2
$\begingroup$

I came accross the following statement in a book:

If one wants to switch on a magnetic field, one must first choose the appropriate complex unperturbed wave functions (that are "adapted" to the perturbation by the field).

I do not really understand what they mean here, though. Could anyone enlighten me ?

Actually this question would help me understand another problem, which I will sum up here. In magnetic optical activity, there are 3 terms of interest called the Faraday terms $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{C}$. In short, these terms are related to the observed rotation of plane polarization of an electromagnetic radiation (i.e. light) directed towards a molecule placed in a static uniform magnetic field. I am concerned about the $\mathcal{A}$ term, in particular, which reads: $$ \mathcal{A}(a\rightarrow j)= \left(\langle j|\mu_z| j\rangle - \langle a|\mu_z|a\rangle\right) \Im\left\{ \langle a|m_x| j\rangle \langle j|m_y| a\rangle \right\} \,, $$ where $\mu$ and $m$ are the magnetic and electric dipole moment operators, respectively. The states $a$ and $j$ are the eigenstates of the unperturbed Hamiltonian, i.e. without the external fields. Up to now, I thought that in the absence of magnetic field, the wave functions could be chosen to be real. If this is the case, than the imaginary part appearing in $\mathcal{A}$ should vanish, since we have real operators. Similarly, the mean value of the magnetic dipole moment operator, $\langle j|\mu_z| j\rangle$ (or $\langle a|\mu_z| a\rangle$), which corresponds to the first-order change in the energy (from perturbation theory), should vanish, since we have an imaginary operator between real states. However, I know that this $\mathcal{A}$ term is nonzero in a lot of cases. Actually, in textbooks, it is said that this term does not vanish only if either the ground or excited state is degenerate. After some researches, I found that in general, when a state is degenerate, the wavefunction was complex, which would then lead to a non-vanishing $\mathcal{A}$. But doesn't this contradict the fact that without magnetic field, the wavefunctions can be chosen to be real ? I am really confused about this. Also, if I choose real wavefunctions for the unperturbed Hamiltonian, does that mean I will never get a finite $\mathcal{A}$ term ? Then I guess I would be missing some physics...

Thanks in advance !

$\endgroup$
  • $\begingroup$ Alright, I think I understand now the statement of the book, thanks to this topic. So very roughly: the eigenstate of the system in the presence of an external magnetic field, let us say $|d\rangle$ will reduce to $|d^0\rangle$ when the perturbation is switched off. However, $|d^0\rangle$ may not be the same eigenstate as the one found "naturally" in the absence of field, but rather a linear combination of these states. The other part of my question still holds, though. $\endgroup$ – Leinahtan May 31 '16 at 13:42

Your Answer

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

Browse other questions tagged or ask your own question.