# Isotropic harmonic oscillator in 3-d – use of spherical harmonics

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.11.

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We’ve solved the 3-d isotropic harmonic oscillator before, so we’ve already solved most of Shankar’s exercise 12.6.11. We can quote the results here. The solution has the form

$\displaystyle \psi_{Elm}=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

The earlier solution uses notation from Griffiths’s book, but as the end result is the same, it’s not worth going through the derivation again using Shankar’s notation.

The potential is

$\displaystyle V(r)=\frac{1}{2}m\omega^{2}r^{2} \ \ \ \ \ (2)$

The radial equation to be solved is

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\left(-1+\frac{l(l+1)}{\rho^{2}}+\rho_{0}^{2}\rho^{2}\right)u \ \ \ \ \ (3)$

If we define

 $\displaystyle \kappa^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle \kappa r\ \ \ \ \ (5)$ $\displaystyle \rho_{0}$ $\displaystyle \equiv$ $\displaystyle \frac{\mu\omega}{\hbar\kappa^{2}}=\frac{\hbar\omega}{2E} \ \ \ \ \ (6)$

Taking the asymptotic behaviour of the radial function for small and large ${r}$ into account leads us to a solution of form

$\displaystyle u(\rho)=\rho^{l+1}e^{-\rho_{0}\rho^{2}/2}v(\rho) \ \ \ \ \ (7)$

Note that Griffiths’s ${v}$ is not the same as Shankar’s ${v}$, the latter of which is defined by Shankar’s equation 12.6.49.

This gives a differential equation for Griffiths’s ${v}$

$\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2\left(l+1-\rho_{0}\rho^{2}\right)\frac{dv}{d\rho}+\rho(1-\rho_{0}(2l+3))v=0 \ \ \ \ \ (8)$

The function ${v}$ can be solved as a power series, giving

$\displaystyle v(\rho)=\sum c_{j}\rho^{j} \ \ \ \ \ (9)$

Substituting into 8 leads to the recursion relation

$\displaystyle c_{q+2}=\frac{\rho_{0}(2q+2l+3)-1}{(q+2)(q+2l+3)}c_{q} \ \ \ \ \ (10)$

with ${c_{1}=0}$, so that ${c_{q}=0}$ for all odd ${q}$. The requirement that the series terminates at some finite value of ${j}$ leads to the quantization condition on ${E}$:

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \hbar\omega\left(q_{max}+l+\frac{3}{2}\right) \ \ \ \ \ (11)$

or, defining ${n=q_{max}+l}$,

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right) \ \ \ \ \ (12)$

We worked out the degeneracies in the earlier post as well, so that the degeneracy of ${E_{n}}$ is

$\displaystyle d\left(n\right)=\frac{1}{2}(n+1)(n+2) \ \ \ \ \ (13)$

To complete Shankar’s exercise, we need to work out the eigenfunctions for ${n=0}$ and ${n=1}$. For ${n=0}$, ${q_{max}=l=0}$, so only ${c_{0}\ne0}$ and we have

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (14)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (15)$ $\displaystyle \psi_{000}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{0}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa e^{-\rho_{0}\rho^{2}/2}Y_{0}^{0}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\sqrt{\frac{2\mu3\omega}{4\pi\hbar}}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (18)$

where in the fourth line we used

 $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2\mu E}}{\hbar}=\frac{\sqrt{2\mu\frac{3}{2}\hbar\omega}}{\hbar}=\sqrt{\frac{3\mu\omega}{\hbar}}\ \ \ \ \ (19)$ $\displaystyle Y_{0}^{0}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{4\pi}} \ \ \ \ \ (20)$

Normalizing this requires that

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{000}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{6\mu\omega}{\hbar}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{2}dr\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (22)$

This is a standard Gaussian integral and can be done using software or tables so we get

$\displaystyle c_{0}=\frac{\sqrt{6}}{3}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (23)$

This gives a wave function of

$\displaystyle \psi_{000}=\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (24)$

which agrees with the earlier result.

For ${n=1}$, the degeneracy is, from 13

$\displaystyle d\left(1\right)=3 \ \ \ \ \ (25)$

The three possibilities are ${m=0,\pm1}$ which are reflected in the three spherical harmonics ${Y_{1}^{0,\pm1}}$. The radial function is the same in all cases, and is obtained from ${q_{max}=0}$, ${l=1}$. From 7, this gives

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (26)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho^{2}e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (27)$ $\displaystyle \psi_{11m}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{1}^{m}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa^{2}re^{-\rho_{0}\rho^{2}/2}Y_{1}^{m}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{m} \ \ \ \ \ (30)$

Again, we work out ${c_{0}}$ by imposing normalization. For example

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{1}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (32)$

The normalization integral is

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{111}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\left(\frac{5\mu\omega}{\hbar}\right)^{2}\frac{3}{8\pi}2\pi\int_{0}^{\pi}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{4}\sin^{3}\theta dr\;d\theta\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{75}{8}\sqrt{\frac{\pi\hbar}{\mu\omega}}=1\ \ \ \ \ (34)$ $\displaystyle c_{0}$ $\displaystyle =$ $\displaystyle \frac{2\sqrt{6}}{15}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (35)$

I used Maple to do the integrals. This gives a wave function of

$\displaystyle \psi_{111}=-\sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{i\phi} \ \ \ \ \ (36)$

We can work out the other two wave functions the same way (I used Maple, so I won’t go into the details):

 $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{-i\phi}\ \ \ \ \ (37)$ $\displaystyle \psi_{110}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\cos\theta \ \ \ \ \ (38)$

The ${\psi_{110}}$ here is the same as ${\psi_{001}}$ in our rectangular solution set. The other two are linear combinations of ${\psi_{100}}$ and ${\psi_{010}}$ from our rectangular set, which were (the suffixes in these 2 equations stand for ${x}$, ${y}$ and ${z}$, and not ${n}$, ${l}$ and ${m}$):

 $\displaystyle \psi_{100}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (39)$ $\displaystyle \psi_{010}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi \ \ \ \ \ (40)$

We have

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}+i\psi_{010}\right)\ \ \ \ \ (41)$ $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}-i\psi_{010}\right) \ \ \ \ \ (42)$