Physics 443 Homework 5 Solutions

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1 Physics 3 Homework 5 Solutions Problem P&S Problem. a p S p lim T iɛ p exp i T T dt d 3 xe ψγ µ ψa µ p. Ignoring the trivial identity contribution and working to the lowest order in e we find p it p ie d x p ψγ µ ψ p A µ. Since both p correspond to p correspond to electrons ψ p e ipx up and p ψ e ip xūp. Therefore, p it p ieūp γ µ up d xe ip px A µ x 3 or using the Fourier transform õq d xa µ xe iqx p it p ieūp γ µ upãµp p. b For A µ x A µ x, à µ q πδq 0 õ q and As stated in the problem we define p it p ieūp γ µ upãµ p f p i πδe f E i. 5 im ieūp γ µ upãµ p f p i. 6 Also, as in P&S we assume that the particles are sent in narrow packets φ in φ pe i b. p π 3 p in, 7 Ei where φ p is narrowly peaked around a momentum p i and b is taken to be orthogonal to p i. Then, the probability for scattering into a single particle state with momentum p f is Pφ i p f d3 p f π 3 E f out p f φ in. 8 Plugging in the expression for φ in and integrating over the impact parameter b we get the cross section as d3 p f π 3 d d 3 k φ k d 3 l φ b l E f π 3 Ek π 3 e i b. l k Mk p f M l p f π δe f E k δe f E l. El

2 First we perform the integral over b. d be i b. l k π δ l k. 9 Then, integral over l is taken so that other integrands are restricted to l k surface. At this point we have E l m + l + k. Then, for l integral δe f E l δ l E f m k l /E l + δ l + E f m k l /E l 0 However, only one of those delta functions contribute to the integral since we are assuming that φ l is narrowly peaked around momentum p i and that restricts us to l k since there is also a φ k term among the integrands. Once l integrals are performed we are left with d3 p f π 3 E f d 3 k π 3 φ k Mk p f π δe f E k k. We now perform the k integral using that φ k is peaked around p i and restricts k to p i and remembering that p i p i we get d 3 k π 3 φ k. This d3 p f π 3 E f p i Mp i p f πδe f E i. Note that p i E i v i and that confirms the form given in the problem. f d p f p f dω and using to integrate over p f, we finally obtain δe f E i δ p f p i p f /E f 3 dω Mp i p f 6π pi p. f c For A 0 Ze πr and Ai 0 we have à 0 q d 3 ra 0 re i q. r π 0 drr dcos θ Ze πr e i q r cos θ 5 Ze q Im dre i q r Ze 0 q. 6 Integral in the last line is divergent and the result above is obtained by adding a small, positive imaginary part to q. To lowest order in p we have ū s p γ 0 u s p mδ s s. Then, using the expression we have found in part b: dω m δ s s Z e 6π p f p i. 7

3 Since we are evaluating this expression at p f p i, p f p i p i sin θ where we take the angle between p i and p f as θ. In the non-relativistic limit we further have p i mv i. Finally, summing over the outgoing electron s spin and averaging over the incoming electron s spin we get dω Z α m vi sin θ. 8 Problem P&S Problem 5. In the previous problem we found that dω Mp p p p 6π, 9 where im ieūp γ µ upãµ p p. 0 We will be working with time independent Coulomb potential where à µ p p Ã0, 0, 0, 0, and Ã0 Ze p p. Summing over the states of the final electron and averaging over the of the initial electron we get M e õ p pã ν p ptr /p + mγ µ /p + mγ ν e à µ p pã ν p p p µ p ν p.pg µν + p ν p µ + m g µν 3 e à 0 p pã 0 p p p 0 p 0 p.p + m, where we used Tr γ µ γ ν γ ρ γ σ g µν g ρσ g µρ g νσ + g µσ g νρ. Taking the angle between p and p as θ and remembering that we evaluate im at p p we have p p p sin θ. M Z e 8 p sin θ Z e p sin θ p + m + p cos θ + m 5 p + m p sin θ 6 7 Since p + m p /β where β is the initial electron s speed and α e /π we finally get dω Z α p β sin θ β sin θ. 8 Now, let us study the same setting by working through e µ e µ scattering amplitude. im ūp ieγ µ up ig µν q ūp ieγ ν up. 9 3

4 Summing and averaging over we get M e p µ q p ν p.p g µν + p ν p µ + m eg µν] p µ p ν p.p g µν + p ν p µ + m ] µg µν 30 8e p q.p p.p + p.p p.p m µp.p m ep.p + m em ] µ. 3 In the limit m µ and in the rest frame of muon: This gives us the leading order behavior of M 8e 6 p sin θ M, at order Om µ as E m µ m µe p cos θ + m em µ] 3 m µe E p sin θ + p cos θ + m e] 33 m µe β p sin θ β sin θ. 3 Using P&S equation.8 in the m µ limit, we find the cross section as dω p /β m µ β α β p sin θ p π m µ β sin θ M Problem 3 P&S Problem 5. The two tree level diagrams that contribute to the scattering amplitude are

5 We take momentum orientations as in for m e 0: s p + p p.p k.k, 37 t p k p.k p.k, 38 u p k p.k p.k. 39 The first diagram s contribution is proportional to ūkγ µ vk vp γ ν up and the second diagram s contribution is proportional to ūkγ µ up vp γ ν vk. As can be deduced from these two expressions there is a relative minus sign between the two contributions. Therefore, im ūk ieγ µ vk ig µν p + p vp ieγ ν up ūk ieγ µ up ig µν ie ūkγ µ vk vp γ µ up s ūkγµ up vp γ µ vk t p + p vp ieγ ν vk 0 ]. Summing and averaging over final and initial states, respectively, we get M e s Tr/kγµ /k γ ν Tr/p γ µ/pγ ν + t Tr/kγµ /pγ ν Tr/p γ µ /k γ ν st Re {Tr/p γ µ/pγ ν /kγ µ /k γ ν } ]. We have Tr/kγ µ /k γ ν Tr/p γ µ/pγ ν 6 k µ k ν k.k g µν + k ν k µ p µp ν p.pg µν + p νp µ 3 3 p.kk.p + p.kp.k ] 8u + t, 5 Tr/kγ µ /pγ ν Tr/p γ µ /k γ ν 3 p.kk.p + k.kp.p] 6 8u + s, 7 5

6 and Tr/p γ µ/pγ ν /kγ µ /k γ ν Tr/p /kγ ν/p/k γ ν 8 8k.pTr/p /k 9 3k.pp.k 50 8u, 5 where we used γ µ γ ν γ ρ γ σ γ µ γ σ γ ρ γ ν and γ µ γ ν γ ρ γ µ g νρ. Using these three expressions we get M e u s + + t ] t s +. 5 s t We now go to a center of mass frame where p E, Eẑ, p E, Eẑ k E, Eˆn, k E, Eˆn, E CM E and ˆn points in a direction with coordinates θ, φ on the sphere. Note that even after we restrict to this center of mass frame we are free to rotate the system around ẑ axis and set φ 0 or, in other words, the differential cross section has no φ dependence. Since we obtain dω 6π ECM M 6π s π d cos θ dω πα u s s + + t M, 53 ] t s +. 5 s t Finally, s E, t p.k E E ˆn.ẑ E cos θ and u p.k E + E ˆn.ẑ E + cos θ gives ] d cos θ πα + cos θ cos θ E + + cos θ cos θ. 55 It diverges as θ 0 because of denominators proportional to t 0. These come from the propagator in the second Feynman diagram, i.e., it is due to the virtual photon going on-shell in the second diagram. Below we plot d cos θ E πα. 6

7 Problem P&S Problem 5. a For e + e γ process, the two contributing tree level diagrams are im vp ieγ µ ɛ µk i /p /k + m p k m ɛ νk ieγ ν up + vp ieγ ν ɛ νk i /p /k + m p k m ɛ µk ieγ µ up 56 ] γ µ /p ie ɛ µk ɛ νk vp /k + mγ ν + γν /p /k + mγ µ up. 57 p.k p.k Using /p + mγ ρ up p ρ γ ρ /p mup p ρ up this can be simplified as γ im ie ɛ µk ɛ µ /k νk vp γ ν γ µ p ν + γν /k γ µ γ ν p µ ] up 58 p.k p.k We now switch to a center of mass frame and choose the following polarization vectors. Note that most importantly a photon polarization vector ɛ µ k should satisfy k µ ɛ µ k 0. Moreover, by gauge invariance shifting the polarization vector by a vector proportional to k µ should not change the amplitude.the representatives we choose below correspond to two circular polarizations and are chosen so that they have ɛ 0 0. p µ m + p, 0, 0, p, p µ m + p, 0, 0, p, p i 0, 0, p, 59 k µ m + p, sin θ, 0, cos θ, k µ m + p, sin θ, 0, cos θ, ˆki sin θ, 0, cos θ, 60 ɛ µ +k ɛ µ k 0, cos θ, i, sin θ, ɛ µ k ɛ µ +k 0, cos θ, i, sin θ. 6 We will employ Latin letter superscripts for 3-vectors. We will now expand the amplitude to the order Op terms of order Op will be necessary for the coming sections of the problem. 7

8 Firstly, m + p m + Op. That further gives k µ m, sin θ, 0, cos θ + Op and k µ m, sin θ, 0, cos θ + Op. Since we choose representatives with ɛ 0 0, γ µ p ν and γ ν p µ Op. terms contribute only to order and p.k m mˆk i p i + Op m + ˆk i p i + Op 6 m p.k m + mˆk i p i + Op m ˆk i p i + Op. 63 m As worked out in problem set, in Weyl basis where γ 0 0 and γ i 0 σ i 0 σ i 0 6 we have and Defining we can write u s p p 0 + m ξ s σ p p 0 +mξs σ p v s p p 0 + m p 0 +mξs ξ s u s 0 ξ s 0 m ξ s σ n p n m ξs σ n p n m m ξs ξ s and v 0 s 0 ξ s u s p m pn γ n u s 0 and v s p m v 0 s m + Op 65 + Op. 66 pn γ n m Note that we could have obtained the same result in a basis independent way by solving Dirac equation order by order in p. Then the amplitude, Mp, s; p, s k, r; k, r where s, s, r, r denotes and polarizations, is im ie m ɛi r k ɛ j r k v 0 s pn γ n γ i /k m γ j γ i p j + ˆk f p f m + γ j /k γ i γ j p i ˆk ] d p d pe γ e u s 0 + Op. m m We define q d p d /m and use /k mγ 0 mˆk n γ n + Op, /k mγ 0 + mˆk n γ n + Op to get im ie ɛ i r k ɛ j r k v 0 s qn γ n γ i γ 0 γ j ˆk n γ i γ n γ j γ i q j + ˆk f q f 69 + γ j γ 0 γ i + ˆk n γ j γ n γ i γ j q i ˆk ] f q f qe γ e u s 0 + Oq. 70 8

9 im ie ɛ i r k ɛ j r k v 0 s γ i γ 0 γ j + γ j γ 0 γ i ˆk n γ i γ n γ j + ˆk n γ j γ n γ i + q d γ i δ jd γ j δ id + q f ˆkf γ i γ 0 γ j γ j γ 0 γ i ˆk n γ i γ n γ j ˆk n γ j γ n γ i { qe γ e, γ i γ 0 γ j + γ j γ 0 γ i ˆk n γ i γ n γ j + ˆk n γ j γ n γ i}] u s 0 + Oq. 7 Any term with an even number of γ k s between v 0 s and u s 0 give zero. Accordingly, we simplify our expression as im ie ɛ i r k ɛ j r k v 0 s ˆk n γ j γ n γ i γ i γ n γ j q f γ i δ jf + γ j δ if q f ˆkf ˆkn γ i γ n γ j + γ j γ n γ i { qe γ e, γ i γ 0 γ j + γ j γ 0 γ i}] u s 0 + Oq. 7 Since γ i γ 0 γ j + γ j γ 0 γ i γ 0 {γ i, γ j } γ 0 δ ij the last term linear in q has vanishing contribution this is the correction due to subleading terms in up and vp. Also, we note that for the first and the third terms in the brackets we can freely anti-commute γ n s through γ i and γ j, because any δ in or δ jn term that will be left will have no contribution since ˆk n ɛ n 0 is satisfied by the polarization vectors. Therefore, im ie ɛ i r k ɛ j r k v 0 ˆkn s γ i, γ j] γ n q f γ i δ jf + γ j δ if + q f ˆkf ˆkn {γ i, γ j }γ n] u s 0 + Oq 73 and using {γ i, γ j } δ ij im ie ɛ i r k ɛ j r k v 0 ˆkn s γ i, γ j] γ n q f δ in δ jf + δ jn δ if + δ ijˆkf ˆkn γ n] u s 0 + Oq. 7 Since v 0 s γ n u s 0 ξ s σn ξ s and v 0 s γ i, γ j] γ n u s 0 iɛ ijk ξ s σk σ n ξ s iɛ ijk ξ s δ kn + iɛ knf σ f ξ s note that ɛ ijk ɛ knf δ in δ jf δ if δ jn give zero contribution due to δ in and δ jn terms giving ˆk n ɛ n 0 we finally get ] im ie ɛ i r k ɛ j r k iɛ ijnˆkn ξ s ξs + q f δ in δ jf + δ jn δ if + δ ijˆkf ˆkn ξ s σn ξ s + Oq. 75 Therefore, to zeroth order in q im 0 e ɛ i r k ɛ j r k ˆk n ɛ ijn ξ s ξs. 76 ξ s ξs Tr ξξ s s and we replace the term in the parenthesis with / to compute the amplitude for a spin-0 configuration and with n i σ i / where n i is a unit vector, for a spin- configuration. That immediately tells us for spin- states M 0 to this order. S 0 configurations, on the other hand, give im 0 S 0 r, r e ɛ i r k ɛ j r k ˆk n ɛ ijn. 77 Moreover, ɛ i r k ɛ j r k ˆk n ɛ ijn is zero for r, r +, or, +, it is + i for r, r +, + and it is i for r, r,. Therefore, for the L 0, S 0 positronium B S0 m d 3 k π ψ 3 k k, k k, k, 78 m m 9

10 where ψ k is the Fourier transform of S wavefunction ψ r π a 3/ e r/a 79 for a /mα. See, for example, Landau&Lifshitz - Quantum Mechanics: Non-relativistic Theory, section 36, and remember to use reduced mass m/. Then, Then, MB S0 ±± m m e d 3 k π ψ 3 km k, k, S 0 ±± 80 d 3 k π ψ 8 3 k e ψ0. 8 m MB S0 ±± 8e m ψ0 8e mα 3 m 8 8 6πm α Summing over polarizations of the photon final states and dividing by since the final state consists of two identical photons we get dγ d 3 k d 3 k 6πm m π 3 k π 3 α 5 + 6πm α 5 π δm k k δ 3 k k + k. 8 Integrating over k The integral gives π and we finally get Γ mα5 π d 3 k δm k. 85 k Γ mα5. 86 Also, we should note that Γ for L 0 states do not get any contribution from Op terms in im since the explicit factor of p f there requires an odd parity spatial wavefunction to give nonvanishing contributions. b For L bound states ψ0 0 and therefore Op 0 terms in the amplitude does not give any contribution. For that purpose we will need Op contribution which we have computed in part a. where we define M f ɛ i r k ɛ j r k ɛ f r k ɛ n r k + ɛ n r im ie m pf M f 87 δ in δ jf + δ jn δ if + δ ijˆkf ˆkn ξ s k ɛ f r k + ɛ d r k ɛ d r k ˆk f ˆkn Tr σn ξ s 88 ] σ n ξξ s s. 89 This gives zero for S 0 configurations since we can compute the amplitude in this case by replacing ξξ s s by a matrix proportional to identity matrix and Tr σ n ] 0. For S case on the other hand we will replace ξξ s s by a matrix proportional to a Pauli matrix and this will have nonzero contributions since Tr σ n σ e ] δ ne. 0

11 c We will consider bound states with momentum k as B k M π ψ 3 i pa p+ k/ Σi b p+ 0, 90 k/ M π ψ b 3 i p a p+ a k/ p+ Σ i p+ k/ k/ b 0. 9 p+ k/ Then, B k B k M π 3 π ψ 3 i p ψ j p 0 a p + k / a p + k / b p + k / a p+ k/ b p + k / a p+ k/ Σ j Σ i b p+ k/ b p+ k/ 0 9 Commuting the piece with a s and a s, for the nonvanishing contributions we get an identity matrix in between and a delta function π 3 δ p p + k k/. We use this to perform the p integral. B k B k M π ψ 3 i p ψ j p + k k 0 b p+ k k/ b p+ k k/ Σ j Σ i b p+ k/ b p+ k/ 0. Let A Σ j Σ i, then 0 b q b q A A A A b l b l 0 0 b q A b + b l q A b 0 93 l π 3 δ 3 q l TrA. 9 Therefore, Since B k B k Mπ 3 δ 3 k k Tr Σ j Σ i] π 3 ψ i p ψ j p δ ij, for Tr Σ i Σ i] this gives π 3 ψ i p ψ j p. 95 B k B k Mπ 3 δ 3 k k. 96 d Note that properly normalized np wavefunctions are ψ i x x i f n r where f n r 3 3/π n mα 5/ + Or 97 n near r 0. Again this could be found in Landau&Lifshitz - Quantum Mechanics: Non-relativistic Theory section 36. Therefore, amplitude for a positronium in np state at its rest frame and with Σ total angular momentum configuration is where we will put M m imσ r, r m π ψ 3 i p ie m m pf M f Σ i r, r, 98

12 where M f Σ i r, r ɛ f r k ɛ n r k + ɛ n r k ɛ f r k + ɛ d r k ɛ d r k ˆk f ˆkn Tr σ n Σ i]. 99 Also, notice the normalization factor m due to the difference between, for example, p and a p 0. Since M f is independent of p we have imσ r, r ie m 3/ Mf Σ i r, r d 3 p π ψ 3 i pp f d 3 xf n rx i d 3 p π 3 e i p. x p f i d 3 xf n rx i f i d 3 x f nr xf r xi + f n rδ if where we used integration by parts on the last line. Since That gives π 3 ψ i pp f. 00 d 3 p i p. x e π 3 0 π 3 e i p. x 0 π 3 e i p. x δ 3 x, π 3 ψ i pp f if n 0δ if. 03 imσ r, r e m 3/ Mi Σ i r, r f n 0 0 and plugging in the f n 0 expression above imσ r, r M i Σ i r, r πn m α 7 3n Our last task will be to compute M i Σ i N r, r for J 0,,, where M i Σ i N r, r ɛ i r k ɛ n r k + ɛ n r k ɛ i r k + ɛ d r k ɛ d r k ˆk iˆkn Tr σ n Σ i N]. 06 Note that for each J there will be J + spin configurations Σ i N where N,..., J +. To get the decay rate we will average over positronium spin configurations and sum over photon polarizations. Then, as in part a we have Γ J d 3 k d 3 k m π 3 k π 3 MΣ N r, r π δm k k δ 3 k J + k + k N,r,r πn m α 7 d 3 k 6m 3n 5 π k M i Σ i N r, r δm k J + N,r,r n mα 7 96n 5 dω k M i Σ i N r, r. 07 π J + N,r,r J In this case Σ i N ɛijk n j N σk / and Tr σ n Σ i ] N ɛijk n j N Tr σ n σ k] ɛ ijn n j N. 08 Since ɛ i r k ɛ n r k + ɛ n r k ɛ i r k + ɛ d r k ɛ d r k ˆk iˆk n is symmetric in the indices i and n, we get zero contribution for any n N and Γ J 0. 09

13 J0 In this case Σ i σ i / 6 for which it is easy to check Tr Σ i Σ i]. Then, Tr σ n Σ i] Tr σ n σ i] 6 3 δin, 0 and Nonzero amplitudes are which gives Plugging this in Γ J above, we get M i Σ i r, r 6ɛ i r k ɛ i r k. M i Σ i +, + M i Σ i, i 6, J + N,r,r M i Σ i N r, r. 3 Γ J0 n mα 7 8n 5. J Now there are five different initial angular momentum states with Σ i N 3 h ij N σj where h N s are symmetric and traceless which we can choose to be real. Proper normalization requires ] δ MN Σ i M Σi N 3 hij M hij N. 5 A possible choice is then h ij 3 δ i δ j + δ i δ j, h ij 3 δ i δ j3 + δ i3 δ j, h ij 3 3 δ i δ j3 + δ i3 δ j, h ij 3 δ i δ j δ i δ j, h ij 5 δ i δ j + δ i δ j δ i3 δ j3. and M i Σ i N r, r Tr σ n Σ i ] N 3 h ij N Tr σ n σ j] 3 hin N, 6 ɛ i r k ɛ j 3 r k + ɛ k r k ɛ k r k ˆk iˆkj h ij N. 7 First of all, M i Σ i N +, + M i Σ i N, and M i Σ i N +, M i Σ i N, +. 8 Also, note that for r, r +, + ɛ i + k ɛ j + k + ɛ k + k ɛ k + k ˆk iˆkj δ ij + δ i δ j δ j δ i i cos θ + δ i δ j3 δ j δ i3 i sin θ, 9 so M i Σ i N +, + Mi Σ i N, 0 for all N. For the rest we find M i Σ i +, i cos θ, M i Σ i +, i sin θ, M i Σ i 3 +, cos θ sin θ, 0 3

14 This gives J + M i Σ i +, + cos θ, M i Σ i 5 +, 3 sin θ. N,r,r M i Σ i N r, r 5 cos θ + sin θ + cos θ sin θ + + cos θ + 3 sin θ 6 5 and consequently Γ J n mα 7 30n 5. Problem 5 P&S Problem 6. We will work in the rest frame of the initial proton and in the limit where electron mass is zero. im ūk ieγ ρ uk ig ] ρµ q + iɛūp ie γ µ F + iσµν q ν m F up. 3 By Gordon identity ūp iσ µν q ν + p µ + p µ ] up mūp γ µ up and hence im ie q ūk γ µ ukūp F + F γ µ p µ + p µ ] m F up. Summing and averaging over we get M e q D µνtr /p + m F + F γ µ p µ + p µ m F where D µν Trγ µ /kγ ν /k / k µ k ν k.k g µν + k ν k µ. M e q D µν F + F Tr /p + mγ µ /p + mγ ν + F F + F F p ν + p ν m Tr /p + mγ µ /p + m + p µ + p µ /p + m F + F γ ν p ν + p ν ] m F, p µ + p µ p ν + p ν m Tr /p + m/p + m m Tr /p + m/p + mγ ν ], and M e q D µν F + F p µ p ν p.pg µν + p ν p µ + m g µν + F m p µ + p µ p ν + p ν p.p + m ] F + F F p µ + p µ p ν + p ν.

15 We will now eliminate p from these expressions and then simplify them. We first find and D µν p µ p ν p.pg µν + p ν p µ + m g µν p.kp.k + p.k p.k m k.k ] 5 p.k + p.k + m q ] 6 D µν p µ + p µ p ν + p ν p.k + p.k m + p.p k.k ] 7 ] p.k + p.k + m q q m, 8 where we used m + p.p m p p m m q /. Secondly, F p m.p + m F + F F F Combining these we get M 8e p.q F q + F + q F + F F F 9 + F q m F. 30 m + q + ] F q 8 m F p.k + p.k + m q q m. Since p.q p + q p q p p q q and p.k + p.k p.kp.k + p.q, M 8e q q F + F + F q m F p.kp.k + m q ]. According to the equation right after P&S equation 5.90 we know that d cos θ E 3πm M. 3 E Using p.k me, p.k me and q k.k EE sin θ/ we get d cos θ πα EE sin θ E E F q m F θ q cos m F + F sin θ ] 3 Lastly, using E/E + E/m cos θ πα d cos θ F q m F ] θ q cos m F + F sin θ E + E m sin θ sin θ. 33 5