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I am given an exercise on perturbation theory involving an electron in a hydrogen atom in the presence of a constant magnetic field $\vec{B} = B_z \hat{z}$. Due to Zeeman effect and Spin-Orbit coupling, the term \begin{equation} \Delta H = H_{SO} + H_Z = \frac{1}{2m^2 c^2} \frac{1}{r} \frac{\mathrm{d}V}{\mathrm{d}r} \vec{L} \cdot \vec{S} + \mu_B B_z (L_z + 2S_z) \end{equation} must be added to the unperturbed Hamiltonian $H_0$, with $V(r) = -e^2/r$ and $\mu_B$ is the Bohr magneton. Supposing that the second term dominates (Paschen-Back effect), I am asked to determine the first order perturbation of the energy spectrum, thus treating only the spin-orbit coupling as a perturbation.

To this end, I have to evaluate the perturbation matrix with entries \begin{equation} \langle l', \frac{1}{2},m', m_s' | H_{SO}| l, \frac{1}{2},m, m_s \rangle. \end{equation} In the solution, only the diagonal entries are computed. I suppose that this might be due to the operator $\vec{L} \cdot \vec{S}$ commuting with $L^2$, $L_z$ and $S_z$, that is to say \begin{equation} [\vec{L}\cdot \vec{S}, L^2] = 0, \quad [\vec{L}\cdot \vec{S}, L_z] = 0, \quad [\vec{L}\cdot \vec{S}, S_z] = 0. \end{equation} This would indeed require that \begin{equation} l'= l, \quad m'=m, \quad m_s' = m_s, \end{equation} leaving only the diagonal terms to compute. I re-expressed $\vec{L}\cdot \vec{S}$ as \begin{equation} \vec{L} \cdot \vec{S} = \frac{1}{2} \left( (\vec{L}+\vec{S})^2 - \vec{L}^2 - \vec{S}^2 \right) \end{equation} but still failed to show that the above commutation relations hold. Am I missing something? Could/Should I use (rotational) symmetry instead?

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  • $\begingroup$ what commutation relations are we talking about? $\endgroup$ Jan 5, 2022 at 13:37
  • $\begingroup$ @ZeroTheHero I was talking about the commutators between $\vec{L}\cdot \vec{S}$ and $L^2$, $L_z$ and $S_z$. $\endgroup$
    – Xavier
    Jan 5, 2022 at 16:25
  • $\begingroup$ sorry I don’t see any commutator so I can’t make sense of “the above commutation relations hold”… $\endgroup$ Jan 5, 2022 at 17:38
  • $\begingroup$ @ZeroTheHero as I understand the sentence "the operator $\vec{L}\cdot\vec{S}$ commuting with $L^2$, $L_z$ and $S_z$" can be read as "$[\vec{L}\cdot\vec{S}, L^2]=0$, $[\vec{L}\cdot\vec{S}, S_z]=0$, $[\vec{L}\cdot\vec{S}, L_z]=0$" (with the latter two being not true, of course) $\endgroup$
    – user275556
    Jan 5, 2022 at 17:52
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    $\begingroup$ you might care to make this clear through an edit. $\endgroup$ Jan 5, 2022 at 19:18

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the operator $\vec{J}^2=(\vec{L}+\vec{S})^2$ (and likewise $\vec{L}\cdot\vec{S}$) does not commute with $L_z$ and $S_z$ separately, but only with the combined $J_z = L_z + S_z$. You can see this also from symmetry considerations: rotating just the spatial angular momentum or just the spin will not preserve the scalar product $\vec{L}\cdot\vec{S}$. In order for this term to be a scalar under rotations, the rotations must be both in the coordinate space (angular momentum) and the spin space, together.

The reason that just the diagonal elements are computed is due to the nature of perturbation theory. In the first order in perturbation theory, we are just concerned with the degenerates states within the subspace that share the same energy. Once you incorporated the Zeeman term $\mu_B B(L_z+2S_z)$ into $H$, it lifted (some of) the degeneracy between states with different $m, m_s$. It has not completely disappeared, though: the states with $|m_s = 1/2, m\rangle$ and $|m_s=-1/2, m+2\rangle$ (if $m$ allows such a change) would still have the same energy. However, you can work out that $\vec{L}\cdot\vec{S}$ cannot connect between these states, because $J_z$ is still preserved, and they belong to different values of $J_z$. So you are left only with the diagonal terms which can play a role to first order in perturbation theory.

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  • $\begingroup$ Thank you very much. I'm not sure of what you mean with "$\vec{L} \cdot \vec{S}$ cannot connect between these states". So to be sure I understand correctly: I can identify the basis $|l, \frac{1}{2}, m, m_s\rangle$ with the basis $|n, l, j, m_j\rangle$, knowing that $j = l \pm \frac{1}{2}$. Then, the fact that $\vec{L}\cdot \vec{S}$ commutes with $L^2$, $S^2$ and $J_z$ ensures that the perturbation matrix we consider in order to determine the first order corrections to the energy is diagonal in the basis of states $|n, l, j, m_j \rangle$. Am I right? $\endgroup$
    – Xavier
    Jan 5, 2022 at 16:26
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    $\begingroup$ yes. when I wrote "cannot connect" I meant that the matrix elements of $\vec{L}\cdot\vec{S}$ between these states will be zero. You are exactly right that in the basis of $|n,l,j,m_j\rangle$ the perturbation $\vec{L}\cdot\vec{S}$ is diagonal. Just note that these are not eigenstates of the unperturbted Hamiltonian, because of the $(L_z+2S_z)$ term $\endgroup$
    – user275556
    Jan 5, 2022 at 16:56

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