3D Symmetric HO in Spherical Coordinates *

We have already solved the problem of a 3D harmonic oscillator by separation of variables in Cartesian coordinates. It is instructive to solve the same problem in spherical coordinatesand compare the results. The potential is

\begin{displaymath}\bgroup\color{black} V(r)={1\over 2}\mu\omega^2r^2 .\egroup\end{displaymath}

Our radial equation is

\left({d^2\over dr^2}+{2\over r}{d\over dr}\right)R_{E\ell}(r)...
-{\ell(\ell+1)\over r^2}R+{2\mu E\over\hbar^2}R&=&0

Write the equation in terms of the dimensionless variable

y&=&{r\over\rho}. \\
\rho&=&\sqrt{\hbar\over\mu\omega} \\
...ho}{d\over dy} \\
{d^2\over dr^2}&=&{1\over\rho^2}{d\over dy^2}

Plugging these into the radial equation, we get

{1\over\rho^2}{d^2R\over dy^2}+{1\over\rho^2}{2\over y}{dR\ove...
... dy}-y^2R
-{\ell(\ell+1)\over y^2}R+{2 E\over\hbar\omega}R&=&0.

Now find the behavior for large \bgroup\color{black}$y$\egroup.

{d^2R\over dy^2}-y^2R=0 \\
R\approx e^{-y^2/2} \\

Also, find the behavior for small \bgroup\color{black}$y$\egroup.

{d^2R\over dy^2}+{2\over y}{dR\over dy}-{\ell(\ell+1)\over y^2...
...l(\ell+1)y^{s-2} \\
s(s+1)=\ell(\ell+1) \\
R\approx y^\ell \\

Explicitly put in this behavior and use a power series expansion to solve the full equation.

\begin{displaymath}\bgroup\color{black} R=y^\ell\sum\limits_{k=0}^\infty a_k y^k...
=\sum\limits_{k=0}^\infty a_k y^{\ell+k} e^{-y^2/2} \egroup\end{displaymath}

We'll need to compute the derivatives.

{dR\over dy}=\sum\limits_{k=0}^\infty a_k
...ell+k-2} \\
-(2\ell+2k+1)y^{\ell+k}+y^{\ell+k+2}]e^{-y^2/2} \\

We can now plug these into the radial equation.

\begin{displaymath}\bgroup\color{black} {d^2R\over dy^2}+{2\over y}{dR\over dy}-y^2R
-{\ell(\ell+1)\over y^2}R+{2E\over\hbar\omega}R=0 \egroup\end{displaymath}

Each term will contain the exponential \bgroup\color{black}$e^{-y^2/2}$\egroup, so we can factor that out. We can also run a single sum over all the terms.

\sum\limits_{k=0}^\infty a_k \left[
\right]=0 \\

The terms for large \bgroup\color{black}$y$\egroup which go like \bgroup\color{black}$y^{\ell+k+2}$\egroup and some of the terms for small \bgroup\color{black}$y$\egroup which go like \bgroup\color{black}$y^{\ell+k-2}$\egroup should cancel if we did our job right.

\sum\limits_{k=0}^\infty a_k \left[
\right]=0 \\

Now as usual, the coefficient for each power of \bgroup\color{black}$y$\egroup must be zero for this sum to be zero for all \bgroup\color{black}$y$\egroup. Before shifting terms, we must examine the first few terms of this sum to learn about conditions on \bgroup\color{black}$a_0$\egroup and \bgroup\color{black}$a_1$\egroup. The first term in the sum runs the risk of giving us a power of \bgroup\color{black}$y$\egroup which cannot be canceled by the second term if \bgroup\color{black}$k<2$\egroup. For \bgroup\color{black}$k=0$\egroup, there is no problem because the term is zero. For \bgroup\color{black}$k=1$\egroup the term is \bgroup\color{black}$(2\ell+2)y^{\ell-1}$\egroup which cannot be made zero unless

\begin{displaymath}\bgroup\color{black} a_1=0 .\egroup\end{displaymath}

This indicates that all the odd terms in the sum will be zero, as we will see from the recursion relation.

Now we will do the usual shift of the first term of the sum so that everything has a \bgroup\color{black}$y^{\ell+k}$\egroup in it.

k\rightarrow k+2 \\
\sum\limits_{k=0}^\infty \left[
...{{2E\over\hbar\omega}-(2\ell+2k+3)\over (k+2)(2\ell+k+3)}a_k \\

For large \bgroup\color{black}$k$\egroup,

\begin{displaymath}\bgroup\color{black} a_{k+2}\approx {2\over k}a_k ,\egroup\end{displaymath}

Which will cause the wave function to diverge. We must terminate the series for some \bgroup\color{black}$k=n_r=0,2,4...$\egroup, by requiring

{2E\over\hbar\omega}-(2\ell+2n_r+3)=0 \\
E=\left(n_r+\ell+{3\over 2}\right)\hbar\omega \\

These are the same energies as we found in Cartesian coordinates. Lets plug this back into the recursion relation.

a_{k+2}=-{(2\ell+2n_r+3)-(2\ell+2k+3)\over(k+2)(2\ell+k+3)}a_k \\
a_{k+2}={2(k-n_r)\over(k+2)(2\ell+k+3)}a_k \\

To rewrite the series in terms of \bgroup\color{black}$y^2$\egroup and let \bgroup\color{black}$k$\egroup take on every integer value, we make the substitutions \bgroup\color{black}$n_r\rightarrow 2n_r$\egroup and \bgroup\color{black}$k\rightarrow 2k$\egroup in the recursion relation for \bgroup\color{black}$a_{k+1}$\egroup in terms of \bgroup\color{black}$a_k$\egroup.

\bgroup\color{black}$\displaystyle a_{k+1}={(k-n_r)\over(k+1)(\ell+k+3/2)}a_k $\egroup
\bgroup\color{black}$\displaystyle R_{n_r\ell}=\sum\limits_{k=0}^\infty a_ky^{\ell+2k}e^{-y^2/2} $\egroup
\bgroup\color{black}$\displaystyle E=\left(2n_r+\ell+{3\over 2}\right)\hbar\omega $\egroup

The table shows the quantum numbers for the states of each energy for our separation in spherical coordinates, and for separation in Cartesian coordinates. Remember that there are \bgroup\color{black}$2\ell +1$\egroup states with different \bgroup\color{black}$z$\egroup components of angular momentum for the spherical coordinate states.

\bgroup\color{black}$E$\egroup \bgroup\color{black}$n_r\ell$\egroup \bgroup\color{black}$n_xn_yn_z$\egroup \bgroup\color{black}$N_{Spherical}$\egroup \bgroup\color{black}$N_{Cartesian}$\egroup
\bgroup\color{black}${3\over 2}\hbar\omega$\egroup 00 000 1 1
\bgroup\color{black}${5\over 2}\hbar\omega$\egroup 01 001(3 perm) 3 3
\bgroup\color{black}${7\over 2}\hbar\omega$\egroup 10, 02 002(3 perm), 011(3 perm) 6 6
\bgroup\color{black}${9\over 2}\hbar\omega$\egroup 11, 03 003(3 perm), 210(6 perm), 111 10 10
\bgroup\color{black}${11\over 2}\hbar\omega$\egroup 20, 12, 04 004(3), 310(6), 220(3), 211(3) 15 15
The number of states at each energy matches exactly. The parities of the states also match. Remember that the parity is \bgroup\color{black}$(-1)^\ell$\egroup for the angular momentum states and that it is \bgroup\color{black}$(-1)^{n_x+n_y+n_z}$\egroup for the Cartesian states. If we were more industrious, we could verify that the wavefunctions in spherical coordinates are just linear combinations of the solutions in Cartesian coordinates.

Jim Branson 2013-04-22