Quantum Electrodynamics

Transcript

1 Quantum Electrodynamics Ling-Fong Li Institute Slide_06 QED / 35

2 Quantum Electrodynamics Lagrangian density for QED, Equations of motion are Quantization Write L= L 0 + L int L = ψ x γ µ i µ ea µ ψ x mψ x ψ x 4 F µνf µν i γ µ µ m ψ x = ea µ γ µ ψ non-linear coupled equations υ F µν = eψγ µ ψ L 0 = ψ i γ µ µ m ψ 4 F µνf µν L int = eψγ µ ψa µ where L 0,free field Lagrangian, L int is interaction part. Conjugate momenta for fermion L 0 ψ α = i ψ α x For em fields choose the gauge A = 0 Conjugate mometa Institute Slide_06 QED 2 / 35

3 From equation of motion π i = L 0 A i = F 0i = E i ν F 0ν = eψ ψ = 2 A 0 = eψ ψ A 0 is not an independent field, Commutation relations {ψ α x, t, ψ β x, t [Ȧi x, t, A j x, t] A 0 = e d 3 x ψ x, t ψ x, t d 4π x = 3 x ρ x, t e x x x } = δ αβ δ 3 x x {ψ α x, t, ψ β x, t = i δ tr x ij x } =... = 0 where δ tr ij x y = d 3 k 2π 3 e i k x y δ ij k i k j k 2 Commutators involving A 0 [ A 0 ] x, t, ψ α x, t = e Hamiltonian density d 3 x 4π x x x [ ψ x, t ψ x ], t, ψ α x, t = e ψ α, t 4π x x Institute Slide_06 QED 3 / 35

4 L H = 0 ψ α ψ α + L 0 A k Ȧk L = ψ i α + βm ψ + E 2 + B 2 + E A 0 + eψγ µ ψa µ 2 and [ H = d 3 x H = d 3 x {ψ ] α i e A + βm ψ + E 2 + B 2 } 2 A 0 does not appear in the interaction, But if we write Then longitudinal part is E = El + E t where El = A 0, E t = A t E d 3 x 2 + B 2 = d 3 x E 2 l + d 3 Et x 2 + B d 3 x E 2 l = e d 3 xd 3 y ρ x, t ρ y, t 2 4π x y Coulomb interaction Without classical solutions, can not do mode expansion to get creation and annihilation operators We can only do perturbation theory. Institute Slide_06 QED 4 / 35

5 Recall that the free field part A 0 satisfy massless Klein-Gordon equation The solution is A 0 = 0 A 0 x, t = d 3 k 2ω2π 3 ɛ k, λ[ak, λe ikx + a + k, λe ikx ] w = k 0 = k λ ɛk, λ, λ =, 2 with k ɛk, λ = 0 Standard choice ɛk, λ ɛk, λ = δ λλ, ɛ k, = ɛk,, ɛ k, 2 = ɛ k, 2 It is convienent to write the mode expansion as, A µ x, t = d 3 k 2ω2π 3 ɛ µ k, λ[ak, λe ikx + a + k, λe ikx ] λ where ɛ µ k, λ = 0, ɛk, λ Institute Slide_06 QED 5 / 35

6 Photon Propagator Feynman propagatpr for photon is From mode expansion, id µν x, x = 0 T A µ x A ν x 0 = θ t t 0 Aµ x A ν x 0 + θ t t 0 A ν x A µ x 0 0 Aµ x A ν x 0 = = = d 3 kd 3 k 2π 3 2ω k 2ω ɛ µ k, λɛ ν k, λ 0 [ak, λe ikx ]a + k, λ e ik x 0 k λ,λ d 3 kd 3 k 2π 3 2ω k ɛ µ k, λɛ ν k, λ δ 3 k k e ikx +ik x λ,λ d 3 k 2π 3 2ω k ɛ µ k, λɛ ν k, λ e ikx x λ,λ Note that We then get 2π dk 0 k 2 0 ω2 + i ε e ik 0t t = { i 2ω e i ωt t for t > t i 2ω e i ωt t for t > t d 4 k e ik x x 2π 4 = k 2 i + i ε d 3 k [ 2π 3 θt t e ikx x + θt t e ikx x ] 2ω k Institute Slide_06 QED 6 / 35

7 and 0 T A µ x A ν x 0 = = i d 3 k 2π 3 2ω k ɛ µ k, λɛ ν k, λ [θt t e ikx x + θt t e ikx x ] λ,λ d 4 k e ik x x 2π 4 k 2 + i ε 2 λ= ɛ ν k, λɛ µ k, λ = id µν x, x Or D µν x, x = d 4 k e ik x x 2π 4 k 2 + i ε 2 λ= ɛ ν k, λɛ µ k, λ polarization vectors ɛ µ k, λ, λ =, 2 are perpendicular to each other. Add 2 more unit vectors to form a complete set η µ =, 0, 0, 0, ˆk µ = k µ k η η µ k η 2 k 2 completeness relation is then, 2 ɛ ν k, λɛ µ k, λ = g µν η µ η ν ˆk µ ˆk ν λ= k µ k ν = g µν k η 2 k k η k µ η ν + η µ k ν + 2 k η 2 k 2 η µ η ν k 2 k η 2 k 2 Institute Slide_06 QED 7 / 35

8 If we define propagator in momentum space as D µν x, x = d 4 k 2π 4 e ik x xdµν k then D µν k = g k 2 µν + i ε k µ k ν k η k µ η ν + η µ k ν k η 2 + k 2 k η 2 k 2 η µ η ν k 2 k η 2 k 2 terms proportional to k µ will not contribute to physical processes and the last term is of the form δ µ0 δ ν0 will be cancelled by the Coulom interaction.. Institute Slide_06 QED 8 / 35

9 Feynman rule in QED The interaction Hamiltonian is, H int = e d 3 x ψγ _ µ ψa µ The Feynman propagators, vertices and external wave functions are given below. Institute Slide_06 QED 9 / 35

10 p p p p Institute Slide_06 QED 0 / 35

11 e + e µ + µ Total Cross Section momenta for this reaction e + p + e p µ + k + µ k Use Feynman rule to write the matrix element as M e + e µ + µ = v _ p, s ieγ µ igµν _uk u p, s q 2, r ieγ ν v k, r = ie 2 _ q 2 v p, s γ µ u p, s _ uk, r γ µ v k, r where q = p + p. Notet that electron vertex have property, q µ _ v p γ µ u p = p + p µ _ v p γ µ u p = _ v p /p + /p u p = 0 Institute Slide_06 QED / 35

12 This shows the term proportional to photon momentum q µ will not contribute in the physical processes. For cross section, we need M which contains factor _ v γ µ u _ v γ µ u = u γ µ γ 0 v = u γ 0 γ µ v = _ uγ µ v More generally, _ v Γu = _ u _ Γv, with _ Γ = γ 0 Γ γ 0 It is easy to see _ γµ = γ µ unpolarized cross section which requires the spin sum, γ µ γ 5 = γ µ γ 5 /a /b /p = /p /b /a u α p, s _ u β p, s = /p + m αβ s v α p, s v _ β p, s = /p m αβ s This can be seen as follows. u α p, s _ u β p, s = E + m s = σ p E +m E + m σ p σ p E + m χ s χ s s = /p + m σ p E +m σ p E +m = E + m σ p E +m p 2 E +m 2 Institute Slide_06 QED 2 / 35

13 Similarly for the v spinor, v α p, s v _ β p, s = E + m s = σ p E +m E m σ p σ p E + m χ s χ s σ p E +m = /p m = E + m p 2 E +m 2 σ p E +m σ p E +m Institute Slide_06 QED 3 / 35

14 A typical calculation is, trace of product of γ matrices. = s _ v α p, s γ µ αβ u β p, s _ u ρ p, s γ ν ρσ v σ p, s s,s _ v α p, s γ µ αβ /p + m βρ γ ν ρσ v σ p, s = γ µ αβ /p + m βρ γ ν ρσ /p m σα = Tr [γ µ /p + m γ ν /p m] Tr γ µ = 0 Tr γ µ γ ν = 4g µν Tr γ µ γ ν γ α γ β = 4 g µν g αβ g µα g νβ + g µβ g να Tr /a /a 2 /a n = a a 2 Tr /a 3 /a n a a 3 Tr /a 2 /a n + + a a n Tr /a 2 /a 3 /a n, n even = 0 n odd With these tools 4 M e + e µ + µ 2 = e 4 spin q 4 Tr [ /p m e γ µ /p + m e γ ν] [ Tr /k ] + m µ γµ /k + mµ γ ν Institute Slide_06 QED 4 / 35

15 for energies m µ. In center of mass, If we set m µ = 0, E = k and 4 M e + e µ + µ 2 = 8 e 4 [ spin q 4 p k p k + p k p k ] p µ = E, 0, 0, E, p µ = E, 0, 0, E k µ = E, k, k µ = E, k, with k ẑ = E cos θ q 2 = p + p 2 = 4E 2, p k = p k = E 2 cos θ, Then 4 spin M 2 = p k = p k = E 2 + cos θ 8e 4 [ 6E 4 E 4 cos θ 2 + E 4 + cos θ 2] = e 4 + cos 2 θ Note that under the parity θ π θ. this matrix element conserves the parity Institute Slide_06 QED 5 / 35

16 The cross section is d σ = I 2E 2E 2π4 δ 4 p + p k k 4 spin M 2 d 3 k 2π 3 2ω d 3 k 2π 3 2ω use the δ function to carry out integrations. introduce the quantity ρ, called the phase space, given by ρ = = 2π 4 δ 4 p + p k k d 3 k d 3 k 2π 3 2ω 2π 3 2ω d Ω 32π 2 The flux factor is The differential crossection is then Or I = E E 2 p p 2 2 m 2 m2 2 = E 2 2E 2 = 2 d σ = 2 4E 2 4 M 2 d Ω spin 32π 2 d σ d Ω = α2 + cos 2 6E 2 θ where α = e 2 is the fine structure constant. The total cross section is 4π σ e + e µ + µ = α2 π 3E 2 Institute Slide_06 QED 6 / 35

17 Or σ e + e µ + µ = 4α2 π 3s with s = p + p 2 2 = 4E 2 Institute Slide_06 QED 7 / 35

18 e + e hadrons One of the interesting procesess in e + e collider is the reaction e + e hadrons According to QCD, theory of strong interaciton, this processes will go through e + e q _ q and then q _ q trun into hadrons. Since coupling of γ to q _ q differs from the coupling to µ + µ only in their charges cross section for q _ q as σ e + e q _ q = 3 Qq 2 4α 2 π = 3 Q 2 q σ e + e µ + µ 3s Q q is electric charge of quark q. The factor of 3 because each quark has 3 colors. Then σ e + e hadrons σ e + e µ + µ = 3 Qi 2 i Summation is over quarks which are allowed by the avaliable energies. e. g., for energy below the the charm quark only u, d, and s quarks should be included, which is not far from the reality. σ e + e hadrons σ e + e µ + µ = 3 [ ] 2 = 2 3 Institute Slide_06 QED 8 / 35

19 Institute Slide_06 QED 9 / 35

20 ep ep, e k + p p e k + p p Proton has strong interaction. First consider proton has no strong interaction and include strong interaction later. The lowest order contribution is, M e + p e + p = _ up, s ieγ µ igµν _uk u p, s q 2, r ieγ ν u k, r = ie 2 _ up q 2, s γ µ u p, s _ uk, r γ µ u k, r Institute Slide_06 QED 20 / 35

21 where q = k k. For unploarized cross section, sum over the spins, 4 M e + p e + p 2 = e 4 spin q 4 Tr [ /p + M γ µ /p + M γ ν] [ Tr /k + m e γµ /k + m e γ ν] Again neglect m e. Compute the traces [ Tr /k γ µ /k γ ν] = 4 [k µ k ν g µν k k + k µ k ν ] Tr [ /p + M γ µ /p + M γ ν] = 4 [p µ p ν g µν p p + p µ p ν ] + 4M 2 g µν Then 4 M e + p e + p 2 = e 4 { [ spin q 4 8 p k p k + p k p k ] } 8M 2 k k Institute Slide_06 QED 2 / 35

22 Use laboratoy frame p µ = M, 0, 0, 0, k µ = E, k, k µ = E, k Then p k = ME, p.k = ME, k k = EE cos θ p k = p + k k.k = p.k + k k, p k = p + k k.k = p.k k k q 2 = k k 2 = 2k k = 2EE cos θ Differential cross section is The phase space is d σ = I 2π 4 δ 4 p + k p k 2p 0 2k 0 4 M 2 d 3p spin 2π 3 2p 0 d 3 k 2π 3 2k 0 ρ = = 2π 4 δ 4 p + k p k d 3 p d 3 k 2π 3 2p 0 2π 3 2k 0 4π 2 δ p 0 + k 0 p 0 k 0 d 3 k 2p 0 2k 0 where p 0 = M k p + k k = M k Institute Slide_06 QED 22 / 35

23 Use the momenta in lab frame, ρ = = 4π 2 δ M + E p 0 E k 2 dk d Ω 2p 0 2E 4π 2 δ M + E p 0 E d ΩE de p 0 Let Then and x = E + p 0 + E dx = de + dp 0 p de = de 0 + E E cos θ p 0 ρ = d ΩE 4π 2 δ x dx M p 0 + E E cos θ = d ΩE 4π 2 M + E cos θ From the argument of the δ function we get the relation, M = x = E + p 0 + E From momentum conservation k p 0 2 = M k = M 2 + E 2 + E 2 2EE cos θ and from energy conservation p 2 0 = M + E E 2 = M 2 + E 2 + E 2 2EE + 2ME 2ME Institute Slide_06 QED 23 / 35

24 Comparing these 2 equations we can solve for E, The phase space is then The flux factor is E = The differential cross section is then Or d σ = I d σ d Ω = E 2 4ME 4π 2 ME 4 M 2 = spin ME E cos θ + M = + E 2E sin 2 θ M 2 ρ = d Ω ME 4π 2 M + E cos θ 2 = d Ω E 2 4π 2 ME I = ME p k = 2π 4 δ 4 p + k p k 2p 0 2k 0 4 M 2 d 3p spin 2π 3 2p 0 E E d 3 k 2π 3 2k 0 2 e 4 { [ 6π 2 M 2 q 4 8 p k p k + p k p k ] } 8M 2 k k Institute Slide_06 QED 24 / 35

25 It is staightforward to get [ p k p k + p k p k ] M 2 k k [ = 2EE M 2 cos 2 θ 2 q2 θ ] 2M 2 sin2 2 d σ d Ω = = E α 2 4 E 2 α 2 [ M 2 4EE sin 2 θ 2 2EE M 2 cos 2 θ 2 q2 θ ] 2M 2 sin2 2 2 E 2 θ2 q2 [cos E 3 sin 4 θ θ ] 2M 2 sin2 2 2 Or d σ d Ω = α2 4E 2 sin 4 θ 2 2 θ2 q2 [cos [ + 2E M ] θ 2M 2 sin2 2 sin 2 θ ] 2 Strong interaction. Use the fact that the γpp interaction is local to parametrize the γpp matrix element as p _ Jµ p = up, s [γ µ F q 2 + i σ µνq ν 2M F 2 q 2 ] u p, s with q = p p 2 Institute Slide_06 QED 25 / 35

26 Lorentz covariance and current conservation have been used. Another useful relation is the Gordon decomposition [ _ up γ µ u p = _ p + p up µ + i σµν p ] p ν u p 2m 2m This can be derived as follows. From Dirac equation /p mup = 0, _ up /p m = 0 and _ up γ µ /p mup = 0, _ up /p mγ µ up = 0 Adding these equations, 2m _ up γ µ u p = _ up γ µ /p + /p γ µ u p = _ up p ν γ µ γ ν + p ν γ ν γ µ u p = _ up p ν } {γ 2 µ, γ ν + ] [γ 2 µ, γ ν + p ν } {γ 2 µ, γ ν ] [γ 2 µ, γ ν u p From this we get _ up γ µ u p = _ up [ p + p µ 2m + i σµν p ] p ν u p 2m F q 2, charge form factor F 2 q 2, magnetic form factor. Note that F q 2 = and F 2 q 2 = 0 correspond to point particle. The charge form factor satifies the condition F 0 =. From Q p = p Institute Slide_06 QED 26 / 35

27 we get p Q p = p p = 2E 2π 3 δ 3 p p Institute Slide_06 QED 27 / 35

28 On the other hand from Eq2 we see that p Q p = d 3 x p J 0 x p = d 3 x p J 0 0 p e ip p x = 2π 3 δ 3 p p _ up, s γ 0 u p, s F 0 = 2E 2π 3 δ 3 p p F 0 compare two equations = F 0 =. To gain more insight, write Q in terms of charge density Q = d 3 x ρ x = d 3 xj 0 x Then p J 0 x p = e iq x p J 0 0 p = e iq x F q 2 _ up, s γ 0 u p, s F q 2 is the Fourier transform of charge density distribution i.e. F q 2 d 3 x ρ x e i q x Expand F q 2 in powers of q 2, F q 2 = F 0 + q 2 F 0 + F 0 is total charge and F 0 is related to the charge radius. Calulate cross section as before, d σ d Ω = α2 4E 2 [ cos 2 θ 2 q 2 /4M 2 [G 2 E q 2 /4M 2 ] GM 2 q 2 ] sin 4 θ 2 [ + 2E M sin 2 θ 2 2M 2 sin2 θ 2 G 2 M Institute Slide_06 QED 28 / 35 ]

29 where G E = F + q2 4M 2 F 2 Experimentally, G E and G M have the form, G M = F + F 2 G E q 2 G M q 2 κ p q 2 /0.7Gev where κ p = 2.79 magnetic moment of the proton. If proton were point like, we would have G E q 2 =G M q 2 = Dependence of q 2 in Eq3 = proton has a structure. For large q 2 the elastic cross section falls off rapidly as G E G M q 4. Institute Slide_06 QED 29 / 35

30 Compton Scattering γ k + e p γ k + e p Two diagrams contribute, The amplitude is given by M γe γe = _ up ieγ µ ε µ k + _ up ieγ µ ε µ k i /p + /k m ieγν ε ν k u p i /p /k m ieγν ε ν k u p Note that if write the amplitude as M = ε µ k M µ Institute Slide_06 QED 30 / 35

31 Then we have k µm µ = ie 2 [ _ up /k /p + /k m γν ε ν k u p + _ up γ µ i ε µ k /p /k m /k u p] Using the relation we get /k = /p + /k m /p m = /p m /p /k m k µm µ = ie 2 [ _ up γ ν ε ν k u p _ up γ µ ε µ k u p] = 0 Similarly, we can show that if we repalce the polarization ε µ k by k µ the amplitude also vanishes. Using this relation we can simplify the polarization sum for the photon as follows. Let us take polarizations to be ε µ k, = 0,, 0, 0, ε µ k, 2 = 0, 0,, 0, with k µ = k, 0, 0, k, Then the polarization sum is But from k µ M µ = 0, we get Thus This is the same as the repalcement ε µ k, λ M µ 2 = M 2 + M 2 2 λ M 0 = M 3, = M 0 2 = M 3 2 εµ k, λ M µ 2 = M 2 + M M 3 2 M 0 2 = g µν M µ M ν λ ε µ k, λ ε µ k, λ g µν λ Institute Slide_06 QED 3 / 35

32 Put the γ- matrices in the numerator, [ _up M = ie 2 ε µε µ /p + /k + m ν γ γ ν u p + _ up γ ν /p /k ] + m 2p k 2p k γ µ u p Using the relations, we get /p + m γ ν u p = 2p ν u p, [ M = ie 2 _ /ε up /k /ε + 2 p ε /ε 2p k + /ε /k /ε + 2 p ε /ε ] 2p k u p Institute Slide_06 QED 32 / 35

33 The photon polarizations are, ε µ = 0, ε, with ε k = 0, ε µ = 0, ε, with ε k = 0, Lab frame, p µ = m, 0, 0, 0, = p ε = p ε = 0 and Summing over spin of the electron [ M = ie 2 _ /ε up /k /ε 2p k + /ε /k /ε ] 2p k u p { 2 M 2 = e 4 Tr /p + m [ /ε /k /ε spin 2p k + /ε /k /ε ] [ /ε /k /ε 2p k /p + m 2p k + /ε /k /ε ]} 2p k The cross section is given by d σ = I 2π 4 δ 4 p + k p k 2p 0 2k 0 4 M 2 d 3p spin 2π 3 2p 0 d 3 k 2π 3 2k 0 phase space ρ = 2π 4 δ 4 p + k p k d 3 p d 3 k 2π 3 2p 0 2π 3 2k 0 is exactly the same as the case for ep scattering and the result is ρ = d Ω 4π 2 ω 2 mω Institute Slide_06 QED 33 / 35

34 It is straightforward to compute the trace with result, This is Klein-Nishima relation. In the limit ω 0, d σ d Ω = α2 ω 2 [ ω 4m 2 ω ω + ω ] ω + 4 ε ε 2 2 d σ d Ω = α2 m 2 ε ε 2 here α m is classical electron radius. Institute Slide_06 QED 34 / 35

35 For unpolarized cross section, sum over polarization of photon, [ ε k, λ ε [ ε k, λ ]2 = k, λ ε k, λ ] 2 λλ λλ Since ε k,, ε k, 2 and k form basis in 3-dimension, completeness relation is Then ε i k, λ ε j k, λ = δ ij ˆk i ˆk j λ [ ε k, λ ε k, λ ] 2 = δij ˆk i ˆk j δij ˆk ˆk λλ i j = + cos 2 θ where ˆk ˆk = cos θ. The cross section is d σ d Ω = α2 ω 2 [ ω 2m 2 ω ω + ω ] ω sin 2 θ The total cross section, σ = πα2 m 2 dz{ [ + ω ] 3 + m z [ + ω m z ] z 2 [ + ω m z ] 2 } At low energies, ω 0, we and at high energies σ = 8πα2 3m 2 [ σ = πα2 ln 2ω ωm m + m 2 + O ω ln m ] ω Institute Slide_06 QED 35 / 35