Some Notes on Field Theory

Transcription

1 Some Notes on Feld Theory Eef van Beveren Centro de Físca Teórca Departamento de Físca da Faculdade de Cêncas e Tecnologa Unversdade de Combra Portugal May 20, 2014

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3 Contents 1 Introducton to Quantum Feld Theory Huygens prncple versus Schrödnger equaton Proof of formula Free Klen Gordon partcles Green s functon for free Klen-Gordon partcles Second Quantzaton Procedure Proof of formula Self-nteractng Klen-Gordon feld Tme-ordered product of two felds Proof of formula Tme-ordered product of four felds Feynman rules part I Two-ponts Green s functon Vacuum bubbles Two-ponts Green s functon contnuaton Feynman rules part II The second order n λ contrbuton to Gx 1,x The amputed Green s functon PI graphs and the self-energy Full propagator Dvergences Integraton n n dmensons Counterterms Subtracton contrbutons Four-ponts Green s functon The vertex The second order terms The amputed vertex functon Regularzaton of the vertex functon Moldng tme evoluton nto a path ntegral Tme evoluton n Quantum Mechancs

4 5 A path ntegral for felds Green s functons The free feld propagator The free-feld path ntegral The free-feld generatng functonal λφ 4 theory The nteracton term The full generatng functonal Feynman dagrams λφ 3 theory The nteracton term The full generatng functonal Feynman dagrams The Bethe-Salpeter equaton The bubble sum The ladder sum The drvng term Fermons Fermons Drac spnors Propertes of the Drac spnors Drac traces Coulomb scatterng Number of states Transton probablty Flux of ncomng partcles Dfferental cross secton Averagng over spns Dfferental cross secton contnued Postron scatterng The electron propagator The photon propagator Electron-muon scatterng Electron-photon scatterng

5 Chapter 1 Introducton to Quantum Feld Theory Quantum Feld Theory s a general technque for dealng wth systems wth an nfnte number of degrees of freedom. Examples are systems of many nteractng partcles or crtcal phenomena lke second order phase transtons. Here we wll concentrate on the scatterng of partcles, but the general framework can be appled to any doman n physcs. For an ntroducton, we smplfy Nature as much as possble and hence assume that Nature exsts out of only one type of partcles, wthout spn, wthout charge and all wth the same mass, m. Such partcles are moreover ther own antpartcles. The objects of our nterest are n-ponts Green s functons, Gx 1,x 2,...,xn, whch represent n-partcle processes where n k partcles enter the nteracton area before scatterng and k partcles leave the nteracton area after scatterng. On the subject of Quantum Feld Theory exsts a vast amount of lterature. Here we wll just menton some books, but the lst s very ncomplete. Many of the deas behnd the theory have been developed by R.P. Feynman and can be found n hs book enttled Quantum Electrodynamcs [9]. A classc course on the subject s contaned n Relatvstc Quantum Felds by J.D. Bjorken and S.D. Drell [10]. Also the books enttled Quantum Feld Theory by C. Itzykson and J-B Zuber [11] and Gauge theory of elementary partcle physcs by Ta-Pe Cheng and Lng-Fong L [12], whch contan a lot of deas worked out n detal, have become classc works n the mean tme. More modern, and also wth a great deal of detal, s the book of George Sterman enttled An ntroducton to Quantum Feld Theory [14]. But theores develop, some of the stuff becomes obsolete and other new areas enter the game, and therefore new strateges are followed for courses wrtten n a modern language and ntended for those who want to work n the fronter areas of physcs. Good examples are the lectures of Perre Ramond enttled Feld Theory a modern prmer [15] and of R.J. Rvers Path ntegral methods n quantum feld theory [16]. Path ntegral technques form the bass of almost all modern lterature on feld theory. The classc book Quantum Mechancs and Path Integrals s wrtten by R.P. Feynman and A.R. Hbbs [17]. 1

6 The dmensonal regularzaton methods, whch were mportant for the proof that non- Abelan Gauge Theores are renormalzable, are developed by Gerard t Hooft and Tny Veltman, and can be found n ther Cern publcaton [18] or n ther publcaton n Nuclear Physcs [19]. Any modern lecture contans a chapter on the ssue. Not exactly on the subject of ntroducng quantum feld theory, but stll wth everythng necessary to study the subject, s the book of Sdney Coleman enttled Aspects of Symmetry [20], whch s strongly recommended for further readng. 2

7 1.1 Huygens prncple versus Schrödnger equaton In the 17th century Chrstaan Huygens The Hague, formulated the foundatons of modern wave mechancs and the theory of lght. The descrpton of the propagaton of waves n matter s nowadays known by Huygens prncple. Accordng to Huygens prncple one may calculate the wave ampltude of an oscllatory phenomenonateachpontnspaceatacertannstanttwhenonedsposesofthefollowng two nformatons: 1 the wave ampltudes at all ponts n space at an earler nstant t and 2 the way n whch the wave propagates through space. The frst nformaton we denote by ψ x,t, whereas the second nformaton s supposed to be contaned n the Green s functong x,t; x,t. Wth thosedefntons onemay express Huygens prncple by the followng relaton ψ x,t = d 3 x G x,t; x,t ψ x,t for t > t. 1.1 In order to quantfy the condton t > t n formula 1.1, one may ntroduce the step functon, θt t, whch vanshes for negatve argument and equals 1 for postve argument,.e. 0 for t < t θt t = for t > t We obtan then from formula 1.1 the relaton θt t ψ x,t = d 3 x G x,t; x,t ψ x,t 1.3 for Huygens prncple. In ths secton we study the relaton of formula 1.3 wth the Schrödnger equaton. For that purpose, we frst express the step functon 1.2 by an ntegral representaton, gven by + 2π θτ = lm dω e ωτ. 1.4 ε 0 ω +ε The ntegral canbe carred out as follows. For τ < 0 one closes the contour n thecomplex ω-plane by a semcrcle n the upper half plane whch does not contan any sngularty. Consequently, the complex contour ntegral vanshes and we obtan as a result the upper equaton of formula 1.2. For τ > 0 one closes the contour n the complex ω-plane by a semcrcle n the lower half plane whch does contan the sngularty at ε. The resdue of the resultng complex contour ntegral equals 1 n the lmt of ε 0. Hence we obtan 2πθτ = 2π, whch results n the lower equaton of formula 1.2. From the ntegral representaton t s moreover easy to verfy that d + dt θt t = dω e ωt t = δt t π So, by applyng / t to equaton 1.3, we fnd the followng relaton δt t ψ x,t + θt t t ψ x,t = d 3 x t G x,t; x,t ψ x,t

8 Now, we come to the Schrödnger equaton whch we wll consder here, gven by t H 0 x ψ x,t = V x,t ψ x,t, 1.7 where H 0 x mght represent the operator 2 /2m, but could be more complcated, and where V represents the potental whch has to be specfed for each dfferent problem under study. Assocated wth equaton 1.7 we defne the Green s functon for free propagaton, or free propagator, G 0 x,x, gven by t H 0 x G 0 x,x = δ 4 x x, 1.8 where we ntroduced x = x,t. Equaton 1.8 can be solved as we wll assume here. Later on, we wll encounter some examples. The relaton between Huygens prcple 1.3 and the Schrödnger equaton 1.7 can now be formulated as follows Gx,x = G 0 x,x + d 4 x G 0 x,x V x Gx,x, 1.9 whch s an ntegral equaton and can be solved by teraton, a procedure whch we wll study frst. In the remanng part of ths secton we wll outlne a proof of relaton 1.9. When one substtutes Gx,x as defned on the lefthand sde of formula 1.9 nto the expresson of the rghthand sde, then one obtans Gx,x = G 0 x,x + d 4 x 1 G 0 x,x 1 V x 1 { G 0 x 1,x + + } d 4 x 2 G 0 x 1,x 2 V x 2 Gx 2,x 1.10 = G 0 x,x + d 4 x 1 G 0 x,x 1 V x 1 G 0 x 1,x + + d 4 x 1 d 4 x 2 G 0 x,x 1 V x 1 G 0 x 1,x 2 V x 2 Gx 2,x. The substtuton can be repeated. One fnds Gx,x = G 0 x,x + d 4 x 1 G 0 x,x 1 V x 1 G 0 x 1,x d 4 x 1 d 4 x 2 G 0 x,x 1 V x 1 G 0 x 1,x 2 V x 2 G 0 x 2,x + d 4 x 1 d 4 x 2 d 4 x 3 G 0 x,x 1 V x 1 G 0 x 1,x 2 V x 2 G 0 x 2,x 3 V x 3 G 0 x 3,x

9 Each term n the sum 1.11 can be evaluated, once the free propagator s known. Hence, when the sum converges one can determne the full propagator Gx,x. Ths s the case for weak potentals V. A way to memorze formula 1.11 s by means of the followng graphcal representaton for each of the terms. G 0 x,x Gx,x = x x + V x 1 G 0 x,x 1 x 1 G 0 x 1,x + x x + V x 1 G 0 x 1,x 2 V x 2 G 0 x,x 1 x1 x G 0 x 2,x 2 + x x The frst graph at the rghthand sde of formula 1.12 represents the free propagator G 0 x,x. The second graph, where the free propagators connect x to x 1 and x 1 to x and where the potental acts once, at space-tme pont x 1, represents the second term of formula 1.11, gven by d 4 x 1 G 0 x,x 1 V x 1 G 0 x 1,x, The thrd graph, where the free propagators connect x to x 1, x 1 to x 2 and x 2 to x and where the potental acts twce, one tme at space-tme pont x 1 and another tme at space-tme pont x 2, represents the thrd term of formula 1.11, gven by d 4 x 1 d 4 x 2 G 0 x,x 1 V x 1 G 0 x 1,x 2 V x 2 G 0 x 2,x. And so on. One mportant property, causalty, of both, the free propagator G 0 x,x and the full propagator Gx,x, should be mentoned here: No sgnal can travel faster than lght. Consequently, nothng can be observed before t happens. Or n formula G x,t; x,t = G 0 x,t; x,t = 0 for t < t

10 1.1.1 Proof of formula 1.9 Below, we wll study a proof of formula 1.9. We show that by substtutng expresson 1.9 nto formula 1.3 one ends up wth the Schrödnger equaton 1.7. The substtuton results n the followng relaton θt t ψx = 1.14 = Next, we let the operator d 3 x {G 0 x,x + } d 4 x G 0 x,x V x Gx,x t H 0 x ψx. work at both sdes of equaton From the lefthand sde of 1.14, also usng the result 1.5, one fnds δt t ψx + θt t { t H 0 x Whereas, from the rghthand sde, also usng the result 1.8, we obtan } ψx d 3 x { t H 0 x } G 0 x,x ψx + + d 3 x d 4 x { t H 0 x } G 0 x,x V x Gx,x ψx = = d 3 x δ 4 x x ψx + + d 3 x d 4 x δ 4 x x V x Gx,x ψx = δt t ψx + d 3 x V x Gx,x ψx = δt t ψx + V x θt t ψx In the last step of equaton 1.16 we used once more equaton 1.3. Combnng results 1.15 and 1.16 one fnds the Schrödnger equaton

11 1.2 Free Klen Gordon partcles Non-nteractng partcles wthout spn or charge are descrbed by the Klen-Gordon equaton, whch satsfes the wave equaton Here we defne 2 t 2 ψx,t = 2 x 2 m2 µ µ = 2 t 2 2 x 2 2 y 2 2 z 2, n order to wrte the Klen-Gordon equaton n the usual form ψx,t µ µ + m 2 ψx = 0, 1.18 where ψx stands for ψ x,t. Notce, that we assume here that gravtatonal effects can be completely gnored and consequently that our partcles move n a Mnkowskan background for whch we adopted the metrc +. As easly can be verfed, a general soluton to the free Klen-Gordon equaton 1.18 s gven by the followng wave packet ψx = d 3 k 2π 3 2E { α k e kx + α k e kx }, 1.19 provded that k, whch stands for E, k, satsfes the mass-shell relaton E 2 = k 2 + m

12 1.3 Green s functon for free Klen-Gordon partcles TheGreen sfuncton, G 0, forafreeklen-gordonpartcle, whchhasthecorrectboundary condtons, s a soluton of the dfferental equaton gven by x µ + m 2 G 0 x,x = δ 4 x x x µ One may construct the correct soluton by defnng the Fourer transform, G0, of G 0, by G 0 x,x = d 4 p 2π 4 epx d 4 p 2π 4 ep x G0 p,p. For ths Fourer transform one fnds, by applyng the Klen-Gordon dfferental equaton 1.21, the relaton d 4 p 2π 4 p 2 +m 2 e px d 4 p 2π 4 ep x G0 p,p = d 4 p 2π 4 epx x, whch s solved by p 2 +m 2 G0 p,p = 2π 4 δ 4 p+p Graphcally one may represent ths soluton by x x E, p E, p whch graph can be nterpreted as follows: Four momentum propagates from event x to event x. Ths s represented by four momentum p whch flows away from x and four momentum p whch flows away from x. Now, four momentum conservaton demands that p equals p. Ths s expressed by the delta functon n formula One defnes the Feynman propagator, S F, by S F p,m 2 = p 2 m 2 and G 0 p,p = 2π 4 δ 4 p+p S F p,m As we wll see n the followng, t s usually very convenent to do all calculatons wth the Feynman propagators and only at the end to bother about four momentum conservaton. 8

13 1.4 Second Quantzaton Procedure Our goal s to descrbe many nteractng partcles, not just one-partcle states. To that am we defne a Hlbert space of many-partcle states, also called Fock space. The most elementary state of ths space s called the vacuum, symbolzed by 0. It s assumed to be the state wth no partcles at all or just smply the ground state of the system of states one consders. Next n the herarchy come the one-partcle states, for our world, just exstng of Klen- Gordon partcles, denoted by k. It s supposed to descrbe a partcle wth momentum k. The operator, whch creates out of the vacuum a one-partcle state, s denoted by a k. Consequently, we may wrte k = a k Two-partcle states, whch descrbe the stuaton n whch n our world only two partcles are present, one wth momentum k 1 and the other wth momentum k 2, are supposed to be gven by k 1, k 2 = a k1 a k Now, we suppose that the order n whch the partcles are created, whch s not a tmeorder but just an operaton order, does not nfluence n any way the resultng two-partcle state. Hence, we fnd as a property of the creaton operators defned n formula 1.24 that they commute,.e. a k1 a k2 = a k2 a k We also defne annhlaton operators, a k, wth the followng propertes a k 0 = 0, a k1 a k2 = a k2 a k1, and [ a k1, a k2 ] = 2π 3 2E 1 δ 3 k1 k Notce that the commutaton relatons for the creaton and annhlaton operators are the contnuum generalzatons of the commutators for n harmonc oscllators, whch also vansh except for [ a, a j] = δj. The next step n the second quantzaton procedure s the replacement of the free Klen- Gordon wave packet, whch s defned n formula 1.19, by a free Klen-Gordon quantum feld,.e. φx = d 3 k 2π 3 2E { a k e kx + a k e kx }, 1.28 whch s an operator whch acts n the many-partcle state Hlbert space. 9

14 The reason why ths procedure s called second quantzaton stems from the fact that we can also defne a conjugate momentum πx = t φx, 1.29 for whch one has the followng equal tme commutaton relatons [φ x,t, φ x,t] = [π x,t, π x,t] = 0 [π x,t, φ x,t] = δ 3 x x Proof of formula 1.30 Frst, we wrte the explct expresson for the conjugate momentum π x,t of φ x,t, namely π x,t = t φ x,t = = d 3 k { 2π 3 2E E a k x Et k e d 3 k 22π 3 { a k e k x Et + a k e k x Et } + a k e k x Et } Then, we substtute formulas 1.28 for the feld and 1.31 for ts conjugate momentum n the expresson for the equal-tme commutators Ths gves: [φ x,t, φ x,t] = d 3 k 2π 3 2E d 3 k 2π 3 2E { [a k, a k ] e k x+ k x E +E t + + [ a k, a k ] e k x k x E E t + + [ a k, a k ] e k x+ k x E +E t + + [ a k, a k ] e k x k x +E +E t } Next, we nsert expressons 1.26 and 1.27 to fnd [φ x,t, φ x,t] = d 3 k 2π 3 2E d 3 k 2π 3 2E 10

15 { 2π 3 2Eδ 3 k k e k x k x E E t + 2π 3 2Eδ 3 k k e k x+ k x E +E t } Upon ntegraton over k, we obtan k = k and E = k 2 +m2 = k 2 +m2 = E hence [φ x,t, φ x,t] = d 3 { k 2π 3 e k x x e } k x x 2E. In the second term one may perform the substton k k for the ntegraton varable, n order to obtan two equal terms wth opposte sgn and thus [φ x,t, φ x,t] = 0. The proof for the equal-tme commutator of two conjugate momentum felds s very smlar. For the equal-tme commutator of the feld and ts conjugate momentum we obtan [π x,t, φ x,t] = { d 3 k 22π 3 d 3 k 2π 3 2E [ a k, a k ] e k x k x E E t + + [ a k, a k ] e k x+ k x E +E t } = d 3 k 22π 3 d 3 k { 2π 3 2E 2π 3 2Eδ 3 k k e k x k x E E t + 2π 3 2Eδ 3 k k e k x+ k x E +E t } = d 3 { k 22π 3 e k x x e } k x x = d 3 k 2π 3e k x x = δ 3 x x. 11

16 1.5 Self-nteractng Klen-Gordon feld In general, one starts a quantum feld theory by defnng a Lagrangan densty, L, whch s a functonal of a quantum feld, ϕ, and ts dervatves L ϕ x,t, µ ϕ x,t The object µϕ n formula 1.32 stands for the four partal dervatves gven by 0 ϕ = t ϕ, 1ϕ = x ϕ, 2ϕ = y ϕ, and 3ϕ = z ϕ. The total Lagrangan, L, for the system under consderaton s gven by the volume ntegral of the Lagrangan densty over all space L = d 3 x L ϕ x,t, µ ϕ x,t. All dynamcs of the system s contaned n the Lagrangan densty. The feld equatons for the quantum feld can be derved from the Lagrangan densty by the use of the Euler-Lagrange equatons where µ L µϕ = 0 L ϕ = µ L 0 ϕ 1 L µϕ, 1.33 L 1 ϕ 2 L 2 ϕ 3 L 3 ϕ. Now, the Lagrangan densty for the self-nteractng scalar feld, or Klen-Gordon feld, whch we wll consder here, s gven by where L ϕ, µ ϕ = 1 2 µ ϕ m2 ϕ 2 λ 4! ϕ4, 1.34 µ ϕ 2 = 0 ϕ 2 1 ϕ 2 2 ϕ 2 3 ϕ 2. The theory, whch follows from the above Lagrangan densty 1.34, s n the lterature known as ϕ 4 theory. Applyng the Euler-Lagrange equatons 1.33 to the Lagrangan densty 1.34, yelds the followng quantum feld equaton µ µ + m 2 ϕx = λ 3! ϕ3 x When we compare the feld equaton 1.35 to the wave equaton 1.18 for a free Klen- Gordon partcle we may conclude that, for vanshng λ, equaton1.35 may be nterpreted as the feld equaton for a free Klen-Gordon feld. The term on the rghthand sde of equaton 1.35, whch stems from the term λϕ 4 /4! n the Lagrangan densty 1.34, 12

17 may be nterpreted as the source term whch descrbes the devaton of the theory for selfnteractng partcles from the free theory because of the presence of nteracton between the partcles. For ths reason we splt the Lagrangan densty n two parts, the free Lagrangan densty L 0 and the nteracton part L nt, defned by L 0 = 1 µ ϕ m2 ϕ 2 and L nt = λ 4! ϕ The frst term n L 0, whch generates the term µ µ ϕ n the feld equaton and s therefore related to the momentum squared of a free Klen-Gordon partcle, s called the knetc term; the second term n L 0 the mass term. As been observed above, n the absence of the source term the feld equaton 1.35 descrbes a free scalar quantum feld, φ, for whch the expresson 1.28 s a general soluton. As mentoned before, the objects of our nterest are the n-pont Green s functons, whch we are now capable of defnng Gx 1,...,xn = 0 {φx T 1 φxn e d 4 y L nt φy } 0 0 {e T d 4 y L nt φy }, where T stands for tme-orderng, whch means that n all expressons the felds must be permuted n such a way that the tme components of ther arguments are decreasng. 13

18 1.6 Tme-ordered product of two felds In ths secton we determne n all detal the vacuum expectaton value of the tme ordered product of two boson felds, also called propagator, and whch s defned by 0 T {φx 1 φx 2 } When we express the tme-orderng n terms of the θ-functon, defned n 1.2, whch vanshes for negatve argument and equals 1 for postve argument, then we obtan the followng two terms 0 T {φx 1 φx 2 } 0 = 0 φx 1 φx 2 0 θt 1 t φx 2 φx 1 0 θt 2 t e. each term beng characterzed by one of the two permutatons of the numbers one and two. From expresson 1.39 we learn that the frst thng to be calculated, are the smple vacuum expectaton values of two felds 0 φx 1 φx 2 0 and 0 φx 2 φx The full expressons for those objects, after the substtuton of formula 1.28 for the felds, are also qute long, but thngs become more managable by the use of the defntons ax = d 3 k 2π 3 2E e kx a k and φx = ax+a x Substtutng those defntons nto the frst term of formula 1.40, one obtans for the vacuum expectaton value of two felds 0 { ax1 +a x 1 }{ ax 2 +a x 2 } 0, 1.42 whch upon multplcaton leaves us wth the followng four terms 0 ax 1 ax a x 1 ax ax 1 a x a x 1 a x Three of the four terms n the expanson 1.43 vansh, as for example one has from the defnton 1.27 for the annhlaton operators that ax 0 = and hence, for a creaton operator d 3 k 2π 3 2E e kx a k 0 = 0, a x = {ax 0 } = As a consequence of those propertes for the operators defned n formula 1.41, we are then left wth only one nonzero contrbuton to frst of the two vacuum expectaton values 1.40 of two felds,.e. 0 φx 1 φx 2 0 = 0 ax1 a x 2 0,

19 whch, upon nserton of the full expresson 1.41 for the operators ax and a x, reads d 3 k 1 2π 3 2E 1 and hence contans the vacuum expectaton value d 3 k 2 2π 3 e k 1x 1 +k 2 x 2 a 0 k1 a 0 k2 2E 2 0 a k1 a k2 0. The latter expresson can easly be handled by the use of the commutaton relatons 1.27 and the propertes 1.27 for the annhlaton operators, whch leads to 0 a k1 a k2 0 = 0 { [ a k1, a k2 ] + a k2 a k1 } 0 and whch turns expresson 1.46 nto 0 φx 1 φx 2 0 = = 0 0 2π 3 2E 1 δ 3 k1 k a k2 a k1 0 = 2π 3 2E 1 δ 3 k1 k 2 d 3 k 1 2π 3 2E 1, d 3 k 2 2π 3 2E 2 e k 1x 1 +k 2 x 2 2π 3 2E 1 δ 3 k1 k 2. Because of the Drac delta functon, one may perform the k 2 -ntegraton and then rename the dummy k 1 ntegraton varable for k. Ths gves the vacuum expectaton value of formula 1.46 ts fnal form 0 φx 1 φx 2 0 = d 3 k 2π 3 2E e kx 1 x The second term of formula 1.40 equals the frst term wth the numbers one and two exchanged. So, we obtan for the vacuum expectaton value 1.39 of the tme ordered product of two boson felds the expresson 0 T {φx 1 φx 2 } 0 = d 3 k 2π 3 2E e kx 1 x 2 for t 1 t 2 d 3 k 2π 3 2E e kx 2 x 1 for t 1 t Now, n the exponents of 1.48 comes kx, whch n our metrc equals Et k x. Hence, n the above expresson we must take Et 1 t 2 for t 1 t 2 and Et 2 t 1 for t 1 t 2, whch s equvalent to takng E t 1 t 2 rrespectve of the order of t 1 and t 2,.e. 0 T {φx 1 φx 2 } 0 = d 3 k 2π 3 2E e k x 1 x 2 E t 1 t 2 for t 1 t 2 d 3 k 2π 3 2E e k x 2 x 1 E t 1 t 2 for t 1 t

20 Furthermore, by changng the ntegraton varable k to k n the lower of the two expressons n formula 1.49, results 0 T {φx 1 φx 2 } 0 = d 3 k 2π 3 2E e k x 1 x 2 E t 1 t Wth complex functon theory one can easly show the followng dentty + dk 0 2π k 2 e k 0t k 0 2 = e +m2 t 2 k m 2 2 k 2 +m 2, 1.51 whch, upon substtuton n formula 1.50, also rememberng that E actually stands for k 2 +m2, gves 0 T {φx 1 φx 2 } 0 = where k stands for k 0, k and d 4 k for dk 0 d 3 k. d 4 k e kx 1 x 2 2π 4 k 2 m 2, 1.52 Notce, that, snce k 0 s an ntegraton varable, k 2, whch equals k 0 2 k 2, s not dentcal to m 2,.e. s off-mass-shell. A graphcal representaton for the propagator 1.52 s as shown below. x k 1 x 2 One mght moreover recognze n the fnal expresson 1.52 for the vacuum expectaton value of the tme ordered product of two boson felds the Feynman propagator whch s gven n formula Proof of formula 1.51 For the proof of formula 1.51, whch we cast here n the form + dk 0 2π e k 0t k 0 2 M 2 = e M t 2M, 1.53 we ntroduce a small postve real number ǫ, such that the rghthand sde of equaton 1.53 gves e M t ǫ t, M +Oǫ whch vanshes n the lmts t ±. At the end of the calculatons we take ǫ 0. By comparson of formulae 1.53 and 1.54, we conclude that we must choose the substtuton M M ǫ

21 The ntegral whch consequently has to be calculated s then + dk 0 2π e k 0t k 0 M +ǫk 0 +M ǫ In the lterature one often fnds the form + dk 0 2π e k 0t k 0 2 M 2 +ǫ Ths can be acheved from expresson 1.56 by the substtuton 2Mǫ ǫ, 1.58 moreover gnorng the term quadratc n ǫ. The ntegrand of formula 1.56 has two sngulartes, or poles, n the complex k 0 plane at M +ǫ and at M ǫ, as ndcated n fgure 1.1. Imk 0 M + ǫ M +M M ǫ Rek 0 Fgure 1.1: The two poles of the ntegrand of formula 1.56 n the complex k 0 plane at M +ǫ and at M ǫ. Next, we concentrate on the numerator of the ntegrand of expresson 1.56,.e. e k 0t When k 0 s complex, then t has a real part and an magnary part Hence, formula 1.59 turns nto k 0 = Rek 0 + Imk e Rek 0t + Imk 0 t In the followng, we study what happens to the expresson 1.61 when we take k 0. Except for the cases where k 0 s real, hence Imk 0 = 0, we fnd and Imk 0 + for k 0 n the upper half complex k 0 plane, 1.62 Imk 0 for k 0 n the lower half complex k 0 plane

22 Consequently, for t < 0, we obtan e Rek 0t + Imk 0 t 0 for k0 n the upper half complex k 0 plane, 1.64 whereas, for t > 0, we obtan e Rek 0t + Imk 0 t 0 for k0 n the lower half complex k 0 plane, 1.65 Let us consder for t < 0 the followng contour n the complex k 0 plane. Imk 0 C u M + ǫ M ǫ Rek 0 When we let the radus of the half crcle of contour C u go to nfnty, then we have C u dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = = dk 0 2π e k 0t k 0 2 M 2 +2Mǫ + half crcle dk 0 2π e k 0t k 0 2 M 2 +2Mǫ. However, the ntegrand of the ntegral over the half crcle vanshes when ts radus approaches nfnty, accordng to formula Accordngly, for nfnte radus one has C u dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = + dk 0 2π e k 0t k 0 2 M 2 +2Mǫ Now, accordng to complex functon theory, the counterclockwse ntegral of a closed contour n the complex plane equals 2π tmes the sum of the resdues on the poles contaned n the closed contour. Here, we have one pole at k 0 = M + ǫ, where the resdue may easly be determned usng formula 1.56, to gve 2π e Mt+ǫt 2M +2ǫ Hence, for t < 0 we obtan + dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = emt+ǫt 2M 2ǫ = e M t ǫ t 2M 2ǫ

23 For t > 0 we consder the followng contour n the complex k 0 plane. M + ǫ Rek 0 M ǫ C l Imk 0 The ntegrand of the ntegral over the half crcle vanshes when ts radus approaches nfnty, accordng to formula Accordngly, for nfnte radus one has here C l dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = + dk 0 2π e k 0t k 0 2 M 2 +2Mǫ Furthermore, accordng to complex functon theory, the clockwse ntegral of a closed contour n the complex plane equals 2π tmes the sum of the resdues on the poles contaned n the closed contour. Here, we have one pole at k 0 = M ǫ, where the resdue may easly be determned usng formula 1.56, to gve 2π e Mt ǫt 2M 2ǫ Hence, for t > 0 we obtan + dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = e Mt ǫt 2M 2ǫ = e M t ǫ t 2M 2ǫ By comparson of formulae 1.69 and 1.72, we fnd for any sgn of t the result + dk 0 2π e k 0t k 0 2 M 2 +2Mǫ = e M t ǫ t 2M 2ǫ Takng the lmt ǫ 0, one fnds formula

24 1.7 Tme-ordered product of four felds In ths secton we determne n all detal the vacuum expectaton value of the tme ordered product of four boson felds, whch s defned by 0 T {φx 1 φx 2 φx 3 φx 4 } When we express the tme-orderng n terms of the θ-functon, then we obtan the followng twenty-four terms 0 T {φx 1 φx 2 φx 3 φx 4 } 0 = 1.75 = 0 φx 1 φx 2 φx 3 φx 4 0 θt 1 t 2 θt 2 t 3 θt 3 t φx 1 φx 2 φx 4 φx 3 0 θt 1 t 2 θt 2 t 4 θt 4 t φx 1 φx 3 φx 2 φx 4 0 θt 1 t 3 θt 3 t 2 θt 2 t 4 +,.e. each term beng characterzed by one of the twenty-four permutatons of the numbers one to four. From expresson 1.75 we learn that the frst thng to be calculated, s the smple vacuum expectaton value of four felds. There are twenty-four of them, whch are all just permutatons of the frst, gven by 0 φx 1 φx 2 φx 3 φx The full expresson for ths object s also qute long, but thngs become more managable by the use of the defntons gven n formula Substtutng those defntons nto the expresson of formula 1.76 for the smple vacuum expectaton value of four felds, one obtans 0 { ax1 +a x 1 }{ ax 2 +a x 2 }{ ax 3 +a x 3 }{ ax 4 +a x 4 } Here we perform the varous multplcatons, to end up wth sxteen terms gven by 0 ax 1 ax 2 ax 3 ax a x 1 ax 2 ax 3 ax ax 1 a x 2 ax 3 ax Several of the terms n the expanson 1.78 vansh because of the propertes 1.44 and 1.45 for the operators defned n formula More complcated cases, lke 0 ax1 ax 2 ax 3 a x 4 0, whch, accordng to the defntons 1.41, equals 20

25 d 3 k 1 2π 3 2E 1 d 3 k 2 2π 3 2E 2 d 3 k 3 2π 3 2E 3 d 3 k 4 2π 3 2E 4 e k 1x 1 k 2 x 2 k 3 x 3 +k 4 x 4 0 a k1 a k2 a k3 a k and hence contans the vacuum expectaton value 0 a k1 a k2 a k3 a k4 0, can be handled by the use of the commutaton relatons 1.27, whch leads to 0 a k1 a k2 a k3 a k4 0 = = 0 a { [ ] k1 a k2 a k3, a } k4 + a k4 a k3 0 = 0 a k1 a k2 0 2π 3 2E 3 δ 3 k3 a k k1 a k2 a k4 a k3 0 = 0. Inspecton of all sxteen terms of 1.78 gves as a result that fourteen of those vansh. We are then left wth only two nonzero contrbutons 0 φx 1 φx 2 φx 3 φx 4 0 = 1.80 = 0 ax1 ax 2 a x 3 a x ax1 a x 2 ax 3 a x 4 0. Ths can easly be seen, snce, frst, a vacuum expectaton value for an operator whch does not have an equal number of creaton and annhlaton operators, lke the one gven n formula 1.79, allways ends up wth an annhlaton operator actng on 0 or a creaton operator actng on 0, by the use of commutaton relatons 1.27 whenever necessary. Moreover, a vacuum expectaton value also vanshes when a creaton operator stands on the lefthand sde or when an annhlaton operator stands on the rghthand sde. Consequently, forx 1 we must have anannhlatonoperatorandforx 4 a creatonoperator. Ths then mples that for x 2 and x 3 we must have one annhlaton and one creaton operator. There are only two possbltes, whch are shown n formula The frst term of 1.80, whch, n a way smlar to formula 1.79, contans the vacuum expectaton value 0 a k1 a k2 a k3 a k4 0, can be handled by the use of the commutaton relatons Frst, we commute a k2 and a k3, whch leads to 0 a k1 a k4 0 2π 3 2E 2 δ 3 k2 k a k1 a k3 a k2 a k

26 Then, we commute n the frst of the above two terms a k1 and a k4, whereas n the second of the above two terms we commute as well a k1 wth a k3, as a k2 wth a k4. The result of those operatons s gven by 42π 6 E 1 E 2 δ 3 k1 k 4 δ 3 k2 k π 6 E 1 E 2 δ 3 k1 k 3 δ 3 k2 k In the second term of 1.80, whch, n a way smlar to formula 1.79, contans the vacuum expectaton value 0 a k1 a k2 a k3 a k4 0, we commute as well a k1 wth a k2, as a k3 wth a k4. The result of those operatons s gven by 42π 6 E 1 E 3 δ 3 k1 k 2 δ 3 k3 k When we sum the two expressons 1.81 and 1.82 and also nclude the ntegratons and the correspondng exponentals, we fnd for the vacuum expectaton value of formula 1.80 the result 0 φx 1 φx 2 φx 3 φx 4 0 = = d 3 k 1 2π 3 2E 1 d 3 k 2 2π 3 2E 2 d 3 k 3 2π 3 2E 3 d 3 k 4 2π 3 2E 4 { e k [ 1x 1 k 2 x 2 +k 3 x 3 +k 4 x 4 42π 6 E 1 E 2 δ 3 k1 k 4 δ 3 k2 k π 6 E 1 E 2 δ 3 k1 k 3 δ 3 k2 k 4 ] + + e k 1x 1 +k 2 x 2 k 3 x 3 +k 4 x 4 42π 6 E 1 E 3 δ 3 k1 k 2 δ 3 k3 k 4 }. Because of the Drac delta functons, one may perform two of the four k-ntegratons n each of the three above terms. In the frst two terms we perform the k 3 and the k 4 ntegratons. Inthethrdtermweperformthe k 2 andthe k 4 ntegratons, andthenrename the k 3 ntegraton varable for k 2. Ths gves the vacuum expectaton value of formula 1.80 ts fnal form 0 φx 1 φx 2 φx 3 φx 4 0 = = d 3 k 1 2π 3 2E 1 d 3 { k 2 2π 3 e k 1x 1 x 4 k 2 x 2 x 3 + 2E 2 + e k 1x 1 x 3 k 2 x 2 x 4 + e k 1 x 1 x 2 k 2 x 3 x 4 }

27 Not a very terrble result, but remember that the vacuum expectaton value 1.75 of the tme ordered product of four boson felds contans twenty-four of such terms, whch now has to be multpled by three. So, we have ended up wth seventy-two terms, hence some bookkeepng s n order. For convenence we defne Ax 1 x 2 = d 3 k 2π 3 2E e kx 1 x Usng ths defnton and the result 1.83, we obtan for the vacuum expectaton value 1.75 of the tme ordered product of four boson felds the expresson 0 T {φx 1 φx 2 φx 3 φx 4 } 0 = = {Ax 1 x 4 Ax 2 x 3 + Ax 1 x 3 Ax 2 x Ax 1 x 2 Ax 3 x 4 }θt 1 t 2 θt 2 t 3 θt 3 t all possble permutatons of 1,2,3 and Indeed 24 3 = 72 terms! However, as we wll see n the followng, ther number can be reduced to three. By nspecton of all twenty-four permutatons of 1.85, we fnd that there are several terms whch contan the same combnaton of A s. Notce, from ther defnton 1.84, that the order of the A s n a product of A s does not matter, but that the order of the coordnate varables x nsde one A do matter. In the followng table, where we denote t 1 t 2 t 3 t 4 by 1234 and smlar for the other tme-orderngs, we have collected all twenty-four possble tme-orderngs whch contrbute to 1.85 and the A s to whch they are multpled. 23

28 tmeorderng Ampltudes nvolved 1234 A1 2 A3 4 A1 3 A2 4 A1 4 A A1 2 A4 3 A1 4 A2 3 A1 3 A A1 3 A2 4 A1 2 A3 4 A1 4 A A1 3 A4 2 A1 4 A3 2 A1 2 A A1 4 A3 2 A1 3 A4 2 A1 2 A A1 4 A2 3 A1 2 A4 3 A1 3 A A2 1 A3 4 A2 3 A1 4 A2 4 A A2 1 A4 3 A2 4 A1 3 A2 3 A A2 3 A1 4 A2 1 A3 4 A2 4 A A2 3 A4 1 A2 4 A3 1 A2 1 A A2 4 A3 1 A2 3 A4 1 A2 1 A A2 4 A1 3 A2 1 A4 3 A2 3 A A3 2 A1 4 A3 1 A2 4 A3 4 A A3 2 A4 1 A3 4 A2 1 A3 1 A A3 1 A2 4 A3 2 A1 4 A3 4 A A3 1 A4 2 A3 4 A1 2 A3 2 A A3 4 A1 2 A3 1 A4 2 A3 2 A A3 4 A2 1 A3 2 A4 1 A3 1 A A4 2 A3 1 A4 3 A2 1 A4 1 A A4 2 A1 3 A4 1 A2 3 A4 3 A A4 3 A2 1 A4 2 A3 1 A4 1 A A4 3 A1 2 A4 1 A3 2 A4 2 A A4 1 A3 2 A4 3 A1 2 A4 2 A A4 1 A2 3 A4 2 A1 3 A4 3 A1 2 For example, when we collect all terms, underlned n the prevous table, whch contan the product of Ax 1 x 2 and Ax 3 x 4, then we fnd sx such terms,.e. Ax 1 x 2 Ax 3 x 4 {θt 1 t 2 θt 2 t 3 θt 3 t θt 1 t 3 θt 3 t 2 θt 2 t 4 + θt 1 t 3 θt 3 t 4 θt 4 t θt 3 t 1 θt 1 t 2 θt 2 t 4 + θt 3 t 1 θt 1 t 4 θt 4 t θt 3 t 4 θt 4 t 1 θt 1 t 2 }

29 In total the vacuum expectaton value 1.85 of the tme ordered product of four boson felds contans twelve such expressons, smlar to the one n formula 1.86, whch gves agan 12 6 = 72 terms. They are collected n the table below. dstnct terms tme-orderng contrbutons A1 2 A A1 3 A A1 4 A A1 2 A A1 4 A A1 3 A A2 1 A A2 1 A A2 4 A A2 3 A A3 2 A A3 1 A Now, when we nspect carefully the sx combnatons of θ-functons n formula 1.86, then we fnd that they together just bult up the tme nterval gven by t 1 t 2 and t 3 t 4. For nstance, when we denote t 1 t 2 t 3 t 4 by 1234 andsmlar forthe other tme-orderngs, then the tme nterval t 1 t 2 and t 3 t 4 just conssts of 1234, 1324, 1342, 3124, 3142, and 3412, whch tme-orderngs represent precsely the sx combnatons of θ-functons n formula Any other permutaton of 1, 2, 3 and 4 les outsde the referred tme nterval. Consequently, formula 1.86 can be substtuted by Ax 1 x 2 Ax 3 x 4 θt 1 t 2 θt 3 t The seventy-two terms contaned n formula 1.85 have reduced to twelve terms of the form A smlar expresson whch comes from the product of Ax 2 x 1 and Ax 3 x 4, and whch s bult up of the tme-ordered terms 2134, 2314, 2341, 3214, 3241, and 3421, s gven by The sum of 1.87 and 1.88 gves whch equals or Ax 2 x 1 Ax 3 x 4 θt 2 t 1 θt 3 t {Ax 1 x 2 θt 1 t 2 + Ax 2 x 1 θt 2 t 1 } Ax 3 x 4 θt 3 t 4, Ax 1 x 2 Ax 3 x 4 θt 3 t 4 when t 1 t 2, Ax 2 x 1 Ax 3 x 4 θt 3 t 4 when t 1 t 2. 25

30 Now, usng the same procedure whch lead from formula 1.48 to formula 1.52, we fnd for the sum of 1.87 and 1.88 the followng d 3 k 2π 3 2E e k x 1 x 2 e E t 1 t 2 Ax3 x 4 θt 3 t 4 = d 4 k e kx 1 x 2 2π 4 k 2 m 2 Ax 3 x 4 θt 3 t Contnung the above procedure, all seventy-two terms of 1.85 can be summed setwse n sx sets of twelve terms. For example, another such set of twelve terms sums up to d 4 k e kx 1 x 2 2π 4 k 2 m 2 Ax 4 x 3 θt 4 t Now, at ths stage, t mght be clear that, along the same reasonng whch lead to formula 1.89, we obtan for the sum of 1.89 and 1.90 the result 2 d 4 k 1 2π 4 e k 1x 1 x 2 k 1 2 m 2 d 4 k 2 2π 4 e k 2x 3 x 4 k 2 2 m A set of twenty-four terms of 1.85 neatly summed up n a compact expresson. The other two sets of twenty-four terms yeld: and 2 d 4 k 1 2π 4 e k 1x 1 x 3 k 1 2 m 2 d 4 k 2 2π 4 e k 2x 2 x 4 k 2 2 m 2, d 4 k 1 2π 4 e k 1x 1 x 4 k 1 2 m 2 d 4 k 2 2π 4 e k 2x 2 x 3 k 2 2 m The whole vacuum expectaton value of the tme ordered product of four boson felds s just gven by the sum of the three expressons, 1.91, 1.92, and Each of those expressons s just the product of two Feynman propagators as gven n formula A graphcal representaton for the three expressons, 1.91, 1.92, and 1.93, can be constructed as follows: The coordnates x 1, x 2, x 3, and x 4 are represented by four dots, as shown below. x 1 x 2 x 3 x 4 Each possble parwse connecton of those dots represents one of the three above expressons accordng to the combnaton of coordnates n the exponents. The three possble parwse connectons are gven below. 26

31 x 1 k 1 x 2 x 1 x 2 x 2 x 1 k 1 k 1 k 2 k 2 x 3 a x 4 x 3 b x 4 k x 2 3 x 4 c Graph a corresponds to expresson 1.91, graph b to expresson 1.92, and graph c to expresson It mght be clear that for more felds the procedure becomes more tedous. For ths reasonwewllmakeuseofthefeynman rules, whchtakecareforusofalltme-orderngs and leave us wth the fnal ntegrals. 27

32 1.8 Feynman rules part I In order to determne an analytc expresson for the vacuum expecton value of a tmeordered product of n felds, gven by 0 T {φx 1 φxn} 0, 1.94 one proceeds as follows. Each feld φ n 1.94 brngs, followng the defnton 1.28 for the quantum felds as well as the procedure whch lead from formula 1.48 to formula 1.52, a Fourer transform ntegraton of the form d 4 k 2π 4 e k x for = 1,...,n The n events are graphcally represented by n dots, as shown n the fgure below x 5 x 4 k 4 k 5 x 6 k6. x 3 k 3 x k 2 2 kn xn x 1 k 1 From each dot flows momentum away, as also ndcated n the same pcture. Those momenta correspond to the Fourer transform ntegraton varables and are closely related to the creaton and annhlaton operators, a k and a k. Consequently, each momentum flow relates to one feld defned at the correspondng event. Hence, when for an event y more felds are nvolved, as many momenta flow from the related dot as there come felds wth argument y n the expresson for the vacuum expectaton value. Overall momentum conservaton gves moreover a factor 2π 4 δ 4 k 1 +k 2 + +kn So far, the procedure s the same for each contrbuton,.e. 0 T {φx 1 φxn} 0 = d 4 k 1 2π 4e k 1x 1 d 4 k 2 2π 4e k 2x2 d 4 kn 2π 4e k nxn 2π 4 δ 4 k 1 +k 2 + +kn { somethng }.1.97 The somethng contans all possble contrbutons, whch are found by contractng parwse, n all possble combnatons, the momenta. When you do t wth a pencl, then you obtan the Feynman graphs. In the analytc expresson one wrtes for each contracton a Feynman propagator, as defned n formula 1.23, and moreover a Drac delta functon to assure momentum conservaton multpled wth 2π 4 of course, except for one par whch follows already from the overall plus all the other Drac delta functons. 28

33 We gve below three examples, the already known vacuum expectaton values of the tme-ordered products of two and four felds and a new vacuum expectaton value whch also nvolves some combnatorcs. I The vacuum expectaton value of the tme-ordered product of two felds. From expresson 1.38, followng the above outlned procedure, we fnd for the vacuum expectaton value of the tme-ordered product of two felds the followng graphc representaton x 1 x 2 k1 Consequently, one has only one possble contracton, whch leads to the analytc expresson gven by k 2 d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x 2 2π 4 δ 4 k 1 +k 2 k 1 2 m 2, 1.98 for whch t s a smple task just perform the k 2 -ntegraton to convnce oneself that ths equals the prevous expresson II The vacuum expectaton value of the tme-ordered product of four felds. From expresson 1.74, followng the above outlned procedure, we fnd for the vacuum expectaton value of the tme-ordered product of four felds the followng graphc representaton x 1 k x 2 1 k 2 k 3 x 3 k 4 x4 Consequently, the general form of the analytc expresson reads d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x 2 d 4 k 3 2π 4 e k 3x 3 d 4 k 4 2π 4 e k 4x 4 2π 4 δ 4 k 1 +k 2 +k 3 +k 4 {contractons} There are three dfferent possble ways to contract the four momenta n ths case, as we already know from secton

34 Contractng k 1 wth k 2 and k 3 wth k 4 gves 2π 4 δ 4 k 1 +k 2 k 1 2 m 2 k 3 2. m 2 Notce that only one of the two contractons nvolves a Drac delta functon for the momentum conservaton, the other par s then automatcally conserved because of the Drac delta functon n formule 1.99 for the overall momentum conservaton. The other two contrbutons to 1.99, wth comparable expressons to the one above, come from the other two possble ways to contract the momenta. In total, we fnd then 0 T {φx 1 φx 2 φx 3 φx 4 } 0 = d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x 2 d 4 k 3 2π 4 e k 3x 3 d 4 k 4 2π 4 e k 4x 4 2π 4 δ 4 k 1 +k 2 +k 3 +k 4 { 2π 4 δ 4 k 1 +k 2 k 1 2 m 2 k 3 2 m π 4 δ 4 k 1 +k 3 + 2π 4 δ 4 k 1 +k 4 k 1 2 m 2 k 2 2 m + 2 } k 1 2 m 2 k 3 2 m 2. After performng two of the four k-ntegratons one obtans the same result as gven by the sum of the three expressons, 1.91, 1.92, and

35 III The vacuum expectaton value of the tme-ordered product of sx felds, out of whch four are at the same event. The vacuum expectaton value of the tme-ordered product of sx felds, out of whch four are at the same event, s gven by 0 T {φx 1 φx 2 φyφyφyφy} 0, whereas ts general structure s represented by the followng graph q 1 q 4 q 2 y q 3 x 1 x 2 k1 and, moreover, ts correspondng analytc expresson takes the form k 2 d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x 2 d 4 q 1 2π 4 e q 1y d 4 q 2 2π 4 e q 2y d 4 q 3 2π 4 e q 3y d 4 q 4 2π 4 e q 4y 2π 4 δ 4 k 1 +k 2 +q 1 +q 2 +q 3 +q 4 {somethng} The somethng contans ffteen contrbutons: For, one of the sx momenta can be contracted wth each of the fve other momenta. One of the four remanng momenta can be contracted wth one out of three momenta. Whereas, the fnally remanng two momenta can only be contracted amongst each other. Ths gves ndeed = possbltes. There are two types of contractons whch can be dstngushed. The frst type, whch we wll refer to as type A contrbutons, s the result of contractng k 1 wth k 2 and the q s amongst each other. The generc graph s depcted below. type A 31

36 There are three such contrbutons, whch result all three n the same analytc expresson, because oneofthe q scanbecontracted wth eachof theremanng threeq sandmoreover ntegraton varables are dummy. We obtan then for type A the expresson 3 2π 4 δ 4 q 1 +q 4 q 1 2 m 2 2π4 δ 4 q 2 +q 3 q 2 2 m 2 k 1 2. m 2 So, the type A contractons lead to the contrbuton 3 d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x d 4 q 1 2π 4 e q 1y d 4 q 2 2π 4 e q 2y d 4 q 3 2π 4 e q 3y d 4 q 4 2π 4 e q 4y 2π 4 δ 4 k 1 +k 2 +q 1 +q 2 +q 3 +q 4 2π 4 δ 4 q 1 +q 4 q 1 2 m 2 2π4 δ 4 q 2 +q 3 q 2 2 m 2 k 1 2. m 2 When we perform the q 3 and q 4 ntegratons, then, because of the Drac delta functons, we end up wth 3 d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x 2 2π 4 δ 4 k 1 +k 2 k 1 2 m 2 d 4 q 1 2π 4 q 1 2 m 2 d 4 q 2 2π 4 q 2 2 m The latter two ntegrals are so-called loop ntegratons snce each can be assocated wth one of the two loops n the graph for the type A contrbutons. When n formula one substtutes x 1, x 2, x 3, and x 4, by y one obtans exactly three tmes the product of those loop ntegratons. Moreover, by comparson to formula 1.98, one fnds that the frst part of the above expresson equals the vacuum expectaton value of the tmeordered product of two felds. Consequently, one may wrte for the type A contrbutons the followng dentty 0 T {φx 1 φx 2 φyφyφyφy} 0 type A contrbutons = = 0 T {φx 1 φx 2 } 0 0 T {φyφyφyφy} Contrbutons, whch are represented by graphs smlar to the graph for the type A contrbuton,.e. graphs whch have dsconnected parts, are called vacuum bubbles. They do not play any role n real physcs as we wll see furtheron. 32

37 The second type of contrbutons to the somethng of formula 1.102, whch we wll refer to as type B contrbutons, stem from the contractons of k 1 and k 2 each wth one of the q s. The generc graph s depcted below In the lterature ths graph s usually drawn as shown hereafter type B There are twelve such contrbutons, whch result all twelve n the same analytc expresson, because one of the k s can be contracted wth each of the four q s, the other k wth any of the three remanng q s and moreover ntegraton varables are dummy, whch gves contrbutons. We obtan then for type B the expresson 4 3 = π 4 δ 4 k 1 +q 1 k 1 2 m 2 2π4 δ 4 k 2 +q 2 k 2 2 m 2 q 3 2. m 2 So, the type B contractons lead to the contrbuton 12 d 4 k 1 2π 4 e k 1x 1 d 4 k 2 2π 4 e k 2x d 4 q 1 2π 4 e q 1y d 4 q 2 2π 4 e q 2y d 4 q 3 2π 4 e q 3y d 4 q 4 2π 4 e q 4y 2π 4 δ 4 k 1 +k 2 +q 1 +q 2 +q 3 +q 4 2π 4 δ 4 k 1 +q 1 k 1 2 m 2 2π4 δ 4 k 2 +q 2 k 2 2 m 2 q 3 2 m 2 33

38 When we perform the q 1 and q 2 ntegratons, then, because of the Drac delta functons, we end up wth 12 d 4 k 1 2π 4 e k 1x 1 y k 1 2 m 2 d 4 k 2 2π 4 e k 2x 2 y k 2 2 m 2 d 4 q 3 2π 4 e q 3y d 4 q 4 2π 4 e q 4y 2π 4 δ 4 q 3 +q 4 q 3 2 m Next, we may perform the q 4 ntegraton, to end up wth 12 d 4 k 1 2π 4 e k 1x 1 y k 1 2 m 2 d 4 k 2 2π 4 e k 2x 2 y k 2 2 m 2 d 4 q 2π 4 q 2 m For the latter part of ths expresson we recognze agan a loop ntegral, correspondng to the loop n the type B graph. 34

39 Chapter 2 Two-ponts Green s functon Followng formula 1.37, also substtutng expresson 1.36 for the nteracton Lagrangan densty, the two-ponts Green s functon s n φ 4 theory defned by Gx 1,x 2 = [ 0 {φx T 1 φx 2 exp d 4 y 4! λ { [ 0 T exp d 4 y 4! λ φ 4 y]} 0 φ y]} 4 0, 2.1 Whenweexpandtheexponentnthenumeratorof2.1,thenweobtanforthenumerator the followng seres of tme-ordered vacuum expectaton values 0 {φx T 1 φx 2 exp [ d 4 y 4! λ φ y]} 4 0 { { = 0 T φx 1 φx d 4 y λ φ 4 y + 4! = [ d 4 y λ ] 2 φ 4 y + 1 [ d 4 y λ 3 φ y] 4 + 2! 4! 3! 4! 0 = 0 T {φx 1 φx 2 } 0 + λ 4! d 4 y 0 T {φx 1 φx 2 φ 4 y} λ 2 d 4 y 1 2! 4! d 4 y 2 0 T {φx 1 φx 2 φ 4 y 1 φ 4 y 2 } λ 3 d 4 y 1 3! 4! +, d 4 y 2 d 4 y 3 0 T {φx 1 φx 2 φ 4 y 1 φ 4 y 2 φ 4 y 3 } 0 + whch may be consdered as an expanson n the couplng constant λ. For the frst term of the expanson 2.2 we recognze the vacuum expectaton value of the tme-ordered product of two felds, for whch we have the analytc expressons 1.38 or The second term, lnear n λ, contans the vacuum expectaton value 1.101, whch we have determned prevously to be equal to the sum of the expressons 35

40 1.105, referred to as the type A contrbuton, and 1.110, whch we called the type B contrbuton. So, up to the frst order n λ we fnd for the numerator of 2.1 the result 0 {φx T 1 φx 2 exp = 0 T {φx 1 φx 2 } 0 + [ d 4 y 4! λ φ y]} 4 0 = 2.3 λ 4! d 4 y {type A + type B} +. Now, for the type A contrbuton we have the dentty gven n formula Consequently, we may also wrte the numerator of 2.1 lke 0 {φx T 1 φx 2 exp = 0 T {φx 1 φx 2 } 0 { [ d 4 y 1 + 4! λ φ y]} 4 0 λ 4! d 4 y 0 T {φ 4 y} 0 = 2.4 } + + λ 4! d 4 y {type B} +. The denomnator of 2.1, expanded to frst order n λ, reads 0 {exp T [ d 4 y 4! λ φ y]} 4 0 = 1 + λ 4! d 4 y 0 T {φ 4 y} So, by dvdng out the denomnator 2.5 of the two-ponts Green s functon 2.1 from the expresson 2.4 for ts numerator, we obtan to frst order n λ the result Gx 1,x 2 = 0 T {φx 1 φx 2 } 0 + λ 4! d 4 y {type B} The type A contrbuton has dsappeared from the fnal expresson for the two-ponts Green s functon, whch result can be generalzed, as we wll dscuss n the next secton. 36

41 2.1 Vacuum bubbles The events y, whch stem from the nteracton part of the Lagrangan densty, are n general called the nternal ponts of a contrbuton to the n-ponts Green s functon. The other events x, whch come as arguments of the n-ponts Green s functon, are referred as the external ponts. Now, when n a Feynman graph for one or more nternal ponts do not exst any propagators, drectly or ndrectly, whch connect them to the external ponts, then the bubble-lke structures around those nternal ponts are called vacuum bubbles. The type A contrbuton to the vacuum expectaton value gven n formula 1.101, contans such vacuum bubble. Other examples, for whch the graphs are shown below, come from the second order n λ term of the expanson 2.2 for the 2-ponts Green s functon. x 1 x 2, x 1 x 2, x 1 x 2 and x 1 x 2. The sum of the contrbutons represented by the frst three of the here shown graphs s, smlarly to the factorzaton for the type A contrbuton, gven by { 0 T {φx 1 φx 2 } 0 0 T 1 2! [ d 4 y 4! λ } 2 φ y] 4 0, 2.7 whch can be consdered to represent the second order n λ vacuum bubble extenson of the vacuum expectaton value of the tme-ordered product of two felds. One can easly magne how the hgher order extensons look lke. In fact, one can proof that the sum of all possble vacuum bubble extensons of the vacuum expectaton value of the tme-ordered product of two felds s just gven by 0 T {φx 1 φx 2 } 0 0 {exp T [ d 4 y 4! λ φ y]} The latter of the four above second order n λ vacuum bubble graphs reads analytcally {type B} 0 { T d 4 y 37 4! λ φ y} 4 0, 2.9