Phy5645

Welcome to the Quantum Mechanics A PHY5645 Fall2008

Schrodinger equation. The most fundamental equation of quantum mechanics which describes the rule according to which a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.

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

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

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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 , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi\,\!} . 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 struck the electron, the electron gained additional momentum, so then our simultaneous momentum measurement could 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 Catastrophy (Blackbody Radiation)

Imagine a perfect absorber cavity (i.e. it absorbs all radiation at all wavelengths, so that its spectral radiance is only going to depend on the temperature). This emission is called the blackbody radiation. This blackbody radiation experiment shows an important failure of classical mechanics. 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 with nodes at the ends. The result predicts that the spectral intensity will increase quadratically with the increasing frequency, and will diverge to infinite energy or intensity squared at a UV frequency, or so called "Ultraviolet Catastophy."

In 1900, Max Planck offered a successful explanation for blackbody radiation. He also assumed the the radiation was due to oscillations of the electron, but the difference between his assumption and Rayleigh's was that he assumed that the possible energies of an oscillator were not continuous. He proposed that the energy of this oscillator would be proportional to the frequency of a constant, the Planck constant.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E=h\nu=\hbar\omega}

Here E is energy, h is the Planck constant (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h=6.626*10^{-34} Joule-seconds \!} ) and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu\!} 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 catastrophy, and instead the energy approached zero as the frequency is increased. In summary, the energy of the electromagnetic radiation is proportional to frequency instead of the amplitude of the electromagnetic waves, defying the classical physics where the energy is proportional to the intensity.

Photoelectric Effect

Consider a system composed of light hitting a metal plate. From experimental observations, a current can be measured as light is incident on the metal plate. This phenomenon was first studied by Albert Einstein (1905). During this period, the classical point of view is that an electron is bound inside of an atom, and an excitation energy is needed in order to release it from the atom. This energy can be brought forth in the form of light. The classical point of view also includes the idea that the energy of this light is proportional to its intensity. Therefore, if enough energy (light) is absorbed by the electron, the electron eventually will be released. However, this is not the case.

The determining factor here is not the intensity of the light, but the frequency used on the electron. If the frequency of the light is the "specific" frequency, the electron will be released. This specific frequency of light is in resonance with the energy "frequency" of the electron. Einstein made the conclusion that the energy of a single photon is proportional to its frequency, not the intensity.

Einstein realized that the classical view that light is a wave was not true, but instead light must be a particle. If light were a wave, then the mechanism that the wave transmits its energy is due to its perturbation, which is the amplitude of the wave. The photoelectric effect clearly shows that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (amplitude)^2 = intensity } has no affect on the electron energy, instead, only a specific frequency will have an effect on the energy of the electron. Agreeing with the UV catastrophy conclusions, he stated that light is made of particles called "photons," with an energy equal to hv.

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\,\!} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho} of this radiation? and how fast is this rate?

The electron in the Bohr's model involves factors of: radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0\,\!} , angular velocity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega\,\!} , charge of the particle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\,\!} , and the speed of light, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\,\!} : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho=\rho(r_0,\omega,e,c)\,\!}

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:Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(er_0, \omega, c) \!}

What are the dimensions of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho\,\!} ?

Essentially, since light is energy, we are looking for how much energy is passed in a given time: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\rho]=\frac{energy}{time} \!}

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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle energy=\frac{e^2 }{length} \!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e} has to be with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0} , multiply by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r^2} , and divide Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle length^2} .

The angular velocity is in frequency, so to get the above equations in energy/time, just multiply it with the angular velocity, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle energy=\frac{e^2 }{length}\frac{r^2}{length^2}*\omega }

(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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 } \!}

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

Double slit thought experiment with classical bullets

Imagine a rattling gun which is shooting bullets in all directions. 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_1} . 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_2} . When both slits are open, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle peak_{12}} (purple) is observed. This agrees with the classical view, where the bullet is the particle and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{12}} is simply a sum of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_2} .

The equation describing the probability of the bullet arrival if both of the slit are open is therefore

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_{12}=p_1+p_2.}

Classical Waves

Double slit thought experiment with water waves

As waves are passed through the double slit, the intensity of the waves which are proportional to the squares of the height of the wave motion Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H1^2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H2^2} 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 (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H12} ) 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H12 = (H1 +H2 )^2}

Schrodinger equation and motion in one dimension

The time dependent Schrodinger equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar \frac{\partial}{\partial t}|\psi\rangle=\mathcal{H}|\psi\rangle}

where the state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\psi\rangle} evolves in time according to the Hamiltonian operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} .

Motion in 1D

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to -\infty}V(x)=V_-, \lim_{x \to +\infty}V(x)=V_+}

and assuming that: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_-<V_+/!} Schrodinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)=E\psi(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rightarrow \frac{d^2}{dx^2}\psi(x)+\frac{2m}{\hbar^2}(E-V(x))\psi(x)=0}

From this equation we can discuss some general properties of 1-D motion as follows:

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E>V_+\!} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E-V(x)>0\!} at both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} . Therefore, the solution of Schrodinger equation are trigonometric function (sine or cosine). The wave function is oscillating at both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} . The particle is in unbound state. The energy spectrum is continous. Both oscillating solutions are allowed, the energy level are two-fold degenerate.

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_-\le E \le V_+} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E-V(x)>0\!} at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\infty} but $E-V(x)<0$ at $+\infty$. Therefore, the wave function is oscillating at $-\infty$ but decaying exponentially at $+\infty$. The energy spectrum is still continous but no longer degenerate.

If $E<V_{-}\!$:

$E-V(x)\!<0$ at both $-\infty$ and $+\infty$. Therefore, the wave function decays exponentially at both $-\infty$ and $+\infty$. The particle is in bound state. The energy spectrum is discrete and non-degenerate.

Operators, eigenfunctions, symmetry, and time evolution

Commutation relations and simulatneous eigenvalues

COMMUTATORS

The commutator of two operators A and B is defined as follows:

$[A,B]=AB-BA\,\!.$

If $[A,B]=0$, we say that the operators $A$ and $B$ commute. Conversely, if , we say that the operators and do not commute. We can think of the commutator between two operators as a measure of how badly the two fail to commute.

Identities:
$[A,B]+[B,A]=0\!$
$[A,A]=0\!$
$[A,B+C]=[A,B]+[A,C]\!$
$[A+B,C]=[A,C]+[B,C]\!$
$[AB,C]=A[B,C]+[A,C]B\!$
$[A,BC]=[A,B]C+B[A,C]\!$
$[A,[B,C]]+[C,[A,B]]+[B,[C,A]]=0\!$

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. All observables are Hermitian. If two Obersvables have simultaneous eigenkets, meaning for two obervables A and B,

${\hat {A}}|\Psi _{AB}\rangle =a|\Psi _{AB}\rangle \!$
${\hat {B}}|\Psi _{AB}\rangle =b|\Psi _{AB}\rangle \!$

Then we have that ${\hat {A}}{\hat {B}}|\Psi _{AB}\rangle ={\hat {A}}b|\Psi _{AB}\rangle =b{\hat {A}}|\Psi _{AB}\rangle =ba|\Psi _{AB}\rangle \!$
Similarly, ${\hat {B}}{\hat {A}}|\Psi _{AB}\rangle ={\hat {B}}a|\Psi _{AB}\rangle =a{\hat {B}}|\Psi _{AB}\rangle =ba|\Psi _{AB}\rangle \!$

So we can see that, ${\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}=\left[{\hat {A}},{\hat {B}}\right]=0\!$
.

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 $\left[A,B\right]\neq 0$, 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 ${\hat {x}}$ and momentum ${\hat {p}}$. In the position representation, ${\hat {x}}=x$ and ${\hat {p}}={\frac {\hbar }{i}}{\frac {\partial }{\partial x}}$.

Applying ${\hat {x}}$ and ${\hat {p}}$ to an arbitrary state ket we can see that: $\left[{\hat {x}},{\hat {p}}\right]=i\hbar .$

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: $\left[{\hat {r}}_{i},{\hat {p}}_{j}\right]=i\hbar \delta _{ij}$
$\left[{\hat {r}}_{i},{\hat {r}}_{j}\right]=\left[{\hat {p}}_{i},{\hat {p}}_{j}\right]=0,$
where the indices stand for x,y, or z components of the 3-vectors.

CONNECTION BETWEEN CLASSICAL MECHANICS AND QUANTUM MEACHNICS

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: $H=\sum _{k}p_{k}{\dot {q}}_{k}-L(q,{\dot {q}},t).$

There are two possibilities.

1. If the Lagrangian $L(q,{\dot {q}},t)$ does not depend explicitly on time the quantity H is conserved.
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.

It is clear from the above equation that:
${\dot {q}}_{k}={\frac {\partial H}{\partial p_{k}}}$
${\dot {p}}_{k}=-{\frac {\partial H}{\partial q_{k}}}$

This pair of the equations is called Hamilton's equations of motions. The following object

$[A,B]=\sum _{k}\left({\frac {\partial A}{\partial q_{k}}}{\frac {\partial B}{\partial p_{k}}}-{\frac {\partial A}{\partial p_{k}}}{\frac {\partial B}{\partial q_{k}}}\right)$

is called Poisson Bracket, and it has interesting properties. To see, let's calculate commutation relationships between coordinates and momenta.

$[p_{i},p_{j}]=\sum _{k}\left({\frac {\partial p_{i}}{\partial q_{k}}}{\frac {\partial p_{j}}{\partial p_{k}}}-{\frac {\partial p_{i}}{\partial p_{k}}}{\frac {\partial p_{j}}{\partial q_{k}}}\right)=0$

$[q_{i},q_{j}]=\sum _{k}\left({\frac {\partial q_{i}}{\partial q_{k}}}{\frac {\partial q_{j}}{\partial p_{k}}}-{\frac {\partial q_{i}}{\partial p_{k}}}{\frac {\partial q_{j}}{\partial q_{k}}}\right)=0$

$[p_{i},q_{j}]=\sum _{k}\left({\frac {\partial p_{i}}{\partial q_{k}}}{\frac {\partial q_{j}}{\partial p_{k}}}-{\frac {\partial p_{i}}{\partial p_{k}}}{\frac {\partial q_{j}}{\partial q_{k}}}\right)=\sum _{k}-\delta _{ik}\delta _{jk}=-\delta _{ij}.$

This relations clearly shows how close are the quantum commutators with classical world. If we perform the following identification:

$\delta _{ij}\rightarrow i\hbar \delta _{ij}.$

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 that one needs to check on a given operator (let's say ${\hat {O}}$) is if it commutes with the Hamiltonian ${\hat {H}}$. If ${\hat {O}}$ commutes with ${\hat {H}}$, then the eigenfunctions of ${\hat {H}}$ can always be chosen to be simultaneous eigenfunctions of ${\hat {O}}$. If ${\hat {O}}$ commutes with the Hamiltonian and does not explicitly depend on time, then ${\hat {O}}$ is a constant of motion.

COMMUTATORS & SYMMETRY

We can define an operator called the parity operator, ${\hat {P}}$ which does the following:

${\hat {P}}f(x)=f(-x).$

The parity operator commutes with the Hamiltonian ${\hat {H}}$ if the potential is symmetric, ${\hat {V}}(r)={\hat {V}}(-r)$. 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

One can prove that if

$[A,B]=iC\;$
then $\langle \left(\Delta A\right)^{2}\rangle \langle \left(\Delta B\right)^{2}\rangle ={\frac {1}{4}}\langle C\rangle ^{2}$

Discrete eigenvalues and bound states. Harmonic oscillator and WKB approximation

Harmonic oscillator spectrum and eigenstates Coherent states Feynman path integral evaluation of the propagator Motion in magnetic field WKB

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 $\mathbf {L} \!$ as:

$\mathbf {L} =\mathbf {r} \times \mathbf {p}$

Working in the spatial representation, we have $\mathbf {r}$ as our radius vector, while $\mathbf {p}$ is the momentum operator.

$\mathbf {p} =-i\hbar \nabla$

Using the cross product in Cartesian coordinate system, we get component of L in each direction:

$L_{x}=yp_{z}-zp_{y}\!$

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)

$L_{\mu }=\epsilon _{\mu \nu \lambda }r_{\mu }p_{\nu }\!$

with

$\epsilon _{ijk}={\begin{cases}+1&{\mbox{if }}(i,j,k){\mbox{ is }}(1,2,3),(3,1,2){\mbox{ or }}(2,3,1),\\-1&{\mbox{if }}(i,j,k){\mbox{ is }}(3,2,1),(1,3,2){\mbox{ or }}(2,1,3),\\0&{\mbox{otherwise: }}i=j{\mbox{ or }}j=k{\mbox{ or }}k=i,\end{cases}}$

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:

$[L_{\mu },r_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }r_{\lambda }$

$[L_{\mu },p_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }p_{\lambda }$

$[L_{\mu },L_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }L_{\lambda }$

For example,

$[L_{\mu },r_{\nu }]=[\epsilon _{\mu \lambda \rho }r_{\lambda }p_{\rho },r_{\nu }]=\epsilon _{\mu \lambda \rho }[r_{\lambda }p_{\rho },r_{\nu }]=\epsilon _{\mu \lambda \rho }r_{\lambda }[p_{\rho },r_{\nu }]=\epsilon _{\mu \lambda \rho }r_{\lambda }{\frac {\hbar }{i}}\delta _{\rho \nu }=\epsilon _{\mu \lambda \nu }r_{\lambda }{\frac {\hbar }{i}}=i\hbar \epsilon _{\mu \nu \lambda }r_{\lambda }$

Also note that for $L^{2}=L_{x}^{2}+L_{y}^{2}+L_{z}^{2}$,

$[L_{\mu },L^{2}]=[L_{\mu },L_{\mu }^{2}]+[L_{\mu },L_{\nu }^{2}]+[L_{\mu },L_{\rho }^{2}]=0+i\hbar (L_{\nu }L_{\rho }+L_{\rho }L_{\nu }-L_{\rho }L_{\nu }-L_{\nu }L_{\rho })=0$

Angular momentum as generator of rotations in 3D

Let $\alpha \!$ represent an infinitesimally small rotation directed along the axis about which the rotation takes place. The changes $\delta \mathbf {w}$ (in the radius vector $\mathbf {w} \!$ of the particle) due to such a rotation is:

$\delta \mathbf {w} =\alpha \times \mathbf {w}$

so

$\psi (\mathbf {w} +\delta \mathbf {w} )=(1+\alpha \cdot (\mathbf {w} \times \nabla ))\psi (\mathbf {w} )$

The expression

$1+\alpha \cdot (\mathbf {w} \times \nabla )$

is the operator of an infinitesimally small rotation. We recognize the equation

$\mathbf {w} \times \nabla ={\frac {i}{\hbar }}\mathbf {L}$

Therefore, the infinitesimal rotation operator is

$\mathbf {R} _{inf}=1+{\frac {i}{\hbar }}\alpha \cdot \mathbf {L}$

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

$\mathbf {R} =e^{{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }$

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

$\mathbf {w} '=e^{{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }\mathbf {w} e^{-{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }$

This equation also implies that if we have a scalar instead of $\mathbf {w} \!$, it would be invariant. We can also calculate the effect of the unitary operator $e^{{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }$ on the states:

$\langle r_{0}|e^{{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }\mathbf {\hat {\mathbf {r} }} e^{-{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }=\langle r_{0}|{\hat {\mathbf {r'} }}=r_{0}'\langle r_{0}|$

$\Rightarrow \psi '(r_{0})=\langle r_{0}|\psi '\rangle =\langle r_{0}|e^{{\frac {i}{\hbar }}\alpha \cdot \mathbf {L} }|\psi \rangle =\langle r_{0}'|\psi \rangle =\psi (r_{0}')$

This is the wavefunction evaluated at a rotated point.

Eigenvalue quantization

The quantization of angular momentum follows simply from the above commutation relations. Define $\beta$ by:

$\beta =L_{x}^{2}+L_{y}^{2}+L_{z}^{2}$

Since $\beta$ is a scalar, it commutes with each component of angular momentum.

Now Define a change of operators as follows:

$L_{+}=L_{x}+iL_{y}\!$

$L_{-}=L_{x}-iL_{y}\!$

From the commutation relations we get

$L_{+}L_{-}=\beta -L_{z}^{2}+\hbar L_{z}$

Similarly,

$L_{-}L_{+}=\beta -L_{z}^{2}-\hbar L_{z}$

Thus,

$[L_{+},L_{-}]=L_{+}L_{-}-L_{-}L_{+}=2\hbar L_{z}$

And

Also, It is easy to show that:

Let L'z be an eigenvalue of Lz. From 5.1.9:

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

Defining the number k by

the inequality 5.1.13 becomes

Similarly, from equation 5.1.10, we get

This result, combined with 5.1.15 shows that

and

From 5.1.12

Now, if

then,

and is then an eigenket of Lz belonging to the eigenvalue:

Similarly, if

then,

is another eigenvalue of Lz, 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 -k. Similarly, using the complex conjugate of 5.1.12, we get that

are eigenvalues of L'z. Thus we may conclude that 2k is an integral multiple of the Plank's constant, and that the eigenvalues are:

If |m> is an eigenstate of Lz with eigenvalue mh, then

Which means that L+ or L- raise or lower the z component of the angular momentum by ħ.

Orbital angular momentum eigenfunctions

INTRODUCTION

Now we construct our eigenfunctions of the orbital angular momentum explicitly. The eigenvalue equation is

                                                                       5.2.1

in terms of wave functions, becomes:

                                  5.2.2

Solving for the dependence, we find

                                                           5.2.3

We construct the dependence using the differential operator representation of

                                           http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=L%5E2%3D-%5Chbar%5E2(%5Cfrac%7B1%7D%7Bsin%5E2%5Ctheta%7D%5Cfrac%7Bd%5E2%7D%7Bd%5Cphi%5E2%7D%2B%5Cfrac%7B1%7D%7Bsin%5Ctheta%7D%5Cfrac%7Bd%7D%7Bd%5Ctheta%7D(sin%5Ctheta%5Cfrac%7Bd%7D%7Bd%5Ctheta%7D))                                                                                5.2.4

Where the eigenvalues of are:

We proceed by using the property of and , defined by

                                       http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=L_%5Cpm%3D%5Cfrac%7B%5Chbar%7D%7Bi%7D%20e%5E%20%7B%5Cpm%20i%20%5Cphi%7D(%5Cpm%20i%5Cfrac%7Bd%7D%7Bd%5Ctheta%7D-cot%5Ctheta%5Cfrac%7Bd%7D%7Bd%5Cphi%7D)                                                                                                        5.2.5

To find the following equation

                                    5.2.6

Using eq. 5.2.2, we get

                                                             5.2.7

And the solution by using 5.2.3 is

                                                                                    5.2.8

Where f(r) is an arbitrary function of r. We can find the angular part of the solution by using . It turns out to be

                                          5.2.9

And we know that are the spherical harmonics defined by

                                               5.2.10

where the function with cosine argument is the associated Legendre polynomials defined by:

with

                                                                                           5.2.11

And so we then can write:

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

As has been shown in 5.1, angular momentum acts as the generator of rotation.

Central forces

Free particle in spherical coordinates

A free particle is a specific case when $V_{0}=0\!$ of the motion in a uniform potential $V(r)=V_{0}\!$. 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:

$(-{\frac {2m}{\hbar ^{2}}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {2m}{\hbar ^{2}}}{\frac {l(l+1)}{r^{2}}}+V_{0})u_{l}(r)=Eu_{l}(r)$

let $k^{2}={\frac {2m}{\hbar ^{2}}}|E-V|$. Rearranging the equation gives

$(-{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {l(l+1)}{r^{2}}}-k^{2})u_{l}(r)=0$

Letting $\rho =kr\!$ gives the equation:

$(-{\frac {\partial ^{2}}{\partial \rho ^{2}}}+{\frac {l(l+1)}{\rho ^{2}}})u_{l}(\rho )=u_{l}(\rho )=d_{l}d_{l}^{+}u_{l}(\rho )$

where $d_{l}\!$ and $d_{l}^{\dagger }\!$ become the raising and lowering operators:

$d_{l}={\frac {\partial }{\partial \rho }}+{\frac {l(l+1)}{\rho }},d_{l}^{\dagger }=-{\frac {\partial }{\partial \rho }}+{\frac {l(l+1)}{\rho }}$

Being $d_{l}^{\dagger }d_{l}=d_{l+1}d_{l+1}^{\dagger }$, it can be shown that

$d_{l}^{\dagger }u_{l}(\rho )=c_{l}u_{l+1}(\rho )$

For l=0, ${\frac {\partial ^{2}}{\partial \rho ^{2}}}u_{0}(\rho )=u_{0}(\rho )$, giving

$u_{0}(\rho )=A\sin(\rho )+B\cos(\rho )\!$

Now applying the raising operator to the ground state

$d_{0}^{\dagger }u_{0}(\rho )=(-{\frac {\partial }{\partial \rho }}+{\frac {l+1}{\rho }})u_{0}(\rho )=c_{0}u_{1}(\rho )$

Hydrogen atom

The Schrodinger equation for the particle moving in central potential can be represented in a spherical coordinate system as follows:

where $L\!$ is the angular momentum operator and $\mu \!$ is the reduced mass.

In this case, being invariant under the rotation, the Hamiltonian, $H\!$, commutes with both $L^{2}\!$ and $L_{z}\!$. Furthermore, $L^{2}\!$ and $L_{z}\!$ commute with each other. Therefore, the energy eigenstates can be chosen to be simultaneously the eigenstates of $H\!$, $L^{2}\!$ and $L_{z}\!$. Such states can be expressed as the following:

$\psi (r,\theta ,\phi )=R(r)Y_{lm}(\theta ,\phi )={\frac {u_{l}(r)}{r}}Y_{lm}(\theta ,\phi )$

where $Y_{l}^{m}(\theta ,\phi )$ is the spherical harmonic, which is the simultaneous eigenstates of $L^{2}\!$ and $L_{z}\!$ and $u_{l}(r)\!$ substitution is made for simplification.

Substitute (6.2.2) into (6.2.1), and taking into account the fact that:

$L^{2}\psi (r,\theta ,\phi )=L^{2}[R(r)Y_{lm}(\theta ,\phi )]=R(r)L^{2}Y_{lm}(\theta ,\phi )=R(r)\hbar ^{2}l(l+1)Y_{lm}(\theta ,\phi )=\hbar ^{2}l(l+1)\psi (r,\theta ,\phi )$

we have the equation for $u_{l}(r)\!$:

$[-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}}{2m}}{\frac {l(l+1)}{r^{2}}}+V(r)]u_{l}(r)=Eu_{l}(r)$

In the hydrogen atom or single electron ion, the potential is the Coulomb potential between the electron and the nucleus:

$V(r)=-{\frac {Ze^{2}}{r}}$

where $Z=1\!$ for the hydrogen, $Z=2\!$ for helium ion $H_{e}^{+}$, etc.

Since we are only concentrating on the bound states, we can take the limits of r:

where

If we allow http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Ckappa%3Da%5E%7B-1%7D, then the large limit of r can be expressed as

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=U_l(r)%5Csim%20e%5E%7B-kl%7D (6.2.8)

U(r) can be expressed as an expansion of polynomials, each term of which needs to satisfy the Schrodinger equation

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=U_l(r)%3D%5Csum_%7Bm%7D%20a_mr%5Em (6.2.9) The first term of the Schrodinger equation turns into

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cfrac%7Bd%5E2U_l(r)%7D%7Bdr%5E2%7D%3D%5Csum_%7Bm%7Da_mm(m-1)r%5E%7Bm-2%7D (6.2.10)

Taking a look at the first term in the Schrodinger equation, we find that the lowest possible value for m is 2. When we plug this value into our equation it produces a recursion relation. http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=2a_2%3DEa_0 (6.2.11)

Using the limits of U, the wavefunction can be expressed as the following where W(kr) ->1 as r->0.

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=U_l(r)%3D(%5Ckappa%20r)%5E%7Bl%2B1%7De%5E%7B-%20%5Ckappa%20r%7DW(%5Ckappa%20r) (6.2.12) To simplify the equation, make a substitution http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Crho%3D%5Ckappa%20r. The equation now turns into

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=U_l(%5Crho)%3D%5Crho%5E%7Bl%2B1%7De%5E%7B-%5Crho%7DW(%5Crho) (6.2.13)

Plugging eqn into the the Schrodinger and simplifying turns into

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cfrac%7Bd%5E2W(%5Crho)%7D%7Bd%5Crho%5E2%7D%2B2(%5Cfrac%7Bl%2B1%7D%7B%5Crho%7D-1)%5Cfrac%7BdW%7D%7Bd%5Crho%7D%2B(%5Cfrac%7B%5Crho_0%7D%7B%5Crho%7D-%5Cfrac%7B2(l%2B1)%7D%7B%5Crho%7D)W(%5Crho)%3D0 (6.2.14)

W(p) can also be expressed as an expansion of polynomials

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=W(%5Crho)%3Da_0%2Ba_1%5Crho%2Ba_2%5Crho%5E2%2B...%3D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7Da_k%5Crho%5Ek%20 (6.2.15)

The Schrodinger equation is then expressed as

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D(a_kk(k-1)%5Crho%5E%7Bk-2%7D%2B2(l%2B1)k%5Crho%5E%7Bk-2%7Da_k-2%5Crho%5E%7Bk-1%7Da_kk)%2B%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D(%5Crho_0a_k%5Crho%5E%7Bk-1%7D-2(l%2B1)a_k%5Crho%5E%7Bk-1%7D)%3D0 (6.2.16)

And simplified into http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D(a_%7Bk%2B1%7D(k%2B1)k%5Crho%5E%7Bk-1%7D%2B2(l%2B1)(k%2B1)%5Crho%5E%7Bk-1%7Da_%7Bk%2B1%7D-2%5Crho%5E%7Bk-1%7Da_kk%2B(%5Crho_0-2(l%2B1))a_k%5Crho%5E%7Bk-1%7D)%3D0 (6.2.17)

Bring all p's to the same power

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=k(k%2B1)a_%7Bk%2B1%7D%2B2(l%2B1)(k%2B1)a_%7Bk%2B1%7D-2ka_k%2B(%5Crho_0-2(l%2B1))a_k%3D0 (6.2.18)

which can be expressed in the simplest fractional form as

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cfrac%7Ba_%7Bk%2B1%7D%7D%7Ba_k%7D%3D%5Cfrac%7B2(k%2Bl%2B1)-%5Crho_0%7D%7B(k%2B1)(k%2B2l%2B2)%7D (6.2.19)

where http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Crho_0%3D2(N%2Bl%2B1) and N=0,1,2... and l=0,1,2,...

In the limit of large k

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cmathop%7B%5Clim%7D%5Climits_%7Bk%20%5Cto%20%5Cinfty%7D%5Cfrac%7Ba_%7Bk%2B1%7D%7D%7Ba_k%7D%5Crightarrow%20%5Cfrac%7B2%7D%7Bk%7D (6.2.2http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=a_%7Bk%2B1%7D%3D%5Cfrac%7B2%7D%7Bk%7Da_k%5Csim%20%5Cfrac%7B2%5Ek%7D%7Bk!%7D this will make eq=\psi(r) \sim e^kr \rightarrow\infty so we need to break, and make eq= a_{k+1}=0 from this, we get the energy spectrum.

  (6.2.21)The fractional form (eqn 6.2.19) can be expressed as a confluent hypergeometric function

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=_%7CF_%7C(a%2Cc%2Cz)%3D1%2B%5Cfrac%7Ba%7D%7Bc%7D%5Cfrac%7Bz%7D%7B1!%7D%2B%5Cfrac%7Ba(a%2B1)%7D%7Bc(c%2B1)%7D%5Cfrac%7Bz%5E2%7D%7B2!%7D%3D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%5Calpha_kz%5Ek (6.2.22)

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cfrac%7B%5Calpha_k%7D%7B%5Calpha_%7Bk-1%7D%7D%3D%5Cfrac%7B%5Cfrac%7Ba(a%2B1)(a%2B2)...(a%2Bk-1)%7D%7Bc(c%2B1)(c%2B2)...(c%2Bk-1)%20k!%7D%7D%7B%5Cfrac%7Ba(a%2B1)(a%2B2)...(a%2Bk-2)%7D%7Bc(c%2B1)(c%2B2)...(c%2Bk-2)%20(k-1)!%7D%7D%3D%5Cfrac%7Ba%2Bk-1%7D%7B(c%2Bk-1)k%7D (6.2.23)

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cfrac%7B%5Calpha%20_%7Bk%2B1%7D%7D%7B%5Calpha_k%7D%3D%5Cfrac%7Ba%2Bk%7D%7B(c%2Bk)(k%2B1)%7D (6.2.23)

where comparison with eqn 6.2.19, we find that

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=c%3D2(l%2B1) http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=a%3D-N http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=z%3D2%5Crho

So the solution of W(p) is

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=W(%5Crho)%3Dconstant_%7CF_%7C(-N%2C2(l%2B1)%2C%202%5Crho) (6.2.24) where http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Crho%3D%5Ckappa%20r%3D%5Csqrt%7B%5Cfrac%7B-2%5Cmu%20E%7D%7B%5Chbar%5E2%7D%7D%20r

Full wavefunction solution with normalization is

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=%5Cpsi_%7Bn%2Cl%2Cm%7D(r%2C%20%5Ctheta%2C%20%5Cphi)%3D%5Cfrac%7Be%5E%7B-%5Ckappa%20r(2%5Ckappa%20r)%5El%7D%7D%7B(2l%2B1)!%7D%5Csqrt%7B%5Cfrac%7B(2%5Ckappa)%5E3(n%2Bl)!%7D%7B2n(n-l%2B1)%7D%7D_%7CF_%7C(-n%2Bl%2B1%2C%202(l%2B1)%2C%202%5Ckappa%20r)Y_%7Blm%7D(%5Ctheta%2C%20%5Cphi) (6.2.25)

The energy is then found to be

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=E%3D-%5Cfrac%7BZ%5E2e%5E4%5Cmu%7D%7B2%5Chbar%5E2n%5E2%7D%3D-%5Cfrac%7BZ%5E2%7D%7Bn%5E2%7DRy (6.2.26)

where Ry=13.6 eV for the hydrogen atom and n=1,2,3,... and the degeneracy for each level is n^2. Below is a chart depicting the energy levels possible for the hydrogen atom for n=1, 2, 3. The parenthesis indicate the degeneracy.

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 E > 0 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.

If $I_{0}\!$ is the number of particles incident from the left per unit area per time and $Id\Omega \!$ the number of those scattered into the cone per time, then the differential cross section is defined as

${\frac {d\sigma }{d\Omega }}={\frac {I(\theta ,\phi )}{I_{0}}}$

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 E+dE where dE 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:

$(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(r))\psi (r)=E\psi (r)$

$\Rightarrow (\nabla ^{2}+k^{2})\psi (r)={\frac {2mV(r)}{\hbar ^{2}}}\psi (r)$

where $E={\frac {\hbar ^{2}k^{2}}{2m}}$ and $V(r)\!$ will be assumed to be finite in a limited region of space $r<d\!$. This is called the range of the force, e.g. nuclear forces $d\sim 10^{-15}m\!$, atomic forces $d\sim 10^{-10}m\!$. 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:

$\psi _{k}=\psi _{k}^{(0)}+\int d^{3}r'G_{k}(\mathbf {r} ,\mathbf {r} '){\frac {2m}{\hbar ^{2}}}V(r')\psi _{k}(r')$

where the Green's function satisfies

$(\nabla ^{2}+k^{2})G_{k}(\mathbf {r} ,\mathbf {r'} )=\delta (\mathbf {r} -\mathbf {r'} )$

and

$(\nabla ^{2}+k^{2})\psi _{k}^{(0)}(x)=0$

and the solution is chosen such that the second term in Eq.(2) corresponds to an outgoing wave. Then

$G_{k}(\mathbf {r} ,\mathbf {r} ')=-{\frac {1}{4\pi }}{\frac {e^{ik|r-r'|}}{|r-r'|}}$

and in the asymptotic limit of $r\to \infty \!$:

$\lim _{r\to \infty }\psi _{k}(\mathbf {r} )=\psi _{k}^{(0)}(\mathbf {r} )-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )\psi _{k}(\mathbf {r'} ){\frac {e^{ik\mathbf {r} }}{r}}=\psi _{k}^{(0)}(\mathbf {r} )+f_{k}(\theta ,\phi ){\frac {e^{ik\mathbf {r} }}{r}}$

where the scattering amplitude

$f_{k}(\theta ,\phi )=-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )\psi _{k}(\mathbf {r'} )$

and the angles $\theta \!$ and $\phi \!$ are the angles between ${\hat {r}}\!$ (the vector defining the detector) and $\mathbf {k} \!$ (the vector defining the in the incoming wave). Now the differential cross section is written through the ratio of the (outgoing) radial current density $j_{r}\!$ and the incident current density $j_{inc}\!$ as

${\frac {d\sigma }{d\Omega }}={\frac {j_{r}r^{2}}{j_{inc}}}$

The radial current is

$j_{r}={\frac {\hbar }{2mi}}(\psi _{sc}^{*}{\frac {\partial \psi _{sc}}{\partial r}}-\psi _{sc}{\frac {\partial \psi _{sc}^{*}}{\partial r}})={\frac {\hbar k}{mr^{2}}}|f(\theta ,\phi )|^{2}$

$\Rightarrow {\frac {d\sigma }{d\Omega }}=|f(\theta ,\phi )|^{2}$

Finally we have 1st Born approximation. For large $r\!$ we find

$\psi _{k}(\mathbf {r} )\approx \psi _{k}^{(0)}(\mathbf {r} )-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )\psi _{k}^{(0)}(\mathbf {r'} ){\frac {e^{ik\mathbf {r} }}{r}}$

and

$f_{k}(\theta ,\phi )=-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )e^{i\mathbf {k} \cdot \mathbf {r} '}$

Central potential scattering and phase shifts

Recall that for scattering we have Green function method

$\psi _{k}=\psi _{k}^{0}+\int d^{3}r'G(\mathbf {r} ,\mathbf {r} ')V(\mathbf {r} ')\psi _{k}(\mathbf {r} ')$

where the Green's function satisfies that

$(\nabla ^{2}+k^{2})G_{k}(\mathbf {r} ,\mathbf {r'} )=\delta (\mathbf {r} -\mathbf {r'} )$

and the solution is chosen such that the second term in Eq.(1) corresponds to an outgoing wave.

$G_{k}(\mathbf {r} ,\mathbf {r'} )=-{\frac {1}{4\pi }}{\frac {e^{ik|\mathbf {r} -\mathbf {r'} |}}{|\mathbf {r} -\mathbf {r'} |}}$

and in the asymptotic limit of r come to infinity

$\lim _{r\to \infty }|\mathbf {r} -\mathbf {r'} |=r-\mathbf {r} \cdot \mathbf {r'}$

Thus

$\lim _{r\to \infty }\psi _{k}(\mathbf {r} )=\psi _{k}^{0}(\mathbf {r} )-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )\psi _{k}(\mathbf {r'} ){\frac {e^{ik\mathbf {r} }}{r}}=\psi _{k}^{0}(\mathbf {r} )+f_{k}(\theta ,\phi ){\frac {e^{ik\mathbf {r} }}{r}}$

where

$f_{k}(\theta ,\phi )=-{\frac {m}{2\pi \hbar ^{2}}}\int d^{3}\mathbf {r'} e^{-ik\mathbf {\hat {r}} \cdot \mathbf {r'} }V(\mathbf {r'} )\psi _{k}(\mathbf {r'} )$

and the angles $\theta \!$ and $\phi \!$ are the angles between ${\hat {r}}\!$ (the vector defining the detector) and $\mathbf {k} \!$ (the vector defining the in the incoming wave).

For central potentials, i.e. if $V(r)=V(|\mathbf {r} |)$, then $f_{k}(\theta )\!$, i.e. the scattering amplitude does not depend on the azimuthal angle $\phi \!$. To determine $f_{k}(\theta )\!$ we need to find the solution of the Schrodinger equation:

$(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(|\mathbf {r} |))\psi ={\frac {\hbar ^{2}k^{2}}{2m}}\psi$

we use spherical coordinates

$(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {h^{2}l(l+1)}{2mr^{2}}}+V(|\mathbf {r} |))u_{l}(r)={\frac {\hbar ^{2}k^{2}}{2m}}u_{l}(r)$

For V(r) with a finite range d, we have shown that for r >> d we have

$(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}})u_{l}(r)={\frac {\hbar ^{2}k^{2}}{2m}}u_{l}(r)$

and the solution is a combination of spherical Bessel and Neumann functions

${\frac {u(r)}{r}}=A_{l}j^{l}(kr)+B_{l}n_{l}(kr)$

when r is large enough we use approximation of Bessel function and Neumann function.

${\frac {u(r)}{r}}\rightarrow A_{l}{\frac {\sin(kr-l{\frac {\pi }{2}})}{kr}}-B_{l}{\frac {\cos(kr-l{\frac {\pi }{2}})}{kr}}$

Letting

${\frac {B_{l}}{A_{l}}}=-\tan \delta _{l}$

here $\delta _{l}\!$ is called phase shift. we can rewrite the above expression (up to a normalization constant) as

${\frac {u_{l}(r)}{r}}\rightarrow {\frac {\sin(kr-l{\frac {\pi }{2}})}{kr}}$

Now since we are seeking the scattering amplitude with azimuthal symmetry, we can write the solution of the Schrodinger equation as a superposition of $m=0\!$ spherical harmonics only:

$\psi =\sum _{l=0}^{\mathop {\infty } }a_{l}(k)p_{l}(\cos \theta ){\frac {u_{l}(r)}{r}}$

$\psi =\sum _{l=0}^{\mathop {\infty } }a_{l}(k)p_{l}(\cos \theta ){\frac {\sin(kr-l{\frac {\pi }{2}}+\delta _{l}))}{kr}}$

where the Legendre polynomials are

$p_{l}(x)={\frac {1}{2^{l}l!}}{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}$

$p_{0}(x)=1;p_{1}(x)=x;p_{2}(x)={\frac {1}{2}}(3x^{2}-1)$

let us fix the coeffcients $a_{l}(k)\!$ by

$e^{ik\cos \theta }+f_{k}(\theta ,\phi ){\frac {e^{ik\mathbf {r} }}{r}}=\sum _{l=0}^{\mathop {\infty } }a_{l}(k)p_{l}(\cos \theta ){\frac {\sin(kr-l{\frac {\pi }{2}}+\delta _{l}))}{kr}}$

which must hold at large r and where we chose the coordinates by letting the incident wave propagate along z-direction. Note that (due to an entirely separate argument):

$e^{ikr\cos \theta }=\sum _{l=0}^{\mathop {\infty } }(2l+1)i^{l}p_{l}(\cos \theta ){\frac {\sin(kr-l{\frac {\pi }{2}})}{kr}}$

so

$\sum _{l=0}^{\mathop {\infty } }(2l+1)i^{l}p_{l}(\cos \theta ){\frac {\sin(kr-l{\frac {\pi }{2}})}{kr}}+f_{k}(\theta ,\phi ){\frac {e^{-ik\mathbf {r} }}{r}}=\sum _{l=0}^{\mathop {\infty } }a_{l}(k)p_{l}(\cos \theta ){\frac {\sin(kr-l{\frac {\pi }{2}}+\delta _{l}))}{kr}}$

We fix the coefficients $a_{l}(k)\!$ by matching the incoming spherical waves on both sides of the above equation. Note that this does not involve $f_{k}(\theta )\!$ since the scattering amplitude controls the outgoing spherical wave. Since $\cos x=(e^{ix}-e^{-ix})/2\!$ we have

$a_{l}(k)=(2l+1)i^{l}e^{i\delta _{l}}$

Therefore:

$f_{k}(\theta )={\frac {1}{k}}\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}}\sin \delta _{l}p_{l}(\cos \theta )$

Note that $\delta _{l}\!$ is a function of k and therefore a function of the incident energy. If $\delta _{l}\!$ 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 $\delta _{l}<0\!$ for repulsive potentials and $\delta _{l}>0\!$ for attractive potentials. Also, if l/k >> d, then the classical impact parameter is much larger than the range of the potential and in this case we expect $\delta _{l}\!$ to be small.

The differential scattering cross section is

${\frac {d\sigma }{d\Omega }}=|f_{k}(\theta )|^{2}={\frac {1}{k^{2}}}|\sum _{l=0}^{\infty }(2l+1)e^{i\delta _{l}}\sin \delta _{l}p_{l}(\cos \theta )|^{2}$

By integrating ${\frac {d\sigma }{d\Omega }}$ over the solid angle we obtain the total scattering cross section

$\sigma _{tot}={\frac {4\pi }{k^{2}}}\sum _{l=0}^{\mathop {\infty } }(2l+1)\sin ^{2}\delta _{l}$

which follows from the orthogonality of the Legendre polynomials

$\int _{-1}^{1}dxp_{l}(x)p_{l}'(x)={\frac {2}{(2l+2)}}\delta _{ll}$

Finally note that since $p_{l}(1)=1\!$ for all $l\!$, we have

$\sigma _{tot}={\frac {4\pi }{k}}\Im mf(0)$

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 V(r) has a differential cross section of:

The scattering amplitude, fk(r), is the coefficient of the outgoing wave.

The Born approximation, often called the first Born approximation, is a technique for find solutions when V(r) is small.

The scattering amplitude can be approximated by: where -ikr is the scattered portion and ik is the incident portion.

The scattering amplitude is defined as the coefficient of the outgoing wave in the asymptotic solution (for large r)

of the Schrodinger equation:

For a central-force potential, V(r), the Born scattering amplitude reduces to

Which leads to the Born cross section:

Born Approximation for Spherically Symmetric Potentials:

Given spherically symmetry we may define

and align the polar axis for the r integral lie along this quantity. We then have:

Our first Born integration then takes the form:

The phi integral introduces a trivial 2pi. For the theta integral we can use the following identity:

and get:

where the angular dependence of f is carried by kappa:

Example 1:

Consider the scattering amplitude from a Gaussian potential of the form:

Our scattering amplitude then becomes:

Integration by parts gives:

Coulomb potential scattering

Example 1

Lets look at an example of a Screened Coulomb (Yukawa) Potential, where we have a potential:

thus we have the differential cross section:

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 1/r potential.

Use a change of coordinates from Cartesian to parabolic coordinates:

, ,

The following is a picture of parabolic coordinates:

Phi represents rotation about the z-axis, xi represents the parabolas with their vertex at a minimum, and eta represents parabolas with their vertex at a maximum.

So now we can write the Schrodinger equation in parabolic coordinates:

So we will seek solutions which are independent of phi. Recall that the scattering amplitude is a function of only.

Look for solutions of the form:

We can tidy up the notation a little bit by using the following substitution:

Now let: From this we can write:

Recall the confluent hypergeometric function:

We can then write the recursion formula as the following:

This implies that:

where the confluent geometric function is written in terms of three new variables, and A is a c-number.

Now we can write the wavefunction due to Coulomb scattering:

Now we should look at the limit where z is taken to go to infiniti, and our confluent hypergeometric function is rewritten as:

Now we can use this to rewrite our equation for Phi of xi:

Rewriting our wavefunction :

where f(theta) is the following:

We can then get our differential cross section from that by squaring it:

If we normalize the wavefunction to give unit flux at large distances, we must take the following for the constant C:

So the wavefunction at large distances is given by the following:

where

Plugging this in for our wavefunction squared:

Now let's use the following quantity to represent the velocity:

For small incident velocities, we can write:

where the first equation is for an attractive Coulomb potential, and the second equation is for a repulsive Coulomb potential.

Two particle scattering

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

Lets 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 r1 <----> r2, rcm = (r1 + r2)/2 is invariant while r = r1-r2 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 Psi(r) by hand:

Under the transformation:

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:

and 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:

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:

Now if the particles are distinguishable, the cross section for observing a scattered particle at 90 degrees is then:

Where if the particles are indistinguishable, we see above that we will have:

Thus the differential cross-section is exactly twice the distinguishable case when the particles are indistinguishable.