Spin-orbit coupling, degeneracy of eigenvalues

I just read in a book about atomic physics that an important part of the fine structure of hydrogen is spin-orbit coupling. The Hamiltonian of spin-orbit coupling in the hydrogen atom is given by $$H_{SO} = \beta L\cdot S = \frac{1}{2}\left(J^2-L^2-S^2\right),$$ where $L$ is the orbital angular momentum operator, $S$ the spin operator and $J = L + S$.

I want to determine the eigenvalues and degeneracies of $H_{SO}$ and the possible values for the quantum number $j$ of $J$ because the book and other sources just tell me that and don't derive it. This is what I've done so far:

Since $[J^2,H]=[L^2,H]=[S^2,H]=[J_z,H]=0$, let $\psi$ be an eigenstate of $H, J^2, L^2, S^2$ and $J_z$. So we get

$$H_{SO}\psi = \frac{\hbar^2\beta}{2}\left(j(j+1)-l(l+1)-s(s+1)\right)\psi$$

and the eigenvalues of $H_{SO}$ are therefore given by $\alpha_{j,l,s} = \frac{\hbar^2\beta}{2}(j(j+1)-l(l+1)-s(s+1))$.

What I'm struggling with is the degeneracy of the eigenvalues and how to determine the possible values for $j$. Can anybody help?

• $J=L+S$ is a direct application of composition of angular momentum operators. Look that up in any good QM reference: you seem to already know enough to read one of those! – user154997 Jun 17 '17 at 12:09
• @LucJ.Bourhis: No matter which book I look into, they all just tell that for example $|l-s|\le j \le l+s$ but I couldn't find a proof for this. And I simply don't know how to approach the degeneracy problem. That's why I'm asking this question on this forum. – MeMeansMe Jun 17 '17 at 12:49
• Which book do you lear QM from? So that I can hopefully pinpoint the exact section to study… I mean I could summarise the composition of angular momentum in an answer but that would not be very efficient compared to study a proper course. – user154997 Jun 17 '17 at 13:04
• I agree with Luc. The possible values for J are very standard material and they're covered in all suitably advanced QM textbooks. – Emilio Pisanty Jun 17 '17 at 23:57

This is basically a problem of recursive counting. Start with the uncoupled basis state, i.e. the set of states of the form $\vert \ell m_\ell \rangle \vert s m_s\rangle$. There are clearly $(2\ell+1)(2s+1)$ of these, and the job is to reorganize them.
The key counting result is based on the observation that $\vert \ell m_\ell\rangle\vert sm_s\rangle$ is an eigenstate of $\hat J_z=\hat L_z+\hat S_z$ with eigenvalue $M=m_\ell+m_s$. With this in mind organize your $\vert \ell m_\ell\rangle\vert sm_s\rangle$ states so that those with the same value of $M$ are on the same line. Explicitly, for instance, you would have \begin{align} \begin{array}{rlll} M=\ell+s:&\vert \ell \ell\rangle\vert s s\rangle \\ M=\ell+s-1:& \vert \ell,\ell-1\rangle\vert ss\rangle& \vert\ell\ell\rangle \vert s,s-1\rangle\\ M=\ell+s-2:&\vert \ell,\ell-2\rangle \vert ss\rangle & \vert\ell,\ell-1\rangle\vert s,s-1\rangle&\vert \ell \ell\rangle \vert s,s-2\rangle\\ \vdots\qquad& \qquad\vdots \end{array} \end{align} and replace each state with a $\bullet$ to get $$\begin{array}{rlll} \ell+s:&\bullet \\ \ell+s-1:&\bullet & \bullet\\ \ell+s-2:&\bullet&\bullet&\bullet\\ \vdots\qquad&\vdots \end{array}$$ Now, if $M=\ell+s$ is the largest value and it occurs once, the value of $j=\ell+s$ must occur once and also all the states $\vert j=\ell+s,m_j\rangle$ will occur once. There is a linear combination of the two states with $M=\ell+s-1$ that will be the state $\vert j=\ell+s,m_j=\ell+s-1\rangle$, there will be a linear combination of the three states with $M=\ell+s-2$ that will be the $\vert j=\ell+s,m_j=\ell+s-2\rangle$ state etc. Since we are only interested in enumerating the possible resulting values of $j$, and not interested in the actual states per se, we can eliminate from our table the first column since it contains one state with $m_j=\ell+s$, one with $m_j=\ell+s-1$ etc. Eliminating this column yields the reduced table $$\begin{array}{rll} \ell+s-1: & \bullet\\ \ell+s-2:&\bullet&\bullet\\ \vdots\qquad&\vdots \end{array}$$ Since the value of $m_j=\ell+s-1$ occurs once, the value $j=\ell+s-1$ must occur once, and the states $\vert \ell+s-1,m_j\rangle$ will each occur once. We take out those from the list by deleting the first column to obtain a further reduced table $$\begin{array}{rl} \ell+s-2:&\bullet\\ \vdots\qquad &\vdots \end{array}$$ The process so continues until exhaustion. In the examples above we have found $j=\ell+s,\ell+s-1$ and the final reduced table of the example, if not empty, would indicate the value of $j=\ell+s-1$. It is clear this process produces a decreasing sequence of $j$. The last value of $j$ is determined by the width of the original table. It is not hard to convince yourself that the width of the table will stop increasing once we reach $M=\vert \ell-s\vert$, and this is the last value of $j$. Thus by exhaustion you find the possible values of $j$ in the range $$\vert \ell-s\vert\le j\le \ell+s\, .$$
As an example consider $\ell=1$ and $s=2$. The original table then looks like $$\begin{array}{rlll} \frac{3}{2}:&\vert 11\rangle\vert 1/2,1/2\rangle \\ \frac{1}{2}:&\vert 10\rangle\vert 1/2,1/2\rangle & \vert 11\rangle\vert 1/2,-1/2\rangle\\ -\frac{1}{2}:&\vert 1,-1\rangle\vert 1/2,1/2\rangle&\vert 10\rangle\vert 1/2,-1/2\rangle\\ -\frac{3}{2}:&\vert 1,-1\rangle \vert 1/2,-1/2\rangle \end{array}\qquad \to \qquad \begin{array}{rlll} \frac{3}{2}:&\bullet \\ \frac{1}{2}:&\bullet & \bullet \\ -\frac{1}{2}:&\bullet &\bullet \\ -\frac{3}{2}:&\bullet \end{array}$$ It is only $2$ column wide, and the width stop growing at $M=1/2$, indicating the possible $j$ in this case are $3/2$ and $1/2$, and indeed $$\vert 1-1/2\vert \le j\le 1+1/2$$ Finally, note that the absolute value is required on the left because one could write state $\vert sm_s\rangle\vert \ell m_\ell\rangle$ without affecting the possible values of $j$.