3 One-Loop Counterterms in QED 3.1 Fermion Self-energy k µ ν p (p k) − We work in Feynman gauge. Applying the rules of QED we have (in d dimensions) d γ 2 (4 d) d k µ i( (p k) + m) ν igµν Σ(p ;m) = iµ − ( ieγ ) · − ( ieγ )− ; (3.1) Z (2π)d − (p k)2 m2 − k2 − − where we have displayed explicitly the scale dependence of the coupling outside four dimensions. Introducing Feynman parameters, we get d µ d k 1 1 γ (γ (p k) + m)γµ Σ(p2;m) = ie2µ(4 d) dα dβδ( α β) · − − π d 1 2 − Z (2 ) Z0 Z0 − − (k2(α + β) 2p k α + (p2 m2)α) − · − (3.2) Now shift k k + pα (and perform the trivial integral over β absorbing the δ-function) ! d µ d k 1 γ (γ p(1 α) + m)γµ Σ(p2;m) = ie2µ(4 d) dα · − − π d 2 (3.3) − Z (2 ) Z0 (k2 + p2(α(1 α) m2α)) − − We have omitted a term d µ d k 1 γ γ kγµ ie2µ(4 d) dα · ; − π d 2 − Z (2 ) Z0 (k2 + p2(α(1 α) m2α)) − − which vanishes by symmetry since the numerator is odd in k, the denominator is even in k and we must integrate over all directions of the vector k. µ µ Setting d = 4 2ε, using γ γ pγµ = 2(1 ε)γ p, γ γµ = 4 2ε and the result from eq.(2.6) we − · − − · − have ε e2 1 4πµ2 Σ p2 m Γ ε dα ε α γ p ε m ( ; ) = 2 ( ) [2(1 )(1 ) (4 2 ) ] 2 2 −(16π ) Z0 − − · − − m α p α(1 α) − − (3.4) 23 Expanding in ε and keeping only the terms which do not vanish as ε 0, we get ! 2 1 2 e 1 Σ(p ;m) = dα(2(1 α)γ p 4m) + ln(4π) γE −(16π2) Z − · − ε − 0 1 m2α p2α(1 α) + dα 2γ p(1 α) 2m + (2(1 α)γ p 4m)ln − − (3.5) Z · − − − · − µ2 0 Performing the integral over α except in the last term, this reduces to 2 2 e 1 Σ(p ;m) = (γ p 4m) + ln(4π) γE −(16π2) · − ε − 1 m2α p2α(1 α) + γ p 2m + dα(2(1 α)γ p 4m)ln − − (3.6) · − Z − · − µ2 0 In order to obtain the (physical) mass subtraction term, δm, and the wavefunction renormalization constant Z , we must expand this in a power series in (γ p m), making use of the relation 2 · − p2 m2 = (γ p m)(γ p + m) = 2m(γ p m) + ((γ p m)2): − · − · · − O · − This enables us to expand the logarithm about p2 = m2. This gives 2 1 2α2 2 e 1 m Σ(p ;m) = 3m + ln(4π) γE m + 2m dα(1 α)ln (16π2) ε − − Z − µ2 0 e2 1 1 m2α2 1 (1 α2) + 1 ln(4π) + γE + 2 (1 α)ln + 4 dα − (γ p m) (16π2) − ε − Z − µ2 Z α · − 0 0 + ((γ p m)2) (3.7) O · − The terms which are O((γ p m)2 and higher are finite and independent of the scale µ. They make · − 2 up the renormalized self-energy ΣR(p ;m). The last integral over α in eq.(3.7) diverges at α = 0. This is a new type of divergence caused by the fact that the photon is massless - it is called an “infrared divergence”. For the moment we regularize this infrared divergence by assigning a small mass, λ to the photon wherever necessary (i.e. we only keep terms in λ which are not regular as λ 0. When we do this the last integral in eq.(3.7) becomes ! 1 α(1 α2) 1 m2 dα λ2 2 − 2 2 = ln( 2 1 + O( ): Z0 α (1 α)λ =m 2 λ − − − Writing (to this order in perturbation theory), 2 2 Σ(p ;m) = δm + (Z 1)(γ p m) + ΣR(p ;m): 2 − · − we have for the mass renormalization (introducing the fine-structure constant α = e2=(4π)), α 1 1 m2α2 δm = m 3 + ln(4π) γE 1 2 dα(1 α)ln 4π ε − − − Z − µ2 0 α 1 4 m2 = 3m + ln(4π) γE + + ln ; (3.8) 4π ε − 3 µ2 24 and for the wavefunction renormalization constant, α 1 1 m2α2 1 α(1 α2)) Z π γ α dα 2 = 1 + 1 ln(4 ) + E + 2 (1 )ln 2 + 4 2 − 2 2 4π − ε − Z0 − µ Z0 α + (1 α)λ =m − α 1 µ2 m2 = 1 + ln + γE ln(4π) 4 + 2ln : (3.9) 4π − ε − m2 − − λ2 3.2 Photon Self-energy (Vacuum polarization) (k q) − ν q µ k ν The photon self-energy Πµ (q2) is, in general, a two-rank tensor, which is formed from the four- momentum of the photon , qµ and the (invariant) metric tensor. It must therefore have the form ν ν ν Πµ (q2) = A(q2)gµ + B(q2)qµq : ν On the other hand Πµ (q2) obeys a Ward identity µν 2 qµΠ (q ) = 0: This can be seen by writing 1 µ 1 1 1 q γµ = (γ k m) (γ (k q) m) (γ (k q) m) − (γ k m) · − · − − · − − · − u ν qµ µ = - = 0 u 25 Applying this to the one-loop graph representing the photon self-energy, we get the difference between two graphs in which one of the two internal fermion propagates has been killed. But these two graphs are identical and so the difference is zero. We may therefore write ν ν ν Πµ (q2) = gµ q2 + qµq Π(q2) (3.10) − In other words only the transverse part of thephoton propagator acquires a higher order correction. The photon has no mass and therefore no mass renormalization. There is only a photon wavefunc- tion renormalization constant Z3. 2 1 2 Π(q ) = (Z3 1) + ΠR(q ) ; (3.11) Z3 − 2 where ΠR(q ) is the renormalized (finite) part of the self-energy. At the one-loop level the prefactor 1=Z3 in eq.(3.11) may be set to unity. The fact that only the transverse part of the photon-propagator acquires a higher-order correction means that the gauge parameter, ξ is renormalized. If we write the leading order propagator as ν ν µν qµq ξ qµq g q2 ( 1) q2 i − − − − h q2 i The renormalized propagator is ν qµqν µ µ ν Z3 g q2 q q i − i ξ 2 2 + ( 1) 4 : − q (1 ΠR(q )) − q − The transverse part of the propagator is renormalized but not the longitudinal part. Near q2 = 0, the renormalized propagator looks like ν ν µν qµq ξ qµq Z3 g q2 ( R 1) q2 i − − − ; − h q2 i where (ξ 1) (ξR 1) = − : − Z3 Now returning to the one-loop graph and inserting the Feynman rules, we get d γ γ µν 2 4 d d k µ i( (k q) + m) ν i( k + m) Π (q ) = iµ − Tr ( ieγ ) · − ( ieγ ) · : (3.12) − Z (2π)d − ((k q)2 m2) − k2 m2) − − − An overall minus sign has been introduced for a loop of fermions. This arises form the fact that the Wick contraction required to construct the Feynman graph requires an interchange of two fermion fields, thereby introducing a minus sign. 26 Feynman parametrization gives d µ ν ν d k 1 Tr[γ (γ (k q) + m)γ (γ k + m)] Πµ (q2) = ie2µ4 d dαdβδ( α β) : − π d 1 · − · 2 (3.13) − Z (2 ) Z0 − − (k2 2k qα + q2α m2) − · − Performing the trace (and integrating over β) gives d µν 2 µ ν µ ν µ ν ν d k 1 g k (k q) m 2k k + k q + q k Πµ (q2) = ie2µ4 d dα · − − − : 4 − π d 2 (3.14) Z (2 ) Z0 (k2 2k qα + q2α m2) − · − Shifting kµ kµ + qµα we get ! d 1 µν 2 α α 2 2 µ ν α α µ ν µν 2 2 4 d d k g k (1 )q m 2k k + 2 (1 )q q Π (q ) = 4ie µ − dα − − − − − ; Z π d Z 2 2α α 2 2 (2 ) 0 (k + q (1 ) m ) − − (3.15) where once again we have omitted terms linear in k, which vanish by symmetric integration. From eq.(3.10) it is sufficient to extract only the terms in the above integral which are proportional ν to gµ . Using eqs.(2.6), (2.8) and (2.9) we have (setting d = 4 2ε) − ε e2 1 4πµ2 1 q2Π q2 dα Γ ε ε q2α α m2 ( ) = 2 2 2 ( 1) (4 2 ) 1 (1 ) − −4π Z0 m q α(1 α) − 2 − − − − − − Γ(ε) q2α(1 α) + m2 (3.16) − − Using Γ(ε) Γ(ε 1) = ; − −(1 ε) − it can be seen that the RHS of eq.(3.16) becomes proportional to q2, so we have ε e2 1 4πµ2 Π q2 Γ ε dαα α ( ) = 2 ( ) (1 ) 2 2 (3.17) −2π Z0 − m q α(1 α) − − Expanding in ε up to terms which vanish as ε 0, and performing the integral over α where ! appropriate, this gives 2 1 2 2α α 2 e 1 m q (1 ) Π(q ) = + ln(4π) γE 6 dαα(1 α)ln − − (3.18) −12π2 ε − − Z − µ2 0 We define Z3 to be 1 + Π(0), so that we have (in terms of the fine-structure constant, α) α 1 m2 Z = 1 + ln(4π) γE ln ; (3.19) 3 − 3π ε − − µ2 and the renormalized photon self energy 2 2 ΠR(q ) = Π(q ) (Z 1); − 3 − 27 is proportional to q2 so that it vanishes as the photon goes on mass-shell.