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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?

atbug
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  • 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? – ACuriousMind Apr 01 '16 at 11:56
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    @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. – atbug Apr 01 '16 at 12:00
  • "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.? – Luke Pritchett Apr 01 '16 at 12:11
  • @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. – atbug Apr 01 '16 at 12:32
  • Electron eigenstates of what? A Hamiltonian? If the Hamiltonian is invariant under some rotation-like symmetry I would absolutely expect electrons to have an energy degeneracy due to their spin. You'd have, e.g., both spin-up and a spin-down electron states with the same energy. – Luke Pritchett Apr 01 '16 at 13:05
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    @LukePritchett, spin-orbit coupling would generally break the degeneracy I think. – atbug Apr 01 '16 at 15:26
  • @ Chong Wang, yes, by breaking a symmetry in that specific example. I'm just trying to say that spin degeneracies in electron energies are not uncommon. If you just have spin-1/2 electrons interacting in some way that doesn't depend on spin then you will have a spin degeneracy. The rep. $\Gamma_6$ looks like an appropriate description of an electron: spinor rep. with twofold degeneracy as long as the Hamiltonian respects with the full symmetry. – Luke Pritchett Apr 01 '16 at 16:27

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