Indirect Exchange Interaction Mediated by Dirac Fermions

Vladimir Litvinov

doi:10.1007/978-3-030-12053-5_7

Magnetism in Topological Insulators pp 117-142 | Cite as

Indirect Exchange Interaction Mediated by Dirac Fermions

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Vladimir Litvinov

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First Online: 08 May 2019

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Abstract

Based on the model of s-d interaction discussed in Chap. 6, the indirect exchange interaction between a pair of magnetic impurities is calculated. The interaction carries the signature of the topological surface states. The chapter discusses possible magnetic textures induced by the indirect exchange.

Keywords

Indirect exchange interaction Spin–spin interaction via Dirac fermions Magnetic exchange in topological insulators RKKY oscillations Non-RKKY oscillations Bloembergen-Rowland magnetic interaction RKKY-type interaction in an electron gas with inverted bands

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Appendix 1. Spin Texture

The spin configuration, governed by the terms t₁ = C₁(S₁ × R_∥)_z(S₂ × R_∥)_z and t₂ = C₂(S₁R_∥)(S₂R_∥) in pair interaction (7.30) and (7.32), stems from the minimum energy it delivers when one rotates spins in the pair. If one rotates S₁ and S₂ in the geometry of Fig. 7.4, the interaction energy corresponding to t₁ and t₂, depends on angles γ₁ and γ₁ as shown in Fig. 7.15.

The terms, t₁ and t₂, favor different spin configurations with respect to the inter-spin vector (see insets in Fig. 7.15a, b) while they are similar in their antiferromagnetic nature as long as C_1,2 > 0. If any of the coefficients C_1,2 becomes negative, the corresponding ordering turns ferromagnetic.

The actual spin texture in a magnetically doped TI film is determined by the total indirect exchange interaction that simultaneously includes competing t₁ and t₂ contributions as well as the Ising term in (7.30). In addition, the coefficients which are the range functions may be variable in sign. If one considers the magnetic ordering on a small inter-spin distance where sign variations are not relevant, the relative magnitude of the interaction on the surfaces and in the middle of the film make us conclude that surfaces experience out-of-plane ferromagnetism, while the spin glass forms in the inner part of the film.

Appendix 2. Calculation of the RKKY in a Vertically Biased Topological Insulator

In order to calculate the range function (7.33) analytically, we use the conduction band energy spectrum (7.21) expanded up to the k² term (reference energy is in the middle of the surface energy gap):

$$ {\displaystyle \begin{array}{c}{E}_{\mathrm{c},\uparrow \downarrow }(k)=\frac{1}{2}\sqrt{\varDelta_S^2+4{\tilde{V}}_{as}^2} + \tilde{B}{k}^2\pm {\alpha}_Rk,\\ {}\tilde{B}=D-{Bg}_1\left({g}_1^2+4{V}^2\right)+\frac{g_1^2\ {\tilde{\mathrm{A}}}_2^2}{\sqrt{\varDelta_S^2+4{\tilde{V}}_{as}^2}}\equiv \frac{{\mathrm{\hslash}}^2}{2m},\\ {}{\alpha}_{\mathrm{R}}=2{\tilde{A}}_2V,\kern1em {g}_1=\frac{\varDelta_S}{\sqrt{\varDelta_S^2+4{\tilde{V}}_{as}^2}},\kern1em V=\frac{{\tilde{V}}_{as}}{\sqrt{\varDelta_S^2+4{\tilde{V}}_{as}^2}}.\end{array}} $$

(7.42)

In what follows we suppose that $ \tilde{B} $ (inverse effective mass) is positive and then the spectrum as a function of momentum comprises two parabolas shifted with respect to each other by Rashba spin-splitting (α_R ≠ 0).

We use an analytical continuation of Bessel functions to a negative k-half-plane,

$$ {\displaystyle \begin{array}{c}{J}_0\left({kR}_{\Big\Vert}\right)=\frac{1}{2}\left({H}_0^{(1)}\left({kR}_{\Big\Vert}\right)+{H}_0^{(2)}\left({kR}_{\Big\Vert}\right)\right),\\ {}{H}_0^{(1)}\left({ze}^{i\pi}\right)=-{H}_0^{(2)}(z),\end{array}} $$

(7.43)

where $ {H}_0^{\left(1,2\right)}\left({R}_{\Big\Vert }k\right) $ are Hankel functions [30]. The integral over momentum in (7.33) can be represented as an integral along the line shifted in the upper k-half-plane to avoid the branch cut of the function $ {H}_0^{(1)} $ on the negative k-axis:

$$ {I}_{c\uparrow}\left({\omega}_n\right)=\frac{1}{2}{\int}_{-\infty + i\delta}^{\infty + i\delta}\frac{kH_0^{(1)}\left({kR}_{\parallel}\right) dk}{i{\omega}_n-\tilde{B}{k}^2-{\alpha}_{\mathrm{R}}k+{\varepsilon}_{\mathrm{F}}},\kern1em {\varepsilon}_{\mathrm{F}}=\mu -\frac{1}{2}\sqrt{\varDelta_S^2+4{\tilde{V}}_{as}^2\ }. $$

(7.44)

Expression (7.44) accounts for the spin-up band only. To keep the integration over this band, we have to flip α_R → − α_R in the course of transition to a negative k half-axis. Because of Kramers degeneracy, mere reflection k → − k would correspond to transition to the spin-down band.

The Hankel function $ {H}_0^{(1)}(z) $ is analytical in the upper k-half-plane and tends to zero on the arc z → ∞. So, in order to calculate (7.44), we close the integration path in the upper half-plane. The poles of the integrand are

$$ {k}_{1,2}=\frac{1}{2\tilde{B}}\left[-{\alpha}_{\mathrm{R}}\pm \sqrt{\alpha_{\mathrm{R}}^2+4\tilde{B}\left(i{\omega}_n+{\varepsilon}_{\mathrm{F}}\right)}\right]. $$

(7.45)

At this point, it is important to make sure that the poles of the integrand are located on both sides of the real k-axis. Otherwise, the integral would be zero as the path can be closed around the region, where the integrand is an analytical function. Moreover, as we will use (7.44) in a subsequent summation over the Matsubara frequency ω_n, it is important to choose branches of poles to hold their positions on the complex k-plane for ω_n > 0 and ω_n < 0, as shown in Fig. 7.16.

Fig. 7.16
Poles of the integrand in (7.44) on the complex k-plane

Finally, the integral (7.44) is found by calculating the residue

$$ {\displaystyle \begin{array}{c}{I}_{c\uparrow}\left({\omega}_n\right)=-\frac{\pi i}{2\tilde{B}}\cdot \frac{-P+\sqrt{P^2+ ix+F}}{\sqrt{P^2+ ix+F}}{H}_0^{(1)}\left[{R}_{\parallel}\left(-P+\sqrt{P^2+ ix+F}\right)\right],\\ {}P=\frac{\alpha_{\mathrm{R}}}{2\tilde{B}};\kern1em F=\frac{\varepsilon_{\mathrm{F}}}{2\tilde{B}};\kern1em x=\frac{\omega_n}{2\tilde{B}}.\end{array}} $$

(7.46)

The integral I_c↓(ω_n), calculated in a similar way, differs from (7.46) by the sign of coefficient P. The combination that enters the range function (7.33) takes the form:

$$ {\displaystyle \begin{array}{l}{I}_{c\uparrow}\left({\omega}_n\right){I}_{c\downarrow}\left({\omega}_n\right)=-\frac{\pi^2}{4{\tilde{B}}^2}\times \left(1-\frac{P^2}{P^2+ ix+F}\right){H}_0^{(1)}\left[{R}_{\parallel}\left(-P+\sqrt{P^2+ ix+F}\right)\right]\\ {}\times {H}_0^{(1)}\left[{R}_{\parallel}\left(P+\sqrt{P^2+ ix+F}\right)\right],\end{array}} $$

(7.47)

The range function contains the sum over Matsubara frequencies which at T → 0 can be replaced with the integral:

$$ {\displaystyle \begin{array}{c}{S}_{\mathrm{M}}=T\sum \limits_{\omega_n}{I}_{c\uparrow}\left({\omega}_n\right){I}_{c\downarrow}\left({\omega}_n\right)=\frac{1}{2\pi i}{\int}_{\varGamma }{I}_{c\uparrow}\left(\upomega \right){I}_{c\downarrow}\left(\upomega \right) d\omega, \\ {}i{\omega}_n\to \omega +i\ \mathit{\operatorname{sign}}\left(\omega \right).\end{array}} $$

(7.48)

Contour Γ = Γ₁ + Γ₂, shown in Fig. 7.17, avoids the branch cut point x₀ = − R² − F and the branch cut line determined by the square root in (7.47).

The integral over the circle around the pole tends to zero when the radius decreases, and S_M can be expressed via integration over the upper and lower sides of path Γ₂:

$$ {\displaystyle \begin{array}{l}{S}_{\mathrm{M}}=\frac{i\pi}{4\tilde{B}}{\int}_0^{\sqrt{P^2+F}} ydy\left(1-\frac{P^2}{y^2}\right)\Big\{{H}_0^{(1)}\left[{R}_{\parallel}\left(-P+y\right)\right]{H}_0^{(1)}\left[{R}_{\parallel}\left(P+y\right)\right]\\ {}\kern2.2em -{H}_0^{(1)}\left[{R}_{\parallel}\left(-P-y\right)\right]{H}_0^{(1)}\left[{R}_{\parallel}\left(P-y\right)\right]\Big\},\\ {}\kern6em y=\sqrt{P^2+x+F}.\end{array}} $$

(7.49)

The Hankel function $ {H}_0^{(1)} $ has a branch cut on the negative part of the real axis. To avoid the integration of $ {H}_0^{(1)} $ in this region, we use the relation $ {H}_0^{(1)}\left(-y\right)=-{H}_0^{(2)}(y) $, so (7.49) takes the form

$$ {\displaystyle \begin{array}{c}{S}_{\mathrm{M}}=\frac{i\pi}{4\tilde{B}}\times \left\{{\int}_0^P{yh}_1(y)\left(1-\frac{P^2}{y^2}\right) dy+{\int}_P^{\sqrt{P^2+F}}{yh}_2(y)\left(1-\frac{P^2}{y^2}\right)\right\} dy,\\ {}{h}_1(y)={H}_0^{(2)}\left[{R}_{\parallel}\left(P+y\right)\right]{H}_0^{(1)}\left[{R}_{\parallel}\left(P-y\right)\right]-{H}_0^{(2)}\left[{R}_{\parallel}\left(P-y\right)\right]{H}_0^{(1)}\left[{R}_{\parallel}\left(P+y\right)\right],\\ {}{h}_2(y)={H}_0^{(1)}\left[{R}_{\parallel}\left(y-P\right)\right]{H}_0^{(1)}\left[{R}_{\parallel}\left(y+P\right)\right]-{H}_0^{(2)}\left[{R}_{\parallel}\left(y+P\right)\right]{H}_0^{(2)}\left[{R}_{\parallel}\left(y-P\right)\right].\end{array}} $$

(7.50)

Changing the variable to s = yR_∥ and using notations (7.46), we get the final result for the range function (7.34) [31]. In (7.34), we used the identities relating to the Hankel and Newman functions: $ {H}_0^{(1)}(s)={J}_0(s)+{iN}_0(s),{H}_0^{(2)}(s)={J}_0(s)-{iN}_0(s). $

Appendix 3. Calculation of Bloembergen-Rowland Coupling

Below we illustrate the calculation method taking the integral I₂ (7.38) as an example. After integration over angles, we come to the expression

$$ \mathrm{Int}=\sum \limits_{\omega_n}\iint dkdq\frac{kq\ {J}_0\left({kR}_{\Big\Vert}\right){J}_0\left({qR}_{\Big\Vert}\right){\left(i{\omega}_n\right)}^2}{\left({\varepsilon}_k^2+{\tilde{A}}_2^2{k}^2-{\omega}_n^2\right)\left({\varepsilon}_q^2+{\tilde{A}}_2^2{q}^2-{\omega}_n^2\right)}\equiv \sum \limits_{\omega_n}{\left[Q\left({\omega}_n\right)\right]}^2, $$

(7.51)

where J₀ is the Bessel function and

$$ {\displaystyle \begin{array}{l}Q\left({\omega}_n\right)={\int}_0^{\infty } kdk\ \frac{\ i{\omega}_n\kern0.50em {J}_0\left({R}_{\parallel }\ k\right)}{{\left(i{\omega}_n\right)}^2-{B}^2{\left({k}^2-{k_0}^2\right)}^2-{\tilde{A}}_2^2{k}^2}\\ {}=\frac{1}{2}{\int}_0^{\infty } kdk\ \frac{\ i{\omega}_n\ }{{\left(i{\omega}_n\right)}^2-{B}^2{\left({k}^2-{k_0}^2\right)}^2-{\tilde{A}}_2^2{k}^2}\left[{H}_0^{(1)}\left({R}_{\Big\Vert }\ k\right)+{H}_0^{(2)}\left({R}_{\Big\Vert }\ k\right)\right].\end{array}} $$

(7.52)

Using the relation $ {H}_0^{(2)}\left({R}_{\Big\Vert }\ k\right)=-{H}_0^{(1)}\left({R}_{\Big\Vert }\ {ke}^{i\pi}\right) $, the second term in (7.52) can be represented as

$$ {\int}_{-\infty + i\delta}^0 ydy\ \frac{\ i{\omega}_n\ {H}_0^{(1)}\left({R}_{\Big\Vert }\ y\right)}{{\left(i{\omega}_n\right)}^2-{B}^2{\left({y}^2-{k_0}^2\right)}^2-{\tilde{A}}_2^2{y}^2}. $$

(7.53)

Combining both terms in (7.52), we obtain

$$ Q(x)=-\frac{1}{2{B}^2}{\int}_{-\infty +i\infty}^{\infty +i\infty } kdk\ \frac{\kern0.5em x\ {H}_0^{(1)}\left({R}_{\Big\Vert }\ k\right)}{\left(k-{k}_1\right)\left(k-{k}_2\right)\left(k-{k}_3\right)\left(k-{k}_4\right)}, $$

(7.54)

where

$$ {k}_{1,2}=\pm {\left[{k}_{\mathrm{M}}^2+\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2},\kern0.5em {k}_{3,4}=\pm {\left[{k}_{\mathrm{M}}^2-\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2},\kern0.5em x=i{\omega}_n. $$

(7.55)

Calculating the integral (7.54), one obtains

$$ {\displaystyle \begin{array}{l}Q(x)=\kern1em \frac{\pi ix\ \left[{H}_0^{(1)}\left({R}_{\Big\Vert }\ {\left[{k}_{\mathrm{M}}^2+\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)-{H}_0^{(1)}\left(-{R}_{\Big\Vert }{\left[{k}_{\mathrm{M}}^2-\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)\right]}{4B\sqrt{x^2-{\varDelta}_S^2/4}}\\ \kern2.5em=-\frac{\pi ix\ \left[{H}_0^{(1)}\left({R}_{\Big\Vert }\ {\left[{k}_{\mathrm{M}}^2+\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)+{H}_0^{(2)}\left({R}_{\Big\Vert }{\left[{k}_M^2-\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)\right]}{4B\sqrt{x^2-{\varDelta}_S^2/4}}.\end{array}} $$

(7.56)

At T = 0, discrete summation is replaced with integration over x : iω_n → x + i δ sign (x), so the expression (7.56), being used in (7.51), gives

$$ T\sum \limits_{\omega_n}{\left[Q\left({\omega}_n\right)\right]}^2=\frac{1}{2\pi i}{\int}_{\varGamma }{\left[Q(x)\right]}^2 dx. $$

(7.57)

The integration path in (7.57) is shown in Fig. 7.18.

One can deform the contour Γ rotating it by π/2 as the turn goes in the region, where the integrand is an analytical function:

$$ {\displaystyle \begin{array}{c}T\sum \limits_{\omega_n}{\left[Q\left({\omega}_n\right)\right]}^2=\frac{\pi i}{16{B}^2}{\int}_0^{i\infty}\frac{x^2}{x^2-{\varDelta}_S^2/4}\Big[{H}_0^{(1)}\left({R}_{\Big\Vert }\ {\left[{k}_{\mathrm{M}}^2+\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)\\ {}\kern1em +{H}_0^{(2)}\left({R}_{\Big\Vert }{\left[{k}_{\mathrm{M}}^2-\frac{1}{B}\sqrt{x^2-{\varDelta}_S^2/4}\right]}^{1/2}\right)\Big]{}^2 dx.\end{array}} $$

(7.58)

Changing the variable to $ =\sqrt{1-4{x}^2/{\varDelta}_S^2} $, we express (7.58) as

$$ {\displaystyle \begin{array}{c}T\sum \limits_{\omega_n}{\left[Q\left({\omega}_n\right)\right]}^2=-\frac{\pi {\varDelta}_S}{32{B}^2}{\int}_1^{\infty}\frac{\sqrt{z^2-1}}{\mathrm{z}}\Big[{H}_0^{(1)}\left(\frac{R_{\Big\Vert }}{R_0}\ {\left[{R}_0^2{k}_{\mathrm{M}}^2+ iz\right]}^{1/2}\right)\\ {}\kern0.9em +{H}_0^{(2)}\left(\frac{R_{\Big\Vert }}{R_0}\ {\left[{R}_0^2{k}_{\mathrm{M}}^2- iz\right]}^{1/2}\right)\Big]{}^2 dz\\ {}=-\frac{\pi {\varDelta}_S}{16{B}^2}\underset{1}{\overset{\infty }{\int }}\frac{\sqrt{z^2-1}}{\mathrm{z}}{\left\{\operatorname{Re}\left[{H}_0^{(1)}\left(\frac{R_{\Big\Vert }}{R_0}\ {\left[{R}_0^2{k}_{\mathrm{M}}^2+ iz\right]}^{1/2}\right)\right]\right\}}^2 dz.\end{array}} $$

(7.59)

The expression (7.59) enters the final result (7.39). The terms I₁ and I₃ in (7.38) were calculated in a similar way.

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Vladimir Litvinov
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1.Sierra Nevada CorporationIrvineUSA

Indirect Exchange Interaction Mediated by Dirac Fermions

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