The problem:

A particle with spin 3/2 has a hamiltonian $H = L^2/2I$, $I$ is a constant and $L$ is the orbital angular momentum operator. Find the degeneracy of the first excited state.

So what I have done is relate the given H with the energy eigenvalues,
$$ H|\psi\rangle = \frac{1}{2I}L^2|\psi\rangle = \frac{1}{2I}l(l+1)|\psi\rangle = E|\psi\rangle, $$ which gives me the possible values of energy for the particle.

Then, as he asks about the first excited state, this means that quantum number $n = 2$. Considering that $l = 1, 2, 3, \ldots, n-1$ is the azimuthal quantum number, then the possible values for $l$ are 0 or 1.

Using these values for $E = \hbar^2l(l+1)$ gives us two possible energy values: $E_1 = 0$ for $l=0$ and $E_2 = \hbar^2/I$ for $l = 1$.

Now comes my problem. I know that spin = quantum number $s = 3/2$ gives me four values for its projection quantum number $m_s$: $-3/2$, $-1/2$, $1/2$ and $3/2$.

I also know that for a given value of $l$, one has $-l \leq m_l \leq l$ possible values for $m_l$. So in this case, $m_l = -1$, $0$ or $1$.

My confusion here is which of these quantum numbers are relevant to this case?

My first thought was that degeneracy would be 12, since for $l = 1$, there are three states related to the three possible values for $m_l$ and, for each of these, there are four possible values for $m_s$, hence 12.

Then I went check with a friend and he thinks it's 4. For that he just considered the spin and possible values for $m_s$, saying there would be four states with same energy, disregarding $m_l$ since the eigenvalues of $L^2$ does not depend on $m$ or $m_l$. And I couldn't disagree nor agree (he wasn't so sure either).

I think I'm missing some key concepts here and would gladly accept any lights shed on this.

  • 1
    $\begingroup$ Without solving this homework problem, please consider the following question: What is $n$ supposed to be? In the hydrogen-atom, the angular momentum does not specify the energy yet, this is why the hydrogen atom has both $n$ and $l$ as quantum numbers. Does this still make sense for your Hamiltonian? $\endgroup$
    – QuantumAI
    Nov 2 '17 at 16:31
  • $\begingroup$ The eigenvalues of $L^2$ not depending on $m_l$ just means the values are degenerate. Different values of $m_l$ are still different states. $\endgroup$
    – octonion
    Nov 2 '17 at 16:33
  • $\begingroup$ @QuantumAI thank you for the comment, helped visualize the issue I was having understanding this. Thanks for your quick replies, guys. $\endgroup$
    – Guest101
    Nov 3 '17 at 22:10

You are correct and your friend is not.

disregarding $m_l$ since the eigenvalues of $L^2$ does not depend on $m$ or $m_l$

This is precisely the reason for why you do need to include the degeneracy in the orbital sector: you have three orthogonal (orbital) states with the same energy, and you need to include them. Then, in addition, each of these corresponds to four orthogonal orbital$\otimes$spin states, to give a total count of $12$.

  • $\begingroup$ Thanks so much! It was very helpful. I think mostly the confusion was based on the fact that for a major part of my QM course not all the quantum numbers were used simultaneously. But now I get that a state is defined by all it's quantum numbers, (n, l, ml, s, ms), therefore any possible values for projections and so forth "hidden" actually represents other possible states existing with that same energy (thus adding degeneracy to that state). Right? $\endgroup$
    – Guest101
    Nov 3 '17 at 22:24

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