A Lorentz-violating low-energy model for the bilayer graphene

Abstract

In this work, we propose a model with Lorentz symmetry violation which describes the electronic low energy limit of the AA-bilayer graphene (BLG) system. The AA-type bilayer is known to preserve the linear dispersion relation of the graphene layer in the low energy limit. The theoretical model shows that in the BLG system, a time-like vector can be associated with the layer separation and contributes to the energy eigenstates. Based on these properties, we can describe in a $(2+1)$-dimensional space-time the fermionic quasi-particles that emerge in the low-energy limit with the introduction of a Lorentz-violating parameter, in analogy with the $(3 + 1)$-dimensional Standard Model Extension. Moreover, we study the consequences of the coupling of these fermionic quasi-particles with the electromagnetic field, and we show via effective action that the low-energy photon acquires a massive spectrum. Finally, using the hydrodynamic approach in the collisionless limit, one finds that the LSV generates a new kind of anomalous thermal current to the vortexes of the system via coupling of the LSV vector.Kindly check and confirm the postbox is correctly identified for affiliation 1.The affiliations are correct.Please confirm the corresponding author is correctly identified.The correspondent author is correct.

Appendix A: The Feynman integrals

In this appendix, we show briefly the Feynman integral in one-loop that were calculated in Section IV. We start with the Feynman parametrization:

$$\begin{aligned} \frac{1}{AB} = \int ^1_0 \frac{dx}{\left[ A x + B(1-x) \right] ^2}, \end{aligned}$$

(76)

that allow us to join the propagator product

$$\begin{aligned}{} & {} \frac{L_{\mu \nu }(p,q)}{ \left[ (p+q)^2 - M_n^2\right] \left( p^2-M_n^2\right) } \nonumber \\{} & \quad =\int _0^1 \frac{dx L_{\mu \nu }(p,q)}{\left( 2 p \cdot q x + x q^2 + p^2 - M_n^2 \right) ^2} \nonumber \\{} & \quad =\int _0^1 \frac{dx L_{\mu \nu }(p,q)}{\left[ (p+ x q)^2 + x(1-x)q^2 - M_n^2 \right] ^2} \nonumber \\{} & \quad =\int _0^1 dx \frac{L_{\mu \nu }(p-xq,q)}{\left( p^2 - \sigma ^2 \right) ^2}, \end{aligned}$$

(77)

where $L_{\mu \nu }(p,q)$ is a generic tensor that depends on the p and q momenta, and $\sigma ^2:= M_n^2 - x(1-x) q^2$. We have applied the shift $p \rightarrow p + x q$ in the last step of (77). Using the known result from the quantum field theory handbook [35], the D-dimension integrals in the momentum space are:

$$\begin{aligned} \int \frac{d^Dp}{(2\pi )^{D}}\frac{1}{(p^2 - \sigma ^2)^\alpha }= & {} \frac{\Gamma (\alpha - \frac{d}{2})}{(2\pi )^D \Gamma (\alpha )} \frac{(-1)^{\frac{D+1}{2}} \pi ^{\frac{D}{2}} }{(- \sigma ^2)^{\alpha - \frac{D}{2}}}, \end{aligned}$$

(78)

$$\begin{aligned} \int \frac{d^Dp}{(2\pi )^{D}} \frac{p_\mu }{(p^2 - \sigma ^2)^\alpha }= & {} 0, \end{aligned}$$

(79)

$$\begin{aligned} \int \frac{d^Dp}{(2\pi )^{D}} \frac{p_\mu p_\nu }{(p^2 - \sigma ^2)^\alpha }= & {} \frac{\Gamma (\alpha - 1-\frac{D}{2})}{(2\pi )^{D} \Gamma (\alpha )} \frac{(-1)^{\frac{D+1}{2}} \pi ^{\frac{D}{2}} }{(- \sigma ^2)^{\alpha -1- \frac{D}{2}}} \eta _{\mu \nu }, \end{aligned}$$

(80)

where $\sigma ^2 > 0$ is required in these results, and implies that $4M_{n}^2 > q^2$. In $(1+2)$-dimensions, we make $D=3$ and $\alpha =2$, thus the previous results are reduced to

$$\begin{aligned} \int \frac{d^3p}{(2\pi )^3} \frac{1}{(p^2 - \sigma ^2)^2}= & {} -\frac{i}{16\pi } \frac{1}{\sqrt{\sigma ^2}}, \end{aligned}$$

(81)

$$\begin{aligned} \int \frac{d^3p}{(2\pi )^3} \frac{p_\mu p_\nu }{(p^2 - \sigma ^2)^2}= & {} - \frac{i}{16\pi } \eta _{\mu \nu } \sqrt{\sigma ^2}. \end{aligned}$$

(82)

Thereby, the Feynman integral is

$$\begin{aligned}{} & {} \int \frac{d^3p}{(2 \pi )^3} \frac{1}{ \left[ (p+q)^2 - M_n^2\right] \left( p^2-M_n^2\right) } \nonumber \\{} & \quad = \int _0^1 dx \int \frac{d^3p}{(2 \pi )^3} \frac{1}{\left( p^2 - \sigma ^2 \right) ^2} \nonumber \\{} & \quad = \frac{ i }{8\pi } \int _0^1 \frac{dx }{\sqrt{ M_n^2 - x(1-x) q^2} } \nonumber \\{} & \quad = -\frac{1 }{8 \pi |M_n|} \coth ^{-1}\left( \frac{\sqrt{q^2}}{2M_{n}} \right). \end{aligned}$$

(83)

Going further, one has:

$$\begin{aligned}{} & {} \int \frac{d^3p}{(2\pi )^3} \frac{(p+q)_{\mu } p_\nu }{ \left[ (p+q)^2 - M_n^2\right] \left( p^2-M_n^2\right) } \nonumber \\{} & \quad =\int _0^1 dx \int \frac{d^3p}{(2\pi )^3} \frac{ p_\mu p_\nu + x (1-x) q_\mu q_\nu }{\left( p^2 - \sigma ^2 \right) ^2} \nonumber \\{} & \quad = -\frac{i |M_n|}{16 \pi } \left[ \eta _{\mu \nu } g(y)+\frac{ q_\mu q_\nu }{M_n^2} f(y) \right], \end{aligned}$$

(84)

in which $y_{n}:= q^2/M_{n}^2$, and the functions f and g are defined by

$$\begin{aligned} f(y_n)= & {} \int _0^1 dx \frac{x(1-x)}{\sqrt{ 1-x(1-x) y_n} } = \nonumber \\= & {} \frac{1}{4 y_n^{3/2}}\left[ -2 \sqrt{y_n}+(4-3y_n) \coth ^{-1}\left( \frac{2}{\sqrt{y_n}}\right) \right. \nonumber \\{} & {} \left. +4 y_n \text{ csch}^{-1}\left( \sqrt{\frac{4}{y_n}-1}\right) \right], \end{aligned}$$

(85)

and

$$\begin{aligned} g(y_n)= & {} \int _0^1 dx \sqrt{ 1 - x(1-x) y_n } \nonumber \\{} & {} =\frac{2\sqrt{y_n}+(3y_n-4) \coth ^{-1}\left( \frac{2}{\sqrt{y_n}}\right) }{4 y_n^{3/2}}. \end{aligned}$$

(86)

Finally, the dimensionally regularized integral (48) that contributes to the self-energy at one loop is

$$\begin{aligned} \Xi _{I,n}(p,D)= & {} - 3 e^2 (\Lambda )^{3-D} \nonumber \\{} & {} \times \int \frac{d^D k}{(2 \pi )^D} \frac{1}{\sqrt{k^2}}\frac{1}{ \left( k+p+I t\right) ^2-M_n ^2}, \nonumber \\ \end{aligned}$$

(87)

in which the result in (1+2) dimensions is recovered when $D=3$, and $\Lambda$ is an arbitrary energy scale to keep the dimensionless coupling constant in D-dimensions. Using the Feynman parametrization

$$\begin{aligned} \frac{1}{A B^{1/2}} = \int _0^1 dx \frac{(1-x)^{-1/2}}{[ x A + (1-x) B ]^{3/2}}, \end{aligned}$$

(88)

the integral (49) can be rewritten as:

$$\begin{aligned} \Xi _{I,n}(p,D)= & {} - 3 e^2 (\Lambda )^{3-D} \int _0^1 dx (1-x)^{-1/2} \times \nonumber \\{} & {} \times \int \frac{d^D k}{(2 \pi )^D} \frac{1}{\left[ k^2 - \Delta (x)^2 \right] ^{3/2}}, \end{aligned}$$

(89)

where $[\Delta (x)]^2 = x M_n^2 - x(1-x)(p+ It)^2>0$, and imposes the condition $2M_{n}^2>(p+It)^2$. By use of identity A3, we obtain:

$$\begin{aligned} \Xi _{I,n}(p,D)= & {} \frac{3 \alpha }{16\pi } (-1)^{(D+3)/2} \Gamma \left( \frac{3-D}{2}\right) \nonumber \\{} & {} \times \int _0^1 dx (1-x)^{-1/2} \left[ \frac{4\pi \Lambda ^2}{\Delta (x)^2} \right] ^{(3-D)/2}. \end{aligned}$$

(90)

We expand around $D=3-\delta$, when $\delta \rightarrow 0$, the result is

$$\begin{aligned} \Xi _{I,n}(p,\delta )= & {} -\frac{3\alpha }{4\pi }\frac{1}{\delta }+\frac{3\alpha \gamma _{E}}{8\pi } -\frac{3\alpha }{4\pi }-\frac{3\alpha }{8\pi }\ln \left( \frac{\pi \Lambda ^2}{M_{n}^2} \right) \nonumber \\{} & {} -\frac{3\alpha }{16\pi }\int _{0}^{1} \!\frac{dx}{\sqrt{1-x}}\ln \left[ \frac{M_{n}^2}{M_n^2 - (1-x)(p+ It)^2} \right] . \end{aligned}$$

(91)

The finite part is

$$\begin{aligned} \Xi _{I,n}^{(f)}(p)= & {} \frac{3\alpha }{8\pi }\left\{ \frac{2M_{n}}{\sqrt{(p+It)^2}}\tanh ^{-1}\left[ \frac{\sqrt{(p+It)^2}}{M_{n}} \right] \right. \nonumber \\{} & {} \left. -\ln \left[ \frac{M_{n}^2}{M_{n}^2-(p+It)^2} \right] \right\}, \end{aligned}$$

(92)

in which we impose the condition of $M_{n}^{2}>(p+It)^2$. If $M_{n}^{2}<(p+It)^2$, we obtain

$$\begin{aligned} \Xi _{I,n}^{(f)}(p)= & {} \frac{3\alpha }{8\pi }\left\{ \frac{2M_{n}}{\sqrt{(p+It)^2}}\tanh ^{-1}\left[ \frac{\sqrt{(p+It)^2}}{M_{n}} \right] \right. \nonumber \\{} & {} \left. -\ln \left[ \frac{M_{n}^2}{(p+It)^2-M_{n}^2}\right] -i\pi \right\}. \end{aligned}$$

(93)

The limit of $p^{\mu }\rightarrow 0$ in (92) yields

$$\begin{aligned} \Xi _{n}^{(f)}(0) \!=\! \frac{3\alpha }{8\pi }\left[ \frac{2M_{n}}{\sqrt{t^2}}\tanh ^{-1}\left( \frac{\sqrt{t^2}}{M_{n}} \right) -\ln \left( \frac{M_{n}^2}{M_{n}^2-t^2} \right) \right], \end{aligned}$$

(94)

if $M_{n}^2>t^2$, and

$$\begin{aligned} \Xi _{n}^{(f)}(0)= & {} \frac{3\alpha }{8\pi }\left[ \frac{2M_{n}}{\sqrt{t^2}}\tanh ^{-1}\left( \frac{M_{n}}{\sqrt{t^2}} \right) \right. \nonumber \\{} & {} \left. -\ln \left( \frac{M_{n}^2}{t^2-M_{n}^2} \right) -i\pi \left( 1-\frac{M_{n}}{\sqrt{t^2}}\right) \right], \end{aligned}$$

(95)

if $t^2>M_{n}^2$.