Lecture 18 Quantum Electrodynamics

Lecture 18 Quantum Electrodynamics Quantum Electrodynamics (QED) describes a U(1) gauge theory coupled to a fermion, e.g. the electron, muon or any other charged particle. As we saw previously the lagrangian for such theory is 1 L = ¯(iD6 − m) − F F µν ; (18.1) 4 µν where the covariant derivative is defined as Dµ (x) = (@µ + ieAµ(x)) (x) ; (18.2) resulting in 1 L = ¯(i6@ − m) − F F µν − eA ¯γµ ; (18.3) 4 µν µ where the last term is the interaction lagrangian and thus e is the photon-fermion coupling. We would like to derive the Feynman rules for this theory, and then compute the amplitudes and cross sections for some interesting processes. 18.1 Feynman Rules for QED In addition to the rules for the photon and fermion propagator, we now need to derive the Feynman rule for the photon-fermion vertex. One way of doing this is to use the generating functional in presence of linearly coupled external sources Z ¯ R 4 µ ¯ ¯ iS[ ; ;Aµ]+i d xfJµA +¯η + ηg Z[η; η;¯ Jµ] = N D D DAµ e : (18.4) 1 2 LECTURE 18. QUANTUM ELECTRODYNAMICS From it we can derive a three-point function with an external photon, a fermion and an antifermion. This is given by 3 3 (3) (−i) δ Z[η; η;¯ Jµ] G (x1; x2; x3) = ; (18.5) Z[0; 0; 0] δJ (x )δη(x )δη¯(x ) µ 1 2 3 Jµ,η;η¯=0 which, when we expand in the interaction term, results in 1 Z ¯ G(3)(x ; x ; x ) = D ¯ D DA eiS0[ ; ;Aµ]A (x ) ¯(x ) (x ) 1 2 3 Z[0; 0; 0] µ µ 1 2 3 Z 4 ¯ ν × 1 − ie d yAν(y) (y)γ (y) + ::: Z 4 ν = (+ie) d yDF µν(x1 − y)Tr [SF(x3 − y)γ SF(x2 − y)] ; (18.6) where S0 is the free action, obtained from the first two terms in (18.3). To obtain the second equality we used Wick's theorem while carefully keeping the spinor indices. Finally, to obtain the vertex rule from (18.6) we assume that we are computing an amplitude for a process involving a photon, a fermion and an antifermion. Using the LSZ formalism 1 on the three-point function in (18.6) we obtain the amplitude in momentum space as ¯ 4 (4) µ∗ hf(p2)f(p3)jA(p1)i = (−ie)(2π) δ (P1 − P2 − P3)u ¯(p3)γµv(p2) (p1) ; (18.7) where the spinor, antispinor and photon polarization appear as a consequence of the application of the LSZ formula. Since this always implies the same Feynman rules for external spinors, antispinors and polarizations, we conclude that the vertex itself will add a factor of (−ie)γµ, where the Lorentz index in the gamma matrix should be contracted with that of the external photon polarization. This is illustrated in Figure 18.1. The complete Feynman rules for QED are then obtained by having • A factor of µ∗(k) for each incoming photon of momentum k. • A factor of µ(k) for each outgoing photon of momentum k. • A factor of i ; (18.8) 6p − m for each internal fermion line. 1Just as we did for the scalar-fermion theory in Lecture 14, and being just as careful with the signs coming from the i's and the action of the Klein-Gordon and Dirac differential operators on fields. 18.2. E+E− ! µ+µ− IN QED 3 Figure 18.1: Feynman rule for the QED vertex. • A factor of −ig µν ; (18.9) q2 + i for each internal photon line (in the Feynman gauge). Finally, the rules regarding spinors and antispinors for incoming and outgoing fermions and antifermions are the same as for those in Yukawa theory. We are now in a position to compute QED amplitudes. We will study a couple of simple examples in what follows. 18.2 e+e− ! µ+µ− in QED We start with a simple process in QED: electron{positron annihilation into muon pairs. As we will see, its simplicity stems from the fact that there is only one diagram contributing. The relevant lagrangian is 1 L =e ¯(iD6 − m )e +µ ¯(iD6 − m )µ − F F αβ ; (18.10) e µ 4 αβ where it is understood that both covariant derivatives are one and the same Dα = (@α + ieAα) ; (18.11) for = e; µ. Here e(x) and µ(x) are the electron and muon fields, respectively. The lagrangian (18.10) implies that the gauge interactions are universal, i.e. the electron and muon interact with the photon with the same coupling2 e. Another consequence of the form of (18.10) is that electron and muon numbers are separately conserved. This means that there are two separate global U(1)'s, one for electron and one for muons, which forbids terms such as 2Do no confuse with the electron field e(x). 4 LECTURE 18. QUANTUM ELECTRODYNAMICS ei¯ Dµ6 ; (18.12) violating the conservation of these charges. Taking this into account we conclude that the only Feynman diagram contributing to e−e+ ! µ−µ+ is the one shown in Figure 18.2. Figure 18.2: Feynman diagram for e−e+ ! µ−µ+ in QED. Using the Feynman rules for QED from the previous section we can now write the amplitude for this process. It is 0 0 −ig iA = (−ie)2 u¯s (p0)γµ vr (k0) µν v¯r(k)γν us(p) ; (18.13) a ab b q2 + i c cd d where we have shown explicitly all spinor indices and we assume the repeated ones are summed over, and we are still using the Feynman gauge for the photon propagator. To compute the cross section we will need the square of the amplitude in (18.13). For this we need to be careful about complex conjugation of the whole amplitude. We will need to compute the adjoint of the two terms to the sides of the photon propagator separately, since their spinor indices are separately contracted. That is, we need y (¯uγµv)y = uyγ0γµv = vy(γµ)y(γ0)yu : (18.14) But remembering that (γ0)y = γ0, and noticing that 8 µ < γ if µ = 0 (γ)y = ; (18.15) : −γµ if µ = 1; 2; 3 we obtain that (¯uγµv)y = vyγ0γµu =vγ ¯ µu : (18.16) 18.2. E+E− ! µ+µ− IN QED 5 Similarly, (¯vγµu)y = (¯uγµv) : (18.17) Then, the complex conjugate of the amplitude (18.13) is 0 0 ig −iAy = (−ie)2 v¯r (k0)γα us (p0) αβ u¯s(p)γβ vr (k) : (18.18) e ef f q2 + i g gh h We can compute the amplitude squared by keeping all the spinor indices explicitly. We obtain 4 e 0 0 0 0 jAj2 = us (p0)¯us (p0) vr (k0)¯vr (k0) γµ γα us(p)¯us(p) vr (k)¯vr(k)(γ ) (γ ) : q4 f a b e ab ef d g h c µ cd α gh (18.19) If we now assume that all spin indices (s; r; s0r0) are summed over, both for the initial and for the final states, we can use the relations X X us(p)¯us(p) =6p + m ; vs(p)¯vs(p) =6p − m : (18.20) s s Using them in (18.19) we obtain e4 jAj2 = (6p0 + m ) γµ (6k0 − m ) γα (6p + m ) (γ ) (6k − m ) (γ ) ; (18.21) q4 µ fa ab µ be ef e dg α gh e hc µ cd where all the spinor indices are summed. We notice that we can write the amplitude squared as the product of two traces over these indices as in e4 jAj2 = Tr [(6p0 + m )γµ(6k0 − m )γα] × Tr [(6p + m )γ (6k − m )γ ] ; (18.22) q4 µ µ e α e µ In order to continue we need to compute the traces of the product of up to four gamma matrices. Let us start with the second factor in (18.22) 6 LECTURE 18. QUANTUM ELECTRODYNAMICS 2 Tr [(6p + me)γα(6k − me)γµ] = Tr [6pγα6kγµ] − meTr [γαγµ] σ ρ 2 = p k Tr [γσγαγργµ] − 4megµα σ ρ 2 = 4p k fgσαgρµ − gµαgρσ + gµσgραg − 4megµα 2 = 4 pαkµ − gµαp · k + pµkα − megµα : (18.23) In the first equality in (18.23) we use the fact that the trace over an odd number of gamma matrices vanishes. The rest of the computations simply makes repeated use of the Clifford algebra of gamma matrices fγµ; γνg = 2gµν ; (18.24) from which we can derive (exercise) Tr [γσγαγργµ] = 4 fgσαgρµ − gσρgαµ + gσµgαρg : (18.25) It is straightforward to obtain a similar result for the first factor in (18.22). The amplitude squared is e4 jAj2 = 16 p0αk0µ − gµαp0 · k0 + p0µk0α − m2 gµα p k − g p · k + p k − m2g : q4 µ α µ µα µ α e µα (18.26) In order to simplify the result we take the limit me = 0. This is typically appropriate since mµ me, and all energies and momenta must be at least of the order of the muon mass for the process to occur. In this limit we have e4 jAj2 = 32 (p · p0)(k · k0) + (p0 · k)(p · k0) + m2 p · k : (18.27) q4 µ Now that we have the amplitude squared in terms of Lorentz-invariant products of the four-momenta, we can either go to a particular frame or express these products in terms of Mandelstam variables. Center of Momentum Frame: Assuming me = 0, the four-momenta can be written in the CM frame as 18.2.

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