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Polarization of an electron scattered by static potentials

Hao-Hao Peng Shanghai United Imaging Healthcare Advanced Technology Research Institute Co., Ltd., Shanghai 201822, China    Ren-Hong Fang [email protected] College of Intelligent Systems Science and Engineering, Hubei Minzu University, Enshi, Hubei 445000, China
Abstract

We study the polarization of an electron scattered by different static potentials. The initial state of the electron is chosen as a wavepacket to construct the definite orbital angular momentum, and the final polarization of the electron, scattered by different static potentials such as vector, pseudovector, scalar and pseudoscalar potentials, is calculated. Numerical results show that, the sign of the polarization of the electron scattered by the vector potential is opposite to the other three cases, and the magnitude order of the polarization value is consistent with recent experimental result in the collision parameter range 0<b<2fm0<b<2\,\mathrm{fm}.

I Introduction

It has long been known that particles with nonzero spin quantum number can polarise in a rotating system (Barnett, 1915; Einstein and de Haas, 1915). It can be qualitatively explained by spin-orbit coupling through which the orbital angular momentum (OAM) of the macroscopic rotating system transfers into the spin angular momentum of microscopic particles (Liang and Wang, 2005a). In high energy peripheral heavy-ion collisions at STAR in BNL and at LHC in CERN, an extremely hot and dense matter (named quark gluon plasma (QGP)) with huge OAM is created (Liang and Wang, 2005a; Betz et al., 2007; Becattini et al., 2008; Wang, 2017), hence a huge vorticity configuration is formed in this matter (Becattini et al., 2013; Pang et al., 2016; Jiang et al., 2016; Deng and Huang, 2016; Li et al., 2017). It is expected that the final hadrons with nonzero spin quantum number will polarise globally along the direction of the initial OAM at the freeze-out stage of QGP (Liang and Wang, 2005a; Gao et al., 2008; Liang and Wang, 2005b; Abelev et al., 2008). After more than ten years’ effort, the global polarization of Λ\Lambda hyperon had been firstly observed at STAR (Abelev et al., 2007; Adamczyk et al., 2017). The collisional energy dependence of the global polarization of Λ\Lambda hyperon can be well recovered by the numerical simulations of the relativistic hydrodynamics model and the multi-phase transport model (Xie et al., 2016; Li et al., 2017). The spin alignment of vector mesons in heavy-ion collisions are also measured these years (Abelev et al., 2008; Zhou, 2018).

There are many methods to theoretically study the polarization of particles in a fluid with relativistic vorticity. Based on the assumption of local equilibrium of spin, the authors in (Becattini et al., 2013; Fang et al., 2016; Yang et al., 2018) derived the relation between the 4-dimensional spin vector (Pauli-Lubanski pseudovector) and vorticity in relativistic case through Wigner function approach. Recently the 4-dimensional spin vector induced by vorticity is also obtained from the method of spin density matrix but without the factor of Fermi-Dirac thermal distribution (Becattini et al., 2019). The first theoretical calculation of the polarization based on microscopic scattering by a static potential is performed in (Liang and Wang, 2005a), in which the concept of global polarization of particles in high energy heavy-ion collisions is firstly proposed. Based on (Liang and Wang, 2005a), the theoretical calculation of quark polarization through two-body scattering (Gao et al., 2008) is performed, and the transport property of polarization in a fluid is also discussed in (Huang et al., 2011). The statistical model and quark coalescense model of hydron polarization are carried out in recent studies (Liang and Wang, 2005b; Becattini et al., 2017; Yang et al., 2018; Sheng et al., 2020a, b). Recently one of us and his collaborators put forward a systematic formulism to calculate the particle polarization in QGP from the spin-orbit coupling (Zhang et al., 2019), in which all 222\rightarrow 2 processes are considered and one can see clearly how the macroscopic vorticity in the fluid induces the microscopic polarization of a particle.

In order to carefully understand the polarization due to spin-orbit coupling, in this article we consider a simple model to calculate the polarization of a Dirac fermion scattered by four types of static potential. We choose the wavepacket as the initial state of the electron which is the same as (Zhang et al., 2019) and sum over the initial polarization states, so that the initial total angular momentum is just the OAM of the incident electron. The final state of the electron is chosen as the common eigenstate of momentum and spin, then the scattering probability of different final spin can be calculated.

This article is organised as follows. In Sec. II, we set up the theoretical formulism for the scattering probability of different final spin. In Sec. III, the polarization of the scattered electron by four types of static potential is studied. We briefly summarise this article in Sec. IV.

In this article, the natural unit where =c=1\hbar=c=1 is adopted. We choose the metric tensor as gμν=diag (+1,1,1,1)g^{\mu\nu}=\text{diag\,}(+1,-1,-1,-1). Greek indices, such as μ,ν,ρ,σ\mu,\nu,\rho,\sigma, run over 0,1,2,30,1,2,3, or t,x,y,zt,x,y,z, while Roman indices, such as i,j,ki,j,k, run over 1,2,31,2,3 or x,y,zx,y,z. The Heaviside-Lorentz convention is chosen for electromagnetism which is consistent with Peskin and Schroeder (Peskin and Schroeder, 1995).

II A wavepacket is scattered by a static potential

In this section we consider an electron wavepacket which is scattered by a classical static potential (time independent). The classical static potential is denoted as 𝒜\mathcal{A}, the center of which is located at the origin. A very convenient choice for the static potential is the screened potential model (Gyulassy and Wang, 1994), whose explicit form in position space is V(𝒙)=Qed|𝒙|/(4π|𝒙|)V(\boldsymbol{x})=Qe^{-d|\boldsymbol{x}|}/(4\pi|\boldsymbol{x}|) with electric charge source QQ and force distance 1/d1/d. The form of this screened potential in momentum space is V(𝒒)=Q/(𝒒2+d2)V(\boldsymbol{q})=Q/(\boldsymbol{q}^{2}+d^{2}). The electron wavepacket is denoted as \mathcal{B}. The in-state |ϕ,λ;𝒃in|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle_{\text{in}} of the single particle wavepacket for this electron in the remote past with impact parameter 𝒃=(b,0,0)\boldsymbol{b}=(b,0,0) can be represented as

|ϕ,λ;𝒃in=d3k(2π)312Ekϕ(𝒌)ei𝒌𝒃|𝒌,λin,|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle_{\text{in}}=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{k}}}\phi(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{b}}|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle_{\text{in}}, (1)

where |𝒌,λin|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle_{\text{in}} represents the in-state of a plane wave state of an electron with momentum 𝒌\boldsymbol{k} and spin projection component λ\lambda_{\mathcal{B}} along the direction of positive yy-axis, and ϕ(𝒌)\phi(\boldsymbol{k}) is a normalised Guassian wavepacket with momentum center 𝒑i=(0,0,c)\boldsymbol{p}_{i}=(0,0,c) and momentum width aa whose explicit form is

ϕ(𝒌)=((8π)3a6)14exp((𝒌𝒑i)2a2).\phi(\boldsymbol{k})=\bigg{(}\frac{(8\pi)^{3}}{a^{6}}\bigg{)}^{\frac{1}{4}}\exp\bigg{(}-\frac{(\boldsymbol{k}-\boldsymbol{p}_{i})^{2}}{a^{2}}\bigg{)}. (2)

Taking use of 𝒌,λ|𝒌,λinin=2Ek(2π)3δ(3)(𝒌𝒌){}_{\text{in}}\langle\boldsymbol{k}^{\prime},\lambda_{\mathcal{B}}|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle_{\text{in}}=2E_{k}(2\pi)^{3}\delta^{(3)}(\boldsymbol{k}^{\prime}-\boldsymbol{k}), we can see that the in-state in Eq. (1) can be normalised to 11 as

ϕ,λ;𝒃|ϕ,λ;𝒃inin=d3k(2π)3|ϕ(𝒌)|2=1.{}_{\text{in}}\langle\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle_{\text{in}}=\int\frac{d^{3}k}{(2\pi)^{3}}|\phi(\boldsymbol{k})|^{2}=1. (3)

Since the electron moves from z-z-axis to +z+z-axis, we set c>0c>0. We also set b>0b>0, then the initial orbital angular momentum of the electron is along y-y-axis. When this electron moves toward the static potential, it will be scattered by some probability. The sacttering probability with spin component λ\lambda_{\mathcal{B}} summed over in the initial state and with spin projection component λ\lambda along +y+y-axis in the final state is

𝒫(λ,b)\displaystyle\mathcal{P}(\lambda,b) =\displaystyle= 12d3p(2π)32Epλ|𝒑,λ|ϕ,λ;𝒃inout|2\displaystyle\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{p}}\sum_{\lambda_{\mathcal{B}}}\bigg{|}{}_{\text{out}}\langle\boldsymbol{p},\lambda|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle_{\text{in}}\bigg{|}^{2} (4)
=\displaystyle= 12d3p(2π)32Epλ|𝒑,λ|𝒮|ϕ,λ;𝒃|2\displaystyle\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{p}}\sum_{\lambda_{\mathcal{B}}}\bigg{|}\langle\boldsymbol{p},\lambda|\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle\bigg{|}^{2}
=\displaystyle= 12d3p(2π)32Epλ|d3k(2π)312Ekϕ(𝒌)ei𝒌𝒃𝒑,λ|𝒮|𝒌,λ|2\displaystyle\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{p}}\sum_{\lambda_{\mathcal{B}}}\bigg{|}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{k}}}\phi(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{b}}\langle\boldsymbol{p},\lambda|\mathcal{S}|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle\bigg{|}^{2}

where 𝒮\mathcal{S} denotes the SS-matrix for the interaction and we choose the plane wave state |𝒑,λ|\boldsymbol{p},\lambda\rangle as the final state with normalization: 𝒑,λ|𝒑,λ=2Ep(2π)3δ(3)(𝒑𝒑)\langle\boldsymbol{p}^{\prime},\lambda|\boldsymbol{p},\lambda\rangle=2E_{p}(2\pi)^{3}\delta^{(3)}(\boldsymbol{p}^{\prime}-\boldsymbol{p}). Then we can obtain the polarization χ(b)\chi(b) of the final electron along +y+y-axis as

χ(b)=𝒫(+,b)𝒫(,b)𝒫(+,b)+𝒫(,b).\chi(b)=\frac{\mathcal{P}(+,b)-\mathcal{P}(-,b)}{\mathcal{P}(+,b)+\mathcal{P}(-,b)}. (5)

The denominator 𝒫(+,b)+𝒫(,b)\mathcal{P}(+,b)+\mathcal{P}(-,b) in Eq. (5) is the normalization factor. In fact, due to the unitarity of the SS-matrix, 𝒮𝒮=1\mathcal{S}^{\dagger}\mathcal{S}=1, the sum λ𝒫(λ,b)\sum_{\lambda}\mathcal{P}(\lambda,b) automatically equal to one, i.e.

λ𝒫(λ,b)\displaystyle\sum_{\lambda}\mathcal{P}(\lambda,b) =\displaystyle= 12d3p(2π)32Epλλ|𝒑,λ|𝒮|ϕ,λ;𝒃|2\displaystyle\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{p}}\sum_{\lambda}\sum_{\lambda_{\mathcal{B}}}\bigg{|}\langle\boldsymbol{p},\lambda|\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle\bigg{|}^{2} (6)
=\displaystyle= 12d3p(2π)32Epλλϕ,λ;𝒃|𝒮|𝒑,λ𝒑,λ|𝒮|ϕ,λ;𝒃\displaystyle\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{p}}\sum_{\lambda}\sum_{\lambda_{\mathcal{B}}}\langle\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}|\mathcal{S}^{\dagger}|\boldsymbol{p},\lambda\rangle\langle\boldsymbol{p},\lambda|\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle
=\displaystyle= 12λϕ,λ;𝒃|𝒮𝒮|ϕ,λ;𝒃\displaystyle\frac{1}{2}\sum_{\lambda_{\mathcal{B}}}\langle\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}|\mathcal{S}^{\dagger}\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle
=\displaystyle= 12λϕ,λ;𝒃|ϕ,λ;𝒃\displaystyle\frac{1}{2}\sum_{\lambda_{\mathcal{B}}}\langle\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}};\boldsymbol{b}\rangle
=\displaystyle= 1.\displaystyle 1.

But in actual calculation, we only consider the process of tree level and ignore the high order processes, so the sum λ𝒫(λ,b)\sum_{\lambda}\mathcal{P}(\lambda,b) no longer equal to one and the normalization factor in the denominator in Eq. (5) is necessary.

III Polarization of the electron scattered by different static potentials

In this section, we will calculate the polarization of an electron scattered by different static potentials. The electron is described by Dirac field. Firstly, we consider the coupling between Dirac field and static vector field, i.e. the interaction part of the Lagrangian is int=eψ¯γμψAμ\mathcal{L}_{\text{int}}=-e\bar{\psi}\gamma^{\mu}\psi A_{\mu} with Aμ(t,𝒙)=(V(𝒙),𝟎)A^{\mu}(t,\boldsymbol{x})=(V(\boldsymbol{x}),\boldsymbol{0}). Ignoring the identity operator in SS-matrix, the term 𝒑,λ|𝒮|𝒌,λ\langle\boldsymbol{p},\lambda|\mathcal{S}|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle in Eq. (4) at first order in coupling ee is

𝒑,λ|𝒮|𝒌,λ\displaystyle\langle\boldsymbol{p},\lambda|\mathcal{S}|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle =\displaystyle= 𝒑,λ|(ie)d4xψ¯(x)γμψ(x)Aμ(x)|𝒌,λ\displaystyle\langle\boldsymbol{p},\lambda|(-ie)\int d^{4}x\bar{\psi}(x)\gamma^{\mu}\psi(x)A_{\mu}(x)|\boldsymbol{k},\lambda_{\mathcal{B}}\rangle (7)
\displaystyle\equiv (ie)u¯(p,λ)γA(𝒑𝒌)u(k,λ)(2π)δ(EpEk),\displaystyle(-ie)\bar{u}(p,\lambda)\gamma\cdot A(\boldsymbol{p}-\boldsymbol{k})u(k,\lambda_{\mathcal{B}})(2\pi)\delta(E_{p}-E_{k}),

where Aμ(𝒑𝒌)=(V(𝒑𝒌),𝟎)A^{\mu}(\boldsymbol{p}-\boldsymbol{k})=(V(\boldsymbol{p}-\boldsymbol{k}),\boldsymbol{0}) and u(k,λ),u(p,λ)u(k,\lambda_{\mathcal{B}}),u(p,\lambda) are solved in chiral representation of Dirac matrix for free Dirac equation. In the following calculation, we will choose u(k,λ),u(p,λ)u(k,\lambda_{\mathcal{B}}),u(p,\lambda) as

u(k,λ)\displaystyle u(k,\lambda_{\mathcal{B}}) =\displaystyle= 12(m+Ek)((m+Ek𝝈𝒌)ξλ(m+Ek+𝝈𝒌)ξλ)\displaystyle\frac{1}{\sqrt{2(m+E_{k})}}\bigg{(}\begin{array}[]{c}(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})\xi_{\lambda_{\mathcal{B}}}\\ (m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})\xi_{\lambda_{\mathcal{B}}}\end{array}\bigg{)} (10)
u(p,λ)\displaystyle u(p,\lambda) =\displaystyle= 12(m+Ep)((m+Ep𝝈𝒑)ξλ(m+Ep+𝝈𝒑)ξλ),\displaystyle\frac{1}{\sqrt{2(m+E_{p})}}\bigg{(}\begin{array}[]{c}(m+E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p})\xi_{\lambda}\\ (m+E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p})\xi_{\lambda}\end{array}\bigg{)}, (13)

where ξλ\xi_{\lambda} is the eigenstate of spin operator 𝝈\boldsymbol{\sigma} along +y+y-axis with eigenvalue λ\lambda, i.e. ξλ𝝈ξλ=λ𝒚^\xi_{\lambda}^{\dagger}\boldsymbol{\sigma}\xi_{\lambda}=\lambda\hat{\boldsymbol{y}} with λ=±1\lambda=\pm 1. Since λ\lambda_{\mathcal{B}} will be summed over later, the explicit form of ξλ\xi_{\lambda_{\mathcal{B}}} is irrelevant.

From Eq. (4), we can obtain the scattering probability 𝒫V(λ,b)\mathcal{P}^{V}(\lambda,b) with the final spin projection component λ\lambda and collision parameter bb for static vector potential, which is calculated in detail in Appendix A. We list the explicit form of 𝒫V(λ,b)\mathcal{P}^{V}(\lambda,b) as follows,

𝒫V(λ,b)=𝒫0V(b)+λ𝒫1V(b),\mathcal{P}^{V}(\lambda,b)=\mathcal{P}_{0}^{V}(b)+\lambda\mathcal{P}_{1}^{V}(b), (14)

with unpolarised probability 𝒫0V(b)\mathcal{P}_{0}^{V}(b) and polarised probability 𝒫1V(b)\mathcal{P}_{1}^{V}(b) defined as 7-dimensional integrals,

𝒫0V(b)\displaystyle\mathcal{P}_{0}^{V}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)cos[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\cos[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×[(Ep+m)2+(Epm)2𝒌^𝒌^+p2𝒑^(𝒌^+𝒌^)]\displaystyle\times[(E_{p}+m)^{2}+(E_{p}-m)^{2}\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}+p^{2}\hat{\boldsymbol{p}}\cdot(\hat{\boldsymbol{k}}+\hat{\boldsymbol{k}}^{\prime})]
𝒫1V(b)\displaystyle\mathcal{P}_{1}^{V}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)sin[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\sin[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×((Epm)2(2𝒚^𝒑^𝒑^𝒚^)(𝒌^×𝒌^)+p2𝒚^[𝒑^×(𝒌^𝒌^)])\displaystyle\times\bigg{(}(E_{p}-m)^{2}(2\hat{\boldsymbol{y}}\cdot\hat{\boldsymbol{p}}\hat{\boldsymbol{p}}-\hat{\boldsymbol{y}})\cdot(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k}}^{\prime})+p^{2}\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\bigg{)}

where dΩpd\Omega_{p}, dΩd\Omega, dΩd\Omega^{\prime} represent the differential solid angle of 𝒑^\hat{\boldsymbol{p}}, 𝒌^\hat{\boldsymbol{k}}, 𝒌^\hat{\boldsymbol{k}}^{\prime} respectively, and 𝒙^=(1,0,0)\hat{\boldsymbol{x}}=(1,0,0), 𝒚^=(0,1,0)\hat{\boldsymbol{y}}=(0,1,0).

Then the polarization χV(b)\chi^{V}(b) of the electron scattered by the static vector potential in Eq. (5) becomes

χV(b)=𝒫1V(b)𝒫0V(b).\chi^{V}(b)=\frac{\mathcal{P}_{1}^{V}(b)}{\mathcal{P}_{0}^{V}(b)}. (15)

If the electron is scattered by the static pseudovector, scalar and pseudoscalar potentials respectively, then the interaction part of the Lagrangian int\mathcal{L}_{\text{int}} becomes ψ¯γμγ5ψAμ,ψ¯ψϕ,ψ¯γ5ψϕ\bar{\psi}\gamma^{\mu}\gamma^{5}\psi A_{\mu},\bar{\psi}\psi\phi,\bar{\psi}\gamma^{5}\psi\phi, where ϕ\phi is a scalar or pseudoscalar field. In Appendix A, we also calculate the polarization function χ(b)\chi(b) for these three types of static potential.

Figure 1 shows the numerical results of the polarization χ(b)\chi(b) as a fucntion of collision parameter bb for four types of static potential with the parameters, where we take following parameters: initial energy of the incident electron c=1GeVc=1\,\mathrm{GeV}, the width of wavepacket a=0.1GeV,a=0.1\,\mathrm{GeV},Debye screen mass d=0.1GeVd=0.1\,\mathrm{GeV}, and electron mass m=0.000511GeVm=0.000511\,\mathrm{GeV}. Since b>0b>0, c>0c>0, i.e. the initial OAM of the electron is along negative yy-axis, we may expect that the final eletron scattered by the static potentials polarises more preferentially along negative yy-axis, which is consistent with the results for pseudovector, scalar and pseudoscalar potentials as shown in Figure 1. However, for vector potential, the final eletron polarises more preferentially along positive yy-axis, which is opposite to the direction of the initial OAM of the electron. This inconsistence may result from the fact that the virtual photon exchanged by the electron and the static potential also carries spin angular momentum of 11\,\hbar, leading to the opposite polarization of the final electron due to the conservation of angular momentum. In Figure 1, we only plot the curves in the range 0<b<6fm0<b<6\,\mathrm{fm}, where the the absolute value of polarization becomes larger as bb increases. Especially in the range 0<b<2fm0<b<2\,\mathrm{fm}, the magnitude order of the polarization value is the same as the recent experimental result (Adamczyk et al., 2017). For the range b>6fmb>6\,\mathrm{fm}, the numerical result becomes unstable, which is not shown in the plot. It is expected that the polarization χ(b)\chi(b) becomes zero as bb tends to be very large, since in this case the influence of the static potential on the electron is very weak.

Refer to caption
Figure 1: Polarization χ(b)\chi(b) as a fucntion of collision parameter bb

IV Summary

In this article, we calculate the polarization of an electron scattered by four different types of static potential. Through the scattering of static potentials, it is expected that the initial OAM of the incident electron can transfer into the final spin angular momentum, which is consistent with the numerical results for pseudovector, scalar and pseudoscalar potentials. However the electron polarises more preferentially opposite to the initial OAM of the electron, which may result from the spin of the virtual photon exchanged. For the range of the collision parameter 0<b<2fm0<b<2\,\mathrm{fm}, the magnitude order of the polarization value is consistent with the recent experimental result.

V Acknowledgments

We thank Qun Wang for helpful discussion. This work was supported in part by the Science and Technology Research Project of Education Department of Hubei Province under Grant No. D20221901, and the National Natural Science Foundation of China (NSFC) under Grant Nos. 12265013 and 12073008.

Appendix A Calculation for 𝒫(λ,b)\mathcal{P}(\lambda,b)

From Eq. (7) and Eq. (1), we obtain

𝒑,λ|𝒮|ϕ,λ\displaystyle\langle\boldsymbol{p},\lambda|\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}}\rangle (16)
=\displaystyle= d3k(2π)3ϕ(𝒌)2Ekei𝒌𝒃(ie)u¯(p,λ)γA(𝒑𝒌)u(k,λ)(2π)δ(EpEk)\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\phi(\boldsymbol{k})}{\sqrt{2E_{k}}}e^{-i\boldsymbol{k}\cdot\boldsymbol{b}}(-ie)\bar{u}(p,\lambda)\gamma\cdot A(\boldsymbol{p}-\boldsymbol{k})u(k,\lambda_{\mathcal{B}})(2\pi)\delta(E_{p}-E_{k})
=\displaystyle= 1(2π)20d|𝒌|𝒌2𝑑Ωϕ(𝒌)2Ekei𝒌𝒃(ie)u¯(p,λ)γA(𝒑𝒌)u(k,λ)Ep|𝒑|δ(|𝒌||𝒑|)\displaystyle\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}d|\boldsymbol{k}|\boldsymbol{k}^{2}\int d\Omega\frac{\phi(\boldsymbol{k})}{\sqrt{2E_{k}}}e^{-i\boldsymbol{k}\cdot\boldsymbol{b}}(-ie)\bar{u}(p,\lambda)\gamma\cdot A(\boldsymbol{p}-\boldsymbol{k})u(k,\lambda_{\mathcal{B}})\frac{E_{p}}{|\boldsymbol{p}|}\delta(|\boldsymbol{k}|-|\boldsymbol{p}|)
=\displaystyle= ie|𝒑|(2π)2Ep2𝑑ΩA0(𝒑𝒌)ϕ(𝒌)ei𝒌𝒃u(p,λ)u(k,λ),\displaystyle\frac{-ie|\boldsymbol{p}|}{(2\pi)^{2}}\sqrt{\frac{E_{p}}{2}}\int d\Omega A_{0}(\boldsymbol{p}-\boldsymbol{k})\phi(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{b}}u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}}),

where dΩd\Omega represents the solid angle of 𝒌^\hat{\boldsymbol{k}}. Taking square of Eq. (16) and summing over λ\lambda_{\mathcal{B}} gives

λ|𝒑,λ|𝒮|ϕ,λ|2\displaystyle\sum_{\lambda_{\mathcal{B}}}\bigg{|}\langle\boldsymbol{p},\lambda|\mathcal{S}|\phi_{\mathcal{B}},\lambda_{\mathcal{B}}\rangle\bigg{|}^{2} (17)
=\displaystyle= α𝒑2Ep(2π)3𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)ei(𝒌𝒌)𝒃\displaystyle\frac{\alpha\boldsymbol{p}^{2}E_{p}}{(2\pi)^{3}}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})e^{-i(\boldsymbol{k}-\boldsymbol{k}^{\prime})\cdot\boldsymbol{b}}
×λ[u(p,λ)u(k,λ)u(k,λ)u(p,λ)]\displaystyle\times\sum_{\lambda_{\mathcal{B}}}[u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})u(p,\lambda)]

where dΩd\Omega^{\prime} represents the solid angle of 𝒌^\hat{\boldsymbol{k}}^{\prime}. In the following, we will calculate the third line of Eq. (17) in detail. We can see

λ[u(p,λ)u(k,λ)u(k,λ)u(p,λ)]=λtr[u(p,λ)u(p,λ)u(k,λ)u(k,λ)]\sum_{\lambda_{\mathcal{B}}}[u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})u(p,\lambda)]=\sum_{\lambda_{\mathcal{B}}}\text{tr}[u(p,\lambda)u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})] (18)

For u(k,λ)u(k,λ)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}}) part, we have

M\displaystyle\text{M}_{\mathcal{B}} =\displaystyle= λu(k,λ)u(k,λ)\displaystyle\sum_{\lambda_{\mathcal{B}}}u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})
=\displaystyle= 12(m+Ek)(m+Ek)\displaystyle\frac{1}{2\sqrt{(m+E_{k})(m+E_{k^{\prime}})}}
×λ((m+Ek𝝈𝒌)ξ(m+Ek+𝝈𝒌)ξ)(ξ(m+Ek𝝈𝒌),ξ(m+Ek+𝝈𝒌))\displaystyle\times\sum_{\lambda_{\mathcal{B}}}\bigg{(}\begin{array}[]{c}(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})\xi_{\mathcal{B}}\\ (m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})\xi_{\mathcal{B}}\end{array}\bigg{)}\left(\begin{array}[]{cc}\xi_{\mathcal{B}}^{\dagger}(m+E_{k^{\prime}}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime}),\xi_{\mathcal{B}}^{\dagger}(&m+E_{k^{\prime}}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})\end{array}\right)
=\displaystyle= 12(m+Ek)(m+Ek)\displaystyle\frac{1}{2\sqrt{(m+E_{k})(m+E_{k^{\prime}})}}
×((m+Ek𝝈𝒌)(m+Ek𝝈𝒌),(m+Ek𝝈𝒌)(m+Ek+𝝈𝒌)(m+Ek+𝝈𝒌)(m+Ek𝝈𝒌),(m+Ek+𝝈𝒌)(m+Ek+𝝈𝒌))\displaystyle\times\left(\begin{array}[]{cc}(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime}),&(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})\\ (m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime}),&(m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})\end{array}\right)

For u(p,λ)u(p,λ)u(p,\lambda)u^{\dagger}(p,\lambda) part, we have

u(p,λ)u(p,λ)=u(p,λ)u¯(p,λ)γ0=12(1+λγ5γS)(γp+m)γ0u(p,\lambda)u^{\dagger}(p,\lambda)=u(p,\lambda)\bar{u}(p,\lambda)\gamma^{0}=\frac{1}{2}(1+\lambda\gamma^{5}\gamma\cdot S)(\gamma\cdot p+m)\gamma^{0} (21)

where SμS^{\mu} is the 4-dimensional spin vector

Sμ=(S0,𝑺)=(𝒑𝒚^m,𝒚^+(𝒚^𝒑)𝒑m(m+Ep))S^{\mu}=(S^{0},\boldsymbol{S})=\bigg{(}\frac{\boldsymbol{p}\cdot\hat{\boldsymbol{y}}}{m},\hat{\boldsymbol{y}}+\frac{(\hat{\boldsymbol{y}}\cdot\boldsymbol{p})\boldsymbol{p}}{m(m+E_{p})}\bigg{)} (22)

We can see

γp+m=(mEp𝝈𝒑Ep+𝝈𝒑m)\gamma\cdot p+m=\left(\begin{array}[]{cc}m&E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p}\\ E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p}&m\end{array}\right)
γ5γS=(0S0+𝝈𝑺S0+𝝈𝑺0)\gamma^{5}\gamma\cdot S=\left(\begin{array}[]{cc}0&-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S}\\ S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S}&0\end{array}\right)

The term related to λ\lambda in u(p,λ)u(p,λ)u(p,\lambda)u^{\dagger}(p,\lambda) is

M1\displaystyle\text{M}_{1} =\displaystyle= λ2γ5γS(γp+m)γ0\displaystyle\frac{\lambda}{2}\gamma^{5}\gamma\cdot S(\gamma\cdot p+m)\gamma^{0}
=\displaystyle= λ2(0S0+𝝈𝑺S0+𝝈𝑺0)(mEp𝝈𝒑Ep+𝝈𝒑m)(0110)\displaystyle\frac{\lambda}{2}\left(\begin{array}[]{cc}0&-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S}\\ S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S}&0\end{array}\right)\left(\begin{array}[]{cc}m&E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p}\\ E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p}&m\end{array}\right)\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)
=\displaystyle= λ2((S0+𝝈𝑺)m(S0+𝝈𝑺)(Ep+𝝈𝒑)(S0+𝝈𝑺)(Ep𝝈𝒑)(S0+𝝈𝑺)m)\displaystyle\frac{\lambda}{2}\left(\begin{array}[]{cc}(-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})m&(-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})(E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p})\\ (S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})(E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p})&(S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})m\end{array}\right)

The term without λ\lambda in u(p,λ)u(p,λ)u(p,\lambda)u^{\dagger}(p,\lambda) is

M0=12(γp+m)γ0=12(mEp𝝈𝒑Ep+𝝈𝒑m)(0110)=12(Ep𝝈𝒑mmEp+𝝈𝒑)\text{M}_{0}=\frac{1}{2}(\gamma\cdot p+m)\gamma^{0}=\frac{1}{2}\left(\begin{array}[]{cc}m&E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p}\\ E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p}&m\end{array}\right)\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)=\frac{1}{2}\left(\begin{array}[]{cc}E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p}&m\\ m&E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p}\end{array}\right)

The term related to λ\lambda in λ[u(p,λ)u(k,λ)u(k,λ)u(p,λ)]\sum_{\lambda_{\mathcal{B}}}[u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})u(p,\lambda)] is

tr(M1M)\displaystyle\text{tr}\,(\text{M}_{1}\text{M}_{\mathcal{B}}) =\displaystyle= (M1M)11+(M1M)22\displaystyle(\text{M}_{1}\text{M}_{\mathcal{B}})_{11}+(\text{M}_{1}\text{M}_{\mathcal{B}})_{22}
=\displaystyle= (M1)11(M)11+(M1)12(M)21+(M1)21(M)12+(M1)22(M)22\displaystyle(\text{M}_{1})_{11}(\text{M}_{\mathcal{B}})_{11}+(\text{M}_{1})_{12}(\text{M}_{\mathcal{B}})_{21}+(\text{M}_{1})_{21}(\text{M}_{\mathcal{B}})_{12}+(\text{M}_{1})_{22}(\text{M}_{\mathcal{B}})_{22}
\displaystyle\equiv λ(m+Ek)(m+Ek)×14(I1+II1+III1+IV1)\displaystyle\frac{\lambda}{\sqrt{(m+E_{k})(m+E_{k^{\prime}})}}\times\frac{1}{4}(\text{I}_{1}+\text{II}_{1}+\text{III}_{1}+\text{IV}_{1})

where

I1\displaystyle\text{I}_{1} =\displaystyle= tr(S0+𝝈𝑺)m(m+Ek𝝈𝒌)(m+Ek𝝈𝒌)\displaystyle\text{tr}\,(-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})m(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
II1\displaystyle\text{II}_{1} =\displaystyle= tr(S0+𝝈𝑺)(Ep+𝝈𝒑)(m+Ek+𝝈𝒌)(m+Ek𝝈𝒌)\displaystyle\text{tr}\,(-S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})(E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p})(m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
III1\displaystyle\text{III}_{1} =\displaystyle= tr(S0+𝝈𝑺)(Ep𝝈𝒑)(m+Ek𝝈𝒌)(m+Ek+𝝈𝒌)\displaystyle\text{tr}\,(S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})(E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p})(m+E_{k}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
IV1\displaystyle\text{IV}_{1} =\displaystyle= tr(S0+𝝈𝑺)m(m+Ek+𝝈𝒌)(m+Ek+𝝈𝒌)\displaystyle\text{tr}\,(S^{0}+\boldsymbol{\sigma}\cdot\boldsymbol{S})m(m+E_{k}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{k^{\prime}}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})

Taking use of

12tr(𝝈𝒂)(𝝈𝒃)=𝒂𝒃\frac{1}{2}\text{tr}\,(\boldsymbol{\sigma}\cdot\boldsymbol{a})(\boldsymbol{\sigma}\cdot\boldsymbol{b})=\boldsymbol{a}\cdot\boldsymbol{b}
12tr(𝝈𝒂)(𝝈𝒃)(𝝈𝒄)=i𝒂(𝒃×𝒄)\frac{1}{2}\text{tr}\,(\boldsymbol{\sigma}\cdot\boldsymbol{a})(\boldsymbol{\sigma}\cdot\boldsymbol{b})(\boldsymbol{\sigma}\cdot\boldsymbol{c})=i\boldsymbol{a}\cdot(\boldsymbol{b}\times\boldsymbol{c})
12tr(𝝈𝒂)(𝝈𝒃)(𝝈𝒄)(𝝈𝒅)=(𝒂𝒃)(𝒄𝒅)(𝒂𝒄)(𝒃𝒅)+(𝒂𝒅)(𝒃𝒄)\frac{1}{2}\text{tr}\,(\boldsymbol{\sigma}\cdot\boldsymbol{a})(\boldsymbol{\sigma}\cdot\boldsymbol{b})(\boldsymbol{\sigma}\cdot\boldsymbol{c})(\boldsymbol{\sigma}\cdot\boldsymbol{d})=(\boldsymbol{a}\cdot\boldsymbol{b})(\boldsymbol{c}\cdot\boldsymbol{d})-(\boldsymbol{a}\cdot\boldsymbol{c})(\boldsymbol{b}\cdot\boldsymbol{d})+(\boldsymbol{a}\cdot\boldsymbol{d})(\boldsymbol{b}\cdot\boldsymbol{c})

we have

14(I1+IV1)=im𝑺(𝒌×𝒌)\frac{1}{4}(\text{I}_{1}+\text{IV}_{1})=im\boldsymbol{S}\cdot(\boldsymbol{k}\times\boldsymbol{k}^{\prime})
14(II1+III1)=i(S0𝒑Ep𝑺)(𝒌×𝒌)+i(m+Ep)𝑺[𝒑×(𝒌𝒌)]\frac{1}{4}(\text{II}_{1}+\text{III}_{1})=i(S^{0}\boldsymbol{p}-E_{p}\boldsymbol{S})\cdot(\boldsymbol{k}\times\boldsymbol{k}^{\prime})+i(m+E_{p})\boldsymbol{S}\cdot[\boldsymbol{p}\times(\boldsymbol{k}-\boldsymbol{k}^{\prime})]

Note that Ek=Ek=EpE_{k}=E_{k^{\prime}}=E_{p} due to the delta function δ(EpEk)\delta(E_{p}-E_{k}) in Eq. (7). Finally we have

tr(M1M)\displaystyle\text{tr}\,(\text{M}_{1}\text{M}_{\mathcal{B}}) =\displaystyle= λ(m+Ek)(m+Ek)×14(I1+II1+III1+IV1)\displaystyle\frac{\lambda}{\sqrt{(m+E_{k})(m+E_{k^{\prime}})}}\times\frac{1}{4}(\text{I}_{1}+\text{II}_{1}+\text{III}_{1}+\text{IV}_{1})
=\displaystyle= iλ((Epm)2(2𝒚^𝒑^𝒑^𝒚^)(𝒌^×𝒌^)+p2𝒚^[𝒑^×(𝒌^𝒌^)])\displaystyle i\lambda\bigg{(}(E_{p}-m)^{2}(2\hat{\boldsymbol{y}}\cdot\hat{\boldsymbol{p}}\hat{\boldsymbol{p}}-\hat{\boldsymbol{y}})\cdot(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k}}^{\prime})+p^{2}\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\bigg{)}

Then we can obtain the coefficient of λ\lambda in Eq. (4) as

𝒫1(b)\displaystyle\mathcal{P}_{1}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)sin[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\sin[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×((Epm)2(2𝒚^𝒑^𝒑^𝒚^)(𝒌^×𝒌^)+p2𝒚^[𝒑^×(𝒌^𝒌^)])\displaystyle\times\bigg{(}(E_{p}-m)^{2}(2\hat{\boldsymbol{y}}\cdot\hat{\boldsymbol{p}}\hat{\boldsymbol{p}}-\hat{\boldsymbol{y}})\cdot(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k}}^{\prime})+p^{2}\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\bigg{)}

where α=e2/4π\alpha=e^{2}/4\pi, p=|𝒑|=|𝒌|=|𝒌|p=|\boldsymbol{p}|=|\boldsymbol{k}|=|\boldsymbol{k}^{\prime}| and dΩpd\Omega_{p} represents the solid angle of 𝒑^\hat{\boldsymbol{p}}. The term errelevant to λ\lambda in λ[u(p,λ)u(k,λ)u(k,λ)u(p,λ)]\sum_{\lambda_{\mathcal{B}}}[u^{\dagger}(p,\lambda)u(k,\lambda_{\mathcal{B}})u^{\dagger}(k^{\prime},\lambda_{\mathcal{B}})u(p,\lambda)] is

tr(M0M)\displaystyle\text{tr}\,(\text{M}_{0}\text{M}_{\mathcal{B}}) =\displaystyle= (M0M)11+(M0M)22\displaystyle(\text{M}_{0}\text{M}_{\mathcal{B}})_{11}+(\text{M}_{0}\text{M}_{\mathcal{B}})_{22}
=\displaystyle= (M0)11(M)11+(M0)12(M)21+(M0)21(M)12+(M0)22(M)22\displaystyle(\text{M}_{0})_{11}(\text{M}_{\mathcal{B}})_{11}+(\text{M}_{0})_{12}(\text{M}_{\mathcal{B}})_{21}+(\text{M}_{0})_{21}(\text{M}_{\mathcal{B}})_{12}+(\text{M}_{0})_{22}(\text{M}_{\mathcal{B}})_{22}
\displaystyle\equiv 1m+Ep×14(I0+II0+III0+IV0)\displaystyle\frac{1}{m+E_{p}}\times\frac{1}{4}(\text{I}_{0}+\text{II}_{0}+\text{III}_{0}+\text{IV}_{0})

where

I0\displaystyle\text{I}_{0} =\displaystyle= tr(Ep𝝈𝒑)(m+Ep𝝈𝒌)(m+Ep𝝈𝒌)\displaystyle\text{tr}\,(E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{p})(m+E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
II0\displaystyle\text{II}_{0} =\displaystyle= trm(m+Ep+𝝈𝒌)(m+Ep𝝈𝒌)\displaystyle\text{tr}\,m(m+E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
III0\displaystyle\text{III}_{0} =\displaystyle= trm(m+Ep𝝈𝒌)(m+Ep+𝝈𝒌)\displaystyle\text{tr}\,m(m+E_{p}-\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})
IV0\displaystyle\text{IV}_{0} =\displaystyle= tr(Ep+𝝈𝒑)(m+Ep+𝝈𝒌)(m+Ep+𝝈𝒌)\displaystyle\text{tr}\,(E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{p})(m+E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{k})(m+E_{p}+\boldsymbol{\sigma}\cdot\boldsymbol{k}^{\prime})

We have

14(I0+IV0)=Ep(m+Ep)2+Ep𝒌𝒌+(m+Ep)𝒑𝒌+(m+Ep)𝒑𝒌\frac{1}{4}(\text{I}_{0}+\text{IV}_{0})=E_{p}(m+E_{p})^{2}+E_{p}\boldsymbol{k}\cdot\boldsymbol{k}^{\prime}+(m+E_{p})\boldsymbol{p}\cdot\boldsymbol{k}+(m+E_{p})\boldsymbol{p}\cdot\boldsymbol{k}^{\prime}
14(II0+III0)=m(m+Ep)2m𝒌𝒌\frac{1}{4}(\text{II}_{0}+\text{III}_{0})=m(m+E_{p})^{2}-m\boldsymbol{k}\cdot\boldsymbol{k}^{\prime}

then

14(I0+II0+III0+IV0)=(m+Ep)3+(Epm)𝒌𝒌+(m+Ep)𝒑(𝒌+𝒌)\frac{1}{4}(\text{I}_{0}+\text{II}_{0}+\text{III}_{0}+\text{IV}_{0})=(m+E_{p})^{3}+(E_{p}-m)\boldsymbol{k}\cdot\boldsymbol{k}^{\prime}+(m+E_{p})\boldsymbol{p}\cdot(\boldsymbol{k}+\boldsymbol{k}^{\prime})

Finally we have

tr(M0M)\displaystyle\text{tr}\,(\text{M}_{0}\text{M}_{\mathcal{B}}) =\displaystyle= 1m+Ep×14(I0+II0+III0+IV0)\displaystyle\frac{1}{m+E_{p}}\times\frac{1}{4}(\text{I}_{0}+\text{II}_{0}+\text{III}_{0}+\text{IV}_{0})
=\displaystyle= (Ep+m)2+(Epm)2𝒌^𝒌^+p2𝒑^(𝒌^+𝒌^)\displaystyle(E_{p}+m)^{2}+(E_{p}-m)^{2}\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}+p^{2}\hat{\boldsymbol{p}}\cdot(\hat{\boldsymbol{k}}+\hat{\boldsymbol{k}}^{\prime})

Then we can obtain the term without λ\lambda in Eq. (4) as

𝒫0(b)\displaystyle\mathcal{P}_{0}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)cos[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\cos[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×[(Ep+m)2+(Epm)2𝒌^𝒌^+p2𝒑^(𝒌^+𝒌^)]\displaystyle\times[(E_{p}+m)^{2}+(E_{p}-m)^{2}\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}+p^{2}\hat{\boldsymbol{p}}\cdot(\hat{\boldsymbol{k}}+\hat{\boldsymbol{k}}^{\prime})]

The probability 𝒫(λ,b)\mathcal{P}(\lambda,b) at impact parameter bb becomes

𝒫(λ,b)=𝒫0(b)+λ𝒫1(b)\mathcal{P}(\lambda,b)=\mathcal{P}_{0}(b)+\lambda\mathcal{P}_{1}(b)

For the potentials of pseudovector, scalar and pseudoscalar, we can replace the γμAμ\gamma^{\mu}A_{\mu} factor in the interaction part of the Lagrangian int=eψ¯γμψAμ\mathcal{L}_{\text{int}}=-e\bar{\psi}\gamma^{\mu}\psi A_{\mu} by γμγ5Aμ,ϕ,γ5ϕ\gamma^{\mu}\gamma^{5}A_{\mu},\phi,\gamma^{5}\phi, where ϕ\phi is a scalar or pseudoscalar field. The calculations for the scattering probability by this three types of interaction are similar to the vector case, and we list the results as follows.

For pseudovector potential, we have

𝒫1PV(b)\displaystyle\mathcal{P}_{1}^{PV}(b) =\displaystyle= α4(2π)6𝑑pp6𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)cos[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{6}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\cos[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×{𝒚^(𝒌×𝒌)𝒚^[𝒑^×(𝒌^𝒌^)]}\displaystyle\times\{\hat{\boldsymbol{y}}\cdot(\boldsymbol{k}\times\boldsymbol{k}^{\prime})-\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\}
𝒫0PV(b)\displaystyle\mathcal{P}_{0}^{PV}(b) =\displaystyle= α4(2π)6𝑑pp6𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)sin[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{6}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\sin[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×(1+𝒌^𝒌^+𝒑^𝒌^+𝒌^𝒑^)\displaystyle\times(1+\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}+\hat{\boldsymbol{p}}\cdot\hat{\boldsymbol{k}}^{\prime}+\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{p}})

For scalar potential, we have

𝒫1S(b)\displaystyle\mathcal{P}_{1}^{S}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)sin[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\sin[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×{(Epm)2(2𝒚^𝒑^𝒑^𝒚^)(𝒌^×𝒌^)p2𝒚^[𝒑^×(𝒌^𝒌^)]}\displaystyle\times\{(E_{p}-m)^{2}(2\hat{\boldsymbol{y}}\cdot\hat{\boldsymbol{p}}\hat{\boldsymbol{p}}-\hat{\boldsymbol{y}})\cdot(\hat{\boldsymbol{k}}\times\hat{\boldsymbol{k}}^{\prime})-p^{2}\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\}
𝒫0S(b)\displaystyle\mathcal{P}_{0}^{S}(b) =\displaystyle= α4(2π)6𝑑pp4𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)cos[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{4}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\cos[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×[(Ep+m)2+(Epm)2𝒌^𝒌^p2𝒑^(𝒌^+𝒌^)]\displaystyle\times[(E_{p}+m)^{2}+(E_{p}-m)^{2}\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}-p^{2}\hat{\boldsymbol{p}}\cdot(\hat{\boldsymbol{k}}+\hat{\boldsymbol{k}}^{\prime})]

For pseudoscalar potential, we have

𝒫1PS(b)\displaystyle\mathcal{P}_{1}^{PS}(b) =\displaystyle= α4(2π)6𝑑pp6𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)cos[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{6}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\cos[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×{𝒚^(𝒌×𝒌)+𝒚^[𝒑^×(𝒌^𝒌^)]}\displaystyle\times\{\hat{\boldsymbol{y}}\cdot(\boldsymbol{k}\times\boldsymbol{k}^{\prime})+\hat{\boldsymbol{y}}\cdot[\hat{\boldsymbol{p}}\times(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]\}
𝒫0PS(b)\displaystyle\mathcal{P}_{0}^{PS}(b) =\displaystyle= α4(2π)6𝑑pp6𝑑Ωp𝑑Ω𝑑ΩA0(𝒑𝒌)A0(𝒑𝒌)ϕ(𝒌)ϕ(𝒌)sin[pb𝒙^(𝒌^𝒌^)]\displaystyle\frac{\alpha}{4(2\pi)^{6}}\int dpp^{6}\int d\Omega_{p}\int d\Omega d\Omega^{\prime}A_{0}(\boldsymbol{p}-\boldsymbol{k})A_{0}(\boldsymbol{p}-\boldsymbol{k}^{\prime})\phi(\boldsymbol{k})\phi(\boldsymbol{k}^{\prime})\sin[pb\hat{\boldsymbol{x}}\cdot(\hat{\boldsymbol{k}}-\hat{\boldsymbol{k}}^{\prime})]
×(1+𝒌^𝒌^𝒑^𝒌^𝒌^𝒑^)\displaystyle\times(1+\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{k}}^{\prime}-\hat{\boldsymbol{p}}\cdot\hat{\boldsymbol{k}}^{\prime}-\hat{\boldsymbol{k}}\cdot\hat{\boldsymbol{p}})

References

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