QFT Perturbation Theory

QFT Perturbation Theory Ling-Fong Li (Institute) Slide_04 1 / 43

Interaction Theory As an illustration, take electromagnetic interaction. Lagrangian density is The combination 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 ) is the covariant derivative. Equations of motion ( 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 controlled by the interaction terms. Quantization Write L = L 0 + L int (Institute) Slide_04 2 / 43

L 0 = ψ ( i γ µ µ m ) ψ 1 4 F µνf µν L int = eψγ µ ψa µ 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 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 (Institute) Slide_04 3 / 43

{ψ α ( ( ) x, t), ψ β x, t [Ȧi ( ( x, t), A j x, t)] Commutators with A 0, } = ( δ αβ δ 3 ) x x {ψ α ( ( x, t), ψ β x, t = ( i δ tr x ) ij x ) } =... = 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 relation Hamiltonian No A 0 in the interaction. If we write 2 1 ( 4π x = δ 3 ) x x x [ H = d 3 x H = d 3 x {ψ ] α ( i e A ) + βm ψ + 1 ( E 2 + B 2) } 2 E = El + E t where El = A 0, E t = A t (Institute) Slide_04 4 / 43

Then 1 ( E d 3 x 2 + B 2) = 1 d 3 x ( E 2 l + d 3 Et x 2 + B 2) 2 2 The longitudinal part is 1 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 / 43

Physical states 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 0 0 2 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 pi 2 = mi 2 for each stable particle. 4 vaccum and one particle states form discrete spectra in p µ assume interactions do not violently the spectrum of states. there is no room to describe bound states (Institute) Slide_04 6 / 43

Spectral Representation In the interacting field theories, can get some useful relation independent of the details of the interaction. Here we will discuss 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 of energy and mometum between two field operators, we get where we have used n 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 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, so that we can write (x y ) = = i (2π) 3 i (2π) 3 1 = d 4 qδ 4 (q p n) 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 ) ] (Institute) Slide_04 7 / 43

where 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) since 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 is the commutator function for free scalar field with mass σ 2.In this Eq (1), the commutator function of the interacting field is an integral over free field commutatotor with different masses is called the spectral representation of the commutator function. 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 (Institute) Slide_04 8 / 43

Or in momentum space where F F ( p 2 ) = d σ 2 ρ ( σ 2) 1 0 p 2 σ 2 + i ε ( p 2 ) = d 4 x F ( (x ), F p 2 ) = d 4 x F (x ) (Institute) Slide_04 9 / 43

In-fields and in-states asymptotic conditions Want to study the interaction by scattering processes where the initial and final states propagate as free particles until they are close enough to interact. We now want to find how the transistion from the initial free particles states to final free particles states. Clearly this transistion is controlled by the interaction terms in the Lagrangian. The initial states: In scattering problem, at t =, particles propagate freely. Let φ in (x ) creates such free particle propagating with physical mass µ in the initial states, ( + µ 2 ) φ in (x ) = 0 Here we allow the particle s physical mass µ to be different from µ 0 that appear in the Largrangian. This has to do with the renormalization effect and will be described later. We will assume that φ in (x ) transforms under coordinate displacements and Lorentz transformation in the same way as φ (x ). Since φ in (x ) satisfies free field equation, we can 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 Invert this expansion a in (k) = i d 3 x fk (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 (Institute) Slide_04 10 / 43

The normalization of these states are, k 1, k 2,...k n, in = [ i ] (2π) 3 2w ki a in (k i ) 0 k 2, in k 1, in = (2π) 3 2w 1 δ 3 ( k1 k 2 ) p 1, p 2,..., p m, in k 1, k 2,...k n in = 0 unless m = n and (p 1, p 2,..., p m ) conicides with (k 1, k 2,...k n). (Institute) Slide_04 11 / 43

Relation between φ in (x ) and φ (x ) To set up the perturbation expansion, want to find the relation between interacting field φ (x ) and φ in (x ) which is a free field. The field equations for these fields are ( ) + µ 2 φ 3 0 φ (x ) = j (x ) = 3! or ( + µ 2 ) φ (x ) = j (x ) + δµ 2 φ (x ) j (x ), δµ 2 = µ 2 µ 2 0 ( + µ 2 ) φ in (x ) = 0 Treat the non-linear term φ3 for the interacting field φ (x ) as a inhomogeneous source term. Formally we can 3! relate the solution to the inhomogeneous equation φ (x ) to the solution to homogeneous equation φ in (x ) as, where z φin (x ) = φ (x ) d 4 ( y ret x y, µ 2 ) j (y ) (2) ( x + µ 2) ret ( x y, µ 2 ) = δ 4 (x y ) and 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, (3) because ret 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 we take matrix element of Eq (2), 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 / 43

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 z 0 φin (x ) p = 0 φ (x ) p and we can 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 / 43

The lower limit in the integral corresponds to the threshold for all states other than the one particle state. We can 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 field. Thus we get the relation 1 = Z + d 4µ 2 σ2 ρ ( σ 2) which implies 0 Z < 1 (4) The inequality Z < 1 is due to the positivity of the spectral function ρ ( q 2). Now we will use this inequality to show that the asymptotic condition in Eq (3) 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 ) These can be consistent with the asymptotic relation φ (x ) z φ in (x ) as x 0, in Eq (3) only if Z = 1, which contradicts the inequality in Eq (4) (Institute) Slide_04 14 / 43

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 + µ 2 ) f = 0 f k ( x, t) is an arbitrary normalizable solution to Klein-Gordon equation. The correct asymptotic condition is 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) Out fields and out states Similar procedure applies to φ out ( + µ 2 0) φout (x ) = 0 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) Asymptotic condition 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 p 1, p 2,..., p n, in = α, in (Institute) Slide_04 15 / 43

Final state p 1, p 2,..., p m, out = β, out S-matrix S βα β, out α, in Introduce S-operator β, out β, in S 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 / 43

LSZ reduction set up the framework to compute S βα. Consider we get the reduction formula, β, out α, p, in = N β p, out α, in + S β,αp = β, out α, p, in i e ip x d 4 x ( + µ 2) β, out φ (x ) α, in z 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 (Institute) Slide_04 17 / 43

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

2 remove positron(anti-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

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] (5) 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 (6) Define U (t, t ) U (t) U 1 (t ) time evolution operator Eq(6) 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 20 / 43

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 21 / 43

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 22 / 43

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 23 / 43

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 24 / 43

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) 0... 0 φ in (x n 1) φ in (x n) 0 + permutations] neven [ 0 φ in (x 1) φ in (x 2) 0... 0 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 25 / 43

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 26 / 43

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 27 / 43

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 28 / 43

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 29 / 43

closed loop diagram :graphs with no external lines(lines with open end) disconnected diagram :a subgraph not connected to any external lines connected diagram :graph not disconnected (Institute) Slide_04 30 / 43

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 i(p 1 y 1 +p 2 y 2 ) e i(q 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 31 / 43

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 32 / 43

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 i(k 1 k 2 k 3 ) z 1 = (2π) 4 δ 4 (k 1 k 2 k 3) d 4 z 2 e i(k 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 i(p 1 y 1 +p 2 y 2 ) e i(q 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 33 / 43

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 z ) 4 (2π) 4 δ 4 (p 1 + p 2 q 1 q 2) +... (Institute) Slide_04 34 / 43

Cross section and Decay rate Write the S-matrix elements as S fi = δ fi + i(2π) 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 i(p 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 35 / 43

Decay rates For a general decay processes with kinematics, n a(p) 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 36 / 43

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 37 / 43

Cross section For a scattering processes, a(p 1 ) + b(p 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 2 2 + 2E 1 E 2 p 2 + p 4 (Institute) Slide_04 38 / 43

(p 1 p 2 ) 2 m1m 2 2 2 = ( p 2 + m1)( p 2 2 + m2) 2 + 2E 1 E 2 p 2 + p 4 m1m 2 2 2 = 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 m2 2 1 1 (2π) 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 39 / 43

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. 6. At each vertex, the factors depend on form of interactions. 1 a. λφ 3 ( i λ) 3! (Institute) Slide_04 40 / 43

1 b. λφ 4 ( i λ) 4! ( ) c. eψγ µ ψa µ ieγ µ d. f ψγ 5 ψφ ( if γ 5 ) (Institute) Slide_04 41 / 43

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 42 / 43

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 transfer(scattering 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 43 / 43