In the case of weak field Zeeman effect (anomalous Zeeman effect) in hydrogen atom, the unperturbed Hamiltonian reads as $$ H_0 = \frac{\hat{p}^2}{2m} + \frac{C_1}{r} + f(r)\mathbf{L}\cdot\mathbf{S} $$ and the perturbation term, which comes from the external magnetic field, is $$ H' = C_2(L_z+2S_z) $$ In my problem, the exact form of $C_1$ and $C_2$ are not important. What bothers me is that every literature always uses the eigenkets of $H_0$, which can be denoted as $|n,L,J,m_j\rangle$ as the unperturbed kets. But these kets do not diagonalize the perturbation term $H'$. Aren't the zeroth-order eigenkets supposed to be chosen such that they diagonalize the perturbation in the case of degenerate levels, which is true in this problem?

  • $\begingroup$ What quantum mechanics textbook are you referring to? I remember the treatment of degenerate (time-independent) perturbation theory being somewhat unclear with regard to the problem that you pose (see eg. Shankar). In general, as you seem to know, perturbation theory uses as the zero-th order kets the very same eigenkets of the unperturbed hamiltonian. In the case of a degenerate energy spectrum, should the perturbation be able to remove the degeneration, the standard treatment is first of all to calculate the matrix elements of the perturbation between the unperturbed eigenkets; ... $\endgroup$ Commented Feb 5, 2016 at 10:38
  • $\begingroup$ ...once you've done so, you'll be able to find the states that diagonalize the perturbation; then, and only then, you can go to the basis which diagonalizes the perturbation, and calculate the energy spectrum to first order. You can't really start from kets that diagonalize the perturbation, as you don't know them from the beginning. Once you've found the kets that diagonalize the perturbation, to find the correct (up to first order in the perturbation) eigenkets you don't need to use the formula with the (in this case) infinite denominators anymore: the two sets coincide to first order $\endgroup$ Commented Feb 5, 2016 at 10:41
  • $\begingroup$ Modern QM by Sakurai and Physics of Atoms and Molecules by Bransden and Joachain, as well as some online resources that I have found. All of them do not seem to really care about the fact that $H'$ is not diagonal in $|n,L,J,m_J\rangle$ basis. I don't really get what you mean. Since I know the exact form of $|n,L,J,m_J\rangle$ and how $H'=L_z+2S_z$ acts on these kets, I can easily compute the matrix element of $H'$ for fixed $n$ and $L$, namely $\langle n,L,J',m_J'|H'|n,L,J,m_J\rangle$. Having found this matrix, I can diagonalize it and find the corresponding basis. $\endgroup$
    – nougako
    Commented Feb 5, 2016 at 10:57
  • $\begingroup$ Aren't those steps of diagonalization not so hard to carry out? But why those authors neglect it? $\endgroup$
    – nougako
    Commented Feb 5, 2016 at 11:02
  • $\begingroup$ You indeed understood correctly. The basis that you'll find is, to first order, the perturbed eigenket basis you're looking for. As for the energy levels, the perturbation removes the degeneracy by adding to the former energy of the new kets the energies that correspond to the diagonal entries. $\endgroup$ Commented Feb 5, 2016 at 11:02

1 Answer 1


After some hours of pondering, I finally realize two things:

  1. If I were to diagonalize $H'$, I must do it in the subspace of fixed $n$ and $J$, instead of fixed $n$ and $L$. This is because the energy eigenvalues of the unperturbed Hamiltonian $H_0$ are specified by $(n,J)$ (states with same $n$ and $J$ but different $L$'s can have the same energy).
  2. The matrix elements of $H'=J_z+S_z$ between states of the same $n,J$ but different $L$'s and/or $m_J$'s vanish, $$ \begin{aligned} \langle n,L',J,m_J'|H'| n,L,J,m_J\rangle &= m_J\hbar\delta_{m_Jm_J'}\delta_{LL'} + \langle n,L',J,m_J'|S_z| n,L,J,m_J\rangle\\ &= \delta_{LL'} \Bigg( m_J\hbar\delta_{m_Jm_J'} + \frac{\hbar}{2} \sqrt{\frac{(L'\pm m_J'+1/2)(L\pm m_J+1/2)}{(2L'+1)(2L+1)}} \delta_{m_J-1/2,m_J'-1/2} \ldots \\ & \ldots -\frac{\hbar}{2} \sqrt{\frac{(L'\mp m_J'+1/2)(L\mp m_J+1/2)}{(2L'+1)(2L+1)}} \delta_{m_J+1/2,m_J'+1/2} \Bigg)\\ &= \delta_{LL'} \Bigg( m_J\hbar\delta_{m_Jm_J'} + \frac{\hbar}{2} \frac{(L\pm m_J+1/2)}{(2L+1)} \delta_{m_J-1/2,m_J'-1/2} \ldots \\ & \ldots -\frac{\hbar}{2} \frac{(L\mp m_J+1/2)}{(2L+1)} \delta_{m_J+1/2,m_J'+1/2} \Bigg)\\ &= K\delta_{m_Jm_J'}\delta_{LL'} \end{aligned} $$ with $$ K = m_J\hbar + \frac{\hbar}{2} \frac{(L\pm m_J+1/2)}{(2L+1)}-\frac{\hbar}{2} \frac{(L\mp m_J+1/2)}{(2L+1)} = m_J\hbar \Bigg( 1 \pm \frac{1}{(2L+1)}\Bigg). $$
  • $\begingroup$ can i say that there is no degeneracy of H', i.e. H' is already diagonal in the basis of J and mj. Therefore, I can use nondegenerate pertubation theory safely? plz reply. $\endgroup$
    – Ka Wa Yip
    Commented Nov 24, 2016 at 0:58
  • $\begingroup$ Degeneracy becomes your concern when it's the unperturbed Hamiltonian that is degenerate. You don't want to care too much about the spectrum of the perturbation itself, at least for first order perturbation theory. As I already showed above, in the subspace of fixed $n$ and $J$ in the spectrum of $H_0$, $H'$ is diagonal. $\endgroup$
    – nougako
    Commented Nov 24, 2016 at 7:47

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