When one computes Hall conductivity $\sigma_{xy}$, one can show that the zero temperature Kubo formula gives

\begin{align} \sigma_{xy}(\omega) = -\frac{i}{\omega} \sum_{n\neq 0} \left[\frac{\langle 0|J_y |n\rangle \langle n | J_x|0 \rangle}{\hbar \omega + E_n - E_0} - \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{\hbar \omega + E_0 - E_n}\right] \end{align}

Now, one is in many instances interested in the $\omega\rightarrow 0$ limit, in which case one Taylor expands the denominators:

\begin{align} \frac{1}{\hbar \omega + E_n - E_0} = \frac{1}{E_n - E_0} - \frac{\hbar\omega}{(E_n - E_0)^2} + \mathcal{O}(\omega^2) \end{align}

In particular, in many places (e.g. David Tong's Quantum Hall notes) they argue away the $\omega^0$ term so that there is no zero-frequency divergence. I want to know how this is explicitly done.

Another stackexchange post argued for the use of the trick $J_x \sim v_x = \frac{\partial}{\partial t} x = [H,x]$, which cancels the energy factors in the denominator and result in the desired vanishing of leading term because the commutator $[x,v_y]=0$. The problem, however, is that performing such a trick for the second term on both $J_x$ and $J_y$ gives a term proportional to the commutator $[x,y]=0$. Thus, such an argument also shows that the second order term also vanishes, which is patently false.

To argue that the leading term vanishes, Tong's lecture notes invoke parity and gauge-invariance (without explicit argument). Parity makes sense, but the fact that $\sigma_{xy} = -\sigma_{yx}$ should be a result, not a constraint, so I feel there should be a more direct way of showing that it vanishes. The statement about gauge-invariance is mysterious to me.

Why is the Heisenberg equation of motion approach to this wrong, and what is the proper way to explicitly show that the leading term vanishes?

  • $\begingroup$ @d_b If we use the commutator for both the current matrix elements, we get two factors of $(E_n-E_0)$, thus cancelling the denominator. $\endgroup$
    – Arkya
    Commented Apr 28, 2020 at 5:06
  • $\begingroup$ Are we talking about Ryogo Kubo here? $\endgroup$ Commented May 3, 2020 at 20:00
  • $\begingroup$ Did you eventually find satisfactory answer to this question? I am struggling with it myself now. $\endgroup$
    – Blazej
    Commented Mar 22, 2021 at 15:26
  • $\begingroup$ Unfortunately not entirely. The answer invoking causality to restrict $\sigma(\omega)$ is a good way to do it, but I would've thought there would be a more direct method. $\endgroup$
    – Aaron
    Commented Mar 23, 2021 at 15:58

2 Answers 2


This might not be a complete answer to all of your questions but here's another way to see why the first term vanishes.

Let's inspect the bracketed expression in your first equation at leading order. Under complex conjugation, this term becomes: \begin{align} \left[\frac{\langle 0|J_y |n\rangle \langle n | J_x|0 \rangle}{ E_n - E_0} - \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{ E_0 - E_n}\right]^* &= \frac{\langle 0|J_x^{\dagger} |n\rangle \langle n | J_y^{\dagger}|0 \rangle}{ E_n - E_0} - \frac{\langle 0|J_x^{\dagger} |n\rangle \langle n | J_y^{\dagger}|0 \rangle}{ E_0 - E_n} \\ &= \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{ E_n - E_0} - \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{ E_0 - E_n}\\ &= \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{ E_n - E_0} - \frac{\langle 0|J_x |n\rangle \langle n | J_y|0 \rangle}{ E_0 - E_n} \end{align} which is the original expression we started with.

This means that the leading order term (with the factor of $i$ out front) is pure imaginary. But we know, from general symmetry grounds, that the imaginary part of $\sigma(\omega)$ must be an odd function of $\omega$, and hence it should vanish at $\omega=0$.

You can see why this would break down for the next order term. In fact, we can check that the next order term (the term that actually leads to TKNN's expression for DC conductivity) is pure real, hence doesn't have to vanish at $\omega=0$.


Notice that the imaginary part of your bracketed expression vanishes in the $\omega \rightarrow 0$ limit.

Furthermore, in Matsubara's formalism, the zero wavevector conductivity

$\sigma(i\Omega) = \frac{i}{i\Omega}\left[\Pi(i\Omega) - \Pi(i\Omega = 0)\right]$,

presents no difficulty in the DC case, where $\Pi$ is the current-current polarization bubble.


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