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Thermodynamics for a rotating chiral fermion system in the uniform magnetic field

Ren-Hong Fang fangrh@sdu.edu.cn Key Laboratory of Particle Physics and Particle Irradiation (MOE), Institute of Frontier and Interdisciplinary Science, Shandong University, Qingdao, Shandong 266237, China Theoretical Physics Research and Innovation Team, College of Intelligent Systems Science and Engineering, Hubei Minzu University, Enshi, Hubei 445000, China
Abstract

We study the thermodynamics for a uniformly rotating system of chiral fermions under the uniform magnetic field. Then we obtain the mathematical expressions of some thermodynamic quantities in terms of the series with respect to the external magnetic field BB, the angular velocity Ω\Omega and the chemical potential μ\mu, expanded around B=0B=0, Ω=0\Omega=0 and μ=0\mu=0. Our results given by such series are a generalization of the expressions available in the references simply corresponding to the lower-order terms of our findings. The zero-temperature limit of our results is also discussed.

I Introduction

The properties of Dirac fermion system have been investigated from many aspects for a long time. For a hydrodynamic system consisting of Dirac fermions under the background of electromagnetic fields, Wigner function is an appropriate tool, which can provide a covariant and gauge invariant formalism (Elze et al., 1986; Vasak et al., 1987). It is worth pointing out that, although Wigner function defined in 8-dimensional phase space is not always non-negative, one can always obtain non-negative probability density when the 4-dimensional momentum is integrated out. For massless (or chiral) fermion system with uniform vorticity and electromagnetic fields, the charge current and the energy-momentum tensor up to the second order have been obtained from Wigner function approach, including chiral anomaly equation, chiral magnetic and vortical effects (Gao et al., 2012; Yang et al., 2020). The pair production in parallel electric and magnetic fields with finite temperature and chemical potential from Wigner function approach is also investigated recently (Sheng et al., 2019). Without external electromagnetic fields, the energy-momentum tensor and charge current of the massless fermion system up to second order in vorticity have been obtained from thermal field theory (Buzzegoli et al., 2017; Buzzegoli and Becattini, 2018; Palermo et al., 2021). For a uniformly rotating massless fermion system, the analytic expressions of the charge current and the energy-momentum tensor are obtained (Ambruş and Winstanley, 2014). For the massive and massless fermion systems under the background of a uniform magnetic field, the general expansions with respect to fermion mass, magnetic field and chemical potential are derived by the approaches of proper-time and grand partition function (Cangemi and Dunne, 1996; Zhang et al., 2020a; Fang et al., 2021). There are also some investigations on the system of free fermion gas, quark matter or hadronic matter, with pure rotation (Chen et al., 2021; Fujimoto et al., 2021; Becattini et al., 2021), or with the coexistence of rotation and magnetic field (Chen et al., 2016; Fukushima et al., 2020), or with specific boundary conditions (Chernodub and Gongyo, 2017a, b, c; Zhang et al., 2020b; Yang et al., 2021). The quantum superfluid phenomena of Dirac fermions in the background of magnetic field and rotation are discussed recently (Liu and Zahed, 2018; Mottola and Sadofyev, 2021).

In this article, we consider a uniformly rotating chiral fermion system in a uniform magnetic field, where we ignore the interaction among the fermions and the directions of the angular velocity and the magnetic field are chosen to be parallel. In this article we will adopt the approach of normal ordering and ensemble average to calculate the thermodynamical quantities of the system. Firstly we briefly derive the Dirac equation in a rotating frame under the background of a uniform magnetic field from the Dirac equation in curved space. Then through solving the eigenvalue equation of the Hamiltonian in cylindrical coordinates, we can obtain a series of Landau levels, from which one can calculate the expectation value of corresponding thermodynamical quantities for each eigenstate. From the approach of ensemble average used in (Vilenkin, 1978, 1979, 1980), the macroscopic thermodynamical quantities can be expressed as the summation over the product of the particle number (Fermi-Dirac distribution) and the expectation value in each eigenstate. We expand all thermodynamical quantities as threefold series at B=0B=0, Ω=0\Omega=0 and μ=0\mu=0, where the lower orders are consistent with that from the approaches of thermal field theory and Wigner function respectively (Buzzegoli et al., 2017; Buzzegoli and Becattini, 2018; Yang et al., 2020), and to our knowledge the general orders have not been obtained before. We also calculate all quantities in zero temperature limit, and obtain the equality of partcile/energy density and corresponding currents along zz-axis, which can provide qualitative reference for the thermodynamics of compact stars in astrophysics, such as neutron star and magnetic star, since the magnitudes of the magnetic field and rotational speed are huge compared to the temperature of the compact stars (Felipe et al., 2008; Itokazu et al., 2018). In this article all thermodynamical quantities will be calculated at the rotating axis (r=0r=0), so the boundary condition at the speed-of-light surface will not affect our results.

From the point of view of hydrodynamics, it has been pointed out that the relativistic hydrodynamical equations with only first order term does not obey the causality (Hiscock and Lindblom, 1983, 1985, 1987), i.e., the group speed of some transport coefficients, such as heat conductivity, would exceed the speed of light (Denicol et al., 2008). Therefore, the high order terms in hydrodynamics are necessary, which indeed repair the issue of causality. There have been some earlier work to study the second order terms of transport phenomena, such as Kubo formula from quantum field theory (Jimenez-Alba and Yee, 2015; Hattori and Yin, 2016), thermal field theory (Buzzegoli et al., 2017; Buzzegoli and Becattini, 2018), Wigner function (Yang et al., 2020, 2022), etc. All of these work are perturbation theory essentially, from which the general order terms have not been obtained. In this article, we consider a special configuration for the electromagnetic field and vorticity field in hydrodynamics, i.e. with a pure homogeneous magnetic field parallel to a homogeneous vorticity field, and obtain the general order terms of all thermodynamic quantities, which is important to study the analytic behavior of hydrodynamics in mathematics.

The rest of this article is organized as follows. In Sec. II and III, we briefly derive the Dirac equation in a uniformly rotating frame and list the Landau levels and corresponding eigenfunctions of a single right-handed fermion, which are just reference review. In Sec. IV and V, we obtain the expressions of some thermodynamic quantities in terms of the series with respect to the external magnetic field BB, the angular velocity Ω\Omega and the chemical potential μ\mu, expanded around B=0B=0, Ω=0\Omega=0 and μ=0\mu=0, which are our main results. In Sec. VI, the zero temperature limit of the thermodynamical quantities is discussed. This article is summarized in Sec. VII.

Throughout this article we adopt natural units where =c=kB=1\hbar=c=k_{B}=1. We use the Heaviside-Lorentz convention for electromagnetism and the chiral representation for gamma matrixes where γ5=diag(1,1,+1,+1)\gamma^{5}=\mathrm{diag\,(-1,-1,+1,+1)}, which is the same as Peskin and Schroeder (Peskin and Schroeder, 1995).

II Dirac equation in a uniformly rotating frame

In this section we briefly introduce the Dirac equation in curved spacetime (Parker and Toms, 2009), which is applied to a uniformly rotating frame (Chen et al., 2016).

In curved spacetime, under the background of the electromagnetic field, the Dirac equation for a single chiral fermion is

iγ¯μDμψ(x)=0,i\underline{\gamma}^{\mu}D_{\mu}\psi(x)=0, (1)

where the covariant derivative DμD_{\mu} and gamma matrices γ¯μ\underline{\gamma}^{\mu} are defined as

Dμ=μ+ieAμ+Γμ,γ¯μ=γaeaμ.D_{\mu}=\partial_{\mu}+ieA_{\mu}+\Gamma_{\mu},\ \ \ \underline{\gamma}^{\mu}=\gamma^{a}e_{a}^{\ \mu}. (2)

The underline in γ¯μ\underline{\gamma}^{\mu} is used to distinguish the spacetime-dependent gamma matrices γ¯μ\underline{\gamma}^{\mu} from the constant gamma matrices γa\gamma^{a}, and Γμ=18ωμab[γa,γb]\Gamma_{\mu}=\frac{1}{8}\omega_{\mu ab}[\gamma^{a},\gamma^{b}] is the affine connection. The definitions of vierbein eaμe_{a}^{\ \mu}, metric tensor gμνg_{\mu\nu}, and spin connection ωμab\omega_{\mu ab} are listed as follows,

eaμ=xμXa,eμa=Xaxμ,gμν=ηabeμaeνb,e_{a}^{\ \mu}=\frac{\partial x^{\mu}}{\partial X^{a}},\ \ \ e_{\ \mu}^{a}=\frac{\partial X^{a}}{\partial x^{\mu}},\ \ \ g_{\mu\nu}=\eta_{ab}e_{\ \mu}^{a}e_{\ \nu}^{b}, (3)
ωμab=gαβeaα(μebβ+Γμνβebν),\omega_{\mu ab}=g_{\alpha\beta}e_{a}^{\ \alpha}(\partial_{\mu}e_{b}^{\ \beta}+\Gamma_{\mu\nu}^{\beta}e_{b}^{\ \nu}), (4)
Γμνβ=12gβσ(gσμ,ν+gσν,μgμν,σ),\Gamma_{\mu\nu}^{\beta}=\frac{1}{2}g^{\beta\sigma}(g_{\sigma\mu,\nu}+g_{\sigma\nu,\mu}-g_{\mu\nu,\sigma}), (5)

where ηab=diag(+1,1,1,1)\eta_{ab}=\mathrm{diag}\,(+1,-1,-1,-1) is the metric tensor in Minkowski space, XaX^{a} and xμx^{\mu} are the coordinates in a local Lorentz frame and in a general frame, respectively.

In curved spacetime, the vector JVμJ_{V}^{\mu}, axial vector JAμJ_{A}^{\mu} and symmetric energy-momentum tensor TμνT^{\mu\nu} become

JVμ=ψ¯γ¯μψ,JAμ=ψ¯γ¯μγ5ψ,J_{V}^{\mu}=\overline{\psi}\underline{\gamma}^{\mu}\psi,\ \ \ \ J_{A}^{\mu}=\overline{\psi}\underline{\gamma}^{\mu}\gamma^{5}\psi, (6)
Tμν=14(ψ¯iγ¯μDνψ+ψ¯iγ¯νDμψ+H.C.),T^{\mu\nu}=\frac{1}{4}\left(\overline{\psi}i\underline{\gamma}^{\mu}D^{\nu}\psi+\overline{\psi}i\underline{\gamma}^{\nu}D^{\mu}\psi+\mathrm{H.C.}\right), (7)

where DμD^{\mu}, γ¯μ\underline{\gamma}^{\mu} in curved spacetime have replaced a\partial^{a}, γa\gamma^{a} in flat spacetime.

Now we consider a frame 𝒦\mathcal{K} rotating uniformly with angular velocity 𝛀=Ω𝐞z\boldsymbol{\Omega}=\Omega\mathbf{e}_{z} relative to an inertial frame KK. The coordinates in 𝒦\mathcal{K} and KK are denoted as xμ=(t,x,y,z)x^{\mu}=(t,x,y,z) and Xa=(T,X,Y,Z)X^{a}=(T,X,Y,Z) respectively, which are related to each other by following transformations,

{T=tX=xcosΩtysinΩtY=xsinΩt+ycosΩtZ=z.\left\{\begin{array}[]{ccc}T&=&t\\ X&=&x\cos\Omega t-y\sin\Omega t\\ Y&=&x\sin\Omega t+y\cos\Omega t\\ Z&=&z\end{array}\right.. (8)

It should be pointed out that the rotational angular velocity Ω\Omega can not be too large, otherwise the synchronous condition in Eq. (8) can not be satisfied. According to Eq. (3), the metric tensor gμνg_{\mu\nu} and its inverse are

gμν=(1(x2+y2)Ω2yΩxΩ0yΩ100xΩ0100001),g_{\mu\nu}=\left(\begin{array}[]{cccc}1-(x^{2}+y^{2})\Omega^{2}&y\Omega&-x\Omega&0\\ y\Omega&-1&0&0\\ -x\Omega&0&-1&0\\ 0&0&0&-1\end{array}\right), (9)
gμν=(1yΩxΩ0yΩy2Ω21xyΩ20xΩxyΩ2x2Ω2100001).g^{\mu\nu}=\left(\begin{array}[]{cccc}1&y\Omega&-x\Omega&0\\ y\Omega&y^{2}\Omega^{2}-1&-xy\Omega^{2}&0\\ -x\Omega&-xy\Omega^{2}&x^{2}\Omega^{2}-1&0\\ 0&0&0&-1\end{array}\right). (10)

Keeping gμνg_{\mu\nu} unchanged, the vierbein eμae_{\ \mu}^{a} still has a freedom degree of an arbitrary local Lorentz transformation. We can choose eμae_{\ \mu}^{a} as

e 00=e 11=e 22=e 33=1,e 01=yΩ,e 02=xΩ,e_{\ 0}^{0}=e_{\ 1}^{1}=e_{\ 2}^{2}=e_{\ 3}^{3}=1,\ \ \ \ e_{\ 0}^{1}=-y\Omega,\ \ \ \ e_{\ 0}^{2}=x\Omega, (11)

with zeros for other components.

Now we consider a single chiral fermion in a uniformly rotating frame under the background of a uniform magnetic field 𝐁=B𝐞z\mathbf{B}=B\mathbf{e}_{z}, and we choose the gauge potential in the inertial frame as Aa=(0,𝐀)A^{a}=(0,\mathbf{A}) with 𝐁=×𝐀\mathbf{B}=\nabla\times\mathbf{A}. The covariant derivative DμD_{\mu} and gamma matrices γ¯μ\underline{\gamma}^{\mu} become

Dμ=(ti2ΩΣ3+Ω(yAxxAy),xieAx,yieAy,zieAz),D_{\mu}=\left(\partial_{t}-\frac{i}{2}\Omega\Sigma_{3}+\Omega(yA_{x}-xA_{y}),\partial_{x}-ieA_{x},\partial_{y}-ieA_{y},\partial_{z}-ieA_{z}\right), (12)
γ¯μ=(γ0,yΩγ0+γ1,xΩγ0+γ2,γ3),\underline{\gamma}^{\mu}=\left(\gamma^{0},y\Omega\gamma^{0}+\gamma^{1},-x\Omega\gamma^{0}+\gamma^{2},\gamma^{3}\right), (13)

and in this case the Dirac equation for a single chiral fermion can be written as

itψ(x)=[iγ0𝜸(ie𝐀)ΩJz]ψ(x),i\frac{\partial}{\partial t}\psi(x)=\left[-i\gamma^{0}\boldsymbol{\gamma}\cdot(\nabla-ie\mathbf{A})-\Omega J_{z}\right]\psi(x), (14)

where ee is the charge of the chiral fermion, Jz=12Σ3i(xyyx)J_{z}=\frac{1}{2}\Sigma_{3}-i(x\partial_{y}-y\partial_{x}) is the zz-component of the total angular momentum 𝐉\mathbf{J}, and the term ΩJz-\Omega J_{z} can be naturally explained as the coupling of the angular momentum 𝐉\mathbf{J} and the angular velocity 𝛀\boldsymbol{\Omega}.

III Landau levels for a single right-handed fermion in a rotating frame

In the chiral representation of gamma matrixes, where γ5=diag(1,1,+1,+1)\gamma^{5}=\mathrm{diag\,(-1,-1,+1,+1)}, we can divide the chiral fermion field into left- and right-handed fermion fields, i.e. ψ=(ψL,ψR)T\psi=(\psi_{L},\psi_{R})^{T}. Since the equations of motion for ψL\psi_{L} and ψR\psi_{R} decouple, we only discuss right-handed fermion field in this article. All results can be directly generalised to the left-handed case. In the following, we set eB>0eB>0 for simplicity.

The right-handed part of Eq. (14) is

itψR(x)=HψR(x),i\frac{\partial}{\partial t}\psi_{R}(x)=H\psi_{R}(x), (15)
H=i𝝈(ie𝐀)ΩJR,z,H=-i\boldsymbol{\sigma}\cdot(\nabla-ie\mathbf{A})-\Omega J_{R,z}, (16)

where HH, JR,z=12σ3i(xyyx)J_{R,z}=\frac{1}{2}\sigma_{3}-i(x\partial_{y}-y\partial_{x}) is Hamiltonian and the zz-component of the total angular momentum of the right-handed fermion. In this article we shall choose the symmetric gauge for 𝐀\mathbf{A}, i.e. 𝐀=(12By,12Bx,0)\mathbf{A}=(-\frac{1}{2}By,\frac{1}{2}Bx,0). Then the explicit form of the Hamiltonian is

H=i𝝈+12eB(yσ1xσ2)ΩJR,z.H=-i\boldsymbol{\sigma}\cdot\nabla+\frac{1}{2}eB\left(y\sigma_{1}-x\sigma_{2}\right)-\Omega J_{R,z}. (17)

It can be proved that, these three Hermitian operators, H,p^z=iz,JR,zH,\hat{p}_{z}=-i\partial_{z},J_{R,z}, are commutative with each other, then we can construct the common eigenfunctions of them. According to the calculations for Landau levels in Appendix A, we list the common eigenfunctions and corresponding energy in cylindrical coordinate system (where the three coordinate variables are z,r,ϕz,r,\phi) as follows:

When m=12,32,52,m=\frac{1}{2},\frac{3}{2},\frac{5}{2},\cdots,

ψλnmpz=n!(n+m12)!(eB(E+pz+mΩ)2(E+mΩ)eρ2ρm214Lnm12(ρ)ei(m12)ϕiλeB(E+mΩ)(E+pz+mΩ)eρ2ρm2+14Ln1m+12(ρ)ei(m+12)ϕ)eiEt+izpz2π,\psi_{\lambda nmp_{z}}=\sqrt{\frac{n!}{(n+m-\frac{1}{2})!}}\left(\begin{array}[]{c}\sqrt{\frac{eB(E+p_{z}+m\Omega)}{2(E+m\Omega)}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}-\frac{1}{4}}L_{n}^{m-\frac{1}{2}}(\rho)e^{i(m-\frac{1}{2})\phi}\\ \frac{i\lambda eB}{\sqrt{(E+m\Omega)(E+p_{z}+m\Omega)}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}+\frac{1}{4}}L_{n-1}^{m+\frac{1}{2}}(\rho)e^{i(m+\frac{1}{2})\phi}\end{array}\right)\frac{e^{-iEt+izp_{z}}}{2\pi}, (18)
E={λpz2+2eBnmΩ,n>0pzmΩ,n=0.E=\left\{\begin{array}[]{cc}\lambda\sqrt{p_{z}^{2}+2eBn}-m\Omega,&n>0\\ p_{z}-m\Omega,&n=0\end{array}\right.. (19)

When m=12,32,52,m=-\frac{1}{2},-\frac{3}{2},-\frac{5}{2},\cdots,

ψλnmpz=n!(nm+12)!(eB(E+pz+mΩ)2(E+mΩ)eρ2ρ14m2Ln12m(ρ)ei(m12)ϕiλeB(nm+12)(E+mΩ)(E+pz+mΩ)eρ2ρ14m2Ln12m(ρ)ei(m+12)ϕ)eiEt+izpz2π,\psi_{\lambda nmp_{z}}=\sqrt{\frac{n!}{(n-m+\frac{1}{2})!}}\left(\begin{array}[]{c}\sqrt{\frac{eB(E+p_{z}+m\Omega)}{2(E+m\Omega)}}e^{-\frac{\rho}{2}}\rho^{\frac{1}{4}-\frac{m}{2}}L_{n}^{\frac{1}{2}-m}(\rho)e^{i(m-\frac{1}{2})\phi}\\ -\frac{i\lambda eB(n-m+\frac{1}{2})}{\sqrt{(E+m\Omega)(E+p_{z}+m\Omega)}}e^{-\frac{\rho}{2}}\rho^{-\frac{1}{4}-\frac{m}{2}}L_{n}^{-\frac{1}{2}-m}(\rho)e^{i(m+\frac{1}{2})\phi}\end{array}\right)\frac{e^{-iEt+izp_{z}}}{2\pi}, (20)
E=λpz2+2eB(nm+12)mΩ,E=\lambda\sqrt{p_{z}^{2}+2eB\left(n-m+\frac{1}{2}\right)}-m\Omega, (21)

where ρ=12eBr2\rho=\frac{1}{2}eBr^{2}, Lnμ(ρ)L_{n}^{\mu}(\rho) is general Laguerre polynomial as introduced in Appendix B, mm is the magnetic quantum number, λ=±1\lambda=\pm 1 represent the states with positive and negative energy respectively, and n=0,1,2,n=0,1,2,\cdots represent different Landau levels. The eigenfunctions ψλnmpz\psi_{\lambda nmp_{z}} are denoted by the group of good quantum numbers (λ,n,m,pz\lambda,n,m,p_{z}), which are normalized according to

𝑑Vψλnmpzψλnmpz=δλλδnnδmmδ(pzpz).\int dV\psi_{\lambda^{\prime}n^{\prime}m^{\prime}p_{z}^{\prime}}^{\dagger}\psi_{\lambda nmp_{z}}=\delta_{\lambda^{\prime}\lambda}\delta_{n^{\prime}n}\delta_{m^{\prime}m}\delta(p_{z}^{\prime}-p_{z}). (22)

IV Particle current

In this section we consider a right-handed fermion system under the background of a uniform magnetic field 𝐁=B𝐞z\mathbf{B}=B\mathbf{e}_{z}, and the system is rotating uniformly with angular velocity 𝛀=Ω𝐞z\boldsymbol{\Omega}=\Omega\mathbf{e}_{z}. The interaction among the fermions in this system is ignored. We assume that this rotating system is in equilibrium with a reservoir, which keeps constant temperature T=1/βT=1/\beta and constant chemical potential μ\mu.

IV.1 Ensemble average

We will calculate the macroscopic particle current of the system at the rotation axis (i.e. at r=0r=0) through ensemble average approach, in which all macroscopic thermodynamical quantities are the ensemble average of the normal ordering of the corresponding field operators.

The forms of the eigenfunctions in Eqs. (18, 20) at r=0r=0 or ρ=0\rho=0 are simplified to

ψλnmpz=eiEt+izpz2π(eB(E+pz+Ω/2)2(E+Ω/2)δm,1/20)+eiEt+izpz2π(0iλeBn+1(EΩ/2)(E+pzΩ/2)δm,1/2),\psi_{\lambda nmp_{z}}=\frac{e^{-iEt+izp_{z}}}{2\pi}\left(\begin{array}[]{c}\sqrt{\frac{eB(E+p_{z}+\Omega/2)}{2(E+\Omega/2)}}\delta_{m,1/2}\\ 0\end{array}\right)+\frac{e^{-iEt+izp_{z}}}{2\pi}\left(\begin{array}[]{c}0\\ -\frac{i\lambda eB\sqrt{n+1}}{\sqrt{(E-\Omega/2)(E+p_{z}-\Omega/2)}}\delta_{m,-1/2}\end{array}\right), (23)

which are to be used in the following calculations of ensemble average. We find that the zz-component mm of the total angular momentum can only take values ±1/2\pm 1/2 due to the absence of the orbital angular momentum at r=0r=0.

For the right-handed fermion system, the field operator of the particle current at r=0r=0 is

Jμ=ψRσμψR,J^{\mu}=\psi_{R}^{\dagger}\sigma^{\mu}\psi_{R}, (24)

with σμ=(1,𝝈)\sigma^{\mu}=(1,\boldsymbol{\sigma}). From the approach of ensemble average used in (Vilenkin, 1978, 1979, 1980), the macroscopic particle current 𝒥μ\mathcal{J}^{\mu} can be calculated from JμJ^{\mu} as follows,

𝒥μ\displaystyle\mathcal{J}^{\mu} =\displaystyle= :Jμ:\displaystyle\langle:J^{\mu}:\rangle (25)
=\displaystyle= n=1λ𝑑pzλeβ(EnλΩ/2λμ)+1ψλ,n,1/2,pzσμψλ,n,1/2,pz\displaystyle\sum_{n=1}^{\infty}\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda}{e^{\beta\left(E_{n}-\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,n,1/2,p_{z}}^{\dagger}\sigma^{\mu}\psi_{\lambda,n,1/2,p_{z}}
+n=0λ𝑑pzλeβ(En+1+λΩ/2λμ)+1ψλ,n,1/2,pzσμψλ,n,1/2,pz\displaystyle+\sum_{n=0}^{\infty}\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda}{e^{\beta\left(E_{n+1}+\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,n,-1/2,p_{z}}^{\dagger}\sigma^{\mu}\psi_{\lambda,n,-1/2,p_{z}}
+λ𝑑pzλθ(λpz)eβ(|pz|λΩ/2λμ)+1ψλ,0,1/2,pzσμψλ,0,1/2,pz,\displaystyle+\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda\theta(\lambda p_{z})}{e^{\beta\left(|p_{z}|-\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,0,1/2,p_{z}}^{\dagger}\sigma^{\mu}\psi_{\lambda,0,1/2,p_{z}},

where ::\langle:\cdots:\rangle means normal ordering and ensemble average of corresponding field operator (Vasak et al., 1987; Dong et al., 2020), θ(x)\theta(x) is the step function, and we have defined En=pz2+2eBnE_{n}=\sqrt{p_{z}^{2}+2eBn}. The second, third, and fourth lines of Eq. (25) represent the contributions of high Landau levels with m=1/2m=1/2, all Landau levels with m=1/2m=-1/2, and the lowest Landau level with m=1/2m=1/2, respectively. We can see that the macroscopic particle current 𝒥μ\mathcal{J}^{\mu} consists of the summation over the product of the particle number (Fermi-Dirac distribution) and the expectation value in each mode described by the quantum numbers (λ,n,m,pz\lambda,n,m,p_{z}).

IV.2 Particle number density

Firstly we calculate the particle number density ρ𝒥0\rho\equiv\mathcal{J}^{0} of the system. Making use of

ψλnmpzψλnmpz={eB4π2E+Ω/2+pzE+Ω/2,m=12eB4π2EΩ/2pzEΩ/2,m=12\psi_{\lambda nmp_{z}}^{\dagger}\psi_{\lambda nmp_{z}}=\left\{\begin{array}[]{cc}\frac{eB}{4\pi^{2}}\frac{E+\Omega/2+p_{z}}{E+\Omega/2},&m=\frac{1}{2}\\ \frac{eB}{4\pi^{2}}\frac{E-\Omega/2-p_{z}}{E-\Omega/2},&m=-\frac{1}{2}\end{array}\right. (26)

and from Eq. (25) one can obtain

ρβ3=bω16π2+12s=±1ag(a+12sω,b),\rho\beta^{3}=\frac{b\omega}{16\pi^{2}}+\frac{1}{2}\sum_{s=\pm 1}\frac{\partial}{\partial a}g\left(a+\frac{1}{2}s\omega,b\right), (27)

where we have defined three dimensionless quantities a=βμa=\beta\mu, b=2eBβ2b=2eB\beta^{2}, ω=βΩ\omega=\beta\Omega, and have defined g(x,b)g(x,b) as

g(x,b)=b4π20𝑑yn=0s=±1(112δn,0)ln(1+esxnb+y2).g(x,b)=\frac{b}{4\pi^{2}}\int_{0}^{\infty}dy\sum_{n=0}^{\infty}\sum_{s=\pm 1}\left(1-\frac{1}{2}\delta_{n,0}\right)\ln\left(1+e^{sx-\sqrt{nb+y^{2}}}\right). (28)

In a recent article (Zhang et al., 2020a), making use of Abel-Plana formula, the authors obtained the asymptotic expansion of g(x,b)g(x,b) at b=0b=0 as follows

g(x,b)\displaystyle g(x,b) =\displaystyle= (7π2360+x212+x424π2)b2lnb2384π2b296π2ln(e2G6)\displaystyle\left(\frac{7\pi^{2}}{360}+\frac{x^{2}}{12}+\frac{x^{4}}{24\pi^{2}}\right)-\frac{b^{2}\ln b^{2}}{384\pi^{2}}-\frac{b^{2}}{96\pi^{2}}\ln\left(\frac{e}{2G^{6}}\right) (29)
12π2n=0(4n+1)!!(4n+4)!!2n+2C2n+1(x)b2n+2,\displaystyle-\frac{1}{2\pi^{2}}\sum_{n=0}^{\infty}\frac{(4n+1)!!}{(4n+4)!!}\mathscr{B}_{2n+2}C_{2n+1}(x)b^{2n+2},

where G=1.28242G=1.28242... is the Glaisher number, n\mathscr{B}_{n} are Bernoulli numbers, and C2n+1(x)C_{2n+1}(x) is defined and expanded at x=0x=0 in the following,

C2n+1(x)\displaystyle C_{2n+1}(x) =\displaystyle= δn,0+1(4n+1)!0𝑑ylnyd4n+1dy4n+1(1ey+x+1+1eyx+1)\displaystyle-\delta_{n,0}+\frac{1}{(4n+1)!}\int_{0}^{\infty}dy\ln y\frac{d^{4n+1}}{dy^{4n+1}}\left(\frac{1}{e^{y+x}+1}+\frac{1}{e^{y-x}+1}\right) (30)
=\displaystyle= (ln4+γ1)δn,0+2(4n+1)!k=0(24n+2k+11)ζ(4n2k)x2k(2k)!.\displaystyle(\ln 4+\gamma-1)\delta_{n,0}+\frac{2}{(4n+1)!}\sum_{k=0}^{\infty}\left(2^{4n+2k+1}-1\right)\zeta^{\prime}(-4n-2k)\frac{x^{2k}}{(2k)!}.

Plugging Eqs. (29, 30) into Eq. (27), one can get the threefold series expansion of the particle number density at a=0a=0, b=0b=0, ω=0\omega=0 or μ=0\mu=0, B=0B=0, Ω=0\Omega=0 as follows,

ρβ3\displaystyle\rho\beta^{3} =\displaystyle= a6+a36π2+aω28π2+bω16π2\displaystyle\frac{a}{6}+\frac{a^{3}}{6\pi^{2}}+\frac{a\omega^{2}}{8\pi^{2}}+\frac{b\omega}{16\pi^{2}} (31)
1π2n=02n+2b2n+2(4n+4)!!(4n)!!j=0ω2j(2j)!22j\displaystyle-\frac{1}{\pi^{2}}\sum_{n=0}^{\infty}\frac{\mathscr{B}_{2n+2}b^{2n+2}}{(4n+4)!!(4n)!!}\sum_{j=0}^{\infty}\frac{\omega^{2j}}{(2j)!2^{2j}}
×k=0(24n+2k+2j+31)ζ(4n2k2j2)a2k+1(2k+1)!.\displaystyle\times\sum_{k=0}^{\infty}\left(2^{4n+2k+2j+3}-1\right)\zeta^{\prime}(-4n-2k-2j-2)\frac{a^{2k+1}}{(2k+1)!}.

The lower orders O(b2,ω2,bω)O(b^{2},\omega^{2},b\omega) in Eq. (31) are consistent with the perturbative results in (Buzzegoli et al., 2017; Buzzegoli and Becattini, 2018; Yang et al., 2020), where the authors used the approaches of thermal field theory and Wigner function respectively.

IV.3 Particle current along zz-axis

Next we calculate the space components of the particle current 𝒥μ\mathcal{J}^{\mu}. According to the rotation symmetry along zz-axis of the system, the xx- and yy-components of 𝒥μ\mathcal{J}^{\mu} vanish. The unique nonzero component is 𝒥z\mathcal{J}^{z}. Making use of

ψλnmpzσ3ψλnmpz={eB4π2E+Ω/2+pzE+Ω/2,m=12eB4π2EΩ/2pzEΩ/2,m=12\psi_{\lambda nmp_{z}}^{\dagger}\sigma_{3}\psi_{\lambda nmp_{z}}=\left\{\begin{array}[]{cc}\frac{eB}{4\pi^{2}}\frac{E+\Omega/2+p_{z}}{E+\Omega/2},&m=\frac{1}{2}\\ -\frac{eB}{4\pi^{2}}\frac{E-\Omega/2-p_{z}}{E-\Omega/2},&m=-\frac{1}{2}\end{array}\right. (32)

and from Eq. (25) one can obtain

𝒥zβ3=ab8π2+12s=±1sag(a+12sω,b),\mathcal{J}^{z}\beta^{3}=\frac{ab}{8\pi^{2}}+\frac{1}{2}\sum_{s=\pm 1}s\frac{\partial}{\partial a}g\left(a+\frac{1}{2}s\omega,b\right), (33)

which can be expanded as the threefold series at a=0a=0, b=0b=0, ω=0\omega=0 or μ=0\mu=0, B=0B=0, Ω=0\Omega=0 as follows,

𝒥zβ3\displaystyle\mathcal{J}^{z}\beta^{3} =\displaystyle= ab8π2+ω12+ω348π2+ωa24π2\displaystyle\frac{ab}{8\pi^{2}}+\frac{\omega}{12}+\frac{\omega^{3}}{48\pi^{2}}+\frac{\omega a^{2}}{4\pi^{2}} (34)
1π2n=02n+2b2n+2(4n+4)!!(4n)!!j=0ω2j+1(2j+1)!22j+1\displaystyle-\frac{1}{\pi^{2}}\sum_{n=0}^{\infty}\frac{\mathscr{B}_{2n+2}b^{2n+2}}{(4n+4)!!(4n)!!}\sum_{j=0}^{\infty}\frac{\omega^{2j+1}}{(2j+1)!2^{2j+1}}
×k=0(24n+2k+2j+31)ζ(4n2k2j2)a2k(2k)!.\displaystyle\times\sum_{k=0}^{\infty}\left(2^{4n+2k+2j+3}-1\right)\zeta^{\prime}(-4n-2k-2j-2)\frac{a^{2k}}{(2k)!}.

When ω=0\omega=0 or Ω=0\Omega=0 in Eq. (34), one can obtain 𝒥zβ3=ab8π2\mathcal{J}^{z}\beta^{3}=\frac{ab}{8\pi^{2}}, which is the chiral magnetic effect (Kharzeev et al., 2008; Fukushima et al., 2008; Son and Surowka, 2009; Kharzeev and Son, 2011; Son and Yamamoto, 2012); When b=0b=0 or B=0B=0 and keeping the leading order of ω\omega in Eq. (34), one can obtain 𝒥zβ3=ω12(1+3a2π2)\mathcal{J}^{z}\beta^{3}=\frac{\omega}{12}\left(1+\frac{3a^{2}}{\pi^{2}}\right), which is the chiral vortical effect (Landsteiner et al., 2011; Golkar and Son, 2015; Hou et al., 2012; Lin and Yang, 2018; Gao et al., 2019; Shitade et al., 2020).

V Energy-momentum tensor

In this section, we will calculate the energy-momentum tensor 𝒯μν\mathcal{T}^{\mu\nu} (at r=0r=0) of the right-handed fermion system as described in Sec. IV. According to the rotation symmetry along zz-axis, the energy-momentum tensor at r=0r=0 are unchanged under the rotation along zz-axis, which leads to following constraints on 𝒯μν\mathcal{T}^{\mu\nu}:

𝒯01=𝒯02=𝒯12=𝒯13=𝒯23=0,𝒯11=𝒯22.\mathcal{T}^{01}=\mathcal{T}^{02}=\mathcal{T}^{12}=\mathcal{T}^{13}=\mathcal{T}^{23}=0,\ \ \ \mathcal{T}^{11}=\mathcal{T}^{22}. (35)

The possible nonzero components of 𝒯μν\mathcal{T}^{\mu\nu} are 𝒯00\mathcal{T}^{00}, 𝒯11=𝒯22\mathcal{T}^{11}=\mathcal{T}^{22}, 𝒯33\mathcal{T}^{33}, and 𝒯03\mathcal{T}^{03}.

For the right-handed fermion system, the field operator of the symmetric energy-momentum tensor at r=0r=0 is

Tμν=14(ψRiσμDRνψR+ψRiσνDRμψR+H.C.),T^{\mu\nu}=\frac{1}{4}\left(\psi_{R}^{\dagger}i\sigma^{\mu}D_{R}^{\nu}\psi_{R}+\psi_{R}^{\dagger}i\sigma^{\nu}D_{R}^{\mu}\psi_{R}+\mathrm{H.C.}\right), (36)

with σμ=(1,𝝈)\sigma^{\mu}=(1,\boldsymbol{\sigma}) and the right-handed covariant derivative DRμD_{R}^{\mu} defined as

DRμ=(ti2Ωσ3,x,y,z).D_{R}^{\mu}=\left(\partial_{t}-\frac{i}{2}\Omega\sigma_{3},-\partial_{x},-\partial_{y},-\partial_{z}\right). (37)

The macroscopic energy-momentum tensor 𝒯μν\mathcal{T}^{\mu\nu} can be calculated from TμνT^{\mu\nu} as follows,

𝒯μν\displaystyle\mathcal{T}^{\mu\nu} =\displaystyle= :Tμν:\displaystyle\langle:T^{\mu\nu}:\rangle
=\displaystyle= 14n=1λ𝑑pzλeβ(EnλΩ/2λμ)+1ψλ,n,1/2,pz(iσμDRν+iσνDRμ)ψλ,n,1/2,pz\displaystyle\frac{1}{4}\sum_{n=1}^{\infty}\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda}{e^{\beta\left(E_{n}-\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,n,1/2,p_{z}}^{\dagger}\left(i\sigma^{\mu}D_{R}^{\nu}+i\sigma^{\nu}D_{R}^{\mu}\right)\psi_{\lambda,n,1/2,p_{z}}
+14n=0λ𝑑pzλeβ(En+1+λΩ/2λμ)+1ψλ,n,1/2,pz(iσμDRν+iσνDRμ)ψλ,n,1/2,pz\displaystyle+\frac{1}{4}\sum_{n=0}^{\infty}\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda}{e^{\beta\left(E_{n+1}+\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,n,-1/2,p_{z}}^{\dagger}\left(i\sigma^{\mu}D_{R}^{\nu}+i\sigma^{\nu}D_{R}^{\mu}\right)\psi_{\lambda,n,-1/2,p_{z}}
+14λ𝑑pzλθ(λpz)eβ(|pz|λΩ/2λμ)+1ψλ,0,1/2,pz(iσμDRν+iσνDRμ)ψλ,0,1/2,pz+H.C.\displaystyle+\frac{1}{4}\sum_{\lambda}\int_{-\infty}^{\infty}dp_{z}\frac{\lambda\theta(\lambda p_{z})}{e^{\beta\left(|p_{z}|-\lambda\Omega/2-\lambda\mu\right)}+1}\psi_{\lambda,0,1/2,p_{z}}^{\dagger}\left(i\sigma^{\mu}D_{R}^{\nu}+i\sigma^{\nu}D_{R}^{\mu}\right)\psi_{\lambda,0,1/2,p_{z}}+\mathrm{H.C.}

V.1 Energy density

Firstly we calculate the energy density ε𝒯00\varepsilon\equiv\mathcal{T}^{00} of the system. Making use of

ψλnmpz(it+12Ωσ3)ψλnmpz={eB8π2(E+pz+Ω/2),m=12eB8π2(EpzΩ/2),m=12\psi_{\lambda nmp_{z}}^{\dagger}\left(i\partial_{t}+\frac{1}{2}\Omega\sigma_{3}\right)\psi_{\lambda nmp_{z}}=\left\{\begin{array}[]{cc}\frac{eB}{8\pi^{2}}(E+p_{z}+\Omega/2),&m=\frac{1}{2}\\ \frac{eB}{8\pi^{2}}(E-p_{z}-\Omega/2),&m=-\frac{1}{2}\end{array}\right. (39)

and from Eq. (25) one can obtain

εβ4=abω16π2+s=±1(32bb)g(a+12sω,b),\varepsilon\beta^{4}=\frac{ab\omega}{16\pi^{2}}+\sum_{s=\pm 1}\left(\frac{3}{2}-b\frac{\partial}{\partial b}\right)g\left(a+\frac{1}{2}s\omega,b\right), (40)

which can be expanded as the threefold series at a=0a=0, b=0b=0, ω=0\omega=0 or μ=0\mu=0, B=0B=0, Ω=0\Omega=0 as follows,

εβ4\displaystyle\varepsilon\beta^{4} =\displaystyle= 7π2120+a24+ω216+a48π2+3a2ω216π2+ω4128π2\displaystyle\frac{7\pi^{2}}{120}+\frac{a^{2}}{4}+\frac{\omega^{2}}{16}+\frac{a^{4}}{8\pi^{2}}+\frac{3a^{2}\omega^{2}}{16\pi^{2}}+\frac{\omega^{4}}{128\pi^{2}} (41)
+abω16π2+b2lnb2384π2+b296π2ln(2eγ+1G6)\displaystyle+\frac{ab\omega}{16\pi^{2}}+\frac{b^{2}\ln b^{2}}{384\pi^{2}}+\frac{b^{2}}{96\pi^{2}}\ln\left(\frac{2e^{\gamma+1}}{G^{6}}\right)
+1π2n=0(4n+1)2n+2b2n+2(4n+4)!!(4n)!!j=0ω2j(2j)!22j\displaystyle+\frac{1}{\pi^{2}}\sum_{n=0}^{\infty}\frac{(4n+1)\mathscr{B}_{2n+2}b^{2n+2}}{(4n+4)!!(4n)!!}\sum_{j=0}^{\infty}\frac{\omega^{2j}}{(2j)!2^{2j}}
×k=0(24n+2k+2j+11)ζ(4n2k2j)a2k(2k)!,\displaystyle\times\sum_{k=0}^{\infty}\left(2^{4n+2k+2j+1}-1\right)\zeta^{\prime}(-4n-2k-2j)\frac{a^{2k}}{(2k)!},

where the logarithmic term b2lnb2b^{2}\ln b^{2} has been discussed in detail in (Zhang et al., 2020a), and its coefficient is independent of ω\omega in this work. It is worth noting that there would be no such logarithmic term if the un-normal ordering description of field operators was adopted (Sheng et al., 2018; Yang et al., 2020).

V.2 Pressure

The pressure PP of the system is 𝒯33\mathcal{T}^{33}. Making use of

ψλnmpzσ3(iz)ψλnmpz={eB8π2(E+pz+Ω/2)pzE+Ω/2,m=12eB8π2(EpzΩ/2)pzEΩ/2,m=12\psi_{\lambda nmp_{z}}^{\dagger}\sigma_{3}(-i\partial_{z})\psi_{\lambda nmp_{z}}=\left\{\begin{array}[]{cc}\frac{eB}{8\pi^{2}}\frac{(E+p_{z}+\Omega/2)p_{z}}{E+\Omega/2},&m=\frac{1}{2}\\ -\frac{eB}{8\pi^{2}}\frac{(E-p_{z}-\Omega/2)p_{z}}{E-\Omega/2},&m=-\frac{1}{2}\end{array}\right. (42)

and from Eq. (25) one can obtain

Pβ4=abω16π2+12s=±1g(a+12sω,b),P\beta^{4}=\frac{ab\omega}{16\pi^{2}}+\frac{1}{2}\sum_{s=\pm 1}g\left(a+\frac{1}{2}s\omega,b\right), (43)

which can be expanded as the threefold series at a=0a=0, b=0b=0, ω=0\omega=0 or μ=0\mu=0, B=0B=0, Ω=0\Omega=0 as follows,

Pβ4\displaystyle P\beta^{4} =\displaystyle= 7π2360+a212+ω248+a424π2+a2ω216π2+ω4384π2\displaystyle\frac{7\pi^{2}}{360}+\frac{a^{2}}{12}+\frac{\omega^{2}}{48}+\frac{a^{4}}{24\pi^{2}}+\frac{a^{2}\omega^{2}}{16\pi^{2}}+\frac{\omega^{4}}{384\pi^{2}} (44)
+abω16π2b2lnb2384π2b296π2ln(2eγG6)\displaystyle+\frac{ab\omega}{16\pi^{2}}-\frac{b^{2}\ln b^{2}}{384\pi^{2}}-\frac{b^{2}}{96\pi^{2}}\ln\left(\frac{2e^{\gamma}}{G^{6}}\right)
1π2n=02n+2b2n+2(4n+4)!!(4n)!!j=0ω2j(2j)!22j\displaystyle-\frac{1}{\pi^{2}}\sum_{n=0}^{\infty}\frac{\mathscr{B}_{2n+2}b^{2n+2}}{(4n+4)!!(4n)!!}\sum_{j=0}^{\infty}\frac{\omega^{2j}}{(2j)!2^{2j}}
×k=0(24n+2k+2j+11)ζ(4n2k2j)a2k(2k)!.\displaystyle\times\sum_{k=0}^{\infty}\left(2^{4n+2k+2j+1}-1\right)\zeta^{\prime}(-4n-2k-2j)\frac{a^{2k}}{(2k)!}.

One can obtain 𝒯11\mathcal{T}^{11} from the traceless condition for energy-momentum tensor, 𝒯00=2𝒯11+𝒯33\mathcal{T}^{00}=2\mathcal{T}^{11}+\mathcal{T}^{33}.

V.3 Energy current

The energy current along zz-axis is 𝒯03\mathcal{T}^{03}. Making use of

ψλnmpz(iz+σ3it+12Ω)ψλnmpz={eB8π2(E+pz+Ω/2)2E+Ω/2,m=12eB8π2(EpzΩ/2)2EΩ/2,m=12\psi_{\lambda nmp_{z}}^{\dagger}\left(-i\partial_{z}+\sigma_{3}i\partial_{t}+\frac{1}{2}\Omega\right)\psi_{\lambda nmp_{z}}=\left\{\begin{array}[]{cc}\frac{eB}{8\pi^{2}}\frac{(E+p_{z}+\Omega/2)^{2}}{E+\Omega/2},&m=\frac{1}{2}\\ -\frac{eB}{8\pi^{2}}\frac{(E-p_{z}-\Omega/2)^{2}}{E-\Omega/2},&m=-\frac{1}{2}\end{array}\right. (45)

and from Eq. (25) one can obtain

𝒯03β4=b8π2(π26+ω28+a22)+s=±1s(1b2b)g(a+12sω,b),\mathcal{T}^{03}\beta^{4}=\frac{b}{8\pi^{2}}\left(\frac{\pi^{2}}{6}+\frac{\omega^{2}}{8}+\frac{a^{2}}{2}\right)+\sum_{s=\pm 1}s\left(1-\frac{b}{2}\frac{\partial}{\partial b}\right)g\left(a+\frac{1}{2}s\omega,b\right), (46)

which can be expanded as the threefold series at a=0a=0, b=0b=0, ω=0\omega=0 or μ=0\mu=0, B=0B=0, Ω=0\Omega=0 as follows,

𝒯03β4\displaystyle\mathcal{T}^{03}\beta^{4} =\displaystyle= b8π2(π26+ω28+a22)+(aω6+a3ω6π2+aω324π2)\displaystyle\frac{b}{8\pi^{2}}\left(\frac{\pi^{2}}{6}+\frac{\omega^{2}}{8}+\frac{a^{2}}{2}\right)+\left(\frac{a\omega}{6}+\frac{a^{3}\omega}{6\pi^{2}}+\frac{a\omega^{3}}{24\pi^{2}}\right) (47)
+2π2n=1n2n+2b2n+2(4n+4)!!(4n)!!j=0ω2j+1(2j+1)!22j+1\displaystyle+\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{n\mathscr{B}_{2n+2}b^{2n+2}}{(4n+4)!!(4n)!!}\sum_{j=0}^{\infty}\frac{\omega^{2j+1}}{(2j+1)!2^{2j+1}}
×k=0(24n+2k+2j+31)ζ(4n2k2j2)a2k+1(2k+1)!.\displaystyle\times\sum_{k=0}^{\infty}\left(2^{4n+2k+2j+3}-1\right)\zeta^{\prime}(-4n-2k-2j-2)\frac{a^{2k+1}}{(2k+1)!}.

Up to now, we have obtained all thermodynamical quantities of the right-handed fermion system. For left-handed fermion system, one can derive corresponding quantities from the right-handed case through space inversion: ρRρL\rho_{R}\rightarrow\rho_{L}, 𝒥Rz𝒥Lz\mathcal{J}_{R}^{z}\rightarrow-\mathcal{J}_{L}^{z}, εRεL\varepsilon_{R}\rightarrow\varepsilon_{L}, PRPLP_{R}\rightarrow P_{L}, 𝒯R03𝒯L03\mathcal{T}_{R}^{03}\rightarrow-\mathcal{T}_{L}^{03}, μRμL\mu_{R}\rightarrow\mu_{L}, BBB\rightarrow B, ΩΩ\Omega\rightarrow\Omega, where the subscripts R,LR,L are used to distinguish the quantities in right-handed case from that in left-handed case.

VI Zero temperature limit

Now we turn to the thermodynamics of the system at zero temperature limit. When the temperature tends to be zero, with chemical potential μ\mu, magnetic field BB, and angular velocity Ω\Omega fixed, then the three dimensionless quantities a=βμa=\beta\mu, b=2eBβ2b=2eB\beta^{2}, ω=βΩ\omega=\beta\Omega all tend to be infinity. The asymptotic behavior of g(x,b)g(x,b) as xx\rightarrow\infty and bb\rightarrow\infty has been obtained in (Zhang et al., 2020a),

limx,bg(x,b)=x2b16π2.\lim_{x,b\rightarrow\infty}g(x,b)=\frac{x^{2}b}{16\pi^{2}}. (48)

From Eqs. (27, 33), one can derive the expressions of the particle density ρ\rho and the current 𝒥z\mathcal{J}^{z} at zero temperature limit as follows,

ρ=𝒥z=eB4π2(μ+Ω2).\rho=\mathcal{J}^{z}=\frac{eB}{4\pi^{2}}\left(\mu+\frac{\Omega}{2}\right). (49)

At zero temperature limit, due to the coupling of the spin with the magnetic field and the angular velocity, the spin alignment of all particles and antiparticles will be along zz-axis of the system. Since these particles are right-handed, they will move along zz-axis with the speed of light cc (c=1c=1 in natural unit), so it is reasonable that the particle density ρ\rho equals to the zz-component current 𝒥z\mathcal{J}^{z} at zero temperature limit.

From Eqs. (40, 43, 46), the expressions of energy density ε\varepsilon, pressure PP and energy current 𝒯03\mathcal{T}^{03} at zero temperature limit are

ε=P=𝒯03=eB8π2(μ+Ω2)2.\varepsilon=P=\mathcal{T}^{03}=\frac{eB}{8\pi^{2}}\left(\mu+\frac{\Omega}{2}\right)^{2}. (50)

The movements of the particles and antiparticles with the speed of light along zz-axis leads to the equality of the energy density ε\varepsilon and the energy current 𝒯03\mathcal{T}^{03}. Since there is no energy current along the direction of the xx- and yy-axis, then 𝒯11\mathcal{T}^{11} and 𝒯22\mathcal{T}^{22} vanish in this system, which results in the equality of the energy density ε\varepsilon and the pressure PP.

VII Summary

In this article, we have investigated the thermodynamics of the uniformly rotating right-handed fermion system under the background of a uniform magnetic field through the approach of normal ordering and ensemble average, where all thermodynamical quantities are expanded as threefold series at B=0B=0, Ω=0\Omega=0 and μ=0\mu=0. For these threefold series, our results at lower orders are consistent with previous ones by other authors. It is worth pointing out that the general orders of BB and Ω\Omega in the expressions of the thermodynamical quantities are obtained for the first time and can provide a useful reference for the high order calculations from several different approaches, such as thermal field theory and Wigner function. We also calculate all quantities in zero temperature limit, and obtain the equality of partcile/energy density and corresponding currents along zz-axis. Since for the chiral fermion the right-handed part decouples from the left-handed part, in this article we only considered the case of the right-handed fermion system, which can be directly generalized to the left-handed case through space inversion. In this article, the currents and energy-momentum tensor are calculated at the rotating axis (r=0r=0), so the boundary condition at the speed-of-light surface will not affect our results. The calculations for these quantities off or far from the rotating axis (r0r\neq 0) as well as with the boundary condition at the speed-of-light surface may be investigated in the future.

VIII Acknowledgments

I thank De-Fu Hou for helpful discussion. This work was supported by the National Natural Science Foundation of China under Grants No. 11890713, and No. 12073008.

Appendix A Landau levels for a single right-handed fermion

The Hamiltonian for a right-handed fermion under the background of the uniform magnetic field 𝐁=B𝐞z\mathbf{B}=B\mathbf{e}_{z} is

H=i𝝈(ie𝐀)=i𝝈+12eB(yσ1xσ2),H=-i\boldsymbol{\sigma}\cdot(\nabla-ie\mathbf{A})=-i\boldsymbol{\sigma}\cdot\nabla+\frac{1}{2}eB(y\sigma_{1}-x\sigma_{2}), (51)

where we have chosen 𝐀=(12By,12Bx,0)\mathbf{A}=(-\frac{1}{2}By,\frac{1}{2}Bx,0) for the gauge potential. One can refer to (Sheng et al., 2018; Sheng, 2019; Dong et al., 2020) for other choices of the gauge potential.

In the following, we will solve the eigenvalue equation of HH in cylindrical coordinates,

Hψ=Eψ.H\psi=E\psi. (52)

We can see that the three Hermitian operators, HH, p^z=iz\hat{p}_{z}=-i\partial_{z}, J^z=12σ3+(xp^yyp^x)\hat{J}_{z}=\frac{1}{2}\sigma_{3}+(x\hat{p}_{y}-y\hat{p}_{x}) are commutative with each other, so the eigenfunction ψ\psi can be chosen as

ψ=(f(r)ei(m12)ϕig(r)ei(m+12)ϕ)eizpz,\psi=\left(\begin{array}[]{c}f(r)e^{i(m-\frac{1}{2})\phi}\\ ig(r)e^{i(m+\frac{1}{2})\phi}\end{array}\right)e^{izp_{z}}, (53)

where <pz<-\infty<p_{z}<\infty and m=±1/2,±3/2,±5/2,m=\pm 1/2,\pm 3/2,\pm 5/2,... are the eigenvalues of p^z\hat{p}_{z} and J^z\hat{J}_{z} respectively. The explicit form of the Hamiltonian HH in cylindrical coordinates is

H=(izeiϕ(ir1rϕ+i2eBr)eiϕ(ir+1rϕi2eBr)iz),H=\left(\begin{array}[]{cc}-i\frac{\partial}{\partial z}&e^{-i\phi}\left(-i\frac{\partial}{\partial r}-\frac{1}{r}\frac{\partial}{\partial\phi}+\frac{i}{2}eBr\right)\\ e^{i\phi}\left(-i\frac{\partial}{\partial r}+\frac{1}{r}\frac{\partial}{\partial\phi}-\frac{i}{2}eBr\right)&i\frac{\partial}{\partial z}\end{array}\right), (54)

then from Eq. (52) we can obtain two differential equations for f(r),g(r)f(r),g(r) as follows,

(pzE)f(r)+(r+m+12r12eBr)g(r)\displaystyle(p_{z}-E)f(r)+\left(\frac{\partial}{\partial r}+\frac{m+\frac{1}{2}}{r}-\frac{1}{2}eBr\right)g(r) =\displaystyle= 0,\displaystyle 0, (55)
(r+m12r12eBr)f(r)+(pzE)g(r)\displaystyle\left(-\frac{\partial}{\partial r}+\frac{m-\frac{1}{2}}{r}-\frac{1}{2}eBr\right)f(r)+(-p_{z}-E)g(r) =\displaystyle= 0,\displaystyle 0, (56)

which are equivalent to

{2r2+1rr(m12)2r2[pz2E2eB(m+12)]14e2B2r2}f(r)=0\left\{\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{\left(m-\frac{1}{2}\right)^{2}}{r^{2}}-\left[p_{z}^{2}-E^{2}-eB\left(m+\frac{1}{2}\right)\right]-\frac{1}{4}e^{2}B^{2}r^{2}\right\}f(r)=0 (57)
g(r)=1pz+E(r+m12r12eBr)f(r)g(r)=\frac{1}{p_{z}+E}\left(-\frac{\partial}{\partial r}+\frac{m-\frac{1}{2}}{r}-\frac{1}{2}eBr\right)f(r) (58)

We can define a dimensionless variable ρ=12eBr2\rho=\frac{1}{2}eBr^{2}, then

ddr=eBrddρ,d2dr2=eBddρ+2eBρd2dρ2.\frac{d}{dr}=eBr\frac{d}{d\rho},\ \ \ \ \frac{d^{2}}{dr^{2}}=eB\frac{d}{d\rho}+2eB\rho\frac{d^{2}}{d\rho^{2}}. (59)

Now Eq. (57) becomes

{ρd2dρ2+ddρ(m12)24ρ12eB[pz2E2eB(m+12)]14ρ}f=0\left\{\rho\frac{d^{2}}{d\rho^{2}}+\frac{d}{d\rho}-\frac{\left(m-\frac{1}{2}\right)^{2}}{4\rho}-\frac{1}{2eB}\left[p_{z}^{2}-E^{2}-eB\left(m+\frac{1}{2}\right)\right]-\frac{1}{4}\rho\right\}f=0 (60)

Next, we choose

f=eρ2ρm214G(ρ),f=e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}-\frac{1}{4}}G(\rho), (61)

then Eqs. (58, 60) become

g=2eBE+pzeρ2ρm2+14G(ρ),g=-\frac{\sqrt{2eB}}{E+p_{z}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}+\frac{1}{4}}G^{\prime}(\rho), (62)
ρG′′+[(m+12)ρ]G12eB(pz2E2)G=0.\rho G^{\prime\prime}+\left[\left(m+\frac{1}{2}\right)-\rho\right]G^{\prime}-\frac{1}{2eB}\left(p_{z}^{2}-E^{2}\right)G=0. (63)

Define following two quantities,

γ=m+12,α=12eB(pz2E2),\gamma=m+\frac{1}{2},\ \ \ \ \alpha=\frac{1}{2eB}\left(p_{z}^{2}-E^{2}\right), (64)

then Eq. (63) becomes

ρG′′+(γρ)GαG=0,\rho G^{\prime\prime}+(\gamma-\rho)G^{\prime}-\alpha G=0, (65)

which is the confluent hypergeometric equation (Zeng, 2007). With the boundary conditions, |f(0)|,|f()|<|f(0)|,|f(\infty)|<\infty, the solutions for G(ρ)G(\rho), f(ρ)f(\rho), g(ρ)g(\rho) can be chosen as:

(1) When γ=0,1,2,\gamma=0,-1,-2,..., i.e. m=1/2,3/2,5/2m=-1/2,-3/2,-5/2, …, the boundary condition |f(0)|<|f(0)|<\infty requires that

G(ρ)=ρ1γF(αγ+1,2γ,ρ)=ρ12mF(αm+12,32m,ρ),G(\rho)=\rho^{1-\gamma}F(\alpha-\gamma+1,2-\gamma,\rho)=\rho^{\frac{1}{2}-m}F\left(\alpha-m+\frac{1}{2},\frac{3}{2}-m,\rho\right), (66)

where F(α,γ,ρ)F(\alpha,\gamma,\rho) is the confluent hypergeometric function as discussed in Appendix B. In addition, the boundary condition |f()|<|f(\infty)|<\infty requires that

αm+12=n(n),E=λpz2+2eB(nm+12)(λ=±1),\alpha-m+\frac{1}{2}=-n\ \ (n\in\mathbb{N}),\ \ \ E=\lambda\sqrt{p_{z}^{2}+2eB\left(n-m+\frac{1}{2}\right)}\ \ \ (\lambda=\pm 1), (67)
G(ρ)=ρ12mF(n,32m,ρ)ρ12mLn12m(ρ),G(\rho)=\rho^{\frac{1}{2}-m}F\left(-n,\frac{3}{2}-m,\rho\right)\sim\rho^{\frac{1}{2}-m}L_{n}^{\frac{1}{2}-m}(\rho), (68)

where Lnk(ρ)L_{n}^{k}(\rho) is the general Laguerre polynomial as discussed in Appendix B. Then one obtain

f(ρ)eρ2ρ14m2Ln12m(ρ),g(ρ)2eBE+pz(nm+12)eρ2ρ14m2Ln12m(ρ).f(\rho)\sim e^{-\frac{\rho}{2}}\rho^{\frac{1}{4}-\frac{m}{2}}L_{n}^{\frac{1}{2}-m}(\rho),\ \ \ \ g(\rho)\sim-\frac{\sqrt{2eB}}{E+p_{z}}\left(n-m+\frac{1}{2}\right)e^{-\frac{\rho}{2}}\rho^{-\frac{1}{4}-\frac{m}{2}}L_{n}^{-\frac{1}{2}-m}(\rho). (69)

(2) When γ=1,2,3,\gamma=1,2,3,..., i.e. m=1/2,3/2,5/2,m=1/2,3/2,5/2,..., the boundary condition |f(0)|<|f(0)|<\infty requires that

G(ρ)=F(α,γ,ρ)=F(α,12+m,ρ).G(\rho)=F(\alpha,\gamma,\rho)=F\left(\alpha,\frac{1}{2}+m,\rho\right). (70)

In addition, the boundary condition |f()|<|f(\infty)|<\infty requires that

α=n(n),E=λpz2+2eBn(λ=±1),\alpha=-n\ \ (n\in\mathbb{N}),\ \ \ E=\lambda\sqrt{p_{z}^{2}+2eBn}\ \ \ (\lambda=\pm 1), (71)
G(ρ)=F(n,12+m,ρ)Lnm12(ρ).G(\rho)=F\left(-n,\frac{1}{2}+m,\rho\right)\sim L_{n}^{m-\frac{1}{2}}(\rho). (72)

Then one obtain

f(ρ)eρ2ρm214Lnm12(ρ),g(ρ)2eBE+pzeρ2ρm2+14Ln1m+12(ρ).f(\rho)\sim e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}-\frac{1}{4}}L_{n}^{m-\frac{1}{2}}(\rho),\ \ \ \ g(\rho)\sim\frac{\sqrt{2eB}}{E+p_{z}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}+\frac{1}{4}}L_{n-1}^{m+\frac{1}{2}}(\rho). (73)

There is a special case we must point out here: When m>0m>0, n=0n=0, we must choose E=pzE=p_{z}, in which case we have f(ρ)=eρ2ρm214f(\rho)=e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}-\frac{1}{4}}, g(ρ)=0g(\rho)=0. There is no physical solution for m>0m>0, n=0n=0, E=pzE=-p_{z}.

Making use of the orthonormal relation of the general Laguerre polynomials,

0𝑑xexxγLmγ(x)Lnγ(x)=Γ(n+γ+1)n!δmn,\int_{0}^{\infty}dxe^{-x}x^{\gamma}L_{m}^{\gamma}(x)L_{n}^{\gamma}(x)=\frac{\Gamma(n+\gamma+1)}{n!}\delta_{mn}, (74)

we can obtain the normalized eigenfunctions as follows:

When m<0m<0,

ψλnmpz=n!(nm+12)!(eB(E+pz)2Eeρ2ρ14m2Ln12mei(m12)ϕiλeB(nm+12)E(E+pz)eρ2ρ14m2Ln12mei(m+12)ϕ)eizpz2π,\psi_{\lambda nmp_{z}}=\sqrt{\frac{n!}{(n-m+\frac{1}{2})!}}\left(\begin{array}[]{c}\sqrt{\frac{eB(E+p_{z})}{2E}}e^{-\frac{\rho}{2}}\rho^{\frac{1}{4}-\frac{m}{2}}L_{n}^{\frac{1}{2}-m}e^{i(m-\frac{1}{2})\phi}\\ -\frac{i\lambda eB(n-m+\frac{1}{2})}{\sqrt{E(E+p_{z})}}e^{-\frac{\rho}{2}}\rho^{-\frac{1}{4}-\frac{m}{2}}L_{n}^{-\frac{1}{2}-m}e^{i(m+\frac{1}{2})\phi}\end{array}\right)\frac{e^{izp_{z}}}{2\pi}, (75)
E=λpz2+2eB(nm+12).E=\lambda\sqrt{p_{z}^{2}+2eB\left(n-m+\frac{1}{2}\right)}. (76)

When m>0m>0,

ψλnmpz=n!(n+m12)!(eB(E+pz)2Eeρ2ρm214Lnm12ei(m12)ϕiλeBE(E+pz)eρ2ρm2+14Ln1m+12ei(m+12)ϕ)eizpz2π,\psi_{\lambda nmp_{z}}=\sqrt{\frac{n!}{(n+m-\frac{1}{2})!}}\left(\begin{array}[]{c}\sqrt{\frac{eB(E+p_{z})}{2E}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}-\frac{1}{4}}L_{n}^{m-\frac{1}{2}}e^{i(m-\frac{1}{2})\phi}\\ \frac{i\lambda eB}{\sqrt{E(E+p_{z})}}e^{-\frac{\rho}{2}}\rho^{\frac{m}{2}+\frac{1}{4}}L_{n-1}^{m+\frac{1}{2}}e^{i(m+\frac{1}{2})\phi}\end{array}\right)\frac{e^{izp_{z}}}{2\pi}, (77)
E=λpz2+2eBn.E=\lambda\sqrt{p_{z}^{2}+2eBn}. (78)

All normalized eigenfunctions are orthogonal with each other,

𝑑Vψλnmpzψλnmpz=δλλδnnδmmδ(pzpz).\int dV\psi_{\lambda^{\prime}n^{\prime}m^{\prime}p_{z}^{\prime}}^{\dagger}\psi_{\lambda nmp_{z}}=\delta_{\lambda^{\prime}\lambda}\delta_{n^{\prime}n}\delta_{m^{\prime}m}\delta(p_{z}^{\prime}-p_{z}). (79)

Appendix B Confluent hypergeometric function and Laguerre polynomial

The confluent hypergeometric equation is (Zeng, 2007)

zy′′+(γz)yαy=0.zy^{\prime\prime}+(\gamma-z)y^{\prime}-\alpha y=0. (80)

When γ\gamma\notin\mathbb{Z}, there are two independent solutions as follows,

y1\displaystyle y_{1} =\displaystyle= F(α,γ,z),\displaystyle F(\alpha,\gamma,z),
y2\displaystyle y_{2} =\displaystyle= z1γF(αγ+1,2γ,z),\displaystyle z^{1-\gamma}F(\alpha-\gamma+1,2-\gamma,z), (81)

where F(α,γ,z)F(\alpha,\gamma,z) is the confluent hypergeometric function defined as

F(α,γ,z)=k=0(α)k(γ)kzkk!1+αγz+α(α+1)γ(γ+1)z22!+α(α+1)(α+2)γ(γ+1)(γ+2)z33!+.F(\alpha,\gamma,z)=\sum_{k=0}^{\infty}\frac{(\alpha)_{k}}{(\gamma)_{k}}\frac{z^{k}}{k!}\equiv 1+\frac{\alpha}{\gamma}z+\frac{\alpha(\alpha+1)}{\gamma(\gamma+1)}\frac{z^{2}}{2!}+\frac{\alpha(\alpha+1)(\alpha+2)}{\gamma(\gamma+1)(\gamma+2)}\frac{z^{3}}{3!}+\cdots. (82)

The asymptotic behavior of F(α,γ,z)F(\alpha,\gamma,z) as zz\rightarrow\infty is the same as eze^{z}. When α\alpha is a non-positive integer, then F(α,γ,z)F(\alpha,\gamma,z) becomes a polynomial.

The general Laguerre polynomial Lnγ(z)L_{n}^{\gamma}(z) is defined from F(α,γ,z)F(\alpha,\gamma,z) as follows (Gradshteyn and Ryzhik, 2014),

Lnγ(z)=Γ(γ+n+1)n!Γ(γ+1)F(n,γ+1,z)=(γ+nn)F(n,γ+1,z),L_{n}^{\gamma}(z)=\frac{\Gamma(\gamma+n+1)}{n!\Gamma(\gamma+1)}F(-n,\gamma+1,z)=\left(\begin{array}[]{c}\gamma+n\\ n\end{array}\right)F(-n,\gamma+1,z), (83)

where γ\gamma\in\mathbb{R} and nn\in\mathbb{N}. Laguerre polynomial Lnγ(z)L_{n}^{\gamma}(z) satisfies following differential equation

zy′′+(γ+1z)y+ny=0zy^{\prime\prime}+(\gamma+1-z)y^{\prime}+ny=0 (84)

We can rewrite Eq. (84) as a type of Sturm-Liouvelle equation,

ddz(zγ+1ezdydz)+nzγezy=0,\frac{d}{dz}\bigg{(}z^{\gamma+1}e^{-z}\frac{dy}{dz}\bigg{)}+nz^{\gamma}e^{-z}y=0, (85)

which gives the orthogonality of Lnγ(z)L_{n}^{\gamma}(z),

0𝑑zezzγLmγ(z)Lnγ(z)=Γ(n+γ+1)n!δmn.\int_{0}^{\infty}dze^{-z}z^{\gamma}L_{m}^{\gamma}(z)L_{n}^{\gamma}(z)=\frac{\Gamma(n+\gamma+1)}{n!}\delta_{mn}. (86)

When γ=0\gamma=0, then Lnγ(z)L_{n}^{\gamma}(z) becomes the normal Laguerre polynomial Ln(z)L_{n}(z),

Ln(z)=Ln0(z)=F(n,1,z).L_{n}(z)=L_{n}^{0}(z)=F(-n,1,z). (87)

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