Daily Archives: Tue, 4 July 2017

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)