Daily Archives: Sat, 4 March 2017

Harmonic oscillator in 2-d and 3-d, and in polar and spherical coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.2.2 – 10.2.3.

We’ve seen that the 3-d isotropic harmonic oscillator can be solved in rectangular coordinates using separation of variables. The Hamiltonian is

\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+\frac{m\omega^{2}}{2}\left(x^{2}+y^{2}+z^{2}\right) \ \ \ \ \ (1)

The solution to the Schrödinger equation is just the product of three one-dimensional oscillator eigenfunctions, one for each coordinate. That is

\displaystyle \psi_{n}\left(x,y,z\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right)\psi_{n_{z}}\left(z\right) \ \ \ \ \ (2)

Each one-dimensional eigenfunction can be expressed in terms of Hermite polynomials as

\displaystyle \psi_{n_{x}}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n_{x}}n_{x}!}}H_{n_{x}}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)

with the functions for {y} and {z} obtained by replacing {x} by {y} or {z} and {n_{x}} by {n_{y}} or {n_{z}}. We also saw earlier that in the 3-d oscillator, the total energy for state {\psi_{n}\left(x,y,z\right)} is given in terms of the quantum numbers of the three 1-d oscillators as

\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right)=\hbar\omega\left(n_{x}+n_{y}+n_{z}+\frac{3}{2}\right) \ \ \ \ \ (4)

and that the degeneracy of level {n} is {\frac{1}{2}\left(n+1\right)\left(n+2\right)}.

Since the Hermite polynomial {H_{n_{x}}} has parity {\left(-1\right)^{n_{x}}} (that is, odd (even) polynomials are odd (even) functions), the 3-d wave function {\psi_{n}} has parity {\left(-1\right)^{n_{x}}\left(-1\right)^{n_{y}}\left(-1\right)^{n_{z}}=\left(-1\right)^{n}}.

We can write the one {n=0} state and three {n=1} states in spherical coordinates using the standard transformation

\displaystyle x \displaystyle = \displaystyle r\sin\theta\cos\phi\ \ \ \ \ (5)
\displaystyle y \displaystyle = \displaystyle r\sin\theta\sin\phi\ \ \ \ \ (6)
\displaystyle z \displaystyle = \displaystyle r\cos\phi \ \ \ \ \ (7)

Using the notation {\psi_{n}=\psi_{n_{x}n_{y}n_{z}}=\psi_{n_{x}}\psi_{n_{y}}\psi_{n_{z}}}, we have, using {H_{0}\left(y\right)=1} and {H_{1}\left(y\right)=2y}:

\displaystyle \psi_{000} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}\ \ \ \ \ (8)
\displaystyle \psi_{100} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (9)
\displaystyle \psi_{010} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi\ \ \ \ \ (10)
\displaystyle \psi_{001} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\cos\theta \ \ \ \ \ (11)

We can check that these are the correct spherical versions of the eigenfunctions by using the Schrödinger equation in spherical coordinates, which is

\displaystyle H\psi=\left[-\frac{\hbar^{2}\nabla^{2}}{2m}+\frac{m\omega^{2}}{2}r^{2}\right]\psi=E\psi \ \ \ \ \ (12)

The spherical laplacian operator is

\displaystyle \nabla^{2}\psi=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}} \ \ \ \ \ (13)

You can grind through the derivatives by hand if you like, but I just used Maple to do it, giving the results

\displaystyle H\psi_{000} \displaystyle = \displaystyle \frac{3}{2}\hbar\omega\psi_{000}\ \ \ \ \ (14)
\displaystyle H\psi_{100} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{100}\ \ \ \ \ (15)
\displaystyle H\psi_{010} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{010}\ \ \ \ \ (16)
\displaystyle H\psi_{001} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{001} \ \ \ \ \ (17)

In two dimensions, the analysis is pretty much the same. In the more general case where the masses are equal, but {\omega_{x}\ne\omega_{y}}, the Hamiltonian is

\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}}{2m}+\frac{m}{2}\left(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}\right) \ \ \ \ \ (18)

A solution by separation of variables still works, with the result

\displaystyle \psi_{n}\left(x,y\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right) \ \ \ \ \ (19)

The total energy is

\displaystyle E_{n}=E_{n_{x}}+E_{n_{y}}=\hbar\omega\left(n_{x}+\frac{1}{2}+n_{y}+\frac{1}{2}\right)=\hbar\omega\left(n+1\right) \ \ \ \ \ (20)

For a given energy level {n=n_{x}+n_{y}}, there are {n+1} ways of forming {n} out of a sum of 2 non-negative integers, so the degeneracy of level {n} is {n+1}.

The one {n=0} state and two {n=1} states are

\displaystyle \psi_{00} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}\ \ \ \ \ (21)
\displaystyle \psi_{10} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}x\ \ \ \ \ (22)
\displaystyle \psi_{01} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}y \ \ \ \ \ (23)

To translate to polar coordinates, we use the transformations

\displaystyle x \displaystyle = \displaystyle \rho\cos\phi\ \ \ \ \ (24)
\displaystyle y \displaystyle = \displaystyle \rho\sin\phi \ \ \ \ \ (25)

so we have

\displaystyle \psi_{00} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\ \ \ \ \ (26)
\displaystyle \psi_{10} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\cos\phi\ \ \ \ \ (27)
\displaystyle \psi_{01} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\sin\phi \ \ \ \ \ (28)

Again, we can check this by plugging these polar formulas into the polar Schrödinger equation, where the 2-d Laplacian is

\displaystyle \nabla^{2}=\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}} \ \ \ \ \ (29)

The results are

\displaystyle H\psi_{00} \displaystyle = \displaystyle \hbar\omega\psi_{00}\ \ \ \ \ (30)
\displaystyle H\psi_{10} \displaystyle = \displaystyle 2\hbar\omega\psi_{10}\ \ \ \ \ (31)
\displaystyle H\psi_{01} \displaystyle = \displaystyle 2\hbar\omega\psi_{01} \ \ \ \ \ (32)