In group theory, to account for electron spin, double group is introduced. The key difference between an ordinary point group and a double group is an extra element $\bar{E}$ with the meaning of a $2\pi$ rotation. Since only a $4\pi$ rotation restores a spin state, $\bar{E}^2 = E$, where $E$ is the identity.

Also, a $2\pi$ rotation always results in a $\pi$ phase in spinor. To put it another way, $\bar{E}=-1$. From this point of view, double group representation character of $\bar{E}$ should always be $-D$, where $D$ is the dimension of the representation.

However, character table of double groups does not follow this rule. For example, for two double groups $T_d$ and $O$, the character table is: enter image description here

The character of $\bar{E}$ is positive in most representations. Does this mean that these representations are not physical? If so, it seems crystals with $T_d$ or $O$ symmetry have degeneracies everywhere, which is hardly acceptable to me.

Can anyone help me resolve this issue?

  • $\begingroup$ Please explain the notation. What are $\Gamma_i, \phi, O, T_d$? Also, do you mean the double cover or covering group with "double group" (see also this question)? What exactly do you want to know about it? $\endgroup$ – ACuriousMind Apr 1 '16 at 11:56
  • 1
    $\begingroup$ @ACuriousMind, no, double group is a well established concept in solid state physics. $T_d$ and $O$ are double groups and $\Gamma_i$ are their representations. $\phi$ has nothing to do with my question. $\endgroup$ – Chong Wang Apr 1 '16 at 12:00
  • $\begingroup$ "Also, a 2π rotation always results in a π phase in spinor." Do you have a source for this claim? Is this your intuition, or are you taking this as the definition of a spinor? My instinct is that the answer is pretty simple: the reps. where $\overline{E}$ does not map to $-I$ are not spinor reps. Is there a reason you expect all of the reps. of these groups to be spinors reps.? $\endgroup$ – Luke Pritchett Apr 1 '16 at 12:11
  • $\begingroup$ @Luke Pritchett, see, for example, "modern quantum mechanics" by sakurai, eq (3.2.15). Electrons are all spinors. So if none of the spinor reps. are 1D, electron eigenstates would all have degeneracy. I don't think this is the case. $\endgroup$ – Chong Wang Apr 1 '16 at 12:32
  • 1
    $\begingroup$ @LukePritchett, spin-orbit coupling would generally break the degeneracy I think. $\endgroup$ – Chong Wang Apr 1 '16 at 15:26

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.