Phy5645: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(269 intermediate revisions by 24 users not shown)
Line 2: Line 2:


''' Welcome to the Quantum Mechanics A PHY5645 Fall2008/2009'''
''' Welcome to the Quantum Mechanics A PHY5645 Fall2008/2009'''
[[Image:SchrodEq.png|thumb|550px|Schrodinger equation. The most fundamental equation of quantum mechanics which describes the rule according to which a state <math>|\Psi\rangle</math> evolves in time.
[[Image:SchrodEq.png|thumb|550px|<b>[[Schrödinger Equation]]</b><br/>The most fundamental equation of quantum mechanics; given a Hamiltonian <math>\mathcal{H}</math>, it describes how a state <math>|\Psi\rangle</math> evolves in time.]]
]]  
 
This is the first semester of a two-semester graduate level sequence, the second being [[phy5646|PHY5646 Quantum B]]. Its goal is to explain the concepts and mathematical methods of Quantum Mechanics, and to prepare a student to solve quantum mechanics problems arising in different physical applications. The emphasis of the courses is equally on conceptual grasp of the subject as well as on problem solving. This sequence of courses builds the foundation for more advanced courses and graduate research in experimental or theoretical physics.  
This is the first semester of a two-semester graduate level sequence, the second being [[phy5646|PHY5646 Quantum B]]. Its goal is to explain the concepts and mathematical methods of Quantum Mechanics, and to prepare a student to solve quantum mechanics problems arising in different physical applications. The emphasis of the courses is equally on conceptual grasp of the subject as well as on problem solving. This sequence of courses builds the foundation for more advanced courses and graduate research in experimental or theoretical physics.  
   
   
Line 12: Line 12:
'''Fall 2009 Midterm is October 15'''
'''Fall 2009 Midterm is October 15'''


----
== Outline of the Course ==
'''Outline of the course:'''
 
 
== Physical Basis of Quantum Mechanics ==
 
=== Basic concepts and theory of motion in QM ===
In Quantum Mechanics, all of the information of the system of interest is contained in a wavefunction , <math>\Psi\,\!</math>. Physical properties of the system such as position, linear and angular momentum, energy, etc. can be represented via linear operators, called observables. These observables are a complete set of commuting Hermitian operators, which means that the common eigenstates (in the case of quantum mechanics, the wavefunctions) of these Hermitian operators form an orthonormal basis. Through these mathematical observables, a set of corresponding physical values can be calculated.
 
In order to clarify the paragraph above, consider an analogous example: Suppose that the system is a book, and we characterize this book by taking measurements of the dimensions of this book and its mass (The volume and mass are enough to characterize this system). A ruler is used to measure the dimensions of the book, and this ruler is the observable operator. The length, width, and height (values) from the measurements are the physical values corresponding to that operator (ruler). For measuring the weight of the book, a balance is used as the operator. The measured mass of the book is the physical value for the corresponding observable. The two observable operators (the ruler and the mass scale) have to commute with each other, otherwise the system can not be characterized at the same time, and the two observables can not be measured with infinite precision.
 
In quantum mechanics, there are some measurements that cannot be done at the same time. For example, suppose we want to measure the position of an electron. What we would do is send a signal (a gamma ray, for example), which would strike the electron and return to our detectors. We have, then, the position of the electron. But as the photon strikes the electron, the electron gains additional momentum, and our simultaneous momentum measurement can not be precise. Therefore both momentum and position cannot be measured at the same time. These measurement are often called "incompatible observables." This is explained in the Heisenberg uncertainty principle and implies, mathematically, that the two operators do not commute.
 
This concept contrasts with classical mechanics, where the two observables that do not commute with each other can still be measured with infinite precision. This is because of the difference in dimension of the object: macroscopic (classical mechanics) and microscopic scale (quantum mechanics). However, the prediction of quantum mechanics must be equivalent to that of the classical mechanics when the energy is very large (classical region). This is known as the Correspondence Principle, formally expressed by Bohr in 1923.
 
We can explain this principle by the following:
In quantum mechanics, particles cannot have arbitrary values of energy, only certain discrete values of energy. There are quantum numbers corresponding to specific values of energy and states of the particle. As the energy gets larger, the spacing between these discrete values becomes relatively small and we can regard the energy levels as a continuum. The region where the energy can be treated as a continuum is what is called the classical region.
 
=== UV Catastrophe (Black Body Radiation) ===
To begin an overview of the evolution of Quantum Mechanics, one must first examine its birthplace, i.e. the black body radiation problem. It is simple to understand that emission of radiation from an object occurs for all temperatures greater then absolute zero. As the temperature of the object rises the energy concentration of the emitted radiation (the spectral distribution) shifts away from the long wavelength, i.e. infrared regions, to the shorter wavelength regions, including the visible spectrum and finally the UV and X-ray regions. Coherently, the total power radiated increases with temperature.
 
Imagine a perfect absorber cavity (i.e. it absorbs all radiation at all wavelengths, so that its spectral radiance only depends on temperature). From Kirchoff's law it follows that such a body would not only be a perfect absorber, but also a perfect ''emitter'' of radiation. "Blackbody" is a kind of material, not only it absorbs all of the radiation when the radiation falls on it, but also it seems black. This emission is called the black body radiation. Lord Rayleigh (John William Strutt) and Sir James Jeans applied classical physics and assumed that the radiation in this perfect absorber could be represented by standing waves. Although the Rayleigh-Jeans result does approach the experimentally recorded values for large values of wavelength, the trend line vastly differs as the wavelength is allowed to tend towards zero. The result predicts that the spectral intensity will increase quadratically with increasing frequency, and would diverge to infinite energy as the wavelength went to zero. For short wavelengths, this became known as the so called "Ultraviolet Catastrophe." This black body radiation experiment shows an important failure of classical mechanics.  The Rayleigh-Jeans law is as follows:
 
<math>\rho_\lambda(T) = \frac{2 c k T}{\lambda^4}</math>
 
where c is the speed of light, k is Boltzmann's constant and T is the temperature in kelvins.
 
Based on a thermodynamic argument, Wien noted that under adiabatic expansion, the energy of a mode of light, the frequency of the mode, and the total temperature of the light change together in the same way, so that their ratios are constant. This implies that in each mode at thermal equilibrium, the adiabatic invariant energy/frequency should only be a function of the adiabatic invariant frequency/temperature:
<math>
\rho _{wien} (T) = \frac{{2c^2 h}}{{\lambda ^5 }}e^{ - \frac{{hc}}{{\lambda kT}}}
</math>
 
 
In 1900, Max Planck offered a successful explanation for black body radiation.  He too postulated that the radiation was due to oscillations of the electron, but the difference between his assumption and Rayleigh's was that he argued that the possible energies of an oscillator were not continuous. He proposed that the energy of an oscillator would be proportional to a constant of the frequency.
 
<math>E=h\nu=\hbar\omega</math>
 
Here E is energy, h is the Planck constant (<math> h=6.626*10^{-34} Joule-seconds \!</math>
) and <math>\nu\!</math> is the frequency of the oscillator. With the concept of energy being discrete in mind, the result is that Planck's calculation avoided the UV catastrophe, and instead the energy approached zero as the frequency tends to infinity increased. Planck's law of black body radiation is as follows:
 
<math>\rho_\lambda(T) = \frac{2 c^2}{\lambda^5}~\frac{h}{e^\frac{hc}{\lambda kT}-1}</math>
 
When <math>\lambda  \to \infty </math> and <math>\lambda  \to 0 </math>, we can easily get Rayleigh-Jeans formula and Wien formula.
 
Before leaving the subject of Black Body Radiation it is important to look at one fundamental realization that has come out of the mathematics. In 1964, A. Penzias and R. Wilson discovered a radio signal of suspected cosmic origin, with an intensity corresponding to approximately 3 K. Upon application of Planck's theorem for said radiation, it soon became evident that the spectrum seen corresponded to that of a black body at 3 K, and since this radiation was incident on Earth evenly from all directions, space itself was deemed to be the emitting black body. This ''cosmic background radiation'' gave credence to the Big Bang theory, and upon analysis of an expanding system, allowed for proof that Planck's theorem holds for black bodies of changing size. The results of this particular proof even allow for a fair estimation into the rate of expansion of the universe since the time the black body radiation was emitted.
 
Homework Problem 1 :[http://wiki.physics.fsu.edu/wiki/index.php?title=Talk:Phy5645&oldid=5851]
 
Homework Problem 2 :[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/UV_catastrophe_problem2]
 
=== Photoelectric Effect ===
Another contributing factor to the emergence of the theory of Quantum Mechanics came with the realization of the particle nature of light through explanation of the photoelectric effect.
 
Consider a system composed of light hitting a metal plate. From experimental observations, first observed by Hertz in 1887, and later by Hallwachs, Stoletov, and Lenard in 1900, a current can be measured when light is incident on the metal plate. During this period, the classical point of view was that an electron was bound inside of an atom, and an excitation energy was needed in order to release it from the atom. This energy could be brought forth in the form of light.  The classical point of view also included the idea that the energy of the light was proportional to its intensity. Therefore, if enough energy (light) is absorbed by the electron, the electron would eventually be released. However, this was not the case. Several odd results came from these studies. First it was noted that, while the current did appear to be proportional to the intensity of the incident light, there were certain minimum frequencies of light below which no current could be produced, regardless of the intensity of the incident beam. Also, the stopping potential of the emitted electrons appeared to depend upon the frequency of the radiation, and not the intensity at all. Finally, the emission appeared to take place instantaneously for any intensity so long as the minimum frequency condition was satisfied.
 
In 1905, Einstein began offering possible explanations for the odd observations made regarding the photoelectric effect. Einstein realized that the classical view of light as a wave was not entirely true, that light must also behave like a particle. This allowed him to postulate that the energy of the incident radiation was not continuous, but was rather composed of quantized packets, proportional to the frequency of the wavelength of incident light. These ''corpuscles'' could then be seen to be completely absorbed by an atom, rather then spreading out over the structure like a wave would, so that the absorption/emission would happen instantly. He commented that since electrons were inherently bound to the atom, a certain minimum energy would be required to remove them, and thus if a corpuscle did not have enough energy, i.e. its frequency was too low, the atom would merely absorb and release it, rather then kicking out an electron as well. From this result, Millikan was able to confirm Einstein's theory a few years later by showing that the stopping potential did indeed depend linearly with respect to the frequency, with an additive term corresponding to the minimum energy required to remove the electron, its ''work function''.
 
The equation describing the kinetic energy of the emitted electron is:
 
<math>\frac{1}{2}mv^2 = h\nu - W</math>
 
Where W is the work function and <math>\nu</math> is the frequency of the incident photon.
 
 
From these results it was clearly evident that light was behaving in a particle-like manner, however the existence of various interference and diffraction experiments still gave evidence for a wave-like nature as well, and thus the dual nature of light was exposed, in stark contrast to classical physics.
 
=== Stability of Matter ===
One of the most important problems to inspire the creation of Quantum Mechanics was the model of the Hydrogen Atom. After Thompson discovered the electron, and Rutherford, the nucleus (or Kern, as he called it), the model of the Hydrogen atom was refined to one of the lighter electron of unit negative elementary charge orbiting the larger proton, of unit positive elementary charge. However, it was well known that classical electrodynamics required that charges accelerated must radiate EM waves, and therefore lose energy. For an electron that moves in circular orbit about the more massive nucleus under the influence of the Coulomb attractive force, here is a simple non-relativistic model of this classical system:
Where <math>r\,\!</math> is the orbital radius, and we neglect the motion of the proton by assuming it is much much more massive than the electron.
 
'''So the question is:  What determines the rate <math>\rho</math> of this radiation?  and how fast is this rate?'''
 
The electron in the Bohr's model involves factors of:  radius  <math>r_0\,\!</math>, angular velocity <math>\omega\,\!</math>, charge of the particle <math>e\,\!</math>, and the speed of light, <math>c\,\!</math>: <math>\rho=\rho(r_0,\omega,e,c)\,\!</math>
           
The radius and charge will not enter separately, this is because if the electron is far from the proton, then the result can only depend on the dipole moment, which is .
 
Therefore the above parameters is now:<math> \rho(er_0, \omega, c) \!</math>
 
'''What is the dimension of <math>\rho\,\!</math>?'''
 
Essentially, since light is energy, we are looking for how much energy is passed in a given time: <math>[\rho]=\frac{energy}{time} \!</math>
 
Knowing this much already imposes certain constraints on the possible dimensions. By using dimensional analysis, let's construct something with units of energy.
 
From potential energy for coulombic electrostatic attractions: <math>energy=\frac{e^2 }{length} \!</math>
 
<math>e\!</math> has to be with respect to <math>r_0\!</math> , multiply by <math>r^2\!</math>  , and divide <math>length^2\!</math>.
 
The angular velocity is in frequency, so to get the above equations in energy/time, just multiply it with the angular velocity, <math>energy=\frac{e^2 }{length}\frac{r^2}{length^2}*\omega </math>
 
(Here, it is seen that the acceleration of the electron will increase with decreasing orbital radius. The radiation due to the acceleration a is given by the Larmor Formula: <math>energy \sim \frac{e^2r_0^2 }{(c/w)^3} w = \frac{e^2r_0^2 }{c^3}w^4\sim\frac{1}{r_0^4 } \!</math>
It was known that the hydrogen atom had a certain radius on the order of .5 angstroms. Given this fact it can easily be seen that the electron will rapidly spiral into the nucleus, in the nanosecond scale. Clearly, the model depicts an unstable atom which would result in an unstable universe.  A better representation of an electron in an atom is needed.
 
Homework Problem :[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/stability_of_matter_hw1]
 
===Double Slit Experiment===
 
'''Bullet'''[[Image:Gun.png|thumb|125px|Double slit thought experiment with classical bullets]] 
 
Imagine a gun which is spraying bullets randomly toward a wall with two slits in it separated by a distance, d.  The slits are about the size of a bullet.  A histogram of the bullet's location after it passes through the two slits is plotted. If slit 2 is closed, but the slit 1 is open, then the green peak is observed which is given by the distribution function <math>p_1</math>.  Similarly, if the slit 1 is closed, but he slit 2 is open, the pink peak is observed which is given by the distribution function <math>p_2</math>.  When both slits are open, <math>peak_{12}</math> (purple) is observed.  This agrees with the classical view, where the bullet is the particle and <math>p_{12}</math> is simply a sum of <math>p_1</math> and <math>p_2</math>.  The bullets do not follow purely linear trajectories because they are allowed to hit the edges of the slits they pass through and be deflected.  It is because the bullets can be deflected that the result of this experiment is a probability distribution rather than the bullets going to just the two locations that are along straight line trajectories from the gun through the slits. 
 
The equation describing the probability of the bullet arrival if both of the slit are open is therefore
 
<math>p_{12}=p_1+p_2.\!</math>
 
'''Classical Waves'''[[Image:waves.png|thumb|125px|right|Double slit thought experiment with water waves]]
 
 
As waves are passed through the double slit, they are diffracted so that the waves emerge from the slit as circular waves, this effect can only occur when the size of the slits is comparable to the wavelength.  The intensity of the waves which are proportional to the squares of the height of the wave motion <math>H_1^2</math> and <math>H_2^2</math> are observed when slit 1 and 2 are closed respectively. These intensities are similar to the histograms for the bullets in the previous demonstration. However, an interference pattern of the intensity (<math>H_{12}</math>) is observed when both slits are opened.  This is due to constructive and destructive interferences of the two waves. The resultant interference is the square of the sum of the two individual wave heights
 
<math>H_{12} = (H_{1} +H_{2} )^2\!</math>
 
'''Hot Tungsten Wire (thermal emission of electrons)'''
A high current is passed through a tungsten wire, resulting in electrons being emitted from the wire which then enter the double slits one at a time, arriving in the same manner as the bullet arrives from the gun. However, after plotting a histogram of the locations where the electron landed, it looks like H_{12} for the double slit wave experiment. This shows that electrons exhibit both the wave and the particle-like character. The probability distribution of the electron's landing on the screen thus exhibits the interference patterns. It is the laws obeyed by these probability "amplitudes" that Quantum Mechanics describes.
[1] R.P. Feynman, R.B. Leighton and M.L.Sands The Feynman Lectures on Physics, vol 3, Addison-Wesley, (1989), Chapter 1.
 
 
=== Stern-Gerlach Experiment ===
[[Image:Stern-Gerlach_experiment.jpg|thumb|450px]]
 
Preformed in 1922 by Walter Stern and Otto Gerlach, this experiment demonstrated that particles have intrinsic spin.
A collimated monochromatic beam of silver atoms (Ag) is subjected to an inhomogeneous magnetic field. Silver was chosen because it has all of its shells full except for one additional electron in the 5s shell. It is necessary to use a non-uniform magnetic field because, if the field was uniform, the trajectory of the silver atoms would be unaffected. In a non-uniform field the force on one end of the dipole is greater than the other.
 
Classical theory would predict that there would be a continuous line at the collector plate because the spin would be random valued.
However, at the collector there was only two spots. The atoms were deflected in the vertical direction by specific amount equal to <math>\pm \hbar/2</math>, thus showing that spin is quantized. Because there were only two spots at the collecter the electorn is a spin ½ particle.
 
It is important to note here that this spin does not arise because the particle is spinning. If this were the case then this would mean that the electron would be spinning faster than the speed of light.
 
A sample problem: [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Double_pinhole_experiment The double pinhole experiment]
 
=== The principle of complementarity ===
The idea of wave-particle duality has no classical counterpart. In classical physics, a given entity must be exclusively one or the other. But this has come at the expense of great conceptual difficulty. We must somehow accommodate the classically irreconcilable wave and particle concepts. This accommodation involves what is known as the principle of complementarity, first enunciated by Bohr. The wave-particle duality is just one of many examples of complementarity. The idea is the following. Objects in nature are neither particles nor waves; a given experiment or measurement which emphasizes one of these properties necessarily does so at the expense of the other. An experiment properly designed to isolate the particle properties, such as Compton scattering or the observation of cloud chamber tracks, provides no information about the wave aspects. Conversely, an experiment properly designed to isolate the wave properties, for example the diffraction, provides no information about its particle aspects. The conflict is thus resolved in the sense that irreconciable aspects are not simultaneously observable in principle. Other examples of complementary aspects are the position and linear momentum of a particle, the energy of a given state and the length of time for which the state exists, the angular orientation of a system and its angular momentum, and so on. The quantum mechanical description of the properties of a physical system is expressed in terms of pairs of mutually complementary variables or properties. Increasing precision in the determination of one such variable necessarily implies decreasing precision in the determination of the other.
 
 
=== The Correspondence Principle ===
 
Thus far we have been concentrating our attention on experiments which defy explanation in terms of classical mechanics and which, at the same time, isolate certain aspects of the laws of quantum mechanics. We must not lose sight, however, of the fact that there exists an enormous domain,the domain of the macroscopic physics, for which classical physics works and works extremely well. There is thus an obvious requirement which quantum mechanics must satisfy - namely, that in the appropriate or classical limit, it must lead to the same predictions as does classical mechanics. Mathematically this limit is that in which <math> \hbar </math>may be regarded as small. For the electromagnetic field, for example, this means that the number of quanta in the field must be very large. For particles it means that the de Broglie wavelengths must be very small compared to all relevant wavelengths. Of course, the statement of quantum mechanics are probabilistic in nature, we have argued, while those of classical mechanics are completely deterministic. Thus, in the classical limit,the quantum mechanical probabilities must become practical certainties; fluctuations must become negligible.
This principle, that in the classical limit the predictions of the laws of quantum mechanics must be in one-to-one correspondence with the predictions of classical mechanics, is called the correspondence principle.
 
For example,in classical mechanics, physical quantities are functions A(r, p) of the position and momentum variables. The correspondence principle consists of choosing in quantum mechanics the same functions of the position and momentum observables.
To the quantity <math>A\left ( r,p \right )</math>, there corresponds the observables <math>\hat{A}=A\left (\hat{r} ,\hat{p} \right )</math>. For instance,
 
<math>\hat{E_{c}}=-\frac{\hbar}{2m}\triangle</math>, <math>\hat{L}=\frac{\hbar}{i} \vec{r }\times\vec{\triangledown }</math>,
 
that is,
                                                     
   
<math>\hat{L_{z}}=\frac{\hbar}{i}\left ( x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x} \right )</math>
 
==  Schrödinger equation ==
 
Imagine a particle constrained to move along a the x-axis, subject to some force <math>F(x,t)\!</math>.  Classically, we would investigate this system by applying Newton's second law, <math>F = ma\!</math>.  Assuming the force is conservative, it could also be expressed as the partial derivative with respect to <math>x</math>, and Newton's second law then reads:
 
<math>m\frac{d^2x}{dt^2}=-\frac{\partial V}{\partial x}</math>
 
The energy for the particle in this regime is given by the addition of its kinetic and potential energies:
 
<math>E = T + V = \frac{p^2}{2m} + V</math>
 
Now by applying the appropriate initial conditions for our particle, we then have a solution for the trajectory of the particle.  As we will see, the above relation is only an approximation to actual physical reality. As we attempt to describe increasingly smaller objects we enter the quantum mechanical regime, where we cannot neglect the particles' wave properties. Allowing <math>\displaystyle{p \rightarrow \frac{\hbar}{i}\frac{d}{dx}}</math> and <math>\displaystyle{E \rightarrow i\hbar \frac{d}{dt}}</math>, we can use the energy equation for a classical particle above to find an equation that describes this wave nature. Thus, we find that the complex amplitude satisfies the Schrodinger Equation for a scalar potential <math>V(x,t)\!</math> in one dimension:
 
<math> i\hbar\frac{\partial}{\partial t}\psi(x,t)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x,t)\right]\psi(x,t) </math>
               
While in 3D: 
 
<math> i\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r},t)\right]\psi(\textbf{r},t)</math>                                                             
Given a solution which satisfies the above Schrodinger equation, Quantum Mechanics provides a mathematical description of the laws obeyed by the probability amplitudes associated with quantum motion.
 
We can also generalize the Schrödinger equation to a system which contains <math> N </math> particles. We assume that the wave function is <math> \psi(\textbf{r}_1,\textbf{r}_2, \ldots, \textbf{r}_N, t) </math> and the Hamiltonian operator of the system can be expressed as:
 
<math>  H= \sum_{k=1}^N \frac{\textbf{p}^2_i}{2m_i}+V(\textbf{r}_1,\textbf{r}_2, \ldots, \textbf{r}_N) </math>
 
So the Schrödinger equation for a many-particle system is:
 
<math>  i\hbar\frac{\partial}{\partial t}\psi(\textbf{r}_1,\textbf{r}_2, \ldots, \textbf{r}_N, t)=\left[\sum_{k=1}^N \frac{\textbf{p}^2_i}{2m_i}+V(\textbf{r}_1,\textbf{r}_2, \ldots, \textbf{r}_N)\right]\psi(\textbf{r}_1,\textbf{r}_2, \ldots, \textbf{r}_N, t) </math>
 
 
=== Stationary states ===
Stationary states are the energy eigenstates of the Hamiltonian operator. These states are called "stationary" because their probability distributions are independent of time.
 
For a conservative system with a time independent potential, <math>V(\textbf{r})</math>, the Schrödinger equation takes the form:
 
<math> i\hbar\frac{\partial \psi(\textbf{r},t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\psi(\textbf{r},t)</math>
 
Since the potential and the Hamiltonian do not depend on time, solutions to this equation can be written as
 
<math>\psi(\textbf{r},t)=e^{-iEt/\hbar}\psi(\textbf{r})</math>.
 
Obviously, for such state the probability density is 
 
<math>|\psi(\textbf{r},t)|^2=|\psi(\textbf{r})|^2</math>
 
which is independent of time. Hence the name "stationary state". The Schrodinger equation now becomes
 
<math>\left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\psi(\textbf{r},t)=E\psi(\textbf{r})</math>
 
which is an eigenvalue equation with eigenfunction <math>\psi(\textbf{r})</math> and eigenvalue <math>E</math>. This equation is known as the time-independent Schrödinger equation.
 
 
A sample problem:  [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Free_particle_SE_problem A free particle Schrodinger equation].
 
=== Conservation of probability ===
 
Next, we wish to show that for any fixed time t, the probability to find a particle in space is equal to one.
 
The quantity <math>|\psi(\textbf{r},t)|^2</math> can be interpreted as probability density. To show that this is true, two conditions must be met. First, the probability amplitude must be positive semi-definite (equal to or greater than zero). This condition is trivial because <math>|\psi(\textbf{r},t)|\!</math> is always a positive function. Second, the probability amplitude must be conserved. This condition can be shown by proving that if the wavefunction is normalized at some time <math>t_0</math> then it must be normalized for any time <math>t</math>:
 
<math>\int_{-\infty}^{\infty}d^3r|\psi(\textbf{r},t_0)|^2=1 \Rightarrow \int dr^3 |\psi(\textbf{r},t)|=1 \;\;\forall t</math>
 
 
The solution to the Schrodinger equation conserves probability, i.e. the probability to find the particle somewhere in the space does not change with time.
To see that it does, consider
 
<math>i\hbar\frac{\partial}{\partial t}\psi(\textbf{r},t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi(\textbf{r},t)</math>
 
Now multiply both sides by the complex conjugate of <math>\psi(\textbf{r},t) \!</math>:
 
<math>i\hbar\psi^*(\textbf{r},t)\frac{\partial}{\partial t}\psi(\textbf{r},t)=\psi^*(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi(\textbf{r},t)</math> <br/>
Now, take the complex conjugate of this entire expression: <br/>
<math>-i\hbar\psi(\textbf{r},t)\frac{\partial}{\partial t}\psi^*(\textbf{r},t)=\psi(\textbf{r},t)\left[-\frac{\hbar^2}{2m}\nabla^2+V(\textbf{r})\right]\psi^*(\textbf{r},t)</math> <br/>
and taking the difference of the above equations we finally find
 
<math>\frac{\partial}{\partial t}\psi^*(\textbf{r},t)\psi(\textbf{r},t)+\nabla\cdot\frac{\hbar}{2im}
\left[\psi^*(\textbf{r},t)\nabla \psi(\textbf{r},t)-(\nabla\psi^*(\textbf{r},t)) \psi(\textbf{r},t)\right]=0</math>
 
Note that this is in the form of a continuity equation
 
<math>\frac{\partial}{\partial t} \rho(\textbf{r},t) + \nabla \cdot \textbf{j}(\textbf{r},t)=0</math>
where
 
<math>\rho(\textbf{r},t)=\psi^*(\textbf{r},t) \psi(\textbf{r},t)\!</math>
 
is the probability density, and 
 
<math>\textbf{j}(\textbf{r},t)=\frac{\hbar}{2im}\left[\psi^*(\textbf{r},t)\nabla \psi(\textbf{r},t)-(\nabla\psi^*(\textbf{r},t)) \psi(\textbf{r},t)\right]</math>
 
is the probability current.
 
Once we know that the densities and currents constructed from the solution of the Schrodinger equation satisfy the continuity equation, it is easy to show that the probability is conserved.
To see that note:
 
<math>\frac{\partial}{\partial t}\int d^3r |\psi(\textbf{r},t)|^2=-\int d^3r(\nabla\cdot \textbf{j})=-\oint d\textbf{A} \cdot \textbf{j} =0</math>
 
where we used the divergence theorem which relates the volume integrals to surface integrals of a vector field.  Since the wavefunction is assumed to vanish outside of the boundary, the current vanishes as well. So, this proved the probability of finding the particle in the whole space is independent of time.
 
=== States, Dirac bra-ket notation ===
The physical state of a system is represented by a set of probability amplitudes (wave functions), which form a linear vector space. This linear vector space is a particular type of space called a Hilbert Space. Another way to think about the Hilbert space is as an infinite dimensional space of square normalizable functions. This is analogous to 3-dimensional space, where the basis is <math> \left( \hat{i}, \hat{j}, \hat{k}\right)</math> in a generalized coordinate system. In the Hilbert space, the basis is formed by an infinite set of complex functions. The basis for a Hilbert space is written like <math> \left( |\psi_0\rangle, |\psi_1\rangle, |\psi_2\rangle, ... , |\psi_j\rangle, ... \right) </math>, where each <math>|\psi_i\rangle </math> is a complex vector function.
 
We denote a state vector <math> \psi\ </math> in Hilbert space with Dirac notation as a “ket” <math>| \psi \rangle</math>, and its complex conjugate (or dual vector) <math> \psi\ </math>* is denoted by a “bra” <math>\langle\psi |</math>.
 
Therefore, in the space of wavefunctions that belong to the Hilbert space, any wavefunction can be written as a linear combination of the basis function: <math> | \phi \rangle = \sum_n c_n|\psi_n\rangle </math>, where <math> c_n </math> denotes a complex number.
 
 
By projecting the state vector <math>|\psi\rangle</math> onto different basis, we can obtain the wavefunctions of the system in different basis. For example, if we project <math>|\psi\rangle</math> onto position basis <math>\langle \textbf{r}|</math>, we would get <math><\textbf{r}|\psi> \equiv  \psi(\textbf{r})</math>, and if we project <math>|\psi\rangle</math> onto momentum basis <math>\langle \textbf{p}|</math>, we would get <math><\textbf{p}|\psi> \equiv  \psi(\textbf{p})</math>, whereas <math>|\psi( \textbf{r} )|^{2}</math> is the probability of finding the system in the position <math>\textbf{r}</math> and and <math>|\psi( \textbf{p} )|^{2}</math> is the probability of finding the system having the momentum <math>\textbf{p}</math>.
 
 
In Dirac notation, the scalar product of two state vectors (<math> \phi\ </math>, <math> \psi\ </math>) is denoted by a “bra-ket” <math>\langle\phi|\psi\rangle </math>. In coordinate representation the scalar product is given by:
 
<math>\langle\phi|\psi\rangle = \int \phi^*(r,t)\psi(r,t)d^3r </math>
 
And so, the normalization condition may now be written:
 
<math>\langle\psi_m|\psi_n\rangle = \delta_{mn} </math>
 
Which additionally shows that any wavevector is determined to within a phase factor, <math> e^{i\gamma}</math>, where <math>\gamma</math> is some real number.
 
The vectors in this space also obey some useful rules following from the fact that the Hilbert space is linear and complete:
 
 
<math> \langle\phi|c\psi\rangle = c\langle\phi|\psi\rangle </math>
 
<math> \langle c\phi|\psi\rangle = c^*\langle\phi|\psi\rangle </math> where c is some c-number.
 
<math> \langle\phi|\psi_1 + \psi_2\rangle = \langle\phi|\psi_1\rangle + \langle\phi|\psi_2\rangle </math>
 
 
In Dirac's notation, Schrödinger's equation is written as
 
<math>i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle=\mathcal{H}|\psi(t)\rangle </math>
 
By projecting the equation in position space, we can obtain the previous form of Schrödinger's equation:
 
<math> i\hbar\frac{\partial \psi(\textbf{r},t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})\right]\psi(\textbf{r},t)</math>
 
On the other hand, we can also project the Schrödinger's equation in other space like momentum space and obtain:
 
<math> i\hbar\frac{\partial \psi(\textbf{p},t)}{\partial t} = \left[ \frac{\textbf {p}^{2}}{2m} + V\left ( i\hbar \frac{\partial}{\partial \textbf{p}}\right)\right]\psi(\textbf{p},t)</math>
 
where <math>\psi(\textbf{p},t)</math> and <math>\psi(\textbf{r},t)</math> are related through Fourier transform as described in the next section.
 
 
For time independent Hamiltonians, the Schrödinger's equation separates and we can seek the solution in the form of stationary states.
 
<math>|\psi_n(t)\rangle=e^{-iE_n t/\hbar}|\psi_n\rangle</math>.
 
The equation for stationary states in the Dirac notation is then
 
<math>E_n|\psi_n\rangle=\mathcal{H}|\psi_n\rangle.</math>
 
The eigenfunctions (now also referred to as eigenvectors) are replaced by eigenkets.
Use of this notation makes solution of the Schrodinger equation much simpler for some problems, where the Hamiltonian can be re-written in the form of matrix operators having some algebra (defined set of operations on the basis vectors) over the Hilbert space of the eigenvectors of that Hamiltonian. (See the section on Operators.)
 
If we prepare an arbitrary state <math>|\phi\rangle </math> at t=0, how does it evolve in time? <math>|\phi\rangle </math> can be expressed as the linear superposition of the energy eignstates:
 
<math> | \phi \rangle=\sum_{n}c_n| \psi_n \rangle </math>
 
Then, we can get:
 
<math> | \phi(t) \rangle=e^{-i\mathcal{H} t/\hbar}| \phi \rangle=e^{-i\mathcal{H} t/\hbar}\sum_{n}c_n| \psi_n \rangle=\sum_{n}c_ne^{-iE_n t/\hbar}|\psi_n\rangle </math>.
 
=== Heisenberg Uncertainty relations ===
 
Consider a long string which contains a wave that moves with a fairly well-defined wavelength across the whole legnth of the string.  The question, "where is the wave" does not seem to make much sense, since it is spread thoughout the length of string.  A quick snap of the wrist and the string produces a small bump-like wave which has a well defined position.  Now the question, "what is the wavelength" does not make sense, since there is no well defined period.  This example illisturates the limitation on measuring the wavelength and the position simultaneously.  Relating the wavelength to momentum yields the de Broglie equation, which is applicable to any wave phenomenon, including the wave equation:
 
<math>p=\frac{h}{\lambda}=\frac{2\pi \hbar}{\lambda}</math>
 
Now that there is a relation between momentum and position, the uncertainty of the measurement of either momentum or position takes mathematical form in the Heisenberg Uncertainty relation:
 
<math>\Delta x \Delta p \geq \frac{\hbar}{2}</math>
 
where the <math>\Delta\ A </math> of each operator represents the positive square root of the variance, given generally by:
 
<math>\langle(\Delta A)^2\rangle=\langle A^2\rangle-\langle A\rangle^2.</math>
 
Although both momentum and position are measurable quantities that will yield precise values when measured, the uncertainty principle states that the deviation in one quantity is directly related to the other quantity.  This deviation in the uncertainty principle is the result of identically prepared systems not yielding identical results. 
 
A generalized expression for the uncertainty of any two operators A and B is:
 
<math>\Delta A\Delta B=\frac{1}{2i}\langle [A,B]\rangle.</math>
 
And thus, there exists an uncertainty relation between any two observables which do not commute.
 
More generally the uncertainty principle states that two canonically conjugated variables cannot be determined simultaneously with a precision exceeding the relation:
 
<math> \Delta A\Delta B = \hbar </math>
 
Canonically conjugated variables are those which are related by the Fourier Transform. More specifically, they are variables that when you take the Fourier Transform of a function that is dependent on one, you get a function that depends on the other. For example, position and momentum are canonically conjugated variables:
 
<math>\Phi(p,t) = \frac{1}{2\pi\hbar}\int^{+\infty}_{-\infty}e^{\frac{-ipx}{\hbar}}\Psi(x,t)dx</math>;
<math>\Psi(x,t) = \frac{1}{2\pi\hbar}\int^{+\infty}_{-\infty}e^{\frac{ipx}{\hbar}}\Phi(p,t)dp</math>.
 
Another example of canonically conjugated variables are energy and time.
It is precisely this relationship that leads to the uncertainty principle. The reader has probably noticed that the relation above – i.e. <math>\Delta A\Delta B = \hbar </math> – is not the familiar uncertainty principle we all know, the one where <math>\hbar  </math> is divided by two. It turns out that the above relation is more general; we only get the more familiar version when the wave-packet is Gaussian.
 
See also
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645#Generalized_Heisenberg_uncertainty_relation Generalized Heisenberg Uncertainty Relation]
 
A worked problem showing the uncertainty in the position of different objects over the lifetime of the universe: [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/uncertainty_relations_problem1 Uncertainty Relations Problem 1]
 
A problem about how to find kinetic energy of a particle, a nucleon specifically, using the uncertanity principle : [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Uncertainty_Relations_Problem_2 Uncertainty Relations Problem 2]
 
Another problem verifying Uncertainty relation:
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HeisenbergUnRelation Verifying the Relation]
 
== Motion in one dimension ==
 
We study the one dimensional problems in quantum theory, not only because the interest of study the simplest cases to learn about the general properties. Actually there are many cases in two and three dimensions that can be reduced to one-dimensional problem like the cases in central potentials.
 
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Problem_1D_sample Example]
 
 
'''Overview'''
 
Let's consider the motion in 1 direction of a particle in the potential V(x). Supposing that V(x) has finite values when x goes to infinity:
 
:<math>\lim_{x \to -\infty}V(x)=V_-,  \lim_{x \to +\infty}V(x)=V_+</math>
 
and assuming that:    <math>V_-<V_+ \!</math>
 
The Schrodinger equation becomes:
 
:<math>\left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right]\psi(x)=E\psi(x)</math>
<math>\rightarrow \frac{d^2}{dx^2}\psi(x)+\frac{2m}{\hbar^2}(E-V(x))\psi(x)=0</math>
 
From this equation we can discuss some general properties of 1-D motion as follows:
 
 
If <math>E>V_+\!</math>:
 
<math>E-V(x)>0\!</math> at both <math>-\infty</math> and <math>+\infty</math>. Therefore, the solutions of Schrodinger equation are linear combinations of trigonometric functions (sine or cosine). The wave function is oscillating at both <math>-\infty</math> and <math>+\infty</math>. The particle is in an unbound state. The energy spectrum is continous. Both oscillating solutions are allowed, and the energy levels are two-fold degenerate.
 
 
If <math>V_-\le E \le V_+</math>:
 
<math>E-V(x)>0\!</math> at <math>-\infty</math> but <math>E-V(x)<0</math> at <math>+\infty</math>. Therefore, the wave function is oscillating at  <math>-\infty</math> but decaying exponentially at <math>+\infty</math>. The energy spectrum is still continous but no longer degenerate.
 
 
If <math>E<V_-\!</math>:
 
<math>E-V(x)\!<0</math> at both <math>-\infty</math> and <math>+\infty</math>. Therefore, the wave function decays exponentially at both <math>-\infty</math> and <math>+\infty</math>. The particle is in a bound state. The energy spectrum is discrete and non-degenerate.
 
 
=== 1D bound states ===
When the energy of a particle is less than the potential at both positive and negative infinity, the particle is '''trapped''' in a well and it goes back and forth between the turning points in the potential with its kinetic energy; however, when the energy is larger than the potential at either infinity, the particle is said to be "unbound". For a bound state, the wavefunction must decay at least exponentially at both infinities.
 
====Basic Properties ====
 
'''1. Non-degeneracy of the bound states in 1D'''
 
Let's consider a more general property that is the [http://en.wikipedia.org/wiki/Wronskian Wronskian] for 2 solutions of the 1-D Schrodinger equation must equal to a constant.
   
Schrodinger equation :
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi</math>
 
is a second-order differential equation. Such equation has 2 linearly independent
<math>\psi_E^{(1)}</math> and <math>\psi_E^{(2)}</math> for each value of <math>E\!</math>:
 
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(1)}}{dx^2}+V(x)\psi_E^{(1)}=E\psi_E^{(1)}</math>
 
<math>-\frac{\hbar^2}{2m}\frac{d^2\psi_E^{(2)}}{dx^2}+V(x)\psi_E^{(2)}=E\psi_E^{(2)}</math>
 
By definition in mathematics, the Wronskian of these functions is:
<math>W=\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}</math>
 
Multiplying equation (2) by <math>\psi_E^{(2)}</math>, equation (3) by <math>\psi_E^{(1)}</math>,
then subtracting one equation from the other, we get:
 
<math>\frac{d^2\psi_E^{(1)}}{dx^2}\psi_E^{(2)}-\frac{d^2\psi_E^{(2)}}{dx^2}\psi_E^{(1)}=0</math>
 
<math>\rightarrow \frac{d}{dx}\left(\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)}\right)=0</math>
 
<math>\rightarrow \frac{dW}{dx}=0</math>
 
<math>\rightarrow W=C</math>
 
where <math>C\!</math> is constant.
So, the [http://en.wikipedia.org/wiki/Wronskian Wronskian] for 2 solutions of the 1-D Schrodinger equation must equal to a constant.
   
For the bound states, the wave function vanish at infinity, i.e:
<math>\psi_E^{(1)}(\infty)=\psi_E^{(2)}(\infty)=0</math>
 
From (4), (5) and (6), it follows that <math>W=0\!</math>
From (4) and (7), we get:
<math>\psi_E^{(1)}\frac{d\psi_E^{(2)}}{dx}-\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}=0</math>
 
<math>\rightarrow \frac{1}{\psi_E^{(1)}}\frac{d\psi_E^{(1)}}{dx}-\frac{1}{\psi_E^{(2)}}\frac{d\psi_E^{(2)}}{dx}=0</math>
 
<math>\rightarrow \frac{d}{dx}\left[\ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})\right]=0</math>
 
<math>\rightarrow \ln(\psi_E^{(1)})-\ln(\psi_E^{(2)})=\mbox{const.}</math>
 
<math>\rightarrow \psi_E^{(1)}=\mbox{const.}\psi_E^{(2)}</math>
 
From (8) it follows that <math>\psi_E^{(1)}</math> and <math>\psi_E^{(2)}</math> describe the same state. Therefore, the bound states in 1D are non-degenerate.
 
 
'''2. The wave function for a real potential V(x) can be chosen real.'''
 
<math> E\psi(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(\textbf{r})}{\partial x^2}+V(\textbf{r})\psi(\textbf{r})</math>
 
so:
<math> E\psi^*(\textbf{r})=-\frac{\hbar^2}{2m}\frac{\partial^2 \psi^*(\textbf{r})}{\partial x^2}+V(\textbf{r})</math>
<math>\rightarrow\psi^*=C\psi</math>
<math>\rightarrow\psi=C^*\psi^*</math>
<math>\rightarrow\psi^*=|C|^2\psi^*</math>
<math>\rightarrow|C|^2=1</math>
 
so:
<math>C=e^{i\theta}\!</math>
<math>\rightarrow\psi^*=e^{i\theta}\psi</math>
 
let <math>\theta=0\!</math>
<math>\rightarrow\psi^*=\psi</math>
So, the wave function is real.
 
 
'''3. The nth excited state has n nodes.'''
 
See the ''Oscillation theorem'' below.
 
==== Parity operator and the symmetry of the wavefunctions ====
 
 
In the above problem, two basic solutions of Schrodinger equation are either odd or even. The general wavefunctions are combinations of odd and even functions. This property originates from the fact that the potential is symmetric or invariant under the inversion <math>x \rightarrow -x</math>, and so does the Hamiltonian. Therefore, the Hamiltonian commutes with the parity operator <math>\hat{P}</math> where <math>\hat{P} \psi(x)=\psi(-x)</math>, or <math> \hat{P}\left |{\mathbf{r}} \right \rangle=\left |{-\mathbf{r}} \right \rangle</math> In this case, the wavefunctions themselves do not need to be odd or even, but they must be some combinations of odd and even functions.
 
One useful law about parity is that, for 1D bound state, if the potential V(x) is inversion symmetric, its wave function is either even or odd.
 
We will prove it step by step:
 
1. If potential V(x) is inversion symmetric, <math>\psi (\mathbf{r}))</math> and <math>\psi (-\mathbf{r})</math> are wave functions with the same eigenvalue.
 
Prove:
Just change the sign of x, and V(x)=V(-x).
 
2. If potential V(x) is inversion symmetric, as for every eigenvalue, '''we can find''' a complete set of eigenfunctions, which are either even or odd.
 
Prove:
If  <math>\psi (\mathbf{r})</math> is a solution of one stationary Schrodinger function, then <math>\psi (-\mathbf{r})</math> is another solution.
Let's formalize
:<math>\text{u(r)}=\psi (\mathbf{r})+\psi (-\mathbf{r})</math> and <math>\text{v(r)=}\psi (\mathbf{r})-\psi (-\mathbf{r})</math>
 
3. So <math>\text{u(r)}</math> is even and <math>\text{v(r)}</math> is odd. Because they only have one eigenfunction.
 
4. For 1D bound state, if the potential V('''r''') is inversion symmetric, its wave function is either even or odd.
 
====Infinite square well====
Let's consider the motion of a particle in an infinite and symmetric square well: <math>V(x)=+\infty</math> for <math>x \ge |L|/2</math>,  otherwise  <math>V(x)=0\!</math>
 
A particle subject to this potential is free everywhere except at the two ends (<math>x = \pm L/2</math>), where the infinite potential keeps the particle confined to the well.  Within the well the Schrodinger equation takes the form:
:<math>-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi</math>
or equivalently,
:<math>\frac{d^2\psi}{dx^2}=-k^2\psi</math>
where     
:<math>k=\frac{\sqrt{2mE}}{\hbar}</math>
Writing the Schrodinger equation in this form, we see that our solution are simply
 
<math>\psi(x) = A \sin (kx) + B \cos (kx)\!</math>:
                                 
Now we impose that the solution must vanish at <math>x = \pm L/2</math>:
:<math>-A\sin(kL/2)+B\cos(kL/2)=0\!</math>
 
:<math>A\sin(kL/2)+B\cos(kL/2)=0\!</math>
Adding the two equations, we get:
:<math>2B\cos(kL/2)=0\!</math>
   
It follows that either <math>B=0\!</math> or <math>\cos(kL/2)=0\!</math>.
 
Case 1: <math>B=0\!</math>.
In this case <math>A\ne0\!</math>, otherwise the wavefunction vanishes every where. Furthermore, it is required that:
<math>\sin(kL/2)=0\!</math>
 
<math>\rightarrow k=2n\pi/L\!</math>
where <math>n=1,2,3,...\!</math>
 
And the wave functions are odd:
<math>\psi(x)=A\sin(2n\pi x/L)\!</math>
 
Case 2: <math>\cos(kL/2)=0\!</math>  <math>\rightarrow k=(2n+1)\pi/L\!</math>
where <math>n=0,1,2,...\!</math>
 
In this case <math>A=0\!</math> and <math>B\ne0\!</math>, and the wavefunctions are even:
<math>\psi(x)=B\cos[(2n+1)\pi x/L]\!</math>
 
The two solutions give the eigenenergies <math> E_n = \frac{\hbar^2}{2m} \left(\frac{\pi n}{L} \right )^2 \!</math>, where <math>n = 1,2,3,...\!</math>  These wave numbers are quantized as a result of the boundary conditions, thus making the energy quantized as well.  The lowest energy state is the ground state, and it has a non-zero energy, which is due to quantum zero point motion.  This ground state is also nodeless, the first excited state has one node, the second excited state has two nodes, and so on.  The wavefunctions are also orthogonal.
 
=== The Dirac Delta function potential ===
 
A delta potential, eg. <math>V_0\delta(x-a)\!</math>, is a special case of the finite square well, where the width of the well goes to zero and the depth of the well goes to infinity, while the produce of the height and depth remains constant. Additional information on the dirac delta function can be found here: [http://en.wikipedia.org/wiki/Dirac_delta_function Dirac Delta Function]. For a delta potential, the wavefunction is still continuous across the potential, ie. <math>x=a\!</math>. However, the first derivative of the wavefunction is discontinous across the potential.
 
For a particle subject to an attractive delta potential <math> V(x) = -V_0\delta(x)\!</math> the Schrodinger equation is
 
:<math>\frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x)</math>
 
For <math> x \neq \!0 </math> the potential term vanishes, and all that is left is
 
:<math>\frac{d^2 \psi(x)}{dx^2} + \frac{2mE}{\hbar^2}\psi(x) = 0</math>
 
A bound state(s) may exist when <math> E < 0 \! </math>, and <math> \psi(x) \! </math> vanishes at <math> x = \pm \infty \!</math>.  The bound state solutions are therefore given by:
 
:<math> \psi_{1}(x) = Ae^{kx} x < 0  \!</math>
 
:<math> \psi_{2}(x) = Be^{-kx} x > 0 \!</math>
 
where
 
The first boundary condition, the continuity of <math> \psi(x) \!</math> at <math> x = 0 \!</math>, yields <math> A = B \! </math>.
 
The second boundary condition, the discontinuity of <math> \frac{d\psi(x)}{dx} \! </math> at <math> x = 0 \!</math>, can be obtained by integrating the Schrodinger equation from <math> -\epsilon \!</math> to <math>\epsilon \!</math> and then letting <math> \epsilon \rightarrow 0 \!</math>
 
Integrating the whole equation across the potential gives
 
:<math>\int_{-\epsilon}^{\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx</math>
 
In the limit <math>\epsilon \rightarrow 0 \!</math>, we have
 
:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]+V_0\psi(0)=0</math>
 
which yields the relation: <math> 2kA = \frac{2mV_0}{\hbar^2}A \!</math>.
 
Since we defined  <math> k = \sqrt{\frac{2m|E|}{\hbar^2}} \!</math>, we have <math> \sqrt{\frac{2m|E|}{\hbar^2}} = \frac{mV_0}{\hbar^2} \!</math>. Then, the energy is
<math> E = -\frac{mV_0^2}{2\hbar^2} \!</math>
 
Finally, we normalize <math> \psi(x) \!</math>:
 
:<math>\int_{-\infty}^{\infty} |\psi(x)|dx=2|B|^2\int_{0}^{\infty} e^{-2kx}dx=\frac{|B|^2}{k}=1 </math>
 
so,
 
:<math>B=\sqrt{k}=\frac{\sqrt{mV_{0}}}{\hbar} </math>
 
Evidently, the delta function well, regardless of its "strength" <math> V_{0} \!</math>, has one bound state:
 
:<math> \psi(x)=\frac{\sqrt{mV_{0}}}{\hbar}e^{\frac{-mV_{0}|x|}{\hbar^{2}}} </math>
 
Similarly, for a delta potential of the form <math>V_0\delta(x-a)\!</math>, the discontinuity of the first derivative can be shown as follows:
 
The Schrodinger equation is
 
:<math>\frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V_0\delta(x-a)\psi(x)=E\psi(x)</math>
 
Integrating the whole equation across the potential gives
 
:<math>\int_{a-\epsilon}^{a+\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{a-\epsilon}^{a+\epsilon} V_0\delta(x-a)\psi(x)dx=\int_{a-\epsilon}^{a+\epsilon} E \psi(x)dx</math>
 
In the limit <math>\epsilon \rightarrow 0 \!</math>, we have
 
:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}\right]+V_0\psi(a)=0</math>
 
Hence the first derivative of the wave function across a delta potential is discontinuous by an amount:
 
:<math>\frac{d \psi(a^+)}{dx}-\frac{d \psi(a^-)}{dx}=\frac{2m V_0}{\hbar^2}\psi(a)</math>
 
=== Scattering states ===
 
The scattering states are those not bound, where the energy spectrum is a continuous band.  Unlike the bound case, the wave-function does not have to vanish at infinity, though a particle can not reflect from infinity often giving a useful boundary condition.  At any changes in the potentials, the wave-function must still be continuous and differentiable as for the bound states.
 
We have to know wave functions and discrete energy levels <math>E_{n}\!</math> for bound state problems; but, for scattering states (unbound states) the energy E isn't discrete. We are interested in obtaining related wave functions in order to use and determine the transmission and reflection coefficients T and R respectively.
 
=== Oscillation theorem ===
 
Let us concentrate on the bound states of a set of wavefunctions. Let <math> \frac{\hbar^2}{2m}=1</math>. Let <math>\psi_1\!</math> be an eigenstate with energy <math>E_1\!</math> and <math>\psi_2\!</math> an eigenstate with energy <math>E_2\!</math>, and <math>E_2>E_1\!</math>. We also can set boundary conditions, where both <math>\psi_1\!</math> and <math>\psi_2\!</math> vanish at <math>x_0\!</math>.This implies that
:<math>-\psi_2\frac{\partial^2 \psi_1}{\partial x^2}+V(x)\psi_2\psi_1=E_1\psi_2\psi_1\!</math>
:<math>-\psi_1\frac{\partial^2 \psi_2}{\partial x^2}+V(x)\psi_1\psi_2=E_2\psi_1\psi_2\!</math>
 
Subtracting the second of these from the first and simplifying, we see that
:<math>\frac{\partial}{\partial x}\left(-\psi_2\frac{\partial \psi_1}{\partial x}+\psi_1\frac{\partial \psi_2}{\partial x}\right)=\left(E_1-E_2\right)\psi_1\psi_2\!</math>
 
If we now integrate both sides of this equation from <math>x_0\!</math> to any position <math>x'\!</math> and simplify, we see that
:<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}+\psi_1(x')\frac{\partial \psi_2(x')}{\partial x}=(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!</math>
 
The key is to now let <math>x'\!</math> be the first position to the right of <math>x_0\!</math> where <math>\psi_1\!</math> vanishes.
:<math>-\psi_2(x')\frac{\partial \psi_1(x')}{\partial x}=(E_1-E_2)\int_{x_0}^{x'}\psi_1\psi_2dx\!\!</math>
Now, if we assume that <math>\psi_2\!</math> does not vanish at or between <math>x'\!</math> and <math>x_0\!</math>, then it is easy to see that the left hand side of the previous equation has a different sign from that of the right hand side, and thus it must be true that <math>\psi_2\!</math> must vanish at least once between <math>x'\!</math> and <math>x_0\!</math> if <math>E_2>E_1\!</math>.
 
=== Transmission-Reflection, S-matrix ===
 
 
''' The step potential '''
 
 
Let's consider one dimensional potential step with an energy <math> E > V_0 \!</math>. That is, we have a potential
:<math> V(x) =
\begin{cases}
0, & x < 0,  \\
V_0, & x > 0.
\end{cases}
</math>
 
The Schroedinger equation is
:<math>
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E \psi(x).
</math>
 
If we divide the region I and the region II for each <math> x < 0 \!</math> and <math> x > 0 \! </math>, the Schroedinger equations for each region are
 
:<math>
- \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(x),
</math>
 
:<math>
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V_0 \right) \psi_{II}(x) = E \psi_{II}(x).
</math>
 
The general wave functions for each region are
:<math> \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x}, </math>
:<math> \psi_{II}(x) = C e^{i k x} + D e^{-i k x}, \!</math>
where
:<math> k_0 = \sqrt{\frac{2mE}{\hbar^2}} \mbox{    and    } k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}. </math>
 
The boundary conditions at <math> x=0 \!</math> require
:<math> \psi_I(0) = \psi_{II}(0) \mbox{      and      } \left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0} </math>
and we have
:<math> A + B = C + D, \!</math>
:<math> k_0\left(A-B\right) = k \left(C-D\right). </math>
 
If we assume the waves incident from the left to the right, we can set <math> D = 0 \! </math>. In this case, reflection occurs at the potential step, and there is transmission to the right. We then have
:<math> A + B = C, \!</math>
:<math> \left(A-B\right) = \frac{k}{k_0} C. </math>
 
From the above equations, we can get
:<math> \frac{B}{A} = \frac{k_0-k}{k_0+k}  \mbox{  and  }
\frac{C}{A} = \frac{2k}{k_0+k}. </math>
 
The current density which is defined by
:<math> j = \frac{\hbar}{2m i} \left[ \psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right] </math>
can be written by
:<math> j =
\begin{cases}
\displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right) & \left( x < 0 \right), \\
{} & {} \\
\displaystyle \frac{\hbar k}{m} \left|C\right|^2 & \left( x < 0 \right).
\end{cases}
</math>
 
The continuity of waves and the current density implies the relation
:<math>
\frac{\left|B\right|^2}{\left|A\right|^2} + \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = 1.
</math>
 
The first is called the ''reflection coefficient'' and the second term is called the ''transmission coefficient'' which are defined by
:<math>
  R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^2},
</math>
 
:<math>
  T = \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = \frac{4k_0k}{\left(k_0 + k \right)^2}.
</math>
 
Thus, we ensure that <math> R + T = 1 \! </math>.
 
 
Now, let's consider <math> 0 < E < V_0 \! </math> case. In this case, <math> k_0 = \frac{\sqrt{2mE}}{\hbar} \! </math> is still real, but <math> k = \frac{\sqrt{2m\left(E-V_0\right)}}{\hbar} \!</math> is imaginary. If we define <math> \kappa \! </math> as a real value following as
:<math> \kappa = \frac{\sqrt{2m\left(V_0-E\right)}}{\hbar},
</math>
then we have
:<math> k = i\kappa . \!</math>
 
Therefore, we have
:<math>
\psi_I = e^{ik_0x} + \frac{k_0-i\kappa}{k_0+i\kappa} e^{-ik_0 x} ,
</math>
:<math>
\psi_{II} = \frac{2k_0}{k_0 + i\kappa} e^{-\kappa x}.
</math>
 
From the second wave function, <math> \psi_{II} \! </math>, we know that the transmitted waves decrease exponentially with relaxation length <math> \frac{1}{\kappa} \!</math>.
 
If we calculate the reflection coefficient and the transmission coefficient in this case,
:<math> R = \left|\frac{j_{ref}}{j_{inc}}\right| = \left| \frac{k_0 - i\kappa}{k_0 + i\kappa} \right|^2 = 1,
</math>
:<math> T = \left|\frac{j_{tr}}{j_{inc}}\right| = 0, </math>
because <math> j_{tr} = 0 \! </math>. That is, the incident waves are totally reflected.
 
The reflected waves have a phase difference from the incident waves. If we rewrite the wave function <math> \psi_I \! </math>,
:<math>
\begin{align}
\psi_I (x) &= e^{ik_0 x} + \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2 + \kappa^2} e^{-ik_0 x} \\
&= e^{ik_0x} + e^{i\theta} e^{-ik_0x} \\
&= 2 e^{i \frac{\theta}{2}} \cos \left( k_0 x - \frac{\theta}{2} \right),
\end{align}
</math>
where the phase difference of the reflected waves with respect to the incident waves defined by
:<math> e^{i \theta} = \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2+\kappa^2}.</math>
Therefore,
:<math> \theta = \tan^{-1} \left( \frac{2\kappa k_0}{\kappa^2 - k_0^2} \right). </math>
 
 
''' The square potential barrier '''
 
For the square potential barrier with
:<math>
V(x) =
\begin{cases}
0, & x < -a ,  \\
V_0, & -a < x < 0, \\
0, & a < a ,
\end{cases}
</math>
we can write the general solution of the Schroedinger equation for <math> 0 < E < V_0 \! </math> :
:<math> \psi(x) =
\begin{cases}
A e^{i k_0x} + B e^{-ik_0x} \equiv \psi_{I}(x) & \left( x < -a \right), \\
C e^{-\kappa x} + D e^{\kappa x} \equiv \psi_{II}(x) & \left( -a < x < 0 \right), \\
F e^{i k_0x} + G e^{-ik_0x} \equiv \psi_{III}(x) & \left( a < x \right),
\end{cases}
</math>
where <math> k_0 = \frac{\sqrt{2mE}}{\hbar} \!</math> and <math> \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}\!</math>.
 
Whit the boundary conditions at <math> x = -a \! </math>, we have
:<math>
A e^{-ik_0 a} + B e^{ik_0a} = C e^{\kappa a} + D e^{-\kappa a }, </math>
:<math>
A e^{-ik_0 a} - B e^{ik_0a} = \frac{i\kappa}{k_0} \left( C e^{\kappa a} - D e^{-\kappa a } \right). </math>
 
Also, there are another boundary conditions at <math> x = a \! </math> and it requires
:<math>
C e^{-\kappa a} + D e^{\kappa a} = F e^{i k_0 a} + G e^{- i k_0 a} , </math>
:<math>
C e^{-\kappa a} - D e^{\kappa a} = - \frac{ik_0}{\kappa} \left( F e^{i k_0 a} - G e^{- i k_0 a} \right). </math>
 
For the convenience, let's express the coefficients of these linear homogeneous relations in terms of matrices:
:<math>
\left( \begin{align} A \\ B \end{align}\right)
= \frac{1}{2} \left(\begin{align}
\left( 1 + \frac{i\kappa}{k_0} \right) e^{\kappa a + i k_0 a} \\
\left( 1 - \frac{i\kappa}{k_0} \right) e^{\kappa a - i k_0 a}
\end{align} \right.
\left. \begin{align}
\left( 1 - \frac{i\kappa}{k_0} \right) e^{-\kappa a + i k_0 a} \\
\left( 1 + \frac{i\kappa}{k_0} \right) e^{-\kappa a - i k_0 a}
\end{align} \right)
\left( \begin{align} C \\ D \end{align} \right)
</math>
 
:<math>
\left( \begin{align} C \\ D \end{align} \right)
= \frac{1}{2} \left( \begin{align}
\left( 1 - \frac{ik_0}{\kappa} \right) e^{\kappa a + i k_0 a} \\
\left( 1 + \frac{ik_0}{\kappa} \right) e^{-\kappa a + i k_0 a}
\end{align} \right.
\left. \begin{align}
\left( 1 + \frac{ik_0}{\kappa} \right) e^{\kappa a - i k_0 a} \\
\left( 1 - \frac{ik_0}{\kappa} \right) e^{-\kappa a - i k_0 a}
\end{align} \right)
\left( \begin{align} F \\ G \end{align} \right)
</math>
 
If we combine these two equations, we have
:<math>
\left( \begin{align} A \\ B \end{align} \right)
= \left( \begin{align}
\left( \cosh 2\kappa a + \frac{i\varepsilon}{2} \sinh 2\kappa a \right) e^{2i k_0 a} \\
-\frac{i\eta}{2} \sinh 2 \kappa a
\end{align} \right.
\left. \begin{align}
\frac{i\eta}{2} \sinh 2\kappa a \\
\left( \cosh 2 \kappa a - \frac{i\varepsilon}{2} \sinh 2 \kappa a \right) e^{- 2i k_0 a}
\end{align} \right)
\left( \begin{align} F \\ G \end{align} \right)
</math>
where <math> \varepsilon = \frac{\kappa}{k_0} - \frac{k_0}{\kappa} \!</math> and <math> \eta = \frac{\kappa}{k_0} + \frac{k_0}{\kappa} \! </math>.
 
Note that <math> \eta^2 - \varepsilon^2 = 4 \! </math>.
 
 
''' The Dirac delta function potential '''
 
Consider the previous [http://wiki.physics.fsu.edu/wiki/index.php?title=Phy5645#The_Dirac_Delta_function_potential Dirac delta function potential], but this time we have scattering states with <math> E>0 </math>. For <math> x<0 </math> the Schrodinger equation reads
 
:<math>\frac{d^2 \psi(x)}{dx^2}=-\frac{2mE}{\hbar^2}\psi(x) = -k^{2}\psi</math>
 
where
 
:<math> k = \sqrt{\frac{2mE}{\hbar^2}} </math>.
 
The general solution is
 
:<math> \psi_{1}(x)=Ae^{ikx}+Be^{-ikx} \!</math>.
 
Similarly, for <math> x>0 \!</math>,
 
:<math> \psi_{2}(x)=Fe^{ikx}+Ge^{-ikx} \!</math>.
 
The continuity of <math> \psi(x)\!</math> at <math>x=0\!</math> requires that
 
:<math> F+G=A+B \!</math>
 
And the other boundary condition, the discontinuity of <math> \frac{d\psi(x)}{dx} \!</math> at <math> x = 0 \!</math>, can be obtained by integrating the Schrodinger equation from <math> -\epsilon \!</math> to <math>\epsilon \!</math> and then letting <math> \epsilon \rightarrow 0 \!</math>
 
Integrating the whole equation across the potential gives
 
:<math>
\int_{-\epsilon}^{\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx
</math>
in the limit <math>\epsilon \rightarrow 0 \!</math>, we have
 
:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]+V_0\psi(0)=0</math>
 
which yields the relation:
 
:<math> ik(F-G-A+B) = -\frac{2mV_0}{\hbar^2}(A+B) </math>.
 
or, more compactly,
 
:<math> F-G = A(1+2i\beta)-B(1-2i\beta), \!</math>
 
where
 
:<math> \beta= \frac{mV_{0}}{\hbar^{2}k} \!</math>.
 
In atypical scattering experiment particles are fired in from one direction-let's say, from the left. In that case the amplitude of the wave coming in from the right will be zero:
 
:<math> G=0 \!</math> (for scattering from the left).
 
<math>A\!</math> is then the amplitude of the '''incident wave''', <math>B\!</math> is the amplitude of the '''reflected wave''', and <math>f\!</math> is the amplitude of the '''transmitted wave'''. Solving the equations of boundary conditions, we find
 
:<math> B= \frac{i\beta}{1-i\beta}A,</math>.
 
and
 
:<math> F= \frac{1}{1-i\beta}A,</math>.
 
Now, the reflection coefficient:
 
:<math> R= \frac{|B|^{2}}{|A|^{2}}=\frac{\beta^{2}}{1+\beta^{2}},</math>
 
meanwhile, the transmission coefficient:
 
:<math> T= \frac{|F|^{2}}{|A|^{2}}=\frac{1}{1+\beta^{2}}</math>.
 
Of course, the sum of these two coefficients should be 1, and it is:
 
:<math> R+T=1\!</math>.
 
Notice that <math> R\!</math> and <math> T\!</math> are functions of <math>\beta\!</math>, and hence of <math>E\!</math>:
 
:<math> R=\frac{1}{1+(2\hbar^{2}E/mV_{0}^{2})} </math>, 
:<math> T=\frac{1}{1+(mV_{0}^{2}/2\hbar^{2}E)}</math>.
 
=== Motion in a periodic potential ===
 
An example of a periodic potential is given in Figure 1 which consists of a series of continuous repeating form of potentials. In other words, the potential is translationally symmetric over a certain period (in Figure 1 it is over period of a).
 
[[Image:periodic potential.jpg]]
 
'''Figure 1.'''
 
V(x)=V(x + a)
 
In this case a is the period.
 
The Hamiltonian of system under periodic potential commutes with the Translation Operator defined as:
:<math>\hat T_a\psi(x)=\psi(x+a)\!</math>
 
This means that there is a simultaneous eigenstate of the Hamiltonian and the Translation Operator. The eigenfunction to the Schrodinger Equation,
:<math>\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi(x)=E\psi(x)</math>
has the form of the following,
:<math>\psi(x)=e^{ikx}u_k(x)\!</math>
where
:<math>u_k(x+a)=u_k(x)\!</math>
 
This result is also known as the Bloch Theorem.
     
Also, by operating the <math>\hat T_a\!</math> operator on the wavefunction (also known as the Bloch wave), it can be seen that this waveform is also an eigenfunction of the <math>\hat T_a\!</math> operator, as shown in the following,
:<math>
\begin{align}
\hat T_a\psi(x)&=\hat T_a \left(e^{ikx}u_k(x)\right)\\
&=\left(e^{ik(x+a)}u_k(x+a)\right)\\
&=e^{ika}\left(e^{ikx}u_k(x)\right)\\
&=e^{ika}\psi(x)
\end{align}</math>
Using the same argument, it is clear that,
:<math>(\hat T_a)^n\psi(x)=e^{ikna}\psi(x)</math>
 
Also, note that if k is complex, then after multiple <math>\hat T_a\!</math> operations, the exponential will "blow-up". Thus, k has to be real. Applying the Bloch Theorem in solving Schrodinger Equation with known periodic potential will reveal interesting and important results such as a band gap opening in the Energy vs k spectrum. For materials with weak electron-electron interaction, given the Fermi energy of the system, one can then deduce whether such a system is metallic, semiconducting, or insulating.
 
[[Image:Insulator-metal.svg.png]]
 
'''Figure 2. Energy band illustration showing the condition for metal, semiconductor, and insulator.'''
 
 
Consider for example the periodic potential and the resulting Schrodinger equation,
:<math>V(x)=V_0\sum_{n=-\infty}^{\infty}\delta(x-na)</math>
:<math>(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V_0\sum_{n=-\infty}^{\infty}\delta(x-na))\psi(x)=E\psi(x)</math>
 
Focusing the attention for case when 0 < x < a, the solution to the Schrodinger equation is of the form:
:<math>E=-\frac{\hbar^2q^2}{2m}</math>
:<math>\psi(x)=Ae^{iqx}+Be^{-iqx}\!</math>
:<math>\psi(x)=e^{ikx}(Ae^{i(q-k)x}+Be^{-i(q+k)x})\!</math>
:<math>\psi(x)=e^{ikx}u_k(x)\!</math>
From periodicity,
:<math>u_k(\epsilon)=u_k(-\epsilon)\!</math>
:<math>u_k(\epsilon)=u_k(a-\epsilon)\!</math>
 
Thus, the wavefunction from x < 0 (left) and x > 0 (right) can be written as:
:<math>\psi_r=e^{ikx}u_k(x)=e^{ikx}(Ae^{i(q-k)x}+Be^{-i(q+k)x})\!</math>
:<math>\psi_r=e^{ikx}u_k(x+a)=e^{ikx}(Ae^{i(q-k)(x+a)}+Be^{-i(q+k)(x+a)})\!</math>
When the continuity requirement at x = 0 is also being imposed, the following relation is found:
:<math>\psi_l(0)=\psi_r(0)\!</math>
:<math>A+B=Ae^{i(q-k)a}+Be^{-i(q+k)a}\!</math>    (1)
 
From differentiability and periodicity, the Schrodinger equation can be solved as the following:
:<math>\int_{-\epsilon}^\epsilon -\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x)}{\partial x^2}dx+\int_{-\epsilon}^\epsilon V_0\sum_{n=-\infty}^{\infty}\delta(x-na)\psi(x)dx=\int_{-\epsilon}^\epsilon E \psi(x)dx</math>
where <math>\epsilon\!</math> is small, approaching zero. In this case, the term on the right hand side can be taken to be 0. Thus,
:<math>-\frac{\hbar^2}{2m}(\frac{\partial\psi_r(\epsilon)}{\partial x}-\frac{\partial\psi_l(-\epsilon)}{\partial x})=-V_0\psi_l(0)</math>
:<math>\lim_{\epsilon\to 0}(\frac{\partial\psi_r(\epsilon)}{\partial x}-\frac{\partial\psi_l(-\epsilon)}{\partial x})=\frac{2mV_0}{\hbar^2}\psi_l(0)</math>
where,
:<math>\lim_{\epsilon\to 0}\frac{\partial\psi_r(\epsilon)}{\partial x}=iq(A-B)</math>
:<math>\lim_{\epsilon\to 0}\frac{\partial\psi_l(-\epsilon)}{\partial x}=iq(Ae^{i(q-k)a}-Be^{-i(q+k)a})</math>
 
Evaluating further, the following condition is found:
:<math>iq(A-B-Ae^{i(q-k)a}+Be^{-i(q+k)a})=\frac{2mV_0}{\hbar^2}(A+B)</math>  (2)
 
By simultaneously solving equation (1) and (2), the relationship between q and k is found to be:
:<math>\cos(ka)=\cos(qa)+\frac{mV_0a}{\hbar^2}\frac{\sin(qa)}{qa}</math> (3)
 
Worked Problem on Periodic Delta Function Potentials: [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Problem2]
 
Where q is the energy band of the system, and k is the accessible energy band. By noting that the maximum value of the LHS is < the maximum value of the RHS of the relation above, it is clearly seen that there are some energy level that are not accessible as shown in Figure 3.
 
[[Image:k_q relation graph.jpg]]
 
Figure 3. Graph of Eq.(3). Wave is representing the LHS function, gray box representing the range of RHS function. Red lines is the forbidden solution, black line is the allowed solution.
As k increases from 0 to  , there will be many solutions. Focusing only to two of the allowed solutions, it is seen that as k increases, there will be two solutions (one solution gives increasing q value, the other solution gives decreasing q value). Using the fact that the energy of the system is  , the dispersion relation (E vs. k) can be plotted as shown in Figure 4.
[[Image:dispersion relation small.jpg]]
 
Figure 4. Energy vs k showing the existence of band gap for system of an electron under periodic potential.
 
Thus, in conjunction with Pauli exclusion principle, the single particle banspectrum such as the one we discussed here constitutes a simple description of band insulators and band metals.
 
If we take a more general case, where the potential is also periodic and always finite. The potential can be expressed as follows:
 
:<math> V(x)=\begin{cases}
 
0, na<x<na+c;\\
 
V_0, na+c<x<(n+1)a.
 
\end{cases}
</math>
 
Further, we assume that the eigenfunction is:
 
:<math>\psi_r=e^{ikx}u_k(x)\! </math>
 
Then this problem is converted to seeking for <math>u_k(x)\!</math> in the region -b<x<c and connecting the wave function in different periods.
 
:<math>\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi(x)=E_k\psi(x), 0<x<c</math>
 
:<math>\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V_0\right)\psi(x)=E_k\psi(x), -b<x<0</math>
 
Further we assume that
 
:<math> k_1^2=\frac{2mE_k}{\hbar^2}, k_2^2=\frac{2m(V_0-E_k)}{\hbar^2} </math>
 
Then we can get the equation for <math> u_k(x)\!</math>:
 
:<math> \frac{d^2u_k(x)}{x^2}+2ik\frac{du_(x)}{dx}+(k_1^2-k^2)u(x)=0, 0<x<c </math>
 
:<math> \frac{d^2u_k(x)}{x^2}+2ik\frac{du_(x)}{dx}-(k_2^2+k^2)u(x)=0, -b<x<0 </math>
 
Solve for <math> u_k(x) \!</math>, we get:
 
:<math> u_k(x)=A_0e^{i(k_1-k)x}+B_0e^{-i(k_1+k)x}, 0<x<c  </math>
 
:<math> u_k(x)=C_0e^{(k_2-ik)x}+D_0e^{-(k_2+ik)x}, -b<x<o </math>
 
By the continuity of <math> \psi(x) \!</math> and <math> \frac{d\psi(x)}{dx}</math> at <math> x=0 </math>:
 
:<math> A_0+B_0=C_0+D_0 \!</math>
 
:<math> i(k_1-k)A_0-i(k_1+k)B_0=(k_2-ik)C_0-(k_2+ik)D_0 \!</math>
 
By periodicity of <math> u_k(x) \!</math>, we get:
 
:<math> u_k(x)=u_k(x+a) \!</math>
 
:<math>C_0e^{(k_2-ik)x}+D_0e^{-(k_2+ik)x}=C_1e^{(k_2-ik)(x+a)}+D_1e^{-(k_2+ik)(x+a)}</math>
 
:<math>C_1=C_0e^{-(k_1-ik)a}</math>
 
:<math>D_1=D_0e^{(k_2+ik)a}</math>
 
By the continuity of <math> \psi(x) \!</math> and <math> \frac{d\psi(x)}{dx}</math> at <math> x=c </math>:
 
:<math>A_0e^{i(k_1-k)x}+B_0e^{-i(k_2+k)x}=C_1e^{(k_2-ik)c}+D_1e^{-(k_2+ik)c}=C_0e^{-(k_2-ik)b}+D_0e^{(k_2+ik)b} </math>
 
:<math> i(k_1-k)e^{i(k_1-k)c}A_0-i(k_1+k)e^{i(k_1+k)c}B_0=(k_2-ik)e^{(k_2-ik)c}C_1-(k_2+ik)e^{-(k_2+ik)c}D_1= (k_2-ik)e^{-(k_2-ik)b}C_0-(k_2+ik)e^{(k_2+ik)b}D_0</math>
 
Now, we get four linear equations of <math>A_0, B_0, C_0 </math> and <math> D_0 </math>, to get the nontrivial solution, we have to make the determinant be zero:
 
:<math> \frac{k_2^2-k_1^2}{2k_1k_2}sinh(k_2b)sin(k_1c)+cosh(k_2b)cos(k_1c)=cos(ka) </math>
 
This is the equation that the energy follows, where
 
:<math> k_1^2=\frac{2mE_k}{\hbar^2}, k_2^2=\frac{2m(V_0-E_k)}{\hbar^2} </math>
 
If we take the limit <math> V_0 \rightarrow 0 and b \rightarrow 0 </math>and mantain <math>V_0b</math> finite, then we can obtain:
 
:<math> k_2=\sqrt{\frac{2m}{\hbar^2}(V_0-E_k)b^2} \simeq\sqrt{(V_0b)\frac{2mb}{\hbar^2}}\ll 1</math>
 
== Operators, eigenfunctions, symmetry, and time evolution ==
 
 
=== Linear Vector Space and Operators ===
Quantum Mechanics can be conveniently formulated in the language of abstract state vectors, from which the various representations (wave mechanics, matrix mechanics, Schrodinger, Heisenberg and interaction pictures, etc.) can be derived. A formulation of quantum mechanics in terms of LVS (Linear Vector Space) hinges on the fact that the Schrodinger equation is linear.An operator is a mathematical operation performed on a vector belonging to a Linear Vector Space, the result of which is another vector belonging to the same Linear Vector Space.
 
==== Ket Space ====
 
In quantum mechanics a physical state is represented by a state vector in a complex vector space and have all the properties described above. Following Dirac we call such a vector a ket denoted by <math>|\alpha ></math> and have become an essential part of quantum mechanics. This state ket is postulated to contain complete information about the physical state i.e. everything we are allowed to ask about the state is contained in the ket. The complex linear vector space that we work in quantum mechanics are usually infinite dimensional. In such case the vector space in question is known as
a Hilbert space after D. Hilbert, who studied vector spaces in infinite dimensions.
 
One of the postulate of quantum mechanics is that <math>|\alpha ></math> and <math>c|\alpha ></math>, with c = 0 represent the same physical state. In other words, only the ”direction” in vector space is of significance.
 
==== Bra Space ====
 
The vector space we have been dealing with is a ket space. We now introduce the notion of a bra space, a vector space ”dual to” the ket space. We postulate that corresponding to every ket <math>|\alpha ></math> there exists a bra denoted by <math><\alpha |</math> in this dual or bra space. The bra space is spanned by a set of bra vectors {<math><\alpha |</math>} which correspond to the set of kets {<math>|\alpha ></math>}. There is one to one correspondence between a ket space and a bra space:
 
<math>|\alpha >\leftrightarrow <\alpha |</math>
 
<math>|a_{1}>,|a_{2}>...,...\leftrightarrow <a_{1}|<a_{2}...,...</math>
 
<math>|\alpha >+|\beta >\leftrightarrow <\alpha |+<\beta |,</math>
 
where <math>\leftrightarrow</math>  stands for dual correspondence. Roughly speaking, we can regard the bra space as some
kind of mirror image of the ket space.
 
The bra dual to <math>c|\alpha ></math> is postulated to be <math>c\ast <\alpha |</math> not <math>c<\alpha |</math>, which is a very important point
to note. More generally we have
 
<math>c_{\alpha }|\alpha >+c_{\beta }|\beta >\leftrightarrow c\ast _{\alpha }<\alpha |+c\ast _{\beta  }<\beta </math>
 
 
 
====Linear Independence, Basis and Orthonormal Basis====
Consider <math> \mathit{N}\!</math> vectors(bra states) <math> |1\rangle </math>,<math> |2\rangle </math>,...<math> |N\rangle </math>. They are linearly independent if the relation
<math> \sum_{i=1}^{N}\mathit{e}_i|1\rangle=0 </math>
necessarily implies <math> \mathit{e}_i=0 \!</math> for <math>i = 1,2,...,N\!</math>. They can be used as basis in a vector space, and decomposition of any vector in terms of basis vectors in unique.
 
While any set of Linearly independent vectors can be used as a basis, normally the discussion is greatly simplified if the basis is an orthonormal one, i.e, <math> \langle \mathit{e_i}|\mathit{e_j} \rangle=\delta_{ij}</math> ,<math>1\leq i\leq j\leq n </math> (<math>\mathit{n}\to \infty </math> for an infinite-dimensional vector space). The action of an operator is completely known once its action on each of the basis vectors of <math> \mathit{V}</math> is given.The Linear Vector Space used in quantum mechanics is known by the name of Hilbert Space.
 
==== The Hilbert Space ====
A Hilbert Space <math>H\!</math> consisting of a set of vectors  <math>\psi,\phi,\chi\!</math>  and a set of scalars <math>a,b,c\!</math> obeys the following properties.
 
 
(a) <math>H\!</math> is a linear vector space.
 
(It obeys all the properties of a linear vector space as mentioned in the previous section.)
 
 
(b) The scalar product defined in <math>H\!</math> is strictly positive.
 
(The scalar product of one element <math>\phi\!</math> with another element <math>\psi\!</math> is a complex number, denoted by <math>\langle\phi|\psi\rangle</math>.This scalar satisfies the following properties
 
1. The scalar product of <math>\phi\!</math> with that of <math>\psi\!</math> is same as the complex conjugate of the scalar product of <math>\psi\!</math> with <math>\phi\!</math>.
              <math>\langle\phi|\psi\rangle</math>=<math>\langle\psi|\phi\rangle^*</math>
 
2. The scalar product of <math>\psi\!</math> with <math>\phi\!</math> is linear with respect to <math>\phi\!</math>.
            <math>\langle\psi|a\psi_{1} + b\psi_{2}\rangle</math> = <math>a\langle\psi|\phi_{1}\rangle + b\langle\psi|\phi_{2}\rangle</math>
 
3. The scalar product of a state vector <math>\psi\!</math> with itself is a positive real number.
              <math>\langle\psi|\psi\rangle</math> = <math>|\psi|^{2}  \geq 0 </math>
 
==== Schwartz Inequality====
 
For two states <math>|\psi\rangle</math> and <math>|\phi\rangle</math> belonging to the Hilbert Space,the Schwartz Inequality:
                          <math>|\langle\psi|\phi\rangle|^{2}</math> <math>\leq</math> <math>\langle\psi|\psi\rangle</math><math>\langle\phi|\phi\rangle</math>
 
If the vectors <math>|\psi\rangle</math>and <math>|\phi\rangle</math>are linearly dependent such that <math>|\psi\rangle</math>=<math>\alpha|\phi\rangle</math>, then the above relation reduces to an equality.
 
====Dual Spaces and Adjoint Operators====
Let <math> \mathit{V}</math> be a Linear Vector Space. From <math> \mathit{V}\!</math>, we can construct another Linear Vector Space <math> \mathit{V^'}</math> by the following rule:
For each element in <math> \mathit{V}</math> of the form <math>|z\rangle \equiv [\alpha |x\rangle + \beta |y\rangle] \in \mathit{V}</math>, we associate it with <math>\langle z| \in [\alpha^* \langle x| + \beta^* \langle y| \in \mathit{V}</math>. <math> \mathit{V^*}\!</math> is called the dual of <math> \mathit{V}\!</math> (the bra space). together, they form a bracket.
Let <math> \mathit{A}</math> be an operator.It is completely known if <math> \langle y|A|x\rangle</math> is known for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math>.
We define a new operator <math>A^\dagger</math> (the adjoint) such that <math> \langle x|A^\dagger|y\rangle=\langle y|A|x\rangle</math> for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math>.
 
====Special Linear Operators in Quantum Mechanics====
 
'''Hermitian Operator:''' An operator <math>H\!</math>is called hermitian iff <math>H = H^\dagger</math>. all physical observables in quantum mechanics are represented by hermitian operators.
 
'''Unitary Operator:''' An operator <math>U\!</math>is called unitary if there exits an unique <math>U^{-1}\!</math> and is equal to <math>U^\dagger</math>, i.e., <math>UU^\dagger = U^\dagger U = 1</math>. An important property of unitary operators is that it preserves the norm of a vector,which in quantum mechanics refers to the conservation of probability under physical operations.
 
'''Antihermitian operator:''' An operator <math>A\!</math> is called antihermitian iff <math>A^\dagger = -A</math>. Every operator <math>C\!</math> can be decomposed uniquely in terms of a hermitian and an antihermitian part:
<math>B = \frac{B+B^\dagger}{2}+\frac{B-B^\dagger}{2}\equiv H+A</math>.
 
'''Antilinear operator:''' An operator <math>A\!</math> is called antilinear if for all <math>|x\rangle</math> and <math>|y\rangle</math> <math> \in V</math> and for every complex numbers <math>\alpha</math> and  <math>\beta</math>,
<math>A[\alpha |x\rangle + \beta |y\rangle]=[\alpha^* |x\rangle + \beta^* |y\rangle]</math>.
In quantum mechanics, we need only linear operators but for one important exception: The operator <math>T\!</math> that represents time reversal of states is an "antilinear operator".
 
====Theorem on Hermitian Operator====
Eigenvalues of a hermitian operators are real and eigenvectors of a hermitian operator and corresponding to different eigenvalues are orthogonal.
 
''Proof:'' Consider the eigenvalue equation:
 
<math>H|x\rangle = \lambda|x\rangle</math>
where <math>|x\rangle</math> is an eigenvector corresponding to an eigenvalue <math>\lambda\!</math> of <math>H\!</math>. Taking the dual of the equation, and using the fact that <math>H^\dagger = H</math>, we get
 
<math>\langle x|H = \lambda\langle x|</math>
 
Taking the scalar product of the first equation with <math>\langle x|</math> and of the second with <math>|x\rangle</math>, we get
 
<math>\langle x|H|x\rangle = \lambda\langle x|x\rangle = \lambda^*\langle x|x\rangle</math>
 
As <math>|x\rangle</math> is not a null vector, <math> \lambda^* = \lambda\!</math>,i.e, <math>\lambda</math> is real. To prove the second part of the theorem, consider another eigenstate <math>|y\rangle</math> with a distinct (real)eigenvalue <math>\lambda^'\neq \lambda</math>. Taking scalar product of the first equation with <math>\langle y|</math> and of the second equation withn <math>\langle x|</math>, we get
 
<math>\langle|H|x\rangle = \lambda\langle y|x\rangle = \lambda^'\langle y|x\rangle</math>.
 
As <math>\lambda \neq \lambda^'</math>, <math>\langle y|x\rangle = 0</math>, i.e., <math>|x\rangle</math> and <math>|y\rangle</math> are mutually orthogonal.
 
====Hermitian adjoint====
Suppose that there exists an operator <math> \hat{A}^{\dagger}</math>, for which <math>\langle \hat{A}^{\dagger} \psi_1|\psi_2\rangle=\langle\psi_1|\hat{A}|\psi_2\rangle </math>. If this holds for all wavefunction, then we construct a Hermitian adjoint.
 
An operator is called Hermitian if: <math> \hat{O}^{\dagger}=\hat{O}</math>.
 
====Relations of Hermitian adjoint====
<math> (\hat{F}^{\dagger})^{\dagger}=\hat{F} </math><br/>
<math> (c\hat{F})^{\dagger}=c^* \hat{F}^{\dagger} </math><br/>
<math> (\hat{F}+\hat{G})^{\dagger}=\hat{F}^{\dagger}+\hat{G}^{\dagger} </math><br/>
<math> (\hat{F}\hat{G})^{\dagger}=\hat{G}^{\dagger}\hat{F}^{\dagger}</math><br/>
<math> c^{\dagger}=c^* </math><br/>
<math> |a\rangle ^{\dagger}=\langle a| </math><br/>
<math> \langle a|b \rangle ^{\dagger}=\langle b|a \rangle </math><br/>
<math> (|a\rangle\langle b|)^{\dagger}=|b\rangle\langle a|</math>
 
====The Law of Calculation for Operators====
(1) If <math> \hat{F} |\psi\rangle=\hat{G} |\psi\rangle </math> for every <math>|\psi\rangle</math>, then <math> \hat{F}</math> is equal to <math> \hat{G}</math>.
 
(2) Commutative Law: <math>\hat{F}+\hat{G}=\hat{G}+\hat{F} \!</math>
 
(3) Associative Law: <math>(\hat{F}+\hat{G})+\hat{H}=\hat{F}+(\hat{G}+\hat{H}) \!</math>
 
(4) Multiply: <math> \hat{F}\hat{G} |\psi\rangle=\hat{F}(\hat{G} |\psi\rangle) </math>
 
(5) Identity operator: <math> \hat{I} |\psi\rangle= |\psi\rangle </math>
 
(6) Zero operator: <math> \hat{0} |\psi\rangle= |0\rangle </math>
 
====Projection Operator====
 
An operator<math>\hat{P}</math> is said to be a Projection operator , if it is Hermitian and equal to its own square.
Thus,<math>\hat{P}</math>= <math>\hat{P}^{\dagger}</math>    and <math>\hat{P}</math>  = <math>\hat{P}^2</math>
Properties of Projection operator:
 
1. The Product of two commuting projection operators <math>\hat{P}_1</math> and <math>\hat{P}_2</math> is also a projection operator.
    <math>(\hat{P}_1\hat{P}_2)^{\dagger} =\hat{P}_2^{\dagger}\hat{P}_1^{\dagger} = \hat{P}_2\hat{P}_1 =\hat{P}_1\hat{P}_2 </math>
and  <math>(\hat{P}_1\hat{P}_2)^2 = \hat{P}_1\hat{P}_2 \hat{P}_1\hat{P}_2 =\hat{P}_1^2\hat{P}_2^2= \hat{P}_1\hat{P}_2</math>
 
2. The sum of two or more Projection operators is not a Projection operator in general.The sum is a Projection operator iff the operators are mutually orthogonal.
 
3. Two Projection operators are said to be orthogonal if the product of the two operators is zero.
 
The Projection operator corresponding to the ket <math>|x \rangle </math> can be written as  <math>\hat{P} = |x\rangle\langle x|</math>.
The completion relation is given by  <math>\Sigma |x\rangle\langle x| = I </math>  or,  <math>\Sigma P_x \!= I</math>
 
For any vector V,
  <math>P_x|V\rangle = |x\rangle\langle x|V\rangle = |x\rangle v_x</math>
 
The Projection operator selects the component of <math>|V\rangle</math> in the direction of    <math>|x\rangle</math>
 
=== Commutation relations and simultaneous eigenvalues ===
 
====Commutators====
 
The commutator of two operators A and B is defined as follows:
 
<math>[A,B]=AB-BA\,\!.</math>
 
When 2 operators <math>A</math> and <math>B</math> commute, then <math>\left[A,B\right]=0</math>.  Conversely, if <math>\left[A,B\right]\neq 0</math>, the operators do not commute, and we can think of the commutator between two operators as a quantization of how badly they fail to commute. 
 
'''Some Identities:'''<br/>
<math> [A,B]+[B,A]=0 \!</math><br/>
<math>[A,A]= 0 \!</math><br/>
<math>[A,B+C]=[A,B]+[A,C]\!</math><br/>
<math>[A+B,C]=[A,C]+[B,C]\!</math><br/>
<math>[AB,C]=A[B,C]+[A,C]B\!</math><br/>
<math>[A,BC]=[A,B]C+B[A,C]\!</math><br/>
<math>[A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0\!</math><br/>
<math> [A,B]=-[B,A]\!</math><br/>
<math> [A,B]^{\dagger} = [\hat{B}^{\dagger},\hat{A}^{\dagger}]</math><br/>
 
 
In addition, if any two operators are Hermitian and their product is Hermitian, then the operators commute because <br/>
<math> (\hat{A}\hat{B})^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger} = \hat{B}\hat{A} </math> <br/>
and <br/>
 
<math>(\hat{A}\hat{B})^{\dagger} = \hat{A}\hat{B}</math><br/>
so we have that <math>\hat{A}\hat{B} = \hat{B}\hat{A}</math>, which means the commutator is zero.
 
It should also be noted that any operator will commute with a constant scalar. 
 
Also it should be noted that:
<math>[A^n,B]=nA^{n-1}[A,B]\!</math><br/>
However this is only true if: <math>[A,[A,B]]=[B,[A,B]]=0\!</math><br/>
 
 
'''Problem on communtators''' : [http://wiki.physics.fsu.edu/wiki/index.php/User_talk:ChorChan]
 
====Compatible observables====
 
An operator which corresponds to some physically measurable property of a system is called an observable. The following is a list of common physical observables and their corresponding operators given in coordinate representation: <br/>
<math> \text{Position, } \textbf{r}:  \textbf{r}  </math><br/>
<math>\text{Momentum, } \textbf{p}:  \frac{\hbar}{i}\nabla </math><br/>
<math>\text{Kinetic Energy, T}=\frac{p^2}{2m}\text{: } - \frac{\hbar^2}{2m}\nabla^2 </math><br/>
<math>\text{Potential Energy, V: } V(\textbf{r}) </math><br/>
<math>\text{Total Energy, E = T+V: } - \frac{\hbar^2}{2m}\nabla^2 + V(\textbf{r})</math><br/>
 
All observables are Hermitian.  If two Obersvables have simultaneous eigenkets, meaning for two obervables A and B,   
 
<math>\hat{A}|\Psi_{AB}\rangle=a|\Psi_{AB}\rangle\!</math><br/>
<math>\hat{B}|\Psi_{AB}\rangle=b|\Psi_{AB}\rangle\!</math><br/>   
 
Then we have that
<math>\hat{A}\hat{B}|\Psi_{AB}\rangle=
\hat{A}b|\Psi_{AB}\rangle=b\hat{A}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!</math><br/>
Similarly,
<math>\hat{B}\hat{A}|\Psi_{AB}\rangle=
\hat{B}a|\Psi_{AB}\rangle=a\hat{B}|\Psi_{AB}\rangle=ba|\Psi_{AB}\rangle\!</math><br/>
 
So we can see that,
<math>\hat{A}\hat{B}-\hat{B}\hat{A}=\left[\hat{A},\hat{B}\right]=0\!</math><br/>.
 
The same logic works in reverse.  So if two operators, A & B commute, i.e. if , then they have simultaneous eigenkets, and they are said to correspond to compatible observables.  Conversely, if  <math>\left[A,B\right]\neq 0</math>,  we say that the operators  and  do not commute and correspond to incompatible observables.
 
====Functions of Operators====
 
It is often useful to consider functions of operators, as the Hamiltonian can be written as such.
The exponential function of an operator is represented as a power series:
 
<math> e^\lambda = 1 + \frac{\hat{A}}{1!} + \frac{\hat{A}^2}{2!}+ \frac{\hat{A}^3}{3!} + ... </math>
 
Which leads to the identity that for some <math>f(\lambda) = e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}}</math>, where <math>\hat{A}</math> and <math>\hat{B}</math> are operators:
 
<math>  e^{\lambda\hat{A}}\hat{B}e^{-\lambda\hat{A}} = \hat{B} + \frac{\lambda}{1!}[\hat{A},\hat{B}] + \frac{\lambda^2}{2!}[\hat{A},[\hat{A},\hat{B}] + \frac{\lambda^3}{3!}[,\hat{A},[\hat{A},[\hat{A},\hat{B}] + ... </math>
 
This follows easily when one considers a Taylor series expansion of <math>f(\lambda)</math>.
It is also useful to consider the commutator of an operator function with an operator. Given operators <math>\hat{l}</math> and <math>\hat{m}</math> where
 
<math>[\hat{l},\hat{m}] = 1 </math>
 
it can be shown that, for arbitrary <math>f(\hat{l})</math>,
 
<math>[f(\hat{l}),\hat{m}] = \frac{d}{d\hat{l}}f(\hat{l})</math>
 
by use of the power series expansion of <math>f(\hat{l})</math> and the commutator identities in the above section.
 
====Physical meaning of various representations====
 
In a given Hilbert basis, it is obvious that the state vector <math>|\psi \rangle</math> is completely determined by the set of its components {Cn },
 
          <math>|\psi \rangle\leftrightarrow \left \{ c_{n}=\langle\phi _{n}| \psi _{n}\right \}</math>
 
which we can write as a column vector, the corresponding bra being the conjugate line vector.
This representation of the state vector is completely equivalent to the wave function <math>\psi \left ( r,t \right )</math>.
 
Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning?
 
In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the <math>C_{n}</math>’s are the amplitudes to find <math>E_{n}</math> in an energy measurement. Therefore,
 
• The representation <math>\psi \left ( r,t \right )</math> is more convenient if we are interested in the properties of the particle in space,
 
• Its Fourier transform <math>\phi \left ( p,t \right )</math> is more convenient if we are interested in its momentum properties,
 
• And the components {Cn} in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle.
 
But, owing to Riesz’s theorem, this can be done with any physical quantity, for instance, the angular momentum, which we examine later on and which
also has discrete eigenvalues. This can be thought of as a “generalization” of the properties of the Fourier transform.   
 
 
====Position and momentum operators====
An extremely useful example is the commutation relation of the position operator <math>\hat{x}</math> and momentum <math>\hat{p}</math>. 
In the position representation, position operator <math>\hat{x}= x</math> and momentum operator <math>\hat{p}= \frac{\hbar}{i}\frac{\partial}{\partial x}</math>. On the other hand, in the momentum representation, momentum operator <math>\hat{p}= p</math> and position operator <math>\hat{x}= i\hbar\frac{\partial}{\partial p}</math>.
 
Applying <math>\hat{x}</math> and <math>\hat{p}</math>  to an arbitrary state ket we can see that:
<math>\left[\hat{x},\hat{p}\right]= i\hbar.</math>
 
The position and momentum operators are incompatible.  This provides a fundamental contrast to classical mechanics in which x and p obviously commute.
 
In three dimensions the canonical commutation relations are:
<math>\left[\hat{r}_i,\hat{p}_j\right]= i\hbar\delta_{ij}</math><br/>
<math>\left[\hat{r}_i,\hat{r}_j\right]=
\left[\hat{p}_i,\hat{p}_j\right]=0,</math><br/>
where the indices stand for x,y, or z components of the 3-vectors.
 
It is again interesting to consider functions of these operators. (Note: <math>\hat{q}</math> and <math>\hat{p}</math> are the vector canonical position and momentum operators, respectively.)
For some function <math>f(\hat{q})</math>, it is easily shown that
 
<math>[\hat{p},f(\hat{q})] = -i\hbar\nabla f(\hat{q})</math>.
 
Additionally, for some <math>F(\hat{q})</math>, the expression <math>e^{\frac{i\lambda}{\hbar}\hat{p}}F(\hat{q})e^{\frac{-i\lambda}{\hbar}\hat{p}}</math> is equivalent to <math>F(\hat{q} + \lambda)</math> where <math>\lambda</math> is some c-number.
 
====Connection between classical and quantum mechanics====
 
There is a wonderful connection between Classical mechanics and Quantum Mechanics. The Hamiltonian is a concept in the frame of classical mechanics. In this frame, the Hamiltonian is defined as:
<math>H=\sum_kp_{k}\dot{q}_{k}-L(q,\dot{q},t).</math>
 
There are two possibilities.
 
1. If the Lagrangian <math>L(q,\dot{q},t)</math> does not depend explicitly on time the quantity H is conserved.<br/>
2. If the Potential and the constraints of the system are time independent, then H is conserved and H is the energy of the system.<br/>
 
It is clear from the above equation that:<br/>
<math>\dot{q}_k=\frac{\partial H}{\partial p_{k}}</math><br/>
<math>\dot{p}_k=-\frac{\partial H}{\partial q_{k}}</math><br/>
 
This pair of the equations is called Hamilton's equations of motions. The following object
 
<math>[A,B]=\sum_{k}\left(\frac{\partial A}{\partial q_{k}}\frac{\partial B}{\partial p_{k}}-
\frac{\partial A}{\partial p_{k}}\frac{\partial B}{\partial q_{k}}\right)
</math>
 
is called Poisson Bracket, and it has interesting properties. To see, let's calculate commutation relationships between coordinates and momenta.
 
<math>[p_i,p_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial p_j}{\partial p_{k}}-
\frac{\partial p_i}{\partial p_{k}}\frac{\partial p_j}{\partial q_{k}}\right)=0
</math>
 
<math>[q_i,q_j]=\sum_{k}\left(\frac{\partial q_i}{\partial q_{k}}\frac{\partial q_j}{\partial p_{k}}-
\frac{\partial q_i}{\partial p_{k}}\frac{\partial q_j}{\partial q_{k}}\right)=0
</math>
 
<math>[p_i,q_j]=\sum_{k}\left(\frac{\partial p_i}{\partial q_{k}}\frac{\partial q_j}{\partial p_{k}}-
\frac{\partial p_i}{\partial p_{k}}\frac{\partial q_j}{\partial q_{k}}\right)=\sum_{k}-\delta_{ik}\delta_{jk}=-\delta_{ij}.
</math>
 
This relations clearly shows how close are the quantum commutators with classical world. If we perform the following identification:
 
<math>\delta_{ij}\rightarrow i\hbar \delta_{ij}.</math>
 
Then we get quantum commutators. This identifications is called canonical quantization. As a final an important remark, the fact that we have classical commutators doesn't mean that we will have Heisenberg uncertainty relation for conjugate classical variables. This is because in classical mechanics the object of study are points (or body as a collection of points). In quantum mechanics we object of study is the state of a particle or system of particles - which describes the probability of finding a particle, and not it's exact, point-like, location of momentum.
 
====Hamiltonian====
 
In Quantum Mechanics an important property is the commutation of a given operator (let's say <math>\hat{O}</math>) and the Hamiltonian <math>\hat{H}</math>.  If  <math>\hat{O}</math> commutes with <math>\hat{H}</math>, then the eigenfunctions of <math>\hat{H}</math> can always be chosen to be simultaneous eigenfunctions of <math>\hat{O}</math>.  If <math>\hat{O}</math> commutes with the Hamiltonian and does not explicitly depend on time, then  <math>\hat{O}</math> is a constant of motion.
 
Prove:
If there is an operator <math>\hat{O}</math>, which <math>[\hat{H},\hat{O}]=0 </math>,
 
<math>\hat{H}|\psi_i\rangle=E_i|\psi_i\rangle </math>
 
<math>\hat{O}\hat{H}|\psi_i\rangle=E_i\hat{O}|\psi_i\rangle </math>
 
so, <math>\hat{H}\hat{O}|\psi_i\rangle=E_i\hat{O}|\psi_i\rangle </math>, eg. <math>\hat{O}|\psi_i\rangle </math> becomes another eigenfunction of energy <math>E_i</math>.
 
Case I: Non-degenerate energy spectrum.
 
<math>\hat{O}|\psi_i\rangle=a|\psi_i\rangle </math>, eg. <math>|\psi_i\rangle </math> is an eigenstate of <math>\hat{O}</math>.
 
Case II: degenerate energy spectrum.
 
<math>|\psi_i\rangle </math> and <math> \hat{O}|\psi_i\rangle </math> have the same energy.
 
So, there is always a linear combination of <math>|\psi_i\rangle </math>, such that the linear combination is an eigenstate of <math> \hat{O} </math>.
 
 
===Symmetry and Its Role in Quantum Mechanics===
 
Symmetry of any quantum mechanical state implies its invariance under certain mathematical transformations, examples being translation and rotation.
====Commutators & symmetry ====
 
We can define an operator called the parity operator, <math>\hat{P}</math> which does the following:
 
<math>\hat{P}f(x)=f(-x).</math>
 
The parity operator commutes with the Hamiltonian <math>\hat{H}</math> if the potential is symmetric, <math>\hat{V}(r)=\hat{V}(-r)</math>.  Since the two commute, the eigenfunctions of the Hamiltonian can be chosen to be eigenfunctions of the parity operator.  This means that if the potential is symmetric, the solutions can be chosen to have definite parity (even and odd functions).
 
===Ehrenfest's Theorem===
 
It is reasonable to expect the motion of a wave packet to agree with the motion of the corresponding classical particle whenever the potential energy changes by a negligible amount over the dimensions of the packet. If we mean by the position and momentum vectors of the packet the weighted averages or expectation values of these quantities, we can show that the classical and quantum mechanics always agree. A component of the velocity of the packet will be the time rate of change of the expectation value of that component of the position; since < x > depends only on the time and the x in the integrand
 
<math>\frac{\mathrm{d} }{\mathrm{d} t}< x > =  \frac{\mathrm{d} }{\mathrm{d} t}\int \psi \ast \left ( r \right )x\psi \left ( r \right )d^{3}r=\int \psi \ast\left ( r \right ) x\frac{\mathrm{d} }{\mathrm{d} x}\psi \left ( r \right )d^{3}r+\int \frac{\mathrm{d} }{\mathrm{d} t}\psi \ast \left ( r \right )x\psi \left ( r \right )d^{3}r</math>
 
Note that
                                   
 
<math>\frac{\partial }{\partial t\ }\psi =-\frac{\hbar}{2im}\triangledown ^{2}\Psi +\frac{V}{i\hbar}\psi</math>
 
and
 
<math>\frac{\partial }{\partial t\ }\psi\ast  =\frac{\hbar}{2im}\triangledown ^{2}\Psi\ast  -\frac{V}{i\hbar}\psi\ast</math> 
 
This may be simplified by substituting for the time derivatives of the wave function and its complex conjugate and canceling the term involving potential V where we continue to assume that V is real:
 
<math>\frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle=\int \psi \ast \left ( r \right )x\left (  \left ( \frac{-\hbar}{2im} \right )\triangledown ^{2}\psi +\frac{V}{i\hbar}\psi \right )d^{3}r+\int \left (  \left ( \frac{\hbar}{2im}\right )\triangledown ^{2 }\psi -\frac{V}{i\hbar}\psi\ast  \right )x\psi d^{3}r</math>
 
<math>\frac{i\hbar}{2m}\int \left \{ \psi \ast \left ( r \right )x\left ( \triangledown ^{2} \psi \right ) -\left ( \triangledown ^{2}\psi \ast  \right )x\psi \right \}d^{3}r</math>
 
We shall now use the identity
<math> \vec{\triangledown }.\left ( x\psi \vec{\triangledown  }\psi \ast \right )=\vec{\triangledown }\left ( x\psi  \right ).\vec{\triangledown }\psi \ast +x\psi \left ( \triangledown ^{2} \psi \ast \right )</math>
 
Consider two scalar function <math>x\psi</math> and <math>x\psi</math> that are continuous and differentiable in some volume V
bounded bounded by a surface S. Applying the divergence theorem to the vector field <math>x\psi\psi  \ast </math>(the
left hand side of the identity) we obtain
 
<math>\int_{S}x\psi \vec{\triangledown }\psi \ast.d\vec{S} =\int_{V}\left \{ \left ( x\psi  \right )\triangledown ^{2}\psi \ast +\left ( \vec{\triangledown }x\psi  \right ).\left ( \bar{\triangledown }\psi \ast  \right ) \right \}d^{3}r</math>
 
Therefore the second integral of first equation can be written as
<math>\int_{V}\left ( \triangledown ^{2}\psi \ast  \right )x\psi d^{3}r=-\int \left ( \vec{\triangledown }\psi \ast  \right ).\left ( \vec{\triangledown } x\psi \right )d^{3}r+\int_{S}\left ( x\psi \vec{\triangledown }\psi \ast  \right ).dS</math>
 
where the second integral of the normal component of xψ ψ ∗ over the infinite bounding surface A is zero because a wave packet ψ vanishes at great distances and hence
 
<math>\int_{V}\left ( \triangledown ^{2}\psi \ast  \right )x\psi d^{3}r=\int \psi \ast \triangledown ^{2}\left ( x\psi  \right )d^{3}r</math>
 
We again used the fact that the surface integral again vanishes, to obtain
 
<math>\int_{V}\left ( \triangledown ^{2}\psi\ast  \right )x\psi d^{3}r=\int_{V}\left ( \psi \ast \triangledown ^{2}\left ( x\psi  \right ) \right )d^{3}r</math>
Thus,
 
<math>\frac{\mathrm{d} }{\mathrm{d} t}\left \langle x \right \rangle=\frac{i\hbar}{2m}\int_{V}\psi \ast \left \{ x\triangledown ^{2}\psi -\triangledown ^{2}\left ( x\psi  \right ) \right \}d^{3}r</math>
<math>=\frac{1}{m}\left \langle p_{x} \right \rangle</math>
 
Since <math>\left \langle x \right \rangle</math> is seen always to be real number from the inherent structure of its definition. The
above equation shows quite incidentally that <math>\left \langle px \right \rangle</math> is real.
 
In similar fashion we can calculate the time rate of change of a component of the momentum of the particle as
 
<math>\frac{\mathrm{d} }{\mathrm{d} t}\left \langle p_{x} \right \rangle=-\frac{\hbar^{2}}{2m}\int \left \{ \left ( \triangledown ^{2}\psi\ast  \right )\frac{\partial \psi }{\partial x}d^{3}r-\psi \ast \triangledown ^{2}\left ( \frac{\partial \psi }{\partial x} \right )+\int V\psi \ast \frac{\partial \psi }{\partial x} d^{3}r-\psi \ast \frac{\partial V\psi }{\partial x} \right \}d^{3}r</math>
 
The integral containing laplacins can be transofrmed into a surface integral by Green’s theorem and write
 
<math>\int_{V}\left \{ \left ( \triangledown ^{2}\psi \ast  \right )\frac{\partial \psi }{\partial x} -\psi \ast \triangledown ^{2}\left ( \frac{\partial \psi }{\partial x} \right )\right \}d^{3}r=\int_{S}\left \{ \frac{\partial \psi }{\partial x} \vec{\triangledown }\psi \ast -\psi \ast \vec{\triangledown }\left ( \frac{\partial \psi }{\partial x} \right )\right \}.d\vec{S}
</math>
 
It is assumed that the last integral vanishes when taken over a very large surface S. (One can also show that volume integral involving the laplacian vanishes by doing integration by parts twice. For instance, we use the identity
 
<math>\vec{\triangledown }.\vec{\triangledown }\psi \ast \left ( \frac{\partial \psi }{\partial x} \right )=\triangledown ^{2}\psi \ast \frac{\partial \psi }{\partial x}+\vec{\triangledown }\psi \ast.\vec{\triangledown }\frac{\partial \psi }{\partial x}</math>
 
Therefore,
 
<math>\int_{V}\triangledown ^{2}\psi\ast  \frac{\partial \psi }{\partial x}d^{3}r=\int_{V}\vec{\triangledown }.\vec{\triangledown }\psi \ast \left ( \frac{\partial \psi }{\partial x} \right )-\int_{V}\vec{\triangledown }\psi \ast.\vec{\triangledown }\frac{\partial \psi }{\partial x}d^{3}r</math>
 
Using the Gauss’s theorem the first on he right hand side can be converted in to a surface integral over the surface that enclose the volume and hence it vanishes. Therefore,
 
<math>\int_{V}\triangledown ^{2}\psi \ast \frac{\partial \psi }{\partial x}d^{3}r=-\int_{V}\vec{\triangledown }\psi .\vec{\triangledown \frac{\partial \psi }{\partial x}}d^{3}r</math>
 
<math>=\int_{V}\psi \ast \triangledown ^{2}\frac{\partial \psi }{\partial x}d^{3}r</math>
 
Thus the Laplacian term vanishes resulting in
 
<math>\frac{\mathrm{d} }{\mathrm{d} t}\left \langle p_{x} \right \rangle=\int \left \{ V\psi \ast \frac{\partial \psi }{\partial x}d^{3}r-\psi \ast \frac{\partial V\psi }{\partial x} \right \}d^{3}r</math>
 
We can establish a general formula for the time derivative of the expectation value < F > of any operator F.
 
The time derivative of the expectation value of any operator F which may be explicitly time dependent, can be written as
 
<math>i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\left \langle \hat{F} \right \rangle=i\hbar\int \psi \ast \hat{F}\frac{\partial }{\partial t}\psi d^{3}r+i\hbar\int \frac{\partial }{\partial t}\psi \ast \hat{F}\psi d^{3}r+i\hbar\int \psi \ast \frac{\partial }{\partial t}\hat{F}\psi d^{3}r</math>
 
In the last step, we have used the Green’s theorem, and vanishing boundary surface terms. The potential energy is, of course, assumed to be real. Compactly, we may write the last result as
 
<math>i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\left \langle \hat{F} \right \rangle=\left \langle \hat{F}\hat{H}-\hat{H} \hat{F}\right \rangle+i\hbar\left \langle \frac{\partial }{\partial t}\hat{F} \right \rangle</math>
 
This formula is of the utmost importance in all facets of quantum mechanics.
 
====Generalized Heisenberg uncertainty relation====
 
If two opperators <math>\hat{A},\hat{B}</math> are Hermitian and
 
:<math>[\hat{A},\hat{B}]=i\hat{C}\;</math><br/>
then
<math>\frac{1}{4}\langle \hat{C}\rangle^2\leq\langle \left(\Delta \hat{A}\right)^2\rangle\langle \left(\Delta \hat{B}\right)^2\rangle</math>
 
Proof:
 
First recall <math>\Delta \hat{O} = \hat{O} - \langle \hat{O} \rangle </math> and note that <math>\Delta \hat{O} </math> is Hermitian if <math>\hat{O} </math> is.
 
Let <math>\alpha</math> be a real scalar and define <math> f(\alpha)</math> as such:
 
:<math> f(\alpha) = |( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle|^2 </math>.
 
So <math> f(\alpha) </math> is the norm squared of some arbitrary state vector after operating <math>(\alpha \Delta \hat{A} - i\Delta \hat{B})</math> on it. Hence by the positive semidefinite property of the norm:
:<math> f(\alpha) \geq 0 </math>
 
Proceeding to calculate this norm squared:
 
:<math>
\begin{align}
f(\alpha)&=\langle \psi |( \alpha \Delta \hat{A} - i\Delta \hat{B})^\dagger( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle\\
&=\langle \psi |( \alpha \Delta \hat{A} + i\Delta \hat{B})( \alpha \Delta \hat{A} - i\Delta \hat{B})|\psi \rangle\\
&=\langle \psi | \alpha^2 \left(\Delta \hat{A}\right)^2 -i\alpha [\Delta \hat{A},\Delta \hat{B}] + \left(\Delta \hat{B}\right)^2  |\psi \rangle\\
&=\langle \psi |\alpha^2 \left(\Delta \hat{A}\right)^2 |\psi \rangle + \langle \psi |\alpha \hat{C} |\psi \rangle + \langle \psi |\left(\Delta \hat{B}\right)^2  |\psi \rangle\\
&=\alpha^2\langle  \left(\Delta \hat{A}\right)^2  \rangle + \alpha \langle \hat{C} \rangle + \langle \left(\Delta \hat{B}\right)^2  \rangle\\
\end{align}</math>
 
Notice that <math>f(\alpha)</math> is a real valued ( expectation values of Hermitian operators are always real ) quadratic in real  <math>\alpha</math> which is always greater than or equal to zero. This implies that there are no real solutions for <math>\alpha</math> or there is exactly 1. This can be seen by attempting to solve for <math>\alpha</math> by using the "quadratic formula" :
 
:<math>
\alpha=\frac{-\langle  \hat{C} \rangle \pm \sqrt{ (\langle  \hat{C} \rangle)^2 -4 \langle  (\Delta \hat{A})^2  \rangle  \langle (\Delta \hat{B})^2  \rangle }}{2\langle  (\Delta \hat{A})^2  \rangle}
</math>
 
Now we exploit our insight about the number of real roots and see that the discriminant, the term under the square root, must be either 0 ( yielding 1 real solution for <math>\alpha</math>) or negative ( yielding 0 real solutions <math>\alpha</math>). Stated more succinctly:
 
:<math>
(\langle  \hat{C} \rangle)^2 -4 \langle  \left(\Delta \hat{A}\right)^2  \rangle  \langle \left(\Delta \hat{B}\right)^2  \rangle \leq 0
</math>
 
<math>\frac{(\left \langle  \hat{C} \right \rangle )^{2}}{4} \leq \left \langle {(\Delta \hat{A})^{2}} \right \rangle \left \langle {(\Delta \hat{B})^{2}} \right \rangle </math>
 
which immediately implies what was to be proved.
 
=== Heisenberg and interaction picture: Equations of motion for operators ===
There are several ways to mathematically approach the change of a quantum mechanical system with time. The Schrodinger picture considers wave functions which change with time while the Heisenberg picture places the time dependence on the operators and deals with wave functions that do not change in time.  The interaction picture, also known as the Dirac picture, is somewhere between the other two, placing time dependence on both operators and wave functions. The two "pictures" correspond, theoretically, to the time evolution of two different things.  Mathematically, all methods should produce the same result.
 
==== Definition of the Heisenberg Picture ====
 
The time evolution operator is at the heart of the "evolution" of a state, and is the same in each picture.  The evolution operator is obtained from the time dependent Schrödinger equation after separating the time and spatial parts of the wave equation:
 
<math>
i\hbar \frac{\partial}{\partial t} U(t,t_0) = H U(t,t_0)
</math>
 
 
The solution to this differential equation depends on the form of .
 
If we know the time evolution operator, <math>U</math>, and the initial state of a particular system, all that is needed is to apply  to the initial state ket.  We then obtain the ket for some later time.
 
<math>
U|\alpha(0)\rangle= |\alpha(t)\rangle.
</math>
 
Therefore, if we know the initial state of a system, we can obtain the expectation value of an operator, A, at some later time:
 
<math>
\langle A \rangle(t) = \langle \alpha(t)|A|\alpha(t)\rangle= \langle \alpha(0)|U^{\dagger}AU|\alpha(0)\rangle.</math>
 
We can make a redefinition by claiming that
 
<math>
A_H(t) = U^{\dagger}AU</math>
 
and taking as our state kets the time independent, initial valued state ket <math>|\alpha(0)\rangle</math>.
 
This formulation of quantum mechanics is called the Heisenberg picture. In this picture, operators evolve in time and state kets do not.  (Note that the difference between the two pictures only lies in the way we write them down).
 
In classical physics we obviously have time evolution of a system - our observable (position or angular momentum or whatever variable we are considering) changes in a way dictated by the classical equations of motion.  We do not talk about state kets in classical mechanics.  Therefore, the Heisenberg, where the operator changes in time, is useful because we can see a closer connection to classical physics than with the Schrödinger picture.
 
==== Comparing the Heisenberg Picture and the Schrodinger Picture ====
 
As mentioned above, both the Heisenberg representation and the Schrodinger representation give the same results for the time dependent expectation values of operators. 
 
In the Schrodinger picture, in which we are most accustomed, the states change over time while the operators remain constant. In other words, the time dependence is carried by the state operators. When the Hamiltonian is independent of time, it is possible to write:
 
<math>
|\Psi(0)\rangle
</math>
 
as the state at t = 0. 
 
At another time t, this becomes
 
<math>
|\Psi(t)\rangle = e^{-iHt/\hbar} |\Psi(0)\rangle
</math>,
 
which in this form solves the Schrodinger equation
 
<math>
i\hbar\frac{\partial|\Psi(t)\rangle}{\partial t}=H|\Psi(t)\rangle
</math>.
Conversely, in the Heisenberg picture to find the time dependent expectation of a given operator, the states remain constant while the operators
change in time.  In this case, you can write the time dependent operator as:
 
<math>
A(t) = e^{iHt/\hbar}Ae^{-iHt\hbar}
</math>,
 
which means the expectation value is,
 
<math>
\langle A \rangle_t = \langle\Psi(0)| A(t)| \Psi(0)\rangle
</math>.
 
==== The Heisenberg Equation of Motion ====
In the Heisenberg picture, the quantum mechanical observables change in time as dictated by the Heisenberg equations of motion.  We can study the evolution of a Heisenberg operator by differentiating equation 2 with respect to time:
 
:<math>
\begin{align}
\frac{d}{dt} A_H  &= \frac{\partial U^{\dagger}}{\partial t} A U + U^{\dagger}\frac{\partial A}{\partial t}  U + U^{\dagger} A \frac{\partial U}{\partial t} \\
&= \frac{-1}{i\hbar} U^{\dagger} HUU^{\dagger}A U + \left(\frac{\partial A}{\partial t}\right)_H + U^{\dagger} AUU^{\dagger}HU \frac{1}{i\hbar} \\
&= \frac{1}{i\hbar}\left(\left[A,H\right]\right)_H+\left(\frac{\partial A}{\partial t}\right)_H.
\end{align}
</math>
 
The last equation is known as the Ehrenfest Theorem.
 
For example, if we have a hamiltonian of the form,
 
<math>H=\frac{p^2}{2m}+V(r),</math>
 
then we can find the Heisenberg equations of motion for p and r. 
 
The position operator in 3D is:
<math>i\hbar\frac{d \hat{r}(t)}{dt} = \left[ \hat{r}(t),H\right] </math> 
 
Since the Hamiltonian as given above has a V(r,t) term and a momentum term the two commuators are as follows:
 
<math> \left[ \hat{r}(t),V(r,t)\right] = 0 </math>
 
<math> \left[ \hat{r}(t), \frac{p(t)^{2}}{2m}\right] = \frac{i\hbar p(t)}{m} </math>
 
this yields
<math> \frac{d \hat{r}(t)_{H}}{dt} = \frac{ \hat{p}(t)_{H}}{m} </math>.
 
 
To find the equations of motion for the momentum you need to evaluate  <math> \left[ \hat{p}(t), V(r(t))\right] </math>,
 
which equals, <math> \left[\hat{p}, V(r)\right] = -i\hbar \nabla V(r) </math>.
 
This yields <math>\frac{d \hat{p}_H}{dt}=-\nabla V(\hat{r}_H).</math>
 
These results are, of course, what we would expect if we only knew classical physics, providing another example of how the Heisenberg picture is more transparently similar to classical physics.
 
In particular, if we apply these equations to the Harmonic oscilator with natural frequency <math>\omega=\sqrt{\frac{k}{m}}\!</math>
 
<math>
\frac{d\hat x_H }{dt}=\frac{\hat p_H }{m}</math><br/>
 
<math>
\frac{d\hat p_H }{dt}=-k{\hat x_H}.
</math>
 
we can solve the above equations of motion and find
 
<math>
\hat x_H(t)=\hat x_H(0)\cos(\omega t)+\frac{\hat p_H(0)}{m\omega}\sin(\omega t)</math><br/>
 
<math>
\hat p_H(t)=\hat p_H(0)\cos(\omega t)-\hat x_H(0)m\omega\sin(\omega t).\!</math>
 
It is important to stress that the above oscillatory solution is for the position and momentum ''operators''.
Also, note that <math>\hat x_H(0)\!</math> and <math>\hat p_H(0)\!</math> correspond to the time independent operators <math> \hat x\!</math> and <math> \hat p\!</math>.
 
==== The Interaction Picture ====
 
The interaction picture is a hybrid between the Schrödinger and Heisenberg pictures.  In this picture both the operators and the state kets are time dependent.  The time dependence is split between the kets and the operators - this is achieved by first splitting the Hamiltonian into two parts:  an exactly soluble, well known part, and a less known, more messy "peturbation".
 
 
If we want to look at this splitting process, we can say that <math>\text{H=H}_{o}+V(t)</math>.
 
<math>\text{A}_{H}=e^{\frac{i}{\hbar }Ht}A_{s}e^{\frac{-i}{\hbar }Ht}</math>
 
<math>\text{A}_{H}=e^{\frac{i}{\hbar }Ht}A_{s}e^{\frac{-i}{\hbar }Ht}</math>
 
 
<math>\to\!</math> Equation of motion :
 
<math>\text{i}\hbar \frac{\partial }{\partial t}\left |{\alpha ,t} \right \rangle_{I}=-H_{o}e^{\frac{i}{\hbar }H_{o}t}\left |{\alpha ,t} \right \rangle_{S}+e^{\frac{i}{\hbar }H_{o}t}(H_{0}+V)\left |{\alpha ,t} \right \rangle_{S}</math>
 
 
<math>e^{\frac{i}{\hbar }H_{o}t}Ve^{\frac{-i}{\hbar }H_{o}t}\text{ . }e^{\frac{i}{\hbar }H_{o}t}\left |{\alpha ,t} \right \rangle_{S}</math>
 
 
If we call firstpart "<math>V_{I}\!</math>" and second part "<math>\left |{\alpha ,t} \right \rangle_{I}</math>"  ,
 
it turns out :
 
<math>\text{=}V_{I}\left |{\alpha ,t} \right \rangle_{I}</math>
 
so;
 
<math>\text{i}\hbar \frac{\partial }{\partial t}\left |{\alpha ,t} \right \rangle_{I}=V_{I}\left |{\alpha ,t} \right \rangle_{I}</math>
 
<math>\frac{d}{dt}A_{I}(t)=\frac{1}{i\hbar }\left [{A_{I},H_{o}} \right ]+\frac{\partial A_{I}}{\partial t}</math>
 
 
and this equation of motion evolves with <math>H_{o}\!</math>.
 
=== Feynman path integrals ===
 
[[Image:pathintegral.gif|thumb|650px|]] 
 
The path integral formulation was developed in 1948 by Richard Feynman.  The path integral formulation of quantum mechanics is a description of quantum theory which generalizes the action principle of classical mechanics. It replaces the classical notion of a single, unique trajectory for a system with a sum, or functional integral, over an infinity of possible trajectories to compute a quantum amplitude.
 
The classical path is the path that minimizes the action.
 
For simplicity, the formalism is developed here in one dimension.
 
Using the path integral method, the propagator, <math>U(t)</math>, is found directly.
The amplitude for a particle to start at <math>x_0</math> at <math>t=0</math>  and end at <math>x</math> at t can be expressed as a path integral
 
<math>
\langle x|\hat{U}(t)|x_0\rangle=\int_{x_0}^{x}D x(t') e^{iS[x(t')]}
</math>
 
Where <math>S[x(t)]\!</math> is the action for the the path <math>x(t')</math>.
 
The action is given by the time integral of the Lagrangian, just as in classical mechanics
<math>
S[x(t')]=\int_0^t dt' \mathcal{L}[x(t'),\dot{x}(t'),t']
</math>
 
Where
<math>
\mathcal{L}[x(t'),\dot{x}(t'),t']=\frac{1}{2}m\dot{x}^2(t')-V(x(t'),t')
</math>
is the Lagrangian.
Knowing the propagator, we can calculate the probability that a particle in state <math>x_0\!</math> at t=0 will be in state <math>x\!</math> at time t by taking the absolute value squared.
 
Explicit evaluation of the path integral for the harmonic oscillator can be found here [[Image:FeynmanHibbs_H_O_Amplitude.pdf]]
 
=== Problems ===
4.1) Prove that there is a unitary operator <math>\tilde{U}(a)</math>, which is a function of <math>\hat p =\frac{\hbar}{i}\frac{d}{dx}</math>, such that for some wavefunction <math>\psi(x)\!</math>, <math>\tilde{U}(a)\psi(x) = \psi(x + a)</math>.
 
[[Phy5645:Problem 4.1 Solution|Solution to 4.1]]
 
== Discrete eigenvalues and bound states. Harmonic oscillator and WKB approximation ==
=== Harmonic oscillator spectrum and eigenstates===
[[Image:chp_energylevels.jpg|thumb|650px|]]
 
1-D harmonic oscillator is a particle moving under the harmonic oscillator potential with the form:
<math>V(x)=\frac{1}{2}k x^2</math> where <math>k\!</math> is the "spring constant".
 
We can see that <math>V(x)\rightarrow \infty </math> as <math>x\rightarrow \pm\infty </math>, therefore, the wave functions must vanish at infinity for any value of the energy. Consequently, all stationary states are bound, the energy spectrum is discrete and non-degenerate. Furthermore, because the potential is an even function, the parity operator commutes with Hamiltonian, hence the wave functions will be even or odd.
 
The energy spectrum and the energy eigenstates can be found by either algebraic method using raising and lowering operators or by analytic method.
 
The Hamiltonian of a 1-D harmonic oscillator is: 
 
<math>H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}k x^2 </math>
 
<math>H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}m\omega^2 x^2</math> , where <math>\omega \!</math> is the natural frequency of the oscillator, and <math>\omega = \sqrt{\frac{k}{m}}</math>.
 
It will be easy to memorize how to construct lowering and raising operator by factorizing  and rewriting  as follows:
<math>
H=\hbar\omega\left(-\frac{\hbar}{2m\omega}\frac{\partial^2}{\partial x^2}+\frac{1}{2\hbar}m\omega x^2\right)
</math>
 
Then we define:
 
<math>\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}x+\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}</math> as the lowering operator, and <br/>
<math>\hat{a}^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}</math> as the raising operator. The Hamiltonian can now be written as:
 
<math>H=\hbar\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)</math>
 
One way to distinguish <math>\hat{a}\!</math> from <math>\hat{a}^{\dagger}</math> is to remember that the ground state wave function is a Gaussian function and  <math>\hat{a}\!</math> will annihilate this state.
Given the lowering and raising operators, the following commutation relation can easily be shown:
<math>[\hat{a},\hat{a}^{\dagger}]=1.</math>
 
Now let  <math>\hat{a}\hat{a}^{\dagger} = \hat{n}</math>, and write the eigenstate equation of <math>\hat{n}</math> as <math>\hat{n} |n\rangle = n |n\rangle </math>.
Now, let's see how <math>\hat{a}\!</math> and <math>\hat{a}^{\dagger}\!</math> act on an energy eigenstate <math>|\Psi\rangle\!</math>:
For <math>\hat{a}\!</math>:
 
<math>H\hat{a} |\Psi\rangle = \left(E-\hbar\omega\right)|\Psi\rangle </math>
 
This means that <math>\hat{a} |\Psi\rangle</math> is also an energy eigenstate but correspoding to a lower energy,
<math>E-\hbar\omega\!</math>. Therefore <math>\hat{a}\!</math> is so-called the lowering operator.
 
Similarly,
 
<math>H\hat{a}^{\dagger}|\Psi\rangle = \left(E+\hbar\omega\right)|\Psi\rangle </math>
 
and <math>\hat{a}^{\dagger}\!</math> is so-called the raising operator.
 
So, starting from any energy eigenstates, we can construct all other energy eigenstates by applying <math>\hat{a}\!</math> or <math>\hat{a}^{\dagger}\!</math> repeatedly. Although there is no limit in applying <math>\hat{a}^{\dagger}\!</math>, there is a limit in applying <math>\hat{a}\!</math>. The process of lowering energy must stop at some point, since <math>E>0</math>. For the eigenstate of lowest energy  <math>|\Psi_0\rangle</math> (the ground state),
we have:
 
<math>\hat{a}|\Psi_0\rangle=0\!</math><br/>
<math>\Rightarrow \left( \sqrt{\frac{m\omega}{2\hbar}}x+\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\right) \Psi_0(x)=0\!</math>
 
This is a first order ordinary differential equation, which can be easily solved, and the result is as follows:
 
<math>\Psi_0(x)=A e^{-\frac{m\omega}{2\hbar}x^2}</math>
 
where <math>A\!</math> is a constant, which can be determined from the normalization condition:
 
<math>1=\int_{-\infty}^{\infty}|\Psi_0(x)|^2 dx = \int_{-\infty}^{\infty}dx A^2 e^{-\frac{m\omega}{\hbar}x^2}=
A^2\sqrt{\frac{\pi\hbar}{m\omega}}\Rightarrow A=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}</math>
 
Thus, normalized ground state wave function is: 
 
<math>\Psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^2}</math>
 
The energy spectrum of 1-D harmonic oscillator is: 
 
<math>\ {E}_{n}=\hbar\omega\left(n+\frac{1}{2}\right); n=0,1,2,\ldots</math> And it is clear that the ground state energy is:
<math>\ {E}_{0}=\frac{1}{2}\hbar\omega </math>
 
'''''Excited state wave function'''''<br/>
Energy eigenstates with <math>E>0\!</math> are called excited states. By applying <math>a^{\dagger}</math> repeatedly
and after normalization process we obtain the wave function for excited states as follows (more details about the normalization process can be found  in Griffiths, Introduction to Quantum Mechanics, 2nd Ed. pg 47): 
 
<math>|\Psi_n\rangle=\frac{(\hat{a}^{\dagger})^n}{\sqrt{n!}}|\Psi_0\rangle\!</math>
 
In the position representation
 
<math>\langle x|\Psi_n\rangle=\left(\frac{m\omega}{\pi\hbar}\right)^{\frac{1}{4}}\frac{1}{\sqrt{2^nn!}}H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-\frac{m\omega}{2\hbar}x^2}\!</math><br/>
 
 
where <math>H_n(\xi)=(-1)^n e^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2}</math> is the Hermite polynomial.
 
 
In the momentum representation the solution looks similar. The raising and lowering operators are defined:
 
<math>\hat{a}^{\dagger}=i\sqrt{\frac{m\omega \hbar}{2}}\frac{d}{dp}-\frac{ip}{\sqrt{2\hbar m \omega}}</math>
 
<math>\hat{a}=i\sqrt{\frac{m\omega \hbar}{2}}\frac{d}{dp}+\frac{ip}{\sqrt{2\hbar m \omega}}</math>
 
The ground state is found by requiring the lowering operator acting on a state to annihilate it:
 
<math>\left( i\sqrt{\frac{m\omega \hbar}{2}}\frac{d}{dp}+\frac{ip}{\sqrt{2m \omega \hbar}}\right) \Psi_0(p)=0</math>
 
We obtain the ground state by solving the differential equation and normalizing it:
<math>\Psi_0(p)=\frac{1}{(m\omega \hbar \pi)^{1/4}}e^{\frac{-p^2}{2m\omega\hbar}}</math>
 
Applying the raising operator repeatedly to this ground state and applying the correct additional normalization factor produces all of the excited states. In terms of the Hermite polynomials these are:
 
<math>\Psi_n(p)=\frac{(-i)^n}{\sqrt{2^n n!}}\left(\frac{1}{\pi m\omega \hbar}\right)^{1/4}H_n\left(\frac{p}{\sqrt{m\omega\hbar }}\right)e^{\frac{-p^2}{2m\omega \hbar}}</math>
 
Note the appearance of the imaginary unit <math>i</math> which is not present in the position representation of these states.
 
 
There are two parts in the wave function of excited states: the Gaussian function part and the Hermite polynomial part. The former accounts for the behavior of the wave function at long distances, while the later accounts for the behavior of the wave function at short distance and the number of nodes of the wave function.
 
Homework Problem 1 :[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HO_problem1]
 
Homework Problem 2 :[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HO_problem2]
 
=== Coherent states ===
 
The general states of an harmonic oscillator can be expressed as a superpostion of the energy eigenstates <math>|n\rangle\!</math>. A class of states that is of particular importance consists of the eigenstates of non-Hermitian lowering operator <math>a\!</math>, with eigenvalue <math>\alpha\!</math>:
 
<math>a|\alpha\rangle=\alpha|\alpha\rangle\!</math>
where <math>\alpha\!</math> can be any complex number.
Such states are called coherent states. The term coherent reflects their important role in optics and quantum electronics.
The following are some properties of coherent states.
Note that it is not possible to construct an eigenstate of <math>a^{\dagger}</math> because
<math>a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle</math>.
I. Coherent states construction.
 
:<math>|\alpha\rangle=\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle=e^{\alpha a^+}|0\rangle</math>
 
:<math>a|\alpha\rangle=\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}a|n\rangle=\sum_{n=1}^{+\infty}
\frac{\alpha^n}{\sqrt{n!}}\sqrt{n}|n-1\rangle=\sum_{n=1}^{+\infty}\frac{\alpha^n}{\sqrt{(n-1)!}}|n-1\rangle=
\alpha\left(\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle\right)=\alpha|\alpha\rangle</math>
II. Coherent states normalization.
<math>|\alpha\rangle=Ne^{\alpha a^{\dagger}} |0\rangle </math>
where <math>N</math> is normalization constant.
 
:<math>1=\langle\alpha|\alpha\rangle=\langle 0|Ne^{\alpha^*a} Ne^{\alpha a^{\dagger}} |0\rangle = N^2\langle 0|e^{\alpha^*a} e^{\alpha a^{\dagger}} |0\rangle </math>
 
For any operators A and B which both commute with their commutator, we have: 
                                               
<math>e^A e^B = e^{A+B} e^{\frac{1}{2}[A,B]} </math>
 
 
and similarly,  <math>e^B e^A = e^{B+A} e^{\frac{1}{2}[B,A]} = e^{A+B} e^{-\frac{1}{2}[A,B]}</math>
 
 
therefore:    <math>e^A e^B = e^B e^A e^{[A,B]}\!</math>
 
Apply this result for <math>A=\alpha ^* a\!</math> and <math>B=\alpha a^{\dagger}\!</math> ( A and B both commute with their commutator because
<math>[A,B]=|\alpha|^2)\!</math>, we have:
 
:<math>1=\langle\alpha|\alpha\rangle = N^2\langle 0|e^{\alpha^*a} e^{\alpha a^{\dagger}} |0\rangle </math>
 
:<math>
\begin{align}
N^2\langle 0|e^{\alpha a^{\dagger}} e^{\alpha^* a} e^{[\alpha^*a,\alpha a^{\dagger}]} |0\rangle
&= N^2e^{|\alpha|^2}\langle 0|e^{\alpha a^{\dagger}} e^{\alpha^* a} |0\rangle \\
&= N^2e^{|\alpha|^2}\langle 0|e^{\alpha a^{\dagger}} |0\rangle \\
&= N^2e^{|\alpha|^2}\langle 0|0\rangle \\
&= N^2e^{|\alpha|^2}
\end{align}
</math>
 
:<math>\rightarrow N=e^{-\frac{1}{2}|\alpha|^2}</math>
 
:<math>\rightarrow \mbox{Normalized coherent states:} |\alpha \rangle = e^{-\frac{1}{2}|\alpha |^2 } e^ {\alpha a^{\dagger} }|0 \rangle </math>
III. Inner product of two coherent states
There is an eigenstate <math>|\alpha\rangle\!</math>
of lowering operator <math>a\!</math> for any complex number <math>\alpha\!</math>. Therefore, we have a set of coherent states. This is NOT an orthogonal set.
Indeed, the inner product of two coherent states <math>|\alpha\rangle\!</math> and <math>|\beta\rangle\!</math> can be calculated as follows:
 
:<math>
\begin{align}
\langle \beta|\alpha \rangle &= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}\langle 0|e^{\beta^*a} e^{\alpha a^+} |0\rangle \\
&= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}\langle 0|e^{\alpha a^+} e^{\beta^* a} e^{[\beta^*a,\alpha a^+]} |0\rangle \\
&= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*}\langle 0|e^{\alpha a^+} e^{\beta^* a} |0 \rangle \\
&= e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*}
\end{align}
</math>
 
<math>\rightarrow |\langle \beta|\alpha \rangle |^2 = e^{-|\alpha-\beta|^2}</math>
 
Hence, the set of coherent states is not orthogonal and the distance <math>|\alpha-\beta|\!</math> in a complex plane measures the degree to which the two eigenstates are 'approximately orthogonal'.
 
=== Feynman path integral evaluation of the propagator ===
 
The propagator for harmonic oscillator can be evaluated as follows:
 
<math><x|\hat{U}(t,0)|x_0>=e^{\frac{i}{\hbar}S}\int_{y(0)=0}^{y(t)=0}D[y(t')]e^{\frac{i}{\hbar}\int_{0}^{t}(\frac{1}{2}my'^2-\frac{1}{2}ky^2)dt'}                                              </math>
 
where <math>y(t')</math> is the deviation of possible trajectories about the classical trajectory.
==== Saddle point action====
The classical action <math>S</math> can be evaluated as follows:
 
<math>S=\int_{0}^{t}(KE-PE)dt </math>
 
Where <math>KE\!</math> is the kinetic engergy and <math>PE\!</math> is the potential energy.
Equation of motion for harmonic oscillator:
<math>x_{cl}(t')=A\cos(\omega t')+B\sin(\omega t')\!</math><br/>
<math>A\!</math> and <math>B\!</math> are constants.
At <math>t'=0\!</math> (starting point),<math>x_{cl}(0)=x_0\rightarrow A=x_0</math>.
 
At  <math>t'=t\!</math> (final point), <math>x_{cl}(t)=x\rightarrow B=\frac{x-x_0\cos(\omega t)}{\sin(\omega t)}                                                                                . </math>
 
Substitute:
 
<math>x_{cl}(t')= x_0\cos(\omega t')+\frac{x-x_0\cos(\omega t)}{\sin(\omega t)}\sin(\omega t')
\Rightarrow \frac{dx_{cl}(t')}{dt'}= -\omega x_0\sin(\omega t')+\omega \frac{x-x_0\cos(\omega t)}{\sin(\omega t)}\cos(\omega t')</math>                                                        <br/>
<math>KE= \frac{1}{2}m\left(\frac{dx_{cl}}{dt}\right)^2=\frac{1}{2}m\left[-\omega x_0\sin(\omega t')+\omega \frac{x-x_0\cos(\omega t)}{\sin(\omega t)}\cos(\omega t')\right]^2</math><br/>
<math>PE= \frac{1}{2}k(x_{cl}(t'))^2=\frac{1}{2}k\left[x_0\cos(\omega t')+\frac{x-x_0\cos(\omega t)}{\sin(\omega t)}\sin(\omega t')\right]^2</math>                                        <br/> 
Substituting, integrating from time 0 to time <math> t \! </math> and simplifying, we get:
 
<math>S=S(t,x,x_0)=\frac{m\omega}{2\sin(\omega t)}((x^2+x_0^2)\cos(\omega t)-2xx_0)</math><br/>
 
==== Harmonic fluctuations ====
Now, let's evaluate the path integral:
<math>A=A(t)=\int_{y(0)=0}^{y(t)=0}D[y(t')]e^{\frac{i}{\hbar}\int_{0}^{t}(\frac{1}{2}my'^2-\frac{1}{2}ky^2)dt'}</math>                                                                      <br/>
 
Note that the integrand is taken over all possible trajectory starting at point <math>x_0</math> at time <math>t'=0</math>,
ending at point <math>x</math> at time <math>t'=t</math>.
Expanding this integral,
 
<math>A(t)=\left(\frac{m}{2\pi i \hbar}\right)^{\frac{N}{2}}\int_{-\infty}^{\infty} dy_1\ldots dy_{N-1}
e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}y^2_{N-1}-
\frac{\Delta t}{2}ky^2_{N-1}\right)\right]}e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}(y_{N-1}-y_{N-2})^2-
\frac{\Delta t}{2}ky^2_{N-2}\right)\right]}\ldots
e^{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}y^2_{1}-
\frac{\Delta t}{2}ky^2_{1}\right)\right]}
</math>
where <math>N\Delta t=t\!</math>.
Expanding the path trajectory in Fourier series, we have
<math>
y(t')=\sum_n a_n \sin\left(\frac{n\pi t'}{t}\right)
</math>
we may express <math>A(t)\!</math> in the form
 
<math>A(t)=C\int_{-\infty}^{\infty} da_1\ldots da_{N-1}
\exp{\left[\sum_{n=1}^{N-1}\frac{im}{2\hbar}\left(\left(\frac{n\pi}{t}\right)^2-
\omega^2\right)a^2_n\right]}
</math>
 
where C is a constant independent of the frequency which comes from the Jacobian of the transformation.
The important point is that it does not depend on the frequency <math>\omega\!</math>.
Thus, evaluating the integral of,
 
<math>A(t)=C'\prod_{n=1}^{N-1}\left[\left(\frac{n\pi}{t}\right)^2-\omega^2\right]^{-\frac{1}{2}}=
C'\prod_{n=1}^{N-1}\left[\left(\frac{n\pi}{t}\right)^2\right]^{-\frac{1}{2}}
\prod_{n=1}^{N-1}\left[1-\left(\frac{\omega t}{n\pi}\right)^2\right]^{-\frac{1}{2}}
</math>
where C' is a constant directly related to C and still independent of the frequency of motion.
Since the first product series in this final expression is also independent of the frequency of motion,
we can absorb it into our constant C' to have a new constant, C''. Simplifying further,
 
<math>A(t)=C''\sqrt{\frac{\omega t}{\sin(\omega t)}}
</math>
 
In the limit <math>\omega\rightarrow 0</math>, we already know that
 
<math>C''=\sqrt{\frac{m}{2\pi i \hbar t}}
</math>
                                           
Thus,
 
<math>A(t)=\sqrt{\frac{m}{2\pi i \hbar t}}\sqrt{\frac{\omega t}{\sin(\omega t)}}=
\sqrt{\frac{m}{2\pi i \hbar \sin(\omega t)}}
</math>
and
 
<math><x|\hat{U}(t,0)|x_0>=\sqrt{\frac{m}{2\pi i \hbar \sin(\omega t)}}
e^{\frac{i}{\hbar}\left(\frac{m\omega}{2sin(\omega t)}((x^2+x_0^2)cos(\omega t)-2xx_0)\right)}
</math>
For a more detailed evaluation of this problem, please see Barone, F. A.; Boschi-Filho, H.; Farina, C. 2002. "Three methods for calculating the Feynman propagator". American Association of Physics Teachers, 2003. Am. J. Phys. 71 (5), May 2003. pp 483-491.
 
=== Motion in electromagnetic field ===
====Gauge====
Gauge theory is a type of field theory in which the Lagrangian is invariant under a certain continuous group of local transformations.
 
Given a distribution of charges and current, and appropriate boundary conditions, the electromagnetic field is unique. However, the electromagnetic potential <math> A^{\mu}\!</math> is not unique. The Maxwell equations can be expressed by electromagnetic field tensor <math> F^{\mu\nu} \!</math>, which is defined by
:<math> F^{\mu\nu}=\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu} </math>.
 
If we set
:<math> A'^{\mu}=A^{\mu}+\partial^{\mu}\chi </math>,
 
then
:<math>
\begin{align}
F'^{\mu \nu } &= \partial ^\mu A'^\nu  - \partial ^\nu A'^\mu \\
&= \partial^{\mu}(A^{\nu}+\partial^{nu}\chi)-\partial^{\nu}(A^{\mu}+\partial^{\mu}\chi) \\
&= \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}+(\partial^{\mu}\partial^{\nu}-\partial^{\nu}\partial^{\mu})\chi \\
&= F^{\mu \nu }
\end{align}
</math>,
eg. the form of Maxwell equations will not change. So, we have a freedom <math> \partial\chi \!</math>, which is called Gauge Freedom here.
 
For the magnetic field case, we can check for gauge invariance:
 
:<math> \left. \left. \frac{{(p - \frac{e}{c}A - \partial \chi )^2 }}{{2m}}\right|\varphi \right\rangle = E|\varphi \rangle </math>,
 
Let <math>
|\varphi \rangle = e^{\frac{i}{\hbar }\frac{e}{c}\chi } |\phi \rangle
</math>,
 
the form <math>
\frac{{(p - \frac{e}{c}A)^2 }}{{2m}}|\phi \rangle = E|\phi \rangle
</math> will not change.
 
Usually, we use two gauges in magnetic field. One is the Laudau Gauge: <math>
A(r) = ( - yB,0,0)\!</math>, the other is the Symmetric Gauge: <math>
A(r) = \frac{1}{2}( - yB,xB,0)\!</math>.
 
We choose Laudau Gauge in the following calculation.
 
====Motion in electromagnetic field====
 
 
 
The Hamiltonian of a particle of charge <math>e\!</math> and mass <math>m\!</math>
in an external electromagnetic field, which may be time-dependent, is given as follows:
 
:<math> H=\frac{1}{2m}\left(\mathbf{p}-\frac{e}{c}\bold A(\bold r,t)\right)^2+e\phi(\bold r,t)</math>
 
 
where <math> \bold{A(\bold r,t)} \!</math> is the vector potential and <math>{\phi(\bold r,t)}\!</math> is the Coulomb potential of the electromagnetic field. This is precisely the classical Hamiltonian for such a problem, with the transformation from canonical momentum to the momentum operator.
 
Let's find out the Heisenberg equations of motion for the position and velocity operators.
For position operator<math>\bold r\!</math>, we have:
 
:<math>
\begin{align}
\frac{d\bold r}{dt} &= \frac{1}{i\hbar} \left[\bold r,H \right] \\
&= \frac{1}{i\hbar} \left[ \bold r, \frac{1}{2m} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2 + e\phi(\bold r,t)\right] \\
&= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)^2\right] \\
&= \frac{1}{2im\hbar} \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right]\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)\right] \\
&= \frac{1}{2im\hbar} \left[\bold r, \bold p\right] \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p - \frac{e}{c}\bold A(\bold r,t)\right) \left[\bold r, \bold p\right] \\
&= \frac{1}{2im\hbar}i\hbar \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right) + \frac{1}{2im\hbar} \left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)i\hbar \\
&= \frac{1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right),
\end{align}
</math>
where (<math>\bold r \!</math> does not depend on <math>t \!</math> explicitly)
is the equation of motion for the position operator <math>\bold r</math>.
This equation also defines the velocity operator <math>\bold v</math>:
                                               
:<math>\bold v= \frac {1}{m}\left(\bold p-\frac{e}{c}\bold A(\bold r,t)\right)</math>                                                                             
 
The Hamiltonian can be rewritten as:
 
:<math>H=\frac {m}{2}\bold v \cdot \bold v+e\phi</math>
 
Therefore, the Heisenberg equation of motion for the velocity operator is:
 
:<math>
\begin{align}
\frac{d\bold v}{dt} &=\frac {1}{i\hbar}\left[\bold v,H\right]+\frac{\partial \bold v}{\partial t} \\
&= \frac {1}{i\hbar}\left[\bold v,\frac{m}{2}\bold v \cdot \bold v\right]+\frac {1}{i\hbar}\left[\bold v,e\phi\right]-\frac{e}{mc} \frac{\partial \bold A}{\partial t}
\end{align}
</math>
(Note that <math>\bold p\!</math> does not depend on <math>t\!</math>  expicitly)
 
Let's use the following commutator identity:
 
:<math>\left[\bold v,\bold v \cdot \bold v\right]=\bold v \times \left(\bold v \times \bold v\right)-\left(\bold v \times \bold v\right) \times \bold v </math> 
 
Substituting, we get:
 
:<math>
\frac{d\bold v}{dt} = \frac{1}{i\hbar} \frac{m}{2}
\left(\bold v \times (\bold v \times \bold v) - (\bold v \times \bold v) \times \bold v \right) + \frac{1}{i\hbar} e[\bold v,\phi] - \frac{e}{mc} \frac{\partial \bold A}{\partial t}</math>
 
Now let's evaluate <math>\bold v \times \bold v \!</math> and <math>[\bold v,\phi] \!</math>:
 
:<math>
\begin{align}
(\bold v \times \bold v)_i &= \epsilon_{ijk} v_j v_k \\
&= \epsilon_{ijk}\frac{1}{m} \left(p_j-\frac{e}{c}A_j(\bold r,t)\right)
\frac{1}{m}\left(p_k-\frac{e}{c}A_k(\bold r,t)\right) \\
&= -\frac{e}{m^2c} \epsilon_{ijk}\left(p_j A_k(\bold r,t) +
A_j(\bold r,t)p_k\right) \\
&= -\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\bold r,t) - \frac{e}{m^2c}
\epsilon_{ijk} A_j(\bold r,t) p_k \\
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\bold r,t)-\frac{e}{m^2c}
\epsilon_{ikj} A_k(\bold r,t) p_j \mbox{(Switching indices in the second terms)} \\
&= -\frac{e}{m^2c}\epsilon_{ijk} p_j A_k(\bold r,t) + \frac{e}{m^2c}
\epsilon_{ijk} A_k(\bold r,t) p_j \\
&= -\frac{e}{m^2c}\epsilon_{ijk}\left[p_j,A_k(\bold r,t)\right] \\
&= -\frac{e}{m^2c}\epsilon_{ijk}\frac{\hbar}{i} \nabla_j A_k(\bold r,t) \\
&= i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right)_i
\end{align}
</math>
 
:<math>
\rightarrow \left[\bold v \times \bold v\right]=i\hbar\frac{e}{m^2c}\left(\nabla \times \bold A\right) = i\hbar\frac{e}{m^2c}\bold B
</math>
 
and
 
:<math>
\begin{align}
\left[\bold v,\phi\right] &= \frac{1}{m} \left[\bold p-\frac{e}{c}\bold A(\bold r, t),\phi(\bold r,t)\right] \\
&= \frac{1}{m} \left[\bold p,\phi(\bold r,t) \right] \\
&= \frac{1}{m} \frac{\hbar}{i}\nabla\phi
\end{align}
</math>
 
Substituting and rearranging, we get:
 
:<math>
m\frac{d\bold v}{dt} = \frac{e}{2c}
\left(\bold v \times \bold B-\bold B \times \bold v \right) + e\bold E
</math>
where
:<math>
\bold E = -\nabla \phi - \frac{1}{c} \frac{\partial \bold A}{\partial t}
</math>                                                                                       
Above is the quantum mechanical version of the equation for the acceleration of the particle in terms of the Lorentz force.
These results can also be deduced in Hamiltonian dynamics due to the similarity between the Hamiltonian dynamics and quantum mechanics.
 
[Problem about [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/harmonicoscinmagneticfield/ a particle moving in a magnetic field]]
 
=== WKB Approximation===
 
WKB method (Wentzel-Kramers-Brillouin method) is a technique for finding approximations to certain differential equations, including the one dimensional Schrodinger equation. 
It was developed in 1926 by Wenzel, Kramers, and Brillouin, for whom it was named.
The logic is that as <math>\hbar\rightarrow 0\!</math>, the wavelength, <math>\lambda=2\pi\hbar/ p\!</math>, tends to zero where the potential is smooth and slowly varying. Therefore, <math>\lambda\!</math> can be thought of as a local quantity <math>\lambda(x)\!</math>. This is a quasi-classical method of solving the Schrodinger equation.
 
In WKB, for a certain turning point, where <math>\lambda\!</math> is infinite, we cannot say that the potential changes slowly. Here, the whole theory is bound to fail. A proper handling of the turning points is the most difficult aspect of the WKB approximation. The potential at the turning point is approximated as linear and slowly varying (almost constant).
 
With this approximation,
:1. We can solve the Schrodinger equation exactly, and match the exact solutions to the WKB regions at either side of this exactly solveable region.
:2. We need to know the region where the WKB wave functions are valid.
 
For <math>V(x) = 0\!</math> (or constant) the solutions to the Schrodinger equation are simply plane waves of the form <math> e^{\pm ikx} </math>. If the potential varies smoothly, and the energy of the particle is fixed, the wave function can be described locally by writing its plane wave form <math> e^{iu(x)} \!</math>.
 
The WKB solution to the Schrodinger equation for a particle in a smoothly varying potential is given by:
 
:<math>\psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left[\frac{i}{\hbar}\int_{x_0}^x p(x')dx'\right]</math>
 
for a classically accessible region where E > V(x) and p(x) is real,
 
and is written as:
 
:<math> \psi(x)\approx \frac{C'}{\sqrt{p(x)}}\exp\left[\frac{1}{\hbar}\int_{x_0}^x p(x')dx'\right]</math>
 
for a classically inaccessible region where <math> E < V(x)\!</math>.
 
In both cases <math>p(x)\!</math> is the classical formula for the momentum of a particle with total energy
<math>E\!</math> and potential energy <math>V(x)\!</math> given by:
 
:<math>p(x)=\sqrt{2m(E-V(x))}\!</math>
 
This is an exact solution if x is constant, otherwise it's a local solution for a locally defined wavelength. There must be a condition on the region in space where the wavelength is locally defined to be sure the wavelength does not vary too much and the locally defined wave function is a valid approximation. This condition is that <math>\delta \lambda (x) \ll 1</math>, which is equivalent to writing <math> \left| \frac{d\lambda}{dx}\right|  = \left|\frac{d}{dx}\left(\frac{\hbar}{p(x)}\right)\right| \ll 1 </math>.
 
 
For example, suppose there is a point, <math> x = a \!</math> which is a classical turning point at a given value of E which separates the regions where <math> E > V(x)\!</math> and <math> E < V(x)\!</math>.  More specifically the classically inaccessible region is to the right of the classical turning point.  In the region appropriately close to the turning point the wave functions can be written as:
 
 
:<math> \psi(x)\approx \frac{A}{\sqrt{p(x)}}\exp\left[\frac{-1}{\hbar}\int_{a}^x p(x')dx'\right]+ \frac{B}{\sqrt{p(x)}}\exp\left[\frac{1}{\hbar}\int_{a}^x p(x')dx'\right] </math>
 
for <math> x \gg a \! </math>,  and
 
:<math>  \psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left[\frac{-i}{\hbar}\int_{a}^x p(x')dx'\right]+ \frac{D}{\sqrt{p(x)}}\exp\left[\frac{i}{\hbar}\int_{a}^x p(x')dx'\right] </math>
 
for <math> x \ll a \!</math>.
 
Note that at the classical turning point <math>p(x_{turning})\rightarrow 0</math> and the WKB solution diverges, which means it is no longer a valid approximation because the true wave function will not exhibit such divergent behavior at the turning points.  Thus, around each turning point we need to splice the two WKB solutions on either side of the turning point with a "patching" function that will straddle each turning point.  Because we only need a solution for this function in the vicinity of the turning points, we can approximate the potential as being linear, which allows us to solve the Schrodinger equation exactly (in the linear region only).  If we center the turning point at the origin the we have:
 
:<math>
V(x)\approx E+V'(0)x\!</math>
 
Solving the Schrodinger equation with our now linearized potential leads to the Airy equation whose solutions are Airy functions.
Our patching wave function is then:
 
:<math>
\psi_p(x)=a Ai\left(\alpha x\right)+bBi\left(\alpha x\right)\!</math>
 
where <math>a,b\!</math> are c-number coefficients and
 
:<math>\alpha=\left(\frac{2m}{\hbar^2}V'(0)\right)^{\frac{1}{3}}</math>
The key to patching the wavefunction in the region of the turning point is to asymptotically match the patching wavefunction to the WKB wavefunctions outside the region of the classical turning point. In the vicinity of the classical turning point,
 
:<math>p^2=2m(-V'(0)x)\Rightarrow 2p\frac{dp}{dx}=-2mV'(0)\Rightarrow \frac{dp}{dx}=-\frac{m}{p}V'(0) </math> <br/>
 
Since the region of applicability of the WKB approximation is
<math>1\gg\frac{1}{2\pi}\left|\frac{d\lambda}{dx}\right|</math>
 
near the turning point
 
:<math>p^3=\hbar m|V'(0)|\Rightarrow |x|\gg \frac{\hbar^{\frac{2}{3}}}{2}|mV'(0)|^{-\frac{1}{3}} </math> <br/>
This implies that the width of the region around the classical turning point vanishes as <math>\hbar^{\frac{2}{3}}</math>.
Thus, we can come as close to the turning point as we wish with the WKB approximations by taking a limit as <math>\hbar\!</math> approaches zero, as long as the distance from the classical turning point is much less than <math>\hbar^{\frac{2}{3}}</math>. Thus, by extending the patching function towards singularity in the direction of the WKB approximated wavefunction, while simultaneously extending the WKB approximated wavefunction toward the classical turning point, it is possible to match the asymptotic forms of the wavefunctions from the two regions, which are then used to patch the wavefunctions together.
This means that it would be useful to have a form of the Airy functions as they approach positive or negative infinity:
:<math>z\rightarrow \infty: Ai(z)\rightarrow \frac{1}{2\sqrt{\pi}}z^{-\frac{1}{4}}e^{-\frac{2}{3}|z|^{\frac{3}{2}}}</math><br/>
:<math>z\rightarrow \infty: Bi(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}e^{\frac{2}{3}|z|^{\frac{3}{2}}}</math><br/>
:<math>z\rightarrow -\infty: Ai(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\cos\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)</math><br/>
:<math>z\rightarrow -\infty: Bi(z)\rightarrow -\frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\sin\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)</math>
 
And noticing that (for negative <math>x\!</math>)
 
:<math>\frac{1}{\hbar}\int_{x}^0p(x')dx'=\sqrt{\frac{2mV'(0)}{\hbar^2}} \int_x^0\sqrt{-x'}dx'=\frac{2}{3}\sqrt{\frac{2mV'(0)}{\hbar^2}}|x|^{\frac{3}{2}}
</math>
 
and
:<math>\frac{1}{\sqrt{p(x)}}=\left(2mV'(0)\right)^{-\frac{1}{4}}|x|^{-\frac{1}{4}}
</math>
it becomes apparent that our WKB approximation of the wavefunction is the same as the patching function in the asymptotic limit.
This must be the case, since as <math>\hbar\rightarrow 0</math> the region of invalidity of the semiclassical wavefunction in the vicinity of the turning point shrinks, while the solution of the linarized potential problem depends only on the accuracy of the linearity of the potential, and not on <math>\hbar</math>. The two regions must therefore overlap.
 
For example, one can take  <math>x\approx \hbar^{\frac{1}{3}}</math> and then take the limit <math>\hbar\rightarrow 0</math>. The semiclassical solution must hold as we are always in the region of its validity and so must the solution of the linearized potential problem. Note that the argument of the Airy functions at <math>x\approx \hbar^{\frac{1}{3}}</math> goes to <math>\pm\infty</math>, which is why we need their asymptotic expansion.
 
====Bohr-Sommerfeld Quantization Rule ====
 
The quantized energy levels of a bound state can be approximated by the WKB method with an expression known as the Bohr-Sommerfeld quantization rule. A particle in a potential well is subject to bound states. This common example of the WKB method can be found in most undergraduate level quanum texts.
 
For a potential with no rigid walls the Bohn-Sommerfeld Quantization rule is:
 
:<math>\oint p(x)dx = 2\int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = (n+\frac{1}{2})2\pi\hbar</math>
 
For a potential with one rigid wall:
 
:<math> \int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = (n+\frac{3}{4})\pi\hbar </math>
 
For a potential with two rigid walls:
 
:<math> \int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = n\pi\hbar </math>
 
For a central potential:
:<math> p_r^2 = E - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}</math>
:<math>
\begin{align}
\int_{r_1}^{r_2}p_r(r)dr &= \int_{0}^{\infty}\sqrt{2m(E_n - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2})}dr \\
&= (n + \frac{1}{2})\pi\hbar
\end{align}
</math>
 
====WKB method for the Coulomb Potential ====
 
For the coulomb potential, the potential is given by:
:<math> V(r) = -\frac{-Ze^2}{r} </math>
 
Since the electron is bound to the nucleus, it can be veiwed as moving between two rigid walls at <math> r = 0 \!</math> and <math> r = a \!</math> with energy <math> E = V(a), a = -\frac{-Ze^2}{E}\!</math>. Thus, the energy of the electron is negative.
 
The energies of the s-state (<math> \ell = 0 \!</math>) can be obtained from:
 
:<math> \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr = n\pi\hbar </math>
 
Using the change of variable: <math> x = \frac{a}{r} </math>
 
:<math>
\begin{align}
\int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr &= \sqrt{-2mE} \int_0^a dr \sqrt{\frac{a}{r} - 1} \\
&= a\sqrt{-2mE} \int_0^1 dx\sqrt{\frac{1}{x} - 1} \\
&= \frac{\pi}{2}a\sqrt{-2mE} \\
&= -Ze^2\pi\sqrt{-\frac{2m}{E}}
\end{align}
</math>
 
Where I have used the integal
:<math> \int_0^1\sqrt{\frac{1}{x} -1} = \frac{\pi}{2} </math>
 
Thus we have the expression:
:<math>-Ze^2\pi\sqrt{-\frac{2m}{E}} = n\pi\hbar </math>
:<math>\Rightarrow E_n = -\frac{mZ^2e^4}{\hbar^2} = -\frac{Z^2e^2}{2a_0}</math>
 
Where <math> a_0\!</math> is the Bohr radius. Notice that this is the correct expression for the energy levels of a Coulomb potential.
 
===Calculation of Gamow Factor for Alpha decay of Nuclei ===
Since the <math>\alpha</math>-decay happens in the nulcie then we can assume that an <math>\alpha</math>-decay is formed in the nucleus just before its emission (although <math>\alpha</math> particle doesnot exist in the nucleus).
Inside the nucleus the particle will experience nuclear interaction which mostly attractive and outside the nucleus the inetraction would be coulombic(replusive).
 
Since the mathematical form of the nuclear interaction is not known we can model it by a square well type potential for the present purpose.
Outside the range of the nuclear interaction would be coulombic. So the coulomb interaction is,
 
<math>V_{coul} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{r}</math>
 
where <math>Z_{1}</math> is the atomic number of the rest of the nucleus(after decay).
 
From the WKB apporximation we know that at the turning point, <math>E= V(x)= V_{coul} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{R_{c}}</math>
 
<math>R_{c} = \frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{E}</math>
 
Now the Transition probabilty
<math>T\cong \Theta ^{2}</math>,
where <math>\Theta = e^{-\int_{b}^{a}q(x)dx}</math>
 
and <math>q(x)= \frac{1}{\hbar}\sqrt{2m\left(V(x)-E\right)}</math>
 
<math>\Theta ^{2} = e^{-2\int_{b}^{a} q(x)dx}</math>
 
In the present problem <math>b= R</math> and <math>a = R_{c}</math>
 
Now,
<math>\int_{R}^{R_{c}} \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}(V(x)-E)^{\frac{1}{2}} dr =  \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}\int_{R}^{R_{c}}
\left(\frac{1}{4\pi\epsilon_{0}}\frac{2z_{1}e^{2}}{r}-E\right)^\frac{1}{2}dr</math>
 
<math>= \left(\frac{2m}{\hbar^{2}}\right)^{\frac{1}{2}}\left(\frac{2z_{1}e^{2}}{4\pi\epsilon_{0}}\right)^{\frac{1}{2}}\int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</math>
 
let, <math>I = \int_{R}^{R_{c}} \left [ \frac{1}{r} - \frac{1}{R_{c}}\right ]^{\frac{1}{2}}dr</math>
 
 
Put,
<math>r= R_{0}cos^{2}\theta</math>
and
 
<math>dr= -R_{0}2cos\theta sin\theta</math>
 
<math>I= 2\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} \left( \frac{R_{c}sin^{2}\theta}{cos^{2}\theta}\right)^{\frac{1}{2}} cos\theta sin\theta d\theta</math>
 
<math>2R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} sin^{2}\theta d\theta  </math>
 
<math>I= R_{c}^{\frac{1}{2}}\int_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}} ( 1-{cos2\theta}) d\theta</math>
 
<math>I= R_{c}^{\frac{1}{2}}\left [ \theta - sin\theta cos\theta  \right ]_{0}^{cos^{-1}\sqrt{\frac{R}{R_{c}}}}</math>
 
<math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - sin \left(cos^{-1}\sqrt{\frac{R}{R_{c}}}\right) cos\left(cos^{-1}\sqrt{\frac{R}{R_{c}}}\right)  \right ]</math>
 
<math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - \sqrt{\frac{R}{R_{c}}}\sqrt{1- \frac{R}{R_{c}}}  \right ]</math>
 
<math>I= R_{c}^{\frac{1}{2}}\left [ cos^{-1}\sqrt{\frac{R}{R_{c}}} - \sqrt{\frac{R}{R_{c}}- \left(\frac{R}{R_{c}}\right)^{2}}  \right ]</math>
 
Let us consider <math>R_{c} \gg R</math>
 
Then we have
 
<math>I\cong \sqrt{R_{c}}\left(cos^{-1}\sqrt{\frac{R}{R_{c}}}-\sqrt{\frac{R}{R_{c}}} \right)</math>
 
where <math>cos^{-1}\sqrt{\frac{R}{R_{c}}} \cong \frac{\pi}{2} - \left(\frac{R}{R_{c}}\right)^{\frac{1}{2}}</math>
 
Setting, charge of <math>\alpha</math>particle = 2= <math>Z_{2}</math>(in general)
 
<math>\int q(x)dx = \left ( \frac{2Mz_{1}z_{2}e^{2}R_{c}}{\hbar^{2}4\pi\epsilon_0} \right )^{\frac{1}{2}}\left [\frac{\pi}{2} - 2\left(\frac{R}{R_{c}}\right)^{\frac{1}{2}}  \right ]</math>
 
Now <math> T\cong e^{-2\int q(x)dx} = exp\left [ -\frac{\pi z_{1}z_{2}e^{2}}{\hbar 4\pi\epsilon_0} \left (\frac{2M}{e}  \right )^{2} + \frac{4}{\hbar} \left ( \frac{2z_{1}z_{2}e^{2}MR}{4\pi\epsilon_0} \right )^{\frac{1}{2}}\right ]</math>
 
Now putting <math>E= \frac{1}{2}mv^{2}</math>, veloctiy of the particle
 
<math> T\cong exp\left ( \frac{-2\pi z_{1}z_{2}e^{2}}{4\pi\epsilon_0\hbar v} \right )exp \left ( \frac{32z_{1}z_{2}e^{2}MR}{4\pi\epsilon_0\hbar^{2} } \right )^{\frac{1}{2}}</math>
 
The 1st exponential term is known as the Gamow factor. The Gamow factor determines the dependence of the probability on the speed (or energy) of the alpha particle.
 
== Angular momentum ==
===Commutation relations===
Multidimensional problems entail the possibility of having rotation as a part of solution. Just like in classical mechanics where we can calculate the angular momentum using vector cross product, we have a very similar form of equation. However, just like any observable in quantum mechanics, this angular momentum is expressed by a Hermitian operator. Similar to classical mechanics we write angular momentum operator <math>\mathbf L\!</math> as:
:<math>\mathbf{L}=\mathbf{r}\times\mathbf{p}</math>
 
Working in the spatial representation, we have <math>\mathbf{r}</math> as our radius vector, while <math>\mathbf{p}</math> is the momentum operator.
:<math>\mathbf{p}=-i\hbar\nabla</math>


Using the cross product in Cartesian coordinate system, we get component of <math>\bold L\!</math> in each direction:
<b>Chapter 1: [[Physical Basis of Quantum Mechanics]]</b>
:<math>L_x=yp_z-zp_y=\frac{\hbar}{i}\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\!</math>


Similarly, using cyclic permutation on the coordinates x, y, z, we get the other two components of the angular momentum operator. All of these can be written in a more compact form using Levi-Civita symbol as (the Einstein summation convention is understood here)
* [[Basic Concepts and Theory of Motion]]
:<math>L_{\mu}=\epsilon_{\mu\nu\lambda}r_\nu p_\lambda\!</math>
* [[UV Catastrophe (Black-Body Radiation)]]
with
* [[Photoelectric Effect]]
:<math>\epsilon_{\mu\nu\lambda} =
* [[Stability of Matter]]
\begin{cases}
* [[Double Slit Experiment]]
+1, & \mbox{if } (\mu,\nu,\lambda) \mbox{ is } (1,2,3), (3,1,2) \mbox{ or } (2,3,1), \\
* [[Stern-Gerlach Experiment]]
-1, & \mbox{if } (\mu,\nu,\lambda) \mbox{ is } (3,2,1), (1,3,2) \mbox{ or } (2,1,3), \\
* [[The Principle of Complementarity]]
0, & \mbox{otherwise: }\mu=\nu \mbox{ or } \nu=\lambda \mbox{ or } \lambda=\mu.
* [[The Correspondence Principle]]
\end{cases}
* [[The Philosophy of Quantum Theory]]
</math>
Or we simply say that the even permutation gives 1, odd permutation gives -1, otherwise, we get 0.


We can immediately verify the following commutation relations:
:<math>[L_\mu,r_\nu]=i\hbar\epsilon_{\mu\nu\lambda}r_\lambda</math>
:<math>[L_\mu,p_\nu]=i\hbar\epsilon_{\mu\nu\lambda}p_\lambda</math>
:<math>[L_\mu,L_\nu]=i\hbar\epsilon_{\mu\nu\lambda}L_\lambda</math> (this relation tells us :<math>\mathbf{L}\times\mathbf{L}=i\hbar\mathbf{L}</math>)
and
:<math>[\hat{\mathbf{n}}\cdot\mathbf{L},\mathbf{r}]=i\hbar(\mathbf{r}\times\hat{\mathbf{n}})</math>
:<math>[\hat{\mathbf{n}}\cdot\mathbf{L},\mathbf{p}]=i\hbar(\mathbf{p}\times\hat{\mathbf{n}})</math>
:<math>[\hat{\mathbf{n}}\cdot\mathbf{L},\mathbf{L}]=i\hbar(\mathbf{L}\times\hat{\mathbf{n}})</math>


For example,
<b>Chapter 2: [[Schrödinger Equation]]</b>  
:<math>
\begin{align}
\left[L_\mu,r_\nu\right] &= [\epsilon_{\mu\lambda\rho}r_\lambda p_\rho,r_\nu] = \epsilon_{\mu\lambda\rho}[r_\lambda p_\rho,r_\nu] = \epsilon_{\mu\lambda\rho}r_\lambda[ p_\rho,r_\nu] \\
&= \epsilon_{\mu\lambda\rho}r_\lambda\frac{\hbar}{i}\delta_{\rho\nu} = \epsilon_{\mu\lambda\nu}r_\lambda\frac{\hbar}{i} \\
&= i\hbar\epsilon_{\mu\nu\lambda}r_\lambda
\end{align}
</math>
   
   
Also, note that for <math>L^2=L_x^2+L_y^2+L_z^2=L_{\mu} L_{\mu}</math>,
* [[Brief Derivation of Schrödinger Equation]]
 
* [[Relation Between the Wave Function and Probability Density]]
:<math>
* [[Stationary States]]
\begin{align}
* [[Heisenberg Uncertainty Principle]]
\left[L_{\mu},L^2\right] &= \left[L_{\mu},L_{\nu} L_{\nu}\right] \\
* [[Some Consequences of the Uncertainty Principle]]
&= L_{\nu}\left[L_{\mu},L_{\nu}\right]+\left[L_{\mu},L_{\nu}\right]L_{\nu} \\
&= L_{\nu} i\hbar \epsilon_{\mu\nu\lambda} L_{\lambda} + i\hbar \epsilon_{\mu\nu\lambda} L_{\lambda} L_{\nu} \\
&= i\hbar \epsilon_{\mu\nu\lambda} L_{\nu} L_{\lambda} - i\hbar \epsilon_{\mu\lambda\nu} L_{\lambda} L_{\nu} \\
&= i\hbar \epsilon_{\mu\nu\lambda} L_{\nu} L_{\lambda} - i\hbar \epsilon_{\mu\nu\lambda} L_{\nu} L_{\lambda} \\
&= 0.
\end{align}
</math>
 
===Angular momentum as a generator of rotations in 3D===
 
Let us consider an infinitesimal rotation <math> \mathbf{\alpha} \!</math> directed along the axis about which the rotation takes place.We then have
:<math> \mathbf{w}' = \mathbf{w} + \mathbf{\alpha} \times \mathbf{w} = \mathbf{w} + \delta\mathbf{w} </math>
The changes <math>\delta \mathbf w</math> (in the radial vector <math>\mathbf w\!</math> of the particle) due to such a rotation is:
 
:<math>\delta \mathbf w=\mathbf{\alpha}\times \mathbf w</math>
so
:<math>\psi\left(\mathbf{w}+\mathbf{\delta} \mathbf{w}\right)=\left(1+\mathbf{\alpha}\cdot\left(\mathbf{w}\times\mathbf{\nabla}\right)\right)\psi\left(\mathbf{w}\right)</math>
 
[[Image:Rotation.jpg]]
 
 
The expression
:<math>1+\mathbf{\alpha}\cdot(\mathbf{w}\times\mathbf{\nabla})</math>
is the operator of an infinitesimally small rotation. We recognize the equation
:<math>\mathbf{w}\times\mathbf{\nabla}=\frac{i}{\hbar}\mathbf{L}</math>
 
Therefore, the infinitesimal rotation operator is
:<math>\mathbf{R}_{inf}=1+\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}</math>
 
This expression is only until the first order correction. The actual rotation operator is calculated by applying this operator N times where N goes to infinity. Doing so, we get the rotation operator for finite angle
:<math>\mathbf{R} = e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}</math>
 
In this form, we recognize that angular momentum is the generator of rotation. And we can write the equation relating the initial vector before rotation with the transformed vector as
:<math>\mathbf{w}'=e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}\mathbf{w} e^{-\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}</math>
This expression of the rotation operator is also valid when the rotation angle is not infinitesimal. What's more, this equation also implies that if we have a scalar instead of <math>\mathbf{w}\!</math>, it would be invariant. We can also calculate the effect of the unitary operator <math>e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}</math> on the states:
:<math>\langle r_0|e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}\mathbf{ \hat{\mathbf{r}}} e^{-\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}=\langle r_0|\hat{\mathbf{r'}}=r_0'\langle r_0|</math>
:<math>\Rightarrow \psi'(r_0)=\langle r_0|\psi'\rangle=\langle r_0|e^{\frac{i}{\hbar}\mathbf{\alpha}\cdot\mathbf{L}}|\psi\rangle=\langle r_0'|\psi\rangle=\psi(r_0')</math>
 
This is the wavefunction evaluated at a rotated point.
 
A sample problem:  [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/Angular_Momentum_Problem_1 the infinitesimal rotation].
 
===Spherical Coordinates ===
Since angular momentum can be represented as a generator of rotations you can use the equation for an infinitesmial rotation to construct the coordinates of angular momentum in spherical coordinates.
 
The cartesian coordinates x,y,z can be written in spherical as follows:
 
<math>x=r\sin\theta\cos\phi \! </math>
<math>y=r\sin\theta\sin\phi \! </math>
<math>z=r\cos\theta \! </math>
 
 
Denote the state <math> \langle \mathbf{r} \!\, | = \langle \mathbf{r}\! \,  \theta \phi |  </math>
 
If you choose <math> \alpha \! </math> along the z-axis then the only coordinate that will change is phi such that <math> \phi \rightarrow \phi + \alpha </math>.  Now the state is written as:
 
<math> \langle \mathbf{r}\! \, \theta \phi | \left(1 + \frac{i}{\hbar} \alpha L_{z}\right) = \langle \mathbf{r} \! \, \theta \phi + \alpha | </math>
 
Working to first order in alpha the right hand side becomes:
 
<math> \langle \mathbf{r}\! \, \theta \phi | + \alpha \frac{\partial}{\partial \phi} \langle \mathbf{r}\! \, \theta \phi | </math>
 
Therefore <math> L_{z} = \frac{\hbar}{i} \frac{\partial}{\partial \phi} </math>
 
 
Now choose <math> \alpha \! </math> along the x-axis then the cartesian coordinates are changed such that:
<math> x \rightarrow x  \!  </math> ,
<math> y \rightarrow y - \alpha z  \!</math>, and
<math> z \rightarrow z + \alpha y \! </math>,
 
from these transformations it can be determined that <math> \delta \theta = -\alpha\sin\phi \! </math> since <math> \delta z = \alpha y \! </math>
and since x does not change it can be determined that <math> \delta \phi  = -\alpha\cot\theta \cos\phi \! </math>.
 
This means that the original state is now written as:
<math> \langle \mathbf{r}\! \, \theta \phi | \left(1 + \frac{i}{\hbar} \alpha L_{x}\right) = \langle \mathbf{r} \! \, \,\theta - \alpha\sin\phi \,  \phi - \alpha\cot\theta\cos\phi | </math>
 
Expanding the right hand side of the above equation as before to the first order of alpha the whole equation becomes:
<math> \langle \mathbf{r}\! \, \theta \phi | L_{x} = \frac{\hbar}{i} \left( -\sin\phi \frac{\partial}{\partial \theta} \, -\cot\theta\cos\phi \frac{\partial}{\partial\phi} \! \right)  \langle \mathbf{r}\! \, \theta \phi | </math>
 
Therfore <math> L_{x} = \frac{\hbar}{i}\left( -\sin\phi \frac{\partial}{\partial \theta} \, -\cot\theta\cos\phi \frac{\partial}{\partial\phi} \! \right) </math>
 
Using the same techinque, choose <math> \alpha \! </math> along the y-axis and the coordinates will change in a similar fashion such that it can be shown that <math> L_{y} = \frac{\hbar}{i} \left(\cos\phi \frac{\partial}{\partial\theta} - \cot\theta\sin\phi \frac{\partial}{\partial\phi} \! \right) </math>
 
===Eigenvalue quantization===
 
The motivation for exploring eigenvalue quantization comes form wanting to solve the energy eigenvalue problem.  It is not possible, in general, to specify and measure more than one component <math> \hat{\mathbf{n}}\cdot\mathbf{L} </math>  of orbital angular momentum.  Yet it is possible to specify <math> \mathbf{L}^2 </math> simulataneously with any one component of <math> \mathbf{L} </math>. Normally <math>\ L_z </math> is choosen.  A central force potential has a Hamiltonian that commutes with <math> \mathbf{L}  </math>, and in that case one can require the energy eigenstates of the system to also be eigenvectors of <math>  \mathbf{L}^2 </math> and one component of <math>  \mathbf{L}  (L_z) </math>.
 
The quantization of angular momentum follows simply from the above commutation relations. Define <math>\beta\!</math> by:
:<math>\beta=L_x^2+L_y^2+L_z^2</math>
 
Since <math>\beta\!</math> is a scalar, it commutes with each component of angular momentum.
 
Now Define a change of operators as follows:
:<math>L_+=L_x+iL_y\!</math>
:<math>L_-=L_x-iL_y\!</math>
 
The choice of these two operators are choosen because their commutation relations with <math> L_z \!</math> and one another only result in soultions with <math>  L_- \!</math>, <math>  L_+ \!</math>, and <math>  L_z \!</math>.
 
From the commutation relations we get
:<math>L_+L_-=\beta-L_z^2+\hbar L_z</math>
 
Similarly,
:<math>L_-L_+=\beta-L_z^2-\hbar L_z</math>
                                                           
Thus,
:<math>[L_+,L_-]=L_+L_--L_-L_+=2\hbar L_z</math>
 
And
:<math>[L_z,L_-]=L_zL_--L_-L_z=-\hbar L_-</math>
 
Also, It is easy to show that:
:<math>[L^2,L_\pm]=0</math>
 
Let <math>L'_z\!</math> be an eigenvalue of <math>L_z\!</math>.
:<math>\langle L_z'|L_+L_-|L_z'\rangle=(\beta-L_z'^2+\hbar L_z')\langle L_z'|L_z'\rangle</math>
                                                             
Since the left hand side of the above equation is the square of the length of a ket, it has to be non-negative. Therefore
:<math>\beta-L_z'^2+\hbar L_z'\ge0</math>
:<math>\Rightarrow \beta+\frac{\hbar^2}{4}\ge (L_z'-\frac{\hbar}{2})^2</math>
:<math>\Rightarrow \beta+\frac{\hbar^2}{4}\ge 0</math>
 
Defining the number <math>k\!</math> by
:<math>k+\frac{\hbar}{2}=\sqrt{\beta+\frac{\hbar^2}{4}}</math>
:<math>\Rightarrow k\ge -\frac{\hbar}{2}</math>
 
The inequality 5.1.13 becomes
:<math>k+\frac{\hbar}{2}\ge |L_z'-\frac{\hbar}{2}|</math>
:<math>\Rightarrow k+\hbar\ge L_z'\ge -k</math>
 
Similarly, from equation 5.1.10, we get
:<math>\langle L_z'|L_-L_+|L_z'\rangle=(\beta-L_z'^2-\hbar L_z')\langle L_z'|L_z'\rangle</math>
:<math>\Rightarrow \beta-L_z'^2-\hbar L_z'\ge0</math>
:<math>\Rightarrow k\ge L_z'\ge -k-\hbar</math>
 
This result, combined with 5.1.15 shows that
:<math>k\ge 0</math>
and                               
:<math>k\ge L_z'\ge -k</math>
 
From 5.1.12
:<math>L_z L_-|L_z'\rangle=(L_- L_z-\hbar L_-)|L_z'\rangle=(L_z'-\hbar)L_-|L_z'\rangle</math>
 
Now, if <math>L_z'\ne 0</math>, then <math>L_-|L_z'\rangle</math> is an eigenket of <math>L_z\!</math> belonging to the eigenvalue <math>L_z'-\hbar</math>.
Similarly, if <math>L_z'-\hbar\ne -k</math>, then <math>L_z'-2\hbar</math> is another eigenvalue of <math>L_z\!</math>, and so on. In this way, we get a series of eigenvalues which must terminate from 5.1.16, and can terminate only with the value <math>\!-k</math>.
Similarly, using the complex conjugate of 5.1.12, we get that <math>L_z',L_z'+\hbar,L_z'+2\hbar,...\!</math> are eigenvalues of L'z. Thus we may conclude that <math>2k\!</math> is an integral multiple of the Planck's constant, and that the eigenvalues are:
:<math>k, k-\hbar,k-2\hbar,...,-k+\hbar,-k</math>
 
If <math>|m\rangle\!</math> is an eigenstate of <math>L_z\!</math> with eigenvalue <math>m\hbar\!</math>, then
:<math>
\begin{align}
L_z L_\pm |m\rangle &= ([L_z,L_\pm]+L_\pm L_z)|m\rangle=(\pm\hbar L_\pm+L_\pm m)|m\rangle \\
&=(m\pm 1)\hbar(L_\pm |m\rangle)
\end{align}
</math>
 
Which means that <math>L_+\!</math> or <math>L_-\!</math> raises or lowers the <math>z\!</math> component of the angular momentum by <math>\hbar\!</math>.
 
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/angularmomcommutation/ Calculation of two angular momentum expressions]
 
===Orbital angular momentum eigenfunctions===
 
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/AngularMomentumProblem Worked Problem] about angular momentum.
 
Now we construct our eigenfunctions of the orbital angular momentum explicitly. The eigenvalue equation is
:<math>L_z|l,m\rangle=m\hbar|l,m\rangle</math>
in terms of wave functions, becomes:
:<math>\langle r,\theta,\phi|L_z|l,m\rangle=-i\hbar\frac{\partial}{\partial \phi}\langle r,\theta,\phi|l,m\rangle=m\hbar \langle r,\theta,\phi|l,m\rangle</math>
Solving for the <math>\phi\!</math> dependence, we find
:<math>\langle r,\theta,\phi|l,m\rangle=e^{im\phi}\langle r,\theta,0|l,m\rangle</math>
 
We construct the <math>\theta\!</math> dependence using the differential operator representation of <math>L^2\!</math>
:<math>L^2=-\hbar^2\left(\frac{1}{\sin \theta^2}\frac{d^2}{d\phi^2}+\frac{1}{\sin \theta}\frac{d}{d \theta}\left(\sin\theta\frac{d}{d\theta}\right)\right)</math>
 
Where the eigenvalues of <math>L^2\!</math> are:
:<math>L^2|l,m\rangle= \hbar^2 l(l+1)|l,m\rangle</math>
 
We proceed by using the property of <math>L_+\!</math> and <math>L_-\!</math>, defined by
:<math>L_\pm=\frac{\hbar}{i}e^{\pm i\phi}\left(\pm i\frac{d}{d\theta}-\cot \theta \frac{d}{d\phi}\right)</math>
to find the following equation
:<math>\langle r,\theta,\phi|L_+|l,l\rangle=-i\hbar e^{i\phi}\left(i\frac{\partial}{\partial\theta}-\cot \theta \frac{\partial}{\partial\phi}\right)\langle r,\theta,\phi|l,l\rangle=0</math>
 
Using the above equations, we get
:<math>\left(\frac{\partial}{\partial \theta}-l\cot\theta\right)\langle r,\theta,\phi|l,l\rangle=0</math>
 
And the solution is
:<math>\langle r,\theta,\phi|l,l\rangle=f(r)e^{il\phi}(\sin\theta)^l</math>
 
where <math>f(r)\!</math> is an arbitrary function of <math>r\!</math>. We can find the angular part of the solution by using <math>L_-\!</math>. It turns out to be
:<math>P_l^m(x)=\frac{(-1)^m}{2^l l!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l</math>
 
And we know that <math>Y_l^m(\theta, \phi)\!</math> are the spherical harmonics defined by
:<math>Y_l^m(\theta, \phi)=(-1)^l \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P_l^m(\cos \theta)e^{im\phi}</math>
where the function with cosine argument is the associated Legendre polynomials defined by:
:<math>P_l^m(x)=(-1)^m (1-x^2)^{m/2}\frac{d^m}{dx^m}P_l(x)</math>
with
:<math>P_l(x)=\frac{1}{2^l l!}\frac{d^l}{dx^l}(x^2-1)^l</math>
 
And so we then can write:
:<math>P_l^m(x)=\frac{(-1)^m}{2^l l!}(1-x^2)^{m/2}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l</math>
 
Central forces are derived from a potential that depends only on the distance r of the moving particle from a fixed point, usually the coordinate origin. Since such forces produce no torque, the orbital angular momentum is conserved.
 
We can rewrite the angular momentum as
:<math>\mathbf L=\mathbf r\times \frac{\hbar}{i}\nabla</math>
As has been shown, angular momentum acts as the generator of rotation.
 
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/AngularMomentumExercise An exercise] with angular momentum.
 
== Central forces  ==
 
===Generalized derivation===
A central potential does not depend on time, but rather depends only on the absolute value of the distance away from the potential's center. A central potential is rotationally invariant, not depending on the orientation. These properties can effectively reduce a three dimensional problem into a one dimensional problem.
 
:<math>H=\frac{p^2}{2m}+V(|r|)</math>
 
Due to the rotational symmetry, <math>[H,L_z]=0\!</math> and <math>[H,L^2]=0\!</math>, and the eigenstates of <math>L^2\!</math>, <math>L_z\!</math> are non-degerate. This allows us to find a complete set of states that are simultaneous eigenfunctions of <math>H\!</math>, <math>L_z\!</math>, and <math>L^2\!</math>. We can label these states by their eigenvalues of <math>|Elm\rangle\!</math>.
 
From this we can get a state of the same energy for a given <math>l\!</math> with a degeneracy of <math>2l+1\!</math>.
We can rewrite the Laplacian as
:<math>\nabla^2=\frac{1}{r}r-\frac{L^2}{\hbar^2 r^2}</math>
 
This makes the Schroedinger equation
:<math>\left(\frac{-\hbar^2}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{L^2}{2mr^2}+V(r)\right)\psi(r,\theta,\phi)=E\psi(r,\theta,\phi)</math>
 
Using separation of variables, <math>\psi(r,\theta,\phi)=f_l(r)Y_{lm}(\theta,\phi)\!</math>, we get:
:<math>\left(\frac{-\hbar^2}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{\hbar^2 l(l+1)}{2mr^2}+V(r)\right)f_l(r)Y_{lm}(\theta,\phi)=Ef_l(r)Y_{lm}(\theta,\phi)</math>
 
The term <math> \frac{\hbar^2 l(l+1)}{2mr^2} </math> is referred to as the centrifugal barrier, which is associated with the motion of the particle. The classical analogue is <math> \frac{l^2}{2mr^2} </math>. The centrifugal barrier prevents the particle from reaching the center of force, causing the wave function to vanish at this point. Multiplying both sides by <math>Y_{l^\prime m'}\!</math> and integrating over the angular dependence reduces the equation to merely a function of <math>r\!</math>.
 
Now if we let <math>u_l(r)=rf_l(r)\!</math>, this gives the radial Schroedinger equation:
:<math>\left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2 l(l+1)}{2mr^2}+V(r)\right)u_l(r)=Eu_l(r)</math>
 
Due to the boundary condition that <math>f_l(r)\!</math> must be finite the origin, <math>u_l(r)\!</math> must vanish.
 
Often looking at the asymptotic behavior of <math>u_l(r)\!</math> can be quite helpful.
 
As <math>r\rightarrow 0\!</math> and <math>V(r)\ll\frac{1}{r^2}\!</math> the dominating term becomes the centrifugal barrier giving the approximate Hamiltonian:
:<math>\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2 l(l+1)}{2mr^2}</math>
which has the solutions <math>u_l(r)\sim r^{l+1},r^{-l}\!</math> where only the first term is physically possible because the second blows up at the origin.
 
As <math>r\rightarrow\infty\!</math> and <math>rV(r)\rightarrow 0</math>(which does not include the monopole <math>\frac{1}{r}</math> coulomb potential) the Hamiltonian approximately becomes
:<math>\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}u_l(r)=Eu_l(r)</math>
letting <math>k=-i\sqrt{\frac{2mE}{\hbar^2}}</math>
gives a solution of <math>u_l(r)=Ae^{kr}+Be^{-kr}\!</math>,  where when <math>k\!</math> is real, <math>B=0\!</math>, but both terms are needed when <math>k\!</math> is imaginary.
 
'''Nomenclature'''
 
Historically, the first four (previously five) values of <math>l\!</math> have taken on names, and additional values of <math>l\!</math> are referred to alphabetically:
:<math>
\begin{cases}
l = 0  & \mbox{s-wave (sharp)}\\
l = 1  & \mbox{p-wave (principal)}\\
l = 2  & \mbox{d-wave (diffuse)}\\
l = 3  & \mbox{f-wave (fundamental)}\\
l = 4  & \mbox{g-wave (previously called t-wave for thick)}\\
l = 5  & \mbox{h-wave}\\
\end{cases} </math>
 
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HydrogenAtomProblem2 Worked Problem] involving the energy levels in a central potential.
 
===Free particle in spherical coordinates===
A free particle is a specific case when <math>V_0=0\!</math> of the motion in a uniform potential <math>V(r)=V_0\!</math>.  So it's more useful to consider a particle moving in a uniform potential. The Schrodinger equation for the radial part of the wave function is:
 
:<math>\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V_0\right)u_l(r)=Eu_l(r)</math>
let <math>k^2=\frac{2m}{\hbar^2}|E-V|</math>.  Rearranging the equation gives
:<math>\left(-\frac{\partial^2}{\partial r^2}+\frac{l(l+1)}{r^2}-k^2\right)u_l(r)=0</math>
Letting <math>\rho=kr\!</math> gives the terms that <math>\frac{1}{r^{2}}=\frac{k^{2}}{\rho ^{2}}</math> and <math>\frac{\partial ^{2}}{\partial r^{2}}=k^{2}\frac{\partial ^{2}}{\partial \rho ^{2}}</math>. Then the equation becomes:
:<math>\left(-\frac{\partial^2}{\partial\rho^2}+\frac{l(l+1)}{\rho^2}\right)u_l(\rho)=u_l(\rho)=d_ld_l^+u_l(\rho)</math>
where <math>d_l\!</math> and <math>d_l^{\dagger}\!</math> become the raising and lowering operators:
 
:<math>d_l=\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}, d_l^\dagger=-\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}</math>
Being <math>d_l^{\dagger}d_l=d_{l+1}d_{l+1}^{\dagger}</math>, it can be shown that
:<math>d_l^\dagger u_l(\rho)=c_l u_{l+1}(\rho)</math>
For <math>\ell =0</math>,  <math>-\frac{\partial^2}{\partial \rho^2} u_0(\rho)=u_0(\rho)</math>, gives the solution as:
:<math>u_0(\rho)=A\sin(\rho)-B\cos(\rho)\!</math>
The raising operator can be applied to the ground state in order to find high orders of <math>\ u_0(\rho)</math>;
:<math>d_0^\dagger u_0(\rho)=\left(-\frac{\partial}{\partial\rho}+\frac{l+1}{\rho}\right)u_0(\rho)=c_0 u_1(\rho)</math>
By this way, we can get the general expression:
:<math>f_l(\rho)=\frac{u_l(\rho)}{\rho}=A_lj_l(\rho)+B_ln_l(\rho)</math>, where j is spherical Bessel function and n is spherical Neumann function.


===Spherical well===
Dividing the potential into two regions, <math>0<r<a\!</math> and <math>r>a\!</math>,
:<math>
\begin{cases}
0<r<a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0)u_l(r)=Eu_l(r), \mbox{where the general solution is a Bessel function}\\
r>a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2l(l+1)}{2mr^2})u_l(r)=Eu_l(r), \mbox{where the general solution is a Hankel function}
\end{cases}
</math>
<math>Aj_l(kr) +Bn_l(kr), r\rightarrow0,n_l(kr)\rightarrow\infty\!</math>


For the <math>l=0\!</math> term, the centrifugal barrier drops out and the equations become the following
<b>Chapter 3: [[Operators, Eigenfunctions, and Symmetry]]</b>
:<math>
\begin{cases}
0<r<a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}-V_0)u_0(r)=Eu_0(r)\\
r>a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2})u_0(r)=Eu_0(r)
\end{cases}
</math>


The generalized solutions are
* [[Linear Vector Spaces and Operators]]
:<math>
* [[Commutation Relations and Simultaneous Eigenvalues]]
\begin{cases}
* [[The Schrödinger Equation in Dirac Notation]]
0<r<a & u_0(r)=Ae^{ikr}+Be^{-ikr}\\
* [[Transformations of Operators and Symmetry]]
r>a & u_0(r)=Ce^{ik'r}+De^{-ik'r}
* [[Time Evolution of Expectation Values and Ehrenfest's Theorem]]
\end{cases}
</math>


Using the boundary condition, <math>u(r=0)=0\!</math>, we find that <math>A=-B\!</math>. The second equation can then be reduced to sinusoidal function where <math>\alpha=2iA\!</math>.
:<math>u_0(r)=2iAsin(kr)=\alpha\sin(kr)=\alpha\sin\left(\frac{r}{\hbar}\sqrt{2m(E+V_0)}\right)</math>
for <math>r>a\!</math>, we know that <math>D=0\!</math> since as <math>r\!</math> approaches infinity, the wavefunction does not go to zero.
:<math>u_0(r)=Ce^{ik'r}+De^{-ik'r}=Ce^{-\frac{r}{\hbar}\sqrt{-2mE}}</math>


Matching the conditions that at <math>r=a\!</math>, the wavefunctions and their derivatives must be continuous which results in 2 equations
<b>Chapter 4: [[Motion in One Dimension]]</b>  
:<math>\alpha\sin\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
:<math>\alpha\frac{\sqrt{2m(E+V_0)}}{\hbar}\cos\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=-\frac{\sqrt{-2mE}}{\hbar}Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
 
Dividing the above equations, we find
:<math>-\cot\left(\sqrt{\frac{2m}{\hbar^2}(V_0-|E|)a^2}\right)=\frac{\sqrt{\frac{2m|E|}{\hbar^2}}}{\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}}</math>, which is the solution for the odd state in 1D square well.
 
Solving for <math>V_0\!</math>, we know that there is no bound state for
:<math>V_0<\frac{\pi^2\hbar^2}{8ma^2}</math>
 
 
===Isotropic Harmonic Oscillator===
 
The radial part of the Schrodinger's Equation for a particle of mass M in an isotropic harmonic oscillator potential  <math>V(r)=\frac{1}{2}Mw^{2}r^2</math> is given by:
 
<math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{nl}(r)}{\partial r^2}+\left(\frac{1}{2}Mw^{2}r^2+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}\right)u_{nl}(r)=Eu_{nl}(r)</math>
 
We look at the solutions <math>u_{nl}</math> in the asymptotic limits of r.
 
As  <math>r\rightarrow 0</math>, the equation reduces to:  <math>-\frac{\hbar^2}{2M}\frac{\partial^2u_{l}(r)}{\partial r^2}+\frac{\hbar^2}{2M}\frac{l(l+1)}{r^2}u_{l}(r)=Eu_{l}(r)</math>
whose nondivergent solution is given by <math>u_l(r)\simeq r^{^{l+1}}</math>.
 
On the otherhand, as <math>  r\rightarrow \infty</math>, the equation becomes:  <math>-\frac{\hbar^2}{2M}\frac{\partial^2u(r)}{\partial r^2}+\frac{1}{2}Mw^{2}r^2u(r)=Eu(r)</math>
whose solution is given by  <math>u(r)\simeq e^-{\frac{Mwr^2}{2\hbar}}</math>.
 
Combining the aymptotic limit solutions we choose the general solution to th e equation as: <math>u_l(r)=f_l(r)r^{l+1}e^-\frac{Mwr^2}{2\hbar}</math>
 
Substituting this expression into the original equation,
<math>\frac{\partial^2f_l(r) }{\partial r^2}+2\left(\frac{l+1}{r}-\frac{Mw}{\hbar}r\right)\frac{\partial f_l(r) }{\partial r}+\left[\frac{2ME}{\hbar^2}-(2l+3)\frac{Mw}{\hbar}\right]f_l(r) =0</math>
 
Now we try the power series solution <math>f_l(r)=\sum_{n=0}^{\infty}a_{n}r^n= a_{0}+a_{1}r+a_{2}r^2+a_{3}r^3+. . . +a_{n}r^n+...</math>
   
   
Subsituting this solution into the reduced form of the equation,
* [[One-Dimensional Bound States]]
<math>\sum_{n=0}^{\infty}[{n(n-1)a_{n}r^{n-2}+2(\frac{l+1}{r}-\frac{Mw}{\hbar}}t)na_nr^{n-1}+[\frac{2ME}{\hbar^2}-(2l+3)\frac{Mw}{\hbar}]a_nr^n]=0</math>
* [[Oscillation Theorem]]
which reduces to the equation  <math>\sum_{n=0}^{\infty}[{n(n+2l+1)a_{n}r^{n-2}+(-\frac{2Mw}{\hbar}n}+\frac{2ME}{\hbar^2}-(2l+3)\frac{Mw}{\hbar})a_nr^n]=0</math>
* [[The Dirac Delta Function Potential]]
* [[Scattering States, Transmission and Reflection]]
* [[Motion in a Periodic Potential]]
* [[Summary of One-Dimensional Systems]]


For this equation to hold, the coefficients of each of the powers of r must vanish seperately.


So,when n =0 the coefficient of <math> r^{-2}</math> is zero,              <math>0.(2l+1)a_0=0</math>    implying that  <math>a_0</math> need not be zero.
<b>Chapter 5: [[Discrete Eigenvalues and Bound States - The Harmonic Oscillator and the WKB Approximation]]</b>


Equating the coefficient of <math> r^{-1}</math>to be zero,                  <math>1.(2l+2)a_1=0</math>    implying that  <math>a_1</math> must be zero.
* [[Harmonic Oscillator Spectrum and Eigenstates]]
* [[Analytical Method for Solving the Simple Harmonic Oscillator]]
* [[Coherent States]]
* [[Charged Particles in an Electromagnetic Field]]
* [[WKB Approximation]]


Equating the coefficient of <math> r^{-n}</math> to be zero, we get the recursion relation which is:
 
                <math>\sum_{n=0}^{\infty}[(n+2)(n+2l+3)a_{n+2}=[-\frac{2ME}{\hbar^2}+(2n+2l+3)\frac{Mw}{\hbar}]a_n</math>


<b>Chapter 6: [[Time Evolution and the Pictures of Quantum Mechanics]]</b>


The function <math>f_l(r)</math> contains only even powers in n and is given by:
* [[The Heisenberg Picture: Equations of Motion for Operators]]
* [[The Interaction Picture]]
* [[The Virial Theorem]]


<math>f_l(r)=\sum_{n=0}^{\infty }a_{2n}r^{2n}=\sum_{n^{'}=0,2,4}^{\infty  }a_{n^{'}}r^{n^{'}}</math>


Now as  <math> n\rightarrow \infty</math> , <math>f_l(r)</math> diverges so that for finite solution, the series should stop after  <math>r^{n^{'}+2}</math> leading to the quantization condition:
<b>Chapter 7: [[Angular Momentum]]</b>
 
                                                  <math>\frac{2M}{\hbar^2}E_{n^{'}l}-\frac{Mw}{\hbar}(2n^{'}+2l+3)=0</math>
 
                                                  <math>E_{n^{'}l}=(n^{'}+l+\frac{3}{2})\hbar w,    n^{'}=0,1,2,3,...</math>
 
As a result, the energy of the isotropic harmonic oscillator is given by:
 
                                              <math>E_{n}=(n+\frac{3}{2})\hbar w,  n=0,1,2,3,...  with  n=n^{'}+l</math>
 
The degeneracy corresponding to the nth level is:
 
                                            <math>g=\frac{1}{2}(n+1)(n+2)</math>
 
 
The total wavefunction of the isotropic Harmonic Oscilaator is given by:
       
                                        <math>\psi_{nlm}(r,\theta ,\phi )=r^{l+1}f_l(r)Y_{lm}(\theta ,\phi)e^-\frac{Mw}{2\hbar}r^2=R_{nl}(r)Y_{lm}(\theta ,\phi )</math>
 
===Hydrogen atom===
[http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HydrogenAtomProblem Worked Problem] involving the hydrogen atom.
 
Another [http://wiki.physics.fsu.edu/wiki/index.php/Phy5645/HydrogenAtomProblem3 worked problem]
 
[[Image:H_atom.jpg]]
 
The Schrodinger equation for the particle moving in central potential can be represented in a spherical coordinate system as follows:
:<math>\left( -\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{L^2}{2\mu r^2}+V(r)\right) \psi(r,\theta,\phi)=E\psi(r,\theta,\phi)</math>
where <math>L\!</math> is the angular momentum operator and <math>\mu\!</math> is the reduced mass.
 
In this case, being invariant under the rotation, the Hamiltonian, <math>H\!</math>, commutes with both <math>L^2\!</math> and <math>L_z\!</math>. Furthermore, <math>L^2\!</math> and <math>L_z\!</math> commute with each other.  Therefore, the energy eigenstates can be chosen to be simultaneously the eigenstates of <math>H\!</math>, <math>L^2\!</math> and <math>L_z\!</math>. Such states can be expressed as the following:
:<math>\psi(r,\theta,\phi)=R(r)Y_{lm}(\theta,\phi)=\frac{u_l(r)}{r}Y_{lm}(\theta,\phi)</math>
where <math>Y_l^m(\theta,\phi)</math> is the spherical harmonic, which is the simultaneous eigenstates of <math>L^2\!</math> and <math>L_z\!</math>, and the <math>\frac{u_l(r)}{r}</math> substitution is made for simplification.
 
Substituting <math>\psi(r,\theta,\phi)=\frac{u_l(r)}{r}Y_{lm}(\theta,\phi) </math> into the Schrodinger equation, and taking into account the fact that:
:<math>L^2\psi(r,\theta,\phi)=L^2[R(r)Y_{lm}(\theta,\phi)]=R(r)L^2Y_{lm}(\theta,\phi)=R(r)\hbar^2l(l+1)Y_{lm}(\theta,\phi)
=\hbar^2l(l+1)\psi(r,\theta,\phi)</math>
we have the equation for <math>u_l(r)\!</math>:
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V(r)\right] u_l(r)=Eu_l(r)</math>
 
In the hydrogen atom or single electron ion, the potential is the Coulomb potential between the electron and the nucleus:
:<math>V(r)=-\frac{Ze^2}{r}</math>
where <math>Z=1\!</math> for the hydrogen, <math>Z=2\!</math> for helium ion <math>He^+\!</math>, etc.
 
Plugging this potential into the Schrodinger equation we obtain
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}-\frac{Ze^2}{r}\right] u_l(r)=Eu_l(r)</math>
 
Since we are only concentrating on the bound states, we can write down the solutions to this equation in the asymptotic limits.
 
In the <math>r \to 0 </math> limit the equation becomes
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right] u_l(r)=0</math>
 
and the solution in this limit is <math>u_l(r)\rightarrow r^{l+1}</math>.
 
In the <math>r \to \infty </math> limit the equation becomes
:<math>\frac{\partial^2 u_l(r)}{\partial r^2}+\frac{2mE}{\hbar^2}u_l(r)=0</math>
 
and the solution in this limit is <math>u_l(r)\rightarrow e^{-\frac{r}{a}}</math>.
 
where
:<math>a=\sqrt{\frac{-\hbar^2}{2m E}}</math>
 
If we allow <math>\kappa=a^{-1}\!</math>, then the large limit of <math>r\!</math> can be expressed as
:<math>u_l(r)\sim e^{-\kappa r}\!</math>
 
Using the limits of <math>u(r)\!</math>, the wavefunction can be expressed as the following
:<math>u_l(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r)\!</math>
 
To simplify the equation, make a substitution <math>\rho=\kappa r\!</math>. The equation now turns into
:<math>u_l(\rho)=\rho^{l+1}e^{-\rho}W(\rho)\!</math>
 
Plugging this equation into the Schrodinger equation and simplifying, it turns into
:<math>\frac{d^2W(\rho)}{d\rho^2}+2\left(\frac{l+1}{\rho}-1\right)\frac{dW(\rho)}{d\rho}+\left(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho}\right)W(\rho)=0</math>        where <math>\rho _{0}=Ze^{2}\sqrt {\frac{2\mu }{\hbar ^{2}E}} </math>
 
Here <math>W(\rho)\!</math> can be expressed as an expansion of polynomials;
:<math>W(\rho)=a_0+a_1 \rho+a_2\rho^2+...=\sum_{k=0}^\infty a_k \rho^k</math>
 
The Schrodinger equation is then expressed as
:<math>\sum_{k=0}^\infty (a_{k}k(k-1)\rho^{k-2}+2(l+1)k\rho^{k-2}a_k-2\rho^{k-1}a_k k)+\sum_{k=0}^\infty(\rho_0 a_k\rho^{k-1}-2(l+1)a_k\rho^{k-1})=0</math>
 
And simplified into
:<math>\sum_{k=0}^\infty (a_{k+1}(k+1)k\rho^{k-1}+2(l+1)(k+1)\rho^{k-1}a_{k+1}-2\rho^{k-1}a_k k+(\rho_0-2(l+1))a_k\rho^{k-1})=0</math>
 
Bring all <math>\rho\!</math>'s to the same power
:<math>k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_k+(\rho_0-2(l+1))a_k=0\!</math>
which can be expressed in the simplest fractional form as
:<math>\frac{a_{k+1}}{a_k}=\frac{2(k+l+1)-\rho_0}{(k+1)(k+2l+2)}</math>
 
where <math>\rho_0=2(N+l+1)\!</math> and <math>N=0,1,2...\!</math> and <math>l=0,1,2,...\!</math>
 
In the limit of large k
:<math>\lim_{k\to\infty}\frac{a_{k+1}}{a_k}\rightarrow\frac{2}{k}</math>
:<math>a_{k+1}=\frac{2}{k}a_k\sim\frac{2^k}{k!}</math>
this will make
:<math>\psi(r) \sim e^{kr} \rightarrow\infty</math>
so we need to break, and make
:<math>a_{k+1}=0\!</math>
from this, we get the energy spectrum.
   
   
The fractional form can be expressed as a confluent hypergeometric function
* [[Commutation Relations]]
:<math>_1F_1(a,c,z)=1+\frac{a}{c}\frac{z}{1!}+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+...=\sum_{k=0}^\infty a_k z^k</math>
* [[Angular Momentum as a Generator of Rotations in 3D]]
It should be noted that a confluent hypergeometric function is a solution of Kummer's equation, which is:
* [[Spherical Coordinates]]
:<math>z\frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0.\,\!</math>
* [[Eigenvalue Quantization]]
Derriving the recursion relations we have:
* [[Orbital Angular Momentum Eigenfunctions]]
:<math>\frac{a_k}{a_{k-1}}=\frac{\frac{a(a+1)(a+2)...(a+k-1)}{c(c+1)(c+2)...(c+k-1)k!}}{\frac{a(a+1)(a+2)...(a+k-2)}{c(c+1)(c+2)...(c+k-2)(k-1)!}}=\frac{a+k-1}{(c+k-1)k}</math>
:<math>\frac{a_{k+1}}{a_k}=\frac{a+k}{(c+k)(k+1)}</math>
by comparison, we find that
:<math>c=2(l+1)\!</math>
:<math>a=-N\!</math>
:<math>z=2\rho\!</math>
 
So the solution of <math>W(\rho)\!</math> is
:<math>W(\rho)=constant _1F_1(-N,2(l+1),2\rho)\!</math>
where :<math>\rho=\kappa r=\sqrt{\frac{-2\mu E}{\hbar^2}}r</math>
Full wavefunction solution with normalization is
:<math>\psi_{n,l,m}(r,\theta,\phi)=\frac{e^{-\kappa r}(2\kappa r)^l}{(2l+1)!}\sqrt{\frac{(2\kappa)^3(n+l)!}{2n(n-l+1)}}\  _1F_1(-n+l+1,2(l+1),2\kappa r)Y_{lm}(\theta,\phi)</math>
 
The first couple of normalized wavefunctions for the hydrogen atom are as follows
 
<math>\psi_{100}= \dfrac{e^{-r/a_o}}{\sqrt{\pi a_o^3}}</math>
 
<math>\psi_{200}= \dfrac{e^{-r/2a_o}}{\sqrt{32\pi a_o^3}} \left( 2-\dfrac{r}{a_o} \right)</math>


<math>\psi_{210}= \dfrac{e^{-r/2a_o}}{\sqrt{32\pi a_o^3}} \left( \dfrac{r}{a_o} \right) cos(\theta)</math>


<math>\psi_{2 \pm 10}= \dfrac{e^{-r/2a_o}}{ \sqrt{64\pi a_o^3}} \left( \dfrac{r}{a_o} \right) sin(\theta) e^{\pm i \phi}</math>
<b>Chapter 8: [[Central Forces]]</b>  
 
The energy is then found to be
:<math>E=-\frac{Z^2 e^4 \mu}{2\hbar^2 n^2}=-\frac{Z^2}{n^2}Ry</math>
 
where <math>Ry=13.6 eV\!</math> for the hydrogen atom and <math>n=1,2,3,...\!</math> and the degeneracy for each level is <math>n^2\!</math>.
 
Below is a chart depicting the energy levels for hydrogen atom for <math>n=1, 2, 3\!</math> in units of <math>Ry</math>. (The parenthesis indicates the degeneracy due to possibile values of the magnetic quantum number <math>m</math> from <math>-l</math> to <math>+l</math>).
 
 
[[Image:Energy levels of H atom.jpeg]]
 
== Continuous eigenvalues and collision theory ==
===Differential cross-section and the Green's function formulation of scattering===
 
 
Much of what we know about forces and interactions in atoms and nuclei has been learned from scattering experiments, in which say atoms in the target are bombarded with beams of particles. These particles are scattered by the target atoms and then detected as a function of a scattering angle and energy. From theoretical point of view, we are now concerned with the continuous part of the energy spectrum. We are free to choose the value of the incident particle energy and by a proper choice of the zero of energy, this corresponds to <math>E>0\!</math> an to eigenfunctions of the unbound states. Before, when we were studying the bound states, the focus was on the discrete energy eigenvalues which allow a direct comparison of theory and experiments. In the continuous part of the spectrum, as it comes into play in scattering, the energy is given by the incident beam, and intensities are the object of measurement and prediction. These being the measures of the likelihood of finding a particle at certain places, are of course related to the eigenfunctions, rather than eigenvalues. Relating observed intensities to calculated wave functions is the first problem in scattering theory.
 
[[Image:Scattering.JPG]]
 
Figure 1: Collimated homogeneous beam of monoenergetic particles, long wavepacket which is approximately a planewave, but strictly does not extend to infinity in all directions, is incident on a target and subsequently scattered into the detector subtending a solid angle <math>d\Omega\!</math>. The detector is assumed to be far away from the scattering center.
 
If <math>I_0\!</math> is the number of particles incident from the left per unit area per time and <math>I(\theta,\phi)d\Omega\!</math> the number of those scattered into the cone per time and if the density of particles in the incident beam is so small such that we can neglect the interaction of the particles with each other and consider their collisions like independent events then this two quantities are proportional to each other. With these considerations the differential cross section is defined as:
:<math>\frac{d\sigma}{d\Omega}=\frac{I(\theta,\phi)}{I_0}</math>
 
There exist two different types of scattering; elastic scattering, where the incident energy is equal to the detected energy and inelastic scattering which arises from lattice vibrations within the sample. For inelastic scattering, one would need to tune the detector detect <math>E+dE\!</math> where <math>dE\!</math> results from the quantum lattice vibrations. For simplification purposes, we will only be discussing elastic scattering.
 
To describe this scattering, start with the stationary Schrodinger equation:
:<math>\left(-\frac{\hbar^2}{2m}\nabla^2+V(\mathbf r)\right)\psi(\mathbf r)=E\psi(\mathbf r)</math>
:<math>\Rightarrow (\nabla^2+k^2)\psi(\mathbf r)=\frac{2mV(\mathbf r)}{\hbar^2}\psi(\mathbf r)</math>
where <math>E=\frac{\hbar^2k^2}{2m}</math> and <math>V(\mathbf r)\!</math> will be assumed to be finite in a limited region of space <math>r<d\!</math>. This is called the range of the force, e.g. nuclear forces <math>d\sim10^{-15}m\!</math> and atomic forces <math>d\sim10^{-10}m\!</math>. Outside this range of forces, the particles move essentially freely. Our problem consists in finding those solutions of the above differential equation which can be written as a superposition of an
incoming and an outgoing scattered waves. We found such solutions by writing the Schrodinger differential equation as an integral equation:
:<math>\psi_k(\mathbf r)=\psi_k^{(0)}( \mathbf r ) +\int d^3r'G_k(\mathbf r,\mathbf r')\frac{2m}{\hbar^2}V(\mathbf r')\psi_k(\mathbf r')</math>
where the Green's function satisfies
:<math>(\nabla^2+k^2)G_k(\mathbf{r},\mathbf{r'})=\delta(\mathbf{r}-\mathbf{r'})</math>
and
:<math>(\nabla^2+k^2 )\psi^{(0)}_k(\mathbf{r})=0</math>
and the solution is chosen such that the second term in the above wave function corresponds to an outgoing wave. Then, the outgoing Green's function can be written as
:<math>G_{k}(\mathbf r,\mathbf r')=-\frac{1}{4\pi}\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}.</math>
 
The full wave function solution will become an integral equation
:<math> \psi _{k}(\mathbf{r})=\psi _{k}^{(0)}(\mathbf{r})-\frac{m}{2\pi \hbar ^{2}}\int {d^{3}r}' \text{ }\frac{e^{ik\left|\mathbf{r}-\mathbf{r}' \right |}}{\left |{\mathbf{r-r'}} \right |}V(\mathbf{r'})\psi _{k}(\mathbf{r'}),</math> 
where the first term is incident plane waves and the second term is scattered waves.
 
The detector is located far away from the scattering potential and we need to discuss the asymptotic behaviour in the limit of <math>r\to\infty\!</math>.
In this limit,
:<math> k \left| \mathbf{r}-\mathbf{r}'\right| \sim kr - k\mathbf{r}' \cdot \mathbf{\hat{r}} = kr - \mathbf{k}' \cdot \mathbf{r}',
</math>
where the wave vector, <math> \mathbf{k}' = k \mathbf{\hat{r}} </math>, is seen far from the scattering potential.
 
Therefore we can write
:<math>
\begin{align}
\lim_{r \to \infty} \psi_k(\mathbf{r})
&= \psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3 r'e^ {-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^{ikr}}{r} \\
&= \psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^{ikr}}{r},
\end{align}
</math>
where the scattering amplitude
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3 r'e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})
</math>
and the angles <math>\theta\!</math> and  <math>\phi\!</math> are the angles between <math>\mathbf{\hat{r}}\!</math> (the vector defining the detector)  and <math>\mathbf{k}\!</math> (the vector defining the in the incoming waves).
 
Now the differential cross section is written through the ratio of the (outgoing) radial current density <math>j_r\!</math> and the incident current density <math>j_{inc}\!</math> as
:<math>\frac{d\sigma}{d\Omega}=\frac{j_r r^2}{j_{inc}}</math>
 
The radial current is
:<math>
j_r=\frac{\hbar}{2mi}\left(\psi_{sc}^*\frac{\partial \psi_{sc}}{\partial r}-\psi_{sc}\frac{\partial \psi_{sc}^*}{\partial r}\right)=\frac{\hbar k}{mr^2}|f_k(\theta,\phi)|^2</math>
:<math>\Rightarrow \frac{d\sigma}{d\Omega}=|f_k(\theta,\phi)|^2</math>
 
For <math>V(\mathbf{r})</math> that is small, the full wave function <math>|\psi_k(\mathbf{r})\rangle  </math> can be obtained by substituting <math>|\psi_k(\mathbf{r})\rangle  </math> into RHS of the integral equation iteratively  as:
 
:<math>
\begin{align}
\psi_k(\mathbf{r})  =\psi_k^{(0)}(\mathbf{r}) + \left(\frac{2m}{\hbar^2}\right) \int d^3r'G_k(\mathbf r,\mathbf r')V(\mathbf{r'})\psi_k^{(0)}(\mathbf{r'}) \\
+ \left(\frac{2m}{\hbar^2}\right)^{2} \int d^3r' \int d^3r'' G_k(\mathbf r,\mathbf r') & V(\mathbf{r'}) G_k(\mathbf r',\mathbf r'')V(\mathbf{r''})  \psi_k^{(0)}(\mathbf{r''})+... 
\end{align}
</math>
 
In 1st Born approximation, we keep only one <math>V(\mathbf{r})</math> term. For large <math>r\!</math> we have
:<math>\psi_k(\mathbf{r})\approx\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3r'e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k^{(0)}(\mathbf{r'})\frac{e^{ikr}}{r}</math>
and
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3r'e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})e^{i\mathbf{k}\cdot \mathbf r'}=-\left(\frac{m}{2\pi\hbar^2}\right)|\langle\mathbf k_{sc}|V|\mathbf k_{inc}\rangle|</math>
 
For a central-force potential, <math> V(r) </math>, the Born scattering amplitude reduces to
:<math> f_k(\theta) = \frac{-m}{2 \pi \hbar^2}\int d^3r'V(r') e^{-i\mathbf q \mathbf r'} </math>
 
where
:<math> \mathbf q = \mathbf k' - \mathbf k </math>
 
is known as the momentum transfer (in units of <math> \hbar </math>). The integral over the solid angle is easily carried out and yields the following result.
: <math> f_{born}(\theta) = \frac{-2m}{\hbar^2} \int_0^\infin dr' V(r') \frac{\sin(qr')}{qr'} {r'}^2 </math>
 
Here we have denoted the scattering angle between <math> k </math> and <math> k' </math> by <math> \theta </math> , and note that <math> k' = k </math> for elastic scattering, so that
: <math> q = 2k \sin\left(\frac{\theta}{2}\right) </math>.
 
As an example, consider the screened Coulomb potential
:<math> V(r) = V_0 \frac{e^{-\alpha r}}{\alpha r}. </math>
 
In the Born approximation , after a simple integration over the angles, we find
: <math>
\begin{align}
f_{born}(\theta)
&= \frac{-2m}{\hbar^2} V_0 \int_0^\infin dr' {r'}^2 \frac{e^{-\alpha r}}{\alpha r'} \frac{\sin(q r')}{qr'} \\
&= \frac{-2m}{\alpha \hbar^2} V_0 \frac{1}{q^2 + \alpha^2} \\
&= \frac{-2mV_0}{\hbar^2 \alpha}  \frac{1}{4k^2 \sin^2(\theta/2) + \alpha^2}
\end{align}
</math>
 
The differential scattering cross section is obtained simply by taking the square of this amplitude.
:<math> \frac{d \sigma}{d \Omega} = \left| \frac{-2mV_0}{\hbar^2 \alpha}  \frac{1}{4k^2 \sin^2(\theta/2) + \alpha^2} \right|^2 </math>
 
The Coulomb potential between two charges <math> q_1 </math> and <math> q_2 </math> is a limiting case of the potential
:<math> V(r) = V_0 \frac{e^{-\alpha r}}{\alpha r} </math>
 
for <math> \alpha \rightarrow 0 \!</math> and <math> V_0 \rightarrow 0 \!</math> with <math> \frac{V_0}{\alpha} = q_1 q_2 \!</math>. Thus, in the Born approximation,
 
:<math>
\begin{align}
\frac{d \sigma}{d \Omega}
&= \frac{m^2 {q_1}^2 {q_2}^2 }{4p^4 \sin^4(\frac{\theta}{2})} \\
&= \frac{{q_1}^2 {q_2}^2}{16E^2 \sin^4(\frac{\theta}{2})}
\end{align}
</math>
 
Note, that the exact Coulomb scattering amplitude differs from the Born amplitude by a phase factor.
 
===Central potential scattering and phase shifts===
 
 
Recall that for scattering, we have Green function method
:<math>\psi_k =\psi_k^{(0)} +\int d^3 r' G(\mathbf r,\mathbf r')V(\mathbf r')\psi_k (\mathbf r'),</math>
where the Green's function satisfies that
:<math>(\nabla^2+k^2)G_k(\mathbf{r},\mathbf{r'})=\delta(\mathbf{r}-\mathbf{r'})</math>
and the solution is chosen such that the second term in Eq.(1) corresponds to an outgoing waves.
:<math>G_k(\mathbf{r},\mathbf{r'})=-\frac{1}{4\pi}\frac{e^ { ik| \mathbf{r}-\mathbf{r'}|}}{|\mathbf{r}-\mathbf{r'}|}</math>
and in the asymptotic limit of r goes to infinity, <math>r\to\infty\!</math>,
:<math>\lim_{r \to \infty}k|\mathbf{r}-\mathbf{r'}|= \lim_{r \to \infty} k \sqrt{ r^2 - 2 \mathbf{r} \cdot \mathbf{r}' + r'^2 } \sim kr \left( 1 - \frac{\mathbf{r}\cdot\mathbf{r}'}{r^2} \right) = kr - k\mathbf{\hat{r}} \cdot \mathbf{r}' = kr - \mathbf{k}'\cdot \mathbf{r}',</math>
where <math> \mathbf{k}' = k \mathbf{\hat{r}} </math>.
Thus
:<math>\lim_{r \to \infty} \psi_k(\mathbf{r})=\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3 r'e^ { -i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^{ikr}}{r}
=\psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^{ikr}}{r},</math>
where
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3 r' e^{-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'}),</math>
where <math> \mathbf{k}' = k \mathbf{\hat{r}} </math> and the angles <math>\theta\!</math> and  <math>\phi\!</math> are the angles between <math>\mathbf{\hat{r}}\!</math> (the vector defining the detector)  and <math>\mathbf{k}\!</math>
(the vector defining the in the incoming waves).
 
For central potentials, i.e. if <math>V(r)=V(|\mathbf{r}|)</math>, then <math>f_k(\theta)\!</math>, i.e. the scattering amplitude does
not depend on the azimuthal angle <math>\phi\!</math>.
To determine <math>f_k(\theta)\!</math>, we need to find the solution of the Schrodinger equation:
 
:<math> \left( -\frac{\hbar^2 }{2m}\nabla^2+V(|\mathbf{r}|) \right) \psi=\frac{\hbar^2 k^2 }{2m}\psi .</math>
We use spherical coordinates and the radial equation is:
   
   
:<math>\left( -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial r^2 }+\frac{\hbar^2 l(l+1)}{2mr^2}+V(|\mathbf{r}|) \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r) .</math>
* [[General Formalism]]
 
* [[Free Particle in Spherical Coordinates]]
For <math>V(r)\!</math> with a finite range <math>d\!</math>, we have shown that for <math>r \gg d\!</math> we have
* [[Spherical Well]]
:<math> \left( -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial r^2 }+\frac{\hbar^2 l(l+1)}{2mr^2} \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r)</math>
* [[Isotropic Harmonic Oscillator]]
* [[Hydrogen Atom]]
* [[WKB in Spherical Coordinates]]


and the solution is a combination of the spherical Bessel functions and the spherical Neumann functions
:<math>\frac{u(r)}{r}=A_l j_l(kr) +B_l n_l(kr).</math>
When <math>r\!</math> is large enough, we use approximation of the spherical Bessel functions and the spherical Neumann functions.


:<math>\frac{u(r)}{r}\rightarrow A_l \frac{\sin(kr-l\frac{\pi}{2})}{kr} -B_l \frac{\cos(kr-l\frac{\pi}{2})}{kr} .</math>
<b>Chapter 9: [[The Path Integral Formulation of Quantum Mechanics]]</b>
Letting
:<math>\frac{B_l }{A_l }=-\tan\delta_l ,</math>
here the angle <math>\delta_l\!</math> is called the phase shift of the <math>\ l^{th}</math>wave and it is the difference in phase between radial parts. Thus, we can rewrite the above expression (up to a normalization constant) as


:<math>\frac{u_l(r) }{r}\rightarrow\frac{\sin(kr-l\frac{\pi}{2} + \delta_l )}{kr} .</math>
* [[Feynman Path Integrals]]
* [[The Free-Particle Propagator]]
* [[Propagator for the Harmonic Oscillator]]


Physically, we expect <math>\delta_l < 0\!</math> for repulsive potentials and <math>\delta_l > 0\!</math> for attractive potentials.
Also, if <math>l/k \gg d\!</math>, then the classical impact parameter is much larger than the range of the
potential and in this case we expect <math>\delta_l\!</math>  to be small.


Now since we are seeking the scattering amplitude with azimuthal symmetry, we can write
<b>Chapter 10: [[Continuous Eigenvalues and Collision Theory]]</b>  
the solution of the Schrodinger equation as a superposition of <math>m=0\!</math> spherical harmonics
only:
:<math>\psi=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{u_l(r) }{r},</math>
:<math>\psi=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr} ,</math>
where the Legendre polynomials are
:<math>P_l(x)= \frac{1}{2^ll!}\frac{d^l }{dx^l }(x^2-1 )^l,</math>
:<math>P_0(x)=1;P_1(x)=x;P_2(x)=\frac{1}{2}(3x^2 -1); \dots .</math>
Let us fix the coeffcients <math>a_l(k)\!</math> by
:<math>e^{ikr\cos\theta}+f_k(\theta)\frac{e^{ikr}}{r}=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr}</math>
which must hold at large <math>r\!</math> and where we chose the coordinates by letting the incident waves
propagate along z-direction. Note that (due to an entirely separate argument):
:<math>e^{ikr\cos\theta}=\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2} ) }{kr}</math>
so
:<math>\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2} ) }{kr}+f_k(\theta)\frac{e^{ikr}}{r}=\sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{\sin(kr-l\frac{\pi}{2}+\delta_l )}{kr} .</math>
We fix the coefficients <math>a_l(k)\!</math> by matching the incoming spherical waves on both sides of the above equation.
Note that this does not involve <math>f_k(\theta)\!</math> since the scattering amplitude controls the outgoing spherical waves.
 
Since <math>\sin (x)=\frac{e^{ix}-e^{-ix}}{2i}\!</math>,
:<math>
\begin{align}
\sum_{l=0}^{\mathop{ \infty}}(2l+1)i^l P_l(\cos\theta) \frac{1}{kr} \frac{1}{2i} \left( e^{i\left( kr-l\frac{\pi}{2} \right)} - e^{-i\left( kr-l\frac{\pi}{2} \right)} \right) +f_k(\theta)\frac{e^{ikr}}{r} \\
= \sum_{l=0}^{\mathop{ \infty}}a_l(k)P_l(\cos\theta) \frac{1}{kr} \frac{1}{2i} \left( e^{i\left(kr-l\frac{\pi}{2}+\delta_l \right)} - e^{-i\left(kr-l\frac{\pi}{2}+\delta_l \right)}\right) .
\end{align}
</math>
 
By matching the coefficients of <math>e^{-ikr}\!</math>, we get
:<math>a_l(k)=(2l+1)i^le^{i\delta_l}</math>
and for the coefficients of <math>e^{ikr}\!</math>, we get
:<math>f_k(\theta)=\frac{1}{k}\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lP_l(\cos\theta) .</math>
Note that <math>\delta_l\!</math> is a function of <math>k\!</math> and therefore a function of the incident energy. If <math>\delta_l(k)\!</math> is
known we can reconstruct the entire scattering amplitude and consequently the differential
cross section. The phase shifts must be determined from the solution of the Schrodinger
equation.
 
 
 
The differential scattering cross section is
:<math>\frac{d\sigma}{d\Omega}=|f_k(\theta) |^2=\frac{1}{k^2 }\left|\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lP_l(\cos\theta) \right|^2</math>
By integrating <math>\frac{d\sigma}{d\Omega}\!</math> over the solid angle <math> \Omega \!</math>, we obtain the total scattering cross section
:<math>
\begin{align}
\sigma_{tot} &= \int \frac{d\sigma}{d\Omega} d\Omega \\
&= \sum_{l=0}^{\infty}\sum_{l'=0}^{\infty}(2l+1)(2l'+1)e^{i\delta_l }e^{-i\delta_{l'}}\sin\delta_l\sin\delta_{l'} \int_{0}^{2\pi} d\phi \int_{0}^{\pi}d\theta \sin\theta P_l(\cos\theta)P_{l'}(\cos\theta) \\
&= \frac{4\pi}{k^2 }\sum_{l=0}^{\mathop{ \infty}}(2l+1)\sin^2\delta_l,
\end{align}
</math>
which follows from the orthogonality of the Legendre polynomials
 
:<math>\int_{-1}^{1}dxP_l(x) P_{l'}(x)=\frac{2}{(2l+1)}\delta_{ll'}.</math>
   
   
Finally, note that since <math>P_l(1) = 1\!</math> for all <math>l\!</math>, we have
* [[Differential Cross Section and the Green's Function Formulation of Scattering]]
:<math>f_k(0) = \frac{1}{k}\sum_{l=0}^{\infty} \left(2l+1\right)e^{i\delta_l}\sin\delta_lP_l(1) = \frac{1}{k}\sum_{l=0}^{\infty} \left(2l+1\right)e^{i\delta_l}\sin\delta_l.
* [[Central Potential Scattering and Phase Shifts]]
</math>
* [[Coulomb Potential Scattering]]
If we take the imaginary part of the scattering amplitude,then
:<math>\Im mf_k(0) = \frac{1}{k} \sum_{l=0}^{\infty} \left(2l+1\right) \sin^2 \delta_l = \frac{k}{4\pi} \sigma_{tot}.
</math>
Therefore, we have
:<math>\sigma_{tot}=\frac{4\pi}{k}\Im mf(0).</math>
This relationship is known as the optical theorem.
The optical theorem is a general law of wave scattering theory. You can see that it relates the forward scattering amplitude to the total cross section of the scattering. It was originally discovered independently by Sellmeier and Lord Rayleigh in 1871.
 
Referring back to the formula for the scattering amplitude, one more important quantity can be discussed:
 
<math>S_l(k)=e^{2i\delta_l(k)}</math>
 
This quantity, for now referred to as the partial scattering due to various angular momenta, is the ratio of the coefficients of the outgoing and incoming waves for a wave scattered on a potential of finite range <math> a \!</math>.
 
These ratios can simplify the problem of evaluating the continuity of the waveform at the boundary <math> r = a \!</math>. In general, if the interior wave function is known to be smoothly continuous across the boundary at <math> r = a \!</math>, then the phase shifts can be expressed in terms of the logarithmic derivatives evaluated at the boundary <math> r = a \!</math>:
 
<math>\beta_l=\left(\frac{a}{f_l(r)}\frac{d f_l(r)}{dr}\right)_{r=a}</math>
 
Using the above equations for the form of <math>f_l(r)</math> beyond the region of scattering, the following relation is found:
 
<math>\beta_l(k) = ka\frac{j'_l(ka)\cos(\delta_l) - n'_l(ka)\sin(\delta_l)}{j_l(ka)\cos(\delta_l) - n_l(ka)\sin(\delta_l)}</math>
 
Thus with algebraic manipulation:
 
<math>S_l = e^{2i\delta_l}= -\frac{j_l(ka) -in_l(ka)}{j_l(ka) + in_l(ka)}  \frac{\beta_l - ka\frac{j'_l(ka) - in'_l(ka)}{j_l(ka)-in_l(ka)}}{\beta_l - ka\frac{j'_l(ka)+in'_l(ka)}{j_l(ka)+in_l(ka)}}</math>
 
Note that if <math>\beta_l \to \infty\!</math>, that is to say <math>f_l(a)=0\!</math> then only the first portion of this expression survives. This is a special quantity corresponding to hard sphere scattering, therefore in general the real phase angles <math>\xi_l\!</math> (the ''hard sphere phase shifts'') are expressed such that:
 
<math>e^{2i\xi_l} = -\frac{j_l(ka)-in_l(ka)}{j_l(ka)+in_l(ka)}</math>
 
Note that these phase shifts are present for any potential, not just that of a hard sphere.
 
[[Worked Problem for Scattering on a Delta-Shell Potential]]
 
==== Scattering by Square Well potential ====
Consider a beam of point particles of mass m scattering from a finite spherical attractive well of depth '''''V<sub>o</sub>''''' and radius '''''a'''''
<br />
<math> V(r) = -V_0 </math> for <math>r<a</math> <br />
<math> V(r) = 0 </math> for <math>r>a</math><br />
 
The Schrödinger equation for <math>r<a</math> <br />
 
<math>\frac{d^2 R_l}{dr^2} + \frac{2}{r} \frac{dR_l}{dr} -\frac{l(l+1)}{r^2}R_l + \frac{2m}{\hbar^2}(E+V_0)R_l = 0</math>
 
Hence, the solution for above differential equation is <math>\ {R_{l}}=A_l j_l(Kr)</math> where <math>\kappa^2 = \frac{2m}{\hbar^2}(E+ V_0)</math><br />
Another solution for the region <math>r>a</math>  is
 
<math>\ {R_l}= B j_l (kr)+C \eta_l(kr)</math> <br /> where A,C,B are arbitrary constants and <math>k^2 =  \frac{2m}{\hbar^2}E</math><br />
Now, for large r, <br />
<math>\ R_l \approx  e^{i\delta_l}\sin{{\frac{kr-l\pi/2 + \delta_l}{kr}}}</math><br />
<math>\frac{C}{B}= -\tan{\delta_l(kr)}</math><br />
We'll basically apply 2 boundary conditions now, continuity of <math>R_l</math> and matching the logarithmic derivatives at <math>r=a</math>, as a result we obtain,
 
<math>-\frac{C}{B}= \tan{\delta_l}=\frac{kj_l'(ka)j_l(\kappa a)-\kappa j_l(ka) j_l'(\kappa a)}{k\eta_l'(ka)j_l(\kappa a)-\kappa \eta_l(ka) j_l'(\kappa a)}</math><br /><br />
Lets discuss some limiting cases:<br />
(a) Consider the case when <math>ka<< l</math> and <math>\kappa a << l</math>, then with some simplification we get  <math>\tan{\delta_l}\approx k^{2l+1} \approx E^{l+1/2}</math><br /><br />
This behavior is a result of the centrifugal barrier that keeps waves of energy far below the barrier from feeling the effect of the potential.<br />
(b) When the phase shift is <math>\pi/2</math> then the partial wave cross section  <math>\sigma_l(k)= \frac{4\pi (2l+1)}{k^2}\sin^2{\delta_l}</math> is maximum. Then, we have a resonant scattering.<br />
From(a), we see that the phase shift is small for ka small.But, when ka changes and passes the resonant condition, the phase shift rises rapidly and has a sharp peak at resonant energy <math>E_R</math>. This can be represented as<br /> <math> \tan{\delta_l} \approx \frac{\gamma (ka)^{2l+1}}{E-E_R}</math><br />
So, partial wave cross section is <math>\sigma_l= \frac{4\pi(2l+1)}{k^2}\frac{[\gamma (ka)^{2l+1}]^2}{(E-E_R)^2 +[\gamma (ka)^{2l+1}]^2 }</math>, which is the ''Breit-Wigner formula'' for resonant cross section
 
===Born approximation and examples of cross-section calculations===
 
 
 
In scattering theory, especially in quantum menchanics, the Born approximation consists of taking the incident field in place of the total field as the driving field at each point in the scatterer. It is the perturbation method applied to scattering by an extended body. It is accurate if the scattered field is small, compared to the incident field, in the scatterer.
 
Recall the scattering of a particle in a potential <math>V(r)\!</math> has a differential cross section of:
:<math>\frac{d\sigma}{d\Omega}=|f_{\mathbf{k}}(\mathbf{\hat{r}})|^2 </math>
 
where the scattering amplitude, <math>f_{\mathbf{k}}(\mathbf{\hat r})\!</math>, is the coefficient of the outgoing waves.
 
The scattering amplitude is defined as the coefficient of the outgoing waves in the asymptotic solution (for large <math>r\!</math>)
:<math>\psi_{\mathbf{k}}(\mathbf{r})\approx N\left(e^{i\mathbf{k}\cdot\mathbf{r}}+\frac{e^{ikr}}{r}f_{\mathbf{k}}(\mathbf{\hat{r}})\right)</math>
 
of the Schrodinger equation:
:<math>(\nabla^2+k^2)\psi=\frac{2m}{\hbar^2}V\psi.</math>
 
The scattering amplitude for elastic scattering in the direction <math> \mathbf{\hat k}' \!</math> is written by
:<math>f_{\mathbf{k}}(\mathbf{\hat k}') = -\frac{m}{2\pi\hbar^2 N}\int d^3 r' e^ {-i\mathbf{k}'\cdot\mathbf{r'}}V(\mathbf{r'})\psi_{\mathbf{k}} \left( \mathbf{r}'\right)</math>.
 
However, since <math> \psi_{\mathbf{k}} \!</math> in the integrand is not an explicit form, if we replace the <math> \psi_{\mathbf{k}}\!</math> by the normalized plane waves, the scattering amplitude can be approximated by
 
:<math>f_{\mathbf{k}}(\mathbf{\hat k}')\approx -\frac{m}{2\pi\hbar^2 }\int d^3 r' e^{-i\mathbf{k}'\cdot\mathbf{r'}} V(\mathbf{r'}) e^{i\mathbf{k}\cdot\mathbf{r'}} ,</math>
where <math>-i\mathbf{k}'\!</math> is the scattered portion and <math>i\mathbf k\!</math> is the incident portion.This is known as the scattering amplitude in the first Born approximation which is a technique to find solutions when <math>V(r)\!</math> is small compared to <math> E - [l(l+1)\frac{\hbar^2}{2mr^2}]</math> .
 
 
'''Born Approximation for Spherically Symmetric Potentials:'''
 
 
For a central-force potential <math>V(r)\!</math>, the Born scattering amplitude reduces to
:<math>f_B(\theta)=-\frac{m}{2\pi\hbar^2}\int V(r')e^{-i\mathbf{q}\cdot\mathbf{r}'}d^3r'</math>
 
where <math> \mathbf{q}=\mathbf{k}'-\mathbf{k}\!</math> is known as the momentum transfer in scattering theory. In fact, this amplitude leads to the Born cross section which is:
 
:<math>\left(\frac{d\sigma}{d\Omega}\right)_B=\left(\frac{m}{2\pi\hbar^2}\right)^2\left|\langle \mathbf{k}_s|r|\mathbf{k}_i\rangle\right|^2 .</math>
 
We may align the polar axis for the <math>r'\!</math> integral lie along this momentum transfer <math> \mathbf{q} \! </math>.  Then, we have
:<math>\mathbf{q} \cdot \mathbf{r}'= q r' \cos \theta'</math>.
 
Our first Born integration then takes the form:
:<math>f_B(\theta) = -\frac{m}{\hbar^2}\int e^{-i qr'\cos\theta'}V(r)r'^2\sin\theta' dr'd\theta'</math>.
 
The <math> \phi' \!</math> integral <math>\int\nolimits_{0}^{2\pi } {d}\phi' </math> gives <math>2\pi\!</math>  while <math>\theta'\!</math> integral can be determined by using the following identity:
:<math>\int_0^\pi e^{-iqr'\cos\theta'}\sin\theta' d\theta'=\frac{2\sin(qr')}{qr'}</math>
and we find the scattering amplitude as:
:<math>f_B(\theta) = -\frac{2m}{\hbar^2}\int_0^\infty V(r')\frac{\sin(qr')}{qr'} r'^2dr'</math>
 
where the angular dependence of <math>f_B(\theta)\!</math> is carried by <math>\mathbf{q}\!</math> and note that <math> \left|\mathbf{k}'\right| = \left| \mathbf{k} \right| = k \! </math> for elastic scattering, so that
:<math>q= \left| \mathbf{k}' - \mathbf{k} \right| = 2k\sin\frac{\theta}{2}</math>.
 
 
'''Example 1:  Gaussian potential  '''
 
Consider the scattering amplitude from a Gaussian potential of the form
:<math>V(r)=Ae^{-\alpha r^2}</math>
 
Our scattering amplitude then becomes:
:<math>
\begin{align}
f_B(\theta) &= -\frac{2mA}{\hbar^2q}\int_0^\infty r'e^{-\alpha r'^2}\sin(q r')dr' \\
&= -\frac{2mA}{\hbar^2q}\int_0^\infty \frac{\partial}{\partial r'}\left(-\frac{1}{2\alpha}e^{-\alpha r'^2}\right)\sin(q r')dr' \\
&= \frac{mA}{\alpha\hbar^2 q}\left(0 + q\int_0^\infty e^{-\alpha r'^2}\cos(q r')dr'\right) \\
&= \frac{mA}{\alpha\hbar^2}\frac{\sqrt{\pi}}{2\sqrt{\alpha}}e^{-\frac{q^2}{4\alpha}} \\
&= \frac{mA\sqrt{\pi}}{2\hbar^2\alpha^{\frac{3}{2}}}e^{-\frac{q^2}{4\alpha}}
\end{align}
</math> .
 
 
'''Problem:  The scattering amplitude and the differential cross section for a exponential potential:''' [http://wiki.physics.fsu.edu/wiki/index.php/Problem_exponential_potential  Problem_Exponential_Potential]
 
 
'''Example 2:  Hard sphere potential '''
:<math>
V(\mathbf{r})=
\begin{cases}
\infty & r \leq d \\
0 & r>0
\end{cases}
</math>
 
For scattering state
:<math>u_{l}(r)=
\begin{cases}
A_{l}j_{l}(kr)+B_{l}n_{l}(kr) & r\geq d \\
0 & r \leq d
\end{cases}
</math>
 
For <math> r\rightarrow\infty</math>,
:<math>
u_{l}(r) \longrightarrow \frac{A_{l}}{kr\cos\delta_{l}}\sin(kr-l\frac{\pi}{2}+\delta_{l})
</math>
 
Matching continuity boundary condition at <math>r=d</math>, we get,
:<math>\frac{B_{l}}{A_{l}}=-\frac{j_{l}(kd)}{n_{l}(kd)}=-\tan\delta_{l}</math>
so the scattering phase shift of the <math>l^{th}</math> wave is:
:<math>
\delta_{l}=\tan^{-1}\frac{j_{l}(kd)}{n_{l}(kd)}
</math>
For <math>kd \ll 1  </math>,
:<math>j_{l}\approx \frac{x^{l}}{(2l+1)!!}</math>
:<math>n_{l}\approx \frac{(2l-1)!!}{x^{l+1}}</math>
 
so
:<math> \delta_{l}\approx\tan^{-1}(-\frac{(kd)^{2l+1}}{2l+1})  \approx -\frac{(kd)^{2l+1}}{2l+1}  </math>
The <math>l=0</math> term dominates in the scattering process, so the scattering amplitude and the cross section are:
 
:<math> f_{k}(\theta)\approx  -\frac{1}{k}e^{-ikd} P_{0}(\cos\theta) \sin kd    </math>
:<math> \sigma= \int d\Omega |f_{k}(\theta)|^{2} = 4\pi d </math>
 
 
 
'''Problem''' : [http://wiki.physics.fsu.edu/wiki/index.php/User_talk:WeiChiaChen]
 
===Coulomb potential scattering===
'''Example 1'''
 
Let's look at an example of a Screened Coulomb (Yukawa) Potential, where we have a potential:
:<math>V(r)=V_0e^{-\alpha r}\frac{1}{r}</math>
 
The scattering amplitude can be written by:
:<math>
\begin{align}
f &= -\frac{m}{2\pi\hbar^2}\int_0^\infty dr' r'^2 2\pi\frac{e^{-i|\mathbf{k}'-\mathbf{k}|r'}-e^{i|\mathbf{k}-\mathbf{k}'|r'}}{-i|\mathbf{k}'-\mathbf{k}|r'}\frac{V_0e^{-\alpha r'}}{r'} \\
&= -\frac{2mV_0}{\hbar^2}\int_0^\infty dr' r'^2 \frac{\sin\left(|\mathbf{k}'-\mathbf{k}|r'\right)}{|\mathbf{k}'-\mathbf{k}|r'}\frac{e^{-\alpha r'}}{r'} \\
&= -\frac{2mV_0}{\hbar^2}\frac{1}{\left(\left|\mathbf{k}'-\mathbf{k}\right|\right)^2+\alpha^2}
\end{align} 
</math>
 
For the elastic scattering <math> \left| \mathbf{k} \right| = \left| \mathbf{k'} \right| = k \!</math>. Therefore,
:<math>
(\left|\mathbf{k}'-\mathbf{k}\right|)^2 = 2k\sin\frac{\theta}{2}
</math>
 
Thus, we have
:<math>
f = -\frac{2mV_0}{\hbar^2}\frac{1}{4k^2\sin^2\frac{\theta}{2}+\alpha^2}
</math>
 
 
thus we have the differential cross section:
:<math>\frac{d\sigma}{d\Omega}= \left( \frac{2mV_0}{\hbar^2} \right) ^2 \left( \frac{1}{4k^2\sin^2\frac{\theta}{2}+\alpha^2}\right) ^2</math>
 
 
 
 
'''Comparison the differential cross sections for various potential type'''
 
[[Image:Yukawa_Gaussian_Exp.jpg|500px]]
 
Here we compare the differential cross sections for the Yukawa, Gaussian and exponential potential. We can see that for small <math> q\alpha \!</math> all differential cross sections are similar. The differential cross section of the exponential potential decreases faster than the others and that of the Yukawa potential decreases little bit slower than the others.
 
 
'''Example 2'''
 
When we are considering scattering due to the Coulomb potential, we can not neglect the effect of this potential at large distances because it is only a <math>\frac{1}{r}</math> potential.
 
Use a change of coordinates from Cartesian to parabolic coordinates:
:<math>\xi=\sqrt{x^2+y^2+z^2}-z</math>
:<math>\eta=\sqrt{x^2+y^2+z^2}+z</math>
:<math>\phi=\tan^{-1}(\frac{y}{x})</math>
 
The following is a picture of parabolic coordinates:
 
[[Image:ParabolicCoordinates.png]]
 
<math>\phi\!</math> represents rotation about the z-axis, <math>\xi\!</math> represents the parabolas with their vertex at a minimum, and <math>\eta\!</math> represents parabolas with their vertex at a maximum.
 
So now we can write the Schrodinger equation in parabolic coordinates:
:<math> \left[ - \frac{\hbar^2}{2\mu}\frac{4}{\xi+\eta} \left( \frac{\partial}{\partial \xi}\xi\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta}\eta\frac{\partial}{\partial \eta}+\frac{\xi+\eta}{4\xi\eta}\frac{\partial^2}{\partial \phi^2} \right) -\frac{2Ze^2}{\xi+\eta} \right] \psi=\frac{\hbar^2k^2}{2\mu}\psi</math>
 
So we will seek solutions which are independent of <math>\phi\!</math>. Recall that the scattering amplitude is a function of <math>\theta\!</math> only.
 
Look for solution of the form:
 
:<math>\psi=e^{\frac{i}{2}k(\eta-\xi)}\Phi(\xi)=e^{ikz}\Phi(r-z)</math>
 
Then it arrives:
:<math> \left[ \xi\frac{\partial^2}{\partial \xi^2}+(1-ik\xi)\frac{\partial}{\partial \xi}+\frac{Ze^2\mu}{k\hbar^2}k \right]\Phi(\xi)=0</math>
 
We can tidy up the notation a little bit by using the following substitution:
:<math>\lambda=\frac{Ze^2\mu}{k\hbar^2}</math>
 
Now let:
:<math>\Phi(\xi)=\sum_{n=0}^\infty a_n \xi^n</math>
 
From this we can write:
:<math>\frac{a_{n+1}}{a_{n}}=\frac{in-\lambda}{(n+1)^2}k=\frac{n+i\lambda}{(n+1)^2}ik</math>
 
 
Recall the confluent hypergeometric function:
:<math>_1F_1(a,c,z)=1+\frac{a}{c}z+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+\dots+\frac{a(a+1)\dots(a+n-1)}{c(c+1)\dots(c+n-1)}\frac{z^n}{n!}+\dots</math>
 
We can then write the recursion formula as the following:
:<math>\dfrac{\alpha_{n+1}}{\alpha_n}= \left( \dfrac{a+n}{c+n} \right) \dfrac{1}{n+1}</math>
 
This implies that:
:<math>\Phi(\xi)=A_1F_1(i\lambda,1,ik\xi)\!</math>
where the confluent geometric function is written in terms of three new variables, and <math>A_1\!</math> is a c-number.
 
Now we can write the wavefunction due to Coulomb scattering:
:<math>\psi(r,z)=A_1F_1(i\lambda,1,ik(r-z))e^{ikz}\!</math>
 
Now we should look at the limit where z is taken to go to infinity, and our confluent hypergeometric function is rewritten as:
:<math>_1F_1(a,c,z)=\frac{\Gamma(c)(-z)^a}{\Gamma(c-a)} \left[ 1-\frac{a(a-c+1)}{z} \right] +\frac{\Gamma(c)}{\Gamma(a)}e^zz^{a-c}</math>
 
Now we can use this to rewrite our equation for <math>\Phi\!</math> of <math>\xi\!</math>:
:<math>\Phi(\xi)=Ae^{\frac{-\pi}{2}\lambda} \left[ \frac{e^{i\lambda \ln(k\xi)}}{\Gamma(1-i\lambda)} \left( 1-\frac{\lambda^2}{ik\xi} \right) +\frac{i\lambda}{ik\xi}\frac{e^{ik\xi+i\lambda \ln(k\xi)}}{\Gamma(1+i\lambda)} \right] </math>
 
Rewriting our wavefunction <math>\psi\!</math>:
:<math>\psi(r,\theta)=\frac{Ce^{\frac{-\pi\lambda}{2}}}{\Gamma(1-i\lambda)} \left[ \left( 1-\frac{\lambda^2}{2ikr}\frac{1}{\sin^2\frac{\theta}{2}} \right) e^{ikz}e^{i\lambda \ln(k(r-z))}+\frac{f(\theta)}{r}e^{ikr+i\lambda \ln(2kr)} \right] </math>
where <math>f(\theta)\!</math> is the following:
:<math>f(\theta)=\frac{\lambda\Gamma(1-i\lambda)}{2k\Gamma(1+i\lambda)}\frac{1}{\sin^2\frac{\theta}{2}}e^{i\lambda \ln(\sin^2\frac{\theta}{2})}=\frac{\lambda\Gamma(1-i\lambda)}{2k\Gamma(1+i\lambda)} \left( \sin^2\frac{\theta}{2} \right) ^{i\lambda-1}</math>
 
We can then get our differential cross section from that by squaring it:
:<math>\frac{d\sigma}{d\Omega}=|f(\theta)|^2=\frac{\lambda^2}{4k^2\sin^4(\frac{\theta}{2})}=\frac{(Ze^2)^2}{16E^2}\frac{1}{\sin^4(\frac{\theta}{2})}</math>
 
If we normalize the wavefunction to give unit flux at large distances, we must take the following for the constant <math>C\!</math>:
:<math>C=\sqrt{\frac{\mu}{\hbar k}}\Gamma(1-i\lambda)e^{\frac{\pi\lambda}{2}}</math>
 
So the wavefunction at large distances is given by the following:
:<math>|\psi(0)|^2=|C|^2=\frac{\mu}{\hbar k}|\Gamma(1-i\lambda)|^2e^{\pi\lambda}</math>
where
:<math>|\Gamma(1-i\lambda)|^2=\frac{2\pi|\lambda|e^{-\pi\lambda}}{e^{2\pi\lambda}-1}</math>
 
Plugging this in for our wavefunction squared:
:<math>|\psi(0)|^2=\frac{2\pi|\lambda|}{\frac{\hbar k}{\mu}|1-e^{2\pi\lambda}|}</math>
 
Now let's use the following quantity to represent the velocity:
:<math>\frac{\hbar k}{\mu}=v</math>
 
For small incident velocities, we can write:
:<math>|\psi(0)|^2=\frac{2\pi Ze^2}{\hbar^2 v^2}</math>
:<math>|\psi(0)|^2=\frac{2\pi Ze^2}{\hbar^2 v^2}e^{-2\pi\frac{Ze^2}{\hbar v}}</math>
where the first equation is for an attractive Coulomb potential, and the second equation is for a repulsive Coulomb potential.
 
The factor <math> e^{-2\pi\frac{Ze^2}{\hbar v}} </math> is known as the Gamow factor.
 
The Gamow Factor or Gamow-Sommerfeld Factor, named after its discoverer George Gamow, is a probability factor for two nuclear particles' chance of overcoming the Coulomb barrier in order to undergo nuclear reactions. For example, nuclear fusion. By classical physics, there is almost no possibility for protons to fuse by crossing each other's Coulomb barrier, but when George Gamow instead applied quantum mechanics to the problem, he found that there was a significant chance for the fusion due to tunneling.
 
The Gamow factor for the Coulomb potential can also be obtained via the WKB method.
 
Recall that <math> \theta = e^{\frac{1}{\hbar} \int_a^b |p(x')|dx'}</math>  and <math>p(x')=\sqrt{\frac{2m}{\hbar^2}(E-V(x'))}</math>
 
For a broad, high barrier <math> \theta >> 1 </math>, and
:<math> T \approx \frac{1}{\theta^2} </math>.
 
In this case <math> \theta = e^{\frac{1}{\hbar} \int_{-a}^0 |p(x')|dx'} </math>
 
Using our previous potential <math> V(x')=\frac{Ze^2}{|x'|} </math> and <math> E=\frac{Ze^2}{a} </math> we get
:<math>\int_a^b |p(x')|dx' = \int_{-a}^0 \sqrt{\frac{2m}{\hbar^2}(|E|-\frac{Ze^2}{|x'|})}dx' = \frac{2mE}{\hbar} \int_{-a}^0 \sqrt{|1-\frac{a}{x'}|} dx'</math>
:<math> => \frac{\sqrt{2mE}}{\hbar}a \int_0^1 \sqrt{\frac{1}{u}-1}du = \frac{\sqrt{2mE}}{\hbar}a\frac{\pi}{2} = \frac{\sqrt{2m}}{\hbar}\frac{Ze^2 \sqrt{E}}{E} \frac{\pi}{2}</math>
:<math> = \frac{\sqrt{2m}}{\hbar} \frac{\pi}{2} \frac{Ze^2}{\sqrt{E}} =  \frac{\sqrt{2m}}{\hbar} \frac{\pi}{2} \frac{Ze^2}{\sqrt{\frac{1}{2}mv^2}} = \pi \frac{Ze^2}{{\hbar}v}</math>
 
So, as above, the transmission coefficient for small v is
:<math> T \approx e^{\frac{-2\pi}{\hbar} \frac{Ze^2}{v}} </math>
 
===Two particle scattering===
[[Image:particle scattering.jpg]]
 
Classically, if we wish to consider a collection of identical particles, say billiard balls, it is always possible to label all the balls such that we can follow a single ball throughout interacting with others.  It is not, however, possible to attach such labels to a quantum mechanical systems of, say, electrons.  Quantum mechanical particles are far to small to attach such physical labels and there are not enough degrees of freedom to label each particle differently.  Again considering the classical case, one could imagine simply recording the position of a given particle throughout its trajectory to distinguish it from any other particle.  Quantum mechanically, however, we again fail in following a single particles trajectory since each time we make a measurement of position we disturb the system of particles in some uncontrollable fashion.  If the wave functions of the particles overlap at all, then the hope of following a single particles trajectory is lost.  We now attempt to study the consequences of such indistinguishably between identical quantum particles.
 
'''Scattering of Identical Particles'''
 
Let's look at the case of two identical bosons (spin 0) from their center of mass frame. To describe the system, we must use a symmetrized wave function.  Under the exchange <math>\mathbf{r}_1\leftrightarrow \mathbf{r}_2\!</math>, <math>\mathbf{r}_{cm} = (\mathbf{r}_1 + \mathbf{r}_2)/2\!</math> is invariant while <math>\mathbf{r} = \mathbf{r}_1-\mathbf{r}_2\!</math> changes sign. So the center of mass wave function is already symmetric. Furthermore, the wave function has even parity.  This implies that the only possible eigenstates of angular momentum of the two particles are those with even angular momentum quantum numbers. This is evident from the property of the associated Legendre polynomials.
 
But we have to symmetrize <math>\psi(\mathbf{r})\!</math> by hand:
:<math>Y_{lm}(-\mathbf{\hat{r}})=(-1)^lY_{lm}(\mathbf{\hat{r}})\!</math>
 
Under the transformation:
:<math>\mathbf{\hat{r}}\rightarrow -\mathbf{\hat{r}}, \theta\rightarrow\pi-\theta, \phi\rightarrow\phi+\pi</math>
 
The first two terms of the symmetrized wave function represent the incident waves corresponding to the center of mass frame.  Note that because we are considering identical particles we cannot distinguish the target particle from the incident one.  Thus, each particle has equal amplitude of being either one.
 
The scattering amplitude is:
:<math>f_{sym}(\theta,\phi)=f(\theta,\phi)+f(\pi-\theta,\phi+\pi)\!</math>
<math>\theta\!</math>and <math>\pi-\theta\!</math> can then be associated with the angle through which each particle is scattered.  The total amplitude for particles to emerge at each angle is then exactly the sum of amplitudes for emerging at each angle, which is given above.  The scattering amplitude remains consistent with the fact that we have two identical particles, and this gives us the differential cross section:
:<math>\frac{d\sigma}{d\Omega}=|f(\theta,\phi)+f(\pi-\theta,\phi+\pi)|^2=|f(\theta,\phi)|^2+|f(\pi-\theta,\phi+\pi)|^2+2\Re e[f(\theta,\phi)f^*(\pi-\theta,\phi+\pi)]</math>
 
Note:
 
The first two terms in the differential cross section is what we would get if we had two distinguishable particles, while the third term give the quantum mechanical interference that goes along with identical particles.
 
As an example, consider scattering through a 90 degree angle. We then have:
:<math>f(\theta)=f(\pi-\theta)=f\left(\frac{\pi}{2}\right)</math>
 
Now if the particles are distinguishable, the cross section for observing a scattered particle at 90 degrees is then:
:<math>\left(\frac{d\sigma}{d\Omega}\right)_{dis}=2\left|f\left(\frac{\pi}{2}\right)\right|^2</math>
 
Where if the particles are indistinguishable, we see above that we will have:
:<math>\left(\frac{d\sigma}{d\Omega}\right)_{ind}=4\left|f\left(\frac{\pi}{2}\right)\right|^2</math>
 
Thus the differential cross-section is exactly twice the distinguishable case when the particles are indistinguishable.

Latest revision as of 14:59, 8 April 2014

Quantum.png

Welcome to the Quantum Mechanics A PHY5645 Fall2008/2009

Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian , it describes how a state evolves in time.

This is the first semester of a two-semester graduate level sequence, the second being PHY5646 Quantum B. Its goal is to explain the concepts and mathematical methods of Quantum Mechanics, and to prepare a student to solve quantum mechanics problems arising in different physical applications. The emphasis of the courses is equally on conceptual grasp of the subject as well as on problem solving. This sequence of courses builds the foundation for more advanced courses and graduate research in experimental or theoretical physics.

The key component of the course is the collaborative student contribution to the course Wiki-textbook. Each team of students is responsible for BOTH writing the assigned chapter AND editing chapters of others.

Team assignments: Fall 2009 student teams

Fall 2009 Midterm is October 15

Outline of the Course

Chapter 1: Physical Basis of Quantum Mechanics


Chapter 2: Schrödinger Equation


Chapter 3: Operators, Eigenfunctions, and Symmetry


Chapter 4: Motion in One Dimension


Chapter 5: Discrete Eigenvalues and Bound States - The Harmonic Oscillator and the WKB Approximation


Chapter 6: Time Evolution and the Pictures of Quantum Mechanics


Chapter 7: Angular Momentum


Chapter 8: Central Forces


Chapter 9: The Path Integral Formulation of Quantum Mechanics


Chapter 10: Continuous Eigenvalues and Collision Theory

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=Phy5645&oldid=15857"