When solving the Schrodinger equation in 2D polar coordinates, one has to deal with various Bessel functions. In the most simple example, the infinite circular potential well, the solutions to the radial differential equation are the Bessel functions of first $[J_m(kr)]$ and second $[Y_m(kr)]$ kind. One usually discards the $Y_m(kr)$ functions on account of their asymptotic behavior at $r = 0$, $$Y_m(kr) \sim (kr)^{-m}$$ and so they are not square integrable functions. However, in the case of zero angular momentum, $m=0$, the Neumann function of zeroth order, $$Y_0(kr) \sim \ln (kr),$$ although infinite at the origin, is square integrable! So why do we have to discard it as well? What are the boundary conditions that have to be satisfied by a radial wave function at the origin?
[Physics] 2D Schrodinger equation in polar coordinates – boundary conditions at the origin
boundary conditionsquantum mechanicsschroedinger equationwavefunction
Related Solutions
The most general widely studied examples of potentials are those that are derived from an exactly given ground state, and which have the property that they contain enough parameters to be closed under taking supersymmetric conjugates. These are called "shape invariant" potentials.
Given a real positive ground state, one can ask "Which potential has this ground state?" The answer is that if the ground state is $\exp(-W(x))$, and its energy is exactly zero, then the potential is:
$ V(x) = {1\over 2} |\nabla W|^2 - {1\over 2} \nabla^2 W$
The conjugate potential is defined with a plus sign between the two terms instead of a minus sign. It is more usual to define W as the derivative of what I am calling W, but this convention is terrible in higher than one dimension.
The two potentials taken together define a supersymmetric quantum mechanics, as originally defined by Witten. The supersymmetric quantum mechanics in imaginary time is a stochastic Brownian process with a drift which is an analytic function of the position. The conjugate potentials correspond to reversing the direction of the drift, and their properties are similar because they are related by a stochastic version of time-reversibility.
If W(x) goes to plus infinity at infinity (so that it actually defines a normalizable ground state), then the ground state is unique, and the conjugate potential has the exact same spectrum as the original potential, except it omits the lowest energy state. This, plus the form of the supercharge, gives exact solutions of many classes of quantum potentials.
Here are some simple W's which correspond to usual elementary quantum mechanics examples:
- W(x) = |x|^2 is the Hamonic oscillator in any dimension
- W(x) = |x| is the delta function potential in 1d, and the Coulomb potential in 3d
- W(x) = log(|cos(x)|) this gives the infinite hard wall
http://arxiv.org/abs/hep-th/9405029 has a bunch of more interesting examples. Any quantum mechanical potential which has closed form energy states is in this class.
A completely diffeent class of widely studied potentials are random potentials, as studied by Halperin and others, to understand Anderson localization.
Later Edit: The original paper by Anderson which started the random potential field is http://prola.aps.org/abstract/PR/v109/i5/p1492_1 "Absence of diffusion in certain random lattices", and it's one of the great classics. The setup is a square lattice with a random potential at each site, an independent random number between -V and V. The continuum limit in one dimension, where the potential is a random gaussian at each point is analyzed by Halperin B. I. Halperin, Green ' s Functions for a Particle in a One-Dimensional Random Potential, Phys. Rev. 139 , A104 (1965). The field is enormous--- look up "localization" on google scholar. It includes "weak localization" effects, which were popular in the mid 90s because they imply that resistance can drop sharply in the presence of a magnetic field, because the perturbative precurser to the localization process is hindered.
The solutions are like sines and cosines (oscillating) when the energy of the particle is greater than the energy of the potential. Those regions are regions where a classical particle can exist. The solutions are like exponentials when the the energy of the particle is lower than the potential, regions where a classical particle cannot exist.
For example, if a particle with energy $E$ is in a region of no potential $V(x)=0$, the wave function will be sine-like. Suppose that particle encounters a barrier of height $V>E$. A classical particle does not have enough energy to pass by that barrier. The region of higher potential is classically forbidden. The particle will bounce off the barrier. But in quantum mechanics, the wave function extends into the barrier $\ldots$ into the forbidden region. In the forbidden region the wave function will fall exponentially.
Best Answer
In order to keep the kinetic energy $\langle K\rangle =\frac{\hbar^2}{2m} \int d^2x~ |\nabla \psi|^2$ finite, a logarithmic $\ln(r)$ singularity of the radial wave function $\psi(r)$ at $r=0$ is unacceptable. This conclusion holds even if we take the potential energy $\langle V \rangle$ into account:
A non-negative potential $V\geq 0 $ will only make the total energy $E= \langle K \rangle + \langle V \rangle$ bigger.
A potential $V$ with a power law singularity $V(r) \sim r^p$, $p>-2$, at $r=0$ would only have a finite contribution.
A negative potential $V<0$ with a power law singularity $V(r) \sim r^p$, $p\leq -2$, at $r=0$ would lead to a spectrum for the Hamiltonian $H=K+V$ that is unbounded from below.
The general principles for imposing boundary conditions of the wavefunction $\psi$ at $r=0$ in various dimensions are also e.g. outlined in this related Phys.SE post and links therein.