Electron anomalous magnetic moment Alexander Kupco March 27, 2017 Correction to the electron-photon vertex γµ is described in the order of e3 by single Feynman graph, Fig. 1: Z d4k N µ Γµ = −ie2 ; (1) (2π)4 D where the nominator is of the form µ ν 0 µ N = γ (=p + =k + m)γ (=p + =k + m)γν (2) and denominator is D = (p0 + k)2 − m2 (p + k)2 − m2 k2 ; (3) with k being the 4-momentum of the virtual photon in the inner loop, and p and p0 4-momenta of incoming and outgoing electron. In this case, one vertex factor (ie) was taken out as it present in the tree-level vertex ieγµ. Singling out a factor of e corresponds to taking the same factor in the charged current J µ = euγ¯ µu. In the first step, we find such linear transformation of the inner photon 4-momentum k that the integral in Γµ becomes gaussian. This can be done introducing Swinger (more often labeled as Feynmann) transformation: 1 Z 1 Z 1−α 1 = 2 dα dβ 3 : (4) a1a2a3 0 0 (αa1 + βa2 + (1 − α − β)a3) In our case, 1 Z 1 Z 1−α 1 = 2 dα dβ : (5) 0 2 2 2 2 2 3 D 0 0 (α[(p + k) − m ] + β[(p + k) − m ] + (1 − α − β)k ) The denominator of this denominator simplifies into d = α[(p0 + k)2 − m2] + β[(p + k)2 − m2] + (1 − α − β)k2 = k2 + 2k · (αp0 + βp) : (6) Introducing l ≡ k + (αp0 + βp) (7) we get d = l2 − m2(α2 + β2) − 2αβp0 · p : = l2 − m2(α + β)2 + αβq2 ; (8) where q = p − p0 is external photon 4-momentum. In the limit of q2 ! 0, this simplifies to d = l2 − m2(α + β)2; for q2 ! 0 : (9) With new inner loop momentum transformation, the nomitator takes the form of µ ν 0 µ 0 N = γ (=l + (1 − α)=p − β=p + m)γ (=l + (1 − β)=p − α=p + m)γν : (10) Figure 1: One loop correction to the vertex function that gives the correction to the electron magnetic moment. 1 Terms linear in l disapear in the integration as now the denominator D is even function of l. The m2-term is 2 2 ν µ 2 µ m : m γ γ γν = −2m γ : (11) Term proportional to γµ does not contributute to the anomalous magnetic moment, it comes only from (p + p0)µ term 1 in the general form of spin{ 2 current preserving parity: i J µ = eu¯ A(q2)γµ + B(q2)(p + p0)µ u = eu¯ F (q2)γµ + F (q2)σµν q u (12) 1 2m 2 ν where the second step comes from Gordon decomposition. This is also the reason why other diagrams, like those with virtual photon radiated and absorbed by an electron before the interaction with external real photon, do not contribute 2 2 as well. At the limit of long wave exchange q ! 0, the term F1(q ! 0) describes overall charge of the object while 2 the term F2(q ! 0) ! κ describes additional anomalous magnetic moment κ to the Bohr level magnetic moment of 1 spin{ 2 particle in units of Bohr magneton µB = e=(2m). Therefor, we have just two effective terms for the nominator: µ µ µ Neff = N1eff + N2eff (13) with µ ν µ 0 ν 0 µ N1eff = m [γ γ ((1 − β)=p − α=p )γν ] + m [γ ((1 − α)=p − β=p)γ γν ] (14) and µ ν 0 µ 0 0 µ 0 N2eff = γ [(1 − α)=p − β=p]γ [(1 − β)=p − α=p ]γν = −2[(1 − β)=p − α=p ]γ [(1 − α)=p − β=p] ; (15) as the last effective term is again proportional to γµ: µ ν µ σ ν σ µ σ σ µ N3eff = γ =lγ =lγν ∼ l lσγ γ γ γσγν = −2l lσγ γ γσ : (16) This is an important feature as the divergent part of the loop integral does not contribute to the magnetic moment of the particle. The whole calculation can be than carried without introducing renormalization procedure. For the first effective term we have µ µ 0µ 0µ µ µ 0µ N1eff = 4m (1 − β)p − αp + (1 − α)p − βp = 4m (1 − 2β)p + (1 − 2α)p : (17) µ µ To evaluate the second term N2eff , we need to image that it is part of current:uN ¯ 2eff u. First we evaluate the following parts of the current, droping the irrelevant γµ terms: u¯ [=pγµ=p0] u =u ¯ [2pµ=p0 − γµ=p=p0] u =u ¯ [2mpµ − 2(p · p0)γµ + γµ=p0m] u ∼ u¯ [2m(p + p0)µ] u u¯ [=p0γµ=p] u = m2u¯ [γµ] u ∼ 0 u¯ [=pγµ=p] u =u ¯ [m=pγµ] u ∼ u¯ [2mpµ] u u¯ [=p0γµ=p0] u =u ¯ [mγµ=p0] u ∼ u¯ 2mp0µ u : (18) With this, we get µ µ 0µ 0µ µ N2eff ∼ −4m(1 − β)(1 − α)(p + p ) + 4mα(1 − α)p + 4mβ(1 − β)p = = −4m (1 − 2β − α + αβ + β2)pµ + (1 − 2α − β + αβ + α2)p0µ : (19) Combining the results, we get for the effective value of nominator µ 2 µ 2 0µ Neff = 4m (α − αβ − β )p + (β − αβ − α )p : (20) The effective vertex correction Γµ has thus two contributions proportional to pµ and p0µ. From the conservation of current, these two must be the same after performing the integral. We can therefor symmetrize the result in the limit of q2 ! 0 as 1 N µ = 4m(pµ + p0µ) (α − αβ − β2 + β − αβ − α2) = 2m(pµ + p0µ) [(α + β)(1 − (α + β))] ; (21) eff 2 and Z 1 Z 1−α Z d4l (α + β)(1 − (α + β)) Γµ = −ie24m(pµ + p0µ) dα dβ : (22) eff 4 2 2 2 3 0 0 (2π) [l − m (α + β) ] This symmetrized form will simplify greatly the final intergration over parameters α and β. To solve this integral, we will use the standard technique of Wick rotation Z d3~l Z dl 1 Z d3~l I = 0 = I (a2 = j~lj2 − m2(α + β)2) ; (23) 1 3 h i3 3 11 (2π) 2π 2 ~ 2 2 2 (2π) l0 − jlj − m (α + β) with Z 1 Z 1 2 dl0 1 dl0 1 I11(a ) = = : (24) 2π 2 2 3 π 2 2 3 −∞ (l0 − a ) 0 (l0 − a ) 2 Wick rotation gives I11 = lim [−I111(") − I113(R) − I114(R)] ; (25) "!0 ;R!1 with Z π Z π π i" exp(i') i" exp(−2i') i" exp(−2i') −I111(") = d' ! d' = = 0 ; (26) 2 2 3 3 3 0 π [(a + " exp(i')) − a ] 0 π [2a"] π(2a") −2i 0 Z π iR exp(i') 1 I113(R) = d' 2 2 3 / ! 0 : (27) 0 π(R exp(2i') − a ) R We can therefore write Z 1 Z 1 idλ0 1 idλ0 1 I11 = − 2 2 3 = − 2 2 3 ; (28) 0 π (λ0 + a ) −∞ 2π (λ0 + a ) and Z 4~ Z 1 d l 1 dr S4D(r) I1 = −i 4 = −i 4 2 2 3 ; (29) (2π) (j~lj2 + b2)3 0 (2π) (r + b ) 2 3 with b = m(α + β). Substituting S4D = 2π r for the surface of 4D{sphere of radius r, we get Z 1 3 Z 1 1 i d%% i dx x i 2x + 1 i I1 = − 2 2 2 3 = − 2 2 3 = − 2 2 (−) 2 = − 2 2 : (30) 8π b 0 (% + 1) 16π b 0 (x + 1) 16π b 2(x + 1) 0 32π b For effective anomalous magnetic vertex we then get Z 1 Z 1−α −i(α + β)(1 − (α + β)) e2 Z 1 Z 1−α 1 − (α + β) Γµ = −ie24m(pµ + p0µ) dα dβ = − (pµ + p0µ) dα dβ : eff 2 2 2 2 0 0 32π m (α + β) 8π m 0 0 α + β (31) Integral over parameters α and β gives: Z 1 Z 1−α 1 − (α + β) Z 1 Z 1 1 − γ Z 1 1 1 dα dβ = dα dγ = − dα ln α − = : (32) 0 0 α + β 0 α γ 0 2 2 µ So Γeff becomes, e2 α Γµ = − (pµ + p0µ) = − (pµ + p0µ) : (33) eff 16π2m 4πm Gordon decomposition in momentum space gives (p + p0)µ i uγ¯ µu =u ¯ + Σµν (p0 − p) u0 : (34) 2m 2m ν This means that the term infront of (pµ + p0µ) multiplied by −2m represents anomalous magnetic moment α α κ = −2m(−1) = (35) 4πm 2π in units of Bohr magneton µB. Magnetic moment of Dirac particle is e ~µ = g S;~ (36) 2m with g-factor equals to two. The perturbative QED then gives for the magnetic moment of electron in the first order of α: α µ = 1 + µ ; (37) 2π B or in terms of g-factor α g = 2 + ; (38) π or g − 2 α a = = : (39) 2 2π As α−1 = 137:036, we get a = 0:001 161 4 compared with current experimental value aexp = 0:001 159 652 180 73(28) : (40) Current theoretical prediction of electron anomalous moment includes terms up to order of α5 and gives ath = 0:001 159 652 181 78(77) : (41) 3 Appendix A - Proof of Swinger parametrization Let's start with simple two-term form of Swinger parametrization 1 Z 1 1 = dα 2 : (42) a1a2 0 (αa1 + (1 − α)a2) The proof is straightforward: Z 1 Z 1 Z a1 1 1 dα(a1 − a2) 1 dx 1 1 1 1 dα 2 = 2 = 2 = − = : (43) 0 (αa1 + (1 − α)a2) a1 − a2 0 (α(a1 − a2) + a2) a1 − a2 a2 x a1 − a2 a2 a1 a1a2 In case of three terms, it can be proofed again by a straightforward integration: Z 1 Z 1−α Z 1 Z 1−α 1 1 dβ(a2 − a3) dα dβ 3 = dα 3 = 0 0 (αa1 + βa2 + (1 − α − β)a3) a2 − a3 0 0 (α(a1 − a3) + β(a2 − a3) + a3) 1 Z 1 Z αa1+(1−α)a2 dx 1 Z 1 1 1 = dα 3 = − dα 2 − 2 = a2 − a3 0 αa1+(1−α)a3 x 2(a2 − a3) 0 (αa1 + (1 − α)a2) (αa1 + (1 − α)a3) 1 1 1 1 1 = − − = ; (44) 2(a2 − a3) a1a2 a1a3 2 a1a2a3 q.e.d.