Chinese Physics C Vol. 44, No. 7 (2020) 074106

Chiral magnetic effect for chiral system*

1) 2) Ren-Da Dong(董仁达) Ren-Hong Fang(方仁洪) De-Fu Hou(侯德富) Duan She(佘端) Institute of Particle Physics and Key Laboratory of Quark and Lepton Physics (MOS), Central China Normal University, Wuhan 430079, China

Abstract: The chiral magnetic effect is concisely derived by employing the Wigner function approach in the chiral fermion system. Subsequently, the chiral magnetic effect is derived by solving the Landau levels of chiral in detail. The second quantization and ensemble average leads to the equation of the chiral magnetic effect for righthand and lefthand fermion systems. The chiral magnetic effect arises uniquely from the contribution of the lowest Landau level. We carefully analyze the lowest Landau level and find that all righthand (chirality is +1) fermions move along the direction of the , whereas all lefthand (chirality is −1) fermions move in the opposite direction of the magnetic field. Hence, the chiral magnetic effect can be explained clearly using a microscopic approach. Keywords: CME, Landau levels, chiral fermions DOI: 10.1088/1674-1137/44/7/074106

1 Introduction the thermodynamic potential Ω. The macroscopic elec- tric current jz along the z-axis can be obtained from the thermodynamic potential Ω. Another study on the CME Quark gluon plasma (QGP) is created in high energy addressing Landau levels is related to the second quantiz- heavy ion collisions, constituting extremely hot and dense ation of the Dirac field. In Ref. [21], the authors determ- matter. An enormous magnetic field can be generated by ined the Landau levels and corresponding Landau wave- high energy peripheral collisions [1-3]. One of the predic- functions for the massive Dirac equation in a uniform tions in QGP is that positively and negatively charged magnetic field, likewise with chemical potential µ and particles seperate along the direction of the magnetic chiral chemical potential µ5. Then, they second-quant- field, which is related to chiral magnetic effect (CME) [4- ized the Dirac field and expanded it by these solved 6]. Numerous efforts have been made to determine the Landau wavefunctions and creation/construction operat- CME in experiments [7-9]. However, due to background ors. The density operator ρˆ can then be determined from noise, no definite CME has been revealed to date. Numer- Hamiltonian Hˆ and particle number operator Nˆ of the ous theoretical methods likewise investigated the CME, system. Finally, they derived the macroscopic electric such as AdS/CFT [10, 11], hydrodynamics [12-14], fi- current jz along the z-axis through the trace of density op- nite temperature field theory [15-18], quantum kinetic erator ρˆ and electric current operator ˆjz , which is simply theory [19], lattice method [20], et al. the CME equation. In this article, we study the CME in detail by determ- From the study on CME for massive Dirac fermions ination of Landau levels. For the massive Dirac fermion through Landau levels, we conclude that the contribution system, several studies on CME addressed Landau levels. to CME arises uniquely from the lowest Landau level, In Ref. [15], Fukushima et al. proposed four methods to while the contributions from higher Landau levels cancel derive the CME. One of these methods made use of each other. However, because of the mass m of the Dirac Landau energy levels for the massive Dirac equation with fermion, the physical picture of the CME for the massive chemical potential µ and chiral chemical potential µ5 in a Dirac fermion system is not as clear as in the massless homogeneous magnetic background B = Bez to construct fermion case, as the physical meaning of the chiral chem-

Received 25 January 2020, Published online 12 May 2020 * R.-H. F. is Supported by the National Natural Science Foundation of China (11847220); D.-F. H. is in part Supported by the National Natural Science Foundation of China (11735007, 11890711) 1) E-mail: [email protected] 2) E-mail: [email protected] Content from this work may be used under the terms of the Creative Commons Attribution 3.0 licence. Any further distribution of this work must main- tain attribution to the author(s) and the title of the work, journal citation and DOI. Article funded by SCOAP3 and published under licence by Chinese Physical Society and the Institute of High Energy Physics of the Chinese Academy of Sciences and the Institute of Modern Physics of the Chinese Academy of Sciences and IOP Pub- lishing Ltd

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ical potential µ5 for the massive fermion case is not en- we show a succinct derivation of CME employing the tirely understood. To address this issue, we list the low- Wigner function approach, which we can use to obtain est Landau level as follows (we set the homogeneous the CME as a quantum effect of the first order in the h¯ ex- magnetic background B = Bez along the z-axis and as- pansion. Subsequently, we turn to determine the Landau sume eB > 0, which is also appropriate for following sec- levels for the chiral fermion system. Because chiral fer- tions), mions are massless, the equations of righthand and   lefthand parts of the chiral fermion field decouple with  φ0     0  1 + each other, which allows us to deal with righthand and ψ , =   i(yky zkz), λ =  , 0λ(ky kz; x) c0λ  φ  e ( 1) (1) lefthand fermion fields independently. Taking the  F0λ 0  L 0 righthand fermion field as an example, we first solve the √ √ energy eigenvalue equation of the righthand fermion field =λ 2+ 2 = λ 2+ 2+ / with energy E m kz , where F0λ ( m kz kz) m, in an external uniform magnetic field and obtain a series and φ0 is the zeroth harmonic oscillator wavefunction of Landau levels. Then, we perform the second quantiza- along the x-axis. To simplify the following discussions, tion for righthand fermion field, which can be expanded we set ky = 0. The z-component of the spin operator for by complete wavefunctions of Landau levels. Finally, the z = 1 σ ,σ zψ = the single particle is S 2 diag(√ 3 3), implying S 0λ CME can be derived through the ensemble average, ex- 1 plicitly indicating that the CME uniquely arises from the (+ )ψ λ. When λ = +1, E = m2 + k2 > 0, then ψ + in Eq. 2 0 z 0 lowest Landau level. By analyzing the physical picture (1) describes a particle with momentum k and spin pro- √ z for the lowest Landau level, we conclude that all z = + 1 λ = − = − 2 + 2 < jection S 2. When 1, E m kz 0, then righthand (chirality is +1) fermions move along the posit- ψ0− in Eq. (1) describes an antiparticle with momentum ive z-direction, and all lefthand (chirality is -1) fermions − z = − 1 kz and spin projection S 2. Thus, in the homogen- move along the negative z-direction. This is the main res- eous magnetic background B = Bez, we obtain a picture ult of this study. This result can qualitatively explain why for the lowest Landau level (with ky = 0): All particles a macroscopic electric current occurs along the direction spin along the (+z)-axis, while all antiparticles spin along of the magnetic field in a chiral fermion system, called the (−z)-axis; however, the z-component momentum of the CME. We emphasize that the CME equation is de- particles and anti-particles can be along both the (+z)-ax- rived by determining Landau levels, without the approx- is or the(−z)-axis. A net electric current is difficult to ob- imation of a weak magnetic field. tain along the magnetic field direction from the point of The rest of this article is organized as follows. In Sec. view of the lowest Landau level for the massive fermion 2, we present a succinct derivation for the CME using case. Wigner function approach. In Sec. 3, we determine the In this article, we focus on a massless fermion (also Landau levels for the righthand fermion field. In Secs. 4 referred to as the “chiral fermion ”) system, where we and 5, we perform the second quantization of the show that it is easy to obtain a net electric current along righthand fermion system and obtain CME through the the magnetic field direction, seen from the picture of the ensemble average. In Sec. 6, we discuss the physical pic- lowest Landau level. The chiral fermion field can be di- ture of the lowest Landau level. Finally, we summarize vided into two independent parts, namely the righthand this study in Sec. 7. Some derivation details are presen- and lefthand parts. First, we set up the notation. The elec- ted in the appendixes. tric charge of a fermion/antifermion is e. The chemical Throughout this article, we adopt natural units, where potential for righthand/lefthand fermions is µR/L , which h¯ = c = kB = 1. The convention for the metric tensor is can be employed to express the chiral and ordinary chem- gµν = diag(+1,−1,−1,−1). The totally antisymmetric Levi- µνρσ 0123 ical potentials as µ5 = (µR − µL)/2 and µ = (µR + µL)/2, re- Civita tensor is ϵ with ϵ = +1, which is in agree- spectively. The chemical potential µ describes the imbal- ment with Peskin [25], but not with Bjorken and Drell ance of fermions and anti-fermions, while the chiral [26]. The Greek indices, µ,ν,ρ,σ, run over 0,1,2,3, or chemical potential µ5 describes the imbalance of t, x,y,z, whereas Roman indices, i, j,k, run over 1,2,3 or righthand and lefthand chirality. Notably, the introduc- x,y,z. We use the Heaviside-Lorentz convention for elec- tion of a chemical potential generally corresponds to a tromagnetism. conserved quantity. The conserved quantity correspond- µ ing to the ordinary chemical potential is total electric 2 A succinct derivation of CME using Wigner charge of the system. However, due to [22, 23], there is no conserved quantity corresponding to function approach the chiral chemical potential µ5, which is crucial for the existence of CME [24]. We concisely derive the CME using the Wigner func- To study the CME in the chiral fermion system, first tion approach for a chiral fermion system. Our starting

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˜ = 1 ε ρσ point is the following covariant and gauge invariant where Fµν 2 µνρσF . We explicitly showed the h¯ factor Wigner function, in Eqs. (8), (9). If we expand Vµ and Aµ order-by-order ⟨ ∫ ( ) ( ) 1 4 −ip·y y y y in h¯ as Wαβ(x, p) = : d ye Ψβ x + U x + , x − (2π)4 2 2 2 Vµ = Vµ + Vµ + 2Vµ + ··· , ( ) ⟩ (0) h¯ (1) h¯ (2) (10) y × Ψα x − : , (2) Aµ = Aµ + Aµ + 2Aµ + ··· , 2 (0) h¯ (1) h¯ (2) (11) where ⟨: ··· :⟩ represents the ensemble average, Ψ(x) is the then, at order o(1) and o(h¯), Eqs. (8), (9) become Dirac field operator for chiral fermions, α, β are Dirac µ p2V = 0, (12) spinor indices, and U(x + y/2, x − y/2) is the gauge link of (0) − / + / µ a straight line from (x y 2) to (x y 2). This specific p2A = 0, (13) choice for the path in the gauge link in the definition of (0) the Wigner function was first proposed in Ref. [27], 2V = − ˜ Aν , p (1)µ eh¯ Fµν (0) (14) where the authors argued that this type of gauge link can W , 2A = − ˜ Vν . create the variable p in the Wigner function (x p) to p (1)µ eh¯ Fµν (0) (15) represent the kinetic momentum, although in principle the µ µ The zeroth order solutions V and A can be de- path in the gauge link is arbitrary. The specific choice of (0) (0) the two points (x  y/2) in the integrand in Eq. (2) is rived by directly calculating the Wigner function without based on the consideration of symmetry. We can also re- the gauge link through the ensemble average in Eq. (2), place (x  y/2) by (x + sy) and (x − (1 − s)y), where s is a which was already obtained by one of the authors and his Vµ Aµ real parameter [28]. collaborators [29]. The results for (0) and (0) are µν ∑[ Suppose that the electromagnetic field is homo- µ 2 µ 1 F V = p δ(p2) θ(p0) (0) β p0−µ geneous in space and time, then from the dynamical (2π)3 e ( s) + 1 Ψ s ] equation satisfied by (x), one can obtain the dynamical 1 W , + θ(−p0) , (16) equation for (x p) as follows, β −p0+µ e ( s) + 1 γ · KW(x, p) = 0, (3) ∑ [ µ 2 µ 1 = i ∇ + ∇ = ∂x − ∂ν A = p δ(p2) s θ(p0) where Kµ µ pµ and µ µ eFµν . Because (0) β p0−µ 2 p (2π)3 e ( s) + 1 W(x, p) is a 4 × 4 matrix, we can decompose it into 16 in- s ] 1 dependent Γ-matrices, + θ(−p0) , (17) β −p0+µ e ( s) + 1 1 5 µ 5 µ 1 µν W = (F + iγ P + γ Vµ + γ γ Aµ + σ Sµν). (4) β = / 4 2 where 1 T is the inverse temperature of the system, µ F ,P,V ,A ,S R/L is the chemical potential for righthand/lefthand fer- The 16 coefficient functions µ µ µν are scalar, =  pseudoscalar, vector, pseudovector, and tensor, respect- mions as mentioned in the introduction, and s 1 cor- responds to the chirality of righthand/lefthand fermions. ively, and they are all real functions because µ µ † V A W = γ0Wγ0. Vector current and axial vector current can The zeroth order solutions (0) and (0) satisfy Eqs. (12), be expressed as the four-momentum integration of Vµ (13), which indicates that they are both on shell. From Aµ Eqs. (14), (15) we directly obtain the first order solutions, and , ∫ µ 2 µν ′ 2 µ 4 µ V = ˜ δ = V , (1) eh¯ F pν (p ) JV (x) d p (5) (2π)3 [ ] ∫ ∑ × θ 0 1 + θ − 0 1 , µ µ s (p ) β 0−µ ( p ) β − 0+µ = 4 A . (p s) + ( p s) + JA(x) d p (6) s e 1 e 1 (18) By multiplying Eq. (3) by γ · K from the lefthand side, µ 2 µν ′ 2 we obtain the quadratic form of Eq. (3) as follows, A = eh¯ F˜ pνδ (p ) ( ) (1) (2π)3 2 i µν ∑[ ] K − σ [Kµ, Kν] W = 0. (7) 1 1 2 × θ(p0) + θ(−p0) , (19) β p0−µ β −p0+µ e ( s) + 1 e ( s) + 1 From Eq. (7), we can obtain two off mass-shell equations s ′ for Vµ and Aµ (see Appendix A for details), where we employ δ (p2) = −δ(p2)/p2. Eqs. (18), (19) are ( ) the same as the second term in Eq. (3) of Ref. [30]. 2 1 2 2 ν p − h¯ ∇ Vµ = −eh¯ F˜µνA , (8) 4 Now, we can calculate JA/V based on Eqs. (5) and (6). Vi ,Ai ( ) Because (0) (0) are odd functions of three-momentum 2 1 2 2 ν i Vi p − h¯ ∇ Aµ = −eh¯ F˜µνV , (9) p, the nonzero contribution to JV/A arises uniquely from (1) 4 Ai and (1). We assume that only a uniform magnetic field

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12 21 03 30 exists B = Bez, i.e. F = −F = −B and F˜ = −F˜ = −B ults are obtained for the lefthand fermion field ΨL). (other components of Fµν , F˜µν are zero), which implies To determine the Landau levels, we must solve the ei- x = y = genvalue equation for the righthand fermion field as fol- JV/A JV/A 0. After integration over the z-components of Eqs. (18, 19) we have lows, ∫ σ · ψ = ψ , eh¯µ i D R E R (26) Jz = d4 pVz = 5 B, (20) V (1) 2π2 with D = (−∂ ,−∂ + ieBx,−∂ ). The details for solving ∫ x y z µ Eq. (26) are provided in Appendix B. We list the eigen- z 4 z eh¯ J = d pA = B, (21) functions and eigenvalues in the following : For n = 0 A (1) 2π2 Landau level, the wavefunction with energy E = kz is where µ5 = (µR − µL)/2 and µ = (µR + µL)/2. Eq. (20) indic- ( ) φ ξ 1 + µ , 0( ) i(yky zkz) ates that if 5 0, a current flows along the magnetic dir- ψR0(ky,kz; x) = e . (27) ection. Becauseh¯ appears in the coefficient of the magnet- 0 L ic field B, an enormous magnetic field is required to pro- For n > 0 Landau level, the wavefunction with energy duce a macroscopic current, which may be realized in E = λE (k ) is n z ( ) high energy heavy ion collisions. Thus far, we derived the φ ξ 1 n( ) i(yky+zkz) CME in the chiral fermion system using the Wigner func- ψRnλ(ky,kz; x) = cnλ e , (28) iFnλφn−1(ξ) L tion approach, and we observe that the CME is a first or- √ der quantum effect in h¯. In fact, the Wigner function ap- λ =  = + 2 = − λ / √where 1, En(kz) 2neB kz , Fnλ(kz) [kz En(kz)] proach is a quantum kinetic theory, which implies the | |2 = / + 2 2neB, normalized√ coefficient√ cnλ 1 (1 F λ), and presence of quantum effects of a multi-particle system, n φn(ξ) = φn( eBx − ky/ eB) is the n-th order wavefunc- such as the CME. tion of a harmonic oscillator. For n > 0 Landau levels, the wavefunctions with ener- 3 Landau levels for righthand fermions gies E = En(kz) correspond to fermions and antifermi- ons, respectively. For the lowest Landau level, the wave- = > In this and the following sections, we derive the CME function with energy E kz 0 corresponds to fermions, = < for a chiral fermion system by determining the Landau whereas that with energy E kz 0 corresponds to anti- levels. The Lagrangian for a chiral fermion field is fermions. The wavefunctions of all Landau levels are or- thonormal and complete. For the lefthand fermion field, L = Ψ(x)iγ · DΨ(x), (22) the eigenfunctions of Landau levels are the same as the µ µ µ with the covariant derivative D = ∂ + ieA , and the elec- righthand case, but with the sign of the eigenvalues tric charge e for particles/antiparticles. For a uniform changed. magnetic field B = Bez along the z-axis, we choose the µ = , , , gauge potential as A (0 0 Bx 0). The equation of mo- 4 Second quantization for righthand fermion tion for the field Ψ(x) is iγ · DΨ(x) = 0, (23) field which can be written in the form of a Schrödinger equation, In this section, we second-quantize the righthand fer- ∂ Ψ , = α · Ψ , , mion field Ψ (x), such that it becomes an operator and i∂ (t x) i D (t x) (24) R t satisfies following anticommutative relations, with D = −∇ + ieA, A = (0, Bx,0). In the chiral representa- † ′ ′ {Ψ (x),Ψ (x )} =δ(3)(x − x ), tion of Dirac γ-matrices, where γ5 = diag(−1,1), R R {Ψ ,Ψ ′ } = . α = −σ,σ Ψ Ψ = ΨT ,ΨT T R(x) R(x ) 0 (29) diag( ), we express in the form ( L R ) . Then, Eq. (24) becomes Because all eigenfunctions for the Hamiltonian of the ( ) ( ) ∂ Ψ , − σ · Ψ , righthand fermion field are orthonormal and complete, L(t x) = i D L(t x) , i (25) we decompose the righthand fermion field operator ΨR(x) ∂t ΨR(t, x) iσ · DΨR(t, x) by these eigenfunctions as∑ which indicates that the two fields ΨL/R, which corres- Ψ (x) = [θ(k )a (k ,k )ψ (k ,k ; x) pond to eigenvalues ∓1 of the matrix γ5, decouple with R z 0 y z R0 y z ky,kz each other. The two fields Ψ / are often referred to as L R † + θ(−k )b (k ,k )ψ (k ,k ; x)] lefthand/righthand fermion fields, respectively. Lefthand ∑ z 0 y z R0 y z and righthand fermions are also referred to as chiral fer- + [an(ky,kz)ψRn+(ky,kz; x) mions. n,ky,kz In the following, we focus on solving the eigenvalue + † , ψ , . bn(ky kz) Rn−(ky kz; x)] (30) equation for the righthand fermion field ΨR (similar res-

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In contrast to the general Fourier decomposition for the within an external uniform magnetic field B = Bez is in second quantization, we place two theta functions θ(kz) equilibrium with a reservior with temperature T and , † , chemical potential µ , then the density operator ρˆ for this in front of a0(ky kz) and b0(ky kz) in the decomposition, R which is very important for the subsequent second quant- righthand fermion system is ization procedure. From Eq. (29), we obtain following an- 1 −β H−µ N ρˆ = e ( R ), (34) ticommutative relations, Z ′ † ′ ′ {θ(k )a (k ,k ),θ(k )a (k ,k )} = θ(k )δ ′ δ ′ where β = 1/T is the inverse temperature, and Z is the z 0 y z z 0 y z z kyky kzkz ′ † ′ ′ grand canonical partition function, {θ(−k )b (k ,k ),θ(−k )b (k ,k )} = θ(−k )δ ′ δ ′ z 0 y z z 0 y z z kyky kzkz −β −µ = (H R N). † ′ ′ Z Tre (35) {an(ky,kz),a ′ (k ,k )} = δnn′ δk k′ δk k′ n y z y y z z ˆ † ′ ′ The expectation value of an operator F in the equilibri- {b (k ,k ),b ′ (k ,k )} = δ ′ δ ′ δ ′ . (31) n y z n y z nn kyky kzkz um state can be calculated as

The two theta functions θ(kz) are always attached to the ⟨: Fˆ :⟩ = Tr(ˆρFˆ). (36) , †, , † lowest Landau level operators, such as a0 a0 b0 b0. The In the Appendix C, we calculated the expectation values Hamiltonian and total particle number of the righthand of occupied number operators as fermion system are ∫ † θ(k ) ∑ ⟨ θ k a k ,k a k ,k ⟩ = z † † : ( z) 0( y z) 0( y z): β k −µ H = d3 xΨ (x)iσ · DΨ (x) = [k θ(k )a (k ,k )a (k ,k ) e ( z R) + 1 R R z z 0 y z 0 y z θ − k ,k † ( kz) y z ⟨: θ(−k )b (k ,k )b (k ,k ):⟩ = ∑ z 0 y z 0 y z β(−k +µ ) † † e z R + 1 + (−kz)θ(−kz)b (ky,kz)b0(ky,kz)] + En(kz)[a (ky,kz) 0 n † 1 n,k ,k ⟨ a k ,k a k ,k ⟩ = y z : n( y z) n( y z): β E k −µ e [ n( z) R] + 1 × a (k ,k ) + b†(k ,k )b (k ,k )], n y z n y z n y z ⟨ † , , ⟩ = 1 . : bn(ky kz)bn(ky kz): β +µ (37) (32) [En(kz) R] + ∫ e 1 † The macroscopic electric current for the righthand fermi- N = d3 xΨ (x)Ψ (x) R R on system is ∑ ⟨ ⟩ † † † = [θ(kz)a (ky,kz)a0(ky,kz) − θ(−kz)b (ky,kz)b0(ky,kz)] = Ψ σΨ . 0 0 JR : R(x) R(x): (38) k ,k y ∑z † † According to the rotational invariance of this system × [a (ky,kz)an(ky,kz) − b (ky,kz)bn(ky,kz)], n n (33) x y , , along the z-axis, J = J = 0. In the following, we calcu- n ky kz R R late Jz . Using Eq. (30), we see that where we omitted the infinite vacuum term. This can be R z † 3 renormalized in the physics calculation and does not af- J =⟨: Ψ (x)σ ΨR(x):⟩ R ∑(R fect our result on the CME coefficient. Evidently, † † † = ⟨ θ , , ⟩ : (kz)a0(ky kz)a0(ky kz): θ(kz)a (ky,kz)a0(ky,kz) and an(ky,kz)an(ky,kz) are the occu- 0 k ,k pied number operators of particles for different Landau y z ) θ − † , , † , , + ⟨ θ − , † , ⟩ levels, and ( kz)b0(ky kz)b0(ky kz) and bn(ky kz)bn(ky kz) : ( kz)b0(ky kz)b0(ky kz): are occupied number operators of antiparticles for differ- † × ψ (k ,k ; x)σ3ψ (k ,k ; x) ent Landau levels. Notably, without introduction of the ∑R0 y z R0 y z θ  , † † 3 two theta functions ( kz) in front of a0(ky kz) and + ⟨: a (ky,kz)an(ky,kz):⟩ψ +(ky,kz; x)σ ψRn+(ky,kz; x) † n Rn , , b (k ,k ) in the decomposition of Ψ (x), the second quant- n ky kz 0 y z R ∑ ization procedure could not be performed successfully. † † 3 + ⟨: bn(ky,kz)b (ky,kz):⟩ψ −(ky,kz; x)σ ψRn−(ky,kz; x) This is different from the massive case [21], where the n Rn n,ky,kz authors determined the Landau levels and corresponding ∑( θ θ − ) = (kz) − ( kz) ψ† , σ3ψ , wavefunctions for the massive Dirac equation in a uni- β −µ β − +µ (ky kz; x) R0(ky kz; x) (kz R)+ ( kz R)+ R0 , e 1 e 1 form magnetic field with chemical potential µ and chiral ky kz µ ∑ chemical potential 5. The wave functions for the massive 1 † 3 + ψ k ,k x σ ψ + k ,k x θ β E k −µ Rn+( y z; ) Rn ( y z; ) case are in a four-component Dirac form, and the func- e [ n( z) R] + 1 n,ky,kz tion is not needed for the second quantization. ∑ − 1 ψ† , σ3ψ , . β +µ −(ky kz; x) Rn−(ky kz; x) [En(kz) R] + Rn , , e 1 5 Chiral magnetic effect n ky kz (39) ψ† , σ3ψ , Supposing that the system of the righthand ferimons First, we sum over ky for R0(ky kz; x) R0(ky kz; x) and

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† ψ (k ,k ; x)σ3ψ λ(k ,k ; x) in Eq. (39), the results are magnetic field, in contrast to the weak magnetic field ap- Rnλ y z ∑ Rn y z ψ† , σ3ψ , proximation through Wigner function approach in Sec. 2. R0(ky kz; x) R0(ky kz; x)

ky ∑( )( )( ) 1 1 0 φ (ξ) 6 Physical picture of lowest Landau level = φ (ξ), 0 0 L2 0 0 −1 0 k ∫y We discuss the physical picture of the lowest Landau ∞ √ √ 1 2 level. The wavefunction and energy of the lowest Land- = dky[φ0( eBx − ky/ eB)] 2πL −∞ au level (n = 0) for the righthand fermion field is ( ) eB φ 1 + = , (40) 0 i(yky zkz) π ψR (ky,kz; x) = e , E = kz. (47) 2 L 0 0 L and ∑ Setting ky = 0 in Eq. (47), we calculate the Hamiltonian, ψ† , σ3ψ , Rnλ(ky kz; x) Rnλ(ky kz; x) particle number, z-component of momentum, and z-com- ky ponent of the spin angular momentum of the righthand ∫ ∞ ( ) 1 2neB[φ − (ξ)]2 fermion system for the lowest Landau level as follows, = k c2 k φ ξ 2 − n 1 ∑ d y nλ( z) [ n( )] 2 † 2πL −∞ [kz + λEn(kz)] = θ , , ( ) H [kz (kz)a0(0 kz)a0(0 kz) eB 2neB k = c2 (k ) 1 − z π nλ z + λ 2 † 2 L [kz En(kz)] + (−k )θ(−k )b (0,k )b (0,k )], λ ∑ z z 0 z 0 z = eB 2 − −2 = eB · kz . † cnλ(kz)[2 cnλ (kz)] (41) N = [θ(kz)a (0,kz)a0(0,kz) 2πL 2πL En(kz) 0 kz Second, we sum over kz in the third equal sign of Eq. † + (−1)θ(−k )b (0,k )b (0,k )], (39), ∑ z 0 z 0 z ∑( ) † θ(k ) θ(−k ) eB Pz = [kzθ(kz)a (0,kz)a0(0,kz) Jz = z − z 0 R β k −µ β −k +µ kz e ( z R) + 1 e ( z R) + 1 2πL k z ( ) + − θ − † , , , ∑ ( kz) ( kz)b0(0 kz)b0(0 kz)] + 1 + 1 eB · kz ∑[ β E k −µ β E k +µ 1 † e [ n( z) R] + 1 e [ n( z) R] + 1 2πL E (k ) = θ , , n,k n z S z (kz)a (0 kz)a0(0 kz) ∫z 2 0 ∞ ( ) k eB θ(k ) θ(−k ) z( ) = k z − z + ] d z β k −µ β −k +µ 0 1 † 4π2 −∞ e ( z R) + 1 e ( z R) + 1 + − θ(−k )b (0,k )b (0,k ) , (48) 2 z 0 z 0 z = eB µ . 4π2 R (42) where the definitions of P and S are ∫ z z x y Combining Eq. (42) and J = J = 0 yields † ∂ R R P = −i d3 xΨ (x) Ψ (x), eµ z R ∂z R J = R B. (43) ∫ R π2 4 1 3 † 3 S z = d xΨ (x)σ ΨR(x). (49) From the calculation above, we see that only the lowest 2 R Landau level contributes to Eq. (43). A similar calcula- Thus, we have a picture for the lowest Landau level: The † tion for the lefthand fermion system shows that operator θ(k )a (0,k ) produces a particle with charge e, µ z 0 z e L > > JL = − B. (44) energy kz 0, z-component of momentum kz 0, and z- 4π2 + 1 component of spin angular momentum 2 (helicity We can also obtain Eq. (44) from Eq. (43) under space in- † h = +1); The operator θ(−kz)b (0,kz) produces a particle → − µ → µ → 0 version: JR JL, R L, B B. If the system is with charge −e, energy −k > 0, z-component of mo- composed of righthand and lefthand fermions, then the z mentum −kz > 0, and z-component of spin angular mo- vector current J and axial current J are V A mentum − 1 (helicity h = −1). This picture indicates that eµ 2 J = J + J = 5 B, (45) all righthand fermions/antifermions move along the (+z)- V R L π2 2 axis, with righthand fermions spinning along the (+z)-ax- eµ − J = J − J = B, (46) is and righthand antifermions spinning along the z-axis. A R L π2 2 If µR > 0, which indicates that there are more righthand where µ5 = (µR − µL)/2 is the chiral chemical potential and fermions than righthand anti-fermions, a net electric cur- µ = (µR + µL)/2. Thus far, we derived the CME in the chir- rent will move along the (+z)-axis, which is referred to as al fermion system by determining Landau levels. We em- the CME for the righthand fermion system. phasize that Eqs. (45), (46) are valid for any strength of The analogous analysis can be applied to lefthand fer-

074106-6 Chinese Physics C Vol. 44, No. 7 (2020) 074106 mions. The picture of the lowest Landau level for a the chiral fermion system in a uniform magnetic field, by lefthand fermion is: All lefthand fermions/antifermions performing the second quantization for the chiral fermi- move along the (−z)-axis, with left fermions spinning on field, expanding the field operator by an eigenfunc- along the (+z)-axis and lefthand antifermions spinning tion of Landau levels, and calculating the ensemble aver- along the (−z)-axis. If µL > 0, which indicates that there is age of the vector current operator, we natrually obtain the more lefthand fermions than lefthand anti-fermions, a net equation for the CME. Notably, no approximations were electric current will move along the (−z)-axis, which is made for the strength of magnetic field in the calculation. referred to as the CME for the lefthand fermion system. Further, we introduced two theta functions θ(kz) in front † Because the total electric current JV of the chiral fer- , , Ψ of a0(ky kz) and b0(ky kz) in the decomposition of R(x), mion system is the summation of the electric current JR which is crucial for the successful performance of the of the righthand fermion system and the electric current subsequent procedure of second quantization. When we JL of the lefthand fermion system, whether JV moves carefully analyze the lowest Landau level, we find that all + along the ( z)-axis will only depend on the sign of righthand (chirality is +1) fermions move along the posit- µ − µ ( R L). Thus, the CME for the chiral fermion system is ive z-direction, and all lefthand (chirality is -1) fermions described microscopically. move along the negative z-direction. Thus, the CME is described microscopically within this picture of the low- 7 Summary est Landau level.

CME arises from the lowest Landau level both for the We are grateful to Hai-Cang Ren and Xin-Li Sheng massive Dirac fermion system and the chiral fermion sys- for valuable discussions. R.-H. F. thanks for the hospital- tem. For the massive case, the physical picture of how the ity of Institute of Frontier and Interdisciplinary Science lowest Landau level contributes to CME is not extens- at Shandong University (Qingdao) where he is currently ively clear. When the Landau levels are determined for visiting.

Appendix A: Vlasov equation and off mass-shell equation

{σµν, } = σµν, The quadratic form for the equation of motion of the Wigner 1 2 {σµν, γ5} = −ϵµνρσσ , function W(x, p) is i ρσ ( ) µν ρ µνρσ 5 {σ ,γ } = 2ϵ γ γσ, 2 − i σµν , W = . K [Kµ Kν] 0 (A1) µν 5 ρ µνρσ 2 {σ ,γ γ } = 2ϵ γσ, µν ρσ µ ρ σ ν µνρσ 2 = 2 − 1 ∇2 + · ∇ , = − {σ ,σ } = 2g [ g ] + 2iϵ γ5. (A7) Using K p 4 ip and [Kµ Kν] ieFµν, Eq. (A1) becomes ( ) 2 1 2 1 µν p − ∇ + ip · ∇ − eFµνσ W = 0. (A2) Then, all matrices appearing in Eqs. (A4) (A5) are the 16 inde- 4 2 pendent Γ-matrices, whose coefficients must be zero. These coeffi- µν † µν µν† W and σ satisfy W = γ0W γ0 and σ = γ0σ γ0. Employing the cient equations are the Vlasov equations and the off mass-shell γ0 Hermi conjugation and subsequently multiplying to both sides equations for F ,P,Vµ,Aµ,Sµν. The Vlasov equations are of Eq. (A2) yields · ∇F = , ( ) p 0 2 1 2 1 µν p · ∇P = 0, p − ∇ − ip · ∇ W − eFµνWσ = 0. (A3) ν 4 2 p · ∇Vµ = eFµνV , ν Eq. (A2) minus Eq. (A3) yields the Vlasov equation for W, p · ∇Aµ = eFµνA , ρ 1 µν p · ∇Qµν = eF Qν ρ, (A8) ip · ∇W − eFµν[σ ,W] = 0. (A4) [µ ] 4 and the off mass-shell equations are Eq. (A2) added to Eq. (A3) yields the off mass-shell equation for ( ) 2 1 2 1 µν W, p − ∇ F = eFµνQ , ( ) 4 2 2 1 2 1 µν ( ) p − ∇ W − eFµν{σ ,W} = 0. (A5) 2 1 2 1 µν 4 4 p − ∇ P = eF˜µνQ , 4 2 To calculate [σµν,W] and {σµν,W} in Eqs. (A4) (A5), we list the fol- ( ) 2 1 2 ν p − ∇ Vµ = −eF˜µνA , lowing useful identities, 4 µν ( ) [σ ,1] = 0, 2 1 2 ν µν 5 p − ∇ Aµ = −eF˜µνV , [σ ,iγ ] = 0, 4 µν ρ ρ µ ν ( ) [σ ,γ ] = −2ig [ γ ], 2 − 1 ∇2 Q = F − ˜ P , µν ρ ρ µ ν p µν e(Fµν Fµν ) (A9) [σ ,γ5γ ] = −2ig [ γ5γ ], 4 µν ρσ µ ρ σ ν ν ρ σ µ [σ ,σ ] = 2ig [ σ ] − 2ig [ σ ] , (A6) ˜ = 1 ε ρσ where Fµν 2 µνρσF .

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Appendix B: Landau levels for righthand fermion field

We solve following eigenvalue equation in detail, ergy E = λEn(kz) (λ = 1), we can obtain ϕ2 as √ σ · ψ = ψ , i D R(x) E R(x) (B1) eB(∂ξ + ξ)φ (ξ) i[k − λE (k )] ϕ = n = z √ n z φ ξ , 2(x) + n−1( ) (B11) with D = (−∂ ,−∂ + ieBx,−∂ ). Because the operator iσ · D is com- i(kz E) 2neB x y z √ mutative with pˆ = −i∂ , pˆ = −i∂ , we can choose ψ as the com- (∂ξ + ξ)φ (ξ) = 2nφ − (ξ) F λ(k ) = y y z z R where we used√ n n 1 . Defining n z mom eigenstate of iσ · D, pˆy and pˆz as follows [k − λE (k )]/ 2neB, the eigenfunction with eigenvalue E = λE (k ) ( ) z n z n z ϕ becomes 1(x) 1 i(yky+zkz) ψR(x,y,z) = e , (B2) ( ) ϕ2(x) L φ ξ n( ) 1 i(yky+zkz) ψRnλ(ky,kz; x) = e . (B12) F λ k φ − ξ where L is the length of the system in y- and z- directions. The ex- i n ( z) n 1( ) L plicit form of σ · D is This is very subtle when n = 0 in Eq. (B11). When n = 0, E = kz, the ( ) ϕ = ∂ + ξ φ ξ = −∂ −∂ + i∂ + eBx first equal sign of Eq. (B11) indicates 2 0 due to ( ξ ) 0( ) 0. σ · D = z x y . (B3) −∂ − i∂ − eBx ∂ Then, the corresponding eigenfunction becomes x y z ( ) φ (ξ) 1 + Inserting Eq. (B2) (B3) into Eq. (B1), we obtain the group of dif- ψ (k ,k ; x) = 0 ei(yky zkz). (B13) R0 y z 0 L ferential equations for ϕ1(x) and ϕ2(x) as When n = 0, E = −k , the denominator of the first equal sign of Eq. i(kz − E)ϕ1 + (∂x + ky − eBx)ϕ2 = 0, (B4) z (B11) becomes zero, in which case we must directly deal with Eqs. (∂ − k + eBx)ϕ − i(k + E)ϕ = 0. (B5) x y 1 z 2 (B4) (B5). In this case, Eqs. (B4) (B5) become ϕ ϕ From Eq. (B5), we can express 2 by 1, then Eq. (B4) becomes 2ikzϕ1 + (∂x + ky − eBx)ϕ2 = 0, (B14) ( ( )2) 2 2 2 2 2 ky ∂ ϕ + E + eB− k − e B x − ϕ = 0, (B6) (∂x − ky + eBx)ϕ1 = 0. (B15) x 1 z eB 1 ϕ ∼ − 1 2 + Eq. (B15) gives 1(x) exp[ 2 eBx xky], then Eq. (B14) becomes which is a typical harmonic oscillator equation. Defining a √ dimen- ( ) 1 2 sionless variable ξ = eB(x − ky/eB), and ϕ1(x) = φ(ξ), then (B6) be- 2ik exp − eBx + xk + (∂ + k − eBx)ϕ = 0. (B16) z 2 y x y 2 comes → ∞ ( ) When x , Eq. (B16) tends to 2φ 2 − 2 d E kz 2 + + 1 − ξ φ = 0. (B7) (∂x − eBx)ϕ2 = 0, (B17) dξ2 eB 1 2 whose solution is ϕ2 ∼ exp( eBx ), which is divergent as x → ∞. With the boundary condition φ → 0 as ξ → ∞, we must set 2 Thus, there is no physical solution when n = 0, E = −kz. E2 − k2 z + 1 = 2n + 1, (B8) Thus far, we obtain the eigenfunctions and eigenvalues of the eB Hamiltonian of the righthand fermion field as follows: = , , ,··· with n 0 1 2 . Thus, energy E can only assume the following For n = 0 Landau level, the wavefunction with energy E = kz is ( ) discrete values, φ 0 1 i(yky+zkz) √ ψR0(ky,kz; x) = e . (B18) 2 0 L E = En(kz) ≡  2neB+ kz , (B9) √ For n > 0 Landau level, the wavefunction with energy E = λEn(kz) is E (k ) = 2neB+ k2 ( ) where we define n z z . The corresponding normal- φ n 1 i(yky+zkz) ψRnλ(ky,kz; x) = cnλ e , (B19) ized solution for equation (B6) is iFnλφn−1 L √ −ξ2/2 √ ϕ1(x) = φn(ξ) = Nne Hn(ξ), (B10) 2 where λ = 1, En(kz) = 2neB+ kz , Fnλ(kz) = [kz − λEn(kz)]/ 2neB, 1 − 1 − 1 ξ2 dn −ξ2 N = (eB) 4 π 4 (2nn!) 2 H (ξ) = (−1)ne e | |2 = / + 2 where n , and n dξn . For en- cnλ 1 (1 Fnλ).

Appendix C: Expectation value of occupied number operators

We calculate the expectation values of particle number operat- ,θ † , = θ † , [H (kz)a0(ky kz)] kz (kz)a0(ky kz) ors. From the expression of the Hamiltonian and the total particle ,θ − † , = − θ − † , [H ( kz)b0(ky kz)] ( kz) ( kz)b0(ky kz) number operator in Eqs. (32, 33), we easily obtain following com- † † [H,an(ky,kz)] = En(kz)an(ky,kz) mutative relations, † † [H,bn(ky,kz)] = En(kz)bn(ky,kz), (C2) ,θ † , = θ † , [N (kz)a0(ky kz)] (kz)a0(ky kz) AB,C = A{B,C} − {A,C}B ,θ − † , = −θ − † , where we employ [ ] . Defining [N ( kz)b0(ky kz)] ( kz)b0(ky kz) † −β −µ † β −µ , † , = † , θ , β = (H RN)θ , (H R N) [N an(ky kz)] an(ky kz) (kz)a0(ky kz; ) e (kz)a0(ky kz)e † † † −β −µ † β −µ , , = − , , θ − , β = (H RN)θ − , (H RN) [N bn(ky kz)] bn(ky kz) (C1) ( kz)b0(ky kz; ) e ( kz)b0(ky kz)e

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† † −β(H−µRN) β(H−µR N) an(ky,kz;β) = e an(ky,kz)e see that † −β −µ † β −µ † , β = (H RN) , (H R N). ⟨ θ , , ⟩ bn(ky kz; ) e bn(ky kz)e (C3) : (kz)a0(ky kz)a0(ky kz): † = ρθ † , , For θ(k )a (k ,k ;β), we obtain Tr[ (kz)a (ky kz)a0(ky kz)] z 0 y z ( 0 ) 1 † ∂ −β(H−µRN) † † = Tr θ(kz)a (ky,kz;β)e a0(ky,kz) [θ(kz)a (ky,kz;β)] = −[H − µRN,θ(kz)a (ky,kz;β)] Z 0 ∂β 0 0 ( ) 1 † −β −µ † β −µ −β(H−µR N) = − (H R N) − µ ,θ , (H RN) = Tr θ(kz)a0(ky,kz)a (ky,kz;β)e e [H RN (kz)a0(ky kz)]e Z 0 −β −µ † β −µ = −e (H R N)[(k − µ )θ(k )a (k ,k )]e (H R N) =⟨ θ , † , β ⟩ z R z 0 y z : (kz)a0(ky kz)a0(ky kz; ): † † −β −µ = −(k − µ )[θ(k )a (k ,k ;β)], =⟨ θ , , ⟩ (kz R) z R z 0 y z : (kz)a0(ky kz)a0(ky kz): e (C4) † −β(kz−µR) −β(kz−µR) =θ(kz)e − ⟨: θ(kz)a (ky,kz)a0(ky,kz):⟩e , (C7) † † 0 with the boundary condition θ(kz)a (ky,kz;0) = θ(kz)a (ky,kz), which 0 0 thus, we obtain implies † θ(kz) † † −β(k −µ ) ⟨θ(kz)a (ky,kz)a0(ky,kz)⟩ = . (C8) θ , β = θ , z R . 0 β(kz−µR) + (kz)a0(ky kz; ) (kz)a0(ky kz)e (C5) e 1 Similarly, we obtain Similar calculations obtain † † −β − +µ † θ(−kz) θ − , β = θ − , ( kz R) ⟨θ − , , ⟩ = ( kz)b (ky kz; ) ( kz)b (ky kz)e ( kz)b (ky kz)b0(ky kz) β − +µ 0 0 0 e ( kz R) + 1 † † −β[E (k )−µ ] a (k ,k ;β) = a (k ,k )e n z R † 1 n y z n y z ⟨a (k ,k )a (k ,k )⟩ = † † n y z n y z β[E (k )−µ ] −β[En(kz)+µR] e n z R + 1 bn(ky,kz;β) = bn(ky,kz)e . (C6) † 1 † ⟨bn(ky,kz)bn(ky,kz)⟩ = . (C9) ⟨ θ , , ⟩ β[En(kz)+µR] + We calculate the expectation value of : (kz)a0(ky kz)a0(ky kz): . We e 1

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