NOTES AND SOLUTIONS FOR QUANTUM FIELD THEORY BY MARK SREDNICKI.

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1 NOTES AND SOLUTIONS FOR QUANTUM FIELD THEORY BY MARK SREDNICKI. ERNEST YEUNG AND ERNEST YEUNG Solutons for Quantum Feld Theory. Mark Srednck. Cambrdge Unversty Press. 7. ISBN hardback Contents Part I Spn Zero 3. Attempts at relatvstc quantum mechancs 3 Problems 3. Lorentz nvarance 6 Problems 6 3. Canoncal quantzaton of scalar felds 9 Problems 9 4. The spn-statstcs theorem 5 5. The LSZ reducton formula 5 6. Path ntegrals n quantum mechancs 7 7. The path ntegral for the harmonc oscllator 9 8. The path ntegral for free-feld theory 9. The path ntegral for nteractng feld theory 8. Scatterng ampltudes and the Feynman rules 34. Cross sectons and decay rates 4. Dmensonal analyss wth c The Lehmann-Källén form of the exact propagator Loop correctons to the propagator The one-loop correcton n Lehmann-Källén form Loop correctons to the vertex Other PI vertces Hgher-order correctons and renormalzablty Perturbaton theory to all orders 6. Two-partcle elastc scatterng at one loop 6 Problems 6. The quantum acton 6 Problems 6. Contnuous symmetres and conserved currents 6 3. Dscrete symmetres: P, T, C and Z Nonabelan symmetres Unstable partcles and resonnaces Infrared dvergences Other renormalzaton schemes The renormalzaton group Effectve feld theory 7 3. Spontaneous symmetry breakng 7 3. Broken symmetry and loop correctons 7 Date: 4 avrl Mathematcs Subject Classfcaton. Quantum Feld Theory. Key words and phrases. Quantum Feld Theory, QFT. I wrte notes, revew papers, and code and make calculatons for physcs, math, and engneerng to help wth educaton and research. Wth your support, we can keep educaton and research materal avalable onlne, openly accessble, and free for anyone, anytme. If you lke what I m tryng to do for physcs educaton research, please go to my Tlt/Open or Patreon crowdfundng campagn, read the msson statement, share the page, and contrbute fnancally f you can. ernestyalumn.tlt.com

2 3. Spontaneous breakng of contnuous symmetres 7 Part II Spn One Half Representatons of the Lorentz group 7 Problems Left- and rght-handed spnor felds 73 Problems Manpulatng spnor ndces 74 Problems Lagrangans for spnor felds 75 Problems Canoncal quantzaton of spnor felds I Spnor technology Canoncal quantzaton of spnor felds II Party, tme reversal, and charge conjugaton 89 Problems LSZ reducton for spn-one-half partcles 9 4. The free fermon propagator The path ntegral for fermon felds Formal development of fermonc path ntegrals The Feynman rules for Drac felds Spn Sums Gamma matrx technology Spn-averaged cross sectons The Feynman rules for Majorana felds 9 5. Massless partcles and spnor helcty 9 Problems 9 5. Loop correctons n Yukawa theory 9 5. Beta functons n Yukawa theory Functonal determnants 9 Part III - Spn One Maxwell s equatons Electrodynamcs n Coulomb gauge 9 Problems LSZ reducton for photons The path ntegral for photons Spnor electrodynamcs Scatterng n spnor electrodynamcs Spnor helcty for spnor electrodynamcs Scalar electrodynamcs Loop correctons n spnor electrodynamcs The vertex functon n spnor electrodynamcs The magnetc moment of the electron Loop correctons n scalar electrodynamcs Beta functons n quantum electrodynamcs Ward denttes n quantum electrodynamcs I 68. Ward denttes n quantum electrodynamcs II 69. Nonabelan gauge theory Notes on the materal n secton Group representatons 7. The path ntegral for nonabelan gauge theory 3 7. The Feynman rules for nonabelan gauge theory The beta functon n nonabelan gauge theory BRST symmetry Chral gauge theores and anomales Anomales n global symmetres Anomales and the path ntegral for fermons 4

3 78. Background feld gauge Gervas-Neveu gauge 4 8. The Feynman rules for N N matrx felds 4 8. Scatterng n quantum chromodynamcs 4 8. Wlson loops, lattce theory, and confnement Chral symmetry breakng Spontaneous breakng of gauge symmetres Spontaneously broken abelan gauge theory Spontaneously broken nonabelan gauge theory The Standard Model: gauge and Hggs sector The Standard Model: lepton sector The Standard Model: quark sector 6 9. Electroweak nteractons of hadrons 6 9. Neutrno masses 6 9. Soltons and monopoles Instantons and theta vacua Quarks and theta vacua Supersymmetry 6 Problems. Soluton.. Part I Spn Zero. Attempts at relatvstc quantum mechancs Recall {α j, α k } ab δ jk δ ab, {α j, β} ab β ab δ ab.7 Suppose β ψ λ ψ β ψ λ ψ ψ ψ λ or λ ± Use hnt. T rα β T rα βα T rα β T rα β {α, α } ab α ab δ ab so α T rβ Snce T rβ and λ β ± and T rβ λ β, then β must be even-dmensonal for a set of ± s to sum up and cancel each other. Lkewse, α ψ λ ψ α ψ λα ψ λ ψ ψ λ ± T rα α j T rα α j α T rα α j T rα α j so T rα j So then α j even dmensonal as well. Soluton.. Recall the poston bass Schrödnger equaton for n partcles. n t ψ n j m j + Ux j + V x j x k ψ.3 Recall H j j k d 3 xa x m + Ux ax + d 3 xd 3 yv x ya xa yayax.3 ψ, t d 3 x,..., d 3 x n ψx,... x n, ta x... a x n.33 and ψ, t H ψ, t. t 3

4 Also recall the commutaton and antcommutaton rules: [ax, ax ] [a x, a x ] [ax, a x ] δ 3 x x Usng some shorthand notaton, H ψ, t d 3 xa x m + U x a x + Now.3 {ax, ax } {a x, a x } d 3 xd 3 yv x y a xa ya y a x {ax, a x } δ 3 x x.38 d 3 x... d 3 x n ψx,..., x n, ta x,..., a x n a x a x... a x n ±a x a x + δ x x a x... a x n ±a x ±a x a x + δ x x a x 3... a x n + δ x x a x... a x n If ths term s appled to, note a x. a x a x... a x n n n ± j δ x xj a x k For the knetc and self-potental terms of H ψ, t, d 3 xa n n x m + U x d 3 x... d 3 x n ψx,..., x n, t ± j δ x xj a x k Evaluate d 3 x, n partcular, wth δ x xj, n a x j m j + U xj d 3 x,..., d 3 x n ψx,..., x n, t± j j To antcommmute a x j place. n j Now, on a x a x,..., a x n. Now j k j j n k j k j a x k over to ts ordered place n n k j a x k ± j, j antcommutatons must take m j + U xj d 3 xv x y a xa ya y a x a x,..., a x n a y d 3 x,..., d 3 x n ψx,..., x n, ta x,..., a x n n ± j d 3 xv x y a xa ya y δ x xj j n n ± j V xj ya x j a ya y a x k j n k j a x k n δ xk y m k j So then d 3 xd 3 yv x y a xa ya y a x a x,..., a x n d 3 y n j k j k j n l j,k a x l n ± j V xj ya x j a y j n V xj x k ± j +m a x j a x k n l j,k a x l n k j n δ xk y m k j a x k n l j,k a x l To brng a x k back n ts ordered place wth n l j,k a x l, then m antcommutatons must occur we really are repeatng what we dd before, exactly, agan. So there s another factor of m. Lkewse for a x j, to be back n ts ordered place, another factor of j results. n j k j n V x j x k a x,..., a x n 4

5 Now n n j k j counts each unque par twce so n n n j V x j x k a x,..., a x n V x j x k a x,..., a x n j k j d 3 xd 3 yv x ya xa ya y a x n j V x j x k j k j k d 3 x,..., d 3 x n ψx,..., x n, ta x,..., a x n d 3 x,..., d 3 x n ψx,... x n, ta x,..., a x n ψ, t H ψ, t s satsfed. t Problem.3. Show explctly that [N, H], where H s gven by eq..3 and N by eq..35. Soluton.3. H Now Recall d 3 xa x m + Ux ax + d 3 xd 3 yv x ya xa yayax.3 d xa x m x + U x a x + d xd yv x ya xa ya y a x notaton N d 3 xa xax d xa xa x notaton.33 NH d za za z d xa x m x + U x a x + d za za z d xd yv x ya xa ya y a x HN d xa x m x + U x a x d za za z + d xd yv x ya xa ya y a x d za za z [ax, ax ] ± [a x, a x ] ± [ax, a x ] ± δ 3 x x Treat x, y, z as ndependent coordnates. a x a + z ±a za x + δ 3 x z [a x, a z ] ± [a z, a x] ± δ 3 z x a xa z ±a z a x δ 3 z x [ m x + U x, a z] [H x, a z] [H x, a z ] d zd xa xh x a x a za z d zd x a za z a xh x a x + d za zh z a z d za z a z d zd x a za z a xh x a x Notce even f the antcommutator rules apples, a z swtched wth a x and a x, for a factor of overall. Lkewse for a [ z wth a x and ] a x. V x y, a z Note that [V x y, a z ]. V x ya xa ya y a x a za z V x ya xa ya y ±a z a x + δx z a z V x ya xa ya za y + δz ya x a z + V x ya xa ya y a z δx z d z d za zv x ya xa ya z a y a x + V xya xa ya x a y + V x ya xa ya y a x d za zv x ya x±a z a y δz ya y a x + V x ya xa y{a x, a y } 5

6 d za zv x ya z a xa ya y a x + d za zv x y δz xa ya y a x + d za zv x ya x δz ya y a x + +V x ya xa y{a x, a y } d za za z V x ya xa ya y a x + V x y{a x, a y}a y a x + V x ya xa y{a x, a y } d za za z V xya xa ya y a x + V x ya xa ya x a y a ya xa y a x d za za z V xya xa ya y a x Then ndeed [N, H]. for both commutaton and antcommutator rules. Lorentz nvarance Problems. Problem.. Verfy that eq..8 follows from eq..3. g µν Λ µ ρλ ν σ g ρσ.3 Soluton.. Recall Λ µ ν δ ν µ + δω ν µ.8. Then g µν δ µ ρ + δ µ ρ δ ν σ + δω ν σ g ρσ g µν δ µ ρ δ ν σ + δ ν σδω µ ρ + δ µ ρ δω ν σ + δω µ ρ δω ν σ Neglect nd. order nfntesmals: δω µ ρ δω ν σ g ρσ g ρσ + gµσδω µ ρ + g ρν δω ν σ Recall A k gl A lk ; g k g kl δ l, and g µσ symmetrc. g µσ δω µ ρ g ρν δω ν σ Problem.. Soluton.. g µσ g µl δω lρ g ρν g νm δω mσ δ m ρ δω mσ δω ρσ g σµ g µl δω lρ δ l ρδ lρ δω σρ δω σρ δω ρσ antsymmetrc tensor Verfy that eq..4 follows from UΛ UΛ UΛ UΛ Λ Λ. Λ Λ Λ Λ + δω Λ + Λ δω Λ UΛ Λ Λ + Λ δω Λ ρσ M ρσ + Λ a ρδω ab Λ b σm ρσ Usng UΛ UΛ UΛ UΛ Λ Λ, UΛ U + δω + δω µνm µν UΛ UΛ UΛ UΛ UΛ + UΛ δω µν M µν UΛ UΛ δω µν M µν UΛ δω µν UΛ M µν UΛ δω µν Λ µ ρλ ν σm ρσ Note that δω µν s just an entry n a rank- tensor, a number, and for arbtrary Lorentz transformaton, each a unquelly determned quantty δω µν. 6 of them for antsymmetrc δω µν. UΛ M µν UΛ Λ µ ρλ ν σm ρσ Problem.3. Verfy that eq..6 follows from eq..4. Soluton.3. Recall that UΛ M µν UΛ Λ µ ρλ ν σm ρσ.4. UΛ M µν UΛ δω abm ab M µν + δω cdm cd M µν δω abm ab M µν + δω cdm cd M µν δω abm ab M µν + M µν δω cd M cd M µν + δω cdm µν M cd δω ab M ab M µν We re only concerned wth terms n frst order of δω. Now [M µν, M ρσ ] M µν M ρσ M ρσ M µν So then runnng through all the ndces and pckng out ndces we re nterested n partcularly ρ, σ, δω cd M µν M cd δω ab M ab M µν δω ρσ M µν M ρσ δω ρσ M ρσ M µν + δω σρ M µν M σρ δω σρ M σrho M µν δω ρσ [M µν, M ρσ ] 6

7 where we used δω ρσ δω σρ M ρσ M σρ antsymmetry. Now Λ µ ρλ ν σm ρσ Λ µ aλ ν b M ab + δω µ a + δω ν b M ab + δω µ a M ab + δω ν b M ab M ab + δω lb g νl M ab + δω ma g µm M ab Consderng only the frst order nfntesmal, and note that for M ax, M yb, a, b run from to 3, δω lb g νl M ab + δω ma g µm M ab δω ρσ g νρ M aσ + δω σρ g νσ M aρ + δω ρσ g µρ M σb + δω σρ g µσ M ρb For a µ and b ν, δω ρσ {g µρ M σb g νρ M aσ + g µσ M ρb + g νσ M aρ } [M µν, M ρσ ] {g µρ M bσ g νρ M aσ g µσ M bρ g νσ M aρ } {g µρ M νσ g νρ M µσ g µσ M νρ g νσ M µρ } Note that f a b, we get. If b µ, a ν, then g µρ M µσ g νρ M νσ g µσ M µρ g νσ M νρ M ρσ M ρσ M σρ + M σρ If b ρ, a σ, If vce-versa, modulo a factor g µρ M ρσ g νρ M σσ g µσ M ρρ g νσ M σρ g µρ M ρσ g νσ M ρσ g µρ g νσ M ρσ g µρ M σσ g νρ M ρσ g µσ M σρ g νσ M ρρ g νρ M ρσ g µσ M σρ g µσ g νρ M ρσ Problem.4. Verfy that eq..7 follows from eq..6. Soluton.4. Recall [M µν, M ρσ ] {g µρ M νσ g νρ M µσ g µσ M νρ g νσ M µρ }.6 [J, J j ] ɛ jk J k [J, K j ] ɛ jk K k [K, K j ] ɛ jk J k.7 Recall J ɛ jkm jk. Then ɛ jk J k ɛ kj ɛ klmm lm δ lδ jm δ m δ jl M lm M j M j M j Recall K M. [K, K j ] [M, M j ] g j M g j M g M j + g M j M j ɛ jk K k ɛ jk M k. [J, K j ] [ ɛ klm kl, M j ] ɛ kl [M kl, M j ] ɛ kl ɛ kl M k M k ɛ jk M k ɛ jk K k g k g l for k, l. Problem.5. Verfy that eq..8 follows from eq..5. Soluton.5. Recall UΛ P µ UΛ Λ µ ν P ν.5. Let Λ + δω. Keep frst order nfntesmals. g kj M l g lj M k g k M lj + g l M kj UΛ P µ UΛ δω abm ab P µ δω cdm cd P µ δω abm ab P µ + P µ δω cd M cd a, b, c, d run from to 3. We sought obtanng term δω ρσ 7

8 δω ρσ P µ M ρσ + δω σρ P µ M σρ δω ρσ M ρσ P µ δω σρ M σρ P µ δω ρσ [P µ, M ρσ ] For the rght hand sde of.5, Λ µ ν P ν + δω µ ν P ν P ν + g µk δω kν P ν δω kν g µk P ν ν, k run from to 3. Extract terms nvolvng ρ, σ from the sum. So then g µρ δω ρσ P σ + g µσ δω σρ P ρ δω ρσ g µρ P σ g µσ P ρ [P µ, M µν ] g µσ P ρ g µρ P σ Problem.6. Verfy that eq..9 follows from eq..8. Soluton.6. Usng.8: [P µ, M ρσ ] g µσ P ρ g µρ P σ, [J, H] ɛ [ jk M jk, H ] ɛ jkg k P j g j P k j, k [J, P j ] ɛ jkg jl [ P l, M jk] assume l,, 3 ɛ jkg jl g lk P j g lj P k ɛ jk P k P k ɛ jk P k [K, H] [ M, H ] [ H, M ] g P G P P [K, P j ] [ M ] [, P j gjk M, P k] g jk [P k, M ] g jk g k P g k P δ j H Note, I use the metrc, g µν. Problem.8. a Let Λ + δω n eq..6, and show that where [ϕx, M µν ] L µν ϕx,.9 L µν xµν x νµ,.3 b Show that [[ϕx, M µν ], M ρσ ] L µν L ρσ ϕx. c Prove the Jacob dentty. [[A, B], C] + [[B, C], A] + [[C, A], B]. Hnt: wrte out all the commutators. d Use your results from parts b and c to show that [ϕx, [M µν, M ρσ ]] L µν ρσ L ρσ L µν ϕx..3 e Smplfy the rght-hand sde of eq..3 as much as possble. f Use your results from part e to verfy eq..6, up to the possblty of a term on the rght-hand sde that commutes wth ϕx and ts dervatves. Such a term, called a central charge, n fact does not arse for the Lorentz algebra. Soluton.8. a Gven UΛ ϕxuλ ϕλ x, Λ + δω, UΛ ϕxuλ I δω µνm µν Keep up to st. order terms: ϕx + ϕx δ abm ab ϕx δω µνm µν ϕx + δω µν ϕxm µν + δω µνδω ab M µν ϕxm ab ϕx δω µν[m µν, ϕx] µ, ν run through,... 3 n the summaton. There ll be a term δω νµ M νµ δω µν M µν δω µν M µν to be ncluded. Hence, a factor of. Now, Λ x Λ ρ νx ν Λ ρ νx ν δ ρ ν + δω ρ νx ν 8

9 b c d e ϕλ x ϕx ρ + δω ρ νx ν ϕx ρ + δω ρ νx ν ρ ϕ ϕx + g ρα δω να x ν g ρβ β ϕ ϕx + g ρβ g ρα δω να x ν β ϕ ϕx + δω να x ν α ϕ Let ν, α run through... 3, pckng out terms nvolvng µ, ν δω µν x µ ν + δω νµ x ν µ δω µν x µ ν x ν µ [M µν, ϕx] x µ ν x ν µ ϕ L µν ϕx [[ϕx, M µν ], M ρσ ] [L µν ϕx, M ρσ ] L µν L ρσ ϕx + [L µν, M ρσ ] ϕx L µν L ρσ ϕx Snce M ρσ doesn t depend on x. Recall M µν s a hermtan operator that s a generator of the Lorentz group. [[A, B], C] + [[B, C], A] + [[C, A], B] [A, B] C C[A, B] + [B, C]A A[B, C] + [C, A]B B[C, A] AB BAC CAB BA + BC CBA ABC CB + CA ACB BCA AC ABC BAC CAB + CBA + BCA CBA ABC + ACB + CAB ACB BCA + BAC 3. Canoncal quantzaton of scalar felds Problems. Problem 3.. Derve eq. 3.9 from eqs. 3., 3.4. Soluton 3.. Usng ak d 3 xe kx [ ϕx + ωϕx] d 3 xe kx ϕx 3. Then we want I wll also use [ak, ak ] [a k, a k ] Πx ϕx 3.4 [ak, a k ] π 3 ωδ 3 k k 3.9 [ϕx, t, ϕx, t] [Πx, t, Πx, t] [ϕx, t, Πx, t] δ 3 x x [ak, ak ] d 3 xd 3 x e kx+k x [ ϕx + ωϕx, ϕx + ωϕx ] [ ϕx + ωϕx, ϕx + ωϕx ] [ ϕ x, ϕ x ] + [ωϕ x, ϕ x ] + [ ϕ x, ωϕ x ] + ω [ϕ x, ϕ x ] Now [ϕ x, ϕ x ], and ϕ ϕ/ t ϕ Π, for ths partcular Lagrangan. We also have [ ϕ x, ϕ x ] [ ϕ x, ϕ x ] [ϕ x, ϕ x ] δ 3 x x [ϕ ϕ x, ϕ x ] [ ϕ x, ϕ x ] δ 3 x x δ 3 x x [ak, ak ] [ a k, a k ]? a k d 3 xe kx [ ϕx + ωϕx] recall we had mposed the condton, ϕ ϕ. [ a k, a k ] [a k, a k ] d 3 xd 3 x e kx+k x [ ϕx + ωϕx, ϕx + ωϕx ] d 3 xd 3 x e kx+k x [ ϕ x, ϕ x ] + [ωϕ x, ϕ x ] + [ ϕ x, ωϕ x ] + ω [ϕ x, ϕ x ] 9

10 Agan [ϕ x, ϕ x ] [ ϕ x, ϕ x ]. [ϕ x, ϕ x ] δ 3 x x [ ϕ x, ϕ x ] δ 3 x x δ 3 x x [ a k, a k ] [ ak, a k ] d 3 xd 3 x e kx k x ϕx + ωϕx, ϕx + ωϕx Now [ ϕ x + ωϕ x, ϕ x + ωϕ x ] [ ϕ x, ϕ x ] ω[ϕ x, ϕ x ] + ω[ ϕ x, ϕ x ] + ω [ϕ x, ϕ x ] ωδ 3 x x + ω δ 3 x x ωδ 3 x x [ ak, a k ] d 3 xd 3 x e kx k x ωδ 3 x x d 3 xωe k k x π 3 ωδ 3 k k Problem 3.. Use the commutaton relatons, eq. 3.9, to show explctly that a state of the form k... k n a k... a k n s an egenstate of the hamltonan, eq. 3.3, wth egenvalue ω + + ω n. The vacuum s annhlated by ak, ak, and we take Ω E n eq Soluton 3.. Recall that [ak, ak ] H [a k, a k ] [ak, a k ] π 3 ωδ 3 k k 3.9 dkωa kak + E Ω V EΩ dkωa kak Lettng A π 3 ω, δ 3 k k δ k, H k... k n dkωa kaka k... a k n Consder a kaka k... a k n a k a ka... a n a k a a k + A δ 3 ka... a n A δ k a k a... a n + a k a a a k + A δ k a 3... a n By nducton, A δ k a k a... a n + A δ k a k a a 3... a n + a k a a a 3 a k + A δ k3 a 4... a n... A n j j δ kj a k a k + a k a... a na k Note that a k. So note ω ωk k m / and lettng ωk j ω j, n n H k... k n ω j a... a n l k j ω j k... k n Note that A factor, A π 3 ω canceled by our Lorentz nvarant dfferental, dk Problem 3.3. and hence that Use UΛ ϕxuλ ϕλ x to show that UΛ akuλ aλ k, UΛ a kuλ a Λ k 3.34 UΛ k... k n Λk... Λk, 3.35 where k... k n a k... a k n s a state of n partcles wth momenta k,..., k n. d 3 k π 3 ω n the ntegraton.

11 Soluton 3.3. Incomplete, 95 Gven UΛ ϕxuλ ϕλ x, UΛ akuλ UΛ d 3 xe kx [ ϕx + ωϕx]uλ d xe kx UΛ ϕxuλ + ωuλ ϕxuλ d xe kx UΛ ϕxuλ + ωϕλ x Now UΛ ϕxuλ UΛ ϕxuλ ϕλ x So then for the expresson before, d xe kx ϕλ x + ωϕλ x d xe Λ kλ x ϕλ x + ωϕλ x aλ k Problem 3.4. Recall that T a ϕxt a ϕx a, where T a exp P µ a µ s the space-tme translaton operator, and P s dentfed as the hamltonan H. a Let a µ be nfntesmal, and derve an expresson for [P µ, ϕx]. b Show that the tme component of your result s equvalent to the Hesenberg equaton of moton ϕ [H, ϕ]. c For a free feld, use the Hesenberg equaton to derve the Klen-Gordon equaton. d defne a spatal momentum operator P d 3 xπx ϕx Use the canoncal commutaton relatons to show that P obeys the relaton you derved n part a. e Express P n terms of ak and a k. Soluton 3.4. INCOMPLETE 98. a Now T a ϕxt a ϕx a. T a exp P µ a µ. Let a µ be nfntesmal. Then b c d e T a ϕxt a + P µ a µ ϕx P µ a µ ϕx + P µ a µ ϕx ϕxp µ a µ ϕx + µ ϕa µ Problem 3.5. keep up to Oa µ terms [P µ, ϕx] µ ϕ Consder a complex that s, nonhermtan scalar feld ϕ wth lagrangan densty L µ ϕ µ ϕ m ϕ ϕ + Ω a Show that ϕ obeys the Klen-Gordon equaton. b Treat ϕ and ϕ as ndependent felds, and fnd the conjugate momentum for each. Compute the hamltonan densty n terms of these conjugate momenta and the felds themselves but not ther tme dervatves. c Wrte the mode expanson of ϕ as ϕx dk[ake kx + b ke kx ] d Assumng canoncal commutaton rleatons for the felds and ther conjugate momenta, fnd the commutaton relatons obeyed by ak and bk and ther hermtan conjugates. e Express the hamltonan n terms of ak and bk and ther hermtan conjugates. What value must Ω have n order for the ground state to have zero energy? Soluton 3.5. a The acton s S d 4 L d 4 x µ ϕ µ ϕ m ϕ ϕ + Ω δs d 4 x µ δϕ µ ϕ µ ϕ µ δϕ m δϕ ϕ m ϕ δϕ

12 From ntegraton by parts, d 4 x µ δϕ µ ϕ d 4 x µ ϕ µ δϕ Ω Ω ds δϕ µ ϕ ds δϕ µ ϕ d 4 xδϕ µ µ ϕ d 4 xδϕ µ µ ϕ δs d 4 x{ µ µ ϕ m ϕδϕ + µ µ ϕ m ϕ δϕ} We obtan Klen-Gordon: µ µ m ϕ µ µ m ϕ extremze acton If you wanted to make ths explct, note that δϕ + δϕ Rδϕ, or assume treat ϕ, ϕ as ndependent whch we assume n the next part. b µ ϕ µ ϕ ncludes ϕ ϕ term n the sum. In Srednck s notaton, µ ϕ µ ϕ ϕ ϕ + ϕ ϕ. The conjugate momenta are Π ϕ L ϕ ϕ Π ϕ Recall H Π ϕ L, defnton of Hamltonan. ϕ H Π ϕ ϕ + Π ϕ ϕ µ ϕ µ ϕ m ϕ ϕ + Ω +Πϕ Π ϕ + Π ϕ Π ϕ + µ ϕ µ ϕ + m ϕ ϕ Ω Π ϕ Π ϕ + ϕ ϕ + m ϕ ϕ Ω c Wrtng the mode expanson of ϕ as ϕx dk[ake kx + b ke kx ] dka k e kx + b k e kx Then ϕ ϕ ϕ dka k e kx + b k e kx dka k ωe kx + b k ωe kx dka k ωe kx + b k ωe kx d 3 xe k x ϕ x d 3 xe k x ka k e kx + b k e kx ka t k π 3 δk + k + eω+ω dkb k π 3 δk k a k e ωt + bk π ω d 3 xe k x ϕ x d 3 xe k x ka k ωe kx + b k ωe kx kω d 3 x e xk +k a k exk k b k eωt a k bk Lkewse, d 3 xe kx ϕx ak ω + b ke ωt ω d 3 xe kx ϕx ak d 3 xe kx ϕ x a k ω eωt + bk ω + b ke ωt d 3 xe kx ϕ x a k e ωt + bk ak bk d 3 xe kx ωϕx + ϕ d 3 xe kx ωϕ x + ϕ x

13 d Assume [ϕx, t, ϕx, t] [Πx, t, Πx, t] [ϕx, t, Πx, t] δ 3 x x In our case ϕ, ϕ treated as ndependent felds. Recall Π ϕ ϕ, Π ϕ ϕ. Assume further, treatng ϕ, ϕ as ndependent, [ ϕx, t, ϕ x, t ] [ ak, a k ] [ bk, b k ] [Π ϕ x, t, Π ϕ x, t] [ Π ϕ x, t, Π ϕ x, t ] [ Π ϕ x, t, Π ϕ x, t ] [ϕx, t, Π ϕ x, t] δ 3 x x, [ϕ x, t, Π ϕ x, t] δ 3 x x d 3 x d 3 x e kx e k x [ωϕx + ϕ, ω k ϕ x ϕ ] d 3 xd 3 x e kx k x ω k [ Π ϕ, ϕ ] + ω[ϕx, Π ϕ ] d 3 xd 3 x e kx k x ω k δ 3 x x + ωδ 3 x x d 3 xe kx k x ω k π 3 ωδ 3 k k d 3 x d 3 x e kx e k x [ωϕ dag x + ϕ, ω k ϕx ϕx ] d 3 xd 3 x e kx k x ω[ Π ϕ, ϕ] + ω[ϕ x, Π ϕ ] d 3 xd 3 x e kx k x ω δ 3 x x + ωδ 3 x x d 3 xe kx k x ω π 3 ωδ 3 k k [ak, ak ] d 3 xd 3 x e kx e k x [ωϕx + ϕ, ω k ϕx + ϕx ] snce [ϕx, ϕx ] [ ϕx, ϕ x ] [ ϕx, Πϕ x ] [ a k, a k ] d 3 xd 3 x e kx e k x [ωϕ x ϕ, ω k ϕ x ϕ x ] snce [ ϕ x, ϕ x ] [ ϕ x, ϕ x ] [ ϕ x, Π ϕ x ] Lookng at bk comprsed of ϕ x, ϕ x and b k comprsed of ϕx, ϕx. Then, lkewse wth a, a, [bk, bk ] [ b k, b k ] [ak, bk ] d 3 xd 3 x e kx e k x [ωϕx + ϕx, ω k ϕx + ϕx ] snce [ϕx, ϕx ] [Π ϕ x, Π ϕ x ] Lkewse for [ a k, b k ]. It can be shown that [ϕx, ϕx ] [ ϕx, Π ϕ x ] [ ak, b k ] [ ϕx, ϕx ] [ a k, bk ] because, recall ak d 3 xe kx [ωφx + φ] d 3 xe kx φ d 3 xe kx [ωφx + Π φ ] b k d 3 x e k x [ωφx φx ] d 3 x e k x φx d 3 xe k x [ωφ + Π φ ] 3

14 [ ak, b k ] d 3 x Π φ φ Π φ φ d 3 x e kx e k x [ωφ x Π φ x, ωφ x + Π φ x ] d 3 xd 3 x e kx e k x ω[φx, Π φ x ] ω[π φ x, φx ] must assume treatng φ, φ ndependent: then [φx, Π φ x ] [φ x, Π φ x] e H Π ϕ Π ϕ + ϕ ϕ + m ϕ ϕ Ω dk ω a k e k x + ω b k e k x dk ωake kx + ωb ke kx + + dk{ ka k e kx + kb k e kx } dk {k a k e k x k b k e k x }+ +m dk dk a k e k x + b k e k x a k e kx + b k e kx Ω where x, kx ωt + k j x k, xj e kx k j e kx. H dk dk{ω ωa k a k e k kx ω ωb k a k e k +kx ω ωa k b +kx k + ω ωb e k k b kx k }+ ek + dk dk {k k a k a ke k k x k k b k a k e k+k x k k a k b k e k+k x + k k b k b k e k k x }+ +m dk dk {a k a k e k kx + b k a k e k +kx + a k b +kx k + b e k k b kx k } ek Consder dong dk d 3 x d 3 k ω π d 3 x. 3 d 3 xe k kx e ω ωt π 3 δ 3 k k π 3 δ 3 k k. Note how ω k k + m, for the last step. Then also d 3 xe k +kx e ω +ωt π 3 δ 3 k + k e ωt π 3 δ 3 k + k Remember, δ 3 k + k makes k k. Lkewse, d 3 xe k kx e ω ω t π 3 δ 3 k k π 3 δ 3 k k d 3 xe k +kx e ω +ωt π 3 δ 3 k + k e ωt π 3 δ 3 k + k Note that m ω k H d 3 xh dk{ ω a k a k ω b ka k e ωt ω a k b k eωt + ω b kb k }+ + +m dk{ k ω a k a k + k b k a k e ωt + k a k b k e ωt + k b k b k ω ω ω }+ dk{ a k a k ω + b ka k ω e ωt + a k b k ω eωt + b kb k ω } + Ω V ω + k + m ω + k + ω k ω H dkω{a k a k + b k b k } + Ω V Now b k b k b k b k + π 3 ωδ 3 k k b k b k + π 3 ω. 4

15 H dkω{a k a k + b k b k} + ɛ Ω V where ɛ V dkπ 3 ω For H, then ɛ V Ω V or d 3 kω k Ω V. d 3 k π 3 ω π3 ω d 3 kω k V π 3 δ 3 nterpreted as volume of space V. ɛ π 3 d 3 kω k π 3 d 3 kω Twce the zero-pont energy was found, due to knds of partcles assocated wth the ϕ, ϕ felds. 4. The spn-statstcs theorem Problem 4.. Verfy eq. 4.. Verfy ts lmt as m. Soluton 4.. Recall that we defned a nonhermtan feld ϕ x, ts Hermtan conjugate ϕ x, dke k x ak 4.3 dke k x a k 4.4 Tme-evolved wth H : ϕ + x, t e Ht ϕ + x, e Ht dke kx ak ϕ x, t e Ht ϕ x, e Ht dke kx a k 4.5 [ϕ + x, ϕ x ] dk dk e kx k x [ak, a k ] dk dk e kx k x π 3 ωδ 3 k k by fundamental canoncal commutaton or antcommutaton relatons dke kx x INCOMPLETE 98 Look up how to do modfed Bessel functons for functons wth symmetry about k. 5. The LSZ reducton formula Problem 5.. Work out the LSZ reducton formula for the complex scalar feld that was ntroduced n problem 3.5. Note that we must specfy the type a or b of each ncomng and outgong partcle. Soluton 5.. Recall, a k d 3 xe kx [ωφ x φ ] d 3 xe kx φ b k d 3 xe kx [ωφx φx] d 3 xe kx φx ak d 3 xe kx [ωφx + φx] d 3 xe kx φ Note there are dfferent types of partcles. Defne f k exp [ k k /4σ ] a b d 3 kf ka k d 3 kf kb k 5

16 Guess, for nteractng theory, lm t a ta t or lm t b tb t or lm t a tb t or n general lm t a t... a mtb m+ t... b nt note [a, b ] d 3 kf k d 3 k f k [a k, b k ] Recall a + a dt a t d 3 kf k d 4 x e kx ωφ x + φ x d 3 kf k d 4 xe kx + ω φ x d 3 kf k d 4 xe kx + k + m φ x d 3 kf k d 4 xe kx x + m φ x So, takng the lmt for fk, a + a d 4 xe kx + m φ x For the Hermtan conjugate, a + a dt a t d 3 kf k d 4 x e kx [ωφx φx] d 3 kf k d 4 xe kx ω φ + d 3 kf k d 4 xe kx k + m + φx d 4 xe kx + m φx Lkewse, b + b d 4 xe kx + m φx b + b d 4 xe kx + m φ x Now f lm a ta t or t + f lm b tb t t + f lm a tb t t + f lm a t... a m tb t... b n t t + Note that the only nontrval commutaton relatons are [ak, a k ] [bk, b k ] π 3 ωδ 3 k k. a s, b s commute, essentally. 6

17 f T a +... a m + b m b n + a... a m b m+... b n n +n d 4 x e k x + m a... d 4 x m e k m x m m + m a d 4 x m+ e k m+ x m+ m+ + m b... d 4 x n e k n x n n + m b d 4 x e kx + m a... d 4 x m e kmxm m + m a d 4 x m+ e km+xm+ m+ + m b... d 4 x n e knxn n + m b T φx... φx m φ x m+... φ x n φ x... φ x m φx m+... φx n e.g. A par of a and b partcles come n; a par of a and b partcles come out. f 4 d 4 x e k x + m a d 4 x e k x + m b d 4 x e kx + m a d 4 x e kx + m b T φx φ x φ x φx Problem Path ntegrals n quantum mechancs a Fnd an explct formula for Dq n eq Your formula should be of the form Dq CΠ N j dq j, where C s a constant that you should compute. b For the case of a free partcle, V Q, evaluate the path ntegral of eq. 6.9 explctly. Hnt: ntegrate over q, then q, etc., and look for a pattern. Express your fnal answer n terms of q, t, q, t, and m. Restore by dmensonal analyss. c Compute q, t q, t q e Ht t q by nsertng a complete set of momentum egenstates, and performng the ntegral over the momentum. Compare wth your result n part b. Soluton 6.. a Recall, for δt, use Weyl orderng. q, t q, t N k dq k N j dp j π epjqj+ qj e Hpj,q j δt where q j q j + q j+ q q q N+ q Defne q j qj+ qj δt Now Hp j, q j only quadratc n p j, Hp j, q j p j m + V q j. Consder usng +Bx dxe Ax π A exp / π exp π δt m dpj π epjqj+ qj e δt B A q j+ q j 4δt m 7 p j m m πδt / mqj+ q j δt exp δt

18 q, t q, t N k m N+ So Dq lm N πδt b For V, where µ m δt. N k m πδt m dq k πδt N+ m N+ dq k exp πδt [ N+ N dq k exp k N dq k ; δt k T N+. N m qj+ q j exp δt δtv q δt j j N m { q j V q j }δt j t t dtl qt, qt m N+ N N q, t q, t m qj+ q j dq k exp δt πδt δt k j N+ N µ N dq k exp µq j+ q j π k j Now ths wll be useful: dye αx y +βz y ] / [ ] π αβ exp x z α + β α + β snce exp αx y + βz y exp αx xy + y + βz yz + y exp αx + βz exp α + βy [ ] dy π 4xα + βz α + β exp exp αx + βz 4α + β [ π x α + β exp α + β z + xαβz + α + x + βαx + αβz + β z ] α + β [ π βαx α + β exp xαβz + αβz ] [ ] π αβ α + β α + β exp x z α + β where my formula, +Bx π dxe Ax A exp B 4A was used. Consder the followng: dq exp µq q + µq q / π [ µ exp µ q q ] dq exp µq 3 q / π µ exp µ q q π Suppose n j dq n j exp j µq n+ q j µ / / [ π π µ ] / µ 3µ exp 3µ/ q 3 q / [ µ exp 3 3 q 3 q ] n n π dq j exp µq j+ q j dq n e µqn+ qn µ j j π n [ ] π π µ/n µ n µ n+ exp n+ q n+ q µ n n 8 n π µ n exp [ µ n q n q ]. n+ n exp [ µ n q n q ] n [ ] µ n + exp n + q n+ q

19 Then by nducton, µ N+ N N µ N+ π dq k exp µq j+ q j π π µ k j [ ] π µ exp π N + N + q N+ q q, t q, t m T N [ N + exp µ N + q N+ q [ ] m m exp q N+ q πδt N + δtπ N + [ ] m exp T q N+ q where µ m δt, δt T N+. [ ] T t t ; q N+ q, q q. q, t q, t m t t π exp m t t q q By dmensonal analyss, [ q, t q, t m m t t π exp q q ] t t ] c q, t q, t q e Ht t q q dp p p e Ht t dpdp epq p e Ht t p q e p where q p e pq π π π dpdp e epq p q π p e Ht t p dpdp π dp p p q e pq e p q p t t e m δp p wth H p m for the free partcle Hamltonan. dpe pq e pq e p t t /m dpe pq q e p t t /m π π π mq π m t t exp q m m t t t t π exp q q t t 7. The path ntegral for the harmonc oscllator Problem 7.. Startng wth eq. 7., do the contour ntegral to verfy eq Soluton 7.. Recall de e Et t π E + ω 7. ɛ Now ω ɛ E ω ɛ E ω ɛ + E On the semcrcle C, E Re θ. Et t Rcos θ + sn θ t Rcθ t + Rsθδt. We see for t >, we sought the lower half semcrcle, t <, we sought the upper half semcrcle. For R, C, due to e Rsθ t factor n ntegrand. e ω ɛ t 7. for t > ω ɛ e ω ɛ t for t < ω ɛ ω e ω t t Problem 7.. Startng wth eq. 7.4, verfy eq Soluton INCOMPLETE 9 ɛ { e ω t ω for t > e ω t ω for t <

20 Recall Gt t ω exp ω t t 7.4 If t t, { ω ttgt t e ωt t f t > t ω eωt t f t < t ω G Problem 7.3. a Use the Hesenberg equaton of moton, A [H, A], to fnd explct expressons for Q and P. Solve these to get the Hesenberg-pcture operators Qt and P t n terms of the Schrödnger-pcture operators Q and P. b Wrte the Schrödnger-pcture operators Q and P n terms of the creaton and annhlaton operators a and a, where H ωa a +. Then, usng your result from part a, wrte the Hesenberg-pcture operators Qt and P t n terms of a and a. c Usng your result from part b, and a a, verfy eqs. 7.6 and 7.7. Soluton 7.3. a Recall HP, Q m P + mω Q the Hamltonan for the harmonc osclator. Recall [Q, P ] canoncal commutaton relaton. Lkewse, So then Q [H, Q] { m [P, Q] + mω [Q, Q]} m P P P m P [P, Q] + [P, Q]P m P [H, P ] { mω Q[Q, P ] + [Q, P ]Q} mω Q mω Q Q ω Q So the form of Qt.P t must be P ω P Qt a q e ωt + b q e ωt P t a p e ωt + b p e ωt Note that Q ωa q e ωt b q e ωt I assume the followng ntal condtons for the Hesenberg representaton of Q, P : Q a q + b q Q Q ωa q b q P m or a q b q P mω snce, for the last statement, Q ought to obey the Hesenberg equaton of moton. Then, t s easy to solve for a q, b q : [ ] [ ] [ ] aq Q P/mω So b q Smlarly for P t, Qt [ ] [ ] [ ] ap P b p mω Q P t Q + [ ] [ ] Q P/mω P e ωt + Q mω [ P e ωt mω [ ] [ ] P mωq P mωq e ωt + P + mωq e ωt Q + P Q P mω mω ] P mωq P + mωq [ aq b q ] [ ap b p ]

21 b Recall mω a x + p mω a mω x p mω a + a + a a e ωt + mω mω ae ωt mω Qt a e ωt + mω P t mω a a mω a e ωt + ae ωt c mω a + a e ωt + Q a + a mω mω P a a a + a + a a e ωt mω mω a e ωt + ae ωt mω mω a a + mω a + a e ωt T Qt Qt a e ωt + ae ωt a e ωt + ae ωt mω e ωt t mω Gt t snce Gt t ω exp ω t t 7.4, and Gt t ω exp ω t t ω exp ωt t T Qt Qt Qt 3 Qt 4 Qt Qt a e ωt3 + ae ωt3 e ωt4 mω Qt Qt e ωt3+t4 + e ωt3 t4 Qt mω mω 3/ a e ωt + ae ωt e ωt3+t4 + e ωt3 t4 Qt mω 3/ eωt3+t4 t + e ωt3 t4 t mω eωt3+t4 t t + e ωt3 t4 t+t 8. The path ntegral for free-feld theory Problem 8.. Startng wth eq. 8., verfy eq. 8.. Soluton 8.. Recall µ t, µ e kx k, ke kx µ t, µ e kx k, ke kx xe kx µ µ e kx k k e kx k e kx for k k k. x + m x x d 4 k k + m e kx x π 4 k + m ɛ snce d 4 k π e kx x s one Drac delta functon, representaton. 4 Problem 8.. Startng wth eq. 8., verfy eq Soluton 8.. Now k k k x x d 4 k π 4 e kx x k + m ɛ d 4 k e kx x π 4 ɛ k +m 8. ɛ δ 4 x x k + m ɛ k + m ɛ k k + m ɛ k k + m ɛ + k kx x k t t + k x x k τ + k x x

22 So from e kx x, consder e kτ. On semcrcle C, let k Re θ, exp Re θ τ exp Rcos θ + sn θτ exp Rcθ exp R sn θτ If τ >, then for lm R, C must be n the lower half plane. If τ <, then for lm R, C must be n the upper half plane. By resdue thm., dk π e k τ k + m ɛ e k +m ɛτ k +m ɛ e k +m ɛτ k +m ɛ Now ω k m wth ω ωk by specal relatvty. f τ > f τ < k ɛ e +m t t k + m d x x 3 k e k x x e ω t t π 3 dke k x x ω t t ω θt t dke kx x + θt t dke kx x where for the nd. term of the very last expresson, consder τ <. Then d 3 k π 3 ω e k x x +ωt t. By symmetry about x x, k k leaves d 3 k unchanged, + overall factor. Problem 8.3. Startng wth eq. 8.3, verfy eq. 8.. Note that the tme dervatve n the Klen-Gordon wave operator can act on ether the feld whch obeys the Klen-Gordon equaton or the tme-orderng step functons. Soluton 8.3. Recall x x θt t dke kx x + θt t dke kx x 8.3 Now Recall x x k x tt and kx x k x x ω k t t Srednck t δt t dke kx x + θt t dk ωe kx x + + δt t dke kx x + θt t dke kx x tt δ t t Dke kx x + δt t dk ωe kx x + θt t dk ω e kx x + + δ t t dke kx x δt t dkωe kx x + θt t dke kx x Note that dθt t dt dθt t dt δt t δt t, and by my notaton, δ t t d dt δt t n all cases for ths partcular problem. Observe a ω term. x yelds the terms k + ω m. So m wll cancel these terms. tt has other terms. Note that dk ωδt t d3 k π 3 ω. ωδt t dke kx x ωδt t δ t t d 3 k π 3 ω ekx x δt t d 3 k π 3 ω ekx x δt t dke kx x δt t d 3 k ek x x ωt t π 3 d 3 k π 3 e kx x δt t d 3 k ek x x ωt t π 3 ω d 3 k e k x x +ωt t π 3

23 x + m δt t + δ t t d 3 k + δt t dke kx x δ t t π 3 ekx x d 3 k π 3 e kx x + dke kx x If t t, or x x, then due to δt t, δt t. If x x, then due to δt t, δt t. Surely δ t t δ t t for t t snce slope s zero for δ for t t. Now d 4 xfxδx x fx d 4 xfx? Lkewse dtfxδt t d e ωt t dtfx dt δt t ω d 3 k π 3 e k x x δ 3 x x so d 3 k π 3 ekx x d 4 xfxδt t dfx dt d 3 k π 3 fx, t e k x x fx, t δ 3 x x d 3 x fx d 3 k π 3 e kx x fx e ωt t δt t { dfx dt ω tt ω + fx, t + d 3 xδ 3 x x dfx ω dt d e ωt t dtfx dt δt t ω d 3 k π 3 e k x x δ 3 x x so dfx dt fx xx δt t{ dfx e ωt t dt ω tt ω fx, t d 3 xδ 3 x x + dfx dt xx d 3 xfx x + m x x fx + fx + fx ω fx fx + fx e ωt t } + fxω e ωt t } ω fx d 4 xfxδx x So snce x + m x x obeys the propertes of a Drac delta functon δ 4 x x, t s equal to the Drac delta functon. Problem 8.4. Use eqs. 3.9, 3.9, and 5.3 and ts hermtan conjugate to verfy the last lne of eq Soluton 8.4. Recall that ϕx dk[ake kx + a ke kx ] 3.9 Recall Z J exp [ak, ak ] [a k, a k ] [ak, a k ] π 3 ωδ 3 k k 3.9 [ ] d 4 [ k Jk J k π 4 k + m exp ɛ 3 ak 5.3 ] d 4 xd 3 x Jx x x Jx 8.

24 Then recall, T ϕx ϕx δ δjx x x [ x x + d 4 k e kx x π 4 k + m ɛ δ δjx Z J J [ δ δjx [ ] d 3 x x x Jx 8. ] d 4 x x x Jx ] δ δjx Now what we want s to show the same result but wth the mode expanson of the feld. Z J J Z J J J x x Consder { T ϕx δ δjx Z J. J T ϕx dk[ake kx + a ke kx ] snce ak ; a k. Then we should get the correct result for the quadratc term: T ϕx ϕx dk dkak e k x + a k e k x ake kx + a ke kx dk dk e k x + kx ak ak + e k x kx a k a k + +e k x kx a k a k + e k x +kx a k a k Snce a k a k a k a k + π3 ωδ 3 k k. dke kx x By Tme orderng, we requre dke kx x f x x > dke kx x f x x > So T ϕx ϕx θx x dke kx x + θx x dke kx x x x So we get back the same result from the mode expanson. Problem 8.5. The retarded and advanced Green s functons for the Klen-Gordon wave operator satsfy ret x y for x y and adv x y for x y. Fnd the pole prescrptons on the rght-hand sde of eq. 8. that yeld these Green s functons. Soluton 8.5. Recall x x d 4 k e kx x π 4 k + m 8. ɛ The ratonale for addng ɛ was to avod the poles. Then n dong the contour ntegraton n complex k, we can take the ntegraton across the real axs. Now k k k Srednck s conventon k + m ɛ k + m ɛ k ωk ɛ k ωk ɛ + k w k w + k w n 4th quadrant for ɛ ωk. w n nd. quadrant. Consder e kτ ; τ x y. kx y k x y k x y 4

25 If τ >, for k Re θ R cos θ + R sn θ, e R cos θτ e R sn θτ. Then close the ntegraton n lower half-plane.e. sn θ <. dk e k τ e wτ Γ w k w + k π π w w e wτ dk e k τ w k + C R e CR k τ w dk k R sn θτ e R w Rdθ for sn θ < e k τ C R ω k ɛx y dk πe w k ω k ɛ k ɛ πe +m x y for x > y k + m If τ <, then close the ntegraton n upper half plane. Then resdue k w enclosed. So fnally, ret x y for x y, e k τ e wτ w π πe k +m x y k w k + m ret x y θy x πe k +m x y k + m for ntegraton closng the upper half plane adv x y for x y, adv x y θx y πe k +m x y for ntegraton closng the lower half plane k + m Consder not appendng ɛ. Then there are poles at w k + m, w. There are a number of ssues to consder: We must consder semcrcle contour ntegrals about the poles of nfntesmally small radus. What branch are we ntegratng on, especally on those semcrcle ntegrals around each of the poles? If τ >, for C δ, C δ semcrcle ntegrals around each of the poles enclosng the poles from the upper half, δ w w δ + + π{ e wτ w + ewτ w } for k w + δe θ. for k w + δe θ. Γ C R + dθ Cδ δeθ e w+δe θ τ δe θ w δe θ + C δ dθδe Cδ θ e w+δe θ τ π δe θ w + δe θ π{ e wτ w π cos wτ w w+δ w+δ dθ ewτ e δeθ τ w δe θ δ θe wτ w δ ewτ π π ewτ w w πe wτ w + ewτ } π{ e wτ w w + ewτ w } As you can see, there are ssues wth whch branch to choose that semcrcles C δ, C δ le on. If τ >, for C δ, C δ semcrcle ntegrals around each of the poles not enclosng the poles from the lower half, e ωτ π ω ω e ωτ π ω + πe ωτ eωτ ω 5 π sn ωτ ω

26 If τ <, for C δ, C δ semcrcle ntegrals around each of the poles enclosng the poles from below the poles, semcrcle n the upper half, + πeωτ ω π ω {e ωτ + πe ωτ ω + π{ e ωτ ω + e ωτ + eωτ ω } } π ω e ωτ 3 + e ωτ If τ <, for C δ, C δ semcrcle ntegrals around each of the poles not enclosng the poles from above the poles, semcrcle n the upper half, Soluton 8.6. { π eωτ ω + πe ωτ ω π sn ωτ ω } π{ eωτ ω e ωτ ω } Supposng Z J exp W J, recall, after some algebrac manpulatons, Z J exp d 4 xd 4 x Jx x x Jx W J d 4 xd 4 x Jx x x Jx Suppose Jx, a classcal source, s a real functon. Now kx x dke dkcos kx x + sn kx x. Lkewse, x x θt t IW J RW J + θt t θt t + θt t dk sn kx x + θt t dk cos kx x + θt t dk sn kx x + dk sn kx x + dk cos kx x dk cos kx x d 4 xd 4 x Jx dk cos kx x Jx d 4 xd 4 x Jx θt t + θt t dk sn kx x Jx Problem 8.7. Repeat the analyss of ths secton for the complex scalar feld that was ntroduced n problem 3.5, and further studed n problem 5.. Wrte your source term n the form J ϕ + Jϕ, and fnd an explct formula, analogous to eq. 8., for Z J, J. Wrte down the approprate generalzaton of eq. 8.4, and use t to compute T ϕx ϕx, T ϕ x ϕx, and T ϕ x ϕ x. Then verfy your results by usng the method of problem 8.4. Fnally, gve the approprate generalzaton of eq Soluton 8.7. L µ φ µ φ m φ φ + Ω + J φ + Jφ S d 4 d 4 k [ xl π 4 φ kk + m φk + Ω φ k φk + J kφk + Jkφ ] k where Jx so that For the other terms, d 4 k π 4 e kx Jk, Jxφ x φx φ x d 4 k π 8 ekx Jkd 4 k e k x φ k d 4 x d 4 k π 4 ekx φk d 4 k π 4 e kx φ k e kx e k x kt e k x+k t 6 d 4 k π 4 Jk φ k φk d 4 xe kx φx φ k d 4 xe kx φ x

27 π 8 µ φ d 4 k d µ φ π 4 µ e kx φ 4 k k π 4 µe k x φk d 4 kd 4 k ω, k ω, k φ k φk e k k x d 4 kd 4 k ωω + k k φ k φk e k k π 4 Drop tlde notaton now. Make ths substtute for the ntegraton varable: So χ χ + χ J + J χ k + m + π 8 d 4 k ω + k φ k φk π 4 χk φk Jk k + m χ φ χ J + k + m k + m χ + J J k + m J k + m J k + m + J + Ω χ + χ + +J χ J + k + m k + m + Ω χ χ + χ J + J χ k + m + k + m + Ω χ χ + Ω χ J + J χ + d 4 kk φ k φk J k + m J k + m χ + J J k + m Ω k + m J J + JJ k + m J k + m + + J χ + Jχ + J J + JJ k + m Now Dφ x dφx φ smply shfted by a constant: Dφ Dχ. Suppose the ntegraton over χ smply yelds Z J,J, and neglect terms contanng Ω let Ω [ Z J, J exp ] d 4 k Jk J [ +k π 4 k + m exp ɛ ] d 4 xd 4 x Jx x x J x where x x d 4 k e kx x π 4 k +m ɛ. Now T φ x... φx m... δ Jx... δ J x mz J, J δ J,J δjx... δ δj x m Z J, J Lkewse, T φx φx [ δ δj x d 4 xjx x x ] d 4 xjx x x Z J, J J,J d 4 x Jx x x Z J, J J,J J,J T φ x φ x T φ x φx δ δjx [ ] d 4 xjx x x Z J, J J,J x x Now φx dk[ake kx + b ke kx ]. Immedately, T φx φx T φ x φ x snce ak b k b k ak b k bk bk a k a k a k 7

28 T φ x φx dk dk a k e kx + bk e kx ak e kx + b k e kx dk dk e kx kx bk b k Recall that k x k x k x k x k t k t and [bk, b k ] π 3 ωδ 3 k k so that bk b k π 3 ωδ 3 k k dk e k x k x k t k t dke k x x k t k t Now x x d 3 k e k x x dk π 3 π d 3 k π 3 e k x x e k t t k + m k } dke k x x e ω t t ω { e k +m t t where ω k m. Note, from the form of the Lagrangan, m a m b mass of partcle of type a, s same as mass of partcle of type b. 9. The path ntegral for nteractng feld theory Soluton 9.. a L L + L L µ φ µ φ m φ Recall for the path ntegral, ZJ e d 4 xl L 4 Z λλφ 4 + L ct 4! Z λλφ 4 + L ct δ δjx Z J e d 4 x 4 Z λλ δ Jx 4 P [ P! For V, vertex, d 4 x 4 [ 4 Z λλ δ Jx P! ] P d 4 yd 4 zjy y zjz ] P d 4 yd 4 zjy y zjz Consder only dagrams wth sources for all legs. Then P 4. We count the number of terms that result n a partcular dagram. Rearrange δ 4 Jx by 4! ways. Rearrange V! vertces. V,!. Rearrange sources at end of each propagator, for each δ Jx.!,! 4. Rearrange the propagator themselves. P! 4!. We can rearrange the 4 propagators n 4! ways, all these arrangements duplcated by exchangng the dervatves at the vertces. S 4! symmetry factor. S! 4!. However, for some gven set of coordnates, y, y, y 3, y 4, 4! ways to choose whch propagator s exactly y, y, y 3, or y 4. 4! 4!. 4!! 4 4! 4! 4 4! 4! 4! 8

29 Dong ths problem more carefully, start from the gven Lagrangan. L µ ϕ µ ϕ m ϕ L 4 Z λλϕ 4 + L ct L ct Z ϕ µ ϕ µ ϕ Z m m ϕ A µ ϕ µ ϕ Bm ϕ Assumng Z λ + Oλ, Z e d 4 xl δ Jx Z Z λλ 4! d 4 x d 4 x Z λλ 4! d 4 yj y y x 4 4 δ J x 4! wth 4! 4 countng factor matchng δ Jx s to propagators. For ϕϕ ϕϕ scatterng, use δ J x δ J x δ J x3 δ J x4, T ϕx ϕx ϕx 3 ϕx 4 Z λλ 4! 4! f 4 d 4 x e kx + m d 4 yd 4 zj y y z J z 4 d 4 x x x x x x3 x x4 x 4 4 Z λλ d 4 x x x x x x3 x x4 x d 4 x e kx + m d 4 x 3 e k3x3 3 + m d 4 x 4 e k4x4 4 + m T ϕ ϕ ϕ 3 ϕ 4 Z λ λπ 4 δ 4 k + k k 3 k 4 For A, B counterterms, consder P propagators. d 4 x δ J x A x + Bm δ J x Z d 4 xδ Jx A x + Bm d 4 yj y y x d 4 zd 4 wj z z w J w d 4 x d 4 zj z z x A x + Bm d 4 yj y y x ZJ d 4 xδ Jx A x + Bm δ Jx! d 4 yd 4 zj y y z J z Remember, only consder connected dagrams. Also, note that x y x x d 4 k e ky x π 4 k +m ɛ T ϕx ϕx f d 4 x x x A x + Bm x x d 4 xe kx + m d 4 xe kx d 4 x e kx + m d 4 k k e ky x π 4 k +m ɛ, so order s mportant. for ths partcular case, ths partcular order d 4 x e kx A x + Bm δ 4 x x +... To deal wth x, use ntegraton by parts: d 4 xe kx xδ 4 x x k d 4 xe kx δ 4 x x ke kx Then the rght-hand sde of Eq. 9 s d 4 x x x A x + Bm x x π4 δ 4 k k Ak + Bm π 4 δ 4 k k Ak + Bm π 4 δ 4 k k Because of δ 4 k k, then the results are the same f labels,, swtched. Thus, a countng factor. 9

30 f Ak + Bm π 4 δ 4 k k +... k X Ak + Bm The assocated vertex factor s Z λ λ We also need to sprnkle counterterm vertces for each propagator wth a factor of Ak + Bm. b c A clever way to see ths Arthur Lpsten, Nov. 7, 8, that tadpoles are canceled, s to note that the symmetry ϕ ϕ. Then ϕ. Or, note that E P 4V. For ϕx, E, but P 4V 4V + P whch s not possble, for V, P Z +. So no counterterm lnear n ϕ s needed. Problem 9.3. Consder a complex scalar feld see problems 3.5, 5., and 8.7 wth L L + L, where L µ ϕ µ ϕ m ϕ ϕ, L 4 Z λλϕ ϕ + L ct L ct Z ϕ µ ϕ µ ϕ Z m m ϕ ϕ Ths theory has two knds of sources, J and J, and so we need a way to tell whch s whch when we draw the dagrams. Rather than labelng the source blobs wth a J or J, we wll ndcate whch s whch by puttng an arrow on the attached propagator that ponts towards the source f t s a J, and away from the source f t s a J. a What knd of vertex appears n the dagrams for ths theory, and what s the assocated vertex factor? Hnt: your answer should nvolve those arrows! b Ignorng the counterterms, draw all the connected dagrams wth E 4 and V, and fnd ther symmetry factors. Hnt: the arrows are mportant! Soluton 9.3. a Recall Consder Note that Z J, J J,J DφDφ e d 4 x[l +Jφ +J φ] exp [ d 4 xd 4 x Jx x x J x ] [ exp φ xφxφ xφx ] [ d 4 xl exp δ δjx δ δj x ] d 4 x 4 Z λλ φ φ δ δjx δ δj x notaton δ Jx δ J x δ Jx δ J x Consder, as an exercse, that s not drectly related to the problem at hand, δ J x Z [ d 4 x J x x δ Jx x]z x xz + [ d 4 x x x J x ]Z δ J x [ d 4 x J x x x] Z + x xz + [ d 4 x x x J x ][ d 4 x J x x x]z δ Jx,J,J In a sense, I say I can set to as an overall multplcatve factor, dagrams that s closed as a loop. 3

31 Recall dong the double Taylor seres expanson: [ exp 4 Z λλ d 4 x [ V! 4 Z λλ d 4 δ δ x δjx δj x Z J, J V δ δjx ] δ δj Z J x ] V P [ P! ] P d 4 yd 4 zjy y zj z Consder vertex. 4 Z λλ d 4 δ δ x δjx δj x [ Consder P! d 4 yd 4 zjy y zj z ] P. For P, applyng the vertex gves zero. For P, 4 Z λλ 8 Z λλ s obtaned. Result: closed double loop, fgure eght. For P 3, 3! [ d 4 y d 4 z J y y z J z ][ d 4 y d 4 z J y y z J z ][ d 4 y 3 d 4 z 3 J y3 y3 z 3 J z 3 ] Consder only connected dagrams. δ J x δjx δ J x d 4 y J y y x d 4 z x z J z d 4 δ Jx y 3 J y3 y3 x d 4 z J z d 4 y 3 J y3 x z y3 x Result: closed loop and propagator legs. For P 4, 4 Z λλ d 4 x d 4 z x z J x d 4 y J y y x d 4 z 3 x z3 J z 3 so we get the vertex, wth 4 propagators attached n the followng manner: d 4 y 4 J y4 y4 x J J J k k 3 k k 4 J Note that E P 4V by nducton. Note frst the 4! factor, cancelng 4! from the 4 propagators, due to matchng functonal dervatves to propagators. Then another factor when usng the LSZ formula see Problem. on the remanng propagators amputatng the external legs. The assocated vertex factor s Z λ λ. b See dagram. Soluton 9.4. a Gven exp W g, J π + dx exp [ x + 6 gx3 + Jx ] The trck s to do the followng Arthur Lpsten, Nov. 7, 8. snce Scratchwork: e x exp W g, J exp π 6 g δ J 3 dxe x +Jx dxe x +Jx dx dye x y dxe x J e J / πe J / 3 rdrdϕe r π e r π π

32 Note also, e x π y x J dy dx exp W g, J exp 6 gδ J 3 e J / P V V! 6 gδ J 3 J P! P Recall Srednck, Sec. 9. Follow the development n Sec. 9. It begns wth ϕ 3 theory. Z J exp d 4 yd 4 zjy y zjz [ ] 3 Z g g Z J d 4 δ x V! 6 δjx V [ P d 4 yd 4 zjy y zjz] 9. P! V Most general dagram contrbutng to ZJ conssts of product of several connected dagrams. Let C I partcular connected dagram. Then general dagram s D S D I C I n I. n I Z counts number of C I s n D, S D s addtonal symmetry factor for D part of symmetry factor not already accounted for by symmetry factors ncluded n each of the connected dagrams. Snce propagator and vertex rearrangements wthn each C I accounted for we only need to consder exchanged of propagators and vertces among dfferent connected dagrams. These can leave the entre dagram D unchanged only f surely the followng must be true: exchange made among dfferent but dentcal connected dagrams. exchanges nvolve all of the propagators and vertces n a gven connected dagram exchanged wthn a C I already accounted for n I! S D I n I! P Z J D C I n I S D n I! C I n I {n I } {n I } I {n I } I I So I clam that Z J exp I C I. Then f we dentfy y z, d 4 yd 4 zjyjz J Z g g g dx δ 3 J δ 3 J n I n I! C I n I I exp C I exp C I Then the result of Z e I C I apples here to, snce the combnatorcs are stll the same,.e., we stll exchange dervatves δ J 3 at vertex g, we stll can exchange sources, J. Thus n our case, J Z J exp W g, J exp 6 gδ J 3 [ g exp V! 6 δ J 3] V J P e I C I P! V P so W g, J I C I. A connected dagram s merely a term n the double Taylor expanson C I S I g V I J E I 3

33 So then W g, J g V I J E I S I I For each set of terms of some specfed V vertces, E external sources, there s the same g V J E factor for each of these terms. Sum over all connected dagrams wth such specfed V, E. g V K J E K g V J E S K S K K K As n Problem 9.a, V! ways to arrange V δ J 3 s, 3! ways to arrange 3 δ J s, ways to arrange J s, P! ways to arrange P J s. So the above result, sayng that a connected dagram s smply a term of the form S I g V I J E I s justfed. So we could mpose the requrement of no tadpoles to rearrange our sum so that where W g, J C V,E g V J E V E C V,E K S K and K s the sum over all connected dagrams wth V vertces and E external sources. b c Now recall Addng a counterterm Y, Note J ew g,j+y are equvalent condtons. Interpret e W as ZJ expw g, J π exp W g, J + Y + π [ dx exp x + ] 6 gx3 + Jx 9.7 [ dx exp x + ] 6 gx3 + Jx + Y x e W g,j+y J J W g, J + Y J J W g, J + Y J ZJ e W g,j e 6 gδ J 3 e J / [ g V! 6 δ J 3] J P e I C I P! P V Recall the development n Sec. 9. Analogously, note ZJ + g 6 δ3 J + J + J Og + J + J gj + OJ 3 Thus a counterterm s necessary. Assumng Y Og, Z J + g J + Y 6 δ3 J + + So Y g, assumng Y Og. The no tadpole condton of δ δ J Z J g or δ J W J g J + Y 4 J + Y J + Y gj + Y 4 4 δ J Z J g + Y δ J exp W g, J + Y J + dx x exp π δj W g, J + Y J s equvalent to δ δj ew g,j+y J or [ x + 6 gx3 + Jx + Y x] J analogous to ϕx. Interpretng connected Feynman dagrams as terms of the form C I S I g V I J E I. 33

34 For a W g, J +Y C V E V,E g V J E δ 9.3 s.t. δj W g, J + Y J, then no connected Feynman dagrams wth tadpoles are ncluded n the sum of connected Feynman dagrams W g, J + Y. d W g, J + Y g V I J E I g V J E C V,E g V J E S I S K I V E K V E where the sum K s over all connected dagrams wth no tadpoles of V vertces, E external sources.. Scatterng ampltudes and the Feynman rules Problem.. Wrte down the Feynman rules for the complex scalar feld of problem 9.3. Remember that there are two knds of partcles now whch we can thnk of as postvely and negatvely charged, and that your rules must have a way of dtngushng them. Hnt: the most drect approach requres two knds of arrows: momentum arrows as dscussed n ths secton and what we mght call charge arrows as dscussed n problem 9.3. Try to fnd a more elegant approach that requres only one knd of arrow. Soluton.. Wth an expresson for ZJ exp W J, we can take functonal dervatves to compute vacuum expectaton values of tme-ordered products of felds. Recall from Problem 9.3, [ Z J, J J,J DϕDϕ e ] d 4 x[l +Jϕ +J ϕ] exp d 4 xd 4 x Jx x x J x Keep n mnd the counterterms [ exp ] [ d 4 xl exp ] d 4 x 4 Z λλ ϕ ϕ L ct Z ϕ µ ϕ µ ϕ Z m m ϕ ϕ A µ ϕ µ ϕ Bm ϕ ϕ Recall dong the double Taylor seres expanson: [ ] exp 4 Z λλ d 4 x δ J x δ J Z x J ZJ, J So the path ntegral, wth nteracton, for our complex scalar feld theory s [ ] V Z J, J V! 4 Z λλ d 4 x δ J x δ [ ] P J d 4 yd 4 zjy y zj z x P! P V Recall, from Problem 8.7, snce, for example T ϕ x ϕx δ J x [ T ϕx ϕx T ϕ x ϕ x T ϕx ϕx [ ] δ J d 4 xjxδx x x Z J, J J,J d 4 xjx x x d 4 x Jx x x Z J, J J,J ] d 4 xjxδx x e 4 Z λλ d 4 x To study nteracton n ths theory, consder the followng. T ϕx 3 ϕ x 4 ϕ x ϕx δ J x 3 δ J x4 δ J x δ J x V! 4 Z λλ V δ Jx δ J x δ J x 3 δ J x4 δ J x δ J ZJ, J x V d 4 z δ J z δ J z Remember to restrct ourselves to fully connected dagrams. 34 P Z J, J x x + Oλ P! J,J d 4 xd 4 x J x x x J x P

35 Consder, for Oλ, λ order, δ J x 3 δ J x4 δ J x δ J x 4 Z λλ d 4 z δ J z δ J d 4 xd 4 x J x x x J x 4 z 4! Matchng up δ Jz δ J z wth sources, there are 4! ways, f ncludng only connected dagrams. Matchng up δ Jy δ Jy δ J x δ J x, there are 4 ways. Note that the factors 4 4! match up to exactly cancel the factors n the denomnators. δ J x 3 δ J x4 δ J x δ J Z x λ λ d 4 z Z λ λ d 4 xj x x z d 4 yj y y z d 4 z x4 z z x3 x z z x d 4 x z x J x d 4 y z y J y T ϕx 3 ϕ x 4 ϕ x ϕx Z λ λ d 4 z x4 z z x3 x z z x + Oλ Recall Problem 5.. LSZ formula for complex scalar feld. f n +n d 4 x e k x + m a... d 4 x m e kmx m m + m a d 4 x m+ e k m+ x m+ m+ + m b... d 4 x n e k n x n n + m b d 4 x e kx + m a... d 4 x m e +kmxm m + m a d 4 x m+ e km+xm+ m+ + m b... d 4 x n e knxn n + m b T ϕx... ϕx m ϕ x m+... ϕ x n ϕ x... ϕ x m ϕx m+... ϕx n In ths case, for the vacuum expectaton, J, J, so ths condton forces a -type partcles, arsng from ϕ to be each pared up wth b -type partcles, arsng from ϕ. Then, note that for the LSZ formula, there s the condton m + n m n m + m Also, note that we are attemptng a more elegant approach because the most drect approach requres two knds of arrows: momentum arrows and what we mght call charge arrows, that dstngush the sources J, J, or, n turn, dstngush between ϕ and ϕ. Then I clam that because of the tme orderng operator of the vacuum expectaton, then a partcles could be dstngushed from b partcles. For ncomng partcles, a partcles correspond to ϕ feld and b partcles correspond to ϕ feld. For outgong partcles, a partcles correspond to ϕ feld and b partcles correspond to ϕ feld. The term T ϕx... ϕ x m+... ϕ x... ϕx m+... make t very clear whch partcles are ncomng and whch are outgong, compelled by the tme orderng operator T. If n ths case, y, y come after x, x, then for ths case, plug the above, very last result nto the LSZ formula, f 4 d 4 x 3 e k3x3 3 + m a d 4 x 4 e k4x4 4 + m b d 4 x e kx + m a d 4 x e kx + m b Z λ λ d 4 z x4 z z x3 x z z x Usng + m x y δ 4 x y.e. Klen-Gordon wave operator acts on propagator,.e. propagator s a Green s functon, then for nstance, 4 + m b x4 z δ 4 d x 4 z 4 x 4e k 4 x 4 e k4z f Z λ λ d 4 z exp [ k 3 k 4 + k + k z] Z λ λπ 4 δ 4 k + k k 3 k 4 35