I am trying to derive the result of the TKNN formula but am experiencing difficulty in deriving the Kubo formula. The Kubo formula used in the TKNN paper is, $$ \sigma_{xy}= \frac{ie^2}{\hbar} \sum_{E^a < E_F < E^b} \frac{\langle a|v_x|b \rangle \langle b|v_y|a \rangle - \langle a|v_y|b \rangle \langle b|v_x|a \rangle}{(E^a - E^b)^2} .$$

The following is my work so far. First of all, from linear response theory, for a general observable, $$ O=\langle\hat{O}(t)\rangle-i\int_{t_0}^{t}\mathrm{d} t' \langle[\hat{O}(t),H_{\mathrm{ext}}(t')]\rangle $$ where $H_{\mathrm{ext}}$ is the perturbing Hamiltonian. Since $H_{\mathrm{ext}}=-\int J\cdot A(r,t)\mathrm{d}^3r$, if we consider current density we have $$ \vec{J}_i(r,t)=\langle j_i(t)\rangle+i\int_{-\infty}^{t}\mathrm{d} t'\int\mathrm{d}^3{r'} \langle[j_i(r,t),j_j(r',t')]\rangle A_j(r',t'). $$ Ignoring the first term, we can rewrite this as $$ \vec{J}_i(r,t)=-\int_{-\infty}^\infty\mathrm{d}t\int \mathrm{d}^3r R_{ij}(r-r',t-t')A_j(r',t') $$ where $$ R_{ij}(r-r',t-t')=-i\theta(t-t')\langle[j_i(r,t),j_j(r',t')]\rangle. $$ We can assume that $A(r,t)=A(r)e^{-i\omega t}$ to show that $E(r,t)=i\omega A(r,t)$. Then by Fourier transforming the expression for the expectation value of the current density, we get $$ \vec{J}(k,\omega)=-\frac{R_{ij}(k,\omega)}{i\omega}E_j(k,\omega) $$ which means DC conductivity is $$ \sigma_{ij}=\lim_{\omega\rightarrow0}\frac{iR_{ij}(k,\omega)}{\omega} $$ The Fourier transform of the response function can be shown to be, $$ R_{ij}(k,\omega)=-i\int_0^\infty \mathrm{d} t e^{i\omega t}\langle[j_i(k,t),j_j(-k,0)]\rangle $$ If we evaluate expectation values using the grand canonical ensemble (ie $\langle \hat{O}\rangle = Tr(e^{-\beta H} \hat{O})/Z$ where $H$ includes the chem. pot.) and integrate over time and perform a first order $\omega$ expansion we end up with $$ R_{ij}(k,\omega)=\sum_{n,m}(e^{-E_m \beta}-e^{-E_n\beta})\left(\frac{\langle n|j_i(k,\omega)|m\rangle\langle m|j_j(-k,\omega)|m\rangle}{E_n-E_m+i\epsilon}-\omega\frac{\langle n|j_i(k,\omega)|m\rangle\langle m|j_j(-k,\omega)|m\rangle}{(E_n-E_m+i\epsilon)^2}\right)/Z $$ This is from where I don't know how to proceed. All derivations for the Kubo formula I have found on this site and other resources online do not consider the position dependence of the current density operator. Any advice that will point me in the right direction will be appreciated.

  • $\begingroup$ Possible duplicate of Kubo Formula for Quantum Hall Effect $\endgroup$ Commented May 5, 2017 at 3:35
  • $\begingroup$ Please take a look at physics.stackexchange.com/q/1906. The answer explains why the first term vanishes and the second term gives the Kubo formula. The derivation there is applicable for position-dependent current operators, just replace every $v_i$ operator there by your $j_i(\pm k)$, the algebra still goes through. $\endgroup$ Commented May 5, 2017 at 3:40
  • $\begingroup$ @EverettYou How exactly does this substitution--replacing $j_i(\pm k)\to v_i$--work? Does the TKNN paper assume that the current operator is not position dependent? $\endgroup$ Commented May 5, 2017 at 14:02
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    $\begingroup$ Because the current $j$ is proportional to the velocity $v$ of the electron as $j=-ev$, so they are essentially the same operator. The TKNN paper did not assume that the current operator is position independent. The matrix element $\langle a|v(q)|b\rangle$ makes perfect sense if we include the momentum dependence of $v$, i.e. $v(q)$ will just scatter a state $|b\rangle$ of momentum $k$ to another state $|a\rangle$ of momentum $k+q$. One can still sum over all such pair of states that satisfies $E_a<E_F<E_b$ and obtain the momentum-dependent Hall conductance $\sigma_{xy}(q)$. $\endgroup$ Commented May 5, 2017 at 17:39
  • $\begingroup$ @EverettYou I must be misunderstanding something. Isn't the Hall conductance supposed to be momentum-independent? $\endgroup$ Commented May 5, 2017 at 17:53

1 Answer 1


The Kubo formula applies to the position dependent case and can be used to compute a momentum-dependent Hall conductivity $$\sigma_{xy}(\boldsymbol{q})= \frac{\mathrm{i}e^2}{\hbar} \sum_{E_a < E_F < E_b} \frac{\langle a|v_x(\boldsymbol{q})|b \rangle \langle b|v_y(-\boldsymbol{q})|a \rangle - \langle a|v_y(-\boldsymbol{q})|b \rangle \langle b|v_x(\boldsymbol{q})|a \rangle}{(E_a - E_b)^2}.$$ This formula can be derived from the last formula (the current-current correlation in the many-body basis) in the question $$\begin{split} R_{xy}(\boldsymbol{q},\omega)=\frac{1}{Z}\sum_{n,m}(e^{-\beta E_m}-e^{-\beta E_n})\bigg(&\frac{\langle n|v_x(\boldsymbol{q},\omega)|m\rangle\langle m|v_y(-\boldsymbol{q},\omega)|m\rangle}{E_n-E_m+\mathrm{i}0_+}\\ &-\omega\frac{\langle n|v_x(\boldsymbol{q},\omega)|m\rangle\langle m|v_y(-\boldsymbol{q},\omega)|m\rangle}{(E_n-E_m+\mathrm{i}0_+)^2}\bigg), \end{split}$$ by noticing that the first term vansishes (as explained in Kubo Formula for Quantum Hall Effect) and take the limit of $\sigma_{xy}(\boldsymbol{q})=\lim_{\omega\to0}\mathrm{i}R_{xy}(\boldsymbol{q},\omega)/\omega$ for the second term. Then switch to the single-particle basis (can be done for free fermions) and take the zero temperature limit $\beta\to\infty$ in the end.

From the momentum-dependent Hall conductivity one can restore the explicit position dependence by Fourier transform $$\sigma_{xy}(\boldsymbol{r}-\boldsymbol{r}')=\sum_{\boldsymbol{q}}\sigma_{xy}(\boldsymbol{q}) e^{\mathrm{i}\boldsymbol{q}\cdot(\boldsymbol{r}-\boldsymbol{r}')}.$$ The uniform conductivity is defined as $\sigma_{xy}=\int d^2\boldsymbol{r}' \sigma_{xy}(\boldsymbol{r}-\boldsymbol{r}')$ which picks out the zero momentum component $\sigma_{xy}(\boldsymbol{q}=0)$. TKNN's paper mainly focus on the uniform Hall conductance and its topological significance. But the generalization of the Kubo formula to the non-uniform (inhomogeneous) case is straight forward as described above. However for a generic momentum $\boldsymbol{q}\neq0$, the Hall conductance $\sigma_{xy}(\boldsymbol{q})$ is no longer quantized to an integer and is no longer related to the topological index (Chern number) of the electronic band structure, for this reason, $\sigma_{xy}(\boldsymbol{q})$ is less investigated. But experimentally $\sigma_{xy}(\boldsymbol{q})$ is definite a quantity that can be measured as well.

Most generally, one can define the momentum and frequency dependent conductivity tensor $\sigma_{ij}(q)$ (where $q=(\omega,\boldsymbol{q})$ is the momentum-frequency vector) from the electron Green's function $G(k)$, $$\begin{split} \sigma_{ij}(q)&=-\mathrm{i}\sum_k \text{Tr} G(k)\gamma_0 G(k)\gamma_i G(k+q)\gamma_j,\\ \gamma_\mu&=-\partial_{k_\mu}G^{-1}(k)\quad(\mu=0,1,2).\end{split}$$ This is called the inhomogeneous dynamical conductivity or optical conductivity. The vertex operator $\gamma_\mu$ replace the current/velocity operator in the quantum field thoery (as a generalization of the gamma matrices of Dirac fermions). See Calculating conductivity from Green's functions for derivation of this general form.

  • $\begingroup$ Thanks for the insight on the non-uniform conductivity. Can you recommend any resources that discusses this topic and notes the distinction between the uniform and non-uniform conductivity? It seems like many derivations brush over this even though it feels like a non-trivial fact. $\endgroup$ Commented May 5, 2017 at 18:14
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    $\begingroup$ Just search for "dynamical conductivity", you can see a lot of discussion of all kinds of conductivities away from the static/uniform case. $\endgroup$ Commented May 5, 2017 at 18:41
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    $\begingroup$ Would you mind explaining what did you mean precisely by "switching to the single-particle basis"? Also, how exactly do you convert from current (which I think of as an operator on the full Fock space) into velocity operator (which seems to me to be an operator on the single particle space)? $\endgroup$
    – Blazej
    Commented Mar 18, 2021 at 9:31

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