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Which of the following statements is not correct ?

  • A.In the presence of spin-orbit coupling, the Hamiltonian does not commute with \(\bar{L}\) and \(\bar{S}\) operators
  • B.Even in presence of spin-orbit coupling, the Hamiltonian commutes with \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) operators
  • C.The spin-orbit interaction for an electron is described by \(H_{so} = (e^{2}/8\pi\epsilon_{0}))\,1/(m^{2}c^{2}r^{3})\,\bar{S}.\bar{L}\).
  • D.In the presence of spin-orbit interaction, the physical quantities measured by \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) are not conserved.

Answer: D

The statement that is not correct is that, with spin-orbit interaction, \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) are not conserved.

Spin-orbit coupling is \(\xi(r)\bar{L}\cdot\bar{S} = \frac{\xi}{2}(J^{2} - L^{2} - S^{2})\). This commutes with \(L^{2}\), \(S^{2}\), \(J^{2}\) and the components of \(\bar{J}\), so those quantities remain conserved, while \(\bar{L}\) and \(\bar{S}\) separately are not, since the coupling transfers angular momentum between them.

So statements A and B are correct, and C is the standard form of the spin-orbit Hamiltonian (including the Thomas factor). Statement D claims the opposite of what the commutators show.

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