Quantum Oscillations of Interlayer Conductivity in a Multilayer Topological Insulator

Abstract

Quantum and difference oscillations of interlayer conductivity in a multilayer system of thin films of topological insulators (TIs) are investigated. Due to the linearity of the carrier spectrum in such a system, new features of quantum oscillations arise. In particular, the frequencies of de Haas–van Alfvén and Shubnikov–de Haas oscillations depend quadratically on the chemical potential, rather than linearly as in systems with parabolic carrier spectrum. For the same reason, the temperature damping factor of oscillations contains the chemical potential. This is due to the nonequidistant character of the Landau levels: the higher the chemical potential, the smaller the distance between Landau levels. However, the beat frequencies, as well as the frequencies of slow oscillations, do not depend on the chemical potential; in this sense, the behavior of these systems is similar to that of conventional non-Dirac systems. Finally, in the Born approximation (in the second order cross-diagram technique), we considered the general case when the interlayer conductivity takes into account both intra- and interband transitions. We have shown that the contribution of intraband transitions is insignificant for the conductivity oscillations in the absence of magnetic impurities. However, in the presence of a Dirac point in the spectrum, a linear (in magnetic field) intraband contribution to conductivity arises from the zero Landau level. At low temperatures, this contribution is exponentially small compared to the intraband contribution and vanishes at zero temperature.

Appendices

APPENDIX A

Hamiltonian and Spectrum

The Hamiltonian of a heterostructure consisting of topological insulators and conventional dielectric films can be expressed as (see [58])

$$\begin{gathered} \mathcal{H} = \sum\limits_{{{{\mathbf{k}}}_{ \bot }},i,j}^{} {c_{{{{{\mathbf{k}}}_{ \bot }}i}}^{\dag }\left[ {{{{v}}_{F}}{{\tau }^{z}}(\hat {z} \times {\boldsymbol{\sigma }}){{{\mathbf{k}}}_{ \bot }}{{\delta }_{{i,j}}} + {{\Delta }_{S}}{{\tau }^{x}}{{\delta }_{{i,j}}}_{{_{{_{{_{{_{{_{{_{{_{{_{{}}}}}}}}}}}}}}}}}}} \right.} \\ \left. { + \,\,\frac{1}{2}{{\Delta }_{D}}{{\tau }^{ + }}{{\delta }_{{j,i + 1}}} + \frac{1}{2}{{\Delta }_{D}}{{\tau }^{ - }}{{\delta }_{{j,i - 1}}}} \right]{{c}_{{{{{\mathbf{k}}}_{ \bot }}j}}}. \\ \end{gathered} $$

(A.1)

Here $\hat {z}$ is the unit vector in the Z direction. Then

$$[\hat {z} \times {\boldsymbol{\sigma }}] \cdot {{{\mathbf{k}}}_{ \bot }} = - {{\sigma }_{y}}{{k}_{x}} + {{\sigma }_{x}}{{k}_{y}},$$

which is the Hamiltonian of chiral Dirac electrons on one of the surfaces of the topological insulator. The expression [$\hat {z}$ × σ] ⋅ k_⊥ corresponds to chirality, which can be easily understood by the left-hand rule: place the left hand palm down on a table. Then the four stretched fingers indicate the direction of the momentum, and the thumb bent at 90° shows the direction of the spin. This orientation corresponds to one chirality. The state with opposite chirality can be represented similarly, except instead of the left hand one should take the right hand. The two TI surfaces are taken into account by the Pauli matrix τ^z, which contains two values of chirality. The indices i and j number the layers of the topological insulator. The term with Δ_S corresponds to hopping between the upper and lower surfaces of the same layer. The Hamiltonian is written in the basis of the states (1 0)^T and (0 1)^T, which indicate membership in the upper and lower surfaces of a given layer of the topological insulator. In this basis, the mixing of surfaces (hopping between them) is given by the Pauli matrix τ^x:

$${{\tau }^{x}}\left( \begin{gathered} 1 \\ 0 \\ \end{gathered} \right) = \left( \begin{gathered} 0 \\ 1 \\ \end{gathered} \right),\quad {{\tau }^{x}}\left( \begin{gathered} 0 \\ 1 \\ \end{gathered} \right) = \left( \begin{gathered} 1 \\ 0 \\ \end{gathered} \right).$$

The delta function δ_{i, j} in this term indicates that the hopping occurs within a single layer. The remaining terms describe hopping between adjacent layers of the TI through the dielectric layers. The term with τ⁺ denotes hopping from the lower surface of a given layer to the upper surface of the adjacent lower layer: the symbol δ_{j, i + 1} is different from zero for the transitions j → i = j – 1. This is clear from the explicit form of the matrix

$$\frac{1}{2}{{\tau }^{ + }} = \left( {\begin{array}{*{20}{c}} 0&1 \\ 0&0 \end{array}} \right),$$

which maps the state (0 1)^T (corresponding to the lower surface) to (1 0)^T (corresponding to the upper surface), but not vice versa. The term with τ^– denotes hopping from the upper surface of a given layer to the lower surface of the adjacent upper layer: the symbol δ_{j, i – 1} is different from zero for the transitions j → i = j + 1. The matrix

$$\frac{1}{2}{{\tau }^{ - }} = \left( {\begin{array}{*{20}{c}} 0&0 \\ 1&0 \end{array}} \right)$$

maps the state (1 0)^T to (0 1)^T, but not vice versa.

Let us find the spectrum corresponding to such a Hamiltonian. To this end, we represent all delta symbols in terms of Fourier transforms:

$$\begin{gathered} \mathcal{H} = \frac{1}{N}\sum\limits_{{{{\mathbf{k}}}_{ \bot }},i,j,{{k}_{z}}}^{} {{{e}^{{i{{k}_{z}}({{r}_{i}} - {{r}_{j}})}}}c_{{{{{\mathbf{k}}}_{ \bot }}i}}^{\dag }} \left[ {{{{v}}_{F}}{{\tau }^{z}}(\hat {z} \times {\boldsymbol{\sigma }}){{{\mathbf{k}}}_{ \bot }}_{{_{{_{{_{{_{{_{{_{{_{{_{{}}}}}}}}}}}}}}}}}}} \right. \\ \left. { + \,{{\Delta }_{S}}{{\tau }^{x}} + \frac{1}{2}{{\Delta }_{D}}{{\tau }^{ + }}{{e}^{{i{{k}_{z}}d}}} + \frac{1}{2}{{\Delta }_{D}}{{\tau }^{ - }}{{e}^{{ - i{{k}_{z}}d}}}} \right]{{c}_{{{{{\mathbf{k}}}_{ \bot }}j}}}, \\ \end{gathered} $$

(A.2)

where we used the representation

$${{\delta }_{{i,j}}} = \frac{1}{N}\sum\limits_{{{k}_{z}}}^{} {{{e}^{{i{{k}_{z}}({{r}_{i}} - {{r}_{j}})}}}} .$$

Here r_{i ± 1} = r_i ± d, and N is the number of layers. Accordingly, in the momentum space we have

$${{\mathcal{H}}_{k}} = \left( {\begin{array}{*{20}{c}} {{{{v}}_{F}}(\hat {z} \times {\boldsymbol{\sigma }}){{{\mathbf{k}}}_{ \bot }}}&{{{\Delta }_{S}} + {{\Delta }_{D}}{{e}^{{i{{k}_{z}}d}}}} \\ {{{\Delta }_{S}} + {{\Delta }_{D}}{{e}^{{ - i{{k}_{z}}d}}}}&{ - {{{v}}_{F}}(\hat {z} \times {\boldsymbol{\sigma }}){{{\mathbf{k}}}_{ \bot }}} \end{array}} \right).$$

The spectrum of such a Hamiltonian is easily found:

$${{\varepsilon }_{k}} = \pm \sqrt {{v}_{F}^{2}(k_{x}^{2} + k_{y}^{2}) + {{\Delta }^{2}}({{k}_{z}})} ,$$

(A.3)

where

$${{\Delta }^{2}}({{k}_{z}}) = \Delta _{S}^{2} + \Delta _{D}^{2} - 2{{\Delta }_{S}}{{\Delta }_{D}}\cos {{k}_{z}}d.$$

The Hamiltonian has SO(3) symmetry:

$${{e}^{{i{\boldsymbol{\tau }} \cdot \phi }}}{{\mathcal{H}}_{k}}{{e}^{{ - i{\boldsymbol{\tau }} \cdot \phi }}} = {{\mathcal{H}}_{k}}.$$

Therefore, it can be represented in the block diagonal form ${{\mathcal{H}}_{k}}$ = τ^z${{\tilde {\mathcal{H}}}_{k}}$, where

$${{\tilde {\mathcal{H}}}_{k}} = {{{v}}_{F}}(\hat {z} \times {\boldsymbol{\sigma }}){{{\mathbf{k}}}_{ \bot }} + {{\sigma }_{z}}\Delta ({{k}_{z}}).$$

This means that it is sufficient to deal with one of the blocks of this form.

APPENDIX B

Density of States, Thermodynamic Potential, and Magnetization

We have formula (5) for the density of states. Let us apply the Poisson formula

$$\begin{gathered} f(0) + 2\sum\limits_{n = 1}^{} {f(n)} \\ = 2\int\limits_0^\infty {f(x)dx + 4\operatorname{Re} \sum\limits_{k = 1}^\infty {\int\limits_0^\infty {f(x){{e}^{{2\pi ikx}}}dx.} } } \\ \end{gathered} $$

(B.1)

Consider separately the oscillating and nonoscillating parts. For the nonoscillating part we have

$${{\rho }_{0}}(\varepsilon ) = \frac{L}{\pi }{{N}_{{LL}}}\int\limits_{ - \frac{\pi }{d}}^{\frac{\pi }{d}} {d{{k}_{z}}\sum\limits_{\alpha = \pm }^{} {\int\limits_0^\infty {\delta (\varepsilon - {{\varepsilon }_{x}})dx,} } } $$

(B.2)

where N_LL = 2/π$l_{H}^{2}$ is the degeneracy of the Landau levels. Summing over the band index, we obtain

$$\begin{gathered} {{\rho }_{0}}(\varepsilon ) = \frac{{2L{{N}_{{LL}}}}}{\pi }{\text{|}}\varepsilon {\text{|}}\int\limits_{ - \frac{\pi }{d}}^{\frac{\pi }{d}} {d{{k}_{z}}} \\ \, \times \int\limits_0^\infty {\delta ({{\varepsilon }^{2}} - 2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x - {{\Delta }^{2}}({{k}_{z}}))dx.} \\ \end{gathered} $$

(B.3)

Integrating first with respect to k_z, we obtain

$$\begin{gathered} {{\rho }_{0}}(\varepsilon ) = 2N{{N}_{{LL}}}\frac{{{\text{|}}\varepsilon {\text{|}}\Theta ({{\varepsilon }^{2}} - {{\Delta }^{2}})}}{{\pi {{\Delta }_{S}}{{\Delta }_{D}}}} \\ \times \int\limits_{{{x}_{{\min }}}}^{{{x}_{{\max }}}} {\frac{{dx}}{{\sqrt {1 - \frac{{{{{(2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x + \Delta _{S}^{2} + \Delta _{D}^{2} - {{\varepsilon }^{2}})}}^{2}}}}{{{{{(2{{\Delta }_{S}}{{\Delta }_{D}})}}^{2}}}}} }}} , \\ \end{gathered} $$

where

$${{x}_{{\min }}} = \frac{{{{\varepsilon }^{2}} - {{{({{\Delta }_{S}} + {{\Delta }_{D}})}}^{2}}}}{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}},$$

$${{x}_{{\max }}} = \frac{{{{\varepsilon }^{2}} - {{{({{\Delta }_{S}} - {{\Delta }_{D}})}}^{2}}}}{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}},$$

Δ = |Δ_S – Δ_D|, and N = L/d is the number of layers. The density of states contains singular points, which correspond to the minimum and maximum of the cosine profile of the Landau levels. For example, for the zero level, these are the points $\varepsilon _{ \pm }^{2}$ = (Δ_S ± Δ_D)². These are nothing but van Hove singularities. Indeed, the spectrum becomes flat near its extrema, which leads to a singularity of the density of states due to the degeneracy of the Landau levels. To integrate with respect to x, we pass to a new variable:

$$\frac{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x + \Delta _{S}^{2} + \Delta _{D}^{2} - {{\varepsilon }^{2}}}}{{2{{\Delta }_{S}}{{\Delta }_{D}}}} = \cos y,$$

$$dx = \frac{{ - {{\Delta }_{S}}{{\Delta }_{D}}\sin ydy}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}.$$

After straightforward integration, we obtain formula (7).

Now, consider the oscillating part

$$\begin{gathered} {{\rho }_{{{\text{osc}}}}}(\varepsilon ) = 2N{{N}_{{LL}}}\frac{{{\text{|}}\varepsilon {\text{|}}}}{\pi }\operatorname{Re} \sum\limits_{k = 1}^\infty {\int\limits_{ - \pi }^\pi {d{{k}_{z}}} } \\ \times \,\,\int\limits_0^\infty {\delta ({{\varepsilon }^{2}} - 2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x - {{\Delta }^{2}}({{k}_{z}})){{e}^{{2\pi ikx}}}dx.} \\ \end{gathered} $$

(B.4)

First, we integrate with respect to x:

$$\begin{gathered} {{\rho }_{{{\text{osc}}}}}(\varepsilon ) = 2N{{N}_{{LL}}}\frac{{{\text{|}}\varepsilon {\text{|}}}}{\pi }\operatorname{Re} \sum\limits_{k = 1}^\infty {\int\limits_{ - \pi }^\pi {d{{k}_{z}}} } \\ \times \,\,\exp \left[ {2\pi ik\frac{{{{\varepsilon }^{2}} - {{\Delta }^{2}}({{k}_{z}})}}{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right]\Theta ({{\varepsilon }^{2}} - {{\Delta }^{2}}({{k}_{z}})). \\ \end{gathered} $$

(B.5)

Integration with respect to k_z finally yields expression (8).

Now, we obtain an expression for the oscillating part of the thermodynamic potential. Substituting the expression for the density of states into (10), we obtain

$$\begin{gathered} {{\Omega }_{{{\text{osc}}}}} = - 2N{{N}_{{LL}}}\frac{T}{\pi }\operatorname{Re} \sum\limits_{k = 1}^\infty {\int\limits_0^\infty {\varepsilon d\varepsilon } } \\ \times \,\,\exp \left[ {\pi ik\frac{{{{\varepsilon }^{2}} - \Delta _{S}^{2} - \Delta _{D}^{2}}}{{{v}_{F}^{2}l_{H}^{2}{{\hbar }^{2}}}}} \right]\mathcal{J}(k,\varepsilon )\ln \left( {1 + {{e}^{{\frac{{\mu - \varepsilon }}{T}}}}} \right). \\ \end{gathered} $$

(B.6)

Introduce intermediate notations:

$$\alpha = \frac{{\pi k}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}},\quad \beta = \pi k\frac{{\Delta _{S}^{2} + \Delta _{D}^{2}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}.$$

Integration with respect to energy yields the integral

$$\begin{gathered} I = \int\limits_0^\infty {\varepsilon \exp [i(\alpha {{\varepsilon }^{2}} - \beta )]} \\ \times \,\,\Theta ({{\varepsilon }^{2}} - {{\Delta }^{2}}({{k}_{z}}))\ln \left( {1 + {{e}^{{\frac{{\mu - \varepsilon }}{T}}}}} \right)d\varepsilon . \\ \end{gathered} $$

(B.7)

Let us represent the integral as

$$I = \frac{1}{2}\int\limits_{\Delta ({{k}_{z}})}^\infty {\exp [i(\alpha {{\varepsilon }^{2}} - \beta )]\ln \left( {1 + {{e}^{{\frac{{\mu - \varepsilon }}{T}}}}} \right)d{{\varepsilon }^{2}}} $$

(B.8)

and first integrate by parts:

$$\begin{gathered} I = - \frac{{{{e}^{{ - i\beta }}}}}{{2i\alpha }}\left[ {\frac{{\mu - \Delta ({{k}_{z}})}}{T}{{e}^{{i\alpha {{\Delta }^{2}}({{k}_{z}})}}} - \frac{1}{T}\int\limits_{\Delta ({{k}_{z}})}^\infty {\frac{{{{e}^{{i\alpha {{\varepsilon }^{2}}}}}d\varepsilon }}{{{{e}^{{\frac{{\varepsilon - \mu }}{T}}}} + 1}}} } \right] \\ = - \frac{{{{e}^{{ - i\beta }}}}}{{2i\alpha }}\left[ {\frac{{\mu - \Delta ({{k}_{z}})}}{T}{{e}^{{i\alpha {{\Delta }^{2}}({{k}_{z}})}}} - \int\limits_{ - \infty }^\infty {\frac{{{{e}^{{i\alpha {{{(yT + \mu )}}^{2}}}}}dy}}{{{{e}^{y}} + 1}}} } \right], \\ \end{gathered} $$

(B.9)

where we replaced the lower limit by –∞, because we use the approximation $\frac{{\mu - \Delta }}{T}$ ≫ 1.

First of all we deal with the first term. Integrating with respect to k_z in the thermodynamic potential, we find

$$\begin{gathered} {{e}^{{ - i\beta }}}\int\limits_0^\pi {d{{k}_{z}}\exp \left[ {2\pi ik\frac{{{{\Delta }_{S}}{{\Delta }_{D}}\cos {{k}_{z}}d}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right]} {{e}^{{i\alpha {{\Delta }^{2}}({{k}_{z}})}}} \\ \, \times (\mu - \Delta ({{k}_{z}})) = {{e}^{{ - i\beta }}}\int\limits_0^\pi {d{{k}_{z}}(\mu - \Delta ({{k}_{z}})){{e}^{{i\beta }}}} \\ = \int\limits_0^\pi {d{{k}_{z}}(\mu - \Delta ({{k}_{z}})).} \\ \end{gathered} $$

This integral is real. Hence, the contribution of the first term of (47) to the thermodynamic potential vanishes.

Next, consider the integral

$$\begin{gathered} \int\limits_{ - \infty }^\infty {\frac{{\exp (i\alpha {{{(yT + \mu )}}^{2}})dy}}{{{{e}^{y}} + 1}}} \\ = \, - {\kern 1pt} 2\pi i\sum\limits_{n = 0}^\infty {{{e}^{{ - 2\pi (2n + 1)\alpha T\mu }}}} {\text{exp}}(i\alpha ({{\mu }^{2}}\, + \,{{(2n\, + \,1)}^{2}}{{\pi }^{2}}{{T}^{2}})), \\ \end{gathered} $$

(B.10)

where we applied the method of residues. Hence

$$\begin{gathered} I = - \frac{\pi }{\alpha }\sum\limits_{n = 0}^\infty {{{e}^{{ - 2\pi (2n + 1)\alpha T\mu }}}} \\ \times \exp (i\alpha ({{\mu }^{2}} - \Delta _{S}^{2} - \Delta _{D}^{2} + {{(2n + 1)}^{2}}{{\pi }^{2}}{{T}^{2}})). \\ \end{gathered} $$

Substituting this expression into the thermodynamic potential, integrating with respect to k_z, and taking into account that

$$\int\limits_0^\pi {d{{k}_{z}}\exp \left[ {2\pi ik\frac{{{{\Delta }_{S}}{{\Delta }_{D}}\cos {{k}_{z}}d}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right] = {{J}_{0}}\left( {2\pi k\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} ,$$

we finally obtain

$$\begin{gathered} {{\Omega }_{{{\text{osc}}}}} = 4N{{N}_{{LL}}}T\sum\limits_{k = 1}^\infty {\frac{{{{J}_{0}}\left( {2\pi k\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)}}{k}} \\ \times \,\,\sum\limits_{n = 0}^\infty {\exp \left( { - 2{{\pi }^{2}}\frac{{(2n + 1)kT\mu }}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} \\ \times \,\,\cos \left( {\frac{{\pi k}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}({{\mu }^{2}} - \Delta _{S}^{2} - \Delta _{D}^{2} + {{{(2n + 1)}}^{2}}{{\pi }^{2}}{{T}^{2}})} \right). \\ \end{gathered} $$

(B.11)

Due to the rapid convergence of the series in n, we neglect the dependence of the oscillating part on n. Then the series is easily summed, and we obtain expression (19).

APPENDIX C

Interlayer Transport

Let us analyze the behavior of the function I(ε). For the velocity ${{{v}}_{z}}$ in the spectrum we obtain

$${{{v}}_{z}} = \frac{{\partial {{\varepsilon }_{{n,{{k}_{z}}}}}}}{{\hbar \partial {{k}_{z}}}} = - \frac{{d{{\Delta }_{S}}{{\Delta }_{D}}}}{\hbar }\frac{{\sin {{k}_{z}}d}}{{\sqrt {2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}n + {{\Delta }^{2}}({{k}_{z}})} }}.$$

(C.1)

Using (2), we obtain

$$\begin{gathered} {{{v}}_{z}} = \frac{{\partial {{\varepsilon }_{{n,{{k}_{z}}}}}}}{{\hbar \partial {{k}_{z}}}} \\ = - \frac{d}{{2\hbar }}\frac{{\sqrt {4\Delta _{S}^{2}\Delta _{D}^{2} - {{{({{\epsilon }^{2}} - 2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}n - \Delta _{S}^{2} - \Delta _{D}^{2})}}^{2}}} }}{\epsilon }. \\ \end{gathered} $$

(C.2)

Applying the Poisson formula and summing over α, we obtain

$$\begin{gathered} {{I}_{0}}(\epsilon ) = {{N}_{{LL}}}\frac{L}{{2\pi d}}\frac{{{{d}^{2}}}}{{4{{\hbar }^{2}}}}\int\limits_{ - \pi }^\pi {dz} \int\limits_0^\infty {dx} \\ \times \,\,\frac{{4\Delta _{S}^{2}\Delta _{D}^{2} - {{{({{\epsilon }^{2}} - 2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x - \Delta _{S}^{2} - \Delta _{D}^{2})}}^{2}}}}{\epsilon } \\ \times \,\,2\delta ({{\varepsilon }^{2}} - 2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x - {{\Delta }^{2}}({{k}_{z}})). \\ \end{gathered} $$

(C.3)

Integrating with respect to z, we find

$$\begin{gathered} {{I}_{0}}(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}{{\Delta }_{S}}{{\Delta }_{D}}}}{{2\pi {{\hbar }^{2}}}} \\ \times \,\,\int\limits_{{{x}_{{\min }}}}^{{{x}_{{\max }}}} {dx\frac{{\sqrt {1 - \frac{{{{{(2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x + \Delta _{S}^{2} + \Delta _{D}^{2} - {{\varepsilon }^{2}})}}^{2}}}}{{4\Delta _{S}^{2}\Delta _{D}^{2}}}} }}{\epsilon }} . \\ \end{gathered} $$

(C.4)

Passing to a new variable

$$\frac{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x + \Delta _{S}^{2} + \Delta _{D}^{2} - {{\varepsilon }^{2}}}}{{2{{\Delta }_{S}}{{\Delta }_{D}}}} = \cos y,$$

(C.5)

we obtain

$${{I}_{0}}(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}\Delta _{S}^{2}\Delta _{D}^{2}}}{{4{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{4}}\epsilon }}.$$

(C.6)

Now, consider the oscillating part

$$\begin{gathered} {{I}_{{{\text{osc}}}}}(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}{{\Delta }_{S}}{{\Delta }_{D}}}}{{\pi {{\hbar }^{2}}\epsilon }}\operatorname{Re} \int\limits_{{{x}_{{\min }}}}^{{{x}_{{\max }}}} {dx} \\ \times \,\,{{e}^{{2\pi iikx}}}\sqrt {1 - \frac{{{{{(2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x - \Delta _{S}^{2} - \Delta _{D}^{2} - {{\varepsilon }^{2}})}}^{2}}}}{{4\Delta _{S}^{2}\Delta _{D}^{2}}}} . \\ \end{gathered} $$

(C.7)

Passing to a new variable

$$\frac{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}x + \Delta _{S}^{2} + \Delta _{D}^{2} - {{\varepsilon }^{2}}}}{{2{{\Delta }_{S}}{{\Delta }_{D}}}} = \cos y,$$

(C.8)

we obtain

$$\begin{gathered} {{I}_{{{\text{osc}}}}}(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}\Delta _{S}^{2}\Delta _{D}^{2}}}{{\pi \epsilon {v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{4}}}} \\ \times \,\,\operatorname{Re} \sum\limits_{k = 1}^\infty {\exp \left( {2\pi ik\left( {\frac{{{{\varepsilon }^{2}} - \Delta _{S}^{2} - \Delta _{D}^{2}}}{{{\text{2}}{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} \right)} \\ \times \,\,\int\limits_0^\pi {\exp \left( {2\pi ik\left( {\frac{{{{\Delta }_{S}}{{\Delta }_{D}}\cos y}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} \right)} {{\sin }^{2}}ydy. \\ \end{gathered} $$

(C.9)

Using the formula

$$\begin{gathered} \int\limits_0^\pi {\exp \left( {2\pi ik\left( {\frac{{{{\Delta }_{S}}{{\Delta }_{D}}\cos y}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} \right)} {{\sin }^{2}}ydy \\ = \pi \left[ {{{J}_{0}}(a) + \frac{{{{d}^{2}}{{J}_{0}}(a)}}{{d{{a}^{2}}}}} \right] = \frac{\pi }{a}{{J}_{1}}(a), \\ \end{gathered} $$

(C.10)

$$a = 2\pi k\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}},$$

we obtain

$$\begin{gathered} {{I}_{{{\text{osc}}}}}(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}}}{{2\pi {{\hbar }^{2}}}}\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{\varepsilon }\sum\limits_{k = 1}^\infty {\frac{1}{k}} \\ \times \,\,\cos \left( {2\pi k\frac{{{{\varepsilon }^{2}} - \Delta _{S}^{2} - \Delta _{D}^{2}}}{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right){{J}_{1}}\left( {2\pi k\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right). \\ \end{gathered} $$

(C.11)

Thus,

$$\begin{gathered} I(\epsilon ) = N{{N}_{{LL}}}\frac{{{{d}^{2}}}}{{2\pi {{\hbar }^{2}}}}\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{\varepsilon }\left[ {\frac{{\pi {{\Delta }_{S}}{{\Delta }_{D}}}}{{2{{\hbar }^{2}}l_{H}^{{ - 2}}{v}_{F}^{2}}}} \right. \\ \left. { + \,\sum\limits_{k = 1}^\infty {\frac{1}{k}{\text{cos}}} \left( {2\pi k\frac{{{{\varepsilon }^{2}}\, - \,\Delta _{S}^{2}\, - \,\Delta _{D}^{2}}}{{2{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right){{J}_{1}}\left( {2\pi k\frac{{{{\Delta }_{S}}{{\Delta }_{D}}}}{{{v}_{F}^{2}l_{H}^{{ - 2}}{{\hbar }^{2}}}}} \right)} \right]. \\ \end{gathered} $$

(C.12)

APPENDIX D

Interlayer Conductivity with a Nondiagonal Matrix Element of the Velocity Operator

Let us obtain an expression for the static conductivity with the use of the Kubo formula,

$$\sigma = - \mathop {\lim }\limits_{\omega \to 0} \left( {\frac{1}{\omega }\operatorname{Im} {{P}^{R}}(\omega )} \right).$$

(D.1)

Here

$${{P}^{R}}(\omega ) = P(q = 0,{{\omega }_{n}}){{{\text{|}}}_{{i{{\omega }_{n}} \to \omega + i0}}}$$

(D.2)

is the Fourier component of the two-time retarded Green function, which is expressed in the imaginary time representation as

$$P(q = 0,{{\omega }_{n}}) = - \frac{1}{V}\int\limits_0^\beta {d\tau {{e}^{{i\hbar {{\omega }_{n}}\tau }}}\langle {{T}_{\tau }}[J(\tau ),J(0)]\rangle ,} $$

(D.3)

where the brackets 〈…〉 denote the Gibbs averaging and averaging over impurities.

Expressing the current operator of free particles in terms of the second quantization operators, we obtain

$$\begin{gathered} {{P}^{R}}({{\omega }_{n}}) = \frac{{{{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}k_{1}^{'}k_{2}^{'}{{\sigma }_{1}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{k_{1}^{'}k_{2}^{'}}}}} \\ \, \times K_{\omega }^{R}({{k}_{1}}{{\sigma }_{1}},{{k}_{2}}{{\sigma }_{1}};k_{1}^{'}{{\sigma }_{2}},k_{2}^{'}{{\sigma }_{1}}). \\ \end{gathered} $$

(D.4)

Here ${{K}_{{{{\omega }_{n}}}}}$(k₁σ₁, k₁σ₁; k₂σ₂, k₂σ₂) is the Fourier transform of the two-particle Green function

$${{K}_{\tau }}(1,2;1',2') = - \langle {{T}_{\tau }}[{{\bar {a}}_{1}}(\tau ){{a}_{2}}(\tau ){{\bar {a}}_{{1'}}}(0){{a}_{{2'}}}(0)]\rangle ,$$

(D.5)

where 1 = k₁, σ₁. In the first approximation, we can represent P^R(ω_n) by the diagram shown in Fig. 3. Then we obtain

$${{P}^{R}}({{\omega }_{n}})\, = \,\frac{{2{{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} \int\limits_0^\beta {d\tau {{e}^{{i\hbar {{\omega }_{n}}\tau }}}} G({{k}_{1}}, - \tau )G({{k}_{2}},\tau ),$$

(D.6)

where we took into account the equality

$${{G}_{{12}}}(\tau ) = {{\delta }_{{12}}}G({{k}_{1}},\tau ),$$

which is valid when averaged over impurities. The function G(k₁, τ) is obtained under expansion in the impurity potential over all disconnected diagrams.

Fig. 3.

Diagrammatic representation of P^R(ω_n).

The standard expansion of the Green functions in Fourier series yields

$$\begin{gathered} {{P}^{R}}({{\omega }_{n}}) = \frac{{2{{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} \\ \times \,\,\frac{1}{\beta }\sum\limits_{s = - \infty }^\infty {H({{k}_{1}},{{\zeta }_{s}})G({{k}_{2}},{{\zeta }_{s}} - \hbar {{\omega }_{n}}),} \\ \end{gathered} $$

(D.7)

which allows us to calculate the sum

$$\frac{1}{\beta }\sum\limits_{s = - \infty }^\infty {G(k,{{\zeta }_{s}})G(k,{{\zeta }_{s}} - \hbar {{\omega }_{n}})} .$$

To this end, we calculate

$${{S}_{{12}}}({{\omega }_{n}}) = \frac{1}{\beta }\sum\limits_{s = - \infty }^\infty {{{G}_{1}}({{\zeta }_{s}}){{G}_{2}}({{\zeta }_{s}} - \hbar {{\omega }_{n}}),} $$

(D.8)

where the indices correspond to momenta. Consider the integral in the complex plane,

$${{I}_{{12}}}({{\omega }_{n}}) = \frac{1}{{2\pi {{i}_{C}}}}f(z){{G}_{1}}(z){{G}_{2}}(z - i\hbar {{\omega }_{n}}),$$

(D.9)

where the contour C encloses all the poles of the distribution function f(z), i.e., the points z_s = iζ_s, and the functions G_{1, 2}(z) are analytic continuations of the Green functions from the discrete points z_s = iζ_s to the whole complex plane. Then, by the residue theorem, we have

$${{I}_{{12}}}({{\omega }_{n}}) = - {{S}_{{12}}}({{\omega }_{n}}).$$

(D.10)

On the other hand, we can change the contour, taking into account the singularities of the Green functions G₁ and G₂. Let us consider these singularities.

The Green function G(z) can be represented in terms of the spectral density as

$$G(z) = \int\limits_{ - \infty }^\infty {d\varepsilon \frac{{{{\rho }_{k}}(\varepsilon )}}{{z - \varepsilon }}.} $$

(D.11)

This integral represents a Cauchy integral, where the contour passes along the real axis. Consequently, the real axis is a singularity at which the Green function is not analytic. Moreover, on either side of this axis, the functions G^R(z) (in the upper half-plane) and G^A(z) (in the lower half-plane) are analytic. Similarly, for the function G₂(z – i$\hbar$ω_n), such a singularity is given by the line Imz – $\hbar$ω_n = 0. These singularities should be removed from the complex plane by means of cuts by appropriately changing the integration contour.

Since G(z) ~ z^–1 as z → ∞, the corresponding contributions from the segments of the circle vanish, and the integral is determined only by the banks of the cut.

Then

$$\begin{gathered} \frac{1}{{2\pi i}}\int\limits_{\infty + i\hbar {{\omega }_{n}}}^{ - \infty + i\hbar {{\omega }_{n}}} {d\varepsilon {\kern 1pt} 'f(\varepsilon {\kern 1pt} ')G_{1}^{R}(\varepsilon {\kern 1pt} ')G_{2}^{A}} (\varepsilon {\kern 1pt} '\, - i\hbar {{\omega }_{n}}) \\ + \,\,\frac{1}{{2\pi i}}\int\limits_{ - \infty + i\hbar {{\omega }_{n}}}^{\infty + i\hbar {{\omega }_{n}}} {d\varepsilon {\kern 1pt} 'f(\varepsilon {\kern 1pt} ')G_{1}^{R}(\varepsilon {\kern 1pt} ')G_{2}^{A}} (\varepsilon {\kern 1pt} '\, - i\hbar {{\omega }_{n}}) \\ = \frac{1}{{2\pi i}}\int\limits_\infty ^{ - \infty } {d\varepsilon f(\varepsilon + i\hbar {{\omega }_{n}})G_{1}^{R}(\varepsilon + i\hbar {{\omega }_{n}})G_{2}^{A}} (\varepsilon ) \\ + \,\,\frac{1}{{2\pi i}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon + i\hbar {{\omega }_{n}})G_{1}^{R}(\varepsilon + i\hbar {{\omega }_{n}})G_{2}^{R}} (\varepsilon ). \\ \end{gathered} $$

(D.12)

Taking into account that f(ε + i$\hbar$ω_n) = f(ε), we obtain

$$\begin{gathered} {{S}_{{12}}}({{\omega }_{n}}) \\ = \, - \frac{1}{{2\pi i}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon )[G_{1}^{R}(\varepsilon )\, - \,G_{1}^{A}(\varepsilon )]} G_{2}^{A}(\varepsilon \, - \,i\hbar {{\omega }_{n}}) \\ - \frac{1}{{2\pi i}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon )G_{1}^{R}(\varepsilon + i\hbar {{\omega }_{n}})[G_{2}^{R}(\varepsilon ) - G_{2}^{A}(\varepsilon )].} \\ \end{gathered} $$

(D.13)

Analytic continuation to the upper half-plane yields

$$\begin{gathered} {{S}_{{12}}}(\omega )\, = \, - \frac{1}{{2\pi i}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon )[G_{{{{k}_{1}}}}^{R}(\varepsilon )\, - \,G_{{{{k}_{1}}}}^{A}(\varepsilon )]} G_{{{{k}_{2}}}}^{A}(\varepsilon \, - \,\hbar \omega ) \\ - \,\,\frac{1}{{2\pi i}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon )G_{{{{k}_{1}}}}^{R}(\varepsilon + \hbar \omega )[G_{{{{k}_{2}}}}^{R}(\varepsilon ) - G_{{{{k}_{2}}}}^{A}(\varepsilon )].} \\ \end{gathered} $$

(D.14)

Taking into account the expression for the spectral density

$$ - 2\pi i{{A}_{1}}(\varepsilon ) = G_{1}^{R}(\varepsilon ) - G_{1}^{A}(\varepsilon )$$

(D.15)

and omitting the indices, we substitute this function into the analytic continuation $P_{{(1)}}^{R}$(ω):

$$\begin{gathered} {{P}^{R}}(\omega ) = \frac{{2{{e}^{2}}}}{{3V}}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}{{S}_{{12}}}(\omega )} \\ = \frac{{2{{e}^{2}}}}{{3V}}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {\left[ {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{1}}}}}(\varepsilon )G_{{{{k}_{2}}}}^{A}(\varepsilon - \hbar \omega )} } \right.} \\ \left. { + \,{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}\int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{2}}}}}(\varepsilon )G_{{{{k}_{1}}}}^{A}(\varepsilon + \hbar \omega )} } \right]. \\ \end{gathered} $$

(D.16)

Changing the notation k₁ $\rightleftarrows$ k₂ in the second term in this expression, we find

$$\begin{gathered} {{P}^{R}}(\omega ) = \frac{{2{{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} \\ \times \int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{1}}}}}(\varepsilon )[G_{{{{k}_{2}}}}^{A}(\varepsilon - \hbar \omega ) + G_{{{{k}_{2}}}}^{R}(\varepsilon + \hbar \omega )]} . \\ \end{gathered} $$

(D.17)

For the imaginary part we have

$$\begin{gathered} \operatorname{Im} {{P}^{R}}(\omega ){{\left[ {\frac{{2{{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} } \right]}^{{ - 1}}} \\ = \int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{1}}}}}(\varepsilon )\operatorname{Im} [G_{{{{k}_{2}}}}^{R}(\varepsilon + \hbar \omega ) + G_{{{{k}_{2}}}}^{A}(\varepsilon - \hbar \omega )]} \\ = \int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{1}}}}}(\varepsilon )[A_{{{{k}_{2}}}}^{{}}(\varepsilon + \hbar \omega ) - A_{{{{k}_{2}}}}^{{}}(\varepsilon - \hbar \omega )]} \\ = - \int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon ){{A}_{{{{k}_{1}}}}}(\varepsilon ){{A}_{{{{k}_{2}}}}}(\varepsilon + \hbar \omega )} \\ + \int\limits_{ - \infty }^\infty {d\varepsilon f(\varepsilon + \hbar \omega ){{A}_{{{{k}_{1}}}}}(\varepsilon + \hbar \omega ){{A}_{{{{k}_{2}}}}}(\varepsilon ).} \\ \end{gathered} $$

(D.18)

Changing the notation k₁ $\rightleftharpoons$ k₂ in the second term, we obtain

$$\begin{gathered} \operatorname{Im} {{P}^{R}}(\omega ) = - \frac{{2\pi {{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} \\ \times \int\limits_{ - \infty }^\infty {d\varepsilon [f(\varepsilon ) - f(\varepsilon + \hbar \omega )]{{A}_{{{{k}_{1}}}}}(\varepsilon ){{A}_{{{{k}_{2}}}}}(\varepsilon + \hbar \omega ).} \\ \end{gathered} $$

(D.19)

Thus,

$$\sigma = \frac{{2\pi \hbar {{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {{{V}_{{{{k}_{1}}{{k}_{2}}}}}{{V}_{{{{k}_{2}}{{k}_{1}}}}}} \int\limits_{ - \infty }^\infty {d\varepsilon \left( { - \frac{{\partial f}}{{\partial \varepsilon }}} \right){{A}_{{{{k}_{1}}}}}(\varepsilon ){{A}_{{{{k}_{2}}}}}(\varepsilon )} .$$

(D.20)

Accordingly, for the interlayer conductivity we obtain

$${{\sigma }_{{zz}}} = \int\limits_{ - \infty }^\infty {d\varepsilon \left( { - \frac{{\partial f}}{{\partial \varepsilon }}} \right){{\Sigma }_{{zz}}}(\varepsilon ),} $$

(D.21)

where

$${{\Sigma }_{{zz}}}(\varepsilon ) = \frac{{2\pi \hbar {{e}^{2}}}}{V}\sum\limits_{{{k}_{1}}{{k}_{2}}}^{} {V_{{{{k}_{1}}{{k}_{2}}}}^{z}V_{{{{k}_{2}}{{k}_{1}}}}^{z}{{A}_{{{{k}_{1}}}}}(\varepsilon ){{A}_{{{{k}_{2}}}}}(\varepsilon ).} $$

(D.22)

In the case of diagonal velocity matrix elements, we obtain the expression

$${{\Sigma }_{{zz}}}(\varepsilon ) = \frac{{2\pi \hbar {{e}^{2}}}}{V}\sum\limits_k^{} {{\text{|}}V_{k}^{z}{{{\text{|}}}^{2}}A_{k}^{2}(\varepsilon ).} $$

(D.23)

In the case of a small number of impurities n_i, we can show that

$$\mathop {\lim }\limits_{{{n}_{i}} \to 0} A_{k}^{2}(\varepsilon ) \approx \frac{{\delta (\varepsilon - {{\varepsilon }_{k}})}}{{2\pi }}\tau (\varepsilon ),$$

and we arrive at formula (14).