Wigner-Weyl calculus in Keldysh technique

Abstract

We discuss the non-equilibrium dynamics of condensed matter/quantum field systems in the framework of Keldysh technique. In order to deal with the inhomogeneous systems we use the Wigner-Weyl formalism. Unification of the mentioned two approaches is demonstrated on the example of Hall conductivity. We express Hall conductivity through the Wigner transformed two-point Green’s functions. We demonstrate how this expression is reduced to the topological number in thermal equilibrium at zero temperature. At the same time both at finite temperature and out of equilibrium the topological invariance is lost. Moreover, Hall conductivity becomes sensitive to interaction corrections.

Appendices

Appendix A: Lesser component of a product of operators

Here we consider the static non-interacting system with the one-particle Hamiltonian that does not depend on time. We will show that if the derivative of distribution function may be neglected

$$\begin{aligned} H &\equiv \int d^{{D+1}}\pi \, \mathop {\mathrm{tr}}\nolimits \left( K_1 \star K_2\star \ldots \star K_n\right) ^< \nonumber \\ &=\int d^{{D+1}}\pi \, f(\pi )\left[ \mathop {\mathrm{tr}}\nolimits \left( K_1 \star K_2\star \ldots \star K_n\right) ^\mathrm{A}\nonumber \right. \\&\quad \left. - \mathop {\mathrm{tr}}\nolimits \left( K_1 \star K_2\star \ldots \star K_n\right) ^\mathrm{R}\right] , \end{aligned}$$

(A1)

where each of the $K_i$ is a Green’s function G, its inverse Q or a derivative of one of those. $f(\pi )$ stands for the initial distribution of the system, not necessarily the equilibrium one. See (14), (19) and (20) for the notions of A/R and “lesser” components.

To prove the above, we first note that

$$\begin{aligned}&\left( K_1 \star K_2\star \ldots K_n\right) ^< = \sum _{i=0}^n \left[ K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\star K^<_{i+1} \right. \left. \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right] , \end{aligned}$$

(A2)

so that in every term of this sum there only one factor in the “lesser” form.

In the case of a static system for any $K_i$ that does not contain derivatives of G/Q, we have (see Eqs. 19 and 20)

$$\begin{aligned} K^<_i = (K^\mathrm{A}_i-K^\mathrm{R}_i) f(\pi ). \end{aligned}$$

(A3)

Then H becomes

$$\begin{aligned} \begin{aligned} H&= \sum _{i=0}^n \int d^{{D+1}}\pi \, \mathop {\mathrm{tr}}\nolimits \left[ K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\right. \\&\left. \quad \star \left[ (K^\mathrm{A}_{i+1}-K^\mathrm{R}_{i+1}) f(\pi )\right] \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right] \\&= \sum _{i=0}^n \int d^{{D+1}}\pi \, \mathop {\mathrm{tr}}\nolimits \left[ K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\star \left[ K^\mathrm{A}_{i+1} f(\pi )\right] \right. \\&\left. \quad \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right] \\&\quad - \mathop {\mathrm{tr}}\nolimits \left[ K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\star \left[ K^\mathrm{R}_{i+1} f(\pi )\right] \right. \\&\quad \left. \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right] . \end{aligned} \end{aligned}$$

(A4)

This expression can be simplified in the case when position of $f(\pi )$ is irrelevant, i.e. when $K_i\star f(\pi ) = K_i f(\pi )$. It is possible in two cases: a) when derivatives of $f(\pi )$ can be assumed being small; b) when all $K_i$ do not depend on time, while $f(\pi )$ depends only on the energy. In the former case, the following expression is only approximate, while in the latter it is exact:

$$\begin{aligned} \begin{aligned} H&= \sum _{i=0}^n \int d^{{D+1}}\pi \, f(\pi )\, \\&\quad \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\star K^\mathrm{A}_{i+1} \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right) \\&\quad - f(\pi )\, \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{R}\star \ldots \star K_i^\mathrm{R}\star K^\mathrm{R}_{i+1} \right. \\&\quad \left. \star K_{i+2}^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right) . \end{aligned}\end{aligned}$$

(A5)

We can see now that in the i-th contribution to the sum the first term in the parentheses cancels the second one at $i-1$. Thus, we will be left only with the first and the last terms of the summation:

$$\begin{aligned} \begin{aligned} H&= \int d^{{D+1}}\pi \, f(\pi )\, \left[ \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{A}\star \ldots \star K_n^\mathrm{A}\right) \right. \\&\quad \left. - \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{R}\star \ldots \star K_n^\mathrm{R}\right) \right] . \end{aligned} \end{aligned}$$

(A6)

Very similar consideration is valid when

$$\begin{aligned} K_i = \partial _\pi G \mathrm{\ or\ } K_i = \partial _\pi Q. \end{aligned}$$

(A7)

To show it we recall, that due to the mutually inverse nature of G and Q, we can always write a derivative of one as a sandwiched derivative of the other,

$$\begin{aligned} \partial _\pi Q = -Q\star \partial _\pi G \star Q,\qquad \partial _\pi G = -G\star \partial _\pi Q \star G . \end{aligned}$$

Then, it is sufficient to analyse expressions of the following type entering (A2) in place of one of the $K_i$-s:

$$\begin{aligned} \begin{aligned} G^\mathrm{R}\star \partial _\pi Q^< \star G^\mathrm{A}&= G^\mathrm{R}\star \partial _\pi \left[ (Q^\mathrm{A}-Q^\mathrm{R}) f(\pi )\right] \star G^\mathrm{A}\\&= G^\mathrm{R}\star \left[ \partial _\pi (Q^\mathrm{A}-Q^\mathrm{R})\right] f(\pi )\star G^\mathrm{A}\\&\quad + G^\mathrm{R}\star \left[ \partial _\pi f(\pi )\right] (Q^\mathrm{A}-Q^\mathrm{R})\star G^\mathrm{A}\end{aligned}\end{aligned}$$

(A8)

Provided that the derivative of f may be neglected, the derivatives of G and Q behave similarly to G (Q) themselves in considered construction, and (A1) holds valid. In Appendix B we give the more detailed derivation in an important particular case of Hall conductivity, when last term of the above is important.

Finally, we note that disregarding the derivatives of $f(\pi )$ is justified, in particular, when considering the low temperature limit of the initially equilibrium distribution. Moreover, if $f=f(\pi _0)$ is thermal equilibrium distribution indeed, we can further rewrite (A2) as a sum over the Matsubara frequencies.

Indeed, by using the analytic properties of $K^{R/A}$ (as inherited from those of G/Q) we have

$$\begin{aligned} H= & {} \int _{{\mathrm{Im}\pi _0 = {-} 0 }} d^{{D+1}}\pi \, f(\pi _0) \mathop {\mathrm{tr}}\nolimits \left( K_1 \star K_2\star \ldots \star K_n\right) \nonumber \\&- \int _{{\mathrm{Im}\pi _0 ={+}0 }} d^{{D+1}}\pi \, f(\pi _0) \mathop {\mathrm{tr}}\nolimits \left( K_1 \star K_2\star \ldots \star K_n\right) , \end{aligned}$$

(A9)

Closing now the integration contour over $\pi ^0$ into upper (lower) half-plane in the first (second) terms of the above, we transform the original integrals to the two integrals surrounding poles of $f(\pi _0)$. Those are given by imaginary unit times Matsubara frequency $\omega _j = (2j+1)\pi /{\beta }$. Calculating the integral using residue theorem we obtain:

$$\begin{aligned} H= & {} 2\pi {i T} \int d^{{D}} \pi \sum _{\omega _j} \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{M}\star K_2^\mathrm{M}\star \ldots \star K_n^\mathrm{M}\right) \end{aligned}$$

(A10)

here the superscript M indicates that corresponding function is taken at Matsubara frequency: $K^\mathrm{M}(\omega _j,\pi _1,...,\pi _D) \equiv K(i \omega _j,\pi _1,...,\pi _D)$. In the limit of small temperatures the sum over the Matsubara frequencies is reduced to an integral and we arrive at:

$$\begin{aligned} H ={i} \int d^{{D+1}} {\Pi } \mathop {\mathrm{tr}}\nolimits \left( K_1^\mathrm{M}\star K_2^\mathrm{M}\star \ldots \star K_n^\mathrm{M}\right) . \end{aligned}$$

(A11)

Here $\Pi$ is “Euclidean” $D+1$ - momentum, i.e. $\Pi ^{D+1} = \omega$ is continuous Matsubara frequency, $\Pi ^i = \pi ^i$ for $i=1,...,D$.

Appendix B: Conductivity in a static system without interactions

Here we derive expressions for Hall and symmetric conductivities through Wigner transformed Green’s functions for the case of a noninteracting static system, in which initial non-thermal distribution depends on energy only in $2+1 D$.

We start from tensor ${{\mathcal {K}}}^{\mu \nu \rho }$ defined in Eq. (74):

$$\begin{aligned}&{ {\mathcal {K}}}^{ijk} = \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}\hat{Q}_0 \right. \left. \star \hat{G}_0\star \partial _{\pi _{{j}}}\hat{Q}_0 \star \partial _{\pi _{{k}}}\hat{G}_0 \right) ^< +\mathrm{c.c.} \end{aligned}$$

(B1)

Hall conductivity is given by ${{\mathcal {K}}}^{i0j} - {{\mathcal {K}}}^{ij0}$, so we study both terms separately. We have

$$\begin{aligned} \begin{aligned} { {\mathcal {K}}}^{i0j}&= \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}\hat{Q}_0\star \partial _{\pi _{0}}\hat{G}_0 \star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0 \right) ^< +\mathrm{c.c.}\\&= \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}} {Q}^\mathrm{R}_0 \star \partial _{\pi _{0}} {G}^\mathrm{R}_0\star \partial _{\pi _{j}} {Q}^\mathrm{R}_0 \star \Big ((G^\mathrm{A}_0-G^\mathrm{R}_0) f(\pi _0)\Big ) \right) \\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}} {Q}^\mathrm{R}_0 \star \partial _{\pi _{0}} {G}^\mathrm{R}_0\star \partial _{\pi _{j}}\Big ((Q^\mathrm{A}_0-Q^\mathrm{R}_0) f(\pi _0)\Big ) \star {G}^\mathrm{A}_0 \right) \\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}} {Q}^\mathrm{R}_0 \star \partial _{\pi _{0}} \Big ((G^\mathrm{A}_0-G^\mathrm{R}_0) f(\pi _0)\Big )\star \partial _{\pi _{j}} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) \\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}\Big ((Q^\mathrm{A}_0-Q^\mathrm{R}_0) f(\pi _0)\Big ) \star \partial _{\pi _{0}} {G}^\mathrm{A}_0\star \partial _{\pi _{j}} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) +\mathrm{c.c.} \\&= -\frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}} {Q}^\mathrm{R}_0 \star \partial _{\pi _{0}} {G}^\mathrm{R}_0\star \partial _{\pi _{j}} {Q}^\mathrm{R}_0 \star G^\mathrm{R}_0 \right) f(\pi _0)\\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}Q^\mathrm{A}_0 \star \partial _{\pi _{0}} {G}^\mathrm{A}_0\star \partial _{\pi _{j}} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) f(\pi _0)\\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}Q^\mathrm{R}_0\star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) \star \partial _{\pi _{j}} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) \partial _{\pi _{0}} f(\pi _0) +\mathrm{c.c.} \end{aligned}\end{aligned}$$

(B2)

In the last equality we used that position of $f(\pi _0)$ is irrelevant, see App. A. Very similarly, for ${{\mathcal {K}}}^{ij0}$ we have

$$\begin{aligned} \begin{aligned} { {\mathcal {K}}}^{ij0}&= -\frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}} {Q}^\mathrm{R}_0 \star {G}^\mathrm{R}_0\star \partial _{\pi _{j}} {Q}^\mathrm{R}_0 \star \partial _{\pi _{0}}G^\mathrm{R}_0 \right) f(\pi _0)\\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 }\\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}Q^\mathrm{A}_0 \star {G}^\mathrm{A}_0\star \partial _{\pi _{j}} {Q}^\mathrm{A}_0 \star \partial _{\pi _{0}} {G}^\mathrm{A}_0 \right) f(\pi _0)\\&\quad + \frac{1}{4 \, {{\mathcal {V}}}} \int \frac{d^3\pi d^2 x}{(2\pi )^3 }\\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi _{i}}Q^\mathrm{R}_0\star {G}^\mathrm{R}_0\star \partial _{\pi _{j}} {Q}^\mathrm{R}_0 \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) \right) \partial _{\pi _{0}} f(\pi _0)+\mathrm{c.c.} \end{aligned}\end{aligned}$$

(B3)

Both in (B2) and (B3) one might be tempted to put in the limit $\epsilon \rightarrow 0$

$$\begin{aligned}&\partial _{\pi _{j}} {Q}^\mathrm{R}_0 \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) = - {Q}^\mathrm{R}_0\star \partial _{\pi _{j}} {G}^\mathrm{R}_0\star {Q}^\mathrm{R}_0 \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0)=0, \end{aligned}$$

alleging that ${Q}^\mathrm{R}_0 \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) = {{\mathcal {O}}}(\epsilon )$. While the latter is true, the former is not legitimate since near the pole $\partial _{\pi _{j}} {G}^\mathrm{R}_0 \sim 1/\epsilon ^2$ and the whole expression is not vanishing.

Thus, for the Hall conductivity we obtain

$$\begin{aligned} \begin{aligned} {-}\sigma _H&= \frac{1}{2\pi }\times \frac{1}{48\pi ^2 \, {{\mathcal {V}}}} \epsilon ^{\mu \nu \rho } \int {d^3\pi d^2 x}\\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi ^{\mu }} {Q}^\mathrm{R}_0 \star \partial _{\pi ^{\nu }} {G}^\mathrm{R}_0\star \partial _{\pi ^{\rho }} {Q}^\mathrm{R}_0 \star G^\mathrm{R}_0 \right) f(\pi ^0)\\&\quad - \frac{1}{2\pi }\times \frac{1}{48\pi ^2 \, {{\mathcal {V}}}} \epsilon ^{\mu \nu \rho }\int {d^3\pi d^2 x} \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi ^{\mu }}Q^\mathrm{A}_0 \star \partial _{\pi ^{\nu }} {G}^\mathrm{A}_0\star \partial _{\pi ^{\rho }} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) f(\pi ^0)\\&\quad + \frac{1}{8 \, {{\mathcal {V}}}} \epsilon ^{ij}\int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi ^{i}}Q^\mathrm{R}_0\star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) \star \partial _{\pi ^{j}} {Q}^\mathrm{A}_0 \star {G}^\mathrm{A}_0 \right) \partial _{\pi ^{0}} f(\pi ^0)\\&\quad - \frac{1}{8 \, {{\mathcal {V}}}}\epsilon ^{ij} \int \frac{d^3\pi d^2 x}{(2\pi )^3 } \\&\quad \mathop {\mathrm{tr}}\nolimits \left( \partial _{\pi ^{i}}Q^\mathrm{R}_0\star {G}^\mathrm{R}_0\star \partial _{\pi ^{j}} {Q}^\mathrm{R}_0 \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) \right) \partial _{\pi ^{0}} f(\pi ^0)+\mathrm{c.c.}\\&= \frac{1}{2\pi } {{\mathcal {N}}}_f + \epsilon ^{ij} {{\mathcal {A}}}_{ij}. \end{aligned} \end{aligned}$$

(B4)

Here

$$\begin{aligned} {{\mathcal {A}}}_{ij}= & {} \frac{1}{8 \, {{\mathcal {V}}}} \int \frac{d^3\pi \, d^2 x}{(2\pi )^3 } \mathop {\mathrm{tr}}\nolimits \left( \left( \partial _{\pi ^{{j}}}Q^\mathrm{A}_0\star {G}^\mathrm{A}_0 \star \partial _{\pi ^{{i}}} {Q}^\mathrm{R}_0\nonumber \right. \right. \\&\left. \left. -\partial _{\pi ^{{i}}}Q^\mathrm{R}_0\star {G}^\mathrm{R}_0 \star \partial _{\pi ^{{j}}} {Q}^\mathrm{R}_0 \right) \nonumber \right. \\&\left. \star ( {G}^\mathrm{A}_0- {G}^\mathrm{R}_0) \right) \partial _{\pi _{0}} f(\pi _0) +\mathrm{c.c.}, \end{aligned}$$

(B5)

and

$$\begin{aligned} {{\mathcal {N}}}_f= & {} -\frac{1}{48\pi ^2 \, {{\mathcal {V}}}} \epsilon ^{{ijk}} \oint d\pi _0 \int {d^2 \mathbf {\pi } d^2 x} \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( G_0\star \partial _{\pi ^{{i}}} {Q}_0 \star \partial _{\pi ^{{j}}} {G}_0\star \partial _{\pi ^{{k}}} {Q}_0 \right) f(\pi _0) +\mathrm{c.c.}, \end{aligned}$$

(B6)

where $\oint$ is an integral over the contour encompassing the whole real axis in positive direction, while

$$\begin{aligned} \begin{aligned} Q_0(x,\pi )&= \pi _0- {H}_W(x,\pi ), \\ G_0(x,\pi )&\star Q_0(x,\pi ) =1. \end{aligned}\end{aligned}$$

(B7)

In a similar way for the longitudinal conductivity we obtain

$$\begin{aligned} \sigma _\parallel ^{ij} = {{\mathcal {A}}}_{\{ij\}} + \mathrm{c.c.} \end{aligned}$$

(B8)

Appendix C: Conductivity in terms of the velocity operator

Let us now rewrite (B4) in terms of the matrix elements of the velocity operator, similar to the derivation given in [63]. For this end we shall use that the trace of the Weyl symbols over the phase space is equal to the functional trace of a product of corresponding operators, for instance, given by the trace of their matrix elements over momentum space

$$\begin{aligned} \begin{aligned} \mathop {\mathrm{Tr}}\nolimits (A_W *B_W)&\equiv \int d^3 X \int \frac{d^3 P}{(2\pi )^3}\mathop {\mathrm{tr}}\nolimits (A_W *B_W) \\&= {\mathop {\mathrm{Tr}}\nolimits } \hat{A} \hat{B} = \int d^3 P d^3 Q A(P,Q) B(Q,P), \end{aligned} \end{aligned}$$

(C1)

where $P^i = (p_0,p_1,p_2) = (\omega ,p)$ and $X^i = (t,x_1,x_2) = (t,x)$. Applying this formula to Eq. (B6) we come to

$$\begin{aligned} \begin{aligned} {{\mathcal {N}}}_f&= -\frac{\epsilon ^{\mu \nu \rho }}{48\pi ^2 \, {{\mathcal {V}}}} \oint d \omega ^{(1)} \, f(\omega ^{(1)}) \int d^2 p^{(1)} \prod _{i=2}^4 d^3 P^{(i)} \\&\quad \mathop {\mathrm{tr}}\nolimits \Bigg [ {G}(P^{(1)},P^{(2)}) \left[ \partial _{P^{(2)}_\mu } + \partial _{P^{(3)}_\mu }\right] {Q}(P^{(2)},P^{(3)})\\&\quad \left( \left[ \partial _{P^{(3)}_\nu } + \partial _{P^{(4)}_\nu }\right] {G}(P^{(3)},P^{(4)}) \right) \\&\quad \left[ \partial _{P^{(4)}_\rho } + \partial _{P^{(1)}_\rho }\right] {Q}(P^{(4)},P^{(1)}) \Bigg ] +\mathrm{c.c.} \end{aligned} \end{aligned}$$

(C2)

For the non-interacting fermions described by Hamiltonian ${{\mathcal {H}}}$ with energy eigenstates $|n\rangle$: ${{\mathcal {H}}}|n\rangle = {{\mathcal {E}}}_n |n\rangle$, the matrix elements in the above are given by

$$\begin{gathered} Q\left( {P^{{(1)}} ,P^{{(2)}} } \right) \equiv \left\langle {P^{{(1)}} |\hat{Q}|P^{{(2)}} } \right\rangle \hfill \\ \quad \quad \quad \quad \quad \,\, = \left( {\delta ^{{(2)}} (p^{{(1)}} - p^{{(2)}} )\omega ^{{(1)}} - \langle p^{{(1)}} |{\mathcal{H}}|p^{{(2)}} \rangle } \right)\delta \left( {\omega ^{{(1)}} - \omega ^{{(2)}} } \right) \hfill \\ G\left( {P^{{(1)}} ,P^{{(2)}} } \right) = \delta \left( {\omega ^{{(1)}} - \omega ^{{(2)}} } \right)\sum\limits_{n} {\frac{{\left\langle {p^{{(1)}} |n} \right\rangle \left\langle {n|p^{{(2)}} } \right\rangle }}{{\omega ^{{(1)}} - {\mathcal{E}}_{n} }}} . \hfill \\ \end{gathered}$$

(C3)

here $\sum _n$ may stand both for discrete spectrum summation, and integration $\int dn$ in the case of continuum one.

To perform further simplifications, we note that

$$\begin{aligned} \partial _{p_i} G = - G \partial _{p_i} Q G , \end{aligned}$$

(C4)

and more importantly,

$$\begin{aligned} \left[ \partial _{p^{(4)}_j} + \partial _{p^{(1)}_j} \right] \langle p ^{(4)}| {{\mathcal {H}}} | p ^{(1)} \rangle= & {} \mathrm {i}\langle p ^{(4)}| {{\mathcal {H}}} {{\hat{x}} }_j \\&-{{\hat{x}} }_j{{\mathcal {H}}} | p ^{(1)}\rangle \equiv \langle p ^{(4)}| {\hat{v}}_j | p ^{(1)}\rangle , \end{aligned}$$

where we introduced the velocity operator ${{\hat{v}}}_i = \mathrm {i}[{{\mathcal {H}}}, {{\hat{x}} }_i]$. By operator ${\hat{x}}_i$ we understand $\mathrm {i}\partial _{p_i}$ acting on the wavefunction written in momentum representation:

$$\begin{aligned} \hat{x}_j \Psi (p) = \langle p|\hat{x}_j |\Psi \rangle = \mathrm {i}\partial _{p_j} \langle p|\Psi \rangle = \mathrm {i}\partial _{p_j} \Psi (p). \end{aligned}$$

Then, for example,

$$\begin{aligned} \hat{x}_j \delta ^{(2)}(q-p)= & {} \langle p|\hat{x}_j |q\rangle = \mathrm {i}\partial _{p_j} \langle p|q \rangle \\= & {} \mathrm {i}\partial _{p_j} \delta ^{(2)}(p-q) = -\mathrm {i}\partial _{p_j}\langle q|p \rangle . \end{aligned}$$

Therefore, we can write

$$\begin{aligned} \hat{x}_j |p\rangle = -\mathrm {i}\partial _{p_j} |p \rangle . \end{aligned}$$

Using the above formulae, we derive that

$$\begin{aligned} \begin{aligned} {{\mathcal {N}}}_f&= {+}\frac{\epsilon ^{ij} }{4\, {{\mathcal {V}}}}\,\sum _{n,k} \int \prod _{l=1}^4 d^2 p ^{(l)} \\&\quad \oint \frac{ f(\omega ) d \omega }{(\omega ^{}-{{\mathcal {E}}}_n)^2 (\omega ^{}-{{\mathcal {E}}}_k)} \langle p ^{(1)}| n \rangle \langle n p ^{(2)}\rangle \langle p ^{(2)}| {\hat{v}}_i | p ^{(3)}\rangle \\&\quad \langle p ^{(3)}| k \rangle \langle k | p ^{(4)}\rangle \langle p ^{(4)}| {\hat{v}}_j | p ^{(1)}\rangle + \mathrm{c.c.} \\&= {+}\frac{2\pi \mathrm {i}\, \epsilon ^{ij} }{4 \, {{\mathcal {V}}}} \sum _{n,k } \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f'({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)^2}\\&\quad \langle n | {\hat{v}}_i | k \rangle \langle k | {\hat{v}}_j | n \rangle + \mathrm{c.c.} \end{aligned}\end{aligned}$$

(C5)

We used here that the momentum eigenvalues compose a full set, $\int d^2p\, |p\rangle \langle p|=1$. Note, that in the case of a discreet spectrum, the term $n=k$ should be understood as a limit ${{\mathcal {E}}}_n\rightarrow {{\mathcal {E}}}_k$, which gives a finite result.

For the non-topological contribution to $\sigma _H$ and for $\sigma _\Vert$ we shall similarly analyze ${{\mathcal {A}}}$ given by (B5). Advanced and retarded components needed for its calculation can be obtained from (C3) as

$$\begin{aligned} Q^{\mathrm{A}/\mathrm{R}}(p^{(1)},p^{(2)})= & {} \left( \delta ^{(2)} (p^{(1)}-p^{(2)})\, \left[ \omega ^{(1)}\pm \mathrm {i}\epsilon \right] \nonumber \right. \\&\left. - \langle p^{(1)}| {{\mathcal {H}}} | p^{(2)}\rangle \right) \delta \left( \omega ^{(1)}-\omega ^{(2)}\right) \nonumber \\&G^{\mathrm{A}/\mathrm{R}}(P^{(1)},P^{(2)}) = \delta \left( \omega ^{(1)}-\omega ^{(2)}\right) \nonumber \\&\sum _{n} \frac{\langle p ^{(1)}| n \rangle \langle n | p ^{(2)}\rangle }{\omega ^{(1)}-{{\mathcal {E}}}_n\pm \mathrm {i}\epsilon } . \end{aligned}$$

(C6)

and thus,

$$\begin{aligned}&\left( {G}^\mathrm{A}_0-{G}^\mathrm{R}_0\right) \left( P^{(1)},P^{(2)}\right) =2\pi \mathrm {i}\, \delta (\omega ^{(1)}-\omega ^{(2)}) \nonumber \\&\quad \sum _{n} \delta (\omega ^{(1)}-{{\mathcal {E}}}_n) \langle p ^{(1)} | n \rangle \langle n | p ^{(2)}\rangle . \end{aligned}$$

(C7)

Then

$$\begin{aligned}&\left( \partial _{\pi _{j}}Q^\mathrm{A}_0 \hat{G}^\mathrm{A}_0 \partial _{\pi _{i}}\hat{Q}^\mathrm{R}_0 \right) \left( P^{(1)},P^{(4)}\right) \nonumber \\&\quad = \delta \left( \omega ^{(1)}-\omega ^{(4)}\right) \int dp^{(2)}dp^{(3)} \langle p ^{(1)}| {\hat{v}}_j | p ^{(2)}\rangle \frac{\left\langle p ^{(2)}| n \right\rangle \left\langle n | p^{(3)}\right\rangle }{\omega ^{(1)}-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } \nonumber \\&\qquad \left\langle p ^{(3)}| {\hat{v}}_i | p ^{(4)}\right\rangle , \end{aligned}$$

(C8)

and

$$\begin{aligned}&(\partial _{\pi _{i}}Q^\mathrm{R}_0 \hat{G}^\mathrm{R}_0 \partial _{\pi _{j}}\hat{Q}^\mathrm{R}_0 )(P^{(1)},P^{(4)}) \nonumber \\&\quad = \delta (\omega ^{(1)}-\omega ^{(4)}) \int dp^{(2)}dp^{(3)} \langle p ^{(1)}| {\hat{v}}_i | p ^{(2)}\rangle \nonumber \\&\quad \frac{\langle p ^{(2)}| n \rangle \langle n | p^{(3)}\rangle }{\omega ^{(1)}-{{\mathcal {E}}}_n - \mathrm {i}\epsilon } \langle p ^{(3)}| {\hat{v}}_j | p ^{(4)}\rangle . \end{aligned}$$

(C9)

All together it gives

$$\begin{aligned} {{\mathcal {A}}}_{ij}= & {} {-} \frac{ \mathrm {i}}{8 \, {{\mathcal {V}}}} \sum _{n,k} f' ({{\mathcal {E}}}_k) \nonumber \\&\left[ \frac{\langle k| {\hat{v}}_j | n \rangle \langle n | {\hat{v}}_i | k \rangle }{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon }\nonumber \right. \\&\left. -\frac{\langle k| {\hat{v}}_i | n \rangle \langle n | {\hat{v}}_j | k \rangle }{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon }\right] +\mathrm{c.c.}, \end{aligned}$$

(C10)

So, that

$$\begin{aligned} \begin{aligned} {{\mathcal {A}}}_{\{ij\}}&= {-} \frac{\mathrm {i}}{8 \, {{\mathcal {V}}}} \sum _{n,k} f' ({{\mathcal {E}}}_k) \left[ \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon }\right. \\&\left. \quad -\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon }\right] \left( \langle k| {\hat{v}}_j | n \rangle \langle n | {\hat{v}}_i | k \rangle \right. \\&\quad \left. +\langle k| {\hat{v}}_i | n \rangle \langle n | {\hat{v}}_j | k \rangle \right) +\mathrm{c.c.}, \\ {{\mathcal {A}}}_{[ij]}&= {-} \frac{ \mathrm {i}}{8 \, {{\mathcal {V}}}} \sum _{n,k} f' ({{\mathcal {E}}}_k)\\&\quad \left[ \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } + \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon }\right] \\&\quad \left( \langle k| {\hat{v}}_j | n \rangle \langle n | {\hat{v}}_i | k \rangle - \langle k| {\hat{v}}_i | n \rangle \langle n | {\hat{v}}_j | k \rangle \right) +\mathrm{c.c.}, \end{aligned}\end{aligned}$$

(C11)

The limit $\epsilon \rightarrow 0$ of these expressions depends on the nature of the spectrum. In the continuum case, the Sokhotski-Plemelj formula gives

$$\begin{aligned}&\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } - \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon } = - 2 \pi \mathrm {i}\, \delta ({{\mathcal {E}}}_k-{{\mathcal {E}}}_n), \qquad \nonumber \\&\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } + \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon } = 2 {{\mathcal {P}}}\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n}, \end{aligned}$$

(C12)

while in the discrete case, the expression for ${{\mathcal {A}}}_{\{ij\}}$ (and thus, for symmetric conductivity) will be divergent in $\epsilon \rightarrow 0$

$$\begin{aligned}&\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } - \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon } = \left\{ \begin{array}{ll} 0, &{} n\ne k\\ \frac{2}{\mathrm {i}\epsilon }, &{} n = k \end{array}\right. , \nonumber \\&\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n + \mathrm {i}\epsilon } + \frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n - \mathrm {i}\epsilon } = \left\{ \begin{array}{ll} \frac{2}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n}, &{} n\ne k\\ 0, &{} n = k \end{array}\right. \end{aligned}$$

(C13)

Summarizing, we have

$$\begin{aligned} \sigma _H= {- \sigma _{xy} } &= - \frac{\mathrm {i}\, \epsilon ^{ij} }{4 \, {{\mathcal {V}}}}\,\sum _{n,k} \left( \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f'({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)^2} \nonumber \right. \\& \quad \left. + f'({{\mathcal {E}}}_k) {{\mathcal {P}}}\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n}\right) \langle n | {\hat{v}}_i | k \rangle \langle k | {\hat{v}}_j | n \rangle + \mathrm{c.c.} \end{aligned}$$

(C14)

We can use this expression both for continuum and discreet spectrum if in the latter case we put ${{\mathcal {P}}}\frac{1}{{{\mathcal {E}}}_k-{{\mathcal {E}}}_n}=0$, ${{\mathcal {E}}}_n={{\mathcal {E}}}_k$. One can see, that in the absence of the singularities at ${{\mathcal {E}}}_n = {{\mathcal {E}}}_k$ the term with $f^\prime$ is cancelled. In Eq. (C14) the singularity is isolated in the second term in the brackets while the first term remains finite at ${{\mathcal {E}}}_n = {{\mathcal {E}}}_k$ (it is reduced to $f^{\prime \prime }({{\mathcal {E}}}_n)/2$). It is worth mentioning that one can rewrite the whole expression in the following alternative form:

$$\begin{aligned} \sigma _H= & {} {-}\frac{\mathrm {i}\, \epsilon ^{ij} }{2 \, {{\mathcal {V}}}}\,\sum _{n,k} \frac{ f({{\mathcal {E}}}_k)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )} \nonumber \\&\langle n | {\hat{v}}_i | k \rangle \langle k | {\hat{v}}_j | n \rangle + \mathrm{c.c.} \end{aligned}$$

(C15)

Written in this form it coincides with expression proposed in [93] (see also [94]). In order to show equivalence of Eqs. (C15) and (C14) let us represent the quotient from the former as follows:

$$\begin{aligned} \begin{aligned} \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )}&= \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f^{\prime }({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )} \\&\quad - \frac{ ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k -i\epsilon ) f^{\prime }({{\mathcal {E}}}_n)}{2({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )}\\&\quad -\frac{ ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k+i\epsilon ) f^{\prime }({{\mathcal {E}}}_n)}{2({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )} \\&= \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f^{\prime }({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )} \\&\quad + \frac{ f^{\prime }({{\mathcal {E}}}_n)}{2({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )}+\frac{ f'({{\mathcal {E}}}_n)}{2({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )} \\&= \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f^{\prime }({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n-i\epsilon )({{\mathcal {E}}}_k-{{\mathcal {E}}}_n+i\epsilon )} \\&\quad + f'({{\mathcal {E}}}_n) {{\mathcal {P}}}\frac{1 }{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)} \end{aligned}\end{aligned}$$

(C16)

Appendix D: Hall conductivity for the noninteracting 2D system in the presence of constant magnetic field

Here we demonstrate how the derived expressions allow to obtain final expressions for the conductivity. We take as an example the simplest system of free non-relativistic electrons in the presence of constant magnetic field. The one-particle Hamiltonian is taken in its simplest form

$$\begin{aligned} {{\mathcal {H}}} = \frac{1}{2m}\left( \pi _1^2 + \pi _2^2\right) -\mu \end{aligned}$$

with $\pi _1 = \hat{p}_1$ and $\pi _2 =\hat{p}_2-{{\mathcal {B}}}x_1$. We have the following property specific for this Hamiltonian to be used further:

$$\begin{aligned} \epsilon ^{ij}\pi _i {{\mathcal {H}}} \pi _j = 3i{{\mathcal {B}}} {{\mathcal {H}}} \end{aligned}$$

The average Hall conductivity may be represented as

$$\begin{aligned} \bar{\sigma }_H= & {} -\frac{\mathrm {i}}{4 {{\mathcal {V}}}}\, \epsilon ^{ij} \, \Big (\sum _{n,k|{{\mathcal {E}}}_n\ne {{\mathcal {E}}}_k} \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) }{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)^2} \nonumber \\&+ \frac{1}{2}\sum _{n,k|{{\mathcal {E}}}_n={{\mathcal {E}}}_k}f^{\prime \prime }({{\mathcal {E}}}_n)\Big ) \langle n|{{\hat{v}}}_i| k \rangle \langle k | {{\hat{v}}}_j | n \rangle + \mathrm{c.c.} \end{aligned}$$

(D1)

In order to calculate the value of $\bar{\sigma }_{H}$ we decompose the coordinates $x_1, x_2$ as follows:

$$\begin{aligned} \hat{x}_1= & {} -\frac{\hat{p}_2-{{\mathcal {B}}}x_1}{{\mathcal {B}}} + \hat{{\mathcal {X}}}_1 = \hat{\xi }_1 + \hat{{\mathcal {X}}}_1,\\ \hat{x}_2= & {} \frac{\hat{p}_1}{{\mathcal {B}}} + \hat{{\mathcal {X}}}_2= \hat{\xi }_2 + \hat{{\mathcal {X}}}_2\,. \end{aligned}$$

The commutation relations follow:

$$\begin{aligned}&[\hat{\xi }_1,\hat{\xi }_2] = \frac{i}{{\mathcal {B}}}\,, \\&[\hat{{\mathcal {X}}}_1,\hat{{\mathcal {X}}}_2] =-\frac{i}{{\mathcal {B}}}\,,\\&[{{\mathcal {H}}}, \xi _1] = -i \frac{\partial }{\partial p_1} {{\mathcal {H}}}\,, \quad \\&[{{\mathcal {H}}}, \xi _2] = -i \frac{\partial }{\partial p_2} {{\mathcal {H}}}\,,\\&[{{\mathcal {H}}}, \hat{{\mathcal {X}}}_1] = [{{\mathcal {H}}}, \hat{{\mathcal {X}}}_2] = 0\,. \end{aligned}$$

Here we use that the Hamiltonian contains the following dependence on x:

$$\begin{aligned} {{\mathcal {H}}}(\hat{p}_1, \hat{p}_2-{{\mathcal {B}}} x_1)\, \end{aligned}$$

and $\frac{\partial ^2}{\partial p_1 \partial p_2}{{\mathcal {H}}}=0$. Thus we obtain:

$$\begin{aligned} \bar{\sigma }_H= & {} -\frac{\mathrm {i}}{2 {{\mathcal {V}}}}\, \epsilon ^{ij} \, \Big (\sum _{n,k|{{\mathcal {E}}}_n\ne {{\mathcal {E}}}_k} \frac{ f({{\mathcal {E}}}_k)-f({{\mathcal {E}}}_n) + ({{\mathcal {E}}}_n-{{\mathcal {E}}}_k) f'({{\mathcal {E}}}_n)}{({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)^2} \nonumber \\&+ \frac{1}{2}\sum _{n,k|{{\mathcal {E}}}_n={{\mathcal {E}}}_k}f^{\prime \prime }({{\mathcal {E}}}_n)\Big )\langle n| [{{\mathcal {H}}}, {{\hat{\xi }}}_i] | k \rangle \langle k | [{{\mathcal {H}}}, {{\hat{\xi }}}_j] | n \rangle \nonumber \\= & {} \frac{\mathrm {i}}{2 {{\mathcal {V}}}}\, \epsilon ^{ij} \, \Bigl (- \sum _{n,k|{{\mathcal {E}}}_n\ne {{\mathcal {E}}}_k} 2f({{\mathcal {E}}}_n)\nonumber \\&+ \frac{1}{2}\sum _{n,k|{{\mathcal {E}}}_n={{\mathcal {E}}}_k}f^{\prime \prime }({{\mathcal {E}}}_n)({{\mathcal {E}}}_k-{{\mathcal {E}}}_n)^2 \Bigr )\,\nonumber \\&\Big [ \langle n| {{\hat{\xi }}}_i | k \rangle \langle k | {{\hat{\xi }}}_j | n \rangle \Big ] \nonumber \\= & {} \frac{\mathrm {i}}{2 {{\mathcal {V}}}}\, \epsilon ^{ij} \, \Bigl (-\sum _{n,k} 2f({{\mathcal {E}}}_n) \nonumber \\&+ \sum _{n,k|{{\mathcal {E}}}_n={{\mathcal {E}}}_k} 2f({{\mathcal {E}}}_n)\Bigr ) \Big [ \langle n| {{\hat{\xi }}}_i | k \rangle \langle k | {{\hat{\xi }}}_j | n \rangle \Big ] \nonumber \\= & {} \frac{\mathrm {i}}{2 {{\mathcal {V}}}}\, \sum _{n} \nonumber \\&\Bigl ( -2f({{\mathcal {E}}}_n)\Bigr )\,\Big [ \langle n| [{{\hat{\xi }}}_1, {{\hat{\xi }}}_2] | n \rangle \Big ] \nonumber \\= & {} \frac{1 }{{{\mathcal {B}}} {{\mathcal {V}}}}\, \sum _{n} f({{\mathcal {E}}}_n)\, \langle n| n \rangle = \frac{\rho }{{\mathcal {B}}}. \end{aligned}$$

(D2)

Here $\rho$ is density of electrons. We used here that momentum $p_2$ is a good quantum number, and it enumerates the eigenstates of the Hamiltonian:

$$\begin{aligned}&{{\mathcal {H}}} |n\rangle = {{\mathcal {H}}}(\hat{p}_1, p_y-{{\mathcal {B}}} x_1)|p_2, q\rangle = {{\mathcal {E}}}_{q}|p_2, q\rangle , \quad q\in \mathbb {N}\,. \end{aligned}$$

We assume that the size of the system is $L\times L$. Properties of the eigenstates of Hamiltonian guarantee that $\langle p^\prime _2, q| \hat{p}_1|p_2, q\rangle =0$ for $p^\prime _2 \ne p_2$. This gives

$$\begin{aligned} \bar{\sigma }_H= & {} \,\sum _{q}\int \frac{dp_2 L}{2\pi {{\mathcal {V}}}} \, \frac{ f({{\mathcal {E}}}_q)}{{\mathcal {B}}}. \end{aligned}$$

(D3)

$\langle x_1 \rangle = p_2/{{\mathcal {B}}}$ plays the role of the center of orbit, and this center should belong to the interval $(-L/2, L/2)$ while ${{\mathcal {E}}}_q$ does not depend on momentum. This gives

$$\begin{aligned} \bar{\sigma }_H= & {} \frac{1}{2\pi } \,\sum _{q = 0,1,...} \, f({{\mathcal {E}}}_q). \end{aligned}$$

(D4)

In case of thermal equilibrium this expression receives the form:

$$\begin{aligned} \bar{\sigma }_H= & {} \frac{1}{2\pi } \,\sum _{q=0,1,...} \, \frac{1}{e^{{{\mathcal {E}}}_q/T}+1}. \end{aligned}$$

(D5)

Here

$$\begin{aligned} {{\mathcal {E}}}_q = \frac{{\mathcal {B}}}{2m} (2q+1)-\mu \end{aligned}$$

where $\mu$ is chemical potential. One can see, that at $T \ll \frac{{\mathcal {B}}}{m}$ this expression is reduced to the zero temperature expression $\bar{\sigma }_H = \frac{N}{2\pi }$, where N is the number of occupied Landau Levels.

Appendix E: Robustness of ${{\mathcal {N}}}_f$ with respect to modification of one-particle Hamiltonian

Let us consider the following quantity for the 2+1 dimensional system

$$\begin{aligned}&{{\mathcal {N}}} = \left ( \frac{\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\right ) \end{aligned}$$

(E1)

We consider the static system with distribution function that depends only on $\pi _0$. Variation of ${\mathcal {N}}$ caused by a variation of $\hat{Q}$ (such that $\delta f(\pi _0)=0$) gives:

$$\begin{aligned} {\delta } {{\mathcal {N}}}= & {} \Big ( \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \delta \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big )\nonumber \\&+ \Big ( \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \partial _{\pi _{i}}\delta \hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0\star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big )\nonumber \\= & {} \Big ( -\frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \delta \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \partial _{\pi _{j}}\hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big )\nonumber \\&+ \Big ( -\frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \partial _{\pi _{i}}\delta \hat{Q}_0 \star \partial _{\pi _{j}}\hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big )\nonumber \\= & {} \Big (- \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \delta \hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big )\nonumber \\&+ \Big ( \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \nonumber \\&\mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \hat{G}_0 \star \delta \hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\Big ) \end{aligned}$$

(E2)

Here similar to the case of quantities considered in Appendix A and Appendix 1 the above expressions obey the cyclic property provided that function $f(\pi _0)$ remains unchanged and the contribution proportional to its derivative may be neglected. Further we simplify these expressions and obtain:

$$\begin{aligned} {\delta } {{\mathcal {N}}} &= \left(- \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \delta \hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\right)\nonumber \\& \quad + \left ( \frac{3\epsilon _{ijk}}{48 \pi ^2 \, {{\mathcal {V}}}} \int {d^3\pi d^2 x} \, \mathop {\mathrm{tr}}\nolimits \left( \hat{G}_0 \star \delta \hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{j}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{k}}\hat{Q}_0 \star \hat{G}_0 \star \partial _{\pi _{i}}\hat{Q}_0 \right) ^< +\mathrm{c.c.}\right )=0 \end{aligned}$$

(E3)

Notice that the property proven here is not the complete topological invariance unlike the case of equilibrium at $T=0$. The value of ${\mathcal {N}}$ may be changed smoothly under the change of distribution function $f(\pi _0)$. Moreover, its value is modified also when $f(\pi _0)$ remains unchanged but the terms in ${\mathcal {N}}$ proportional to the derivative of f gives valuable contribution.

Appendix F: Hall conductivity for the system of massive 2D Dirac fermions

The system of massive 2D Dirac fermions in equilibrium at zero temperature has been concerned in Sect. 5.2. It corresponds to

$$\begin{aligned} Q = {\varvec{1}} \omega -v_F (\sigma ^1 \pi _1 + \sigma ^2 \pi _2 + \sigma ^3 m). \end{aligned}$$

Here m is a mass-type parameter, $\sigma _i$ are Pauli matrices, and ${\varvec{1}}$ is a unit $2\times 2$ matrix. In equilibrium at $T=0$ Hall conductivity is given by

$$\begin{aligned} \sigma _H = {-}\frac{{{\mathcal {N}}}}{2\pi }. \end{aligned}$$

with [83]

$$\begin{aligned} {{\mathcal {N}}}^{(0)} = \frac{1}{2} \mathrm{sign} \, m. \end{aligned}$$

Recall that for purely two-dimensional systems these fermions always come in pairs, and the total value of ${\mathcal {N}}$ is integer rather than half-integer.

Now let us calculate corrections to $\sigma _H$ at finite temperatures using the developed formalism. For simplicity we consider the case when Fermi energy is set to zero. Our starting point is Eq. (C14) for Hall conductivity. The Hamiltonian can be written as follows:

$$\begin{aligned} H = v_F \left( \begin{array}{cc}m &{} p_1 - \mathrm {i}p_2 \\ p_1 +\mathrm {i}p_2 &{} - m \end{array}\right) \end{aligned}$$

In the following for simplicity we will consider the case $v_F = 1$ (the nontrivial value of Fermi velocity may be easily restored in the final answer). The eigenvalues of this Hamiltonian are $E_\pm ({p}) =\pm \sqrt{|{p}|^2+m^2}$. The corresponding eigenvectors are

$$\begin{aligned} |n \rangle\equiv & {} |a {p}\rangle = \frac{|{p}|}{\sqrt{2\sqrt{|{ p}|^2+m^2} \left(\sqrt{|{ p}|^2+m^2}+ a m\right)}} &\left( \begin{array}{c} 1 \\ - \frac{m -a \sqrt{|{ p}|^2+m^2}}{p_x +\mathrm {i}p_y} \end{array}\right) |{ p}\rangle \end{aligned}$$

here $a= \pm 1$, with $+1$ corresponding to the conductance band with positive energy while $-1$ marks valence band with negative energy. Momenta eigenstates are normalized to 1 in discrete space, $\langle {q}|{p}\rangle = (2\pi )^2 \delta ^{(2)}({q}-{p})/{{\mathcal {V}}}$.

Since velocity operator $\hat{v}_k = \sigma _k$ does not contain momentum, its matrix elements between the states with definite momenta p and q contain a delta-function $\delta ^{(2)}({p-q})$. In Eq. (C14) each of two sums over the quantum states is to be substituted by ${{\mathcal {V}}}\sum _{a \in \{c,v\}} \int \frac{d^2 p}{(2\pi )^2}$. We also denote

$$\begin{aligned}&\langle a,{p}|\hat{v}_i|b,{ p}^\prime \rangle \langle b,{ p}^\prime |\hat{v}_j|a,{p}\rangle \\&\quad = \langle a,b;i,j;{p} \rangle \frac{(2\pi )^2}{{\mathcal {V}}} \delta ^{(2)}({p}-{p}^\prime ) \frac{(2\pi )^2}{{\mathcal {V}}} \delta ^{(2)}({p}-{p}^\prime ) \\&\quad = \langle a,b;i,j;{p} \rangle \frac{(2\pi )^2}{{\mathcal {V}}} \delta ^{(2)}({p}'-{p}). \end{aligned}$$

In the last equality we used the fact that the factor $\frac{(2\pi )^2}{{\mathcal {V}}} \delta ^{(2)}(0)$ is to be replaced by unity, which becomes clear if we consider the system inside a large but finite rectangular box with periodic boundary conditions and replace the integral over momentum by the sum over its discrete values. Furthermore, it is easy to see that $\langle a,b;i,j; {p}\rangle =\langle a,b;j,i; {p}\rangle$. Using these notations we may rewrite Eq. (C14) as follows

$$\begin{aligned} \sigma _H= & {} {+} \frac{\mathrm {i}}{4}\int \frac{d^2p}{(2\pi )^2} \nonumber \\&\sum _{a,b = \pm 1} \Big [ \frac{f(E_a({p}))-f(E_b({p}))}{(E_a({p})-E_b({p}))^2} \Big ] \nonumber \\&\epsilon _{ij} \langle a,b;i,j;{p}\rangle +c.c., \end{aligned}$$

(F1)

where f is Fermi distribution.

Using the given above explicit expressions for the 2D Dirac spinors, we obtain $\epsilon _{ij} \langle +,-;i,j;{p}\rangle =2\mathrm {i}\mathrm{Im} \langle +,-;1,2;{p}\rangle =2\mathrm {i}m/\sqrt{{p}^2+m^2}$. We represent $\sigma _H = I/(8\pi ^2)$. Here

$$\begin{aligned} I= & {} -\mathrm {i}\int d^2p \sum _{a,b} \frac{f(E_a({p}))-f(E_b({p}))}{(E_a({p})-E_b({p}))^2} \epsilon _{ij} \langle a,b;i,j; {p}\rangle \nonumber \\= & {} -2\mathrm {i}\int d^2p \frac{f(E_+({p}))-f(E_-({p}))}{(E_+({p})-E_-({p}))^2} \epsilon _{ij} \langle c,v;i,j; {p}\rangle \nonumber \\= & {} -2\mathrm {i}\int d^2p \Big (\frac{1}{e^{-\beta \sqrt{{p}^2+m^2}}+1}-\frac{1}{e^{\beta \sqrt{{p}^2+m^2}}+1} \Big )\nonumber \\&\frac{1}{4({p}^2+m^2)}\frac{2\mathrm {i}m}{\sqrt{{p}^2+m^2}} \nonumber \\= & {} \int d^2p \frac{m}{({p}^2+m^2)^{3/2}}\, \mathrm{th}\left( \frac{\beta \sqrt{{p}^2+m^2}}{2} \right) \end{aligned}$$

(F2)

Then, after changing variables we obtain the following expression for Hall conductivity:

$$\begin{aligned} \sigma _H= & {} {-} \frac{\alpha }{4\pi } \int _{|\alpha |}^{+\infty } \frac{du}{u^2 \, \mathrm{th}(u/2)}. \end{aligned}$$

(F3)

where $\alpha =v_F\beta m\equiv v_F m/T$ is dimensionless (we restored here Fermi velocity). This expression tends to ${-}\frac{1}{4\pi }\mathrm{sign}\,m$ at $T\rightarrow 0$.