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''' 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.
]]
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.
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:''' [[Phy5645_Fall09_teams|Fall 2009 student teams]]
 
'''Fall 2009 Midterm is October 15'''
 
----
'''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. 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. 
 
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.
 
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.
 
=== Photoelectric Effect ===
Another contributing factor to the emergence of the theory of Quantum Mechanics came with the realization of the dual 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''.
 
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 by an EM field must radiate, 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 are the dimensions 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 <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 of an electron in an atom is needed.
 
===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.
 
==  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>\frac{d^2x}{dt^2}=-\frac{\partial V}{\partial x}</math>
 
The energy for a 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 (below) subject to a (scalar) potential <math>V(x,t)</math>:    <math> i\hbar\frac{\partial}{\partial t}\psi(x,t)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x,t) </math> 1D Schrodinger equation subject to a (scalar) potential                        2.0.1
While in 3D: 
 
<math> i\hbar\frac{\partial}{\partial t}\psi(r,t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(r)\right]\psi(r,t)</math> 3D Schrodinger equation subject to a (scalar) potential                        2.0.2                                           
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.
 
 
=== Stationary states ===
Stationary states are the energy eigenstates of the Hamiltonian operator. These states are called "stationary states" because their eigenvalues <math>E</math> 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>. Then the eigenvalue equation <math>H\psi(\textbf{r}) = E\psi(\textbf{r})</math> can be written as:
 
<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>
 
where <math>\psi(\textbf{r})</math> is the eigenfunction of the Hamiltonian operator. This equation is known as the time-independent Schrödinger equation.
 
If time t is fixed, evaluated and integrated over all space, the probability will be equal to one. This is a consequence of the wavefunction being normalized over time.
 
 
The probability amplitude, <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 the magnitude of <math>\psi^2</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(r,t_0)|^2=1 \Rightarrow \int dr^3 |\psi(r,t)|=1 \;\;\forall t</math>
 
Does the solution to the Schrodinger Equation conserve the probability, i.e. are we guaranteed that 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(x,t)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x,t)</math> <br/>
<math>i\hbar\psi^*(x,t)\frac{\partial}{\partial t}\psi(x,t)=\psi^*(x,t)\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x,t)</math> <br/>
Taking a complex conjugate <br/>
<math>-i\hbar\psi(x,t)\frac{\partial}{\partial t}\psi^*(x,t)=\psi(x,t)\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi^*(x,t)</math> <br/>
and taking the difference of the above equations we finally find
 
<math>\frac{\partial}{\partial t}\psi^*(x,t)\psi(x,t)+\nabla\cdot\frac{\hbar}{2im}
\left[\psi^*(x,t)\nabla \psi(x,t)-\psi^*(x,t)\nabla \psi(x,t)\right]=0</math>
 
Note that this is in the form of a continuity equation
 
<math>\frac{\partial}{\partial t} \rho(r,t) + \nabla \cdot j(r,t)=0</math>
where
 
<math>\rho(r,t)=\psi^*(r,t) \psi(r,t)\!</math>
 
is the probability density
 
and 
 
<math>j(r,t)=\frac{\hbar}{2im}\left[\psi^*(r,t)\nabla \psi(r,t)-\psi^*(r,t)\nabla \psi(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(r,t)|^2=-\int d^3r(\nabla\cdot j)=-\oint dA\cdot 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.
The time independence of the probability to find particle somewhere in space is what we wished to prove.
 
=== 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 called the 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. So the basis in a Hilbert space looks like <math> \left( |\psi_0\rangle, |\psi_1\rangle, |\psi_2\rangle, ... , |\psi_j\rangle, ... \right) </math>.
 
A state vector <math> \psi\ </math> in Hilbert space is denoted in Dirac notation by a “ket” <math>| \psi \rangle</math>, and its complex conjugate <math> \psi\ </math>* is denoted by a “bra” <math>\langle\psi |</math>.
 
So the basis in a Hilbert space looks like <math> \left( |\psi_0\rangle, |\psi_1\rangle, |\psi_2\rangle, ... , |\psi_j\rangle, ... \right) </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.
 
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>
 
In Dirac's notation, the eigenfunctions are replaced by eigenkets (or simply kets). In this notation Schrödinger's equation is written as
 
<math>i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle=\mathcal{H}|\psi(t)\rangle </math>
 
For time independent Hamiltonians, the above equation separates and we can seek the solution of the form of (stationary states)
 
<math>|\psi(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>
 
=== Heisenberg Uncertainty relations ===
 
Consider a long string which contains a wave that moves with a fairly well-defined wavelength.  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 being held now has a well defined position since the wave's small bump-like wave is noticable.  Now the question, "what is the wavelength" does not make sense, since there is no well defined period.  Obviously there is a limitation on measuring simultaneously the wavelength and the position.  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\mathcal{O}</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 shown to hold in most any undergraduate text:
 
<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.
 
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]
 
== Motion in one dimension ==
 
'''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>
Schrodinger equation:
:<math>[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]\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 solution of Schrodinger equation are trigonometric function (sine or cosine). The wave function is oscillating at both <math>-\infty</math> and <math>+\infty</math>. The particle is in unbound state. The energy spectrum is continous. Both oscillating solutions are allowed, the energy level 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 bound state. The energy spectrum is discrete and non-degenerate.
 
=== 1D bound states ===
 
====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 those of the simple harmonic oscillator, i.e. <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.
 
==== 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 properties 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> (<math>\hat{P} \psi(x)=\psi(-x)</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.
 
==== Non-degeneracy of the bound states in 1D====
 
Let's consider a more general property that is the 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}(\frac{d\psi_E^{(1)}}{dx}\psi_E^{(2)}-\frac{d\psi_E^{(2)}}{dx}\psi_E^{(1)})=0</math>
 
<math>\rightarrow \frac{dW}{dx}=0</math>
 
<math>\rightarrow W=C</math>
 
where <math>C\!</math> is constant.
So, the 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}[ln(\psi_E^{(1)})-ln(\psi_E^{(2)})]=0</math>
 
<math>\rightarrow ln(\psi_E^{(1)})-ln(\psi_E^{(2)})=constant</math>
 
<math>\rightarrow \psi_E^{(1)}=constant.\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.
 
=== 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 plus/minus 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.
 
For the delta function potential the derivative of the wave-function is not differential, but has a step.  Integrating the schrodinger wave equation from just one side of the step to just the other and then taking the limit as the difference between the integral limits becomes zero.
 
 
We have to know wave functions and discrete energy levels <math>\text{E}_{n}</math> for bound state problems; but, for scattering states (unbound states) the energy E isn't discrete. We are interested in the wave functions in order to use and determine the transmission and reflection coefficients T and R.
 
=== Oscillation theorem ===
 
Let us concentrate on the bound states of a set of wavefunctions. 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}(-\psi_2\frac{\partial \psi_1}{\partial x}+\psi_1\frac{\partial \psi_2}{\partial x})=(E_1-E_2)\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 ===
 
=== 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.'''
 
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>(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x))\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 (e^{ikx}u_k(x))\\
&=(e^{ik(x+a)}u_k(x+a))\\
&=e^{ika}(e^{ikx}u_k(x))\\
&=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)
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.
 
== Operators, eigenfunctions, symmetry, and time evolution ==
 
=== Commutation relations and simulatneous eigenvalues ===
 
====Commutators====
 
The commutator of two operators A and B is defined as follows:
 
<math>[A,B]=AB-BA\,\!.</math>
 
When <math>\left[A,B\right]=0</math>, the operators <math>A</math> and <math>B</math> commute.  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. 
 
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^n,B]=nA^{n-1}[A,B]\!</math><br/>
 
In addition, if any two operators are Hermetian and their product is Hermetian, 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. 
 
Some more complicated commutator identities can be found here
http://sites.google.com/site/phy5645fall2008/some-useful-commutator-identites
 
====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.
 
====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,  <math>\hat{x}= x</math> and  <math>\hat{p}= \frac{\hbar}{i}\frac{\partial}{\partial x}</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.
 
====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 commutaion of a give opperator (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.
 
====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).
 
====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 is can be seen by attempting to solve for <math>\alpha</math> by using the "quadratic formula" :
 
:<math>
\alpha=\frac{-\langle \Delta \hat{C} \rangle \pm \sqrt{ (\langle \Delta \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 \Delta \hat{C} \rangle)^2 -4 \langle  \left(\Delta \hat{A}\right)^2  \rangle  \langle \left(\Delta \hat{B}\right)^2  \rangle \leq 0
</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 opperators and deals with wave functions that do not change in time.  The interaction picture, also known as the Dirac picture, is somewhee between the other two, placing time dependence on both opperators 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 with 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. 
 
In the Schrodinger picture, in which we are most accustom to, to find the time dependent expectation values of a given operator, the states change over time while the operator remains constant.  When the Hamiltonian is independent of time it is possible to write:
 
<math>
|\Psi(0)\rangle
</math>
 
as the state at a time 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>
\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} HA U + \left(\frac{\partial A}{\partial t}\right)_H + U^{\dagger} AHU \frac{1}{i\hbar} =
\frac{1}{i\hbar}\left(\left[A,H\right]\right)_H+\left(\frac{\partial A}{\partial t}\right)_H.</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{ \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>
x_H(t)=x_H(0)\cos(\omega t)+\frac{p_H(0)}{m\omega}\sin(\omega t)</math><br/>
 
<math>
p_H(t)=p_H(0)\cos(\omega t)-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''.
 
==== 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>
 
 
- - > 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]]
 
== 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 in the potential of the form:
<math>V(x)=\frac{1}{2}m\omega^2 x^2</math>
 
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 values 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 lowering, raising operators or analysis method.
 
Hamiltonian of 1-D harmonic oscillator: 
 
<math>H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}m\omega^2 x^2</math>
 
It will be easy to memorize how to contruct 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>a=\sqrt{\frac{m\omega}{2\hbar}}x+\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}</math> as the lowering operator, and <br/>
<math>a^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}x-\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}</math> as the raising operator.
 
One way to distinguish <math>a\!</math> from <math>a^{\dagger}</math> is to remember that the ground state wave function is a Gaussian function and  <math>a\!</math> will annihilate this state.
We have the following commutation relation:
<math>[a,a^{\dagger}]=1.</math>
 
<math>H\!</math> can be rewritten in terms of <math>a\!</math> and <math>a^{\dagger}</math> as follows:
 
<math>H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right)</math>
Now, let's see how <math>a\!</math> and <math>a^{\dagger}\!</math> act on an energy eigenstate <math>|\Psi\rangle\!</math>:
For <math>a\!</math>:
 
<math>Ha |\Psi\rangle = \left(E-\hbar\omega\right)|\Psi\rangle </math>
 
This means that <math>a |\Psi\rangle</math> is also an energy eigenstate but correspoding to a lower energy,
<math>E-\hbar\omega\!</math>. <math>a\!</math> is therefore the lowering operator.
 
Similarly,
 
<math>Ha^{\dagger}|\Psi\rangle = \left(E+\hbar\omega\right)|\Psi\rangle </math>
 
and <math>a^{\dagger}\!</math> is the so-called raising operator.
 
So, starting from any energy eigenstates, we can construct all other energy eigenstates by applying <math>a\!</math> or <math>a^{\dagger}\!</math> repeatedly. Although there is no limit in applying <math>a^{\dagger}\!</math>, there is a limit in applying <math>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>a|\Psi_0\rangle=0\!</math><br/>
<math>\Rightarrow \sqrt{\frac{m\omega}{2\hbar}}x+\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx}\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>
 
Normalized ground state wave function: 
 
<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=\hbar\omega\left(n+\frac{1}{2}\right); n=0,1,2,\ldots</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{(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
 
There are two parts in the wave function of excited states: Gaussian function part and 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.
 
=== Coherent states ===
 
The general states of an harmonic oscillator can be expressed as a superpostion of the energy eigenstates eq=|n>. 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>=\alpha|\alpha></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^+|n>=\sqrt{n+1}|n+1></math>.
I. Coherent states construction.
 
<math>|\alpha>=\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}|n>=e^{\alpha a^+}|0></math><br/>
<math>a|\alpha>=\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}a|n>=\sum_{n=1}^{+\infty}
\frac{\alpha^n}{\sqrt{n!}}\sqrt{n}|n-1>=\sum_{n=1}^{+\infty}\frac{\alpha^n}{\sqrt{(n-1)!}}|n-1>=
\alpha(\sum_{n=0}^{+\infty}\frac{\alpha^n}{\sqrt{n!}}|n>)=\alpha|\alpha></math>
II. Coherent states normalization.
<math>|\alpha>=Ne^{\alpha a^+} |0></math>
where <math>N</math> is normalization constant.
 
<math>1=<\alpha|\alpha>=<0|Ne^{\alpha^*a} Ne^{\alpha a^+} |0>=N^2<0|e^{\alpha^*a} e^{\alpha a^+} |0></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^+</math> ( A and B both commute with their commutator because
<math>[A,B]=|\alpha|^2)</math>, we have:
 
<math>1=<\alpha|\alpha>=N^2<0|e^{\alpha^*a} e^{\alpha a^+} |0></math><br/>
<math>N^2<0|e^{\alpha a^+} e^{\alpha^* a} e^{[\alpha^*a,\alpha a^+]} |0>=N^2e^{|\alpha|^2}<0|e^{\alpha a^+} e^{\alpha^* a} |0></math><br/>
<math>=N^2e^{|\alpha|^2}<0|e^{\alpha a^+} |0>=N^2e^{|\alpha|^2}<0|0>=N^2e^{|\alpha|^2}</math>
<math>\rightarrow N=e^{-\frac{1}{2}|\alpha|^2}</math>
 
<math>\rightarrow \mbox{Normalized coherent states:} |\alpha > = e^{-\frac{1}{2}|\alpha |^2 } e^ {\alpha a^{+} }|0></math>
III. Inner product of two coherent states
There is an eigenstate <math>|\alpha></math>
of lowering operator eq=a 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></math> and <math>|\beta></math> can be calculated as follows:
 
<math><\beta|\alpha>=e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}<0|e^{\beta^*a} e^{\alpha a^+} |0></math>
 
<math>=e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}<0|e^{\alpha a^+} e^{\beta^* a} e^{[\beta^*a,\alpha a^+]} |0></math>
 
<math>=e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*}<0|e^{\alpha a^+} e^{\beta^* a} |0></math>
 
<math>=e^{-\frac{1}{2}|\alpha|^2}e^{-\frac{1}{2}|\beta|^2}e^{\alpha \beta^*}</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 deviations 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')=Acos(\omega t')+Bsin(\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_0cos(\omega t)}{sin(\omega t)}                                                                                . </math>
 
Substitute:
 
<math>x_{cl}(t')= x_0cos(\omega t')+\frac{x-x_0cos(\omega t)}{sin(\omega t)}sin(\omega t')
\Rightarrow \frac{dx_{cl}(t')}{dt'}= -\omega x_0sin(\omega t')+\omega \frac{x-x_0cos(\omega t)}{sin(\omega t)}cos(\omega t')</math>                                                        <br/>
<math>KE= \frac{1}{2}m(\frac{dx_{cl}}{dt})^2=\frac{1}{2}m[-\omega x_0sin(\omega t')+\omega \frac{x-x_0cos(\omega t)}{sin(\omega t)}cos(\omega t')]^2</math><br/>
<math>PE= \frac{1}{2}k(x_{cl}(t'))^2=\frac{1}{2}k[x_0cos(\omega t')+\frac{x-x_0cos(\omega t)}{sin(\omega t)}sin(\omega t')]^2</math>                                        <br/> 
Substituting, integrating from time 0 to time t and simplifying, we get:
 
<math>S=S(t,x,x_0)=\frac{m\omega}{2sin(\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}
\exp{\left[\frac{i}{\hbar}\left(\frac{m}{2\Delta t}y^2_{N-1}-
\frac{\Delta t}{2}ky^2_{N-1}\right)\right]}\exp{\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
\exp{\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 ===
 
 
 
Hamiltonian of a particle of charge <math>e\!</math> and mass <math>m\!</math>
in an external electromagetic field, which may be time-dependent is given as follows:
 
<math>H=\frac{1}{2m}(\bold p-\frac{e}{c}\bold A(\bold r,t))^2+e\phi(\bold r,t)</math>
 
 
where <math>\bold A\!</math> and <math>\phi\!</math> are the vector and scalar potentials of the electromagnetic field, respectively.
Let's find out the Heisenberg equation of motion for the position and velocity operators.
For <math>\bold r\!</math> we have:
 
<math>\frac{d\bold r}{dt}=\frac {1}{i\hbar}[\bold r,H] = \frac {1}{i\hbar} [\bold r, \frac{1}{2m}(\bold p-\frac{e}{c}\bold A(\bold r,t))^2+e\phi(\bold r,t)]</math>
(<math>\bold r</math> does not depend on <math>t</math> explicitly)
 
<math>= \frac {1}{2im\hbar} [\bold r, (\bold p-\frac{e}{c}\bold A(\bold r,t))^2]</math><br/>
 
<math>= \frac {1}{2im\hbar}[\bold r, \bold p-\frac{e}{c}\bold A(\bold r,t))](\bold p-\frac{e}{c}\bold A(\bold r,t))+\frac {1}{2im\hbar}(\bold p-\frac{e}{c}\bold A(\bold r,t))[\bold r, \bold p-\frac{e}{c}\bold A(\bold r,t))]</math>
 
<math>= \frac {1}{2im\hbar}[\bold r, \bold p](\bold p-\frac{e}{c}\bold A(\bold r,t))+\frac {1}{2im\hbar}(\bold p-\frac{e}{c}\bold A(\bold r,t))[\bold r, \bold p]</math>
 
<math>= \frac {1}{2im\hbar}ih(\bold p-\frac{e}{c}\bold A(\bold r,t))+\frac {1}{2im\hbar}(\bold p-\frac{e}{c}\bold A(\bold r,t))ih</math>
 
<math>= \frac {1}{m}(\bold p-\frac{e}{c}\bold A(\bold r,t))</math>
 
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}(\bold p-\frac{e}{c}\bold A(\bold r,t))</math>                                                                             
 
The Hamiltonian can be rewritten as:
 
<math>H=\frac {m}{2}\bold v.\bold v+e\phi</math>
 
The Heisenberg equation of motion for velecity operator is:
 
<math>\frac{d\bold v}{dt}=\frac {1}{i\hbar}[\bold v,H]+\frac{\partial \bold v}{\partial t}</math>
<math>=\frac {1}{i\hbar}[\bold v,\frac{m}{2}\bold v.\bold v]+\frac {1}{i\hbar}[\bold v,e\phi]-\frac{e}{mc} \frac{\partial \bold A}{\partial t}</math>                                              (note that <math>\bold p</math> does not denpend on <math>t</math>  expicitly)
 
Let's use the following commutator identity:
 
<math>[\bold v,\bold v.\bold v]=\bold v \times (\bold v \times \bold v)-(\bold v \times \bold v) \times \bold v</math>                                                     
 
Substituting, we get:
 
<math>\frac{d\bold v}{dt}=\frac {1}{i\hbar}\frac{m}{2}(\bold v \times (\bold v \times \bold v)-(\bold v \times \bold v) \times \bold v)+\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>(\bold v \times \bold v)_i=\epsilon_{ijk}v_jv_k=\epsilon_{ijk}\frac{1}{m}(p_j-\frac{e}{c}A_j(\bold r,t))\frac{1}{m}(p_k-\frac{e}{c}A_k(\bold r,t))</math> (sum over all repeated indices)<br/>
<math>=-\frac{e}{m^2c}\epsilon_{ijk}(p_jA_k(\bold r,t)+A_j(\bold r,t)p_k)</math> (because <math>p_j</math> and <math>p_k</math>
commute and so do <math>A_j</math> and <math>A_k</math>)<br/>
<math>=-\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</math><br/>
<math>=-\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\bold r,t)-\frac{e}{m^2c}\epsilon_{ikj}A_k(\bold r,t)p_j</math> (Switching indices in the second tems)<br/>
<math>=-\frac{e}{m^2c}\epsilon_{ijk}p_jA_k(\bold r,t)+\frac{e}{m^2c}\epsilon_{ijk}A_k(\bold r,t)p_j</math> (because <math>\epsilon_{ikj}=-\epsilon_{ijk}</math>)<br/>
<math>=-\frac{e}{m^2c}\epsilon_{ijk}[p_j,A_k(\bold r,t)]</math><br/>
<math>=-\frac{e}{m^2c}\epsilon_{ijk}\frac{\hbar}{i}\nabla_jA_k(\bold r,t) eq==i\hbar\frac{e}{m^2c}(\nabla \times \bold A)_i</math><br/>
<math>\rightarrow [\bold v \times \bold v]=i\hbar\frac{e}{m^2c}(\nabla \times \bold A)=i\hbar\frac{e}{m^2c}\bold B</math>                                                                  <br/>
where
 
<math>[\bold v,\phi]=\frac{1}{m}[\bold p-\frac{e}{c}\bold A(\bold r, t),\phi(\bold r,t)]=\frac{1}{m}[\bold p,\phi(\bold r,t)]=\frac{1}{m}\frac{\hbar}{i}\nabla\phi</math></br>
 
Substituting and rearranging, we get:
 
<math>m\frac{d\bold v}{dt}=\frac {e}{2c}(\bold v \times \bold B-\bold B \times \bold v)+e\bold E</math>
where
<math>\bold E=-\nabla \phi-\frac {1}{c}\frac {\partial \bold A}{\partial t}</math><br/>                                                                                         
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.
 
=== WKB ===
 
WKB 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, and any smooth potential is slowly varying in this limit. 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, the potential at the turning point is approximated as linear and slowly increasing.
 
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>
 
where <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>
 
However, note that at a classical turning point <math>p(x_{turning})\rightarrow 0</math> and the WKB solution diverges which means it is unacceptable. 
The true wave function, of course, will not exhibit such behavior.  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.  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 solution are the Airy functions.
Our patching 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 function to the wavefunctions in the region 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 dopends 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.
 
== 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:
:<math>L_x=yp_z-zp_y=\frac{\hbar}{i}(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y})\!</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)
:<math>L_{\mu}=\epsilon_{\mu\nu\lambda}r_\nu p_\lambda\!</math>
with
:<math>\epsilon_{ijk} =
\begin{cases}
+1 & \mbox{if } (i,j,k) \mbox{ is } (1,2,3), (3,1,2) \mbox{ or } (2,3,1), \\
-1 & \mbox{if } (i,j,k) \mbox{ is } (3,2,1), (1,3,2) \mbox{ or } (2,1,3), \\
0 & \mbox{otherwise: }i=j \mbox{ or } j=k \mbox{ or } k=i,
\end{cases}
</math>
Or we simply say that the even permutation gives 1, odd permutation -1, else 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>
 
For example,
:<math>[L_\mu,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[ 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</math>
Also note that for <math>L^2=L_x^2+L_y^2+L_z^2</math>,
:<math>[L_\mu,L^2]=[L_\mu,L_\mu^2]+[L_\mu,L_\nu^2]+[L_\mu,L_\rho^2]=0+i\hbar(L_\nu L_\rho+L_\rho L_\nu-L_\rho L_\nu-L_\nu L_\rho)=0 </math>
 
===Angular momentum as generator of rotations in 3D===
 
Let <math>\mathbf\alpha\!</math> represent an infinitesimally small rotation directed along the axis about which the rotation takes place. The changes <math>\delta \mathbf w</math> (in the radius vector <math>\mathbf w\!</math> of the particle) due to such a rotation is:
 
[[Image:Rotation.jpg]]
:<math>\delta \mathbf w=\mathbf{\alpha}\times \mathbf w</math>
so
:<math>\psi(\mathbf w+\delta \mathbf w)=(1+\alpha\cdot(\mathbf w\times\nabla))\psi(\mathbf w)</math>
 
The expression
:<math>1+\alpha\cdot(\mathbf w\times\nabla)</math>
is the operator of an infinitesimally small rotation. We recognize the equation
:<math>\mathbf w\times\nabla=\frac{i}{\hbar}\mathbf L</math>
 
Therefore, the infinitesimal rotation operator is
:<math>\mathbf R_{inf}=1+\frac{i}{\hbar}\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}\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}\alpha\cdot\mathbf L}\mathbf w e^{-\frac{i}{\hbar}\alpha\cdot\mathbf L}</math>
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}\alpha\cdot\mathbf L}</math> on the states:
:<math>\langle r_0|e^{\frac{i}{\hbar}\alpha\cdot\mathbf L}\mathbf \hat{\mathbf r} e^{-\frac{i}{\hbar}\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}\alpha\cdot\mathbf L}|\psi\rangle=\langle r_0'|\psi\rangle=\psi(r_0')</math>
 
This is the wavefunction evaluated at a rotated point.
 
===Eigenvalue quantization===
 
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>
 
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>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</math>
:<math>\Rightarrow L_z (L_\pm |m\rangle)=(m\pm 1)\hbar(L_\pm |m\rangle)</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>.
 
===Orbital angular momentum eigenfunctions===
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(\frac{1}{\sin \theta^2}\frac{d^2}{d\phi^2}+\frac{1}{\sin \theta}\frac{d}{d \theta}(\sin\theta\frac{d}{d\theta}))</math>
 
Where the eigenvalues of <math>L^2\!</math> are:
:<math>L^2|l,m\rangle=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}(\pm i\frac{d}{d\theta}-\cot \theta \frac{d}{d\phi})</math>
to find the following equation
:<math>\langle r,\theta,\phi|L_+|l,l\rangle=-i\hbar e^{i\phi}(i\frac{\partial}{\partial\theta}-\cot \theta \frac{\partial}{\partial\phi})\langle r,\theta,\phi|l,l\rangle=0</math>
 
Using eq. 5.2.2, we get
:<math>(\frac{\partial}{\partial \theta}-l\cot\theta)\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 Eq. 5.1 as
:<math>\mathbf L=\mathbf r\times \frac{\hbar}{i}\nabla</math>
As has been shown in 5.1, angular momentum acts as the generator of rotation.
 
 
 
 
== Central forces  ==
 
===Generalized derivation===
A central potential only depends on the distance away from the potential's center and is rotationally invariant, not depending on the orientation.
:<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>.  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}\frac{\partial^2}{\partial r^2}r-\frac{L^2}{\hbar^2 r^2}</math>
 
This makes the Schroedinger equation
:<math>(\frac{-\hbar^2}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{L^2}{2mr^2}+V(r))\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>(\frac{-\hbar^2}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{\hbar^2 l(l+1)}{2mr^2}+V(r))f_l(r)Y_{lm}(\theta,\phi)=Ef_l(r)Y_{lm}(\theta,\phi)</math>
 
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>(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2 l(l+1)}{2mr^2}+V(r))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 need when <math>k\!</math> is imaginary.
 
'''Nomenclature'''
 
Historically, the different values of <math>l\!</math> have taken on names:
:<math>
\begin{cases}
l = 0  & \mbox{s-wave (sharp)}\\
l = 1  & \mbox{p-wave (principle)}\\
l = 2  & \mbox{d-wave (diffuse)}\\
l = 3  & \mbox{f-wave (fundamental)}
\end{cases} </math>
 
===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>(-\frac{2m}{\hbar^2}\frac{\partial^2}{\partial r^2}+\frac{2m}{\hbar^2}\frac{l(l+1)}{r^2}+V_0)u_l(r)=Eu_l(r)</math>
let <math>k^2=\frac{2m}{\hbar^2}|E-V|</math>.  Rearranging the equation gives
:<math>(-\frac{\partial^2}{\partial r^2}+\frac{l(l+1)}{r^2}-k^2)u_l(r)=0</math>
Letting <math>\rho=kr\!</math> gives the equation:
:<math>(-\frac{\partial^2}{\partial\rho^2}+\frac{l(l+1)}{\rho^2})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(l+1)}{\rho}, d_l^\dagger=-\frac{\partial}{\partial\rho}+\frac{l(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 l=0, <math>\frac{\partial^2}{\partial \rho^2} u_0(\rho)=u_0(\rho)</math>, giving
:<math>u_0(\rho)=A\sin(\rho)+B\cos(\rho)\!</math>
Now applying the raising operator to the ground state
:<math>d_0^\dagger u_0(\rho)=(-\frac{\partial}{\partial\rho}+\frac{l+1}{\rho})u_0(\rho)=c_0 u_1(\rho)</math>
===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 equation}\\
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 Henkel 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
:<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
:<math>
\begin{cases}
0<r<a & u_0(r)=Ae^{ikr}+Be^{-ikr}\\
r>a & u_0(r)=Ce^{ik'r}+De^{-ik'r}
\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(\frac{r}{\hbar}\sqrt{2m(E+V_0)})</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
:<math>\alpha\sin(\frac{a}{\hbar}\sqrt{2m(E+V_0)})=Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
:<math>\alpha\frac{\sqrt{2m(E+V_0)}}{\hbar}\cos(\frac{a}{\hbar}\sqrt{2m(E+V_0)})=-\frac{\sqrt{-2mE}}{\hbar}Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}</math>
 
Dividing the above equations, we find
:<math>-\cot(\sqrt{\frac{2m}{\hbar^2}(V_0-|E|)a^2})=\frac{\sqrt{\frac{2m|E|}{\hbar^2}}}{\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}}</math>
 
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>
 
===Hydrogen atom===
[[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 <math>u_l(r)\!</math> substitution is made for simplification.
 
Substitute (6.2.2) into (6.2.1), 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>H_e^+</math>, etc.
 
Since we are only concentrating on the bound states, we can take the limits of <math>r\!</math>:
:<math>\lim_{r \to 0}u_l(r)\rightarrow r^{l+1}</math>
:<math>\lim_{r \to \infty}u_l(r)\rightarrow e^{-\frac{r}{a}}</math>
where
:<math>a=\sqrt{\frac{-\hbar^2}{2\mu 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 eqn into the the Schrodinger and simplifying turns into
:<math>\frac{d^2W(\rho)}{d\rho^2}+2(\frac{l+1}{\rho}-1)\frac{dW}{d\rho}+(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho})W(\rho)</math>
 
<math>W(p)\!</math> can be expressed as an expansion of polynomials
:<math>W(p)=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
:<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>
:<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>
 
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 possible for the hydrogen atom for <math>n=1, 2, 3\!</math>. The parenthesis indicate the degeneracy.
[[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>Id\Omega\!</math> the number of those scattered into the cone per time, then 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 experiment, start with the stationary Schrodinger equation:
:<math>(-\frac{\hbar^2}{2m}\nabla^2+V(r))\psi(r)=E\psi(r)</math>
:<math>\Rightarrow (\nabla^2+k^2)\psi(r)=\frac{2mV(r)}{\hbar^2}\psi(r)</math>
where <math>E=\frac{\hbar^2k^2}{2m}</math> and <math>V(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>, 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=\psi_k^{(0)} +\int d^3r'G_k(\mathbf r,\mathbf r')\frac{2m}{\hbar^2}V(r')\psi_k(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(x) =0</math>
and the solution is chosen such that the second term in Eq.(2) corresponds to an outgoing wave. Then
:<math>G_{k}(\mathbf r,\mathbf r')=-\frac{1}{4\pi}\frac{e^{ik|r-r'|}}{|r-r'|}</math>
and in the asymptotic limit of <math>r\to\infty\!</math>:
:<math>\lim_{r \to \infty} \psi_k(\mathbf{r})=\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ {-ik\mathbf{\hat{r}}\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^ { ik\mathbf{r}}}{r}=\psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^ { ik\mathbf{r}}}{r}</math>
where the scattering amplitude
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ { -ik\mathbf{\hat{r}}\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>\hat{r}\!</math> (the vector defining the detector)  and <math>\mathbf{k}\!</math> (the vector defining the in the incoming wave).
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}(\psi_{sc}^*\frac{\partial \psi_{sc}}{\partial r}-\psi_{sc}\frac{\partial \psi_{sc}^*}{\partial r})=\frac{\hbar k}{mr^2}|f_k(\theta,\phi)|^2</math>
:<math>\Rightarrow \frac{d\sigma}{d\Omega}=|f_k(\theta,\phi)|^2</math>
Finally we have 1st Born approximation. For large <math>r\!</math> we find
:<math>\psi_k(\mathbf{r})\approx\psi_k^{(0)}(\mathbf{r}) -\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ {-ik\mathbf{\hat{r}}\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k^{(0)}(\mathbf{r'})\frac{e^ { ik\mathbf{r}}}{r}</math>
and
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ { -ik\mathbf{\hat{r}}\cdot\mathbf{r'}}V(\mathbf{r'})e^{i\mathbf{k}\cdot \mathbf r'}</math>
 
===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
wave.
:<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 come to infinity
:<math>\lim_{r \to \infty}| \mathbf{r}-\mathbf{r'}|=r-\mathbf{r}\cdot\mathbf{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\mathbf{r'}e^ { -ik\mathbf{\hat{r}}\cdot\mathbf{r'}}V(\mathbf{r'})\psi_k(\mathbf{r'})\frac{e^ { ik\mathbf{r}}}{r}
=\psi_k^{(0)}(\mathbf{r})+f_k(\theta,\phi)\frac{e^ { ik\mathbf{r}}}{r}</math>
where
:<math>f_k(\theta,\phi)=-\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ { -ik\mathbf{\hat{r}}\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>\hat{r}\!</math> (the vector defining the detector)  and <math>\mathbf{k}\!</math>
(the vector defining the in the incoming wave).
 
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
:<math>\left( -\frac{\hbar^2 }{2m}\frac{\partial^2 }{\partial r^2 }+\frac{h^2 l(l+1)}{2mr^2}+V(|\mathbf{r}|) \right) u_l(r) =\frac{\hbar^2 k^2 }{2m}u_l(r)</math>
 
For <math>V(r)\!</math> with a finite range <math>d\!</math>, we have shown that for <math>r \gg d\!</math> we have
:<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>
 
and the solution is a combination of spherical Bessel and 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 Bessel function and Neumann function.
 
:<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>
Letting
:<math>\frac{B_l }{A_l }=-\tan\delta_l</math>
here <math>\delta_l\!</math> is called phase shift.
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})}{kr}</math>
 
Now since we are seeking the scattering amplitude with azimuthal symmetry, we can write
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)</math>
let us fix the coeffcients <math>a_l(k)\!</math> by
:<math>e^{ik\cos\theta}+f_k(\theta,\phi)\frac{e^ { ik\mathbf{r}}}{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 wave
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,\phi)\frac{e^ { -ik\mathbf{r}}}{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 wave. Since <math>\cos x=(e^{ix} -e^{-ix})/2\!</math> we have
:<math>a_l(k)=(2l+1)i^le^{i\delta_l}</math>
Therefore:
:<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 k and therefore a function of the incident energy. If <math>\delta_l\!</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.
 
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.
 
The differential scattering cross section is
:<math>\frac{d\sigma}{d\Omega}=|f_k(\theta) |^2=\frac{1}{k^2 }|\sum_{l=0}^{\infty}(2l+1)e^{i\delta_l }\sin\delta_lp_l(\cos\theta) |^2</math>
By integrating <math>\frac{d\sigma}{d\Omega}</math> over the solid angle we obtain the total scattering cross section
:<math>\sigma_{tot}=\frac{4\pi}{k^2 }\sum_{l=0}^{\mathop{ \infty}}(2l+1)\sin^2\delta_l</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
:<math>\sigma_{tot}=\frac{4\pi}{k}\Im mf(0)</math>
here we take the imaginary part.
This relationship is known as the optical theorem.
 
===Born approximation and examples of cross-section calculations===
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_k(\mathbf{\hat{r}})|^2</math>
 
The scattering amplitude, <math>f_k(\mathbf{\hat r})\!</math>, is the coefficient of the outgoing wave.
 
The Born approximation, often called the first Born approximation, is a technique to find solutions when <math>V(r)\!</math> is small.
 
The scattering amplitude can be approximated by
:<math>f_k(\mathbf{\hat r})\approx -\frac{m}{2\pi\hbar^2 }\int d^3\mathbf{r'}e^ {-i(k\mathbf{\hat{r}}-\mathbf{k})\cdot\mathbf{r'}}V(\mathbf{r'})</math>
 
where <math>-ik\mathbf{\hat r}\!</math> is the scattered portion and <math>i\mathbf k\!</math> is the incident portion.
 
The scattering amplitude is defined as the coefficient of the outgoing wave in the asymptotic solution (for large <math>r\!</math>)
:<math>\psi_k(r)\approx N(e^{ikr}+\frac{e^{ikr}}{r}f_k(r))</math>
 
of the Schrodinger equation:
:<math>(\nabla^2+k^2)\psi=\frac{2m}{\hbar^2}V\psi</math>
 
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^{-iqr}d^3r</math>
 
Which leads to the Born cross section:
:<math>(\frac{d\sigma}{d\Omega})_B=(\frac{m}{2\pi\hbar^2})^2|\langle K_s|r|K_i\rangle|^2</math>
 
 
'''Born Approximation for Spherically Symmetric Potentials:'''
 
Given spherically symmetry we may define
:<math>\mathbf \kappa=\mathbf{k'}-\bold{k}</math>
and align the polar axis for the <math>r\!</math> integral lie along this quantity.  We then have:
:<math>(\mathbf{k'}-\bold{k})\cdot \mathbf{r}=\kappa r \cos \theta'</math>
 
Our first Born integration then takes the form:
:<math>f(\theta)\approx -\frac{2m}{\hbar^2}\int e^{i\kappa r\cos\theta'}V(r)r^2\sin\theta' drd\theta'</math>
 
The phi integral introduces a trivial <math>2\pi\!</math>. For the <math>\theta'\!</math> integral we can use the following identity:
:<math>\int_0^\pi e^{isr\cos\theta'}\sin\theta' d\theta'=-\frac{2\sin(sr)}{sr}</math>
and get:
:<math>f(\theta)\approx -\frac{2m}{\hbar^2\kappa}\int_0^\infty V(r)r\sin(\kappa r) dr</math>
 
where the angular dependence of <math>f(\theta)\!</math> is carried by <math>\kappa\!</math>:
:<math>\kappa=2k\sin\frac{\theta}{2}</math>
 
'''Example 1:'''
 
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>f(\theta)\approx \frac{-2mA}{\hbar^2\kappa}\int_0^\infty re^{-\alpha r^2}\sin(\kappa r)dr</math>
:<math>=\frac{-2mA}{\hbar^2\kappa}\int_0^\infty \frac{d}{dr}(\frac{-1}{2\alpha}e^{-\alpha r^2})\sin(\kappa r)dr</math>
:<math>=\frac{mA}{\alpha\hbar^2\kappa}(0-\kappa\int_0^\infty e^{-\alpha r^2}\cos(\kappa r)dr)</math>
:<math>=\frac{-mA}{\alpha\hbar^2}\frac{\sqrt{\pi}}{2\sqrt{\alpha}}e^{-\frac{\kappa^2}{4\alpha}}</math>
:<math>=-\frac{mA\sqrt{\pi}}{2\hbar^2\alpha^{\frac{3}{2}}}e^{-\frac{\kappa^2}{4\alpha}}</math>
 
===Coulomb potential scattering===
'''Example 1'''
 
Lets 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>
:<math>\Rightarrow f=\frac{-m}{2\pi\hbar^2}\int_0^\infty dr' r'^2 2\pi\frac{e^{-i|k'-k|r'}-e^{i|k-k'|r'}}{-i|k'-k|r'}\frac{V_0e^{-\alpha r'}}{r'}</math>
:<math>\Rightarrow f=\frac{-2mV_0}{\hbar^2}\int_0^\infty dr' r'^2 \frac{\sin(|k'-k|r')}{|k'-k|r'}\frac{e^{-\alpha r'}}{r'}</math>
:<math>\Rightarrow f=\frac{-2mV_0}{\hbar^2}\frac{1}{(k'-k)^2+\alpha^2}=\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>
 
'''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 solutions of the form:
:<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\!</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\xi)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)}{2} \right] +\frac{\Gamma(c)}{\Gamma(n)}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)^4}{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.
 
===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.  We could, for example, label each ball with a different color.  Then, after an arbitrary number of interactions, we can distinguish, say, a red ball from any other.  It is not, however, possible to attach such labels to 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>r_1\leftrightarrow r_2\!</math>, <math>r_{cm} = (r_1 + r_2)/2\!</math> is invariant while <math>r = r_1-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(r)\!</math> by hand:
:<math>Y_{lm}(-r)=(-1)^lY_{lm}(r)\!</math>
 
Under the transformation:
:<math>r\rightarrow -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(\frac{\pi}{2})</math>
 
Now if the particles are distinguishable, the cross section for observing a scattered particle at 90 degrees is then:
:<math>(\frac{d\sigma}{d\Omega})_{dis}=2|f(\frac{\pi}{2})|^2</math>
 
Where if the particles are indistinguishable, we see above that we will have:
:<math>(\frac{d\sigma}{d\Omega})_{ind}=4|f(\frac{\pi}{2})|^2</math>
 
Thus the differential cross-section is exactly twice the distinguishable case when the particles are indistinguishable.

Revision as of 23:05, 5 October 2009

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