Homework 3 on Quantum Field Theory II - Advanced Quantum Mechanics II | PHYSICS 523, Assignments of Physics

Download Homework 3 on Quantum Field Theory II - Advanced Quantum Mechanics II | PHYSICS 523 and more Assignments Physics in PDF only on Docsity! Physics 523, Quantum Field Theory II Homework 3 Due Wednesday, 28st January 2004 Jacob Lewis Bourjaily The Rosenbluth Formula We are to prove the Rosenbluth Formula by considering the elastic scattering of a relativistic electron off of a proton while correcting the vertex function of the proton. The amplitude for this process is, iM =
p kp ′ ←−q k′ p+ e− = u(k′)(−ieγµ)u(k)−iq2 u(p′)(−ieΓµ)u(p). a) Let us simplify the amplitude using the Gordon identity. Recall that we showed in class that the generalized vertex function Γµ may be written in terms of functions F1(q2) and F2(q2) as Γµ = γµF1(q2) + iσµνqν 2m F2(q2). Inserting this into the amplitude and recalling the Gordon identity, we see that iM = i e2 q2 u(k′)γµu(k)u(p′)Γµu(p), = i e2 q2 u(k′)γµu(k)u(p′) ( γµF1 + iσµνqν 2m F2 ) u(p), = i e2 q2 u(k′)γµu(k)u(p′) ( γµF1 + iσµνqν 2m F2 + (p′ + p)µ 2m F2 − (p ′ + p)µ 2m F2 ) u(p), = i e2 q2 u(k′)γµu(k)u(p′) ( γµ(F1 + F2)− (p ′ + p)µ 2m F2 ) u(p), ∴ Γµ = γµ(F1 + F2)− (p ′ + p)µ 2m F2. b) Let us compute the spin-averaged amplitude squared directly. We see that |M |2 = e 4 4q4 ∑ spin u(k′)γµu(k)u(p′) ( γµ(F1 + F2)− (p ′ + p)µ 2m F2 ) u(p)u(p) ( γν(F1 + F2)− (p ′ + p)ν 2m F2 ) u(p′)u(k)γνu(k′), = e4 4q4 Tr [(6k′ + me)γµ(6k + me)γν ]× { (F1 + F2)2Tr [(6p′ + m)γµ(6p + m)γν ]− F2(F1 + F2) (p ′ + p)ν 2m Tr [(6p′ + m)γµ(6p + m)] −F2(F1 + F2) (p ′ + p)µ 2m Tr [(6p′ + m)(6p + m)γν ] + F 22 (p′ + p)µ(p′ + p)ν 4m2 Tr [(6p′ + m)(6p + m)] } , = 4e4 q4 ( k′µkν + k ′ νkµ − gµν(k′ · k −m2e) )× { (F1 + F2)2 ( p′µpν + p′νpµ − gµν(p′ · p−m2))− F2(F1 + F2)(p′ + p)µ(p′ + p)ν + F 22 4m2 (p′ + p)µ(p′ + p)ν(p′ · p + m2) } , = 4e4 q4 ( k′µkν + k ′ νkµ − gµν(k′ · k −m2e) )× { (F1 + F2)2 ( p′µpν + p′νpµ − gµν(p′ · p−m2)) + (p′ + p)µ(p′ + p)ν ( p′ · p + m2 4m2 F 22 − F2(F1 + F2) )} , ∴ |M |2 = 8e 4 q4 [ (F1 + F2)2 ( k′ · p′k · p + k′ · pk · p− k′ · km2 − p′ · pm2e + 2m2m2e ) + ( p′ · p + m2 4m2 F 22 − F2(F1 + F2) )( k′ · (p′ + p)k · (p′ + p)− 1 2 (k′ · k −m2e)(p′ + p)2 )] . 1 2 JACOB LEWIS BOURJAILY c) Let us consider the kinematics of this reaction in the initial rest frame of the proton. In this frame we see that p = (m,~0), k = (E, Eẑ), k′ = (E′, ~k′), p′ = (E − E′ + m,−~k) with |~k′| = E′. We have defined the momentum transfer q such that p′ − p = q = k − k′. Noting that p · p′ = m2 + Em− E′m, let us compute p′2. p′2 = (p+q)2 = p2+2p ·q+q2 = m2+2p ·(p′−p)+q2 = −m2+2p′ ·p+q2 = m2+2Em−2E′m+q2 = m2, =⇒ q2 = 2E′m− 2Em, ∴ E′ = E + q 2 2m . If we write k′ = (E′, 0, E′ sin θ, cos θ) so that q = (E − E′, 0,−E′ sin θ,E − E′ cos θ) we see q2 = E′2−2EE′+E2−E′2−E′2 sin2 θ−E2 +2EE′ cos θ−E′2 cos2 θ = 2EE′(cos θ−1) = −4EE′ sin2 θ 2 . Using our identity derived above that E′ = E + q 2 2m , we may conclude that q2 = −4E2 sin2 θ 2 − q 2 2m 4E sin2 θ 2 , ∴ q2 = − 4E 2 sin2 θ2 1 + 2Em sin 2 θ 2 . Let us now compute all of the required inner products to compute the desired amplitude squared. Noting that p2 = p′2 = m2, k2 = k′2 = 0, and p · k = Em we may derive all of our necessary identities and inner products indirectly (it’s more fun that way). We notice that p′2 = m2 = p2 + 2p · q + q2 = −m2 + 2p′ · p + q2, ∴ p′ · p = m2 − q 2 2 . Similarly, p′2 = m2 = p2 + 2p · q + q2 = m2 + 2p · k − 2p · k′ + q2, but we know that p · k = Em, ∴ p · k′ = Em + q 2 2 . Likewise, k′2 = 0 = k2 − 2k · q + q2 = 2k · k′ + q2 = 0, ∴ k′ · k = −q 2 2 . And k′2 = 0 = k2 − 2k · q + q2 = −2k · p′ + 2k · p + q2, where we know that k · p = Em and ∴ k · p′ = Em + q 2 2 . Similarly, k2 = 0 = k′2 + 2q · k′ + q2 = 2p′ · k′ + q2, ∴ p′ · k′ = Em. Tabulating our results, we have shown that k′ · k = −q 2 2 p′ · p = m2 − q 2 2 k′ · p = Em + q 2 2 p′ · k = Em + q 2 2 k′ · p′ = Em p · k = Em. These imply that k · (p′ + p) = 2Em + q 2 2 , k′ · (p′ + p) = 2Em + q 2 2 , and (p + p′)2 = 4m2 − q2.