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\!}

(hot) Tungsten Wire (electrons)

While a high current is passed through a tungsten wire, electrons are flying off of the wires and entering 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 H12 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 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 F(x,t)\!} . Classically, we would investigate this system by applying Newton's second law, 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 F = ma} . Assuming the force is conservative, it could also be expressed as the partial derivative with respect to 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 x} , and Newton's second law then reads:

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{d^2x}{dt^2}=-\frac{\partial V}{\partial x}}

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. This is most manifest as we attempt to describe increasingly smaller objects, in which case we enter the quantum mechanical regime where we cannot neglect the particles' wave properties. The complex amplitude satisfies the Schrodinger equation (below) subject to a (scalar) potential 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(x,t)} : 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(x,t)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x,t) } 1D Schrodinger equation subject to a (scalar) potential 2.0.1

While in 3D:

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(r,t)=\left[-\frac{\hbar^2}{2m}\nabla^2+V(r)\right]\psi(r,t)} 3D Schrodinger equation subject to a (scalar) potential 2.0.2

Given a solution which satisfies the above Schrodinger equation, Quantum Mechanics provides mathematical descriptions of the laws obeyed by the probability amplitudes associated with quantum motion.

Stationary states

Stationary states are states with definite values of energy 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 therefore eigenvalues of the Hamiltonian operator (see below). Taking 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(x,t)=e^{-iEt/\hbar}\psi(x)} , the stationary states are the solutions of the time-independent Schrödinger 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 E\psi(x)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x)}

If time t is fixed, evaluated and integrated over all space, the probability will end up equaling to one. The wavefunction has to be normalized over time. Let's see 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 |\psi(r,t)|^2} can be interpreted as probability density.We need

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 \int_{-\infty}^{\infty}d^3r|\psi(r,t_0)|^2=1 \Rightarrow \int dr^3 |\psi(r,t)|=1 \;\;\forall t}

i.e. if the wavefunction is normalized at to unity at some 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 t_0} then it must be normalized for any 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 t} .

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

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(x,t)=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x,t)}
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\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)}
Taking a complex conjugate
$-i\hbar \psi (x,t){\frac {\partial }{\partial t}}\psi ^{*}(x,t)=\psi (x,t)\left[-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V(x)\right]\psi ^{*}(x,t)$
and taking the difference of the above equations we finally find

${\frac {\partial }{\partial t}}\psi ^{*}(x,t)\psi (x,t)+\nabla \cdot {\frac {\hbar }{2im}}\left[\psi ^{*}(x,t)\nabla \psi (x,t)-\psi ^{*}(x,t)\nabla \psi (x,t)\right]=0$

Note that this is a form of a continuity equation

${\frac {\partial }{\partial t}}\rho (r,t)+\nabla \cdot j(r,t)=0$

where

$\rho (r,t)=\psi ^{*}(r,t)\psi (r,t)\!$

is the probability density

and

$j(r,t)={\frac {\hbar }{2im}}\left[\psi ^{*}(r,t)\nabla \psi (r,t)-\psi ^{*}(r,t)\nabla \psi (r,t)\right]$ 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:

${\frac {\partial }{\partial t}}\int d^{3}r|\psi (r,t)|^{2}=-\int d^{3}r(\nabla \cdot j)=-\oint dA\cdot j=0$

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 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

In Dirac's notation, the eigenfunctions are replaced by eigenkets (or simply kets) which are represented by . In this notation Schrödinger's equation is written as

$i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle ={\mathcal {H}}|\psi (t)\rangle$

For time independent Hamiltonians, the above equation separates and we can seek the solution of the form of (stationary states)

$|\psi (t)\rangle =e^{-iE_{n}t/\hbar }|\psi _{n}\rangle$.

The equation for stationary states in the Dirac notation is then

$E_{n}|\psi _{n}\rangle ={\mathcal {H}}|\psi _{n}\rangle .$

Heisenberg Uncertainty relations

Given a wave equation, the corresponding wavelength is usually easy to imagine. The momentum of the particle which is represented by the equation is then related to the wavelength by the de Broglie equation:

$p={\frac {h}{\lambda }}={\frac {2\pi \hbar }{\lambda }}$

However, for a given wave equation which has a well-defined wavelength in some region of space, the question 'where is the wave' seems not to make much sense. This question can be cleared up if we now imagine holding a string and giving it a quick snap with the wrist. Now a single pulse traverses the length of the string for which we can easily take measurements of its position. So, there seems to be some limitation on how precise we can make simultaneous measurements of both momentum and position. This uncertainty of measurement of either momentum or position take mathematical form in the Heisenberg Uncertainty relation:

$\Delta x\Delta p\geq {\frac {\hbar }{2}}$

where the $\Delta !/$ of each operator represents the positive square root of the variance, given generally 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 \langle(\Delta A)^2\rangle=\langle A^2\rangle-\langle A\rangle^2.}

A generalized expression for the uncertainty of any two operators A and B is shown to hold in most any undergraduate text:

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 \Delta A\Delta B=\frac{1}{2i}\langle [A,B]\rangle.}

And thus, there exists an uncertainty relation between any two observables which do not commute.

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

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 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} . Therefore, the wave function is oscillating 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 decaying exponentially 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} . The energy spectrum is still continous but no longer 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 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 wave function decays exponentially 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 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: 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(x)=+\infty} for 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 x \ge L/2} , otherwise 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(x)=0\!}

A particle subject to this potential is free everywhere except at the two ends (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 x = \pm L/2} ), where the infinite potential keeps the particle confined to the well. Within the well the Schrodinger equation takes the form:

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\psi}{dx^2}=E\psi}

or equivalently,

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{d^2\psi}{dx^2}=-k^2\psi}

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 k=\frac{\sqrt{2mE}}{\hbar}}

Writing the Schrodinger equation in this form, we see that our solution are those of the simple harmonic oscillator:

Now we impose that the solution must vanish 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 x = \pm L/2} :

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 -A\sin(kL/2)+B\cos(kL/2)=0\!}

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 A\sin(kL/2)+B\cos(kL/2)=0\!}

Adding the two equation, we get:

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 B\cos(kL/2)=0\!}

It follows that either 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 B=0\!} or 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 \cos(kL/2)=0\!} .

Case 1: 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 B=0\!} . In this case 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 A\ne0\!} , otherwise the wavefunction vanishes every where. Furthermore, it is required 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 \sin(kL/2)=0\!}

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 k=2n\pi/L\!} 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 n=1,2,3,...\!}

And the wave functions are odd: 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(x)=A\sin(2n\pi x/L)\!}

Case 2: 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 \cos(kL/2)=0\!} 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 k=(2n+1)\pi/L\!} where $n=0,1,2,...\!$

In this case $A=0\!$ and $B\neq 0\!$, and the wavefunctions are even: $\psi (x)=B\cos[(2n+1)\pi x/L]\!$

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 $x\rightarrow -x$, and so does the Hamiltonian. Therefore, the Hamiltonian commutes with the parity operator ${\hat {P}}$ (${\hat {P}}\psi (x)=\psi (-x)$). 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 : $-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}+V(x)\psi =E\psi$

is a second-order differential equation. Such equation has 2 linearly independent $\psi _{E}^{(1)}$ and $\psi _{E}^{(2)}$ for each value of $E\!$:

$-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi _{E}^{(1)}}{dx^{2}}}+V(x)\psi _{E}^{(1)}=E\psi _{E}^{(1)}$

$-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi _{E}^{(2)}}{dx^{2}}}+V(x)\psi _{E}^{(2)}=E\psi _{E}^{(2)}$

By definition in mathematics, the Wronskian of these functions is: $W=\psi _{E}^{(1)}{\frac {d\psi _{E}^{(2)}}{dx}}-{\frac {d\psi _{E}^{(1)}}{dx}}\psi _{E}^{(2)}$

Multiplying equation (2) by $\psi _{E}^{(2)}$, equation (3) by $\psi _{E}^{(1)}$, then subtracting one equation from the other, we get:

${\frac {d^{2}\psi _{E}^{(1)}}{dx^{2}}}\psi _{E}^{(2)}-{\frac {d^{2}\psi _{E}^{(2)}}{dx^{2}}}\psi _{E}^{(1)}=0$

$\rightarrow {\frac {d}{dx}}({\frac {d\psi _{E}^{(1)}}{dx}}\psi _{E}^{(2)}-{\frac {d\psi _{E}^{(2)}}{dx}}\psi _{E}^{(1)})=0$

$\rightarrow {\frac {dW}{dx}}=0$

$\rightarrow W=C$

where $C\!$ 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: $\psi _{E}^{(1)}(\infty )=\psi _{E}^{(2)}(\infty )=0$

From (4), (5) and (6), it follows that $W=0\!$ From (4) and (7), we get: $\psi _{E}^{(1)}{\frac {d\psi _{E}^{(2)}}{dx}}-{\frac {d\psi _{E}^{(1)}}{dx}}\psi _{E}^{(2)}=0$

$\rightarrow {\frac {1}{\psi _{E}^{(1)}}}{\frac {d\psi _{E}^{(1)}}{dx}}-{\frac {1}{\psi _{E}^{(2)}}}{\frac {d\psi _{E}^{(2)}}{dx}}=0$

$\rightarrow {\frac {d}{dx}}[ln(\psi _{E}^{(1)})-ln(\psi _{E}^{(2)})]=0$

$\rightarrow ln(\psi _{E}^{(1)})-ln(\psi _{E}^{(2)})=constant$

$\rightarrow \psi _{E}^{(1)}=constant.\psi _{E}^{(2)}$

From (8) it follows that $\psi _{E}^{(1)}$ and $\psi _{E}^{(2)}$ 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.

Oscillation theorem

Let us concentrate on the bound states of a set of wavefunctions. Let $\psi _{1}\!$ be an eigenstate with energy $E_{1}\!$ and $psi_{2}\!$ an eigenstate with energy $E_{2}\!$, and $E_{2}>E_{1}\!$. We also can set boundary conditions, where both $\psi _{1}\!$ and $\psi _{2}\!$ vanish at $x_{0}\!$.This implies that

$-\psi _{2}{\frac {\partial ^{2}\psi _{1}}{\partial x^{2}}}+V(x)\psi _{2}\psi _{1}=E_{1}\psi _{2}\psi _{1}\!$

$-\psi _{1}{\frac {\partial ^{2}\psi _{2}}{\partial x^{2}}}+V(x)\psi _{1}\psi _{2}=E_{2}\psi _{1}\psi _{2}\!$

Subtracting the second of these from the first and simplifying, we see that

${\frac {\partial }{\partial x}}(-\psi _{2}{\frac {\partial \psi _{1}}{\partial x}}+\psi _{1}{\frac {\partial \psi _{2}}{\partial x}})=(E_{1}-E_{2})\psi _{1}\psi _{2}\!$

If we now integrate both sides of this equation from $x_{0}\!$ to any position $x'\!$ and simplify, we see that

$-\psi _{2}(x'){\frac {\partial \psi _{1}(x')}{\partial x}}+\psi _{1}(x'){\frac {\partial \psi _{2}(x')}{\partial x}}=(E_{1}-E_{2})\int _{x_{0}}^{x'}\psi _{1}\psi _{2}dx\!$

The key is to now let $x'\!$ be the first position to the right of $x_{0}\!$ where $\psi _{1}\!$ vanishes.

$-\psi _{2}(x'){\frac {\partial \psi _{1}(x')}{\partial x}}=(E_{1}-E_{2})\int _{x_{0}}^{x'}\psi _{1}\psi _{2}dx\!\!$

Now, if we assume that $\psi _{2}\!$ does not vanish at or between $x'\!$ and $x_{0}\!$, 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 $\psi _{2}\!$ must vanish at least once between $x'\!$ and $x_{0}\!$ if $E_{2}>E_{1}\!$.

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).

Figure 1.

The Hamiltonian of system under periodic potential commutes with the Translation Operator defined as:

${\hat {T}}_{a}\psi (x)=\psi (x+a)\!$

This means that there is a simultaneous eigenstate of the Hamiltonian and the Translation Operator. The eigenfunction to the Schrodinger Equation,

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

has the form of the following,

$\psi (x)=e^{ikx}u_{k}(x)\!$

where

$u_{k}(x+a)=u_{k}(x)\!$

This result is also known as the Bloch Theorem.

Also, by operating the ${\hat {T}}_{a}\!$ operator on the wavefunction (also known as the Bloch wave), it can be seen that this waveform is also an eigenfunction of the ${\hat {T}}_{a}\!$ operator, as shown in the following,

${\begin{aligned}{\hat {T}}_{a}\psi (x)&={\hat {T}}_{a}(e^{ikx}u_{k}(x))\\&=(e^{ik(x+a)}u_{k}(x+a))\\&=e^{ika}(e^{ikx}u_{k}(x))\\&=e^{ika}\psi (x)\end{aligned}}$

Using the same argument, it is clear that,

$({\hat {T}}_{a})^{n}\psi (x)=e^{ikna}\psi (x)$

Also, note that if k is complex, then after multiple ${\hat {T}}_{a}\!$ 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.

Figure 2. Energy band illustration showing the condition for metal, semiconductor, and insulator.

Consider for example the periodic potential and the resulting Schrodinger equation,

$V(x)=V_{0}\sum _{n=-\infty }^{\infty }\delta (x-na)$

$(-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+V_{0}\sum _{n=-\infty }^{\infty }\delta (x-na))\psi (x)=E\psi (x)$

Focusing the attention for case when 0 < x < a, the solution to the Schrodinger equation is of the form:

$E=-{\frac {\hbar ^{2}q^{2}}{2m}}$

$\psi (x)=Ae^{iqx}+B{-iqx}\!$

$\psi (x)=e^{ikx}(Ae^{i(q-k)x}+Be^{-i(q+k)x})\!$

$\psi (x)=e^{ikx}u_{k}(x)\!$

From periodicity,

$u_{k}(\epsilon )=u_{k}(-\epsilon )\!$

$u_{k}(\epsilon )=u_{k}(a-\epsilon )\!$

Thus, the wavefunction from x < 0 (left) and x > 0 (right) can be written as:

$\psi _{r}=e^{ikx}u_{k}(x)=e^{ikx}(Ae^{i(q-k)x}+Be^{-i(q+k)x})\!$

$\psi _{r}=e^{ikx}u_{k}(x+a)=e^{ikx}(Ae^{i(q-k)(x+a)}+Be^{-i(q+k)(x+a)})\!$

When the continuity requirement at x = 0 is also being imposed, the following relation is found:

$\psi _{l}(0)=\psi _{r}(0)\!$

$A+B=Ae^{i(q-k)a}+Be^{-i(q+k)a}\!$ (1)

From differentiability and periodicity, the Schrodinger equation can be solved as the following:

$\int _{-\epsilon }^{\epsilon }-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi (x)}{\partial x^{2}}}dx+\int _{-\epsilon }^{\epsilon }V_{0}\sum _{n=-\infty }^{\infty }\delta (x-na)\psi (x)dx=\int _{-\epsilon }^{\epsilon }E\psi (x)dx$

where $\epsilon \!$ is small, approaching zero. In this case, the term on the right hand side can be taken to be 0. Thus,

$-{\frac {\hbar ^{2}}{2m}}({\frac {\partial \psi _{r}(\epsilon )}{\partial x}}-{\frac {\partial \psi _{l}(-\epsilon )}{\partial x}})=-V_{0}\psi _{l}(0)$

$\lim _{\epsilon \to 0}({\frac {\partial \psi _{r}(\epsilon )}{\partial x}}-{\frac {\partial \psi _{l}(-\epsilon )}{\partial x}})={\frac {2mV_{0}}{\hbar ^{2}}}\psi _{l}(0)$

where,

$\lim _{\epsilon \to 0}{\frac {\partial \psi _{r}(\epsilon )}{\partial x}}=iq(A-B)$

$\lim _{\epsilon \to 0}{\frac {\partial \psi _{l}(-\epsilon )}{\partial x}}=iq(Ae^{i(q-k)a}-Be^{-i(q+k)a})$

Evaluating further, the following condition is found:

$iq(A-B-Ae^{i(q-k)a}+Be^{-i(q+k)a})={\frac {2mV_{0}}{\hbar ^{2}}}(A+B)$ (2)

By simultaneously solving equation (1) and (2), the relationship between q and k is found to be:

$\cos(ka)=\cos(qa)+{\frac {mV_{0}a}{\hbar ^{2}}}{\frac {\sin(qa)}{qa}}$ (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.

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.

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:

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

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

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 $\langle \left(\Delta A\right)^{2}\rangle \langle \left(\Delta B\right)^{2}\rangle ={\frac {1}{4}}\langle C\rangle ^{2}$

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. While the Schrodinger picture, in which a wave function changes with time, is the most obvious treatment, the Heisenberg and Interaction pictures are sometimes more convenient. The two "pictures" correspond, theoretically, to the time evolution of two different things. In the Schrodinger picture, the wave functions evolve in time. In the Heisenberg picture, the operators evolve in time. Mathematically, both 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:

$i\hbar {\frac {\partial }{\partial t}}U(t,t_{0})=HU(t,t_{0})$

The solution to this differential equation depends on the form of .

If we know the time evolution operator, $U$, 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.

$U|\alpha (0)\rangle =|\alpha (t)\rangle .$

Therefore, if we know the initial state of a system, we can obtain the expectation value of an operator, A, at some later time:

$\langle A\rangle (t)=\langle \alpha (t)|A|\alpha (t)\rangle =\langle \alpha (0)|U^{\dagger }AU|\alpha (0)\rangle .$

We can make a redefinition by claiming that

$A_{H}(t)=U^{\dagger }AU$

and taking as our state kets the time independent, initial valued state ket $|\alpha (0)\rangle$.

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.

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:

${\frac {d}{dt}}A_{H}={\frac {\partial U^{\dagger }}{\partial t}}AU+U^{\dagger }{\frac {\partial A}{\partial t}}U+U^{\dagger }A{\frac {\partial T}{\partial t}}={\frac {-1}{i\hbar }}U^{\dagger }HAU+\left({\frac {\partial A}{\partial t}}\right)_{H}+U^{\dagger }AHU{\frac {1}{i\hbar }}={\frac {1}{i\hbar }}\left(\left[A,H\right]\right)_{H}+\left({\frac {\partial A}{\partial t}}\right)_{H}.$

The last equation is known as the Ehrenfest Theorem.

For example, if we have a hamiltonian of the form,

$H={\frac {p^{2}}{2m}}+V(r),$

then the Heisenberg equations of motion for p and r are

${\frac {d{\hat {r}}_{H}}{dt}}={\frac {{\hat {p}}_{H}}{m}},$
${\frac {d{\hat {p}}_{H}}{dt}}=-\nabla V({\hat {r}}_{H}).$

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 $\omega ={\sqrt {\frac {k}{m}}}\!$

${\frac {d{\hat {x}}_{H}}{dt}}={\frac {{\hat {p}}_{H}}{m}}$

${\frac {d{\hat {p}}_{H}}{dt}}=-k{{\hat {x}}_{H}}.$

we can solve the above equations of motion and find

$x_{H}(t)=x_{0}(0)\cos(\omega t)+{\frac {p_{H}(0)}{m\omega }}\sin(\omega t)$

$p_{H}(t)=p_{0}(0)\cos(\omega t)-x_{H}(0)m\omega \sin(\omega t).\!$

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".

Feynman path integrals

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, $U(t)$, is found directly. The amplitude for a particle to start at $x_{0}$ at $t=0$ and end at $x$ at t can be expressed as a path integral

$\langle x|{\hat {U}}(t)|x_{0}\rangle =\int _{x_{0}}^{x}Dx(t')e^{iS[x(t')]}$

Where $S[x(t)]$ is the action for the the path $x(t')$.

The action is given by the time integral of the Lagrangian, just as in classical mechanics $S[x(t')]=\int _{0}^{t}dt'{\mathcal {L}}[x(t'),{\dot {x}}(t'),t']$

Where ${\mathcal {L}}[x(t'),{\dot {x}}(t'),t']={\frac {1}{2}}m{\dot {x}}^{2}(t')-V(x[t'],t')$ is the Lagrangian. Knowing the propagator, we can calculate the probability that a particle in state $x_{0}$ at t=0 will be in state $x$ at time t by taking the absolute value squared.

Discrete eigenvalues and bound states. Harmonic oscillator and WKB approximation

Harmonic oscillator spectrum and eigenstates

1-D harmonic oscillator is a particle moving in the potential of the form: $V(x)={\frac {1}{2}}m\omega ^{2}x^{2}$

We can see that $V(x)\rightarrow \infty$ as $x\rightarrow \pm \infty$, 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:

$H=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {1}{2}}m\omega ^{2}x^{2}$

It will be easy to memorize how to contruct lowering and raising operator by factorizing and rewriting as follows: $H=\hbar \omega \left(-{\frac {\hbar }{2m\omega }}{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {1}{2\hbar }}m\omega x^{2}\right)$

Then we define:

$a={\sqrt {\frac {m\omega }{2\hbar }}}x+{\sqrt {\frac {\hbar }{2m\omega }}}{\frac {d}{dx}}$ as the lowering operator, and
$a^{\dagger }={\sqrt {\frac {m\omega }{2\hbar }}}x-{\sqrt {\frac {\hbar }{2m\omega }}}{\frac {d}{dx}}$ as the raising operator.

One way to distinguish $a\!$ from $a^{\dagger }$ is to remember that the ground state wave function is a Gaussian function and $a\!$ will annihilate this state. We have the following commutation relation: $[a,a^{\dagger }]=1.$

$H\!$ can be rewritten in terms of $a\!$ and $a^{\dagger }$ as follows:

$H=\hbar \omega \left(a^{\dagger }a+{\frac {1}{2}}\right)$ 

Now, let's see how $a\!$ and $a^{\dagger }\!$ act on an energy eigenstate $|\Psi \rangle \!$: For $a\!$:

$Ha|\Psi \rangle =\left(E-\hbar \omega \right)|\Psi \rangle$

This means that $a|\Psi \rangle$ is also an energy eigenstate but correspoding to a lower energy, $E-\hbar \omega \!$. $a\!$ is therefore the lowering operator.

Similarly,

$Ha^{\dagger }|\Psi \rangle =\left(E+\hbar \omega \right)|\Psi \rangle$

and $a^{\dagger }\!$ is the so-called raising operator.

So, starting from any energy eigenstates, we can construct all other energy eigenstates by applying $a\!$ or $a^{\dagger }\!$ repeatedly. Although there is no limit in applying $a^{\dagger }\!$, there is a limit in applying $a\!$. The process of lowering energy must stop at some point, since $E>0$. For the eigenstate of lowest energy $|Psi_{0}\rangle$ (the ground state), we have:

$a|\Psi _{0}\rangle =0\!$
$\Rightarrow {\sqrt {\frac {m\omega }{2\hbar }}}x+{\sqrt {\frac {\hbar }{2m\omega }}}{\frac {d}{dx}}\Psi _{0}(x)=0\!$

This is a first order ordinary differential equation, which can be easily solved, and the result is as follows:

$\Psi _{0}(x)=Ae^{-{\frac {m\omega }{2\hbar }}x^{2}}$

where $A\!$ is a constant, which can be determined from the normalization condition:

$1=\int _{-\infty }^{\infty }|\Psi _{0}(x)|^{2}dx=\int _{-\infty }^{\infty }dxA^{2}e^{-{\frac {m\omega }{\hbar }}x^{2}}=A^{2}{\sqrt {\frac {\pi \hbar }{m\omega }}}\Rightarrow A=\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{4}}$

Normalized ground state wave function:

$\Psi _{0}(x)=\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{4}}e^{-{\frac {m\omega }{2\hbar }}x^{2}}$

The energy spectrum of 1-D harmonic oscillator is:

$E=\hbar \omega \left(n+{\frac {1}{2}}\right);n=0,1,2,\ldots$

Excited state wave function
Energy eigenstates with $E>0$ are called excited states. By applying $a^{\dagger }$ 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):

$|\Psi _{n}\rangle ={\frac {(a^{\dagger })^{n}}{\sqrt {n!}}}|\Psi _{0}\rangle \!$

In the position representation

$\langle x|\Psi _{n}\rangle =\left({\frac {m\omega }{\pi \hbar }}\right)^{\frac {1}{4}}{\frac {1}{\sqrt {2^{n}n!}}}H_{n}\left({\sqrt {\frac {m\omega }{\hbar }}}x\right)e^{-{\frac {m\omega }{2\hbar }}x^{2}}\!$

where $H_{n}(\xi )=(-1)^{n}e^{\xi ^{2}}{\frac {d^{n}}{d\xi ^{n}}}e^{-\xi ^{2}}$ 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 $a$, with eigenvalue $\alpha$:

$a|\alpha >=\alpha |\alpha >$ where $\alpha$ 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 $a{\dagger }$ because $a^{+}|n>={\sqrt {n+1}}|n+1>$.

I. Coherent states construction.

$|\alpha >=\sum _{n=0}^{+\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}|n>=e^{\alpha a^{+}}|0>$
$a|\alpha >=\sum _{n=0}^{+\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}a|n>=\sum _{n=1}^{+\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}{\sqrt {n}}|n-1>=\sum _{n=1}^{+\infty }{\frac {\alpha ^{n}}{\sqrt {(n-1)!}}}|n-1>=\alpha (\sum _{n=0}^{+\infty }{\frac {\alpha ^{n}}{\sqrt {n!}}}|n>)=\alpha |\alpha >$

II. Coherent states normalization.

$|\alpha >=Ne^{\alpha a^{+}}|0>$ where $N$ is normalization constant.

$1=<\alpha |\alpha >=<0|Ne^{\alpha ^{*}a}Ne^{\alpha a^{+}}|0>=N^{2}<0|e^{\alpha ^{*}a}e^{\alpha a^{+}}|0>$ For any operators A and B which both commute with their commutator, we have: $e^{A}e^{B}=e^{A+B}e^{{\frac {1}{2}}[A,B]}$

and similarly,

  $e^{B}e^{A}=e^{B+A}e^{{\frac {1}{2}}[B,A]}=e^{A+B}e^{-{\frac {1}{2}}[A,B]}$

therefore:

   $e^{A}e^{B}=e^{B}e^{A}e^{-[A,B]}\!$

Apply this result for $A=\alpha ^{*}a$ and $B=\alpha a^{+}$ ( A and B both commute with their commutator because $[A,B]=|\alpha |^{2})$, we have:

$1=<\alpha |\alpha >=N^{2}<0|e^{\alpha ^{*}a}e^{\alpha a^{+}}|0>$
$N^{2}<0|e^{\alpha a^{+}}e^{\alpha ^{*}a}e^{[\alpha ^{*}a,\alpha a^{+}]}|0>=N^{2}e^{|\alpha |^{2}}<0|e^{\alpha a^{+}}e^{\alpha ^{*}a}|0>$
$=N^{2}e^{|\alpha |^{2}}<0|e^{\alpha a^{+}}|0>=N^{2}e^{|\alpha |^{2}}<0|0>=N^{2}e^{|\alpha |^{2}}$ $\rightarrow N=e^{-{\frac {1}{2}}|\alpha |^{2}}$

 $\rightarrow {\mbox{Normalized coherent states:}}|\alpha >=e^{-{\frac {1}{2}}|\alpha |^{2}}e^{\alpha a^{+}}|0>$

III. Inner product of two coherent states

There is an eigenstate $|\alpha >$ of lowering operator eq=a for any complex number $\alpha$. Therefore, we have a set of coherent states. This is NOT an orthogonal set. Indeed, the inner product of two coherent states $|\alpha >$ and $|\beta >$ can be calculated as follows:

$<\beta |\alpha >=e^{-{\frac {1}{2}}|\alpha |^{2}}e^{-{\frac {1}{2}}|\beta |^{2}}<0|e^{\beta ^{*}a}e^{\alpha a^{+}}|0>$

$=e^{-{\frac {1}{2}}|\alpha |^{2}}e^{-{\frac {1}{2}}|\beta |^{2}}<0|e^{\alpha a^{+}}e^{\beta ^{*}a}e^{[\beta ^{*}a,\alpha a^{+}]}|0>$

$=e^{-{\frac {1}{2}}|\alpha |^{2}}e^{-{\frac {1}{2}}|\beta |^{2}}e^{\alpha \beta ^{*}}<0|e^{\alpha a^{+}}e^{\beta ^{*}a}|0>$

$=e^{-{\frac {1}{2}}|\alpha |^{2}}e^{-{\frac {1}{2}}|\beta |^{2}}e^{\alpha \beta ^{*}}$

$\rightarrow |<\beta |\alpha >|^{2}=e^{-|\alpha -\beta |^{2}}$

Hence, the set of coherent states is not orthogonal and the distance $|\alpha -\beta |$ 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:

$<x|{\hat {U}}(t,0)|x_{0}>=e^{{\frac {i}{\hbar }}S}\int _{y(0)=0}^{y(t)=0}D[y(t')]e^{{\frac {i}{\hbar }}\int _{0}^{t}({\frac {1}{2}}my'^{2}-{\frac {1}{2}}ky^{2})dt'}$

where $y(t')$ is the deviations of possible trajectories about the classical trajectory.

Saddle point action

The classical action $S$ can be evaluated as follows:

$S=\int _{0}^{t}(KE-PE)dt$

Where $KE$ is the kinetic engergy and $PE$ is the potential energy.

Equation of motion for harmonic oscillator: $x_{cl}(t')=Acos(\omega t')+Bsin(\omega t')\!$
$A$ and $B$ are constants.

At $t'=0$ (starting point),$x_{cl}(0)=x_{0}\rightarrow A=x_{0}$.

At $t'=t$ (final point), $x_{cl}(t)=x\rightarrow B={\frac {x-x_{0}cos(\omega t)}{sin(\omega t)}}.$

Substitute:

$x_{cl}(t')=x_{0}cos(\omega t')+{\frac {x-x_{0}cos(\omega t)}{sin(\omega t)}}sin(\omega t')\Rightarrow {\frac {dx_{cl}(t')}{dt'}}=-\omega x_{0}sin(\omega t')+\omega {\frac {x-x_{0}cos(\omega t)}{sin(\omega t)}}cos(\omega t')$
$KE={\frac {1}{2}}m({\frac {dx_{cl}}{dt}})^{2}={\frac {1}{2}}m[-\omega x_{0}sin(\omega t')+\omega {\frac {x-x_{0}cos(\omega t)}{sin(\omega t)}}cos(\omega t')]^{2}$
$PE={\frac {1}{2}}k(x_{cl}(t'))^{2}={\frac {1}{2}}k[x_{0}cos(\omega t')+{\frac {x-x_{0}cos(\omega t)}{sin(\omega t)}}sin(\omega t')]^{2}$
Substituting, integrating from time 0 to time t and simplifying, we get:

$S=S(t,x,x_{0})={\frac {m\omega }{2sin(\omega t)}}((x^{2}+x_{0}^{2})cos(\omega t)-2xx_{0})$

Harmonic fluctuations

Now, let's evaluate the path integral: $A=A(t)=\int _{y(0)=0}^{y(t)=0}D[y(t')]e^{{\frac {i}{\hbar }}\int _{0}^{t}({\frac {1}{2}}my'^{2}-{\frac {1}{2}}ky^{2})dt'}$

Note that the integrand is taken over all possible trajectory starting at point $x_{0}$ at time $t'=0$, ending at point $x$ at time $t'=t$.

Expanding this integral,

$A(t)=\left({\frac {m}{2\pi i\hbar }}\right)^{\frac {N}{2}}\int _{-\infty }^{\infty }dy_{1}\ldots dy_{N-1}\exp {\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{N-1}^{2}-{\frac {\Delta t}{2}}ky_{N-1}^{2}\right)\right]}\exp {\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}(y_{N-1}-y_{N-2})^{2}-{\frac {\Delta t}{2}}ky_{N-2}^{2}\right)\right]}\ldots \exp {\left[{\frac {i}{\hbar }}\left({\frac {m}{2\Delta t}}y_{1}^{2}-{\frac {\Delta t}{2}}ky_{1}^{2}\right)\right]}$

where $N\Delta t=t\!$.

Expanding the path trajectory in Fourier series, we have $y(t')=\sum _{n}a_{n}\sin \left({\frac {n\pi t'}{t}}\right)$

we may express $A(t)\!$ in the form

$A(t)=C\int _{-\infty }^{\infty }da_{1}\ldots da_{N-1}\exp {\left[\sum _{n=1}^{N-1}{\frac {im}{2\hbar }}\left(\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right)a_{n}^{2}\right]}$

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 $\omega \!$. Thus, evaluating the integral of,

$A(t)=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}-\omega ^{2}\right]^{-{\frac {1}{2}}}=C'\prod _{n=1}^{N-1}\left[\left({\frac {n\pi }{t}}\right)^{2}\right]^{-{\frac {1}{2}}}\prod _{n=1}^{N-1}\left[1-\left({\frac {\omega t}{n\pi }}\right)^{2}\right]^{-{\frac {1}{2}}}$

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,

$A(t)=C''{\sqrt {\frac {\omega t}{\sin(\omega t)}}}$

In the limit $\omega \rightarrow 0$, we already know that

$C''={\sqrt {\frac {m}{2\pi i\hbar t}}}$

Thus,

$A(t)={\sqrt {\frac {m}{2\pi i\hbar t}}}{\sqrt {\frac {\omega t}{\sin(\omega t)}}}={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}$

and

$<x|{\hat {U}}(t,0)|x_{0}>={\sqrt {\frac {m}{2\pi i\hbar \sin(\omega t)}}}e^{{\frac {i}{\hbar }}\left({\frac {m\omega }{2sin(\omega t)}}((x^{2}+x_{0}^{2})cos(\omega t)-2xx_{0})\right)}$

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 $e\!$ and mass $m\!$ in an external electromagetic field, which may be time-dependent is given as follows:

$H={\frac {1}{2m}}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))^{2}+e\phi ({\mathbf {r}},t)$

where ${\mathbf {A}}\!$ and $\phi \!$ 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 ${\mathbf {r}}\!$ we have:

${\frac {d{\mathbf {r}}}{dt}}={\frac {1}{i\hbar }}[{\mathbf {r}},H]={\frac {1}{i\hbar }}[{\mathbf {r}},{\frac {1}{2m}}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))^{2}+e\phi ({\mathbf {r}},t)]$ (${\mathbf {r}}$ does not depend on $t$ explicitly)

$={\frac {1}{2im\hbar }}[{\mathbf {r}},({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))^{2}]$

$={\frac {1}{2im\hbar }}[{\mathbf {r}},{\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))]({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))+{\frac {1}{2im\hbar }}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))[{\mathbf {r}},{\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))]$

$={\frac {1}{2im\hbar }}[{\mathbf {r}},{\mathbf {p}}]({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))+{\frac {1}{2im\hbar }}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))[{\mathbf {r}},{\mathbf {p}}]$

$={\frac {1}{2im\hbar }}ih({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))+{\frac {1}{2im\hbar }}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))ih$

$={\frac {1}{m}}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))$

is the equation of motion for the position operator ${\mathbf {r}}$. This equation also defines the velocity operator ${\mathbf {v}}$:

${\mathbf {v}}={\frac {1}{m}}({\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t))$

The Hamiltonian can be rewritten as:

$H={\frac {m}{2}}{\mathbf {v}}.{\mathbf {v}}+e\phi$

The Heisenberg equation of motion for velecity operator is:

${\frac {d{\mathbf {v}}}{dt}}={\frac {1}{i\hbar }}[{\mathbf {v}},H]+{\frac {\partial {\mathbf {v}}}{\partial t}}$ $={\frac {1}{i\hbar }}[{\mathbf {v}},{\frac {m}{2}}{\mathbf {v}}.{\mathbf {v}}]+{\frac {1}{i\hbar }}[{\mathbf {v}},e\phi ]-{\frac {e}{mc}}{\frac {\partial {\mathbf {A}}}{\partial t}}$ (note that ${\mathbf {p}}$ does not denpend on $t$ expicitly)

Let's use the following commutator identity:

$[{\mathbf {v}},{\mathbf {v}}.{\mathbf {v}}]={\mathbf {v}}\times ({\mathbf {v}}\times {\mathbf {v}})-({\mathbf {v}}\times {\mathbf {v}})\times {\mathbf {v}}$

Substituting, we get:

${\frac {d{\mathbf {v}}}{dt}}={\frac {1}{i\hbar }}{\frac {m}{2}}({\mathbf {v}}\times ({\mathbf {v}}\times {\mathbf {v}})-({\mathbf {v}}\times {\mathbf {v}})\times {\mathbf {v}})+{\frac {1}{i\hbar }}e[{\mathbf {v}},\phi ]-{\frac {e}{mc}}{\frac {\partial {\mathbf {A}}}{\partial t}}$

Now let's evaluate ${\mathbf {v}}\times {\mathbf {v}}$ and $[{\mathbf {v}},\phi ]$:

$({\mathbf {v}}\times {\mathbf {v}})_{i}=\epsilon _{ijk}v_{j}v_{k}=\epsilon _{ijk}{\frac {1}{m}}(p_{j}-{\frac {e}{c}}A_{j}({\mathbf {r}},t)){\frac {1}{m}}(p_{k}-{\frac {e}{c}}A_{k}({\mathbf {r}},t))$ (sum over all repeated indices)
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}(p_{j}A_{k}({\mathbf {r}},t)+A_{j}({\mathbf {r}},t)p_{k})$ (because $p_{j}$ and $p_{k}$ commute and so do $A_{j}$ and $A_{k}$)
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)-{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{j}({\mathbf {r}},t)p_{k}$
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)-{\frac {e}{m^{2}c}}\epsilon _{ikj}A_{k}({\mathbf {r}},t)p_{j}$ (Switching indices in the second tems)
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}p_{j}A_{k}({\mathbf {r}},t)+{\frac {e}{m^{2}c}}\epsilon _{ijk}A_{k}({\mathbf {r}},t)p_{j}$ (because $\epsilon _{ikj}=-\epsilon _{ijk}$)
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}[p_{j},A_{k}({\mathbf {r}},t)]$
$=-{\frac {e}{m^{2}c}}\epsilon _{ijk}{\frac {\hbar }{i}}\nabla _{j}A_{k}({\mathbf {r}},t)eq==i\hbar {\frac {e}{m^{2}c}}(\nabla \times {\mathbf {A}})_{i}$
$\rightarrow [{\mathbf {v}}\times {\mathbf {v}}]=i\hbar {\frac {e}{m^{2}c}}(\nabla \times {\mathbf {A}})=i\hbar {\frac {e}{m^{2}c}}{\mathbf {B}}$
where

$[{\mathbf {v}},\phi ]={\frac {1}{m}}[{\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t),\phi ({\mathbf {r}},t)]={\frac {1}{m}}[{\mathbf {p}},\phi ({\mathbf {r}},t)]={\frac {1}{m}}{\frac {\hbar }{i}}\nabla \phi$

Substituting and rearranging, we get:

$m{\frac {d{\mathbf {v}}}{dt}}={\frac {e}{2c}}({\mathbf {v}}\times {\mathbf {B}}-{\mathbf {B}}\times {\mathbf {v}})+e{\mathbf {E}}$

where ${\mathbf {E}}=-\nabla \phi -{\frac {1}{c}}{\frac {\partial {\mathbf {A}}}{\partial t}}$
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 $\hbar \rightarrow 0\!$, the wavelength, $\lambda =2\pi \hbar /p\!$, tends to zero, and any smooth potential is slowly varying in this limit. Therefore, $\lambda \!$ can be thought of as a local quantity $\lambda (x)\!$. 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:

$\psi (x)\approx {\frac {C}{\sqrt {p(x)}}}\exp \left[{\frac {i}{\hbar }}\int _{x_{0}}^{x}p(x')dx'\right]$

where $p(x)\!$ is the classical formula for the momentum of a particle with total energy $E\!$ and potential energy $V(x)\!$ given by:

$p(x)={\sqrt {2m(E-V(x))}}\!$

However, note that at a classical turning point $p(x_{turning})\rightarrow 0$ 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:

$V(x)\approx E+V'(0)x\!$

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:

$\psi _{p}(x)=aAi\left(\alpha x\right)+bBi\left(\alpha x\right)\!$

where $a,b\!$ are c-number coefficients and

$\alpha =\left({\frac {2m}{\hbar ^{2}}}V'(0)\right)^{\frac {1}{3}}$

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,

$p^{2}=2m(-V'(0)x)\Rightarrow 2p{\frac {dp}{dx}}=-2mV'(0)\Rightarrow {\frac {dp}{dx}}=-{\frac {m}{p}}V'(0)$

Since the region of applicability of the WKB approximation is $1\gg {\frac {1}{2\pi }}\left|{\frac {d\lambda }{dx}}\right|$

near the turning point

$p^{3}=\hbar m|V'(0)|\Rightarrow |x|\gg {\frac {\hbar ^{\frac {2}{3}}}{2}}|mV'(0)|^{-{\frac {1}{3}}}$

This implies that the width of the region around the classical turning point vanishes as $\hbar ^{\frac {2}{3}}$. Thus, we can come as close to the turning point as we wish with the WKB approximations by taking a limit as $\hbar \!$ approaches zero, as long as the distance from the classical turning point is much less than $\hbar ^{\frac {2}{3}}$. 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:

$z\rightarrow \infty :Ai(z)\rightarrow {\frac {1}{2{\sqrt {\pi }}}}z^{-{\frac {1}{4}}}e^{-{\frac {2}{3}}|z|^{\frac {3}{2}}}$
$z\rightarrow \infty :Bi(z)\rightarrow {\frac {1}{\sqrt {\pi }}}z^{-{\frac {1}{4}}}e^{{\frac {2}{3}}|z|^{\frac {3}{2}}}$
$z\rightarrow -\infty :Ai(z)\rightarrow {\frac {1}{\sqrt {\pi }}}z^{-{\frac {1}{4}}}\cos \left({\frac {2}{3}}|z|^{\frac {3}{2}}-{\frac {\pi }{4}}\right)$
$z\rightarrow -\infty :Bi(z)\rightarrow -{\frac {1}{\sqrt {\pi }}}z^{-{\frac {1}{4}}}\sin \left({\frac {2}{3}}|z|^{\frac {3}{2}}-{\frac {\pi }{4}}\right)$

And noticing that (for negative $x$)

${\frac {1}{\hbar }}\int _{x}^{0}p(x')dx'={\sqrt {\frac {2mV'(0)}{\hbar ^{2}}}}\int _{x}^{0}{\sqrt {-x'}}dx'={\frac {2}{3}}{\sqrt {\frac {2mV'(0)}{\hbar ^{2}}}}|x|^{\frac {3}{2}}$

and

${\frac {1}{\sqrt {p(x)}}}=\left(2mV'(0)\right)^{-{\frac {1}{4}}}|x|^{-{\frac {1}{4}}}$

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 $\hbar \rightarrow 0$ 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 $\hbar$. The two regions must therefore overlap.

For example, one can take $x\approx \hbar ^{\frac {1}{3}}$ and then take the limit $\hbar \rightarrow 0$. 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 $x\approx \hbar ^{\frac {1}{3}}$ goes to $\pm \infty$, 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 $\mathbf {L} \!$ 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 ${\mathbf {L}}\!$ in each direction:

$L_{x}=yp_{z}-zp_{y}={\frac {\hbar }{i}}(y{\frac {\partial }{\partial z}}-z{\frac {\partial }{\partial y}})\!$

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)

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

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 $\mathbf {\alpha } \!$ 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:

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

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

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

Also, It is easy to show that:

$[L^{2},L_{\pm }]=0$

Let $L'_{z}\!$ be an eigenvalue of $L_{z}\!$.

$\langle L_{z}'|L_{+}L_{-}|L_{z}'\rangle =(\beta -L_{z}'^{2}+\hbar L_{z}')\langle L_{z}'|L_{z}'\rangle$

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

$\beta -L_{z}'^{2}+\hbar L_{z}'\geq 0$

$\Rightarrow \beta +{\frac {\hbar ^{2}}{4}}\geq (L_{z}'-{\frac {\hbar }{2}})^{2}$

$\Rightarrow \beta +{\frac {\hbar ^{2}}{4}}\geq 0$

Defining the number $k\!$ by

$k+{\frac {\hbar }{2}}={\sqrt {\beta +{\frac {\hbar ^{2}}{4}}}}$

$\Rightarrow k\geq -{\frac {\hbar }{2}}$

The inequality 5.1.13 becomes

$k+{\frac {\hbar }{2}}\geq |L_{z}'-{\frac {\hbar }{2}}|$

$\Rightarrow k+\hbar \geq L_{z}'\geq -k$

Similarly, from equation 5.1.10, we get

$\langle L_{z}'|L_{-}L_{+}|L_{z}'\rangle =(\beta -L_{z}'^{2}-\hbar L_{z}')\langle L_{z}'|L_{z}'\rangle$

$\Rightarrow \beta -L_{z}'^{2}-\hbar L_{z}'\geq 0$

$\Rightarrow k\geq L_{z}'\geq -k-\hbar$

This result, combined with 5.1.15 shows that

$k\geq 0$

and

$k\geq L_{z}'\geq -k$

From 5.1.12

$L_{z}L_{-}|L_{z}'\rangle =(L_{-}L_{z}-\hbar L_{-})|L_{z}'\rangle =(L_{z}'-\hbar )L_{-}|L_{z}'\rangle$

Now, if $L_{z}'\neq 0$, then $L_{-}|L_{z}'\rangle$ is an eigenket of $L_{z}\!$ belonging to the eigenvalue $L_{z}'-\hbar$. Similarly, if $L_{z}'-\hbar \neq -k$, then $L_{z}'-2\hbar$ is another eigenvalue of $L_{z}\!$, 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 $L_{z}',L_{z}'+\hbar ,L_{z}'+2\hbar ,...\!$ are eigenvalues of L'z. Thus we may conclude that $2k\!$ is an integral multiple of the Planck's constant, and that the eigenvalues are:

$k,k-\hbar ,k-2\hbar ,...,-k+\hbar ,-k$

If $|m\rangle \!$ is an eigenstate of $L_{z}\!$ with eigenvalue $m\hbar \!$, then

$L_{z}L_{\pm }|m\rangle =([L_{z},L_{\pm }]+L_{\pm }L_{z})|m\rangle =(\pm \hbar L_{\pm }+L_{\pm }m)|m\rangle$

$\Rightarrow L_{z}(L_{\pm }|m\rangle )=(m\pm 1)\hbar (L_{\pm }|m\rangle )$

Which means 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 L_+\!} or 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 L_-\!} raises or lowers the 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 z\!} component of the angular momentum 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 \hbar\!} .

Orbital angular momentum eigenfunctions

Now we construct our eigenfunctions of the orbital angular momentum explicitly. The eigenvalue 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 L_z|l,m\rangle=m\hbar|l,m\rangle}

in terms of wave functions, becomes:

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 \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}

Solving for the 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 \phi\!} dependence, we find

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 \langle r,\theta,\phi|l,m\rangle=e^{im\phi}\langle r,\theta,0|l,m\rangle}

We construct the 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 \theta\!} dependence using the differential operator representation 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 L^2\!}

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 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}))}

Where the eigenvalues 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 L^2\!} are:

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 L^2|l,m\rangle=l(l+1)|l,m\rangle}

We proceed by using the property 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 L_+\!} 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 L_-\!} , defined 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 L_\pm=\frac{\hbar}{i}e^{\pm i\phi}(\pm i\frac{d}{d\theta}-\cot \theta \frac{d}{d\phi})}

to find the following 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 \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}

Using eq. 5.2.2, we get

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{\partial}{\partial \theta}-l\cot\theta)\langle r,\theta,\phi|l,l\rangle=0}

And the solution 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 \langle r,\theta,\phi|l,l\rangle=f(r)e^{il\phi}(\sin\theta)^l}

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

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_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}

And we know 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 Y_l^m(\theta, \phi)\!} are the spherical harmonics defined 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 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}}

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

$P_{l}^{m}(x)=(-1)^{m}(1-x^{2})^{m/2}{\frac {d^{m}}{dx^{m}}}P_{l}(x)$

with

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

And so we then can write:

$P_{l}^{m}(x)={\frac {(-1)^{m}}{2^{l}l!}}(1-x^{2})^{m/2}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}$

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

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

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.

$H={\frac {p^{2}}{2m}}+V(|r|)$

Due to the rotational symmetry, $[H,L_{z}]=0\!$ and $[H,L^{2}]=0\!$. This allows us to find a complete set of states that are simultaneous eigenfunctions of $H\!$, $L_{z}\!$, and $L^{2}\!$. We can label these states by their eigenvalues of $|Elm\rangle \!$.

From this we can get a state of the same energy for a given $l\!$ with a degeneracy of $2l+1\!$. We can rewrite the Laplacian as

$\nabla ^{2}={\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r-{\frac {L^{2}}{\hbar ^{2}r^{2}}}$

This makes the Schroedinger equation

$({\frac {-\hbar ^{2}}{2m}}{\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r+{\frac {L^{2}}{2mr^{2}}}+V(r))\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )$

Using separation of variables, $\psi (r,\theta ,\phi )=f_{l}(r)Y_{lm}(\theta ,\phi )\!$, we get:

$({\frac {-\hbar ^{2}}{2m}}{\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}+V(r))f_{l}(r)Y_{lm}(\theta ,\phi )=Ef_{l}(r)Y_{lm}(\theta ,\phi )$

Multiplying both sides by $Y_{l^{\prime }m'}\!$ and integrating over the angular dependence reduces the equation to merely a function of $r\!$.

Now if we let $u_{l}(r)=rf_{l}(r)\!$, this gives the radial Schroedinger equation:

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

Due to the boundary condition that $f_{l}(r)\!$ must be finite the origin, $u_{l}(r)\!$ must vanish.

Often looking at the asymptotic behavior of $u_{l}(r)\!$ can be quite helpful.

As $r\rightarrow 0\!$ and $V(r)\ll {\frac {1}{r^{2}}}\!$ the dominating term becomes the centrifugal barrier giving the approximate Hamiltonian:

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

which has the solutions $u_{l}(r)\sim r^{l+1},r^{-l}\!$ where only the first term is physically possible because the second blows up at the origin.

As $r\rightarrow \infty \!$ and $rV(r)\rightarrow 0$(which does not include the monopole ${\frac {1}{r}}$ coulomb potential) the Hamiltonian approximately becomes

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

letting $k=-i{\sqrt {\frac {2mE}{\hbar ^{2}}}}$ gives a solution of $u_{l}(r)=Ae^{kr}+Be^{-kr}\!$, where when $k\!$ is real, $B=0\!$, but both terms are need when $k\!$ is imaginary.

Nomenclature

Historically, the different values of $l\!$ have taken on names:

${\begin{cases}l=0&{\mbox{s-wave (sharp)}}\\l=1&{\mbox{p-wave (principle)}}\\l=2&{\mbox{d-wave (diffuse)}}\\l=3&{\mbox{f-wave (fundamental)}}\end{cases}}$

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

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

Spherical well

Dividing the potential into two regions, $0<r<a\!$ and $r>a\!$,

${\begin{cases}0<r<a&({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}-V_{0})u_{l}(r)=Eu_{l}(r),{\mbox{where the general solution is a Bessel equation}}\\r>a&({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}})u_{l}(r)=Eu_{l}(r),{\mbox{where the general solution is a Henkel function}}\end{cases}}$

$Aj_{l}(kr)+Bn_{l}(kr),r\rightarrow 0,n_{l}(kr)\rightarrow \infty \!$

For the $l=0\!$ term, the centrifugal barrier drops out and the equations become the following

${\begin{cases}0<r<a&({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}-V_{0})u_{0}(r)=Eu_{0}(r)\\r>a&({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}})u_{0}(r)=Eu_{0}(r)\end{cases}}$

The generalized solutions are

${\begin{cases}0<r<a&u_{0}(r)=Ae^{ikr}+Be^{-ikr}\\r>a&u_{0}(r)=Ce^{ik'r}+De^{-ik'r}\end{cases}}$

Using the boundary condition, $u(r=0)=0\!$, we find that $A=-B\!$. The second equation can then be reduced to sinusoidal function where $\alpha =2iA\!$.

$u_{0}(r)=2iAsin(kr)=\alpha \sin(kr)=\alpha \sin({\frac {r}{\hbar }}{\sqrt {2m(E+V_{0})}})$

for $r>a\!$, we know that $D=0\!$ since as $r\!$ approaches infinity, the wavefunction does not go to zero.

$u_{0}(r)=Ce^{ik'r}+De^{-ik'r}=Ce^{-{\frac {r}{\hbar }}{\sqrt {-2mE}}}$

Matching the conditions that at $r=a\!$, the wavefunctions and their derivatives must be continuous which results in 2 equations

$\alpha \sin({\frac {a}{\hbar }}{\sqrt {2m(E+V_{0})}})=Ce^{-{\frac {a}{\hbar }}{\sqrt {-2mE}}}$

$\alpha {\frac {\sqrt {2m(E+V_{0})}}{\hbar }}\cos({\frac {a}{\hbar }}{\sqrt {2m(E+V_{0})}})=-{\frac {\sqrt {-2mE}}{\hbar }}Ce^{-{\frac {a}{\hbar }}{\sqrt {-2mE}}}$

Dividing the above equations, we find

$-\cot({\sqrt {{\frac {2m}{\hbar ^{2}}}(V_{0}-|E|)a^{2}}})={\frac {\sqrt {\frac {2m|E|}{\hbar ^{2}}}}{\sqrt {\frac {2m(V_{0}-|E|)}{\hbar ^{2}}}}}$

Solving for $V_{0}\!$, we know that there is no bound state for

$V_{0}<{\frac {\pi ^{2}\hbar ^{2}}{8ma^{2}}}$

Hydrogen atom

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

$(-{\frac {\hbar ^{2}}{2\mu }}{\frac {1}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r+{\frac {L^{2}}{2\mu r^{2}}}+V(r))\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )$

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 $r\!$:

$\lim _{r\to 0}u_{l}(r)\rightarrow r^{l+1}$

$\lim _{r\to \infty }u_{l}(r)\rightarrow e^{-{\frac {r}{a}}}$

where

$a={\sqrt {\frac {-\hbar ^{2}}{2\mu E}}}$

If we allow $\kappa =a^{-1}\!$, then the large limit of $r\!$ can be expressed as

$u_{l}(r)\sim e^{-\kappa r}\!$

Using the limits of $u(r)\!$, the wavefunction can be expressed as the following

$u_{l}(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r)\!$

To simplify the equation, make a substitution $\rho =\kappa r\!$. The equation now turns into

$u_{l}(\rho )=\rho ^{l+1}e^{-\rho }W(\rho )\!$

Plugging eqn into the the Schrodinger and simplifying turns into

${\frac {d^{2}W(\rho )}{d\rho ^{2}}}+2({\frac {l+1}{\rho }}-1){\frac {dW}{d\rho }}+({\frac {\rho _{0}}{\rho }}-{\frac {2(l+1)}{\rho }})W(\rho )$

$W(p)\!$ can be expressed as an expansion of polynomials

$W(p)=a_{0}+a_{1}\rho +a_{2}\rho ^{2}+...=\sum _{k=0}^{\infty }a_{k}\rho ^{k}$

The Schrodinger equation is then expressed as

$\sum _{k=0}^{\infty }(a_{k}k(k-1)\rho ^{k-2}+2(l+1)k\rho ^{k-2}a_{k}-2\rho ^{k-1}a_{k}k)+\sum _{k=0}^{\infty }(\rho _{0}a_{k}\rho ^{k-1}-2(l+1)a_{k}\rho ^{k-1})=0$

And simplified into

$\sum _{k=0}^{\infty }(a_{k+1}(k+1)k\rho ^{k-1}+2(l+1)(k+1)\rho ^{k-1}a_{k+1}-2\rho ^{k-1}a_{k}k+(\rho _{0}-2(l+1))a_{k}\rho ^{k-1})=0$

Bring all $\rho \!$'s to the same power

$k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_{k}+(\rho _{0}-2(l+1))a_{k}=0\!$

which can be expressed in the simplest fractional form as

${\frac {a_{k+1}}{a_{k}}}={\frac {2(k+l+1)-\rho _{0}}{(k+1)(k+2l+2)}}$

where $\rho _{0}=2(N+l+1)\!$ and $N=0,1,2...\!$ and $l=0,1,2,...\!$

In the limit of large k

$\lim _{k\to \infty }{\frac {a_{k+1}}{a_{k}}}\rightarrow {\frac {2}{k}}$

$a_{k+1}={\frac {2}{k}}a_{k}\sim {\frac {2^{k}}{k!}}$

this will make

$\psi (r)\sim e^{kr}\rightarrow \infty$

so we need to break, and make

$a_{k+1}=0\!$

from this, we get the energy spectrum.

The fractional form can be expressed as a confluent hypergeometric function

$_{1}F_{1}(a,c,z)=1+{\frac {a}{c}}{\frac {z}{1!}}+{\frac {a(a+1)}{c(c+1)}}{\frac {z^{2}}{2!}}+...=\sum _{k=0}^{\infty }a_{k}z^{k}$

${\frac {a_{k}}{a_{k-1}}}={\frac {\frac {a(a+1)(a+2)...(a+k-1)}{c(c+1)(c+2)...(c+k-1)k!}}{\frac {a(a+1)(a+2)...(a+k-2)}{c(c+1)(c+2)...(c+k-2)(k-1)!}}}={\frac {a+k-1}{(c+k-1)k}}$

${\frac {a_{k+1}}{a_{k}}}={\frac {a+k}{(c+k)(k+1)}}$

by comparison, we find that

$c=2(l+1)\!$

$a=-N\!$

$z=2\rho \!$

So the solution of $W(\rho )\!$ is

$W(\rho )=constant_{1}F_{1}(-N,2(l+1),2\rho )\!$

where :$\rho =\kappa r={\sqrt {\frac {-2\mu E}{\hbar ^{2}}}}r$ Full wavefunction solution with normalization is

$\psi _{n,l,m}(r,\theta ,\phi )={\frac {e^{-\kappa r}(2\kappa r)^{l}}{(2l+1)!}}{\sqrt {\frac {(2\kappa )^{3}(n+l)!}{2n(n-l+1)}}}\ _{1}F_{1}(-n+l+1,2(l+1),2\kappa r)Y_{lm}(\theta ,\phi )$

The energy is then found to be

$E=-{\frac {Z^{2}e^{4}\mu }{2\hbar ^{2}n^{2}}}=-{\frac {Z^{2}}{n^{2}}}Ry$

where $Ry=13.6eV\!$ 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.

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 $d\Omega \!$. The detector is assumed to be far away from the scattering center.

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

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

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

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 $r\gg d\!$ 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 $l/k\gg d\!$, 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

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

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:

${\frac {d\sigma }{d\Omega }}=|f_{k}(\mathbf {\hat {r}} )|^{2}$

The scattering amplitude, $f_{k}(\mathbf {\hat {r}} )\!$, is the coefficient of the outgoing wave.

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

The scattering amplitude can be approximated by

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

where $-ik\mathbf {\hat {r}} \!$ is the scattered portion and $i\mathbf {k} \!$ is the incident portion.

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

$\psi _{k}(r)\approx N(e^{ikr}+{\frac {e^{ikr}}{r}}f_{k}(r))$

of the Schrodinger equation:

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

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

$f_{B}(\theta )=-{\frac {m}{2\pi \hbar ^{2}}}\int V(r)e^{-iqr}d^{3}r$

Which leads to the Born cross section:

$({\frac {d\sigma }{d\Omega }})_{B}=({\frac {m}{2\pi \hbar ^{2}}})^{2}|\langle K_{s}|r|K_{i}\rangle |^{2}$

Born Approximation for Spherically Symmetric Potentials:

Given spherically symmetry we may define

$\mathbf {\kappa } =\mathbf {k'} -{\mathbf {k}}$

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

$(\mathbf {k'} -{\mathbf {k}})\cdot \mathbf {r} =\kappa r\cos \theta '$

Our first Born integration then takes the form:

$f(\theta )\approx -{\frac {2m}{\hbar ^{2}}}\int e^{i\kappa r\cos \theta '}V(r)r^{2}\sin \theta 'drd\theta '$

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

$\int _{0}^{\pi }e^{isr\cos \theta '}\sin \theta 'd\theta '=-{\frac {2\sin(sr)}{sr}}$

and get:

$f(\theta )\approx -{\frac {2m}{\hbar ^{2}\kappa }}\int _{0}^{\infty }V(r)r\sin(\kappa r)dr$

where the angular dependence of $f(\theta )\!$ is carried by $\kappa \!$:

$\kappa =2k\sin {\frac {\theta }{2}}$

Example 1:

Consider the scattering amplitude from a Gaussian potential of the form

$V(r)=Ae^{-\alpha r^{2}}$

Our scattering amplitude then becomes:

$f(\theta )\approx {\frac {-2mA}{\hbar ^{2}\kappa }}\int _{0}^{\infty }re^{-\alpha r^{2}}\sin(\kappa r)dr$

$={\frac {-2mA}{\hbar ^{2}\kappa }}\int _{0}^{\infty }{\frac {d}{dr}}({\frac {-1}{2\alpha }}e^{-\alpha r^{2}})\sin(\kappa r)dr$

$={\frac {mA}{\alpha \hbar ^{2}\kappa }}(0-\kappa \int _{0}^{\infty }e^{-\alpha r^{2}}\cos(\kappa r)dr)$

$={\frac {-mA}{\alpha \hbar ^{2}}}{\frac {\sqrt {\pi }}{2{\sqrt {\alpha }}}}e^{-{\frac {\kappa ^{2}}{4\alpha }}}$

$=-{\frac {mA{\sqrt {\pi }}}{2\hbar ^{2}\alpha ^{\frac {3}{2}}}}e^{-{\frac {\kappa ^{2}}{4\alpha }}}$

Coulomb potential scattering

Example 1

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

$V(r)=V_{0}e^{-\alpha r}{\frac {1}{r}}$

$\Rightarrow f={\frac {-m}{2\pi \hbar ^{2}}}\int _{0}^{\infty }dr'r'^{2}2\pi {\frac {e^{-i|k'-k|r'}-e^{i|k-k'|r'}}{-i|k'-k|r'}}{\frac {V_{0}e^{-\alpha r'}}{r'}}$

$\Rightarrow f={\frac {-2mV_{0}}{\hbar ^{2}}}\int _{0}^{\infty }dr'r'^{2}{\frac {\sin(|k'-k|r')}{|k'-k|r'}}{\frac {e^{-\alpha r'}}{r'}}$

$\Rightarrow f={\frac {-2mV_{0}}{\hbar ^{2}}}{\frac {1}{(k'-k)^{2}+\alpha ^{2}}}={\frac {-2mV_{0}}{\hbar ^{2}}}{\frac {1}{4k^{2}\sin ^{2}{\frac {\theta }{2}}+\alpha ^{2}}}$

thus we have the differential cross section:

${\frac {d\sigma }{d\Omega }}=({\frac {2mV_{0}}{\hbar ^{2}}})^{2}({\frac {1}{4k^{2}\sin ^{2}{\frac {\theta }{2}}+\alpha ^{2}}})^{2}$

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 ${\frac {1}{r}}$ potential.

Use a change of coordinates from Cartesian to parabolic coordinates:

$\xi ={\sqrt {x^{2}+y^{2}+z^{2}}}-z$

$\eta ={\sqrt {x^{2}+y^{2}+z^{2}}}+z$

$\phi =\tan ^{-1}({\frac {y}{x}})$

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:

$[{\frac {\hbar ^{2}}{2\mu }}{\frac {4}{\xi +\eta }}({\frac {\partial }{\partial \xi }}\xi {\frac {\partial }{\partial \xi }}+{\frac {\partial }{\partial \eta }}\eta {\frac {\partial }{\partial \eta }}+{\frac {\xi +\eta }{4\xi \eta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}})-{\frac {2Ze^{2}}{\xi +\eta }}]\psi ={\frac {\hbar ^{2}k^{2}}{2\mu }}\psi$

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

Look for solutions of the form:

$[\xi {\frac {\partial ^{2}}{\partial \xi ^{2}}}+(1-ik\xi ){\frac {\partial }{\partial \xi }}+{\frac {Ze^{2}\mu }{k\hbar ^{2}}}k]\Phi (\xi )=0$

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

$\lambda ={\frac {Ze^{2}\mu }{k\hbar ^{2}}}$

Now let:

$\Phi (\xi )=\sum _{n=0}^{\infty }a_{n}\xi ^{n}$

From this we can write:

${\frac {a_{n+1}}{a_{n}}}={\frac {in-\lambda }{(n+1)^{2}}}k={\frac {n+i\lambda }{(n+1)^{2}}}ik$

Recall the confluent hypergeometric function:

$_{1}F_{1}(a,c,z)=1+{\frac {a}{c}}z+{\frac {a(a+1)}{c(c+1)}}{\frac {z^{2}}{2!}}+\dots +{\frac {a(a+1)\dots (a+n-1)}{c(c+1)\dots (c+n-1)}}{\frac {z^{n}}{n!}}+\dots$

We can then write the recursion formula as the following:

${\frac {\alpha _{n+1}}{\alpha _{n}}}=({\frac {a+n}{c+n}}){\frac {1}{n+1}}$

This implies that:

$\Phi (\xi )=A_{1}F_{1}(i\lambda ,1,ik\xi )\!$

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:

$\psi (r,z)=A_{1}F_{1}(i\lambda ,1,ik\xi )e^{ikz}\!$

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

$_{1}F_{1}(a,c,z)={\frac {\Gamma (c)(-z)^{a}}{\Gamma (c-a)}}[1-{\frac {a(a-c+1)}{2}}]+{\frac {\Gamma (c)}{\Gamma (n)}}e^{z}z^{a-c}$

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

$\Phi (\xi )=Ae^{{\frac {-\pi }{2}}\lambda }[{\frac {e^{-i\lambda \ln(k\xi )}}{\Gamma (1-i\lambda )}}(1-{\frac {\lambda ^{2}}{ik\xi }})+{\frac {i\lambda }{ik\xi }}{\frac {e^{ik\xi -i\lambda \ln(k\xi )}}{\Gamma (1+i\lambda )}}]$

Rewriting our wavefunction $\psi \!$:

$\psi (r,\theta )={\frac {Ce^{\frac {-\pi \lambda }{2}}}{\Gamma (1-i\lambda )}}[(1-{\frac {\lambda ^{2}}{2ikr}}{\frac {1}{\sin ^{2}{\frac {\theta }{2}}}})e^{ikz}e^{-i\lambda \ln(k(r-z))}+{\frac {f(\theta )}{r}}e^{ikr+i\lambda \ln(2kr)}]$

where $f(\theta )\!$ is the following:

$f(\theta )={\frac {\lambda \Gamma (1-i\lambda )}{2k\Gamma (1+i\lambda )}}{\frac {1}{\sin ^{2}{\frac {\theta }{2}}}}e^{i\lambda \ln(\sin ^{2}{\frac {\theta }{2}})}={\frac {\lambda \Gamma (1-i\lambda )}{2k\Gamma (1+i\lambda )}}(\sin ^{2}{\frac {\theta }{2}})^{i\lambda -1}$

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

${\frac {d\sigma }{d\Omega }}=|f(\theta )|^{2}={\frac {\lambda ^{2}}{4k^{2}\sin ^{4}({\frac {\theta }{2}})}}={\frac {(Ze^{2})^{4}}{16E^{2}}}{\frac {1}{\sin ^{4}({\frac {\theta }{2}})}}$

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

$C={\sqrt {\frac {\mu }{\hbar k}}}\Gamma (1-i\lambda )e^{\frac {\pi \lambda }{2}}$

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

$|\psi (0)|^{2}=|C|^{2}={\frac {\mu }{\hbar k}}|\Gamma (1-i\lambda )|^{2}e^{\pi \lambda }$

where

$|\Gamma (1-i\lambda )|^{2}={\frac {2\pi |\lambda |e^{-\pi \lambda }}{e^{-2\pi \lambda }-1}}$

Plugging this in for our wavefunction squared:

$|\psi (0)|^{2}={\frac {2\pi |\lambda |}{{\frac {\hbar k}{\mu }}|1-e^{-2\pi \lambda }|}}$

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

${\frac {\hbar k}{\mu }}=v$

For small incident velocities, we can write:

$|\psi (0)|^{2}={\frac {2\pi Ze^{2}}{\hbar ^{2}v^{2}}}$

$|\psi (0)|^{2}={\frac {2\pi Ze^{2}}{\hbar ^{2}v^{2}}}e^{-2\pi {\frac {Ze^{2}}{\hbar v}}}$

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

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 $r_{1}\leftrightarrow r_{2}\!$, $r_{cm}=(r_{1}+r_{2})/2\!$ is invariant while $r=r_{1}-r_{2}\!$ 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:

$Y_{lm}(-r)=(-1)^{l}Y_{lm}(r)\!$

Under the transformation:

$r\rightarrow -r,\theta \rightarrow \pi -\theta ,\phi \rightarrow \phi +\pi$

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:

$f_{sym}(\theta ,\phi )=f(\theta ,\phi )+f(\pi -\theta ,\phi +\pi )\!$

$\theta \!$and $\pi -\theta \!$ 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:

${\frac {d\sigma }{d\Omega }}=|f(\theta ,\phi )+f(\pi -\theta ,\phi +\pi )|^{2}=|f(\theta ,\phi )|^{2}+|f(\pi -\theta ,\phi +\pi )|^{2}+2\Re e[f(\theta ,\phi )f^{*}(\pi -\theta ,\phi +\pi )]$

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:

$f(\theta )=f(\pi -\theta )=f({\frac {\pi }{2}})$

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

$({\frac {d\sigma }{d\Omega }})_{dis}=2|f({\frac {\pi }{2}})|^{2}$

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

$({\frac {d\sigma }{d\Omega }})_{ind}=4|f({\frac {\pi }{2}})|^{2}$

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