Solution set #7 Physics 571 Tuesday 3/17/2014. p 1. p 2. Figure 1: Muon production (e + e µ + µ ) γ ν /p 2

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1 Solution set #7 Physics 571 Tuesday 3/17/2014 μ 1. The amplitude is Figure 1: Muon production ( e µ + µ ) it = ie2 s (v 2γ µ u 1 )(u 1 γ µ v 2 ), (1) so the spin averaged squared amplitude is T 2 = e4 4s 2 (v 2 γ µ u 1 )(u 1 γ µ v 2 )(v 2 γ ν u 1 )(u 1 γ ν v 2 ) s 1,s 2,s 1,s 2 = e4 4s 2 Tr γ µ /p 1 γ ν /p 2 Tr γ ν/p 1 γ µ/p 2 = 4 e4 s 2 pµ 1 pν 2 + p ν 1p µ 2 gµν p 1 p 2 p 1 µp 2 ν + p 1 νp 2 µ g µν p 1 p 2 = 8 e4 s 2 (p 1 p 1 )(p 2 p 2 ) + (p 1 p 2 )(p 2 p 1 ) = 2e 4 (t2 + u 2 ) s 2 (2) μ μ θ Figure 2: Definition of the scattering angle θ. In the CM frame we have so we obtain p 1 = E(1, 0, 0, 1), p 2 = E(1, 0, 0, 1), p 1 = E(1, ˆn), p 2 = E(1, ˆn), (3) ˆn = (sin θ cos φ, sin θ sin φ, cos θ), s = 4E 2, (4) T 2 = e 4 (1 + cos 2 θ). (5) 1

2 The differential crosssection is then given by dσ dω = 1 64π 2 s p 1 p 1 T 2 = α2 4s (1 + cos2 θ). (6) Because our amplitude is azimuthally symmetric, we can use dω = 2πd cos θ to arrive at dσ d cos θ = α2 μπ 2s (1 + cos2 θ). μ (7) Integrating 1 d cos θ gives the total cross section 1 σ μ = + 4α2 π 3s 2. In this case two graphs contribute, with opposite signs: θ (8) Figure 3: Bhabha scattering ( e e ); note the relative minus sign. and the amplitude is given by it = ie2 s (v 2γ µ u 1 )(u 1 γ µ v 2 ) ie2 t (u 1 γµ u 1 )(v 2 γ µ v 2 ), (9) The spin averaged cross section is then given by where T 2 = 1 4 as in the previous problem, while spins T 2 = e 4 Φss s 2 Φ st + Φ ts st + Φ tt t 2, (10) Φ ss = 2(t 2 + u 2 ), (11) Φ tt = 1 (u 1 γ µ u 1 )(u 1 γ ν u 1 )(v 2 γ µ v 2 )(v 2 γ ν v 2 ) 4 spins = 1 γ 4 Tr µ /p 1 γ ν /p 1 Tr γ µ/p 2 γ ν/p 2 = 4 p µ 1 pν 1 + pν 1p µ 1 gµν p 1 p 1 p 2µ p 2 ν + p 2ν p 2 µ g µν p 2 p 2 = 8 (p 1 p 1 )(p 2 p 2 ) + (p 1 p 2 )(p 2 p 1 ) = 2(s 2 + u 2 ). (12) and Φ st = 1 (v 2 γ µ u 1 )(u 1 γ µ v 2 )(u 1 γ ν u 1 )(v 2 γ ν v 2 ) 4 spins 2

3 Then since Φ ts = Φ st we also have = 1 4 Tr /p 1 γ ν /p 1 γ µ/p 2 γ ν/p 2 γ µ = 1 2 Tr /p 1 /p 2 γ µ/p 1 /p 2 γ µ = 8(p 1 p 2 )(p 1 p 2 ) = 2u 2. (13) Φ ts = 2u 2. (14) 3. (a) Writing the Lagrangian in terms of Σ(x) = e iθ(x)/ 2f we get L = 1 4 F µνf µν f 2 (D µ Σ) (D µ Σ), D µ Σ = ( µ + iea µ )Σ. (15) Under the gauge transformation A µ A µ + µ Λ we have D µ UD µ U where U = exp( ieλ). Therefore the theory is gauge invariant if we simultaneously transform Σ UΣ, or θ θ ie 2fΛ. Note that this is a shift symmetry as far as θ is concerned, characteristic of a Goldstone boson field. Note the weird feature that if we take U = Σ = Σ 1, then we can remove the Σ field from the Lagrangian completely! This is called unitary gauge ; it makes the theory look like a simple theory of a massive photon, but if you start doing loops in the nonabelian generalization you get massively confused with infinities everywhere (which are gauge artifacts). (b) Gauge symmetry implies that the kinetic operator has a zero eigenvalue, and therefore is not invertible. There are many ways to show this. One way is to treat the coupling of the A µ to θ as an interaction, and then sum it up to all orders in perturbation theory. We write the Lagrangian as where L = 1 2 A µk µν A ν θ 2 θ 2 efa µ µ θ, (16) K µν = ( ( 2 M 2 )g µν µ ν) = ( (k 2 + M 2 )g µν k µ k ν). (17) There is no problem defining the photon propagator ik 1, because of the mass term. It is easy to check that Ag µν + Bk µ k ν 1 = 1 B g µν A A + Bk 2 k µk ν, (18) so that with A = (k 2 + M 2 ) and B = 1 we have K 1 αβ = 1 k 2 + M 2 g αβ + k αk β M 2 (19) and the theta propagator is just i/k 2, while the vertex connecting the θ to A µ is i 2efk µ. Summing the diagrams Figure 4: Diagrams giving the full photon propagator. 3

4 we get the formal expression for the photon propagator ik 1 n=0 AK 1 n, Aµν = 2e2 f 2 k µ k ν k 2 = M 2 k µ k ν k 2. (20) Formally we perform the geometric sum to get ik AK 1 = i(k A) 1. (21) But (K A) µν = (k 2 + M 2 ) (g µν kµ k ν ) k 2 (22) which has no inverse, since (K A) µν k ν = 0. So we see that these diagrams do not converge; we do not have a sensible photon propagator. (c) Adding the gauge fixing term modifies K: ( K µν = (( 2 M 2 )g µν µ ν (1 1 )) ξ with inverse (using formula in eq. (18)) Kµν ξ = k 2 + M 2 g µν k µ k ν k 2 + ξm 2 ( = ((k 2 + M 2 )g µν k µ k ν (1 1 )) ξ ξ 0 1 k 2 + M 2 g µν k µk ν k 2, (23). (24) We see that Landau gauge is nice because K 1 is proportional to the transverse projection operator (it is annihilated by k ν ). Since it is annihilated by k µ, and the interaction with the θ field in the diagrams in Fig. 4 is proportional to k ν, in fact the above expresison in eq. (24) is the exact photon propagator. Note that it has a pole at k 2 = M 2, so we really do have a gauge invariant theory with a massive photon! All we need is spontaneous symmetry breaking to give us the θ field (the Higgs Mechanism). We say that the gauge boson ate θ, which now reappears as the longitudinal component of the massive gauge field. (Recall: we can get in the rest frame of a massive gauge field, and therefore the fact that it has spin 1 means that there must be three physical polarizations instead of just two). (d) We need a θ field for every gauge boson that gets a mass; if we give mass to all of the gauge bosons, we need one θ for every generator in the gauge group. A generalization of Σ then is naturally of the form Σ = e iθata/fa. (25) (The 2 goes away becaus we normalize Tr T a T b = 1 2 ). Then we can take where L =... f 2 Tr (D µ Σ) (D µ Σ) (26) D µ Σ = µ Σ iga a µt a Σ. (27) Note that unlike the Abelian case, this operator has an infinite number of terms as you expand Σ in powers of the θ a fields...it is inherently nonrenormalizable. The Higgs mechanism fixes this. 4

5 4. (a) With δa µ = µ δλ, we have G Λ = 2 + 2βA µ µ (28) and so the determinant of this object is reproduced by introducing the ghost fields c, c with path integral Dc Dc e is gh, L gh = c ( 2 + 2βA µ µ) c. (29) (b) As mentioned, this problem does not use L gh, but only L gf = 1 2ξ ( µa µ + βa µ A µ ) = 1 2ξ ( µa µ ) 2 β ξ µa µ A ν A ν β 2ξ A µa µ A ν A ν. (30) This theory has both a+ 3photon vertex and a 4photon + vertex with the following Feynman rules: +... β, 2β ξ (g αβ p γ + permutations) β, β, i 4β2 ξ (g αβg γδ + g αγ g βδ + g αδ g βγ ) 2β ξ (g αβ p γ + permutations) Figure 5: Feynman rules β, β,for photon vertices. It is important to get the symmetry factors right. Momenta β, β, β, are all taken to be flowing in toward the 3photon vertex. i 4β2 ξ (g αβg γδ + g αγ g βδ + g αδ g βγ ) The graphs we have to sum at tree level for γγ γγ are these: + + β, β, (I) β, β, (II) (III) (IV) (I) (II) (III) (IV) Figure 6: Graphs contributing to γγ γγ at tree level. Recall that for any external line p ɛ(p) = 0, where ɛ(p) is the polarization vector. Therefore we only need to include the part of the 3photon vertex where p acts on the internal propagator. That gives us a factor of ( 2β ξ ) 2 ( q µ q ν ) ( ) i (g µν q 2 qµ q ν ) (1 ξ) = i 4β2 q2 ξ. (31) for each of the graphs (IIII), where q is flowing through the propagator. (We have the minus sign in q µ q ν vertex factors because if q flows through the propagator, at one end it flows into 5

6 the vertex and at the other it flows out.). It follows then that the graphs give: (I + II + III) = i 4β2 ξ (g αβg γδ + g αδ g βγ + g αγ g βδ ) = (IV ) (32)...so the four graphs sum to zero and we have no treelevel photonphoton scattering. To prove a similar result at one loop requires using the ghosts, which I figured was too much work for this assignment. 6