Physics 217 FINAL EXAM SOLUTIONS Fall u(p,λ) by any method of your choosing.

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1 Physics 27 FINAL EXAM SOLUTIONS Fall 206. The helicity spinor u(p, λ satisfies u(p,λu(p,λ = 2m. ( In parts (a and (b, you may assume that m 0. (a Evaluate u(p,λ by any method of your choosing. Using the Dirac equation for the u-spinor, /pu(p,λ = mu(p,λ, u(p,λ/p = mu(p,λ. where the second version of the Dirac equation follows from the first version by hermitian conjugation followed by γ 0 γ µ γ0 = γ µ. Assuming that m 0, it trivially follows that Consequently Thus, we can write, u(p,λ = /p m u(p,λ, u(p,λ = u(p,λ /p m. u(p,λ = u(p,λ /p m γ 5 u(p,λ = u(p,λ = 2 u(p,λ ( /p m γ 5 +γ 5 /p m u(p,λ. /p u(p,λ = 0, (2 m after noting that /pγ 5 +γ 5 /p = p µ { γ µ, γ 5 } = 0. (b Evaluate u(p,λγ µ u(p,λ. Using the same strategy as in part (a, u(p,λγ µ u(p,λ = 2 u(p,λ ( /p m γµ +γ µ /p m after noting that and making use of eq. (. /pγ µ +γ µ /p = p ν { γ ν, γ µ} = 2g µν p ν = p µ, u(p,λ = pµ m u(p,λu(p,λ = 2pµ, (

2 (c Evaluate u(p,λγ µ γ 5 u(p,λ, assuming the the fermion mass m = 0. Recall eq. (20 from Solution Set, We then immediately get after making use of eq. (. γ 5 u(p,λ = 2λu(p,λ. u(p,λγ µ γ 5 u(p,λ = 2λu(p,λγ µ u(p,λ = 4λp µ, (4 (d Evaluate u(p,λγ µ γ 5 u(p,λ, assuming that the fermion mass, m 0. Express your result in terms of the spin vector s µ. How would you use this result in the case of m = 0 to reproduce the answer obtained in part (c? The method used in parts (a and (b will not succeed here. Instead, we modify the method as follows. First recall eq. (4 from Solution Set, γ 5 s/u(p,λ = 2λu(p,λ. Noting that λ = ± 2 so that (2λ2 =, it then follows that u(p,λ = 2λγ 5 s/u(p,λ, u(p,λ = 2λ s/, where the second result above is obtained from the first by hermitian conjugation followed by γ 0 γ µ γ0 = γ µ and γ 0 γ 5 γ0 = γ 5 (the latter is easily obtained from the former by using the definition of γ 5 = iγ 0 γ γ 2 γ and the anticommuting properties of the gamma matrices. In the final step, { γ µ, γ 5 } = 0 is used. Following a similar strategy to the one used in parts (a and (b, u(p,λγ µ γ 5 u(p,λ = 2λ s/γ µ γ 5 u(p,λ = 2λu(p,λγ µ γ 5 γ 5 s/u(p = 2λu(p,λs/γ µ u(p,λ = 2λu(p,λγ µ s/u(p, after using { γ µ, γ 5 } = 0 and (γ5 2 = I to write γ 5 s/γ µ γ 5 = s/γ µ. Hence, it follows that u(p,λγ µ γ 5 u(p,λ = λu(p,λ(γ µ s/+s/γ µ u(p,λ = 2λs µ u(p,λu(p,λ = 4λms µ, (5 after noting that s/γ µ +γ µ s/ = s ν { γ ν, γ µ} = 2g µν s ν = s µ, and making use of eq. (. If m = 0, then the spin vector s µ does not exist. Nevertheless, we can use eq. (9 from Solution Set to write ( s µ = pµ m m +O. E In particular, this method yields u(p,λγ µ γ 5 u(p,λ = i(p ν /mu(p,λσ νµ γ 5 u(p,λ, which simply relates two different spinor product expressions. In the latter expression, Σ νµ 2 iγµ, γ ν. 2

3 Taking the m 0 limit is equivalent to lim m 0 ms µ = p µ. Inserting this result into eq. (5 then reproduces eq. (4. ALTERNATIVE SOLUTION TO PROBLEM First, we consider the case where the mass m 0. In this case, shall make use of eq. (6 of Solution Set, u α (p,λū β (p,λ = 2 (+2λγ5 s/(/p+m αβ, (6 where the four-component spinor indices have been made explicit. Let us denote an arbitrary 4 4 matrix by Γ and consider the quantity, u(p,λγu(p,λ. Using eq. (6 u β (p,λγ βα u α (p,λ = Γ βα u α (p,λu β (p,λ = 2 Γ βα(+2λγ 5 s/(/p+m αβ, where there is an implicit sum over pairs of repeated indices. Hence, it follows that u β (p,λγ βα u α (p,λ = 2 Tr (+2λγ 5 s/(/p+mγ. (7 This is the master formula from which all results of Problem (for the case of m 0 can be derived. First, we derive u(p,λ = 2 Tr (+2λγ 5 s/(/p+mγ 5 = 2 Tr γ 5 (+2λγ 5 s/(/p+m = 2 Tr (γ 5 +2λs/(/p+m = λtr(s//p = 4λs p = 0, after using the fact that the trace of a product of matrices is invariant under the cyclic permutation of the matrices. At the final step, we used s p = 0 (which is an important property of the spin four-vector s µ as discussed in Problem of Solution Set. We have thus successfully rederived eq. (2. Similarly, we can derive, u(p,λ = 2 Tr (+2λγ 5 s/(/p+mγ µ = 2 Tr γ µ (+2λγ 5 s/(/p+m = 2 Tr(γµ /p = 2p µ. u(p,λγ µ γ 5 u(p,λ = 2 Tr (+2λγ 5 s/(/p+mγ µ γ 5 = 2 Tr γ µ γ 5 (+2λγ 5 s/(/p+m = 2 Tr (γ µ γ 5 +2λγ µ s/(/p+m = λmtr(γ µ s/ = 4λms µ, (8 which reproduces eqs. ( and (5. The method presented above relies onthe assumption that m 0. In the case of m = 0, the spin four-vector s µ does not exist. Nevertheless, the above method is easily modified in the case of m = 0. The starting point is ( u α (p,λu β (p,λ = +2λγ5 2 /p (9 αβ

4 which is derived in part (d of problem in Solution Set. We have again made the four-component spinor labels explicit. Thus, in the case of m = 0, the new master formula that supplants eq. (7 is u β (p,λγ βα u α (p,λ = 2 Tr( +2λγ 5 /pγ. (0 We can now repeat the three derivations presented above, but now in the case of m = 0. We then obtain, u(p,λ = 2 Tr( +2λγ 5 /pγ5 = 0. u(p,λγ µ u(p,λ = 2 Tr( +2λγ 5 /pγ µ = 2 Tr(/pγµ = 2p µ. Note that there results are identical to the corresponding results in the case of m 0. This is quite interesting, since strictly speaking the previous derivations relied on the assumption that m 0. Finally, one can derive u(p,λγ µ γ 5 u(p,λ = 2 Tr( +2λγ 5 /pγ µ γ 5 2 Tr γ µ γ 5 ( +2λγ5 /p which reproduces eq. (4. = 2 Tr (γ µ γ 5 +2λγ µ /p = λtr(γ µ /p = 4λp µ, 2. Quantum electrodynamics (QED is a theory of spin-/2 electrons and positrons interacting with photons. Suppose we add to QED a real massive pseudoscalar field Φ that couples to electrons and positrons via L int = igψ(xγ 5 Ψ(xΦ(x ( (a Which, if any, of the discrete symmetries P, C and T are violated if the interaction in eq. ( is added to the QED Lagrangian? You may assume that the only scalar selfinteraction term allowed is Φ 4 (x, although this has no bearing on the rest of the problem. In class, we showed that the bilinear covariant Ψγ 5 Ψ(x is odd under P and T and even under C (in the latter case, we assume that the interaction Lagrangian is normal ordered. However, we are free to assign transformation properties to the scalar field such that Φ is also odd under P and T and even under C. With respect to such an assignment, the contribution of eq. ( to the action is invariant under P, T and C transformations. Note that the scalar kinetic energy term is quadratic in the fields and the scalar selfinteraction is quartic in the fields, so these terms are automatically P, T and C conserving. We conclude that there is a consistent choice for the P, T and C transformation properties of the scalar and fermion fields governed by eq. ( and the interactions of QED such that P, T, and C are conserved. 4

5 (b Draw all distinct Feynman diagrams that contribute to the second-order scattering process γe e Φ. There are two distinct Feynman diagrams for γe e Φ, shown below. γ k p 2 p 2 k e γ k k+p p 2 e e p q Φ e + p q Φ Each line is labeled with a four-momentum. The four-momenta of all fermion lines point in the direction of the arrow. The incoming photon has four-momentum k and the outgoing pseudoscalar has four-momentum q. (c What is the Feynman rule for the Φe + e vertex? Write down the invariant matrix element corresponding to the Feynman diagrams of part (b. Do not assume that the mass of the electron is zero. The Feynman rule is obtained from il int by stripping off the fields and including an appropriate symmetry factor if there are identical particles at the interaction vertex. In the case of eq. (, Ψ and Ψ are treated as being distinct, so there is no extra symmetry factor associated with the Φe + e vertex. The two Feynman rules for the interaction vertices appearing in the Feynman diagrams of part (b are shown below. γ µ p e ieγ µ p e + p e Φ gγ 5 p e + We label the four-momenta as indicated in part (b. Employing the Feynman rules, the invariant matrix element is given by the sum of two terms corresponding to the two 5

6 Feynman diagrams shown in part (b, 2 im = u(p 2,λ 2 ieγ µi(/p 2 /k +m e i(/k +/p +m e gγ t m 2 5 +gγ 5 ieγ u(p µ e s m 2,λ ǫ µ (k,λ, (2 e where we have introduced the Mandelstam invariants, s = (p +k 2 = m 2 e +2p k, t = (k p 2 2 = m 2 e 2k p 2, ( after using k 2 = 0 (corresponding to the massless photon. Note that the relative sign between the two terms in eq. (2 is positive. This sign corresponds to the fact that the order of the spinors that appears after applying the Feynman rules to the two diagrams of part (b is the same for both diagrams (and therefore one is an even permutation of the other. (d Your answer in part (c has the form M = M µ ǫ µ (k,λ, where k is the photon fourmomentum and λ is the photon helicity. Check your answer by verifying that k µ M µ = 0. Using eq. (2, γµ (/p 2 /k +m e γ M µ = iegu(p 2,λ 2 5 t m 2 e + γ 5 (/k +/p +m e γ µ u(p s m 2,λ. e It follows that k µ /k(/p2 /k +m e γ M µ = iegu(p 2,λ 2 5 t m 2 e + γ 5 (/k +/p +m e /k u(p s m 2,λ. e Noting that /k/k = k 2 = 0, and using {γ µ, γ ν } = 2g µν to obtain /k/p 2 = 2k p 2 /p 2 /k, /p /k = 2p k /k/p, it follows that k µ M µ = iegu(p 2,λ 2 2k p2 (/p 2 m e /k γ 5 t m 2 e 2p k /k(/p m e + γ 5 u(p s m 2,λ. e This can now be simplified by employing the Dirac equation, (/p m e u(p,λ = 0, u(p 2 (/p 2 m e = 0. Hence, we obtain, k µ 2k p2 M µ = iegu(p 2,λ 2 γ 5 u(p,λ t m 2 e + 2p k. (4 s m 2 e 2 Remember that the order of the spinors and gamma matrices in eq. (2 are dictated by the rule: traverse a Feynman graph along the fermion lines in the direction opposite to the direction of the arrows. 6

7 Finally, eq. ( implies that 2k p 2 = m 2 e t and 2p k = s m 2 e. Inserting these results back into eq. (4, we see that the expression in brackets above cancels exactly. That is, k µ M µ = 0. This result is a consequence of electromagnetic gauge invariance.. The Z boson is a massive spin particle (of mass m Z. Suppose there exists a light spin-0 particle φ of mass m φ. Then one possible decay mode of the Z boson is Z φγ. Denote the four-momenta of the Z boson, the spin-0 particle φ and the photon (γ by p, q and k, respectively. An explicit computation of the decay matrix element yields the following result for the invariant matrix element for Z φγ, M sλ = A(g µν p k k µ p ν ǫ µ (p,sǫ ν (k,λ, (5 whereǫ µ (p,sisthepolarizationvectorofthez bosonandǫ µ (k,λisthepolarizationvector of the photon. The Z boson and photonhelicities are denoted by s and λ, respectively. The coefficient A, which has units of inverse mass and is a function of m φ, m Z and couplings that govern the decay, will not be specified further. (a Square the invariant matrix element, sum over the final state photon helicities and average over the initial state Z boson helicities. At the end of your calculation, any dot product of four-vectors should be re-expressed in terms of m Z and m φ. Squaring eq. (5, summing over the final state photon helicities and averaging over the initial state Z boson helicities yields M sλ 2 = A 2 (g µν p k k µ p ν (g αβ p k k α p β ǫ µ (p,sǫ α (p,s ǫ ν (k,λ ǫ β (k,λ. s=,0, λ=± (6 The polarization sum for a massive spin- boson is given by eq. (60 of Solution Set 4, s=,0, ǫ µ (p,sǫ α (p,s = g µα + p µp α. (7 For the photon polarization sum, we are instructed to make the substitution, ǫ β (k,λǫ ν (k,λ g βν, (8 λ=± which will significantly reduce the algebra involved in completing this calculation. Prior to the discovery of the Higgs boson, one of the Higgs boson (H search techniques employed at the LEP collider in the 990s was a search for Z Hγ. No evidence was ever found. We now know that the Higgs boson is heavier than the Z boson, so this decay is not permitted. 7

8 Substituting the results of eqs. (7 and (8 into eq. (6, we obtain ( M sλ 2 = A 2 (g µν p k k µ p ν (g αβ p k k α p β g µα p µp α g m 2 βν Z = A 2 g µα (p k 2 p k(k α p µ +k µ p α +p 2 k µ k α( g µα p µp α. (9 Noting that p 2 =, it follows that ( p µ g µα p ( µp α = 0, p α g m 2 µα p µp α Z Hence, eq. (9 simplifies to M sλ 2 = A 2 g µα (p k 2 + kµ k α( g µα p µp α = 0. = 2 A 2 (p k 2, (20 after using k 2 = 0 (corresponding to the massless photon. Finally, we need to evaluate p k. Using four-momentum conservation, q = p k. Squaring both sides and identifying q 2 = m 2 φ then yields m 2 φ = m2 Z 2p k = p k = 2 (m2 Z m φ 2. Inserting this result back into eq. (20, we end up with M sλ 2 = 6 A 2 ( m 2 φ 2. (2 (bexplainwhyitispermissibletomakethesubstitutionshownineq.(8,eventhough the formula for the photon polarization sum given in part (c of problem 4 on Problem Set 4 is a much more complicated expression. The photon polarization sum is given by eq. (69 of Solution Set 4, λ=± ǫ β (k,λǫ ν (k,λ = g βν + k βn ν +k ν n β k n where n µ (;0,0,0. However, note that k βk ν (k n 2, (22 k ν (g µν p k k µ p ν = k µ p k k µ p k = 0, k β (g αβ p k k α p β = k α p k k α p k = 0. Thus, when inserting eq. (22 into eq. (6, the terms proportional to k ν and k β do not contribute to eq. (6. That is, in evaluating eq. (6, we will get the same answer by replacing eq. (22 with eq. (8. This is not an accident in fact, it is consequence of 8

9 electromagnetic gauge invariance! Indeed, as suggested by problem 4, part (c of Problem Set 5, the gauge invariance check of the invariant amplitude guarantees that the photon polarization sum of QED can always be implemented via eq. (8. This is an especially useful simplification. (c Using the result obtained in part (a, derive an expression for the decay width, Γ(Z φγ. Your final result should depend on m Z, m φ and A. Using eq. (5 of the class handout entitled, Two-particle Lorentz invariant phase space, Γ = λ/2 (m 2,m 2 φ,0 6πm Z M sλ 2. (2 where λ(m 2,m 2 φ,0 is the triangle function defined in eq. (9 of the class handout cited above. Using its definition, λ /2 (,m2 φ,0 = m2 Z m2 φ. Hence, inserting eq. (2 into eq. (2 then yields Γ = m Z A 2 96π ( m2 φ. 9