1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $J=L+S$ 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! $\endgroup$ – user154997 Jun 17 '17 at 12:09
  • $\begingroup$ @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. $\endgroup$ – MeMeansMe Jun 17 '17 at 12:49
  • 1
    $\begingroup$ 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. $\endgroup$ – user154997 Jun 17 '17 at 13:04
  • $\begingroup$ I agree with Luc. The possible values for J are very standard material and they're covered in all suitably advanced QM textbooks. $\endgroup$ – Emilio Pisanty Jun 17 '17 at 23:57
0
$\begingroup$

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 $j$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.