Quantum Electrodynamics

Transcription

1 Quantum Electrodynamics Wei Wang and Zhi-Peng Xing SJTU Wei Wang and Zhi-Peng Xing (SJTU) QED 1 / 35

2 Contents 1 QED 2 e + e µ + µ 3 Helicity Amplitudes 4 ep ep 5 Compton Scattering 6 e + e 2γ Wei Wang and Zhi-Peng Xing (SJTU) QED 2 / 35

3 Quantum Electrodynamics Lagrangian density for QED, L = ψ(x)γ µ (i µ ea µ )ψ(x) m ψ(x)ψ(x) 1 4 F µνf µν Equation of motion are (iγ µ µ m)ψ(x) = ea µ γ µ ψ(x) non-linear coupled equations ν F µν = e ψ(x)γ µ ψ(x) Quantization WriteL = L 0 + L int L 0 = ψ(x)(iγ µ µ m)ψ(x) 1 4 F µνf µν L int = e ψ(x)γ µ ψ(x)a µ wherel 0,free field Lagrangian, L int is interaction part. Conjugate momenta for fermion For em fields choose the gauge L ( 0 ψ α) = iψ α(x) Wei Wang and Zhi-Peng Xing (SJTU) QED 3 / 35

4 From equation of motion π i = A 0 is not an independent field, L ( 0 A i ) = F 0i = E i ν F 0ν = eψ ψ 2 A 0 = eψ ψ A 0 = e d 3 x ψ (x,t)ψ(x,t) 4π x x Commutation relations: = e d 3 x ρ(x,t) x x [ψ α ( x, t), ψ β ( x, t)] = δ αβ δ 3 ( x x ), [ψ α ( x, t), ψ β ( x, t)] =... = 0 [ A i ( x, t), A j ( x, t)] = iδij tr( x x ) where δij tr d ( x y) = 3 k e i k ( x y) (δ (2π) 3 ij k ik j ) k 2 Commutators involving A 0 : [A 0 ( x, t), ψ α ( x, t)] = e d 3 x 4π x x [ψ ( x, t)ψ( x, t), ψ α ( x, t)] = e 4π ψ α( x,t) x x Wei Wang and Zhi-Peng Xing (SJTU) QED 4 / 35

5 Hamiltonian density H = L ψ ( 0 ψ α) α + L ( 0 A A k ) k L = ψ ( i α + βm)ψ ( E 2 B 2 ) + E A 0 + e ψγ µ ψa µ and H = d 3 xh = d 3 xψ [ α ( i e A) + βm]ψ ( E 2 B 2 ) A 0 does not appear in the interaction, But if we write then longitudinal part is 1 2 E = E l + E t where El = A 0, E t = A t 1 2 d 3 x( E 2 B 2 ) = 1 2 d 3 xe l 2 + d 3 x( E t 2 B 2 ) d 3 x E 2 l = e 4pi d 3 xd 3 y ρ( x,t)ρ( y,t) x y Coulomb interaction Without classical solutions, can not do mode expansion to get creation and annihilation operators We can only do perturbation theory. Wei Wang and Zhi-Peng Xing (SJTU) QED 5 / 35

6 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 2ω(2π) 3 ɛ(k, λ)[a(k, λ)e ikx + a (k, λ)e ikx ] w = k 0 = λ ɛ(k, λ), λ = 1, 2 with k ɛ(k, λ) = Standard choice ɛ(k, λ) ɛ(k, λ ) = δ λλ, ɛ( k, 1) = ɛ(k, 1), ɛ( k, 2) = ɛ( k, 2) It is convienent to write the mode expansion as, A µ ( x, t) = d 3 k 2ω(2π) 3 ɛ µ (k, λ)[a(k, λ)e ikx + a (k, λ)e ikx ] λ Wei where Wang and Zhi-Peng Xing (SJTU) QED 6 / 35

7 Photon Propagator Feynman propagatpr for photon is 2 ɛν (k, λ)ɛ µ (k, λ) = g µν η µ η ν ˆk µˆkν Wei Wang and Zhi-Peng Xing (SJTU) QED 7 / 35 id µν (x, x ) = 0 T (A µ (x)a ν (x )) 0 = θ(t t ) 0 A µ (x)a ν (x ) 0 + θ(t t ) 0 A ν (x )A µ (x) 0 Using mode expansion, id µν (x, x d 4 e ik(x x) ) = (2π) 4 k 2 + iɛ 2 = 1ɛ ν (k, λ)ɛ µ (k, λ) polarization vectorsɛ µ (k, λ), λ = 1, 2 are perpendicular to each other. Add 2 more unit vectors to form a complete set η µ = (1, 0, 0, 0), ˆkµ = kµ (k η)η µ (k η) 2 k 2 completeness relation is then, λ

8 then D µν (k) = 1 k 2 + iɛ [ g k µ k ν µν (k η) 2 k 2 + (k η)(k µην + ηµk ν ) (k η) 2 k 2 k2 η µ η ν (k η) 2 terms proportional tok µ will not contribute to physical processes and the last term is of the formδ µ0 δ ν0 will be cancelled by the Coulom interaction.. Wei Wang and Zhi-Peng Xing (SJTU) QED 8 / 35

9 Feynman rule in QED The interaction Hamiltonian is, H i nt = e d 3 x ψγ µ ψa µ The Feynman propagators, vertices and external wave functions are given below. Wei Wang and Zhi-Peng Xing (SJTU) QED 9 / 35

10 e + e µ + µ : Total Cross Section The momenta for this reaction is used as e + (p ) + e (p) µ + (k ) + µ (k) and the Feynman diagram is given as e (p) µ (k) e + (p ) µ + (k ) Use Feynman rule to write the matrix element as im = v(p, s )( ieγ µ )u(p, s) ig µν q 2 ū(k, r )( ieγ ν )v(k, r) where q = p + p. = ie2 q 2 v(p, s )γ µ u(p, s)ū(k, r )γ µ v(k, r), Wei Wang and Zhi-Peng Xing (SJTU) QED 10 / 35

11 e + e µ + µ Note 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. This shows the term proportional to photon momentum q µ will not contribute in the physical processes. For cross section, we needm which contains factor( vγ µ u) More generally, ( vγ µ u) = u (γ µ ) (γ 0 ) v = u γ 0 γ µ v ( vγu) = ū Γv, Γ = γ 0 Γ γ 0. Wei Wang and Zhi-Peng Xing (SJTU) QED 11 / 35

12 e + e µ + µ It is easy to see γ µ = γ µ γ µ γ 5 = γ µ γ 5 a b p = p b a unpolarized cross section which requires the spin sum, u α (p, s)ū β (p, s) = ( p + m) αβ s v α (p, s) v β (p, s) = ( p m) αβ s Wei Wang and Zhi-Peng Xing (SJTU) QED 12 / 35

13 A typical calculation is, s,s v α (p, s )(γ µ ) αβ u β (p, s)ū 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)] trace of product of γ matrices: 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/ 1 a/ 2 a/ n ] = 0, n odd With these tools M 2 = e4 q 4 Tr[(p/ m e )γ µ (p/ + m e )γ ν ]Tr[(k/ + m µ )γ µ (k/ + m µ )γ ν ] spin Wei Wang and Zhi-Peng Xing (SJTU) QED 13 / 35

14 e + e µ + µ : cross section Tr[(p/ m e )γ µ (p/ + m e )γ ν ] = Tr[p/ γ µ p/γ ν ] m 2 T r[γ µ γ ν ] = 4[p µ p ν g µν (p p ) + p µ p ν ] 4m 2 eg µν, Tr[(k/ + m µ )γ µ (k/ + m µ )γ ν ] = Tr[k/ γ µ k/γ ν ] m 2 T r[γ µ γ ν ] = 4[k µ k ν g µν (k k ) + k µ k ν ] 4m 2 µg µν for energies m µ. 1 M 2 = 8 e4 4 q 4 [(p k)(p k ) + (p k)(p k )] spin In center of mass, 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 If we set m µ = 0, E = k and q 2 = (p + p ) 2 = 4E 2, p k = p k = E 2 (1 cos θ) Wei Wang and Zhi-Peng Xing (SJTU) p k = p k QED = E 2 (1 + cos θ) 14 / 35

15 e + e µ + µ : cross section We choose the kinematics as follows: Then 1 4 spin M 2 = 8 8e4 16E 4 [E4 (1 cos θ) 2 + E 4 (1 + cos θ) 2 ] = e 4 (1 + cos 2 θ). Note that under the parity θ π θ. this matrix element conserves the parity The cross section is dσ = I 2E 2E (2π)4 δ 4 (p + p k k ) 1 M 2 d3 k d 3 k 4 (2π) 3 2ω (2π) 3 2ω spin Wei Wang and Zhi-Peng Xing (SJTU) QED 15 / 35

16 use the δ-function to carry out integrations. introduce the quantityρ, 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 = 1 4π 2 δ(2e ω ω ) d3 k 4ωω = 1 32π 2 δ(e ω) k2 dkdω ω 2 = dω 32π 2 The flux factor is I = 1 (p 1 p 2 ) E 1 E 2 m 2 1 m2 2 = 1 2 E 2 2E2 = 2 The differential crossection is then Or dσ = E 2 (1 4 spin M 2 ) dω 32π 2 dσ dω = α2 16E 2 (1 + cos2 θ) e 2 Wei Wang and Zhi-Peng Xing (SJTU) QED 16 / 35

17 e + e hadrons One of the interesting procesess ine + e collider is the reaction e + e hadrons According to QCD, theory of strong interaction, 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 toµ + µ only in their charges cross section forq q as σ(e + e q q) = 3(Q 2 q) 4α2 π 3s = 3(Q 2 q)σ(e + e µ + µ ), Q q is the electric charge of quark q. The factor of 3 because each quark has 3 colors. Then R = σ(e+ e hadrons) σ(e + e µ + µ ) This has used the quark-hadron duality. = 3( i Q 2 i ), Wei Wang and Zhi-Peng Xing (SJTU) QED 17 / 35

18 e + e hadrons 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, R = σ(e+ e hadrons) σ(e + e µ + µ ) which is not far from the reality. = 3[( 2 3 )2 + ( 1 3 )2 + ( 1 3 )2 ] = 2 Wei Wang and Zhi-Peng Xing (SJTU) QED 18 / 35

19 e + e hadrons 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, R = σ(e+ e hadrons) σ(e + e µ + µ ) which is not far from the reality. = 3[( 2 3 )2 + ( 1 3 )2 + ( 1 3 )2 ] = 2 Wei Wang and Zhi-Peng Xing (SJTU) QED 19 / 35

20 Helicity Amplitudes In the following, we will give the helicity amplitudes. One can parametrize the momentum as The spinor on the direction (θ, φ) is given as The polarization vector corresponding to the momentum (θ, φ) is given as: Wei Wang and Zhi-Peng Xing (SJTU) QED 20 / 35

21 e µ e µ and ep ep e(k)+p(p) e(k )+p(p ) e (k) e(k ) p(p) 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) = ū(p, s )( ieγ µ )u(p, s)( igµν )ū(k, r )( ieγ ν )u(k, r) q 2 = ie2 ū(p, s )γ µ u(p, a)ū(k, r )( ieγ ν )u(k, r) q 2 where q = k- k. For unploarized cross section, sum over the spins, 1 M(e + p e + p) 2 = e4 4 q 4 T r[( p + M)γ µ ( p + M)γ ν ]T r[( k + M e )γ µ Wei Wang and Zhi-Peng Xing (SJTU) QED 21 / 35

22 Again neglectm e. Compute the traces T r[ k γ µ kγ ν ] = 4[k µ k ν g µν (k k ) + k µ k ν ] T r[( 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 1 4 spin M(e + p e + p) 2 = e4 q 4 {8[(p k)(p k ) + (p k)(p k )] 8M 2 (k Wei Wang and Zhi-Peng Xing (SJTU) QED 22 / 35

23 More useful to use the laboratoy frame Then p µ = (M, 0, 0, 0), k µ = (E, k), k µ = (E, k ) p k = ME, p k = ME, k k = EE (1 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 (1 cos θ) Differential cross section is dσ = (2π) 4 δ 4 (p + k p k ) 1 I 2p 0 2k d3p M spin d 3 k (2π) 3 2p 0 (2π) 3 2k 0 The phase space is ρ = (2π) 4 δ 4 (p + k p k d 3 p d 3 k ) (2π) 3 2p 0 (2π) 3 2k 0 = 1 4π 2 δ(p 0 + k 0 p 0 k 0) d3 k 2p 0 2k 0 where p 0 = M 2 + ( p + k k ) 2 = M 2 + ( k k ) 2 Wei Wang and Zhi-Peng Xing (SJTU) QED 23 / 35

24 Use the momenta in lab frame, ρ = 1 4π 2 δ(m + E p 0 E ) k 2 dk dω 2p 0 2E = 1 4π 2 δ(m + E p 0 E ) E de dω p 0 Let Then and ρ = 1 4π 2 x = E + p 0 + E dx = de (1 + dp 0 de ) = de ( p 0 + E E cos θ ) dωe dx δ(x M) (p 0 + E E cos θ) = 1 dωe 4π 2 M + E(1 cos θ) From the argument of the δ function we get the relation,m = x = E + p 0 + E Solve for E, E ME E = = Wei Wang and Zhi-Peng Xing (SJTU) QED 24 / 35 p 0

25 The differential cross section is then dσ = 1 I Or dσ dω = 1 4ME 1 1 (2π) 4 δ 4 (p + k p k ) 1 2p 0 2k E 2 1 4π 2 ME 4 2 d3p M spin d 3 k (2π) 3 2p 0 (2π) 3 2k 0 spin M 2 = ( E E )2 1 16π 2 M 2 e 4 q 4 {8[(p k)(p k ) + (p Wei Wang and Zhi-Peng Xing (SJTU) QED 25 / 35

26 It is staightforward to get Or [(p k)(p k ) + (p k)(p k )] M 2 (k k ) = [(p k)((p + k k ) k ) + (p k )(p + k k ) k M 2 (k k )] dσ dω = (E E = 2EE M 2 + M 2 EE (1 cos θ)( q2 2M 2 1) = 2EE M 2 [cos 2 θ 2 q2 2M 2 sin2 θ 2 ] )2 α2 M 2 1 (4EE sin 2 θ 2 )2EE M 2 [cos 2 θ 2 = α2 4 E 1 E 3 sin 4 θ [cos 2 θ 2 2 dσ dω = α2 1 [cos 2 θ 2 q2 sin 2 θ 2M 2 2 ] 4E 2 sin 4 θ 2 [1 + ( 2E M ) sin2 θ 2 ] q2 2M 2 sin2 θ 2 ] q2 2M 2 sin2 θ 2 ] Wei Wang and Zhi-Peng Xing (SJTU) QED 26 / 35

27 p Q p = p p = 2E(2π) 3 δ 3 ( p p Wei Wang and Zhi-Peng Xing (SJTU) QED ) 27 / 35 Now include the strong interaction. Use the fact that theγp P interaction is local to parametrize theγp P matrix element as p J µ p = ū(p, s )[γ µ F 1 (q 2 ) + iσ µνq ν 2M F 2(q 2 )]u(p, s) with q = Lorentz covariance and current conservation have been used. Another useful relation is the Gordon decomposition ū(p )γ µ u(p) = ū(p ) = ū(p )[ (p + p ) µ 2m + iσµν (p p) ν ]u(p) 2m F 1 (q 2 ), charge form factor F 2 (q 2 ), magnetic form factor. Note thatf 1 (q 2 ) = 1andF 2 (q 2 ) = 0 correspond to point particle. The charge form factor satifies the conditionf 1 (0) = 1.Form Q p = p we get

28 F 1 (q 2 )is the Fourier transform of charge density distribution i.e. Wei Wang and Zhi-Peng Xing (SJTU) QED 28 / 35 On the other hand from Eq(2) we see that p Q p = d 3 x p J 0 (x) p = d 3 p J 0 (0) p e i(p p) x = 2π) 3 δ 3 ( p p )ū(p, s )γ 0 u(p, s)f 1 (0) = 2E(2π) 3 δ 3 ( p p )F 1 (0) compare two equations F 1 (0) = 1. To gain more insight, write Q in terms of charge density Q = d 3 xρ(x) = d 3 xj 0 (x) then p J 0 (x) p = e iq x p J 0 (0) p = e iq x F 1 (q 2 )ū(p, s )γ 0 u(p, s)

29 where G E = F 1 + G M = F 1 + F 2 Experimentally,G E andg M have the form, q2 4M 2 F 2 G E (q 2 ) G M(q 2 ) χ p 1 (1 q 2 /0.7Gev 2 ) 2 whereχ p = 2.79 magnetic moment of the proton. If proton were point like, we would haveg E (q 2 ) = G M (q 2 ) = 1 Dependence ofq 2 in Eq(3) proton has a structure. For largeq 2 the elastic cross section falls off rapidly asg E G M q 4. Wei Wang and Zhi-Peng Xing (SJTU) QED 29 / 35

30 Compton Scattering M = ie 2 ɛ ɛ [ū(p )γ µ p+ k + m γ ν u(p) + ū(p )γ µ p+ k + m Wei Wang and Zhi-Peng Xing (SJTU) QED γ ν u(p)] 30 / 35 Two diagrams contribute, γ(k) + e(p) γ(k ) + e(p ) The amplitude is given by M(γe γe) = ū(p )( ieγ µ )ɛ µ(k i ) p k m ( ieγν )ɛ ν (k)u(p) +ū(p )( ieγ µ i )ɛ µ (k) p k m ( ieγν )ɛ ν(k )u(p) Put theγ matrices in the numerator,

31 Using the relations, ( p + m)γ ν u(p) = 2p ν u(p) we get M = ie 2 ū(p )[ ɛ k ɛ + 2(p ɛ) ɛ 2p k + ɛ k ɛ + 2(p ɛ) ɛ 2p k ]u(p) Wei Wang and Zhi-Peng Xing (SJTU) QED 31 / 35

32 The photon polarizations are, ɛ µ = (0, ɛ), with ɛ k = 0, ɛ µ = (0, ɛ ), with ɛ k = 0, Lab frame,p µ = (m, 0, 0, 0), (p ɛ) = (p ɛ ) = 0and M = ie 2 ū(p )[ ɛ k ɛ 2p k + ɛ k ɛ 2p k ]u(p) Summing over spin of the electron 1 2 spin M 2 = e 4 T r{( p + m)[ ɛ k ɛ 2p k + ɛ k ɛ 2p k ]( p + m)[ ɛ k ɛ 2p k + ɛ k ɛ 2p k The cross section is given by dσ = 1 I 1 1 (2π) 4 δ 4 (p + k p k ) 1 2p 0 2k d3p M spin d 3 k (2π) 3 2p 0 (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 Wei Wang and Zhi-Peng Xing (SJTU) QED 32 / 35

33 It is straightforward to compute the trace with result, dσ dω = α2 4m 2 (ω ω )2 [ ω ω + ω ω + 4(ɛ ɛ ) 2 2] This is Klein-Nishima relation. In the limit ω 0, here α mis classical electron radius. dσ dω = α2 m 2 (ɛ ɛ ) 2 Wei Wang and Zhi-Peng Xing (SJTU) QED 33 / 35

34 For unpolarized cross section, sum over polarization of photon, [ɛ(k, λ) ɛ (k, λ )] 2 = [ ɛ(k, λ) ɛ (k, λ )] 2 λλ λλ Since ɛ(k, 1), ɛ(k, 2)anf k form basis in 3-dimension, completeness relation is ɛ i (k, λ)ɛ j (k, λ) = δ ij ˆk iˆkj then λ λλ [ɛ(k, λ) ɛ (k, λ )] 2 = (δ ij ˆk iˆkj )(δ ij ˆk iˆk j) = 1 + cos 2 θ whereˆk ˆk = cos θ. The cross section is dσ dω = α2 2m 2 (ω ω )2 [ ω ω + ω ω sin2 θ] The total cross section, σ = πα2 1 1 dz{ [1 + ω m (1 + 1 z)]3 [1 + ω m (1 z)] 1 z 2 } (1 z)]2 m 2 1 [1 + ω m Wei Wang and Zhi-Peng Xing (SJTU) QED 34 / 35

35 e + e 2γ and cross symmetry Wei Wang and Zhi-Peng Xing (SJTU) QED 35 / 35