0
$\begingroup$

The Hamiltonian of spin-orbit coupling in an external magnetic field $\vec{B} = B\vec{e}_z$ is given by $$ H = \beta L\cdot S+\frac{\mu_b}{\hbar}(L_z+2S_z)B.$$ The ladder operators are $$ L_\pm = L_x \pm iL_y\qquad S_\pm = S_x \pm iS_y$$ and operate on the angular momentum eigenstates as follows:

$$ L_\pm|l,m_l\rangle = \hbar\sqrt{(l\pm m_l+1)(l\mp m_l)}|l,m_l\pm 1\rangle$$ $$S_\pm|s,m_s\rangle = \hbar\sqrt{(s\pm m_s+1)(s\mp m_s)}|s,m_s\pm 1\rangle$$

The Hamiltonian can then be expressed in terms of these operators:

$$ H = \frac \beta 4 (L_+S_-+L_-S_+)+L_zS_z+\frac{\mu_B}{\hbar} (L_z+2S_z)$$

My question is how to calculate the matrix elements of $H$ using the uncoupled basis states $|m_l,m_s\rangle$, so $\langle m_l',m_s'|H|m_l,m_s\rangle$, for a $p$-electron ($l=1$). I know the result which I found in a book, but that book doesn't show the explicit computation of the matrix entries. Since I'm quite new to this kind of things, it would be great if someone could show me how to do the calculation. I've been trying things for days now but nothing turned out to be correct.

|cite|improve this question
$\endgroup$
1
$\begingroup$

The angular momentum operators act on the $L$-part of the state only, while the spin operators act on the $S$-part only. Thus, $\hat L_+\hat S_-+\hat L_-\hat S_+$ acting on $\vert \ell=1,m_\ell=0\rangle\vert s=1/2,m_s=1/2\rangle$ will give \begin{align} &\hat L_+\hat S_-\vert 10\rangle\vert 1/2,1/2\rangle+ \hat L_-\hat S_+\vert 10\rangle\vert 1/2,1/2\rangle \\ &=\left[\hat L_+\vert 10\rangle\right]\left[\hat S_-\vert 1/2,1/2\rangle\right]+ \left[\hat L_-\vert 10\rangle\right]\left[\hat S_+\vert 1/2,1/2\rangle\right]\, ,\\ &=\sqrt{2}\hbar\vert 1,1\rangle\,\hbar\vert 1/2,-1/2\rangle+0\, ,\\ &=\sqrt{2}\hbar^2\vert 11\rangle\vert 1/2,-1/2\rangle. \end{align}

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thank you very much. So is it correct that $|m_l,m_s\rangle = |l,m_l\rangle |s,m_s\rangle$? And what about for example $L_z|1,0\rangle |1/2, 1/2\rangle$? Is the $S$-part of the state unchanged here? $\endgroup$ – MeMeansMe Jun 18 '17 at 13:27
  • $\begingroup$ Yes in general $\hat L_z$ on $\vert \ell m_\ell\rangle \vert sm_s\rangle$ simply returns $\hbar m_\ell\vert \ell m_\ell\rangle\vert sm_s\rangle$. In the case of your comment $m_\ell=0$ so $\hat L_z$ will actually kill that specific state. $\endgroup$ – ZeroTheHero Jun 18 '17 at 13:33
  • $\begingroup$ @MeMeansMe in addition to the previous, you really need to keep track of the values of $\ell$ and $s$ in your kets for the matrix elements do depend on these quantum numbers. $\endgroup$ – ZeroTheHero Jun 18 '17 at 14:21

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.