I'm working through Griffiths' Introduction to Quantum Mechanics (2nd edition) and I'm trying to solve problem 4.24 b. In this problem, you're supposed to first find the normalized eigenfunctions to the allowed energies of a rigid rotator, which I correctly realized should be spherical harmonics. Then, you should find the degeneracy of the $n^\text{th}$ energy level and I don't know how to do this. The correct answer says $2n+1$ but they never explain how they found this answer and for me it feels like it's taken out of the blue, but maybe I don't know enough about quantum numbers yet. I would appreciate it a lot if someone could explain how the degeneracies work for spherical harmonics as eigenfunctions for the $L^2$ operator.

  • $\begingroup$ Are you privy to the definition of the degeneracy of an eigenvalue of a linear operator? $\endgroup$ – joshphysics Jan 2 '15 at 18:09
  • $\begingroup$ @Djamillah I don't know that Griffiths book and I saw many complaints of it. But, spherical harmonics are eigenfunctions of two operators, $L^2$, and $L_z$. These functions are denoted as $Y_l^m (\theta, \phi)$, and for each value $l$ there are $2l + 1$ values of $m$, i.e. $m = l, l-1,..., 0,... -(l-1), -l$. You can see all this in the Wikipedia, en.wikipedia.org/wiki/…. But, the typical notation for the spherical harmonics uses the indexes $l$ and $m$ there is no $n$ (which usually has another meaning). $\endgroup$ – Sofia Jan 2 '15 at 22:30
  • $\begingroup$ (continuation) Could it be that Griffiths used the notation $n$ instead of $l$? Let me suppose so. Then, if you have a quantum system which displays a different energy for each value of $l$, but the energy doesn't depend also on $m$, the energy level $l$ is degenerated. That means, more than one single state corresponds to it, in our case $2l+1$ states. $m$ is called magnetic number, because for some systems a magnetic field can split the energy level $l$ into $2l+1$ sub-levels, one sub-level for each value of $m$. $\endgroup$ – Sofia Jan 2 '15 at 22:36
  • $\begingroup$ @Sofia: Long comment. Why not an answer? $\endgroup$ – DanielSank Jan 3 '15 at 1:22

I’m not sure why this was bumped to the community page, since the relevant answer is contained in bits and pieces in the preceding comments and answer.

Anyway, in summary: @Sofia is correct. Griffiths uses a non-standard convention in this question, replacing $l$ with $n$. ‘Degeneracy’ just means that there are multiple states with the same energy.


Since the Hamiltonian for a rigid rotor of length $a$ with two particles of mass $m$ at each end is

$\hat{H} = \frac{\hat{L}^2}{ma^2}$,

and since the spherical harmonics are the eigenfunctions of the $\hat{L}^2$ operator,

$\hat{L}^2Y^m_l(\theta,\phi) = \hbar^2 l(l+1)Y^m_l(\theta,\phi)$

(where the $Y^m_l(\theta,\phi)$ are the spherical harmonics) we have that the (degenerate) energy values are

$E_l = \frac{\hbar^2l(l+1)}{ma^2}$.

Then, purely because of the definition of the spherical harmonics (i.e. that there are $2l+1$ values of $m$ for every value of $l$), we can see that the degeneracy is $2l+1$.

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Griffiths 2nd edition Equation [4.118] and [4.119]:

$$L^2 f_\ell ^m=\hbar ^2\ell (\ell+1)f_\ell ^m$$

and $$ L_z f_\ell ^m=\hbar m f_\ell ^m$$

where $\ell =0, 1/2, 1, 3/2, \ldots $ and $m=-\ell ,\ldots ,\ell$.

There is are degeneracies in the $L^2$ operator since the eigenvalue only depends on the $\ell $ index.

There are a total of $2\ell +1$ indices in the $m=-\ell ,\ldots ,\ell$.

Section 4.3 p.178, the eigenfunctions $f_\ell ^m$ mentioned here in these two equations are the Spherical Harmonics $Y_\ell ^m$.

Footnote 17 references Schiff Quantum Mechanics for a detailed derivation of the wave-equation solutions to the Hydrogen atom, however, this should be the 3rd edition of that book for page 93.

I am still trying to get a sense of why the half-integer solN's of $\ell $ are sometimes dropped, and sometimes not (planning to ask a professor). I believe it has something to do with Spin.

In Exercise #4.24, for example (the massless Rigid-Rotor problem), Griffiths does indeed swap the $\ell $ index symbol for an $n$.

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  • $\begingroup$ Half-integer solutions for orbital angular momentum are dropped because otherwise the wave function (spherical part) wouldn't be continuous on the circle (a rotation of $2\pi$ would get a minus sign otherwise). $\endgroup$ – gented Nov 10 '16 at 23:11
  • $\begingroup$ I thought a -1 from a $2\pi $ rotation was something we wanted with a spin-1/2 particle like an electron? (Perhaps I am confusing orbital momentum and spin). $\endgroup$ – Andrew Dynneson Nov 11 '16 at 0:21
  • $\begingroup$ The spherical harmonics have the form $f(\theta)e^{-i2\pi\cdot m}$, $m$ being the $L_z$ eigenvalue. If half-integer, rotation of $2\pi$ bring back a minus sign. $\endgroup$ – gented Nov 11 '16 at 8:12

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