QFT Perturbation Theory

Transcription

1 QFT Perturbation Theory Ling-Fong Li Institute) Slide_04 1 / 44

2 Interaction Theory As an illustration, take electromagnetic interaction. Lagrangian density is The combination is the covariant derivative. Equations of motion L = ψ x ) γ µ i µ ea µ ) ψ x ) mψ x ) ψ x ) 1 4 F µνf µν D µ = µ + ea µ ) i γ µ µ m ) ψ x ) = ea µ γ µ ψ non-linear coupled equations υ F µν = eψγ µ ψ No exact solutions are known. Expansion in terms of Fourier components, x ) ψ, t = s d 3 p [b p, s, t) u p, s) e i p x + d p, s, t) υ p, s) e i p x ] 2π) 3 2E p Then the operators b and d time dependent and the time evolution is controlled by the interaction terms. Quantization Write L = L 0 + L int L 0 = ψ i γ µ µ m ) ψ 1 4 F µνf µν L int = eψγ µ ψa µ Institute) Slide_04 2 / 44

3 where L 0,free field part, while L int interaction. Conjugate momenta L 0 ψ α ) = i ψ α x ) For electromagnetic fields, choose the gauge A = 0, Radiaiton Gauge then From equation of motion π i = L 0 A i ) = F 0i = E i ν F 0ν = eψ ψ = 2 A 0 = eψ ψ We can t take A 0 to be zero but A 0 can be expressed in terms of other 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 4π x x } = δ αβ δ 3 ) x x {ψ α x, t), ψ β x, t = i δ tr x ) ij x ) } =... = 0 Institute) Slide_04 3 / 44

4 Commutators with A 0, [ A 0 x, t), ψ α ] x, t) = e d 3 x 4π x x x ) = e ψ α, t 4π x x [ ψ x, t)ψ x )], t), ψ α x, t We have used the relationwe have used the relation 2 1 4π x = δ 3 ) x x x This can be seen as follows. In the spherical coordinates the Laplacian is of the form, From this we get 2 = 1 r 2 r 2 ) + 1 r r r 2 sin θ 2 1 r ) = 1 r 2 r r 2 1) r 2 To deal with the singularity at r 0, consider the integral 1 d 3 x 2 r ) = d 3 x [ sin θ ) + θ θ 1 r 2 sin 2 θ ) = 0, if r = 0 ) 1 ] = d S [ r 2 φ 2 ) 1 ] r Institute) Slide_04 4 / 44

5 We can take the surface to) be the surface of a sphere with radius a centered around the origin so that d S = a 2 ˆr d Ω. Using 1 = ˆr, we get r r π x = δ 3 ) x x x Hamiltonian [ H = d 3 x H = d 3 x {ψ ] α i e A ) + βm ψ + 1 E 2 + B 2) } 2 No A 0 in the interaction. If we write Then The longitudinal part is E = El + E t where El = A 0, E t = A t 1 E d 3 x 2 + B 2) = 1 d 3 x E 2 l + d 3 Et x 2 + B 2) d 3 x E 2 l = 1 d 3 xa 0 2 A 0 = e 2 d 3 xd 3 y ρ x, t) ρ y, t) 2 2 4π x y Institute) Slide_04 5 / 44

6 Perturbation Theory Can t solve the classical equations of motion. Can t do mode expansion to introduce a and a.the only approximation we can do in field theory is the perturbation theory. We will now set up the framework for the perturbation. Physical states In high energy physics, we study the scattering processes. Assume interactions all short-range, far away from interaction region, particles propagate as free particles. Choose the physical states to be eigensates of energy momentum operators, Satisfy requirements; 1 eigenvalues p µ all in forward light cone, P µ Ψ = p µ Ψ p 2 = p µ p µ 0, p non-degenerate Lorentz invariant ground state 0 > with zero energy, p 0 0 = 0, = p 0 = 0 3 There exists stable single particle states p i with p 2 i = m 2 i for each stable particle. 4 vaccum and one particle states form discrete spectra in p µ Institute) Slide_04 6 / 44

7 assume interactions do not violently the spectrum of states. there is no room to describe bound states Spectral Representation Eventhough it is diffi cult to get exact results in the interacting field theories, it is still possible to get some useful relation independent of the details of the interaction. Here we will discuss one such example of spectral representation of commutator or propagator of interacting scalar fields. We write the vacuum expectation value of commutator of scalar fields as i x, y ) = 0 [φ x ), φ y )] 0 = 0 φ x ) φ y ) φ y ) φ x )) 0 Inserting a complete set of eigenstates with definite energy and mometum between two field operators, we get for the first term n where we have use the relation The commutator function is then 0 φ x ) n n φ y ) 0 = e ipn x e ipn y 0 φ 0) n n φ 0) 0 n 0 φ x ) n = 0 e ip x φ 0) e ip x n = e ipn x 0 φ 0) n i x, y ) = 0 φ 0) n n φ 0) 0 [e ipn x y ) e ipn x y ) ] = i x y ) n We see that the commutator function is function of the difference x y rather than x, y separately. It is useful to group togather states with same p n by introducing the identity, 1 = d 4 qδ 4 q p n) Institute) Slide_04 7 / 44

8 so that we can write where x y ) = i 2π) 3 = i 2π) 3 d 4 q 2π) 3 δ 4 q p n) 0 φ 0) n 2 [e ipn x y ) e ipn x y ) ] n d 4 qρ q) [e ipn x y ) e ipn x y ) ] ρ q) = 2π) 3 δ 4 q p n) 0 φ 0) n 2 n is usually called the spectral function. It is clear that ρ q) is a Lorentz scalar and can be written as ρ q) = ρ q 2) θ q 0) due to the fact that all physical momenta are non-zero only in the forward light-cone. Thus ρ q 2) vanishes for q 2 < 0 and is positive semi-definite for q 2 > 0.It is convienent to write the commutator function as x y ) = i 2π) 3 d 4 qρ q 2) θ q 0) [e iq x y ) e iq x y ) ] 1) = d σ 2 ρ σ 2) d 4 qδ q 2 σ 2) iq x y ) ε q 0) e = d σ 2 ρ σ 2) x y, σ 2) 0 where x y, σ 2) = d 4 qδ q 2 σ 2) iq x y ) ε q 0) e Institute) Slide_04 8 / 44

9 is the commutator function for free scalar field with mass σ 2.In this Eq 1), the commutator function of the interacting field is experssed as integral over free field commutatotor function with different masses is called the spectral representation of the commutator function. Exactly the same procedure can be applied to the propagator function to get F x y ) = i 0 T φ x ) φ y )) 0 = d σ 2 ρ σ 2) F x y, σ 2 ) 0 Or in momentum space F p 2 ) = d σ 2 ρ σ 2) 1 0 p 2 σ 2 + i ε where F p 2 ) = d 4 x F x ) e ip x, F p 2 ) = d 4 x F x ) e ip x Institute) Slide_04 9 / 44

10 In-fields and in-states asymptotic conditions Consider Eq. of motion L = 1 2 µ φ )2 µ2 0 2 φ2 + λ 4! φ4 conjugate momenta ) + µ 2 λ 0 φ = j x ) = 3! φ3 π x ) = L 0 φ = 0φ Commutation relations [π x, t), φ y, t)] = i δ 3 x y ) [π x, t), π y, t)] = [φ x, t), φ y, t)] = 0 At t =, φ in x ) creates free particle with physical mass µ. + µ 2 ) φ in x ) = 0 we allow physical mass µ to be different from µ 0. Assume that φ in x ) transforms same way as φ x ). In particular, [ pµ, φ in x ) ] = i µ φ in x ) φ in x ) creates one particle state from vacuum. Expand φ in x ) in terms of free solution of Klein-Gordon equation, φ in x ) = d 3 k [ a in k) f k x ) + a in k) fk x ) ] 1 f k x ) = e 2π) ik x 3 2w k Institute) Slide_04 10 / 44

11 Invert this expansion a in k) = i d 3 xfk x ) 0 φ in x ) We also have [p µ, a in k)] = k µ a in k), [ p µ, a in k) ] = k µ a in k) States are defined by k 1, in = 2π) 3 2w k a in k) 0 k 1, k 2,...k n in = [ i ] 2π) 3 2w ki a in k i ) 0 Institute) Slide_04 11 / 44

12 Relation between φ in x ) and φ x ) The field equations for interacting fields Or + µ 2 0) φ x ) = j x ) + µ 2 ) φ x ) = j x ) + δµ 2 φ x ) = j x ) δµ 2 = µ 2 µ 2 0 and for the in-fields, + µ 2 ) φ in x ) = 0 Formally relates φ x ) to φ in x ) by Green s function, z φin x ) = φ x ) d 4 y ret x y, µ 2 ) j y ) where x + µ 2) ret x y, µ 2 ) = δ 4 x y ), ret x y, µ 2 ) = 0 for x 0 < y 0 is the retarded Green s function. The factor z is included to permit the normalization of φ in to unit amplitude for its matrix element to create one-particle state from vacuum. This suggests that φ x ) z φ in x ) as x 0, 2) because the retarded Green s function vanishes as x 0. This relation which relates 2 operators can be viewed as strong convergence relation.it turns out that this leads to contradiction. To see this take matrix element of Eq??), between vacuum and one-particle state z 0 φin x ) p = 0 φ x ) p d 4 y ret x y, µ 2 ) 0 j y ) p Institute) Slide_04 12 / 44

13 The second term vanishes because 0 j y ) p = + µ 2) 0 φ x ) p = + µ 2) e ip x 0 φ 0) p = p 2 + µ 2) e ip x 0 φ 0) p = 0 So we get z 0 φin x ) p = 0 φ x ) p and we interpret z as the probabilty amplitude for the interacting field to produce one-particle from the vacuum. Using the mode expansion for the free field φ in x ), we get 0 φ in x ) p = d 3 k e 2π) ik x 0 a in k) p = 3 2w k d 3 k e 2π) ik x 2π) 3 2w p δ 3 p k) = e ip x 3 2w k and 0 φ 0) p = z 0 φ in 0) p Its contribution to the spectral function is ρ q 2) 1particle = d 3 p 2π) 3 2π) 3 δ 4 p q) 0 φ 0) p 2 1 = Z δ p 0 q 0) = Z δ q 2 µ 2) θ q 0) 2w p 2ω p Put this into the spectral representation in Eq 1) x y ) = Z x y, µ 2) + d 4µ 2 σ2 ρ σ 2) x y, σ 2) Institute) Slide_04 13 / 44

14 The lower limit in the integral corresponds to the threshold for all states other than the one particle state. Differentiate this with respect to x 0 and set x 0 = y 0 to get the canonical equal time commutator, lim i x y ) = 0 [ 0 φ x, x x 0 y 0), φ y, x 0)] 0 = i δ 3 x y ) 0 x 0 = lim [iz x y, µ 2) + x 0 y 0 x d 0 4µ 2 σ2 ρ σ 2) x y, σ 2) ] = i δ 3 x y ) [Z + d 4µ 2 σ2 ρ σ 2) ] where we have used lim x y, µ 2) = i δ 3 x y ) x 0 y 0 x 0 This follows from the fact that x y, σ 2) simply corresponds to commutator for free scalar fields. Thus we get the relation 1 = Z + d 4µ 2 σ2 ρ σ 2) which implies the inequality 0 Z < 1 3) The inequality Z < 1 is due to the positivity of the spectral function ρ q 2). Now we use this inequality to show that the asymptotic condition in Eq 2) will lead to contradiction. Recall that both the interacting field φ x ) and the free field φ in x ) satisfy the same equal time commutation relation, [ 0 φ x, x 0), φ y, x 0)] = i δ 3 x y ), [ 0 φ in x, x 0), φ in y, x 0)] = i δ 3 x y ) Institute) Slide_04 14 / 44

15 These relations can be consistent with the asymptotic relation φ x ) z φ in x ) as x 0, in Eq 2) only if Z = 1, which contradicts the inequality in Eq 3). Correct asymptotic condition Lehmann, Symanzik, and Zimmermann) Let α, β be any two normalizable states, φ f t) is defined φ f t) i d 3 xfk x, t) 0 φ x, t) with f k x, t) is an arbitrary normalizable solution to Klein-Gordon equation. The correct asymptotic condition is + µ 2 ) f = 0 lim x 0 α φf t) β = z α φ f in t) β with φf in t) = i d 3 xf x, t) 0 φ in x, t) Institute) Slide_04 15 / 44

16 Out fields and out states Similar procedure applies to φ out + µ 2 0) φout x ) = 0 Asymptotic condition φ out x ) = d 3 k [ a out k) f k x ) + a out k) fk x ) ], [ p µ, a out k) ] = k µ a out k) lim t α φf t) β = z α φ f out t) β S-matrix Scattering processes: start n non-interacting particles.they interact when close to each other. After interaction, m particles seperate Initial state S-matrix Final state Introduce S-operator p 1, p 2,..., p n, in = α, in p 1, p 2,..., p m, out = β, out β, out β, in S S βα β, out α, in β, out S 1 = β, in then the S matrix element can be written as a matrix element of S operator between 2 in states, Institute) Slide_04 16 / 44

17 LSZ reduction set up the framework to compute S βα. Consider S β,αp = β, out α, p, in Using creation operator for the in state, S β,αp = β, out α, p, in = 2π) 3 2w p β, out a in p) α, in = 2π) 3 2w p β, out a out p) α, in + β, out [ a in p) a out p) ] α, in ] [ = N β p, out α, in i β, out d 3 xf p x ) ] 0 [φ in x ) φ out x )] α, in Here β p, out is state β, out by removing a particle with momentum p and N = 2π) 3 2w p Use the symptotic conditions and the identity lim x 0 α φ in x ) β = 1 lim α φ x ) β, α φ z t out lim x 0 ) d 3 xg 1 x ) 0 g 2 x ) = x ) β = 1 z lim t α φ x ) β d 4 x [ g 1 x ) 2 0g 2 x ) 2 0g 1 x ) g 2 x ) ] Institute) Slide_04 17 / 44

18 d 3 xf p x ) 0 [φ in x ) φ out x )] = d 4 x [ 2 0f p x ) φ x ) f p x ) 2 0φ x ) ] = d 4 xf p x ) + µ 2) φ x ) we get the reduction formula, β, out α, p, in = N β p, out α, in + i e ip x d 4 x + µ 2) β, out φ x ) α, in z where we have used 2 0 f p x ) = 2 i µ 2) f p x ) and carry out the integration by parts. To remove a particle with momentum p from β β, out φ x ) α, in = γp, out φ x ) α, in = N γ, out a out p ) φ x ) α, in = [ γ, out φ x ) a in p ) α, in γ, out a out p ) φ x ) φ x ) a in p ) α, in ] = γ, out φ x ) α p, in i d 3 y γ, out φ out y ) φ x ) φ x ) φ in y )) α, in 0 f p y ) same procedure as before = γ, out φ x ) α p, in i ) d 3 y lim lim γ, out T φ y ) φ x ))) α, in 0 f z y 0 y 0 p y ) Institute) Slide_04 18 / 44

19 β, out φ x ) α, in = { γ, out φ x ) α p, in } + i d 4 y γ, out T φ y ) φ x )) α, in y + µ 2) e ip x z remove all particles from "in" and "out" state p 1,..., p n, out q 1,..., q m, in = i ) m+n m z i=1j=1 n d 4 x i d 4 y j e iq i x i 0 T φ y 1)...φ y m ) φ x 1)...φ x m )) 0 x + µ 2) yj + µ 2) e ip j x j for all p j = q i remove all particles from "in" and "out" state p 1,..., p n, out q 1,..., q m, in = i ) m+n m z i=1j=1 n d 4 x i d 4 y j e iq i x i 0 T φ y 1)...φ y m ) φ x 1)...φ x m )) 0 x + µ 2) yj + µ 2) e ip j x j for all p j = q i Institute) Slide_04 19 / 44

20 In and Out fields for Fermions generalization to fermions. in-field ψ in x ) = where d 3 [ p bin p, s) U p,s x ) + d in p, s) V p,s x ) ] s U p,s x ) = 1 2π) 3 2E p u p, s) e ip x V p,s x ) = 1 2π) 3 2E p υ p, s) e ip x Inversion b in p, s) = b in p, s) = d 3 xu p,s x ) ψ in x ) d in p, s) = d 3 x ψ in x ) V p,s x ) d 3 x ψ in x ) U p,s x ) d in p, s) = d 3 xv p,s x ) ψ in x ) Reduction formula for fermions 1 remove electron from in-state β, out α; ps, in = i z2 d 4 x β, out ψ α x ) α, in i γ µ µ m ) ip x u p, s) e αβ

21 2 remove positronanti-particle) from in-state β, out α; ps, in = i d 4 xe ip x ν α p, s) i γ µ µ m ) z2 αβ β, out ψ β x ) α, in 3 remove electron from out-state β; p s, out α, in = i z2 d 4 xu α p, s ip x ) e i γ µ µ m ) αβ β, out ψ β x ) α, in 4 remove positron from out-state β; p s, out α, in = i d 4 x β, out ψ α x ) α, in i γ µ µ m ) z2 αβ υ p, s ) e ip x

22 U matrix relation between interacting fields φ x ), π x ) andfree fields φ in x ), π in x ). Assume φ x, t) = U 1 t) φ in x, t) U t), π x, t) = U 1 t) π in x, t) U t) In-fields satisfy, 0 φ in x ) = i [H in φ in, π in), φ in ], 0 π in x ) = i [H in φ in, π in), π in] 4) where H in φ in, π in) is free field Hamiltonian with mass µ. Time evolution of φ x ), π x ) is, We find 0 φ x ) = i [H φ, π), φ], 0 π x ) = i [H φ, π), π] i U t) =H I t) U t) t 5) Define U t, t ) U t) U 1 t ) time evolution operator Eq5) becomes, convert this to integral equation i U t, t ) t = H I t) U t, t ) with U t, t) = 1 t U t, t ) = 1 i dt 1 H I t 1) U t 1, t ) t Institute) Slide_04 22 / 44

23 which includes the initial condition. Iterate this equation assuming H I is "small", t t t1 U t, t ) = 1 i dt 1 H I t 1) + i) 2 dt 1 H I t 1) dt 2 H I t 2) +... t t t t t1 tn 1 + i) n dt 1 dt 2... dt n H I t 1) H I t 2)...H I t n) +... t t t The second term can be written as U 2) = i) 2 t = i) 2 t = i) 2 t t1 dt 1 dt 2 H I t 1) H I t 2) t t t dt 2 dt 1 H I t 1) H I t 2) t t 2 t dt 1 dt 2 H I t 2) H I t 1) t t 2 where we have interchange the order of integration. Renaming t 1 and t 2,we get U 2) = i) 2 t t dt 1 t t 2 dt 2 H I t 2) H I t 1) We can use time-ordered product to combine these two equivalent expression so that the t 2 integration goes from t to t U 2) = i)2 t t dt 1 dt 2 T H I t 2) H I t 1)) 2 t t We can generalize these steps to higher terms in U so that This can be written as Institute) Slide_04 23 / 44

24 U t, t ) = 1 + n=1 = T exp i) n t t t dt 1 dt 2... dt n T H I t 1) H I t 2)...H I t n)) n! t t t [ t ]) i d 4 x H I φ t in, π in) Institute) Slide_04 24 / 44

25 Perturbation Expansion of Vaccum expectation value From LSZ reduction, S matrix is of the form, τ x 1, x 2,..., x n) = 0 T φ x 1) φ x 2)...φ x n)) 0 Using U matrix, write this in terms of φ in τ = 0 T U 1 t 1) φ in x 1) U t 1, t 2) φ in x 2) U t 2, t 3)...U t n 1, t n) φ in x n) U t n) ) 0 = 0 T U 1 t) U t, t 1) φ in x 1)...φ in x n) U t n, t ) U t ) ) 0 Let t > t 1...t n > t,then we can pull U 1 t) and U t ) out of the time-ordered product, and combine U s and φ in τ = 0 U 1 t) TU t, t 1) φ in x 1)...φ in x n) U t n, t ))U t ) 0 [ t ] = 0 U 1 t) T φ in x 1)...φ in x n) exp i H I t ) dt )U t ) 0 t Theorem: 0 is an eigenstate ofu t) as t. Then we get the result p, α, in U t) 0 = 0 as t for all in-states.this means This completes the proof. U t) 0 = λ 0 λ some phase as t Institute) Slide_04 25 / 44

26 Similarly we can show that These phases can be written as U t) 0 = λ + 0 λ + some phase as t t λ λ + = [ 0 T exp i H I t ) dt ) 0 ] 1 t Now we have vacuum expectation value τ x 1, x 2,..., x n) completely in terms of φ in, t τ x 1, x 2,..., x n) = 0 U 1 t)t φ in x 1) φ in x 2)...φ in x n) exp i t t = λ λ + 0 T φ in x 1) φ in x 2)...φ in x n) exp i H I t ) dt ) t ) H I t ) dt ) U t) 0 ) 0 or τ x 1, x 2,..., x n) = 0 T φ in x 1) φ in x 2)...φ in x n) exp i ) H I t ) dt ) 0 0 T exp i ) H I t ) dt ) 0 Institute) Slide_04 26 / 44

27 For computation we expand the exponential of H I, to write τ x 1, x 2,..., x n) = i) m m! dy 1...dy m 0 T φ in x 1)...φ in x n) H I y 1) H I y 2)...H I y m )) 0 m=0 i) m m! dy 1...dy m 0 T H I y 1) H I y 2)...H I y m )) 0 m=0 Wick s theorem To compute product of free fields φ in between vacuum, convert to normal ordering. Results are summarized below; T φ in x 1)...φ in x n)) = : φ in x 1)...φ in x n) : + [ 0 φ in x 1) φ in x 2) 0 : φ in x 3) φ in x 4)...φ in x n) : +permutations] +[ 0 φ in x 1) φ in x 2) 0 0 φ in x 3) φ in x 4) 0 : φ in x 5)...φ in x n) : +permutations]... { [ 0 φin x + 1) φ in x 2) 0 0 φ in x 3) φ in x 4) φ in x n 1) φ in x n) 0 + permutations] neven [ 0 φ in x 1) φ in x 2) φ in x n 2) φ in x n 1) 0 φ in x n) + permutations] n odd This can be proved by induction. Illustrate this for n=2. Difference between T ) and :) :is a c-number, T φ in x 1) φ in x 2)) =: φ in x 1) φ in x 2) : + c number ) Institute) Slide_04 27 / 44

28 take matrix element between vacuum state, 0 T φ in x 1) φ in x 2)) 0 = c number ) Then T φ in x 1) φ in x 2)) =: φ in x 1) φ in x 2) : + 0 T φ in x 1) φ in x 2)) 0 Most useful application of Wick s theorem 0 T φ in x 1)...φ in x n)) 0 = 0 n odd 0 T φ in x 1)...φ in x n)) 0 = [ 0 T φ in x 1) φ in x 2)) 0 0 T φ in x 3) φ in x 4)) 0...] permu n even Notation φ in x 1) φ in x 2 ) = 0 T φ in x 1) φ in x 2)) 0 Contraction Example: Institute) Slide_04 28 / 44

29 Feynman Propagators From Wick s theorem most important quantity is vacuum expecation of two free fields, called Feynman propagator. For complex scalar field Fermion field 0 T φ in x ) φ in y )) 0 = i F x y, µ 2 ) = i with i F k) = = i 0 T φ in x ) φ in y )) 0 = i F d 4 k 2π) 4 e ik x y ) i F k) i k 2 µ 2 + i ε x y, µ 2 ) = i d 4 k e ik x y ) 2π) 4 k 2 µ 2 + i ε d 4 k e ik x y ) 2π) 4 k 2 µ 2 + i ε photon field 0 T ψ in α x ) ψin β y )) 0 = is F x y, m) αβ = i d 4 p ip x y ) e 4 2π) γ µ p µ + m ) αβ p 2 m 2 + i ε = d 4 p 2π) 4 e ip x y ) is F p) αβ Institute) Slide_04 29 / 44

30 0 T A in µ x ) A in υ y )) 0 = idf tr x y ) = i d 4 k e ik x y ) 2π) 4 k 2 + i ε ) k µ k ν k η) k µ η ν + k ν η µ g µν k η) 2 + k 2 k η) 2 k 2 η µ η ν k 2 k η) 2 k 2 where η µ = 1, 0, 0, 0) It can be shown that in QED only term contributes is " g µν " as a consequence of the gauge invariance. Institute) Slide_04 30 / 44

31 Graphical representation Each line propagator) represents a contraction in Wick s expansion e.q. y x i F x y, µ 2 ) β y > α x is F x y, m) αβ ν µ Vaccum Amplitude In the denominator of τ function, there are no external lines id tr F x y ) i) m d 4 y 1...d 4 y m 0 T H I φ m=0 m! in y 1))...H I φ in y m ))) 0 e.q. 2nd order term for the case H I = λ 3! : φ 3 in : Institute) Slide_04 31 / 44

32 closed loop diagram :graphs with no external lineslines with open end) disconnected diagram :a subgraph not connected to any external lines connected diagram :graph not disconnected Institute) Slide_04 32 / 44

33 All graphs appearing in the numerator of the τ function can be seperated uniquely into connected and disconnected parts. It turns out that disconnected part is cancelle by those in denominator. Example :H I = λ 3! φ3 in φ q 1) + φ q 2) φ p 1) + φ p 2) S βα = β, out α, in = p 1, p 2, out q 1, q 2, in ) i 4 = d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 y 1 e ip 2 y 2 y1 + µ 2) y2 + µ 2) 0 T φ y 1) φ y 2) φ x 1) φ x 2)) 0 > z = x1 + µ 2) x2 + µ 2) e iq 1 x 1 e iq 2 x 2 ) i 4 d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 µ 2 p 2 ) z 1 µ 2 p2 2 ) µ 2 q1 2 ) µ 2 q2 2 ) τ y 1, y 2, x 1, x 2) e ip 1 y 1 +p 2 y 2 ) e iq 1 x 1 +q 2 x 2 ) Perturbation expansion of τ function i) τ y 1, y 2, x 1, x 2) = n d 4 z 1...d 4 z n 0 T φ n n! in y 1) φ in y 2) φ in x 1) φ in x 2) H I φ in z 1))...H I φ in z n))) 0 Lowest order contribution τ 2) y 1, y 2, x 1, x 2) = i)2 2! Using Wick s theorem, the connected diagrams are, ) λ d 4 z 1 d 4 z 2 0 T φ in y 1) φ in y 2) φ in x 1) φ in x 2) 3! φ3 in z 1) λ ) 3! φ3 in z 2)) 0 Institute) Slide_04 33 / 44

34 Their contribution to τ y 1, y 2, x 1, x 2) is τ 2) y 1, y 2, x 1, x 2) = i λ) 2 2! d 4 z 1 d 4 z 2 i F y 1 z 1) i F y 2 z 1) i F z 2 x 1) i F z 2 x 2) i F z 1 z 2) +... use the propagator in momentum space Then i F x ) = d 4 k 2π) 4 i k 2 µ 2 + i ε e ik x Institute) Slide_04 34 / 44

35 τ 2) y 1, y 2, x 1, x 2) = i λ) 2 2! d 4 z 1 d 4 z 2 d 4 k 1 2π) 4 d 4 k 2 2π) 4... e ik 1 y 1 z 1 ) i F k 1) e ik 2 z 1 x 1 ) i F k 2) d 4 k 5 2π) 4 e ik 3 z 1 z 2 ) i F k 3) e ik 4 y 2 z 2 ) i F k 4) e ik 5 z 2 x 2 ) i F k 5) z 1 integration z 2 integration d 4 z 1 e ik 1 k 2 k 3 ) z 1 = 2π) 4 δ 4 k 1 k 2 k 3) d 4 z 2 e ik 3 +k 4 k 5 ) z 2 = 2π) 4 δ 4 k 3 + k 4 k 5) energy-momentum conservation at each vertex Then τ 2) y 1, y 2, x 1, x 2) = i λ) 2 2! d 4 k 1 2π) 4... d 4 k 4 2π) 4 2π)4 δ 4 k 1 k 2 + k 4 k 5) i F k 1) i F k 2) i F k 4) i F k 5) i F k 1 k 2) e ik 1 y 1 e ik 2 x 1 e ik 4 y 2 e ik 5 x 2 τ 2) y 1, y 2, x 1, x 2) d 4 x 1 d 4 x 2 d 4 y 1 d 4 y 2 e ip 1 y 1 +p 2 y 2 ) e iq 1 x 1 +q 2 x 2 ) e ik 1 y 1 e ik 2 x 1 e ik 4 y 2 e ik 5 x 2 = 2π) 4 δ 4 k 2 q 1) 2π) 4 δ 4 k 5 q 2) 2π) 4 δ 4 p 1 k) 2π) 4 δ 4 p 2 k 4) Institute) Slide_04 35 / 44

36 We see that the external line propagators cancell out and S βα = i λ)2 2! This is rather simple answer in momentum space. ) 1 4 i z [ q 1 + q 2) µ ] 2 2π)4 δ 4 p 1 + p 2 q 1 q 2) +... Institute) Slide_04 36 / 44

37 Cross section and Decay rate Write the S-matrix elements as S fi = δ fi + i2π) 4 δ 4 p f p i )T fi T fi : invariant amplitude for i f. For i = f, the transistion probability is S fi 2 = 2π) 4 δ 4 0)[2π) 4 δ 4 p f p i ) T fi 2 ] To interprete δ 4 0), we write 2π) 4 δ 4 p f p i ) = d 4 xe ip f p i )x The integration is over some large but finite volume V and time interval T. Then we can interprete δ 4 0) as 2π) 4 δ 4 0) = VT and write S fi 2 = VT [2π) 4 δ 4 p f p i ) T fi 2 The transistion rate transistion probability per unit time) is then ω fi = 2π) 4 δ 4 p f p i ) T fi 2 V Institute) Slide_04 37 / 44

38 Decay rates For a general decay processes with kinematics, n ap) c 1 k 1 ) + c 2 k 2 ) c n k n ) p f = k i l=1 p i = p number of states in the volume elements d 3 k 1...d 3 k n in momentum space is n l=1 d 3 k l 2π) 3 2ω kl The transition rate, summing over final states is d ω = 2π) 4 δ 4 p Σ n j=1k j ) T fi 2 V For the invariant normalization of the physical states n l=1 d 3 k l 2π) 3 2ω kl For p = p, which is the number of particle in the initial state. The decay rate per particle is then < p p >= 2π) 3 δ 3 p p )2ω p < p p >= 2π) 3 δ 3 0)2ω p = 2V ω p d ω = d ω = 2π) 4 δ 4 p Σ n 2V ω j=1k j ) T fi 2 n 1 d p 2ω p 3 k l l=1 2π) 3 2ω kl Institute) Slide_04 38 / 44

39 If there are "m" identical particles in the final state, divide this by m! d ω = 1 2ωp T fi 2 d 3 k 1 2π) 3 d... 3 kn 2ω 1 2π) 3 2ωn 2π)4 δ 4 p Σ n j=1 k j )S S = j 1 m j )! Institute) Slide_04 39 / 44

40 Cross section For a scattering processes, ap 1 ) + bp 2 ) c 1 k 1 ) + c 2 k 2 ) c n k n ) the transition rate is, after summing over final states, d ω = 2π) 4 δ 4 p 1 + p 2 Σ n j=1k j ) T fi 2 V n l=1 d 3 k l 2π) 3 2ω kl Normalize this to 1 particle in the beam and 1 particle in the target and divide this by the flux relative velocity divided by the volume, to get differential cross section d σ = 1 2ω p1 V 1 2ω p2 V 2π)4 δ 4 p 1 + p 2 Σ n j=1k j ) T fi 2 V n l=1 d 3 k l 2π) 3 2ω kl V v 1 v 2 Velocity factor can be written as I = v 1 v 2 = p 1 p 2 E 1 E 2 In the C.M. frame p 1 = p 2 = p p 1 = E 1, p), p 2 = E 2, p) I = p E 1 E 2 E 1 + E 2 ) p 1 p 2 ) 2 = E 1 E 2 + p 2 ) 2 = E 2 1 E E 1 E 2 p 2 + p 4 Institute) Slide_04 40 / 44

41 p 1 p 2 ) 2 m1m = p 2 + m1) p m2) 2 + 2E 1 E 2 p 2 + p 4 m1m = p 2 [2 p 2 + m1 2 + m2) 2 + 2E 1 E 2 ] d σ = 1 I = p 2 E 1 + E 2 ) 2 I = 1 E 1 E 2 p 1 p 2 ) 2 m 2 1 m π) 4 δ 4 p 1 + p 2 Σ n 2ω p1 2ω j=1k j ) T fi 2 n d p2 3 k l l=1 2π) 3 2ω l Institute) Slide_04 41 / 44

42 Feynman Rules Since final forms for T fi are quite simple, can use simple rules to sidestep all those tedious intermediate steps. Draw all connected Feynman graphs with appropriate external lines. Label each with momenta and impose momentum conservation for each vertex. 1. For each internal fermion line with momentum p,enter the propagator is F p) = i p µ γ µ m + i ε 2. For each internal boson line of spin 0,with momentum q,enter the propagator i F q) = i q 2 µ 2 + i ε 3. For each internal photon line with momentum k,enter the propagator id F k) µυ = ig µν k 2 + i ε 4. For each internal momentum l not fixed by momentum conservation,enter d 4 l 2π) 4 5. For each closed fermion loop,enter -1). Also a factor of -1) between graphs which differ by an interchange of two external identical fermion lines. Institute) Slide_04 42 / 44

43 6. At each vertex, the factors depend on form of interactions. 1 a. λφ 3 i λ) 3! 1 b. λφ 4 i λ) 4! ) c. eψγ µ ψa µ ieγ µ d. f ψγ 5 ψφ if γ 5 ) Example in λφ 3 theory In λφ 3 theory, consider the processes φ k 1) + φ k 2) φ k 3) + φ k 4) To order λ 2, we have 3 Feynman diagrams We can write down the matrix element for each graph, Institute) Slide_04 43 / 44

44 T a) = i λ) 2 i k 1 k 3) 2 T b) = i λ) 2 i µ 2 k 1 + k 2) 2 T c) = i λ) 2 i µ 2 k 1 k 4) 2 µ 2 Total amplitude T = T a) + T b) + T c) Mandelstam variables s = k 1 + k 2) 2 total energy in c.m. frame t = k 1 k 3) 2 momentum transferscattering angle) u = k 1 k 4) 2 Usually these amplitudes are written as T a) = λ) 2 s + t + u = 4µ 2 i t µ 2 T b) = λ) 2 i s µ 2 T c) = λ) 2 i u µ 2 Institute) Slide_04 44 / 44