8.323 Relativistic Quantum Field Theory I

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1 MIT OpenCourseWare Relativistic Quantum Field Theory I Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 Alan Guth, Lecture, May 13, 2008, p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.323: Relativistic Quantum Field Theory I Interacting Field Theories Tuesday, May 13, 2008 Alan Guth Alan Guth Massachusetts Institute of Technology 8.323, May 13, 2008 Time-Dependent Perturbation Theory Sample theory: λφ 4 Then L = λ ( µ φ) m φ φ ! H = H 0 + H int, where H int = d 3 x λ φ 4 ( x). 4! Goal: to perturbatively calculate matrix elements of the Heisenberg field φ( x,t)= e iht φ( x,0)e iht in thestate Ω, thetrueground stateof theinteracting theory. 1

3 Alan Guth, Lecture, May 13, 2008, p. 2. picture: Interaction Choose any reference time t 0, at which the interaction picture operators and Heisenberg operators will coincide. Define φ I ( x,t) e ih 0 (t t 0 ) φ( x,t 0 )e ih 0 (t t 0 ). At t = t 0, can expand Heisenberg φ and π in creation and annihilation operators: d 3 p ( ) φ( x,t 0 )= (2π) 1 a 2 p e i p x + a 2 p e i p x. π( x,t 0 )=φ ( x,t 0 )= E p d 3 p (2π) 1 ( ) ie 2 p a p e i p x + ie p a 2E p e i p x. p a p creates state of momentum p, but not energy E p not singleparticle. But [ ] [φ( x,t 0 ),π( y,t 0 )] = 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 p. 2Ep x 0 =t t 0 How to express φ( x,t): φ( x,t)=e ih(t t 0 ) e ih 0 (t t 0 ) φ I ( x,t)e ih 0(t t 0 ) e ih(t t 0 ) U (t, t 0 )φ I ( x,t)u(t, t 0 ), where U(t, t 0 )=e ih 0 (t t 0 ) e ih(t t 0 ). Differential equation for U: i t U(t, t 0)=e ih 0(t t 0 ) (H H 0 )e ih(t t 0 ) = e ih 0(t t 0 ) H int e ih(t t 0 ) where = e ih 0(t t 0 ) H int e ih 0 (t t 0 ) e ih 0(t t 0 ) e ih(t t 0 ) = H I (t)u(t, t 0 ), H I (t) =e ih 0 (t t 0 ) H int e ih 0 (t t 0 ) = d 3 x λ 4! φ4 I ( x,t). 3

4 Alan Guth, Lecture, May 13, 2008, p. 3. Solution to differential equation: i U(t, t 0 )=H I (t)u(t, t 0 ) t with U(t 0,t 0 )=I implies the integral equation t U(t, t 0 )=I i dt H I (t )U(t,t 0 ). t 0 To first order in H I, t U(t, t 0 )=I i dt 1 H I (t 1 ). t 0 To second order in H I, t t t 1 U(t, t 0 )=I i dt 1 H I (t 1 )+( i) 2 dt 1 dt 2 H I (t 1 )H I (t 2 ). t 0 t 0 t 0 t 0 t 0 t 0 To third order, t t 1 t 2 U(t, t 0 )=...+( i) 3 dt 1 dt 2 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 0 )=...+( i) 3 t =...+ ( i)3 3! t 0 dt 1 t t 0 dt 1 t1 t 0 dt 2 t t 0 dt 2 t2 t t 0 dt 3 H I (t 1 )H I (t 2 )H I (t 3 ) t 0 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 0 )=I +( i) T { [ exp i dt 1 H I (t 1 )+ ( i)2 t 0 2! t ]} dt H I (t ). t 0 t t 0 dt 1 t t 0 dt 2 T {H I (t 1 )H I (t 2 )}

5 Alan Guth, Lecture, May 13, 2008, p. 4. Generalize to arbitrary t 0 : U(t 2,t 1 ) T { [ t2 ]} exp i dt H I (t ) t 1, where (for t 1 <t 0 <t 2 ) U(t 2,t 1 )= T { [ t2 ]} { [ t0 ]} exp i dt H I (t ) T exp i dt H I (t ) t 0 t 1 = U(t 2,t 0 )U 1 (t 1,t 0 ). Given have U(t, t 0 )= e ih 0 (t t 0 ) e ih(t t 0 ), U(t 2,t 1 )= e ih 0 (t 2 t 0 ) e ih(t 2 t 0 ) e ih(t 1 t 0 ) e ih 0 (t 1 t 0 ) = e ih 0 (t 2 t 0 ) e ih(t 2 t 1 ) e ih 0 (t 1 t 0 ). 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

6 Alan Guth, Lecture, May 13, 2008, p. 5. Construction of true ground state Ω : Assumethat 0 has nonzero overlap with Ω : e iht 0 = e ie n T n n 0. n If T had large negative imaginary part, all other states would be suppressed relative to Ω. ( ) 1 Ω = lim e ih(t +t 0 ) 0 e ie 0 (T +t 0 ) Ω 0 T (1 iɛ) ( ) 1 = lim e ih(t +t 0 ) e ih 0 (T +t 0 ) 0 e ie 0 (T +t 0 ) Ω 0. T (1 iɛ) Recall U(t 2,t 1 )= e ih 0 (t 2 t 0 ) e ih(t 2 t 1 ) e ih 0 (t 1 t 0 ), so ( ) 1 Ω = lim U(t 0, T ) 0 e ie 0 (T +t 0 ) Ω 0. T (1 iɛ) 8 Similarly, Recall Ω = ( ) 1 lim e ie 0 (T t 0 ) Ω 0 0 U(T,t0 ). T (1 iɛ) φ( x,x 0 )= U (x 0,t 0 )φ I ( x,x 0 )U(x 0,t 0 ). So, for x 0 >y 0, Ω φ(x)φ(y) Ω = lim T (1 iɛ) 0 U(T,t 0)U(t 0,x 0 )φ I ( x,x 0 )U(x 0,t 0 ) U(t 0,y 0 )φ I ( y,y 0 )U(y 0,t 0 )U(t 0, T ) 0 Normalization factor = lim T (1 iɛ) 0 U(T,x0 )φ I ( x,x 0 )U(x 0,y 0 ) φ I ( y,y 0 )U(y 0, T ) 0 Normalization factor. But Normalization factor = Ω Ω 1, 9

7 Time-Dependent Perturbation Theory: Alan Guth, Lecture, May 13, 2008, p. 6. so Ω φ(x)φ(y) Ω = lim T (1 iɛ) { [ ]} T 0 T φ I (x)φ I (y)exp i dth I (t) T 0 { [ ]} T. 0 T exp i dth I (t) 0 T Causality: H I (t) = d 3 xh( x,t), so Ω φ(x)φ(y) Ω = { [ ]} 0 T φ I (x)φ I (y)exp i d 4 z H I (z) 0 { [ ]} 0 T. exp i d 4 z H I (z) 0 [ If z 1 and z 2 are spacelike-separated, ] their time ordering is frame-dependent. Need H I (z 1 ), H I (z 2 ) =0 to get same answer in all frames. 10 { [ ]} 0 T φ I (x)φ I (y)exp i d 4 z H I (z) 0 Ω φ(x)φ(y) Ω = { [ ]}. 0 T exp i d 4 z H I (z) 0 11

8 Alan Guth, Lecture, May 13, 2008, p. 7. Wick's Theorem Gian-Carlo Wick October 15, 1909 April 20, 1992 For more information see The National Academies Press Biographical Memoir 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: 0 T {φ(x 1 )φ(x 2 )...φ(x m )} 0 = all possiblefull contractions}. Example: 0 T {φ(x 1 )φ(x 2 )φ(x 3 )φ(x 4 )} 0 = 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

9 Alan Guth, Lecture, May 13, 2008, p. 8. Feynman Diagrams Example: 0 T {φ(x 1 )φ(x 2 )φ(x 3 )φ(x 4 )} 0 = 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 φ 4 (z) 4! { [ ] } Ω T {φ(x)φ(y)} Ω = 0 T φ(x)φ(y)+φ(x)φ(y) i d 4 zh I (z) { [ ]} 0 T φ(x)φ(y) i d 4 zh I (z) 0 ( ) 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! = 15

10 Alan Guth, Lecture, May 13, 2008, p. 9. Very Nontrivial Example: How many identical contractions are there? 1 3! 4! 2 8 symmetry factor Overall factor: 1 ( 1 ) 3 3! 4 3 4! = Symmetry Factors: 17

11 Momentum Space Feynman Rules: Alan Guth, Lecture, May 13, 2008, p. 10. Feynman Rules for λφ 4 Theory { [ ]} ( ) sum of all possiblediagrams 0 T φ(x)φ(y)exp i d4 z H I (z) 0 = with two external points. Rules: 18 d 4 p i ip (x y) D F (x y) = e. (2π) 4 p 2 m 2 + iɛ Vertex: d 4 ze ip 1 z e ip 2 z e ip 3 z e ip 4z =(2π) 4 δ (4) (p 1 + p 2 + p 3 + p 4 ). 19

12 Alan Guth, Lecture, May 13, 2008, p Disconnected Diagrams Consider Momentum conservation at one vertex implies conservation at the other. Graph is proportional to (2π) 4 δ (4) (0) Use Disconnected diagrams: (2π) 4 δ (4) (0) = (volumeof space) 2T. 21

13 Alan Guth, Lecture, May 13, 2008, 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

14 Alan Guth, Lecture, May 13, 2008, p. 13. For our example, Now look at denominator of matrix element: 24 Finally, Summarizing, Vacuum energy density: 25

15 Alan Guth, Lecture, May 13, 2008, p. 14. Final sum for four-point function: 26 Summary of Time-Dependent Perturbation Theory Example: The four-point function: 27

16 Definition of Cross Section: Definition of Decay Rate: Alan Guth, Lecture, May 13, 2008, p. 15. Cross Sections and Decay Rates 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 Number of scattering events of specified type Then σ. 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 = 1.Then Number of events = σ/a. 28 Number of decays per unit time Γ. Number of particles present Number of surviving particles at time t: N(t) =N 0 e Γt. Mean lifetime: ( ) 1 dn τ = dt t = 1/Γ. N 0 0 dt Half-life: 1 e Γt = = t 1/2 = τ ln

17 Unstable Particles as Resonances (Preview): Initial and Final States In- and Out-States: Alan Guth, Lecture, May 13, 2008, p. 16. Unstable particles are not eigenstates of H; they are resonances in scattering experiments. In nonrelativistic quantum mechanics, the Breit-Wigner formula 1 1 f(e) = σ. E E 0 + iγ/2 (E E 0 ) 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 1 p 2 m 2 + imγ 2E p (p 0 E p + i(m/e p )Γ/2), (p 0 ) 2 p 2 m 2 =(p 0 ) 2 2 E p = ( p 0 )( ) ( ) + E p p 0 E p 2E p p 0 E p. 30 Cross Sections and the S-Matrix 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 0,in was then interpreted as the amplitude for producing a set of final particles with momenta p 1... p N. 31

18 Alan Guth, Lecture, May 13, 2008, 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 States: Wavepacket One-particle incoming wave packet: d 3 k φ = (2π) 1 3 φ( k ) k,in, 2E k where φ φ = 1 = d 3 k 2 (2π) 3 φ( k ) =1. Two-particleinitial state: i b k B d 3 k A d 3 k B φ A ( k A ) φ B ( k B )e φ A φ B, b,in = k A k B, in, (2π) 3 (2π) 3 (2E A )(2E B ) 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 d 3 p f φ f ( p f ) φ 1... φ n, out = (2π) p p n, out. 2E f =1 f 33

19 Alan Guth, Lecture, May 13, 2008, p. 18. S-Matrix: The 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 it k A k B (2π) 4 δ (4) 1... p n k A + k B p f im(k A k B {p f }). 35

20 Alan Guth, Lecture, May 13, 2008, 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: ( ) n { } { } d 3 p f 1 P AB, b p p n = p (2π) p n S φ A φ B, 2E b. f f =1 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 k A d 3 k B φ ( A k A ) φ ( B k B ) (2π) 3 (2π) 3 (2 E A )(2 E B ) e i b ( kb k B ) p 1... p n S k A k B p 1... p n S k A k B. 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 ( ) ( k A k B = im k A k B { p f } (2π) 4 δ (4) k A + k B ) p f.,, 37

21 Alan Guth, Lecture, May 13, 2008, p. 20. ( ) Wefirst integrateove r k A and k B using δ (2) k B k B and δ (4) ( ) ( ) ( ) k A + k B p f = δ (2) k + k z A B p f δ ka + k z B p z f ( ) δ Ē A + Ē B E f, where the beam is taken along the z-axis. After integrating over k and k A B,we areleft with ( ) ( ) d ( ) k z d A k z B δ k z A + k z B pz f δ Ē A + Ē B E d z f k F (k z = A δ A), where the first δ-function was used to integrate k z B,and ( ( ) 2 F ( 2 ) z k k A )= k kz + m 2 + k + p z z + m 2 E f. A A B f A Then d ( ) 1 k z = A δ F ( k z z A), evaluated where F ( ka)=0. df d k z A 38 Rewriting ( ( ) 2 2 ) z F ( k )= 2 + k A k kz + m k + p z z + m 2 E f, A A B onefinds z z k z df k p A f A = d k z A E A E B ( ) z A + k z B z Remembering the δ-function constraint δ k p f from theprevious slide, one has z z df k A kb z z = = v A v B. z d ka E A E B z What values of k z satisfy theconstraint F ( k A A)=0? Therearetwo solutions, since F ( z ka )=0can be manipulated into a simple quadratic equation. (To see this, move 2 onesquareroot to therhs of theequation and squareboth sides. The( k z A) 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 k A = 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 f A

22 Alan Guth, Lecture, May 13, 2008, p. 21. Then n dσ = f =1 d 3 p f 1 (2π) 3 2E f d 3 k A (2π) 3 d 3 k B φ A ( k A ) 2 φ B ( k B ) 2 (2π) 3 (2E A )(2E B ) v z A vz B ( ) M k A k B { p f } 2 ( (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 vz B, M k A k B { p f } 2, 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 (2π) 3 d 3 k B (2π) 3 φ A( k A ) 2 φ B ( k B ) 2 =1,,. 40 so finally dσ = M ( p A p B { p f }) 2 (2E A )(2E B ) v z B dπ n(p A + p B ). A vz 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

23 Alan Guth, Lecture, May 13, 2008, 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) 4 δ (4) p A + p B p f (2π) 3 2E f f =1 d 3 p 1 d 3 p 2 1 = (2π) 4 δ (4) (p A + p B p 1 p 2 ) (2π) 3 (2π) 3 (2E 1 )(2E 2 ) d 3 p 1 1 = (2π)δ(E cm E 1 E 2 ) (2π) 3 (2E 1 )(2E 2 ) ( ) 2 p 1 dp 1 1 =dω (2π)δ E cm p2 1 + m2 1 p2 1 + m2 (2π) 3 (2E 2 1 )(2E 2 ) 2 p 1 1 p 1 p 1 1 =dω + (2π) 2 (2E 1 )(2E 2 ) E 1 E 2 p 1 =dω. 16π 2 E cm 42 The two-particle final state, center-of-mass cross section is then ( ) dσ = dω cm p A M ( p A p B p 1 p 2 ) 2 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 ( ) dσ = dω cm v z A v z B = 2 p A E A = 4 p A E cm, M 2 64π 2 E 2 cm (all masses equal). 43

24 Evaluation of the Decay Rate: Alan Guth, Lecture, May 13, 2008, p. 23. 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 z z dπ n (p A + p B ), (2E A )(2E B ) v v A B dγ = M ( p A { p f }) 2 dπ n (p A ). 2E 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 4.6. 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 { [ ]} = 0 T φi (x 1 )...φ I (x )exp i d 4 z H I (z) n 0 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

25 Alan Guth, Lecture, May 13, 2008, 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) 4 δ (4) P p f, (2π) 3 2E f f =1 and the differential cross section is dσ = M ( p A p B { p f }) z z dπ n (p A + p B ). (2E A )(2E B ) v v A 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 B Special case two-particle final states, in the center-of-mass frame: ( ) 2 dσ p A M( p A p B p 1 p 2 ) = dω 64π 2 E z z A E B (E A + E B ) v v cm A B M 2 = (if all masses are equal). 64π 2 E 2 cm Decay rate from S-matrix elements: dγ = M ( p A { p f }) 2 dπ n (p A ). 2E 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

26 Alan Guth, Lecture, May 13, 2008, 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(0)e ip x p = e ip x Ω φ(0) 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, 2 one can show that it contains poles when any p i is on its mass shell, p i = m 2 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

27 Alan Guth, Lecture, May 13, 2008, p. 26. Calculation of 2 2 Scattering: Normalization conventions: p = 2E p a ( p) 0, [ 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 ) 0 a 1 a 2 a A a B 0 [ =(2E A )(2E B )(2π) 6 δ (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. 50 To first order in H I : { p 1 p 2 T i λ I 4! { = p 1 p 2 N i λ 4! I 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 [ 6 φ(x)φ(x)+ 3 4! 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 [ (2π) 3 a I ( ] k)e ik x + a I ( k )e ik x, 2E k so φ + I (x) p I = e ip x 0 I, where φ + I and φ I refer to the parts of φ I (x) containing annihilation and creation operators, respectively. 51 ].

28 Alan Guth, Lecture, May 13, 2008, 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 4! d 4 x I p 1 p 2 p A p B I = i λ 4! d 4 x I p 1 p 2 p A p B I, so this term also contributes only to the uninteresting 1 part of S = 1 + it. 52 The singly contracted term i 6λ 4! d 4 x I p 1 p 2 φ(x)φ(x) p A p B I contains terms where one φ(x) contracts with an incoming particleand theother contracts with an outgoing particle, giving the Feynman diagrams 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

29 Alan Guth, Lecture, May 13, 2008, 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

30 Alan Guth, Lecture, May 13, 2008, p. 29. Amputation: Consider the 2nd order diagram Contribution is 1 d 4 p i d 4 k i 2 (2π) 4 p 2 m 2 (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 1 = p 2 m 2 0, 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 0, 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

31 Alan Guth, Lecture, May 13, 2008, p. 30. 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

32 Time-ordered products of Fermi fields: Alan Guth, Lecture, May 13, 2008, 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 φ...ψ... ψ... Ω { [ ]} = 0 T φ I...ψ I...ψ I...exp i d 4 z H I (z) 0 I = Sum of all connected diagrams with specified external points. I,connected But, to use Wick s theorem, we must define time-ordering and normal ordering for fermion operators. 60 Suppose x and y are spacelike separated, with y 0 >x 0.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

33 The Fermion Propagator: Normal-ordered products of Fermi operators: Alan Guth, Lecture, May 13, 2008, p. 32. By this definition { T {ψ(x) ψ(y)} ψ(x) ψ(y) ψ(y)ψ(x) for x 0 >y 0 for y 0 >x 0. In free field theory we have already learned that 0 T {ψ(x)ψ (y)} d 4 p i( p + 0 = (2π) 4 p 2 m 2 m) + iɛ e ip (x y) S F (x y). 62 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 )} = s a ( p)a s ( q ). If N is to act consistently on both sides, then Generalizing, s N{a s ( q )a ( p)} = a ( p)a s ( q ). s 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

34 Alan Guth, Lecture, May 13, 2008, p. 33. Theorem: Wick's T {ψ 1 ψ 2 ψ 3...} = N{ψ 1 ψ 2 ψ (all possiblecontractions)}. A samplecontraction would be 64