MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department

Alan Guth, 8323 Lecture, May 13, 28, p1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8323: Relativistic Quantum Field Theory I Interacting Field Theories Tuesday, May 13, 28 Alan Guth Alan Guth Massachusetts Institute of Technology 8323, May 13, 28 Time-Dependent Perturbation Theory Sample theory: λφ 4 L = 1 2 ( µφ) 2 1 2 m2 φ 2 λ 4! φ4 Then H = H + H int, where H int = d 3 x λ 4! φ4 ( x) Goal: to perturbatively calculate matrix elements of the Heisenberg field φ( x,t)=e iht φ( x,)e iht in thestate Ω, thetrueground stateof theinteracting theory 1

Alan Guth, 8323 Lecture, May 13, 28, p2 Interaction picture: Choose any reference time t, at which the interaction picture operators and Heisenberg operators will coincide Define φ I ( x,t) e ih (t t ) φ( x,t )e ih (t t ) At t = t, can expand Heisenberg φ and π in creation and annihilation operators: d 3 p 1 φ( x,t )= (a ) (2π) 2 p e i p x + a p e i p x 2Ep π( x,t )= φ( x,t )= d 3 p 1 ( ) ie (2π) 2 p a p e i p x + ie p a p e i p x 2Ep a p creates state of momentum p, but not energy E p not singleparticle But [φ( x,t ),π( y,t )] = iδ 3 ( x y ) = [ ] a p,a q =(2π) 3 δ (3) ( p q ) 2 Can write φ I ( x,t) for all t: d 3 p 1 ) φ I ( x,t)= (a (2π) 2 p e ip x + a eip x x p 2Ep =t t How to express φ( x,t): φ( x,t)=e ih(t t) e ih (t t ) φ I ( x,t)e ih (t t ) e ih(t t ) U (t, t )φ I ( x,t)u(t, t ), where U(t, t )=e ih (t t ) e ih(t t) Differential equation for U: i t U(t, t )=e ih (t t ) (H H )e ih(t t ) = e ih (t t ) H int e ih(t t ) where = e ih (t t ) H int e ih (t t ) e ih (t t ) e ih(t t ) = H I (t)u(t, t ), H I (t) =e ih (t t ) H int e ih (t t ) = d 3 x λ 4! φ4 I ( x,t) 3

Alan Guth, 8323 Lecture, May 13, 28, p3 Solution to differential equation: i t U(t, t )=H I (t)u(t, t ) with U(t,t )=I implies the integral equation To first order in H I, To second order in H I, U(t, t )=I i To third order, U(t, t )=I i t U(t, t )=+( i) 3 t t t dt H I (t )U(t,t ) t U(t, t )=I i dt 1 H I (t 1 ) t t dt 1 H I (t 1 )+( i) 2 t dt 1 t1 t dt 2 t t dt 1 t2 t1 t dt 2 H I (t 1 )H I (t 2 ) t dt 3 H I (t 1 )H I (t 2 )H I (t 3 ) 4 Notethat t 1 t 2 t 3 Can rewrite 3rd order term as U(t, t )=+( i) 3 t = + ( i)3 3! t dt 1 t t dt 1 t1 t dt 2 t t dt 2 t2 t t dt 3 H I (t 1 )H I (t 2 )H I (t 3 ) t dt 3 T {H I (t 1 )H I (t 2 )H I (t 3 )}, where T {} is time-ordered product (earliest time to right) Finally, t U(t, t )=I +( i) T { [ exp i dt 1 H I (t 1 )+ ( i)2 t 2! t ]} dt H I (t ) t t t dt 1 t t dt 2 T {H I (t 1 )H I (t 2 )} + 5

Alan Guth, 8323 Lecture, May 13, 28, p4 Generalize to arbitrary t : U(t 2,t 1 ) T { [ t2 ]} exp i dt H I (t ) t 1, where (for t 1 <t <t 2 ) U(t 2,t 1 )=T { [ t2 ]} { [ t ]} exp i dt H I (t ) T exp i dt H I (t ) t t 1 = U(t 2,t )U 1 (t 1,t ) Given have U(t, t )=e ih (t t ) e ih(t t ), U(t 2,t 1 )=e ih (t 2 t ) e ih(t 2 t ) e ih(t 1 t ) e ih (t 1 t ) = e ih (t 2 t ) e ih(t 2 t 1 ) e ih (t 1 t ) 6 Properties: U is unitary U(t 2,t 1 ) T U(t 3,t 2 )U(t 2,t 1 )=U(t 3,t 1 ) U(t 2,t 1 ) 1 = U(t 1,t 2 ) { [ t2 ]} exp i dt H I (t ) t 1 7

Alan Guth, 8323 Lecture, May 13, 28, p5 Construction of true ground state Ω : Assumethat has nonzero overlap with Ω : e iht = n e ie nt n n If T had large negative imaginary part, all other states would be suppressed relative to Ω Ω = ( ) 1 lim T (1 iɛ) e ih(t +t) e ie (T +t ) Ω = lim T (1 iɛ) e ih(t +t ) e ih (T +t ) ( e ie (T +t ) Ω ) 1 Recall so Ω = U(t 2,t 1 )=e ih (t 2 t ) e ih(t 2 t 1 ) e ih (t 1 t ), ( 1 ) lim, T ) T (1 iɛ) e ie (T +t ) Ω 8 Similarly, Recall Ω = ( 1 lim e ie (T t ) Ω ) U(T,t ) T (1 iɛ) φ( x,x )=U (x,t )φ I ( x,x )U(x,t ) So, for x >y, Ω φ(x)φ(y) Ω = lim T (1 iɛ) U(T,t )U(t,x )φ I ( x,x )U(x,t ) U(t,y )φ I ( y,y )U(y,t )U(t, T ) Normalization factor = lim T (1 iɛ) U(T,x )φ I ( x,x )U(x,y ) φ I ( y,y )U(y, T ) Normalization factor But Normalization factor = Ω Ω 1, 9

Alan Guth, 8323 Lecture, May 13, 28, p6 so Ω φ(x)φ(y) Ω = lim T (1 iɛ) { [ T T φ I (x)φ I (y)exp i dth I (t)]} T { [ T T exp i dth I (t)]} T Causality: H I (t) = d 3 xh( x,t), so Ω φ(x)φ(y) Ω = [ {φ T I (x)φ I (y)exp i [ {exp T i d 4 z H I (z)]} d 4 z H I (z)]} If z 1 and z 2 are spacelike-separated, their time ordering is frame-dependent Need [ H I (z 1 ), H I (z 2 )] =to get same answer in all frames 1 Time-Dependent Perturbation Theory: Ω φ(x)φ(y) Ω = [ {φ T I (x)φ I (y)exp i [ {exp T i d 4 z H I (z)]} d 4 z H I (z)]} 11

Alan Guth, 8323 Lecture, May 13, 28, p7 Wick's Theorem Gian-Carlo Wick October 15, 199 April 2, 1992 http://stillsnapedu/html/biomems/gwickhtml 12 T {φ(x 1 )φ(x 2 ) φ(x m )} = N{φ(x 1 )φ(x 2 ) φ(x m )+ all possiblecontractions} Example: Corollary: T {φ(x 1 )φ(x 2 ) φ(x m )} = all possiblefull contractions} Example: T {φ(x 1 )φ(x 2 )φ(x 3 )φ(x 4 )} = F (x 1 x 2 ) F (x 3 x 4 ) + F (x 1 x 3 ) F (x 2 x 4 )+ F (x 1 x 4 ) F (x 2 x 3 ) 13

Alan Guth, 8323 Lecture, May 13, 28, p8 Feynman Diagrams Example: T {φ(x 1 )φ(x 2 )φ(x 3 )φ(x 4 )} = F (x 1 x 2 ) F (x 3 x 4 ) + F (x 1 x 3 ) F (x 2 x 4 )+ F (x 1 x 4 ) F (x 2 x 3 ) Feynman diagrams: [Note: the diagrams and some equations on this and the next 12 pages were taken from An Introduction to Quantum Field Theory, by Michael Peskin and Daniel Schroeder] 14 Nontrivial Example: H I (z) = λ 4! φ4 (z) Ω T {φ(x)φ(y)} Ω = {φ(x)φ(y)+φ(x)φ(y) T {φ(x)φ(y) T [ i d 4 zh I (z)]} [ i ( ) iλ =3 D F (x y) d 4 zd F (z z)d F (z z) 4! ( ) iλ +12 d 4 zd F (x z)d F (y z)d F (z z) 4! ] d 4 zh I (z) + } = 15

Alan Guth, 8323 Lecture, May 13, 28, p9 Very Nontrivial Example: How many identical contractions are there? Overall factor: 1 3! ( ) 1 3 4! 3! 4 3 4! 4 3 1 2 = 1 8 1 symmetry factor 16 Symmetry Factors: 17

Alan Guth, 8323 Lecture, May 13, 28, p 1 Feynman Rules for λφ 4 Theory [ {φ(x)φ(y)exp T i d 4 zh I (z)]} = ( ) sum of all possiblediagrams with two external points Rules: 18 Momentum Space Feynman Rules: D F (x y) = d 4 p i e ip (x y) (2π) 4 p 2 m 2 + iɛ Vertex: d 4 ze ip 1z e ip 2z e ip 3z e ip 4z =(2π) 4 δ (4) (p 1 + p 2 + p 3 + p 4 ) 19

Alan Guth, 8323 Lecture, May 13, 28, p 11 2 Disconnected Diagrams Consider Momentum conservation at one vertex implies conservation at the other Graph is proportional to (2π) 4 δ (4) () Use Disconnected diagrams: (2π) 4 δ (4) () = (volumeof space) 2T 21

Alan Guth, 8323 Lecture, May 13, 28, p 12 Give each disconnected piece a name: Then Diagram = (value of connected piece) So, thesum of all diagrams is: i 1 n i! (V i) n i Factoring, 22 Factoring even more: 23

Alan Guth, 8323 Lecture, May 13, 28, p 13 For our example, Now look at denominator of matrix element: 24 Finally, Summarizing, Vacuum energy density: 25

Alan Guth, 8323 Lecture, May 13, 28, p 14 Final sum for four-point function: 26 Summary of Time-Dependent Perturbation Theory Example: The four-point function: 27

Alan Guth, 8323 Lecture, May 13, 28, p 15 Cross Sections and Decay Rates Definition of Cross Section: where ρ A and ρ B = number density of particles A = cross sectional area of beams l A and l B = lengths of particle packets Then σ Number of scattering events of specified type ρ A l A ρ B l B A Notethat σ depends on frame Special case: 1 particle in each beam, so ρ A l A A =1and ρ B l B A =1Then Number of events = σ/a 28 Definition of Decay Rate: Γ Number of decays per unit time Number of particles present Number of surviving particles at time t: N(t) =N e Γt Mean lifetime: Half-life: τ = 1 dt N e Γt = 1 2 ( dn dt ) t =1/Γ = t 1/2 = τ ln 2 29

Alan Guth, 8323 Lecture, May 13, 28, p 16 Unstable Particles as Resonances (Preview): Unstable particles are not eigenstates of H; they are resonances in scattering experiments In nonrelativistic quantum mechanics, the Breit-Wigner formula f(e) 1 E E + iγ/2 = σ 1 (E E ) 2 +Γ 2 /4 The full width at half max of theresonance= Γ In the relativistic theory, the Breit-Wigner formula is replaced by a modified (Lorentz-invariant) propagator: which can be seen using 1 p 2 m 2 + imγ 1 2E p (p E p + i(m/e p )Γ/2), (p ) 2 p 2 m 2 =(p ) 2 E 2 p = ( p + E p )( p E p ) 2E p ( p E p ) 3 Cross Sections and the S-Matrix Initial and Final States In- and Out-States: Recall our discussion of particle creation by an external source, ( + m 2 )φ(x) =j(x), where j was assumed to be nonzero only during a finite interval t 1 <t<t 2 In that case, the Fock space of the free theory for t<t 1 defined the in-states, thefock spaceof thefreetheory for t>t 2 defined the out-states, and we could calculate exactly the relationship between the two We started in the in-vacuum and stayed there The amplitude p 1 p 2 p N, out,in was then interpreted as the amplitude for producing a set of final particles with momenta p 1 p N 31

Alan Guth, 8323 Lecture, May 13, 28, p 17 For interacting QFT s, it is more complicated The interactions do not turn off, and affect even the 1-particle states It is still possible to define in- and out-states p 1 p N, in and p 1 p N, out with the following properties: They are exact eigenstates of the full Hamiltonian At asymptotically early times, wavepackets constructed from p 1 p N, in evolve as free wavepackets (The pieces of this ket that describe the scattering vanish in stationary phase approximation at early times) These states are used to describetheinitial stateof thescattering At asymptotically late times, wavepackets constructed from p 1 p N, out evolve as free wavepackets These states are used to describe the final state 32 Wavepacket States: One-particle incoming wave packet: φ = where φ φ =1 = d 3 k (2π) 3 1 2E k φ( k ) k,in d 3 k (2π) 3 φ( k ) 2 =1, Two-particleinitial state: φ A φ B, b,in = d 3 k A (2π) 3 d 3 k B (2π) 3 φ A ( k A ) φ B ( k B )e i b k B (2EA )(2E B ) k A k B, in, where b is a vector which translates particle B orthogonal to thebeam, so that we can construct collisions with different impact parameters Multiparticlefinal state: n φ 1 φ n, out = d 3 p f φ f ( p f ) p (2π) 3 1 p n, out 2Ef f=1 33

Alan Guth, 8323 Lecture, May 13, 28, p 18 The S-Matrix: Definition: S Ψ, out = Ψ, in Therefore Ψ, out Ψ, in = Ψ, out S Ψ, out But S maps a complete set of orthonormal states onto a complete set of orthonormal states, so S is unitary Therefore Ψ, out S Ψ, out = Ψ, out S SS Ψ, out = Ψ, in S Ψ, in, so P&S often do not label the states as in or out 34 S, T,andM: No scattering = final = initial, so separate this part of S: S 1+iT But T must contain a momentum-conserving δ-function, so define p 1 p n it k A k B (2π) 4 δ (4) ( k A + k B p f ) im(k A k B {p f }) 35

Alan Guth, 8323 Lecture, May 13, 28, p 19 Evaluation of the Cross Section: The probability of scattering into the specified final states is just the square of the S-matrix element, summed over the final states: P ( { } { } AB, ) n b p1 p n = f=1 d 3 p f 1 p (2π) 3 1 p n S φ A φ B, 2E 2 b f To relate to the cross section, think of a single particle B scattering off of a particle A, with impact parameter vector b: Remembering that the cross section can be viewed as the cross sectional area blocked off by the target particle, dσ = d 2 b P ( { } { } AB, ) b p1 p n 36 Substituting the expression for P and writing out the wavepacket integrals describing the initial state, n dσ = d 2 b d 3 p f 1 d 3 k A d 3 k B φ A ( k A ) φ B ( k B ) (2π) 3 2E f (2π) 3 (2π) 3 (2EA )(2E B ) f=1 d 3 ka d 3 kb φ A ( ka ) φ B ( kb ) (2π) 3 (2π) 3 (2 Ē A )(2ĒB) ei b ( kb k B ) p 1 p n S k A k B p 1 p n S ka kb This can besimplified by using ( d 2 be i b ( kb k B ) =(2π) 2 δ (2) k B ) k B p 1 p n S ( ) k A k B = im k A ( k B { p f } (2π) 4 δ (4) k A + k B ) p f, ( ) p 1 p n S ka kb = im ka kb { p f } (2π) 4 δ ( ka (4) + k B ) p f, 37

Alan Guth, 8323 Lecture, May 13, 28, p 2 ( Wefirst integrateover ka and kb using δ (2) k B ) k B and δ (4) ( ka + k B ) ( p f = δ (2) k A + k B ) p f δ ( kz A + k z B ) p z f δ (ĒA + ĒB ) E f, where the beam is taken along the z-axis After integrating over k A and k B,we areleft with d k z A d k z B ( kz δ A + k z B ) p z f δ (ĒA + ĒB ) E f = d k z A δ ( F ( k z A )), where the first δ-function was used to integrate k z B,and F ( k z A k )= 2 ) 2 ( A + ( kz A + m2 + k 2 B + p z f k 2 A) z + m2 E f Then d k z A δ ( F ( k z A) ) = 1 df d k z A, evaluated where F ( k z A)= 38 Rewriting F ( k z A k )= 2 ) 2 ( A + ( kz A + m2 + k 2 B + p z f k 2 A) z + m2 E f, onefinds df d k z = k z A p z f k z A A ĒA Ē B Remembering the δ-function constraint δ ( kz A + k z B ) p z f from theprevious slide, one has df d k z = kz A k z B A ĒA ĒB = vz A v z B What values of k z A satisfy theconstraint F ( k z A )=? Therearetwo solutions, since F ( k z A )=can be manipulated into a simple quadratic equation (To see this, move onesquareroot to therhs of theequation and squareboth sides The( k z A )2 term on each side cancels, leaving only linear terms and a square root on the LHS Isolate thesquareroot and squareboth sides again, obtaining a quadratic equation) One solution gives ka = k A, and the other corresponds to A and B approaching each other from opposite directions Assume that the initial wavepacket is too narrow to overlap the 2nd solution 39

Alan Guth, 8323 Lecture, May 13, 28, p 21 Then n dσ = f=1 d 3 p f 1 d 3 k A (2π) 3 2E f (2π) 3 d 3 k B φ A ( k A ) 2 φ B ( k B ) 2 (2π) 3 (2E A )(2E B ) va z vz B ( ) M k A 2 ( k B { p f } (2π) 4 δ (4) k A + k B ) p f Define the relativistically invariant n-body phase space measure n dπ n (P ) f=1 d 3 p f 1 ( (2π) 4 δ (4) P ) p (2π) 3 f 2E f, and assumethat E A ( k A ), E B ( ( k B ), va z M vz B, k A 2 k B { p f }), and dπ n (k A + k B ) are all sufficiently slowly varying that they can be evaluated at the central momenta of the two intial wavepackets, k A = p A and k B = p B Then the normalization of the wavepackets implies that d 3 k A d 3 k B (2π) 3 (2π) φ A( k 3 A ) 2 φ B ( k B ) 2 =1, 4 so finally dσ = M ( p A p B { p f }) 2 (2E A )(2E B ) v z A vz B dπ n(p A + p B ) This formula holds whether the final state particles are distinguishable or not In calculating a total cross section, however, one must not double-count final states If thefinal statecontains n identical particles, one must either restrict the integration or dividetheanswer by n! 41

Alan Guth, 8323 Lecture, May 13, 28, p 22 Special Case: Two-Particle Final State: In the center of mass (CM) frame, p A = p B and E cm = E A + E B,so 2 dπ 2 (p A + p B )= d 3 p f 1 ( (2π) 3 (2π) 4 δ (4) p A + p B ) p f 2E f f=1 = d3 p 1 (2π) 3 d 3 p 2 (2π) 3 1 (2E 1 )(2E 2 ) (2π)4 δ (4) (p A + p B p 1 p 2 ) = d3 p 1 (2π) 3 1 (2E 1 )(2E 2 ) (2π)δ(E cm E 1 E 2 ) =dω p2 1 dp 1 (2π) 3 1 (2E 1 )(2E 2 ) (2π)δ =dω p2 1 1 (2π) 2 (2E 1 )(2E 2 ) p 1 =dω 16π 2 E cm p 1 + p 1 E 1 ( ) E cm p 21 + m21 p 2 1 + m2 2 E 2 1 42 The two-particle final state, center-of-mass cross section is then ( ) dσ = p A M( p A p B p 1 p 2 ) 2 dω cm 64π 2 E A E B (E A + E B ) va z vz B If all four masses are equal, then E A = E B = 1 2 E cm and so v z A v z B = 2 p A E A = 4 p A E cm, ( ) dσ = M 2 dω cm 64π 2 Ecm 2 (all masses equal) 43

Alan Guth, 8323 Lecture, May 13, 28, p 23 Evaluation of the Decay Rate: The formula for decay rates is more difficult to justify, since decaying particles have to be viewed as resonances in a scattering experiment For now we just state the result By analogy with the formula for cross sections, wewrite dσ = M ( p A p B { p f }) 2 (2E A )(2E B ) v z A vz B dπ n(p A + p B ), dγ = M ( p A { p f }) 2 2E A dπ n (p A ) Here M cannot be defined in terms of an S-matrix, since decaying particles cannot be described by wavepackets constructed in the asymptotic past M can be calculated, however, by the Feynman rules that Peskin & Schroeder describe in Section 46 If some or all of the final state particles are identical, then the same comments that were made about cross sections apply here 44 Summary of Key Results So Far Time-dependent perturbation theory: Ω T {φ(x 1 ) φ(x n )} Ω [ = {φ T I (x 1 ) φ I (x n )exp i d 4 z H I (z)]} connected = Sum of all connected diagrams with external points x 1 x n Status: derivation was more or less rigorous, except for ignoring problems connected with renormalization: evaluation of integrals in this expression will lead to divergences These questions will be dealt with next term If the theory is regulated, for example by defining it on a lattice of finite size, the formula above would be exactly true for the regulated theory One finds, however, that the limit as the lattice spacing goes to zero cannot be taken unless the parameters m, λ, etc, are allowed to vary as the limit is taken, and in addition the field operators must be rescaled 45

Alan Guth, 8323 Lecture, May 13, 28, p 24 Cross sections from S-matrix elements: S =1+iT, where ( p 1 p n it k A k B (2π) 4 δ (4) k A + k B ) p f im(k A k B {p f }) Therelativistically invariant n-body phasespacemeasureis n dπ n (P ) d 3 p f 1 ( (2π) 3 (2π) 4 δ (4) P ) p f 2E f f=1 and the differential cross section is dσ = M ( p A p B { p f }) 2 (2E A )(2E B ) v z A vz B dπ n(p A + p B ), Status: this derivation was more or less rigorous, making mild assumptions about in- and out-states These assumptions, and the formula above, will need to be modified slightly when massless particles are present, since the resulting longrange forces modify particle trajectories even in the asymptotic past These modifications arise only in higher order perturbation theory, and are part of the renormalization issue 46 Special case two-particle final states, in the center-of-mass frame: ( ) dσ = p A M( p A p B p 1 p 2 ) 2 dω cm 64π 2 E A E B (E A + E B ) va z vz B = M 2 64π 2 E 2 cm (if all masses are equal) Decay rate from S-matrix elements: dγ = M ( p A { p f }) 2 2E A dπ n (p A ) Status: completely nonrigorous at this point Unstable particles should be treated as resonances, an issue which is discussed in Peskin & Schroeder in Chapter 7 47

Alan Guth, 8323 Lecture, May 13, 28, p 25 Computing the S-Matrix From Feynman Diagrams Complication: p 1 p n S p A p B p 1 p n, out p A p B, in, but the in- and out-states are hard to construct: even single-particle states are modified by interactions Thesolution will makeuseof thefact that Ω φ(x) p = Ω e ip x φ()e ip x p = e ip x Ω φ() p is an exact expression for the interacting fields, with the full operator P µ and the exact eigenstate p By generalizing this to in- and out-states, it will be possibleto manipulatethecorrelation functions Ω T (φ 1 φ n ) Ω by inserting complete sets of in- and out-states at various places When the correlation function is Fourier-transformed in its variables x 1 x n to producea function of p 1 p n, one can show that it contains poles when any p i is on its mass shell, p 2 i = m2 i,and that theresiduewhen all thep i areon mass shell is thes-matrix element 48 A derivation will be given in Chapter 7, but for now we accept the intuitive notion that U(t 2,t 1 ) describes time evolution in the interaction picture, and that the S-matrix describes time evolution from minus infinity to infinity So we write p 1 p n S p A p B { [ = p 1 p n T exp i I ]} d 4 xh I (x) p A p B I connected, amputated, where connected means that the disconnected diagrams will cancel out as before, and the meaning of amputated will be discussed below It will be shown in Chapter 7 that this formula is valid, up to an overall multiplicative factor that arises only in higher-order perturbation theory, and is associated with the rescaling of field operators required by renormalization 49

Alan Guth, 8323 Lecture, May 13, 28, p 26 Calculation of 2 2 Scattering: Normalization conventions: p = 2E p a ( p), [ a( q ),a p ] =(2π) 3 δ (3) ( q p) To zeroth order in H I, p 1 p 2 S p A p B = I p 1 p 2 p A p B I = (2E 1 )(2E 2 )(2E A )(2E B ) a 1 a 2 a A a B =(2E A )(2E B )(2π) [δ 6 (3) ( p A p 1 )δ (3) ( p B p 2 ) ] + δ (3) ( p A p 2 )δ (3) ( p B p 1 ) Graphically, Contributes only to 1 of S =1+iT 5 To first order in H I : p 1 p 2 T I = I { i λ 4! { p 1 p 2 N i λ 4! } d 4 xφ 4 I(x) p A p B I } d 4 xφ 4 I(x)+contractions p A p B I, where contractions = i λ 4! [ 6 φ(x)φ(x)+3 ] Uncontracted fields can destroy particles in initial state or create them in the final state: φ I (x) =φ + I (x)+φ I (x) = d 3 k 1 [a (2π) 3 I ( k)e ik x + a I ( k )e ik x], 2E k so φ + I (x) p I = e ip x I, where φ + I and φ I refer to the parts of φ I(x) containing annihilation and creation operators, respectively 51

Alan Guth, 8323 Lecture, May 13, 28, p 27 This leads to a new type of contraction: We show this kind of contraction in a Feynman diagram as an external line Looking at the contracted terms from the Wick expansion, the fully contracted term produces a multiple of the identity matrix element, i λ d 4 x I p 1 p 2 p A p B = i λ d 4 x I p 1 p 2 p A p B 4! I 4! I, so this term also contributes only to the uninteresting 1 part of S =1+iT 52 The singly contracted term i 6λ d 4 x I p 1 p 2 φ(x)φ(x) 4! p A p B contains terms where one φ(x) contracts with an incoming particleand theother contracts with an outgoing particle, giving the Feynman diagrams I The integration over x gives an energy-momentum conserving δ-function, and the uncontracted inner product produces another, so these diagrams are again a contribution to the 1 of S =1+iT Thecontributions to T comefrom fully connected diagrams, where all external lines are connected to each other 53

Alan Guth, 8323 Lecture, May 13, 28, p 28 Nontrivial contribution to T: There are 4! ways of contracting the 4 fields with the 4 external lines, so the contribution is ( 4! i λ ) d 4 xe i(p A+p B p 1 p 2 ) x 4! = iλ(2π) 4 δ (4) (p A + p B p 1 p 2 ) so im(2π) 4 δ (4) (p A + p B p 1 p 2 ), M = λ 54 Repeating, M = λ By our previous rules, this implies ( ) dσ = dω cm λ 2 64π 2 E 2 cm For σ total oneuses thefact that thetwo final particles areidentical If weintegrate over all final angles we have double-counted, so we divide the answer by 2! σ total = λ 2 64π 2 E 2 cm 4π 1 2! = λ 2 32πE 2 cm 55

Alan Guth, 8323 Lecture, May 13, 28, p 29 Amputation: Consider the 2nd order diagram Contribution is 1 d 4 p i 2 (2π) 4 p 2 m 2 d 4 k i (2π) 4 k 2 m 2 ( iλ)(2π) 4 δ (4) (p A + p B p 1 p 2 ) ( iλ)(2π) 4 δ (4) (p B p ) Notethat δ (4) (p B p ) = p 2 = m 2,so 1 p 2 m = 1 2, which is infinite Any diagram in which all the momentum from one external line is channeled through a single internal line will produce an infinite propagator 56 Notethat δ (4) (p B p ) = p 2 = m 2,so 1 p 2 m 2 = 1, which is infinite Any diagram in which all the momentum from one external line is channeled through a single internal line will produce an infinite propagator Amputation: Eliminateall diagrams for which cutting a singlelineresults in separating a single leg from the rest of the diagram For example, 57

Alan Guth, 8323 Lecture, May 13, 28, p 3 Feynman Rules for λφ 4 in Position Space: im (2π) 4 δ (4) (p A + p B p f ) = (sum of all connected, amputated diagrams), wherethediagrams areconstructed by thefollowing rules: 58 Feynman Rules for λφ 4 in Momentum Space: im = (sum of all connected, amputated diagrams), wherethediagrams areconstructed by thefollowing rules: 59

Alan Guth, 8323 Lecture, May 13, 28, p 31 Feynman Rules for Fermions Time-dependent perturbation theory: Generalizes easily, since H I is bilinear in Fermi fields, so it obeys commutation relations: Ω T { φψ ψ } Ω = I [ {φ T I ψ I ψ I exp i d 4 z H I (z)]} I,connected = Sum of all connected diagrams with specified external points But, to use Wick s theorem, we must define time-ordering and normal ordering for fermion operators 6 Time-ordered products of Fermi fields: Suppose x and y are spacelike separated, with y >x Then ψ(x)ψ(y) = ψ(y)ψ(x) The RHS is already time-ordered, so T { ψ(y)ψ(x)} = ψ(y)ψ(x) If T is to act consistently on both sides, then Generalizing, T {ψ(x)ψ(y)} = ψ(y)ψ(x) T {ψ 1 ψ 2 ψ n } =(product of ψ s ordered by time, earliest to right) ( 1) N, where N is the number of interchanges necessary to bring the ordering on the LHS to theordering on therhs (Hereψ represents a general Fermi field, ψ or ψ) 61

Alan Guth, 8323 Lecture, May 13, 28, p 32 The Fermion Propagator: By this definition { T {ψ(x) ψ(y)} ψ(x) ψ(y) for x >y ψ(y)ψ(x) for y >x In free field theory we have already learned that T {ψ(x) ψ(y)} = d 4 p (2π) 4 i( p + m) p 2 m 2 + iɛ e ip (x y) S F (x y) 62 Normal-ordered products of Fermi operators: For p q, a s ( q )a s ( p) = a s ( p)a s ( q ) The RHS is normal-ordered, so one presumably defines N{ a s ( p)a s ( q )} = a s ( p)a s ( q ) IfN is to act consistently on both sides, then Generalizing, N{a s ( q )a s ( p)} = a s ( p)a s ( q ) N{product of a s and a s} = (product with all a s to theright) ( 1) N, where N is the number of interchanges necessary to bring the ordering on the LHS to theordering on therhs 63

Alan Guth, 8323 Lecture, May 13, 28, p 33 Wick's Theorem: T {ψ 1 ψ2 ψ 3 } = N{ψ 1 ψ2 ψ 3 + (all possiblecontractions)} A samplecontraction would be 64