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Quark gluon plasma (QGP) is created in high energy heavy ion collisions, constituting extremely hot and dense matter. An enormous magnetic field can be generated by high energy peripheral collisions [1-3]. One of the predictions in QGP is that positively and negatively charged particles seperate along the direction of the magnetic field, which is related to chiral magnetic effect (CME) [4-6]. Numerous efforts have been made to determine the CME in experiments [7-9]. However, due to background noise, no definite CME has been revealed to date. Numerous theoretical methods likewise investigated the CME, such as AdS/CFT [10, 11], hydrodynamics [12-14], finite temperature field theory [15-18], quantum kinetic theory [19], lattice method [20], et al.
In this article, we study the CME in detail by determination of Landau levels. For the massive Dirac fermion system, several studies on CME addressed Landau levels. In Ref. [15], Fukushima et al. proposed four methods to derive the CME. One of these methods made use of Landau energy levels for the massive Dirac equation with chemical potential
μ and chiral chemical potentialμ5 in a homogeneous magnetic backgroundB=Bez to construct the thermodynamic potentialΩ . The macroscopic electric currentjz along the z-axis can be obtained from the thermodynamic potentialΩ . Another study on the CME addressing Landau levels is related to the second quantization of the Dirac field. In Ref. [21], the authors determined the Landau levels and corresponding Landau wavefunctions for the massive Dirac equation in a uniform magnetic field, likewise with chemical potentialμ and chiral chemical potentialμ5 . Then, they second-quantized the Dirac field and expanded it by these solved Landau wavefunctions and creation/construction operators. The density operatorˆρ can then be determined from HamiltonianˆH and particle number operatorˆN of the system. Finally, they derived the macroscopic electric currentjz along the z-axis through the trace of density operatorˆρ and electric current operatorˆjz , which is simply the CME equation.From the study on CME for massive Dirac fermions through Landau levels, we conclude that the contribution to CME arises uniquely from the lowest Landau level, while the contributions from higher Landau levels cancel each other. However, because of the mass m of the Dirac fermion, the physical picture of the CME for the massive Dirac fermion system is not as clear as in the massless fermion case, as the physical meaning of the chiral chemical potential
μ5 for the massive fermion case is not entirely understood. To address this issue, we list the lowest Landau level as follows (we set the homogeneous magnetic backgroundB=Bez along the z-axis and assumeeB>0 , which is also appropriate for following sections),ψ0λ(ky,kz;x)=c0λ(φ00F0λφ00)1Lei(yky+zkz), (λ=±1),
(1) with energy
E=λ√m2+k2z , whereF0λ=(λ√m2+k2z+kz)/m , andφ0 is the zeroth harmonic oscillator wavefunction along the x-axis. To simplify the following discussions, we setky=0 . The z-component of the spin operator for the single particle isSz=12diag(σ3,σ3) , implyingSzψ0λ= (+12)ψ0λ . Whenλ=+1 ,E=√m2+k2z>0 , thenψ0+ in Eq. (1) describes a particle with momentumkz and spin projectionSz=+12 . Whenλ=−1 ,E=−√m2+k2z<0 , thenψ0− in Eq. (1) describes an antiparticle with momentum−kz and spin projectionSz=−12 . Thus, in the homogeneous magnetic backgroundB=Bez , we obtain a picture for the lowest Landau level (withky=0 ): All particles spin along the(+z) -axis, while all antiparticles spin along the(−z) -axis; however, the z-component momentum of particles and anti-particles can be along both the(+z) -axis or the(−z) -axis. A net electric current is difficult to obtain along the magnetic field direction from the point of view of the lowest Landau level for the massive fermion case.In this article, we focus on a massless fermion (also referred to as the “chiral fermion”) system, where we show that it is easy to obtain a net electric current along the magnetic field direction, seen from the picture of the lowest Landau level. The chiral fermion field can be divided into two independent parts, namely the righthand and lefthand parts. First, we set up the notation. The electric charge of a fermion/antifermion is
±e . The chemical potential for righthand/lefthand fermions isμR/L , which can be employed to express the chiral and ordinary chemical potentials asμ5=(μR−μL)/2 andμ=(μR+μL)/2 , respectively. The chemical potentialμ describes the imbalance of fermions and anti-fermions, while the chiral chemical potentialμ5 describes the imbalance of righthand and lefthand chirality. Notably, the introduction of a chemical potential generally corresponds to a conserved quantity. The conserved quantity corresponding to the ordinary chemical potentialμ is total electric charge of the system. However, due to chiral anomaly [22, 23], there is no conserved quantity corresponding to the chiral chemical potentialμ5 , which is crucial for the existence of CME [24].To study the CME in the chiral fermion system, first we show a succinct derivation of CME employing the Wigner function approach, which we can use to obtain the CME as a quantum effect of the first order in the
ℏ expansion. Subsequently, we turn to determine the Landau levels for the chiral fermion system. Because chiral fermions are massless, the equations of righthand and lefthand parts of the chiral fermion field decouple with each other, which allows us to deal with righthand and lefthand fermion fields independently. Taking the righthand fermion field as an example, we first solve the energy eigenvalue equation of the righthand fermion field in an external uniform magnetic field and obtain a series of Landau levels. Then, we perform the second quantization for righthand fermion field, which can be expanded by complete wavefunctions of Landau levels. Finally, the CME can be derived through the ensemble average, explicitly indicating that the CME uniquely arises from the lowest Landau level. By analyzing the physical picture for the lowest Landau level, we conclude that all righthand (chirality is +1) fermions move along the positive z-direction, and all lefthand (chirality is -1) fermions move along the negative z-direction. This is the main result of this study. This result can qualitatively explain why a macroscopic electric current occurs along the direction of the magnetic field in a chiral fermion system, called the CME. We emphasize that the CME equation is derived by determining Landau levels, without the approximation of a weak magnetic field.The rest of this article is organized as follows. In Sec. 2, we present a succinct derivation for the CME using Wigner function approach. In Sec. 3, we determine the Landau levels for the righthand fermion field. In Secs. 4 and 5, we perform the second quantization of the righthand fermion system and obtain CME through the ensemble average. In Sec. 6, we discuss the physical picture of the lowest Landau level. Finally, we summarize this study in Sec. 7. Some derivation details are presented in the appendixes.
Throughout this article, we adopt natural units, where
ℏ=c=kB=1 . The convention for the metric tensor isgμν=diag(+1,−1,−1,−1) . The totally antisymmetric Levi-Civita tensor isϵμνρσ withϵ0123=+1 , which is in agreement with Peskin [25], but not with Bjorken and Drell [26]. The Greek indices,μ,ν,ρ,σ , run over0,1,2,3 , ort,x,y,z , whereas Roman indices,i,j,k , run over1,2,3 orx,y,z . We use the Heaviside-Lorentz convention for electromagnetism. -
We concisely derive the CME using the Wigner function approach for a chiral fermion system. Our starting point is the following covariant and gauge invariant Wigner function,
Wαβ(x,p)=⟨:1(2π)4∫d4ye−ip⋅y¯Ψβ(x+y2)U(x+y2,x−y2)×Ψα(x−y2):⟩,
(2) where
⟨:⋯:⟩ represents the ensemble average,Ψ(x) is the Dirac field operator for chiral fermions,α ,β are Dirac spinor indices, andU(x+y/2,x−y/2) is the gauge link of a straight line from(x−y/2) to(x+y/2) . This specific choice for the path in the gauge link in the definition of the Wigner function was first proposed in Ref. [27], where the authors argued that this type of gauge link can create the variable p in the Wigner functionW(x,p) to represent the kinetic momentum, although in principle the path in the gauge link is arbitrary. The specific choice of the two points(x±y/2) in the integrand in Eq. (2) is based on the consideration of symmetry. We can also replace(x±y/2) by(x+sy) and(x−(1−s)y) , where s is a real parameter [28].Suppose that the electromagnetic field
Fμν is homogeneous in space and time, then from the dynamical equation satisfied byΨ(x) , one can obtain the dynamical equation forW(x,p) as follows,γ⋅KW(x,p)=0,
(3) where
Kμ=i2∇μ+pμ and∇μ=∂xμ−eFμν∂νp . BecauseW(x,p) is a4×4 matrix, we can decompose it into 16 independentΓ -matrices,W=14(F+iγ5P+γμVμ+γ5γμAμ+12σμνSμν).
(4) The 16 coefficient functions
F,P,Vμ,Aμ,Sμν are scalar, pseudoscalar, vector, pseudovector, and tensor, respectively, and they are all real functions becauseW†=γ0Wγ0 . Vector current and axial vector current can be expressed as the four-momentum integration ofVμ andAμ ,JμV(x)=∫d4pVμ,
(5) JμA(x)=∫d4pAμ.
(6) By multiplying Eq. (3) by
γ⋅K from the lefthand side, we obtain the quadratic form of Eq. (3) as follows,(K2−i2σμν[Kμ,Kν])W=0.
(7) From Eq. (7), we can obtain two off mass-shell equations for
Vμ andAμ (see Appendix A for details),(p2−14ℏ2∇2)Vμ=−eℏ˜FμνAν,
(8) (p2−14ℏ2∇2)Aμ=−eℏ˜FμνVν,
(9) where
˜Fμν=12εμνρσFρσ . We explicitly showed theℏ factor in Eqs. (8), (9). If we expandVμ andAμ order-by-order inℏ asVμ=Vμ(0)+ℏVμ(1)+ℏ2Vμ(2)+⋯,
(10) Aμ=Aμ(0)+ℏAμ(1)+ℏ2Aμ(2)+⋯,
(11) then, at order
o(1) ando(ℏ) , Eqs. (8), (9) becomep2Vμ(0)=0,
(12) p2Aμ(0)=0,
(13) p2V(1)μ=−eℏ˜FμνAν(0),
(14) p2A(1)μ=−eℏ˜FμνVν(0).
(15) The zeroth order solutions
Vμ(0) andAμ(0) can be derived by directly calculating the Wigner function without the gauge link through the ensemble average in Eq. (2), which was already obtained by one of the authors and his collaborators [29]. The results forVμ(0) andAμ(0) areVμ(0)=2(2π)3pμδ(p2)∑s[θ(p0)1eβ(p0−μs)+1+θ(−p0)1eβ(−p0+μs)+1],
(16) Aμ(0)=2(2π)3pμδ(p2)∑ss[θ(p0)1eβ(p0−μs)+1+θ(−p0)1eβ(−p0+μs)+1],
(17) where
β=1/T is the inverse temperature of the system,μR/L is the chemical potential for righthand/lefthand fermions as mentioned in the introduction, ands=±1 corresponds to the chirality of righthand/lefthand fermions. The zeroth order solutionsVμ(0) andAμ(0) satisfy Eqs. (12), (13), which indicates that they are both on shell. From Eqs. (14), (15) we directly obtain the first order solutions,Vμ(1)=2(2π)3eℏ˜Fμνpνδ′(p2)×∑ss[θ(p0)1eβ(p0−μs)+1+θ(−p0)1eβ(−p0+μs)+1],
(18) Aμ(1)=2(2π)3eℏ˜Fμνpνδ′(p2)×∑s[θ(p0)1eβ(p0−μs)+1+θ(−p0)1eβ(−p0+μs)+1],
(19) where we employ
δ′(p2)=−δ(p2)/p2 . Eqs. (18), (19) are the same as the second term in Eq. (3) of Ref. [30].Now, we can calculate
JA/V based on Eqs. (5) and (6). BecauseVi(0),Ai(0) are odd functions of three-momentump , the nonzero contribution toJiV/A arises uniquely fromVi(1) andAi(1) . We assume that only a uniform magnetic field existsB=Bez , i.e.F12=−F21=−B and˜F03=−˜F30=−B (other components ofFμν ,˜Fμν are zero), which impliesJxV/A=JyV/A=0 . After integration over the z-components of Eqs. (18, 19) we haveJzV=∫d4pVz(1)=eℏμ52π2B,
(20) JzA=∫d4pAz(1)=eℏμ2π2B,
(21) where
μ5=(μR−μL)/2 andμ=(μR+μL)/2 . Eq. (20) indicates that ifμ5≠0 , a current flows along the magnetic direction. Becauseℏ appears in the coefficient of the magnetic field B, an enormous magnetic field is required to produce a macroscopic current, which may be realized in high energy heavy ion collisions. Thus far, we derived the CME in the chiral fermion system using the Wigner function approach, and we observe that the CME is a first order quantum effect inℏ . In fact, the Wigner function approach is a quantum kinetic theory, which implies the presence of quantum effects of a multi-particle system, such as the CME. -
In this and the following sections, we derive the CME for a chiral fermion system by determining the Landau levels. The Lagrangian for a chiral fermion field is
L=¯Ψ(x)iγ⋅DΨ(x),
(22) with the covariant derivative
Dμ=∂μ+ieAμ , and the electric charge±e for particles/antiparticles. For a uniform magnetic fieldB=Bez along the z-axis, we choose the gauge potential asAμ=(0,0,Bx,0) . The equation of motion for the fieldΨ(x) isiγ⋅DΨ(x)=0,
(23) which can be written in the form of a Schrödinger equation,
i∂∂tΨ(t,x)=iα⋅DΨ(t,x),
(24) with
D=−∇+ieA ,A=(0,Bx,0) . In the chiral representation of Diracγ -matrices, whereγ5=diag(−1,1) ,α=diag(−σ,σ) , we expressΨ in the formΨ=(ΨTL,ΨTR)T . Then, Eq. (24) becomesi∂∂t(ΨL(t,x)ΨR(t,x))=(−iσ⋅DΨL(t,x)iσ⋅DΨR(t,x)),
(25) which indicates that the two fields
ΨL/R , which correspond to eigenvalues∓1 of the matrixγ5 , decouple with each other. The two fieldsΨL/R are often referred to as lefthand/righthand fermion fields, respectively. Lefthand and righthand fermions are also referred to as chiral fermions.In the following, we focus on solving the eigenvalue equation for the righthand fermion field
ΨR (similar results are obtained for the lefthand fermion fieldΨL ).To determine the Landau levels, we must solve the eigenvalue equation for the righthand fermion field as follows,
iσ⋅DψR=EψR,
(26) with
D=(−∂x,−∂y+ieBx,−∂z) . The details for solving Eq. (26) are provided in Appendix B. We list the eigenfunctions and eigenvalues in the following : Forn=0 Landau level, the wavefunction with energyE=kz isψR0(ky,kz;x)=(φ0(ξ)0)1Lei(yky+zkz).
(27) For
n>0 Landau level, the wavefunction with energyE=λEn(kz) isψRnλ(ky,kz;x)=cnλ(φn(ξ)iFnλφn−1(ξ))1Lei(yky+zkz),
(28) where
λ=±1 ,En(kz)=√2neB+k2z ,Fnλ(kz)=[kz−λEn(kz)]/ √2neB , normalized coefficient|cnλ|2=1/(1+F2nλ) , andφn(ξ)=φn(√eBx−ky/√eB) is the n-th order wavefunction of a harmonic oscillator.For
n>0 Landau levels, the wavefunctions with energiesE=±En(kz) correspond to fermions and antifermions, respectively. For the lowest Landau level, the wavefunction with energyE=kz>0 corresponds to fermions, whereas that with energyE=kz<0 corresponds to antifermions. The wavefunctions of all Landau levels are orthonormal and complete. For the lefthand fermion field, the eigenfunctions of Landau levels are the same as the righthand case, but with the sign of the eigenvalues changed. -
In this section, we second-quantize the righthand fermion field
ΨR(x) , such that it becomes an operator and satisfies following anticommutative relations,{ΨR(x),Ψ†R(x′)}=δ(3)(x−x′),{ΨR(x),ΨR(x′)}=0.
(29) Because all eigenfunctions for the Hamiltonian of the righthand fermion field are orthonormal and complete, we decompose the righthand fermion field operator
ΨR(x) by these eigenfunctions asΨR(x)=∑ky,kz[θ(kz)a0(ky,kz)ψR0(ky,kz;x)+θ(−kz)b†0(ky,kz)ψR0(ky,kz;x)]+∑n,ky,kz[an(ky,kz)ψRn+(ky,kz;x)+b†n(ky,kz)ψRn−(ky,kz;x)].
(30) In contrast to the general Fourier decomposition for the second quantization, we place two theta functions
θ(±kz) in front ofa0(ky,kz) andb†0(ky,kz) in the decomposition, which is very important for the subsequent second quantization procedure. From Eq. (29), we obtain following anticommutative relations,{θ(kz)a0(ky,kz),θ(k′z)a†0(k′y,k′z)}=θ(kz)δkyk′yδkzk′z{θ(−kz)b0(ky,kz),θ(−k′z)b†0(k′y,k′z)}=θ(−kz)δkyk′yδkzk′z{an(ky,kz),a†n′(k′y,k′z)}=δnn′δkyk′yδkzk′z{bn(ky,kz),b†n′(k′y,k′z)}=δnn′δkyk′yδkzk′z.
(31) The two theta functions
θ(±kz) are always attached to the lowest Landau level operators, such asa0,a†0,b0,b†0 . The Hamiltonian and total particle number of the righthand fermion system areH=∫d3xΨ†R(x)iσ⋅DΨR(x)=∑ky,kz[kzθ(kz)a†0(ky,kz)a0(ky,kz)+(−kz)θ(−kz)b†0(ky,kz)b0(ky,kz)]+∑n,ky,kzEn(kz)[a†n(ky,kz)×an(ky,kz)+b†n(ky,kz)bn(ky,kz)],
(32) N=∫d3xΨ†R(x)ΨR(x)=∑ky,kz[θ(kz)a†0(ky,kz)a0(ky,kz)−θ(−kz)b†0(ky,kz)b0(ky,kz)]×∑n,ky,kz[a†n(ky,kz)an(ky,kz)−b†n(ky,kz)bn(ky,kz)],
(33) where we omitted the infinite vacuum term. This can be renormalized in the physics calculation and does not affect our result on the CME coefficient. Evidently,
θ(kz)a†0(ky,kz)a0(ky,kz) anda†n(ky,kz)an(ky,kz) are the occupied number operators of particles for different Landau levels, andθ(−kz)b†0(ky,kz)b0(ky,kz) andb†n(ky,kz)bn(ky,kz) are occupied number operators of antiparticles for different Landau levels. Notably, without introduction of the two theta functionsθ(±kz) in front ofa0(ky,kz) andb†0(ky,kz) in the decomposition ofΨR(x) , the second quantization procedure could not be performed successfully. This is different from the massive case [21], where the authors determined the Landau levels and corresponding wavefunctions for the massive Dirac equation in a uniform magnetic field with chemical potentialμ and chiral chemical potentialμ5 . The wave functions for the massive case are in a four-component Dirac form, and theθ function is not needed for the second quantization. -
Supposing that the system of the righthand ferimons within an external uniform magnetic field
B=Bez is in equilibrium with a reservior with temperature T and chemical potentialμR , then the density operatorˆρ for this righthand fermion system isˆρ=1Ze−β(H−μRN),
(34) where
β=1/T is the inverse temperature, and Z is the grand canonical partition function,Z=Tre−β(H−μRN).
(35) The expectation value of an operator
ˆF in the equilibrium state can be calculated as⟨:ˆF:⟩=Tr(ˆρˆF).
(36) In the Appendix C, we calculated the expectation values of occupied number operators as
⟨:θ(kz)a†0(ky,kz)a0(ky,kz):⟩=θ(kz)eβ(kz−μR)+1⟨:θ(−kz)b†0(ky,kz)b0(ky,kz):⟩=θ(−kz)eβ(−kz+μR)+1⟨:a†n(ky,kz)an(ky,kz):⟩=1eβ[En(kz)−μR]+1⟨:b†n(ky,kz)bn(ky,kz):⟩=1eβ[En(kz)+μR]+1.
(37) The macroscopic electric current for the righthand fermion system is
JR=⟨:Ψ†R(x)σΨR(x):⟩.
(38) According to the rotational invariance of this system along the z-axis,
JxR=JyR=0 . In the following, we calculateJzR . Using Eq. (30), we see thatJzR=⟨:Ψ†R(x)σ3ΨR(x):⟩=∑ky,kz(⟨:θ(kz)a†0(ky,kz)a0(ky,kz):⟩+⟨:θ(−kz)b0(ky,kz)b†0(ky,kz):⟩)×ψ†R0(ky,kz;x)σ3ψR0(ky,kz;x)+∑n,ky,kz⟨:a†n(ky,kz)an(ky,kz):⟩ψ†Rn+(ky,kz;x)σ3ψRn+(ky,kz;x)+∑n,ky,kz⟨:bn(ky,kz)b†n(ky,kz):⟩ψ†Rn−(ky,kz;x)σ3ψRn−(ky,kz;x)=∑ky,kz(θ(kz)eβ(kz−μR)+1−θ(−kz)eβ(−kz+μR)+1)ψ†R0(ky,kz;x)σ3ψR0(ky,kz;x)+∑n,ky,kz1eβ[En(kz)−μR]+1ψ†Rn+(ky,kz;x)σ3ψRn+(ky,kz;x)−∑n,ky,kz1eβ[En(kz)+μR]+1ψ†Rn−(ky,kz;x)σ3ψRn−(ky,kz;x).
(39) First, we sum over
ky forψ†R0(ky,kz;x)σ3ψR0(ky,kz;x) andψ†Rnλ(ky,kz;x)σ3ψRnλ(ky,kz;x) in Eq. (39), the results are∑kyψ†R0(ky,kz;x)σ3ψR0(ky,kz;x)=1L2∑ky(φ0(ξ),0)(100−1)(φ0(ξ)0)=12πL∫∞−∞dky[φ0(√eBx−ky/√eB)]2=eB2πL,
(40) and
∑kyψ†Rnλ(ky,kz;x)σ3ψRnλ(ky,kz;x)=12πL∫∞−∞dkyc2nλ(kz)([φn(ξ)]2−2neB[φn−1(ξ)]2[kz+λEn(kz)]2)=eB2πLc2nλ(kz)(1−2neB[kz+λEn(kz)]2)=eB2πLc2nλ(kz)[2−c−2nλ(kz)]=eB2πL⋅λkzEn(kz).
(41) Second, we sum over
kz in the third equal sign of Eq. (39),JzR=∑kz(θ(kz)eβ(kz−μR)+1−θ(−kz)eβ(−kz+μR)+1)eB2πL+∑n,kz(1eβ[En(kz)−μR]+1+1eβ[En(kz)+μR]+1)eB2πL⋅kzEn(kz)=eB4π2∫∞−∞dkz(θ(kz)eβ(kz−μR)+1−θ(−kz)eβ(−kz+μR)+1)+0=eB4π2μR.
(42) Combining Eq. (42) and
JxR=JyR=0 yieldsJR=eμR4π2B.
(43) From the calculation above, we see that only the lowest Landau level contributes to Eq. (43). A similar calculation for the lefthand fermion system shows that
JL=−eμL4π2B.
(44) We can also obtain Eq. (44) from Eq. (43) under space inversion:
JR→−JL ,μR→μL ,B→B . If the system is composed of righthand and lefthand fermions, then the vector currentJV and axial currentJA areJV=JR+JL=eμ52π2B,
(45) JA=JR−JL=eμ2π2B,
(46) where
μ5=(μR−μL)/2 is the chiral chemical potential andμ=(μR+μL)/2 . Thus far, we derived the CME in the chiral fermion system by determining Landau levels. We emphasize that Eqs. (45), (46) are valid for any strength of magnetic field, in contrast to the weak magnetic field approximation through Wigner function approach in Sec. 2. -
We discuss the physical picture of the lowest Landau level. The wavefunction and energy of the lowest Landau level (
n=0 ) for the righthand fermion field isψR0(ky,kz;x)=(φ00)1Lei(yky+zkz), E=kz.
(47) Setting
ky=0 in Eq. (47), we calculate the Hamiltonian, particle number, z-component of momentum, and z-component of the spin angular momentum of the righthand fermion system for the lowest Landau level as follows,H=∑kz[kzθ(kz)a†0(0,kz)a0(0,kz)+(−kz)θ(−kz)b†0(0,kz)b0(0,kz)],N=∑kz[θ(kz)a†0(0,kz)a0(0,kz)+(−1)θ(−kz)b†0(0,kz)b0(0,kz)],Pz=∑kz[kzθ(kz)a†0(0,kz)a0(0,kz)+(−kz)θ(−kz)b†0(0,kz)b0(0,kz)],Sz=∑kz[12θ(kz)a†0(0,kz)a0(0,kz)+(−12)θ(−kz)b†0(0,kz)b0(0,kz)],
(48) where the definitions of
Pz andSz arePz=−i∫d3xΨ†R(x)∂∂zΨR(x),Sz=12∫d3xΨ†R(x)σ3ΨR(x).
(49) Thus, we have a picture for the lowest Landau level: The operator
θ(kz)a†0(0,kz) produces a particle with charge e, energykz>0 , z-component of momentumkz>0 , and z-component of spin angular momentum+12 (helicityh=+1 ); The operatorθ(−kz)b†0(0,kz) produces a particle with charge−e , energy−kz>0 , z-component of momentum−kz>0 , and z-component of spin angular momentum−12 (helicityh=−1 ). This picture indicates that all righthand fermions/antifermions move along the(+z) -axis, with righthand fermions spinning along the(+z) -axis and righthand antifermions spinning along the−z -axis. IfμR>0 , which indicates that there are more righthand fermions than righthand anti-fermions, a net electric current will move along the(+z) -axis, which is referred to as the CME for the righthand fermion system.The analogous analysis can be applied to lefthand fermions. The picture of the lowest Landau level for a lefthand fermion is: All lefthand fermions/antifermions move along the
(−z) -axis, with left fermions spinning along the(+z) -axis and lefthand antifermions spinning along the(−z) -axis. IfμL>0 , which indicates that there is more lefthand fermions than lefthand anti-fermions, a net electric current will move along the(−z) -axis, which is referred to as the CME for the lefthand fermion system.Because the total electric current
JV of the chiral fermion system is the summation of the electric currentJR of the righthand fermion system and the electric currentJL of the lefthand fermion system, whetherJV moves along the(+z) -axis will only depend on the sign of(μR−μL) . Thus, the CME for the chiral fermion system is described microscopically. -
The quadratic form for the equation of motion of the Wigner function
W(x,p) is(K2−i2σμν[Kμ,Kν])W=0.
(A1) Using
K2=p2−14∇2+ip⋅∇ and[Kμ,Kν]=−ieFμν , Eq. (A1) becomes(p2−14∇2+ip⋅∇−12eFμνσμν)W=0.
(A2) W andσμν satisfyW=γ0W†γ0 andσμν=γ0σμν†γ0 . Employing the Hermi conjugation and subsequently multiplyingγ0 to both sides of Eq. (A2) yields(p2−14∇2−ip⋅∇)W−12eFμνWσμν=0.
(A3) Eq. (A2) minus Eq. (A3) yields the Vlasov equation for
W ,ip⋅∇W−14eFμν[σμν,W]=0.
(A4) Eq. (A2) added to Eq. (A3) yields the off mass-shell equation for
W ,(p2−14∇2)W−14eFμν{σμν,W}=0.
(A5) To calculate
[σμν,W] and{σμν,W} in Eqs. (A4) (A5), we list the following useful identities,[σμν,1]=0,[σμν,iγ5]=0,[σμν,γρ]=−2igρ[μγν],[σμν,γ5γρ]=−2igρ[μγ5γν],[σμν,σρσ]=2igμ[ρσσ]ν−2igν[ρσσ]μ,
(A6) {σμν,1}=2σμν,{σμν,iγ5}=−ϵμνρσσρσ,{σμν,γρ}=2ϵμνρσγ5γσ,{σμν,γ5γρ}=2ϵμνρσγσ,{σμν,σρσ}=2gμ[ρgσ]ν+2iϵμνρσγ5.
(A7) Then, all matrices appearing in Eqs. (A4) (A5) are the 16 independent
Γ -matrices, whose coefficients must be zero. These coefficient equations are the Vlasov equations and the off mass-shell equations forF,P,Vμ,Aμ,Sμν . The Vlasov equations arep⋅∇F=0,p⋅∇P=0,p⋅∇Vμ=eFμνVν,p⋅∇Aμ=eFμνAν,p⋅∇Qμν=eFρ [μQν]ρ,
(A8) and the off mass-shell equations are
(p2−14∇2)F=12eFμνQμν,(p2−14∇2)P=12e˜FμνQμν,(p2−14∇2)Vμ=−e˜FμνAν,(p2−14∇2)Aμ=−e˜FμνVν,(p2−14∇2)Qμν=e(FμνF−˜FμνP),
(A9) where
˜Fμν=12εμνρσFρσ . -
We solve following eigenvalue equation in detail,
iσ⋅DψR(x)=EψR(x),
(B1) with
D=(−∂x,−∂y+ieBx,−∂z) . Because the operatoriσ⋅D is commutative withˆpy=−i∂y,ˆpz=−i∂z , we can chooseψR as the commom eigenstate ofiσ⋅D ,ˆpy andˆpz as followsψR(x,y,z)=(ϕ1(x)ϕ2(x))1Lei(yky+zkz),
(B2) where L is the length of the system in y- and z- directions. The explicit form of
σ⋅D isσ⋅D=(−∂z−∂x+i∂y+eBx−∂x−i∂y−eBx∂z).
(B3) Inserting Eq. (B2) (B3) into Eq. (B1), we obtain the group of differential equations for
ϕ1(x) andϕ2(x) asi(kz−E)ϕ1+(∂x+ky−eBx)ϕ2=0,
(B4) (∂x−ky+eBx)ϕ1−i(kz+E)ϕ2=0.
(B5) From Eq. (B5), we can express
ϕ2 byϕ1 , then Eq. (B4) becomes∂2xϕ1+(E2+eB−k2z−e2B2(x−kyeB)2)ϕ1=0,
(B6) which is a typical harmonic oscillator equation. Defining a dimensionless variable
ξ=√eB(x−ky/eB) , andϕ1(x)=φ(ξ) , then (B6) becomesd2φdξ2+(E2−k2zeB+1−ξ2)φ=0.
(B7) With the boundary condition
φ→0 asξ→±∞ , we must setE2−k2zeB+1=2n+1,
(B8) with
n=0,1,2,⋯ . Thus, energy E can only assume the following discrete values,E=±En(kz)≡±√2neB+k2z,
(B9) where we define
En(kz)=√2neB+k2z . The corresponding normalized solution for equation (B6) isϕ1(x)=φn(ξ)=Nne−ξ2/2Hn(ξ),
(B10) where
Nn=(eB)14π−14(2nn!)−12 , andHn(ξ)=(−1)neξ2dndξne−ξ2 . For energyE=λEn(kz) (λ=±1 ), we can obtainϕ2 asϕ2(x)=√eB(∂ξ+ξ)φn(ξ)i(kz+E)=i[kz−λEn(kz)]√2neBφn−1(ξ),
(B11) where we used
(∂ξ+ξ)φn(ξ)=√2nφn−1(ξ) . DefiningFnλ(kz)= [kz−λEn(kz)]/√2neB , the eigenfunction with eigenvalueE=λEn(kz) becomesψRnλ(ky,kz;x)=(φn(ξ)iFnλ(kz)φn−1(ξ))1Lei(yky+zkz).
(B12) This is very subtle when
n=0 in Eq. (B11). Whenn=0,E=kz , the first equal sign of Eq. (B11) indicatesϕ2=0 due to(∂ξ+ξ)φ0(ξ)=0 . Then, the corresponding eigenfunction becomesψR0(ky,kz;x)=(φ0(ξ)0)1Lei(yky+zkz).
(B13) When
n=0,E=−kz , the denominator of the first equal sign of Eq. (B11) becomes zero, in which case we must directly deal with Eqs. (B4) (B5). In this case, Eqs. (B4) (B5) become2ikzϕ1+(∂x+ky−eBx)ϕ2=0,
(B14) (∂x−ky+eBx)ϕ1=0.
(B15) Eq. (B15) gives
ϕ1(x)∼exp[−12eBx2+xky] , then Eq. (B14) becomes2ikzexp(−12eBx2+xky)+(∂x+ky−eBx)ϕ2=0.
(B16) When
x→±∞ , Eq. (B16) tends to(∂x−eBx)ϕ2=0,
(B17) whose solution is
ϕ2∼exp(12eBx2) , which is divergent asx→±∞ . Thus, there is no physical solution whenn=0,E=−kz .Thus far, we obtain the eigenfunctions and eigenvalues of the Hamiltonian of the righthand fermion field as follows:
For
n=0 Landau level, the wavefunction with energyE=kz isψR0(ky,kz;x)=(φ00)1Lei(yky+zkz).
(B18) For
n>0 Landau level, the wavefunction with energyE=λEn(kz) isψRnλ(ky,kz;x)=cnλ(φniFnλφn−1)1Lei(yky+zkz),
(B19) where
λ=±1 ,En(kz)=√2neB+k2z ,Fnλ(kz)=[kz−λEn(kz)]/√2neB ,|cnλ|2=1/(1+F2nλ) . -
We calculate the expectation values of particle number operators. From the expression of the Hamiltonian and the total particle number operator in Eqs. (32, 33), we easily obtain following commutative relations,
[N,θ(kz)a†0(ky,kz)]=θ(kz)a†0(ky,kz)[N,θ(−kz)b†0(ky,kz)]=−θ(−kz)b†0(ky,kz)[N,a†n(ky,kz)]=a†n(ky,kz)[N,b†n(ky,kz)]=−b†n(ky,kz),
(C1) [H,θ(kz)a†0(ky,kz)]=kzθ(kz)a†0(ky,kz)[H,θ(−kz)b†0(ky,kz)]=(−kz)θ(−kz)b†0(ky,kz)[H,a†n(ky,kz)]=En(kz)a†n(ky,kz)[H,b†n(ky,kz)]=En(kz)b†n(ky,kz),
(C2) where we employ
[AB,C]=A{B,C}−{A,C}B . Definingθ(kz)a†0(ky,kz;β)=e−β(H−μRN)θ(kz)a†0(ky,kz)eβ(H−μRN)θ(−kz)b†0(ky,kz;β)=e−β(H−μRN)θ(−kz)b†0(ky,kz)eβ(H−μRN)
a†n(ky,kz;β)=e−β(H−μRN)a†n(ky,kz)eβ(H−μRN)b†n(ky,kz;β)=e−β(H−μRN)b†n(ky,kz)eβ(H−μRN).
(C3) For
θ(kz)a†0(ky,kz;β) , we obtain∂∂β[θ(kz)a†0(ky,kz;β)]=−[H−μRN,θ(kz)a†0(ky,kz;β)]=−e−β(H−μRN)[H−μRN,θ(kz)a†0(ky,kz)]eβ(H−μRN)=−e−β(H−μRN)[(kz−μR)θ(kz)a†0(ky,kz)]eβ(H−μRN)=−(kz−μR)[θ(kz)a†0(ky,kz;β)],
(C4) with the boundary condition
θ(kz)a†0(ky,kz;0)=θ(kz)a†0(ky,kz) , which impliesθ(kz)a†0(ky,kz;β)=θ(kz)a†0(ky,kz)e−β(kz−μR).
(C5) Similarly, we obtain
θ(−kz)b†0(ky,kz;β)=θ(−kz)b†0(ky,kz)e−β(−kz+μR)a†n(ky,kz;β)=a†n(ky,kz)e−β[En(kz)−μR]b†n(ky,kz;β)=b†n(ky,kz)e−β[En(kz)+μR].
(C6) We calculate the expectation value of
⟨:θ(kz)a†0(ky,kz) a0(ky,kz):⟩ . We see that⟨:θ(kz)a†0(ky,kz)a0(ky,kz):⟩=Tr[ρθ(kz)a†0(ky,kz)a0(ky,kz)]=1ZTr(θ(kz)a†0(ky,kz;β)e−β(H−μRN)a0(ky,kz))=1ZTr(θ(kz)a0(ky,kz)a†0(ky,kz;β)e−β(H−μRN))=⟨:θ(kz)a0(ky,kz)a†0(ky,kz;β):⟩=⟨:θ(kz)a0(ky,kz)a†0(ky,kz):⟩e−β(kz−μR)=θ(kz)e−β(kz−μR)−⟨:θ(kz)a†0(ky,kz)a0(ky,kz):⟩e−β(kz−μR),
(C7) thus, we obtain
⟨θ(kz)a†0(ky,kz)a0(ky,kz)⟩=θ(kz)eβ(kz−μR)+1.
(C8) Similar calculations obtain
⟨θ(−kz)b†0(ky,kz)b0(ky,kz)⟩=θ(−kz)eβ(−kz+μR)+1⟨a†n(ky,kz)an(ky,kz)⟩=1eβ[En(kz)−μR]+1⟨b†n(ky,kz)bn(ky,kz)⟩=1eβ[En(kz)+μR]+1.
(C9)
Chiral magnetic effect for chiral fermion system
- Received Date: 2020-01-25
- Available Online: 2020-07-01
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 fermions 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 magnetic field, 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.