1 Interactions and Green functions

Transcription

1 Interactng Quantum Feld Theory 1 D. E. Soper 2 Unversty of Oregon Physcs 665, Quantum Feld Theory February Interactons and Green functons In these sectons, we dscuss perturbaton theory for the nteractng theory L = 1 2 ( µφ)( µ φ) 1 2 m2 φ 2 1 4! λφ4. (1) We wll try to fnd the scatterng matrx for ths theory. The dea s to consder that λ 1 and expand n powers of λ. We wll follow the argument n Peskn and Schroeder qute closely. Before we try to fnd the scatterng matrx, we wll nvestgate somethng a lttle smpler, the Green functons G(x 1, x 2,..., x N ) = Ω T φ(x 1 )φ(x 2 ) φ(x N ) Ω. (2) Here Ω s the vacuum state the real nteractng vacuum state. Followng the notaton of Peskn and Schroeder we use 0 for the vacuum of the free theory. The T ndcates that the feld operators are to be tme ordered, those wth the latest tmes go to the left. For example, f t 2 < t 1 < t 3 then T φ(x 1 )φ(x 2 )φ(x 3 ) = φ(x 3 )φ(x 1 )φ(x 2 ). (3) To smplfy thngs a lttle more, let s smply look for the two-pont Green functon G(x, y) = Ω T φ(x)φ(y) Ω. (4) and let s assume that x 0 > y 0 so that G(x, y) = Ω φ(x)φ(y) Ω. (5) Once we have solved ths problem, t wll be easy to generalze to N-pont Green functons wth any orderng for the tmes. 1 Copyrght, 2001, D. E. Soper 2 soper@bovne.uoregon.edu 1

2 2 The nteracton pcture We wrte ( H = H 0 + H nt = H 0 + d x 1 4! λφ4 + E 0 ). (6) Here H 0 s the hamltonan for the free theory and E 0 s whatever constant we have to add to get the energy of the nteractng vacuum to be zero. We wrte the tme dependence of the feld as φ(t, x) = e Ht φ(0, x)e Ht. (7) Let us also defne the nteracton pcture feld φ I (t, x) by φ I (t, x) = e H 0t φ(0, x)e H 0t. (8) Then φ I (t, x) s a free feld and we can wrte t n terms of creaton and destructon operators as φ I (t, x) = (2π) 3 d k 2ω( k) { e k µx µ a( k) + e +kµxµ a ( k) } (9) The nteractng feld s related to the nteracton pcture free feld by where φ(t, x) = U(0, t)φ I (t, x)u(t, 0). (10) U(t, t ) = e H 0t e H(t t ) e H 0t. (11) We can also use U(t, t ) to relate Ω to 0. We let Ω be the ground state of H, wth energy that we have adjusted to be zero, and let n be the egenstates of H wth hgher energes. Then e HT 0 = Ω Ω 0 + n e EnT n n 0 (12) Let T be large wth a large negatve magnary part, T (1 ɛ). In ths lmt, the n terms vansh and we have Ω = lm T (1 ɛ) e HT 0 / Ω 0 = lm U(0, T ) 0 / Ω 0, (13) T (1 ɛ) and, wth a smlar argument, Ω = lm T (1 ɛ) 0 e HT / 0 Ω = lm 0 U(T, 0)/ 0 Ω. (14) T (1 ɛ) 2

3 Wth these results we can wrte G(x, y) = Ω φ(x)φ(y) Ω = Ω U(0, x 0 )φ I (x)u(x 0, 0)U(0, y 0 )φ I (y)u(y 0, 0) Ω = Ω U(0, x 0 )φ I (x)u(x 0, y 0 )φ I (y)u(y 0, 0) Ω 1 = lm T (1 ɛ) 0 Ω Ω 0 0 U(T, 0)U(0, x0 )φ I (x) U(x 0, y 0 )φ I (y)u(y 0, 0)U(0, T ) 0 = lm T (1 ɛ) 0 U(T, x 0 )φ I (x)u(x 0, y 0 )φ I (y)u(y 0, T ) 0 0 Ω Ω 0 (15) We can smplfy ths a lttle more by wrtng the zero-pont Green functon as 0 U(T, T ) 0 1 = Ω Ω = lm (16) T (1 ɛ) 0 Ω Ω 0 If we dvde by 1 wrtten n ths way, we get G(x, y) = lm T (1 ɛ) 0 U(T, x 0 )φ I (x)u(x 0, y 0 )φ I (y)u(y 0, T ) 0. (17) 0 U(T, T ) 0 Ths s a nce compact expresson. To use t, we just have to understand U(t, t ). 3 The evoluton operator Well, f we have to understand U(t, t ), then we should work wth t. From ts defnton, U(t, t ) obeys the dfferental equaton d dt U(t, t ) = e +H 0t (H H 0 )e H 0t e +H 0t e H(t t ) e H 0t (18) wth the boundary condton U(t, t ) = 1. equaton as We can wrte the dfferental d dt U(t, t ) = H I (t)u(t, t ) (19) where H I (t) = e H 0t (H H 0 )e H 0t. (20) 3

4 The soluton of the dfferental equaton wth the boundary condton s ( t ) U(t, t ) = T exp dτ H I (τ) t (21) where the T denotes tme orderng of the operators after expandng the exponental. That s t U(t, t ) = 1 dτ H I (τ) t τ2 t +( ) 2 dτ 2 dτ 1 H I (τ 2 ) H I (τ 1 ) t t t +( ) 3 τ3 τ2 dτ 3 dτ 2 dτ 1 H I (τ 3 ) H I (τ 2 ) H I (τ 1 ) t t t +. (22) Note that the Nth order term n the expanson of the exponental has a 1/N!, whch s cancelled because there are N! tme orderngs of the operators, each gvng a contrbuton of the same form. Thus, n the result G(x, y) = lm T (1 ɛ) the denomnator s The numerator s { ( 0 φ I (y) 0 U(T, x 0 )φ I (x)u(x 0, y 0 )φ I (y)u(y 0, T ) 0, (23) 0 U(T, T ) 0 0 U(T, T ) 0 = 0 T exp T exp { T exp T ( x 0 dτ H I (τ) y 0 T )} dτ H I (τ) ( φ I (x) Snce (we have supposed) x 0 > y 0, ths s 0 T exp ( T T )} dτ H I (τ) T { T T exp dτ H I (τ) Wrtten n ths way, the result works also for y 0 > x 0. 4 ( ) x 0 y 0 0. (24) dτ H I (τ) )} 0 (25) ) φ I (x)φ I (y) 0. (26)

5 4 Result for the Green Functon Thus we arrve at the smple result 0 T exp ( T T G(x, y) = lm dτ H I(τ) ) φ I (x)φ I (y) 0 T (1 ɛ) 0 T exp ( T T dτ H I(τ) ). (27) 0 For Green functons wth N operators, a smlar dervaton gves 0 T exp ( T T G(x 1,..., x N ) = lm dτ H I(τ) ) φ I (x 1 ) φ I (x N ) 0 T (1 ɛ) 0 T exp ( T T dτ H I(τ) ). 0 (28) It s easy. Just add some dots! 5 Wck s theorem If we expand the numerator N n our result (28) for a Green functon we get an expresson of the form N (x 1,..., x N ) = 1 { m λ } dz j 0 T φ(x 1 ) φ(x N ) φ 4 (z 1 ) φ 4 (z m ) 0. m! j=1 4! (29) Here each φ s an nteracton pcture feld φ I but I leave out the subscrpt. The λφ 4 factors come from expandng the exponental. The ntegrand n the exponent contans also a constant term that adjusts the vacuum energy, but ths constant term n the exponent creates a constant factor, whch cancels the correspondng factor n the denomnator. Thus I defne N and the correspondng denomnator D to not have ths factor. We see that we need to be able to evaluate the (bare) vacuum expectaton value of a tme ordered product of free felds, 0 T φ(x 1 ) φ(x n ) 0. (30) Here there are N + 4m felds and some of the x s are what are wrtten as z s n Eq. (29). Wck s theorem tells how evaluate Eq. (30). The frst step s to wrte each φ as φ(x) = φ + (x) + φ (x) (31) 5

6 where φ + (x) = (2π) 3 φ (x) = (2π) 3 d k 2ω( k) e k x a(k) d k 2ω( k) e+k x a (k). (32) Let s start wth n = 2. Defne 0 T φ(x 1 )φ(x 2 ) 0 = D F (x 1 x 2 ) (33) It s easy to work out what D F (x 1 x 2 ) s. For x 0 1 > x 0 2 we have D F (x 1 x 2 ) = 0 φ(x 1 )φ(x 2 ) 0 = 0 φ + (x 1 )φ (x 2 ) 0 = (2π) 6 d k 2ω( e k x 1 k) d p 2ω( k) e+p x 2 0 a(k) a (p) 0 = (2π) 3 d k 2ω( e k (x 1 x 2 ) k) = (2π) 4 d 4 k e k (x 1 x 2 ) 2π 2ω( k) δ(k0 ω( k)) { } = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) k 0 ω( k) + ɛ k 0 ω( k) ɛ { } = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) k 0 ω( k) + ɛ k 0 + ω( k) ɛ = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) 2ω( k) (k 0 ) 2 ω( k) 2 + ɛ = (2π) 4 d 4 k e k (x 1 x 2 ) (34) k 2 m 2 + ɛ The most subtle part of ths dervaton s that wth x 0 1 > x 0 2 the terms wth poles n the upper half k 0 plane gve zero contrbuton and hence can be modfed. 6

7 Exercse: Show that ths works also for x 0 2 > x 0 1. Please justfy the steps. We can wrte the operator product T φ(x 1 )φ(x 2 ) tself n a productve way. Defne the normal ordered product of any number of felds, denoted by N(φ(x 1 ) φ(x N )), as what you get by decomposng each φ nto φ + + φ and expandng the result, then n each term movng the φ + operators to the rght of the φ operators. Wthn the set of φ + operators, the orderng doesn t matter snce these operators commute. Smlarly, wthn the set of φ operators, the orderng doesn t matter snce these operators commute. The man feature of the normal ordered product s that, snce 0 φ = 0 and φ + 0 = 0, we have 0 N(φ(x 1 ) φ(x N )) 0 = 0. For two feld operators we have, f x 0 1 > x 0 2, T φ(x 1 )φ(x 2 ) = φ + (x 1 )φ + (x 2 ) + φ + (x 1 )φ (x 2 ) + φ (x 1 )φ + (x 2 ) + φ (x 1 )φ (x 2 ) = φ + (x 1 )φ + (x 2 ) + φ (x 2 )φ + (x 1 ) + φ (x 1 )φ + (x 2 ) + φ (x 1 )φ (x 2 ) +[φ + (x 1 ), φ (x 2 )] = N(φ(x 1 )φ(x 2 )) + [φ + (x 1 ), φ (x 2 )]. (35) Now we recognze that [φ + (x 1 ), φ (x 2 )] s a number. If we take the vacuum expectaton value of the whole equaton, the vacuum expectaton value of N(φ(x 1 )φ(x 2 )) s zero. Thus the number [φ + (x 1 ), φ (x 2 )] s 0 T φ(x 1 )φ(x 2 ) 0. Thus T φ(x 1 )φ(x 2 ) = N(φ(x 1 )φ(x 2 )) + 0 T φ(x 1 )φ(x 2 ) 0, (36) or T φ(x 1 )φ(x 2 ) = N(φ(x 1 )φ(x 2 )) + D F (x 1 x 2 ). (37) Exercse: Show that ths same result holds for x 0 2 > x 0 1 also. In general, Wck s theorem says that T φ(x 1 ) φ(x n ) = N(φ(x 1 ) φ(x n )) + contractons. (38) Here contractons means that you replace any par of felds {φ(x ), φ(x j )} n the normal product by D F (x x j ) and multply by the normal product 7

8 of all the rest of the felds. Sum over all possble ways of dong ths. Then replace any two pars of felds by a factor D F (x x j ) for each par and multply by the normal product of all the rest of the felds. Sum over all possble ways of dong ths. Next replace all possble combnatons of three pars of felds n the same manner and sum over all the ways to do ths. Contnung n ths way, the last terms have all felds replaced by D F (x x j ) factors f n s even, or all but one f n s odd. The theorem s proved by nducton, but t s a easer to explan than to wrte out the proof. Here are two hnts. Frst, decompose the feld nto φ + + φ and work on the theorem term by term. Second, use [A, B 1 B 2 B 3 B n ] = [A, B 1 ] B 2 B 3 B n + B 1 [A, B 2 ]B 3 B n +. (39) The theorem can now be appled to 0 T φ(x 1 ) φ(x n ) 0 to gve zero of n s odd and, f n s even 0 T φ(x 1 ) φ(x n ) 0 = D F (x 1 x 2 ) D F (x 1 x 2 ) D F (x n 1 x n )+ (40) where you have to sum over all ways of parng the x s. 6 Feynman graphs for Green functons Recall that the numerator N n our result for a Green functon has the form N (x 1,..., x N ) = 1 { m λ } dz j 0 T φ(x 1 ) φ(x N ) φ 4 (z 1 ) φ 4 (z m ) 0. m! j=1 4! (41) For a two-pont functon at order zero n perturbaton theory, we get the bare propagator 0 T φ(x 1 )φ(x 2 ) 0 = D F (x 1 x 2 ) = (2π) 4 d 4 k e k (x 1 x 2 ) k 2 m 2 + ɛ (42) Now we can apply Wck s theorem to Eq. (41) to get rules for expressng the numerator as a sum of ntegrals. Frst, draw a dot, usually refered to as an external pont, labelled x for each φ(x ) n Eq. (41). Draw another dot labelled z, called a vertex, for each φ(z ) 4 n Eq. (41). Connect the dots wth lnes representng propagaton of a partcle from one of the ponts to another. Each φ(x ) external pont should have one lne attached to t; each φ(z ) 4 vertex should have four lnes attached to t. 8

9 To each lne from, say, y to y j assocate a propagator D F (y y j ). To each vertex z assocate a factor λ and an ntegraton d 4 z. To each external pont x assocate a factor 1. For the graph as a whole, dvde by a symmetry factor (explaned below). The dea of the symmetry factor s to compute each dstnct graph only once and multply by the number of tmes t occurs accordng to Wck s theorem, whle dvdng by the m! and 4! factors n Eq. (41). The symmetry factors are straghtforward n that a computer program can fgure them out, but t s a lttle awkard to state the rules such a program should use. Here s a set of rules. Frst, label each vertex, 1,...,m (and recall the 1/m!). Furthermore, regard the 4 ponts that make up each vertex as havng dstnct label (and recall the 1/4!). Next, label the propagators 1,...,p n all possble ways and dvde by 1/p!. Now, frst of all, there s a 1/4! assocated wth each vertex, but there are usually 4! dfferent ways to attach 4 dfferently labelled propagators to the 4 ponts n the vertex. Thus we can normally just count one of these ways and cancel the 1/4!. An excepton occurs f there s a propagator that connects a vertex to tself. If one propagator connects a certan vertex to tself, there are 4 3 dstnct ways to connect the other two propagators, so we have a factor 4 3/4! = 1/2. If two labelled propagators connect a certan vertex to tself, there are 6 ways to do t, so we have a factor 6/4! = 1/4. Thus there s a factor 1/2 for each propagator that connects a vertex to tself. Second, stll countng the vertces as dstnct but now countng the four ponts wthn a vertex as the same, usually the p! ways to relabel the propagators gve dfferent graphs and we get each of these once, so we can just count one of them and cancel the 1/p!. But f S P of the relabellngs gve the same graph, the number of tmes we get ths result s smaller by a factor 1/S P. Fnally, now countng the four ponts wthn a vertex as the same and all of the propagators as the same, usually the m! relabellngs of the vertces gve dstnct graphs, so we can just count one of them and cancel the 1/m!. But f S V of the relabellngs gve the same graph, the number of tmes we get ths result s smaller by a factor 1/S V. Thus our symmetry rules are 9

10 A factor 1/2 for each propagator that connects a vertex to tself. A factor 1/S P where S P s the number of permutatons of labellngs of the propagators that gve the same graph when we count the vertces as dstnct. A factor 1/S V where S V s the number of permutatons of labellngs of the vertces that gve the same graph (when we count the propagators as dentcal). In an actual applcaton, t s a good dea to check the result aganst Wck s theorem. 7 The denomnator There s also a denomnator, D = 1 { m λ } dz j 0 T φ 4 (z 1 ) φ 4 (z m ) 0. (43) m! 4! j=1 We treat ths n the same way. The zero order term s just 1. Then we get all of the Feynman graphs wth no external lnes, whch we can call vacuum graphs. Now the numerator contans subgraphs wth no external lnes. That s, some terms n the numerator are connected graphs (all of the vertces are connected to the external ponts by propagators), whle some have the form In fact (connected graph) (vacuum graph) (44) N = (connected graph) 1 + (vacuum graph) j. (45) j whle the denomnator has the form D = 1 + (vacuum graph) j. (46) j Thus the vacuum graph factor n the numerator just cancels the denomnator. As a result G(x 1,..., x N ) = (connected graph). (47) 10

11 8 Green functons n momentum space Let s defne Fourer transformed Green functons by dx 1 dx N exp( k j x j ) G(x 1,..., x N ) = (2π) 4 δ (4)( kj ) G(k1,..., k N ). (48) We represent G(k 1,..., k N ) by a graph wth external lnes labelled by the momenta k j wth the momenta comng nto the graph. To evaluate G(k 1,..., k N ), replace each propagator by D F (x 1 x 2 ) = (2π) 4 d 4 p e p (x 1 x 2 ) (49) p 2 m 2 + ɛ Then at each external pont, the p µ n the propagator gets changed nto the external momentum k µ j. At each vertex, there s an ntegral over a z that wll produce a ( ) (2π) 4 δ (4) k. (50) where the k µ are the momenta comng nto the vertex. We can use these delta functons to elmnate some of the ntegratons over the momenta p on nternal propagator lnes. We wll be left wth one delta functon for overall momentum conservaton, whch we have already factored out of G(k 1,..., k N ). Then we are left wth an ntegraton over one propagator momentum for each loop n the dagram. Ths gves the rules: Label the lnes by ther momenta, usng momentum conservaton at each vertex. For each loop, there wll be one momentum that s not constraned by momentum conservaton. Supply an ntegraton (2π) 4 d 4 p. (51) To each lne assocate a propagator To each vertex assocate a factor λ. p 2 m 2 + ɛ. (52) For the graph as a whole, dvde by the symmetry factor. 11

12 9 Tme flow and the non-relatvstc lmt It s of some nterest to see what happens f we separate dfferent tme orderngs n our perturbaton theory. Recall what we had for the propagator. If x 0 1 > x 0 2, then D F (x 1 x 2 ) = 0 T φ(x 1 )φ(x 2 ) 0 = 0 φ(x 1 )φ(x 2 ) 0 = 0 φ + (x 1 )φ (x 2 ) 0 = (2π) 3 d k 2ω( e k (x 1 x 2 ) k) = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) k 0 ω( k) + ɛ. (53) If x 0 1 < x 0 2, then D F (x 1 x 2 ) = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) k 0 ω( k) + ɛ. (54) (To get ths, just change k to k.) Now the expresson on the rght hand sde of Eq. (53) s zero f x 0 1 < x 0 2, whle the expresson on the rght hand sde of Eq. (54) s zero f x 0 1 > x 0 2. Thus f we defne we have D ± (x 1 x 2 ) = (2π) 4 d 4 k e k (x 1 x 2 ) 1 2ω( k) ±k 0 ω( k) + ɛ (55) D F (x 1 x 2 ) = D + (x 1 x 2 ) + D (x 1 x 2 ) (56) where D + (x 1 x 2 ) s nonzero for x 0 1 > x 0 2 and D (x 1 x 2 ) s nonzero for x 0 1 < x 0 2. We can break up each propagator n a Feynman graph n ths way. A convenent notaton for ths would be to put an arrow on the lne showng whch way tme flows. Then n momentum space the rule for a propagator s 1 2ω( k) ±k 0 ω( k) + ɛ (57) wth the sgn n ±k 0 chosen accordng to the drecton of the tme flow arrow relatve to the momentum flow arrow. 12

13 Ths analyss makes t easy to see what the non-relatvstc lmt s. Consder a Green functon n whch some of the lnes are ncomng and some are outgong. The ncomng lnes are assocated wth early tmes and for these we choose the momentum sgns so that the momentum comes nto the graph. The outgong lnes are assocated wth late tmes and for these we choose the momentum sgns so that the momentum comes out of the graph. Let us suppose that for each ncomng and outgong lne we have p 0 m. Then we can consder the non-relatvstc lmt n whch m wth p 0 m fnte. Our graph wll survve n ths lmt f, at each vertex, the number of ncomng partcles equals the number of outgong partcles. Ths allows us to have also k 0 m for the nternal lne energes. Then we can approxmate 1 2ω( k) k 0 ω( k) + ɛ 1 2m E NR k 2 /(2m) + ɛ (58) where E NR k 0 m. (59) Ths gves a smple non-relatvstc theory. We redefne all energes by subtractng the rest energy. The knetc energy s k 2 /(2m). For each propagator there s a factor 1/(2m) that comes from our normalzaton of states, whch s rather pecular from a non-relatvstc pont of vew. Assocate a factor q/ 2m wth each external vertex. We can get rd of these by changng normalzatons of the feld. Also, redefne the couplng by λ = (2m) 2 λ. (60) Then all of the 1/(2m) factors go away. Is ths a feld theory? Yes. For a free feld use ψ(t, x) = (2π) 3 d k e Et+ k x b( k), (61) where [b( k), b ( p)] = (2π) 3 δ( k p). (62) That s, we use just the destructon part of the feld and we remove the unnterestng exp( mt) part of the tme dependence. We also change normalzaton factors. The commutaton relatons n poston space are [ψ(t, x), ψ(t, y)] = 0 [ψ(t, x), ψ (t, y)] = δ( x y). (63) 13

14 Ths has the nterpretaton that ψ(t, x) destroys a partcle at poston x at tme t, whle ψ (t, x) creates one. Exercse: Show that Eq. (62) mples [ψ(t, x), ψ (t, y)] = δ( x y). The free hamltonan s H 0 = Perhaps a better way to wrte ths s H 0 = We add nteractons by defnng H nt = d x ψ ( x) 1 2m 2 ψ( x) (64) d x 1 2m ψ (t, x) ψ(t, x) (65) d x λ 4 (ψ (t, x)) 2 (ψ(t, x)) 2. (66) One can check that ths gves the Feynman rules that we have derved by takng the m lmt of the relatvstc theory. Exercse: What equaton of moton for ψ( x, t) follows from the hamltonan H 0 + H nt n Eqs. (65) and (66)? We can extend ths dea to partcles that nteract by means of a potental. We take H nt = d x d y 1 2 ψ (t, x)ψ (t, y) ψ(t, x)ψ(t, y) V( x y). (67) Then the nteracton (66) s a specal case of ths wth a delta functon potental. Exercse: Evaluate H nt x, y, where H nt s evaluated at t = 0 and the state x, y s what you get by applyng ψ (0, x)ψ (0, y) to the vacuum state. Suppose that the Fourer transform of V s Ṽ( q) = d x e q x V( x). (68) Then the Feynman rules for Green functons ncludes a new rule. We can represent the acton of the potental by, say, a dashed lne that couples to 14

15 an ncomng and an outgong partcle lne at each end. The lne carres momentum q determned by momentum conservaton. We wrte For each nteracton, a factor Ṽ( q). What f we take one step backward and look for an acton for our theory. Ths works. The acton s { S[ψ, ψ] = dt d x ψ (t, x)[ 2 tψ(t, x)] [ 2 tψ (t, x)]ψ(t, x) 1 ψ 2m (t, x) ψ(t, } x) dt d x d y 1 2 ψ (t, x)ψ (t, y) ψ(t, x)ψ(t, y) V( x y). (69) Exercse: Wth ths acton, evaluate the equaton of moton δs[ψ, ψ] δψ (t, x) = 0. (70) Havng the acton s useful for dervng the conservaton laws and the correspondng conserved quanttes from Nöther s theorem. It s also useful because the acton comes nto the path ntegral formulaton of the quantum feld theory (to be dscussed later n the course). But the relaton of the commutaton relatons to Posson brackets s dfferent than the prevous cases we have dscussed. That s because the formulaton we used for the relatvstc theory s really desgned for theores wth equatons of moton that are second order n / t. The Schrödnger equaton s frst order n tme dervatves. Exercse: The acton (69) s nvarant under translatons n space and tme. What are the correspondng conserved quanttes? You wll have to derve a new verson of Nöther s theorem to cover ths case. 15

16 10 Charged scalar feld So far we have talked about a feld that destroys and creates partcles that are ther own antpartcles (lke photons). Wth just a few small modfcatons, we can have a feld that destroys and creates partcles that are not ther own antpartcles lke, say, 4 He nucle. Snce we are makng modfcatons to a theory that we already understand, we can be bref. Consder the lagrangan L = ( µ φ )( µ φ) m 2 φ φ λ 4 (φ φ) 2. (71) Here φ s the adjont of φ, but n dong the algebra t s convenent to regard φ and φ as ndependent felds. If we want we can wrte φ = φ 1 + φ 2, where φ 1 and φ 2 are real (e. self-adjont) felds. Then we can vary φ 1 and φ 2 ndependently. The equatons of moton are 0 = δs[φ, φ] δφ (x) = ( µ µ m 2 )φ(x) λ 2 (φ (x)φ(x))φ(x) (72) and 0 = δs[φ, φ] δφ(x) = ( µ µ m 2 )φ (x) λ 2 (φ (x)φ(x))φ (x) (73) whch s just the adjont of the frst equaton. To put ths n the hamltonan formulaton, we defne and Then the hamltonan densty s π( x) = δl[ φ, φ, φ, φ] δ φ( x) π ( x) = δl[ φ, φ, φ, φ] δ φ ( x) = φ ( x) (74) = φ( x). (75) H = π φ + π φ L = π π + ( φ ) ( φ) + m 2 φ φ + λ 4 (φ φ) 2. (76) The commutaton relatons are [φ( x), π( y)] = δ( x y) [φ ( x), π ( y)] = δ( x y). (77) 16

17 One can check that wth these commutaton relatons, the hamltonan generates the rght equatons of moton. Now for a free feld (λ = 0) we can solve the theory exactly and wrte φ(t, x) = (2π) 3 d k 2ω( k) { e k µx µ a( k) + e +kµxµ b ( k) } (78) Here a( k) destroys a partcle and b ( k) creates somethng dfferent an antpartcle. The commutaton relatons are [a( k), a ( p)] = (2π) 3 2ω( k) δ( k p) [b( k), b ( p)] = (2π) 3 2ω( k) δ( k p). (79) These are equvalent to the φ, π commutaton relatons gven above. Now to do perturbaton theory, we mtate the prevous treatment, usng 0 T φ(x 1 )φ (x 2 ) 0 = D F (x 1 x 2 ) = (2π) 4 d 4 k e k (x 1 x 2 ) k 2 m 2 + ɛ (80) We represent ths n graphs as a lne wth an arrow gong from x 2 to x 1. The arrow remnds us that we create a partcle at x 2 and annhlate t at x 1. Or else we create an antpartcle at x 1 and annhlate t at x 2. Thus the arrow represents the drecton of partcle flow. Note the dfference n conventon wth our (rather non-conventonal) use of an arrow to represent the drecton of tme flow n prevous sectons. As before, we can easly derve the Feynman rules for Green functons for ths theory. We draw all connected graphs for the desred Green functon. Now the propagators have arrows representng the drecton of partcle flow. At each vertex, there are two ncomng arrows and two outgong arrows. Label the lnes by ther momenta, usng momentum conservaton at each vertex. For each loop, there wll be one momentum that s not constraned by momentum conservaton. Supply an ntegraton (2π) 4 d 4 p. (81) 17

18 To each lne assocate a propagator To each vertex assocate a factor λ. p 2 m 2 + ɛ. (82) For the graph as a whole, dvde by the symmetry factor. Exercse: What are the Feynman rules for Green functons for a theory wth one real feld φ and one complex feld ψ wth the lagrangan densty L = ( µ ψ )( µ ψ) m 2 ψ ψ ( µφ)( µ φ) gφ ψ ψ (83) 11 Scatterng Let s defne a Green functon for our standard φ 4 theory wth some of the momenta, denoted k, gong nto the graph and some, denoted q, comng out dy 1 dy M dx 1 dx N exp( q j y j k j x j ) Ω T φ(y 1 ) φ(y M )φ(x 1 ), φ(x N ) Ω = (2π) 4 δ (4)( kj q j ) G(q1..., q M ; k 1,..., k N ).(84) We would lke to use G(q 1..., q M ; k 1,..., k N ) to descrbe the scatterng matrx S F I = out q 1..., q M k 1,..., k N n (85) (Please refer to the notes on the S-matrx from last quarter for the defntons.) The relaton s often called the LSZ reducton formula (Lehmann, Symanzk & Zmmermann). To start wth, we note that wth our normalzatons n a free feld theory a feld can annhlate a partcle to gve the vacuum state wth ampltude 1: Ω φ(0) k = 1 (free theory). (86) We suppose that we are near enough to ths stuaton that a feld can annhlate a partcle to gve the vacuum state, just wth some dfferent ampltude, whch we call Z: Ω φ(0) k = Z. (87) 18

19 (We choose the phase of the states so that Z s real and postve.) Wth ths defnton, the LSZ formula s (2π) 4 δ (4)( kj q j ) G(q1..., q M ; k 1,..., k N ) M j=1 Z q 2 M 2 + ɛ out q 1..., q M k 1,..., k N n N j=1 Z k 2 M 2 + ɛ (88) Here M s the physcal mass of the partcles (whch may not be the same as the mass m n the lagrangan). Here the means asymptotc equalty as one approaches the poles at q 2 = M 2, k 2 = M 2. That s, the equaton asserts that both sdes of the equaton have poles at q 2 = M 2, k 2 = M 2 and the resdues of these poles match. Thus out q 1..., q M k 1,..., k N n = (2π) 4 δ (4)( kj ) q j lm { q 2 M2 k 2 M2 M q 2 M 2 + ɛ N j=1 k 2 M 2 + ɛ Z j=1 Z G(q } 1..., q M ; k 1,..., k N ) (89) To see why ths s at least plausble, let s look frst at the two pont functon. G(k) = dx exp( k x) Ω T φ(0)φ(x) Ω. (90) Exercse: Show that ths s the same as (2π) 4 δ 4 (q k) G(k) = dx dy exp( k x + q y) Ω T φ(y)φ(x) Ω. (91) You can use exp(p x)φ(y) exp( P x) = φ(x + y) (92) where P µ s the energy-momentum operator. Ths s equavalent to [P µ, φ(y)] = µ φ(y) (93) whch are the standard commutaton relatons between P µ and the felds. 19

20 We wll look for poles n G(k) at k 2 = M 2 wth k 0 > 0. Such a pole can come only from the ntegraton regon x 0, as wll become apparent from the constructon below. Acceptng ths for the moment, we wrte G(k) T d x dx 0 exp( k x) Ω φ(0)φ(x) Ω. (94) Also f we ntsert ntermedate states between the φ(0) and the φ(x),the pole can come only from sngle partcle states, as wll become apparent from the constructon below. Acceptng ths for the moment, we wrte G(k) T dx 0 d x exp( k x) (2π) 3 Translaton nvarance, Eq. (92), gves Then, usng the defnton of Z, we have G(k) Z d p 2ω( p) Ω φ(0) p p φ(x) Ω. (95) p φ(x) Ω = exp(p x) p φ(0) Ω (96) T dx 0 d x (2π) 3 Performng the x ntegral and usng p 0 = ω( p) we have Ths s G(k) Z G(k) T d p exp( (k p) x). (97) 2ω( p) dx 0 1 2ω( k) exp( (k0 ω( k))x 0 ). (98) Z 2ω( k) k 0 ω( k) + ɛ exp(t (k0 ω( k))). (99) At the pole, we can relplace the exponental by 1 and we can replace 2ω( p) by k 0 + ω( k). Then G(k) Z (k 0 ) 2 ω( k) 2 + ɛ. (100) That s Z G(k) k 2 M 2 + ɛ. (101) Now that we see how ths works, we can go back and notce that the part of the ntegraton over x 0 over any fnte tme nterval, say from T 1 to T 2 does 20

21 not gve a pole. Repeatng essentally the same argument, the part from some fnte tme to + gves a pole of the same form along the negatve energy hyperbola k 2 = M 2 wth k 0 < 0. Repeatng the argument, we get more sngulartes n k 2 startng at the lowest mass squared for a two partcle state, k 2 = (2M) 2. Snce two partcle states can have masses M 2 that are anythng down to 2M, the sngularty at k 2 = (2M) 2 s not solated. (It s a branch pont.) Thus Eq. (101) gves the structure of G(k) as one approaches the one partcle pole only. Eq. (101) s what one uses to calculate Z n perturbaton theory. You calculate G(k) at some order of perturbaton theory and extract the resdue of the one partcle pole. Ths pole wll generally have moved from k 2 = m 2 to somewhere else, k 2 = M 2, so one calculates M 2 too. Now we would lke to make the LSZ formula at least plausble, followng the argument n Peskn and Schroeder Chapter 7.2. We start wth (2π) 4 δ (4)( kj ) q j G(q1..., q M ; k 1,..., k N ) = dy 1 dy M dx 1 dx N exp( q j y j k j x j ) Ω T φ(y 1 ) φ(y M )φ(x 1 ), φ(x N ) Ω (102) We are gong to look for poles at the k 2 j = M 2 and q 2 j = M 2 wth postve energes, k 0 j > 0, q 0 j > 0. As we learned wth the two pont functon, these poles come from the part of the x 0 j ntegraton wth x 0 j and the part of the y 0 j ntegraton wth y 0 j +. Thus we can wrte (2π) 4 δ (4)( kj ) q j G(q1..., q M ; k 1,..., k N ) dy 1 dy M θ(yj 0 > T ) exp( q j y j ) j dx 1 dx N θ(x 0 j < T ) exp( k j x j ) j Ω T {φ(y 1 ) φ(y M )}T {φ(x 1 ), φ(x N )} Ω (103) Now we nsert a sum over a complete set of out state just to the rght of the φ(y) factors and a sum over a complete set of n state just to the left of the φ(x) factors: (2π) 4 δ (4)( kj q j ) G(q1..., q M ; k 1,..., k N ) 21

22 dy 1 dy M θ(yj 0 > T ) exp( q j y j ) j 1 K (2π) 3 d p K K! =1 2ω( p ) Ω T {φ(y 1 ) φ(y M )} p 1,..., p K out 1 L (2π) 3 d l j L L! j=1 2ω( l j ) dx 1 dx N θ(x 0 j < T ) exp( k j x j ) j n l 1,... l L T {φ(x 1 ), φ(x N )} Ω out p 1,... p K l 1,... l L n (104) Now lets look at the matrx element of the φ(y) felds. As we shall see, we can get the requste poles at q 2 = M 2 out of ths. The way t can come about s f each feld operator destroys one partcle. Then we should have K = N and L = M. The essental approxmaton for the fnal state partcles s Ω T {φ(y 1 ) φ(y M )} p 1,..., p K out Ω φ(y 1 ) p 1 out Ω φ(y M ) p M out + (105) where the omtted terms are of the same type, but wth the ndces of the ys matched up n a dfferent order wth the ndces of the ps. There are M! such terms, and each wll gve the same contrbuton to the fnal answer, so we wll just keep the one term dsplayed above and cancel the factor 1/K! n the formula. The dea of ths replacement s that the y 0 are very large and n the out state after a long tme the partcles are all a long way from each other, so each feld operator annhlates one partcle ndependently of what s happenng wth the other partcles. Of course, to really make ths sensble, we should put the partcles nto wave packet states, so that ther wave functons have very lttle overlap n space after a long tme. We could also ntegrate the feld operators aganst wave packet wave functons nstead of plane waves. However, I wll skp further exploratons along these lnes. For the n states we make the same knd of replacement n p 1,..., p N T {φ(x 1 ) φ(x N )} Ω n p 1 φ(x 1 ) Ω n p N φ(x N ) Ω + (106) 22

23 Then we have (2π) 4 δ (4)( kj ) q j G(q1..., q M ; k 1,..., k N ) dy j θ(yj 0 > T ) exp(q j y j ) (2π) 3 j d p j 2ω( p j ) Ω φ(y j) p j out dx θ(x 0 < T ) exp( k x ) (2π) 3 d l 2ω( l ) n p φ(x ) Ω out p 1,... p M l 1,... l N n (107) Performng the ntegrals gves (2π) 4 δ (4)( kj q j ) G(q1..., q M ; k 1,..., k N ) j Z q 2 j M 2 + ɛ Z k 2 M 2 + ɛ out q 1,... q M k 1,... k N n, (108) whch s the LSZ formula. We should note that the Green functon and the S-matx both contan terms n whch one or more partcles go from the ntal state to the fnal state wthout nteractng at all. Ths dervaton does not apply to those no scatterng terms, so let s understand that we have subtracted these terms from both sdes of the equaton. Note that the possblty of havng non-trval factors Z arses because a partcle can nteract wth tself even after t s far from any other partcles. In standard non-relatvstc physcs ths doesn t happen, so Z = The S-matrx and amputated Green functons We can wrte the Green functon dscussed n the prevous secton as G(q 1..., q M ; k 1,..., k N ) = j G 2 (q j ) G 2 (k ) Γ(q 1..., q M ; k 1,..., k N ) (109) Here G 2 s the two partcle Green functon, called the full propagator, and Γ s the amputated Green functon and s defned by ths eequaton. 23

24 Graphcally, we just omt the extermal legs of the dagram, ncludng any nteractons on the external legs. One says that Γ s one partcle rreducble. We have Z G 2 (k ) (110) k 2 M 2 + ɛ at the sngle partcle pole. Snce we calculate the S-matrx by factorng out an Z (111) k 2 M 2 + ɛ for each external partcle, we can restate the prescrpton as tellng us to factor out the full propagators, leavng the amputated Green functon, and multply by Z: out q 1..., q M k 1,..., k N n = (2π) 4 δ (4)( kj q j ) ( Z ) N+M lm q 2 M2 k 2 M2 Γ(q1..., q M ; k 1,..., k N ). (112) 24