Symmetry operators like

  • $\hat{1}$: unity operator
  • $\hat{R}_{x,\pi}$: rotation around $\vec{e}_{x}$ with angle $\pi$
  • $\hat{M}_x$: mirror at a plane with normal in e.g. x direction
  • $\hat{I}$: inversion
  • ...

should be composed of an operator acting in position space and an operator acting in spin space, like $\hat{R}_{\alpha,\beta}=\hat{R}^{pos}_{\alpha,\beta}\hat{R}^{spin}_{\alpha,\beta}$. However, typically just the explicit form in position space is provided: $$\hat{1}^{pos}:\vec{\Psi}(x,y,z)\to \vec{\Psi}(x,y,z)$$ $$\hat{R}^{pos}_{\alpha,\beta}:\vec{\Psi}(x,y,z)\to \vec{\Psi}(x,-y,-z)$$ $$\hat{M}^{pos}_x:\vec{\Psi}(x,y,z)\to \vec{\Psi}(-x,y,z)$$ $$\hat{I}^{pos}:\vec{\Psi}(x,y,z)\to \vec{\Psi}(-x,-y,-z)$$ Lets say $\vec{\Psi}$ is a spinor of the form $\begin{pmatrix}\text{up}\\\text{down}\end{pmatrix}$. Just for the time inversion operator I found the full form $$\hat{T}:\vec{\Psi}(x,y,z)\to -i \begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0 \end{pmatrix} \vec{\Psi}^*(x,y,z)$$ What are the explicit forms for the spin operators for the other symmetry operators?

I guess a rotation can be expressed by $\hat{R}^{spin}_{\alpha,\beta}:\vec{\Psi}(x,y,z)\to e^{-i \beta \vec{\sigma}\cdot \vec{e}_{\alpha}}\vec{\Psi}(x,y,z)$ and $\overline{\hat{R}}^{spin}_{\alpha,\beta}:\vec{\Psi}(x,y,z)\to - e^{-i \beta \vec{\sigma}\cdot \vec{e}_{\alpha}}\vec{\Psi}(x,y,z)$ with the Pauli-matrices $\vec{\sigma}$? Both lead to the same measurable output, thus a system with a certain rotation symmetry (in position space) should be invariant under both transformations? The inversion (without considering antiparticles) is simply $\hat{I}^{\text{spin}}=1$? Can I deduce $\hat{M}^{\text{spin}}_x=\hat{I}^{\text{spin}} \hat{R}^{spin}_{x,\pi}=\hat{R}^{spin}_{x,\pi}$? And am I right, that $\overline{1}=-1$ (this Bethe-symmetry-operator for double-groups) is always a symmetry operator?

Thanks for any comment!

  • $\begingroup$ Perhaps what you are looking for are the Jones matrices used for polarization optics. $\endgroup$ Feb 8, 2020 at 4:34
  • $\begingroup$ @flippifanus: While the Jones formalism is related to this topic, it is just one application. Actually I try to understand some backgrounds of the Luttinger/Kane model and some geometry consideration in QDs. But thanks for the comment! $\endgroup$
    – Matthiasho
    Feb 10, 2020 at 10:53


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.