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)

4 thoughts on “Harmonic oscillator in 2-d and 3-d, and in polar and spherical coordinates

  1. Yasin Şale

    Hi Sir,

    Good job!

    But passage from Eqn. (18) to Eqn. (20) is not clear enough, for people who are unfamiliar with these objects. I think you’d better to remark that $ \omega_x = \omega_y $, before or after Eqn. (20). Unless, you must write total energy of 2d oscillator as $ E=\hbar\omega_x(n_x+\frac{1}{2})+ \hbar\omega_y(n_y+\frac{1}{2}) $.

    Thanks for your clear, brief and instructive articles.

    Reply
    1. gwrowe Post author

      I’ve covered the solution by separation of variables in the post in the link “3-d isotropic harmonic oscillator” in the first line. The solution uses the same method whether or not {\omega_{x}=\omega_{y}}, giving equation 20 for the energies.

      Reply
  2. Pingback: Harmonic oscillator in 2 dimensions: comparison with rectangular coordinates | Physics pages

  3. Pingback: Isotropic harmonic oscillator in 3-d – use of spherical harmonics | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *