In general, what is the time-reversal of Hall conductivity $\sigma_{xy}(q,\omega)$? Is it $\sigma_{yx}^*(-q,\omega)$?

Is it possible to derive it from Kubo formula $$\sigma_{xy}(q,\omega)=\frac{1}{\omega V} \int_0^\infty dt \,e^{i\omega t} \langle\psi|[j_x^\dagger(q,t),j_y(q,0)|\psi\rangle$$ with the time-reversal operation $$\langle\alpha|\hat{O}|\beta\rangle \rightarrow \langle\beta|(T^{-1}\hat{O}T)^\dagger|\alpha\rangle$$ where $T=UK$ with unitary $U$ reversing momenta and so on and complex conjugation $K$?

Is it $T^{-1}j_x^\dagger(q,t)T = j_x^{\dagger*}(-q,-t)$ and how to proceed? Also I remember I read somewhere if $q=0$ it's $\sigma_{yx}(\omega)$.


While you can derive it from the Kubo formula, I don't think that there is a general rule. While you know how the operators and the variables will transform, you don't know in the general case how the Hamiltonian will transform. The behavior of the states and of the density matrix under time-reversal symmetry is model-specific. For example, is there a magnetic field in the setup like in the QHE? or is the setup time-reversal symmetric like in the QSHE?

Once you know how the Hamiltonian, and following that the states and the density matrix will response to time-reversal operation, you can get an expression for the behavior of the Hall conductivity.

edit following discussion in comments: The expectation value is taken with respect to the density matrix $\rho = \exp(-\beta H)$, so under TR

$$ \langle \hat{O} \rangle \to \rm{Tr} \left( T e^{-\beta H} \hat{O} T^{-1} \right)$$ and one has to know how $H$ is transformed. While it is true that when studying $\langle \psi | \hat{O} | \psi \rangle$ you can transform just the operator, once you take the thermal averaging you must enter into particulars of the setup itself. Even when you study the zero temperature limit, and then take the expectation value of the operator at the ground state, you should know that the ground state itself is the same for $H$ and $THT^{-1}$, as these two Hamiltonians might have different ground states.

| cite | improve this answer | |
  • $\begingroup$ The general time-reversal operation I showed (2nd formula) implies that one doesn't need info of the states. We are talking about general transformation here. $\endgroup$ – xiaohuamao Feb 7 at 9:26
  • $\begingroup$ That is correct, but you didn't take into account the density matrix. The true averaging is $$ \langle \hat{O} \rangle = \rm{Tr}\left[ \rho_0 \hat{O} \right]$$ which means that you need to know how the Hamiltonian and its eigenstates / eigenvalues behave. It could be that if you restrict yourself to the ground-state (i.e. $T\to 0$) it is easier $\endgroup$ – yu-v Feb 7 at 9:29
  • $\begingroup$ trace is nothing but many state expectation value in the end. My previous argument still holds. $\endgroup$ – xiaohuamao Feb 7 at 9:31
  • $\begingroup$ It is state expectation value with respect to the Hamiltonian. You need to know how $H$ transforms under $T$ $\endgroup$ – yu-v Feb 7 at 9:31
  • $\begingroup$ Let's stick to zero temperature if you like. $\endgroup$ – xiaohuamao Feb 7 at 9:33

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.