# A particle has hamiltonian $H = L^2/2I$ and spin 3/2. What is the degeneracy of the first excited state?

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.

• 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? – QuantumAI Nov 2 '17 at 16:31
• 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. – octonion Nov 2 '17 at 16:33
• @QuantumAI thank you for the comment, helped visualize the issue I was having understanding this. Thanks for your quick replies, guys. – Guest101 Nov 3 '17 at 22:10

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 $12$.