Elementary Processes. Chapter 5

Transcription

1 Chapter 5 Elementary Processes We want to extend the prevous dscusson to the case where felds nteract among themselves, rather than wth an external source. The am s to gve an expresson for the S-matrx n terms of n felds, as before. We wll see that ut s convenent to express the matrx elements of S n terms of Green s functons, that s, vacuum expectaton values of tme-ordered products of elementary felds, as n h0 T (x ) (x n ) 0. As before the S matrx connects n states to out states, out = S n, and correspondngly n felds to out felds. But we can no longer say that (x)! n(x) as t! (nor (x)! out (x) as t! +) wth n,out (x) freefelds(a free feld s one wthout nteractons, e.g., t satsfes the KG equaton). Before we explan that n more detal let us better understand the role of n states (and n felds). In a collson process we start wth partcles that are wldly separated, so nteractons between them can be ntally neglected. As partcles approach each other the nteractons can no longer be neglected, they turn on. So one could thnk of the stuaton by replacng H 0 (the nteracton part of the Hamltonan) by f(t)h 0, where f(t) s a smooth functon that turns on slowly (adabatcally), then stays on for some long perod over whch the collson takes place, say, f(t) = for T < t < T, and then turns o slowly agan, f(t) =0 t >T 0 and f(t) smoothly decreasng (ncreasng) n T<t<T 0 ( T 0 <t< T ): f(t) T 0 T T T 0 t We want both T and T 0 to be arbtrarly large, and we want T 0 T n the 76

2 77 process to avod sudden changes that can ntroduce extraneous e ects, e.g., par producton. We wll use ths model, but t s not qute general enough. The reason s that f the nteracton s turned o we may not be able to descrbe n and out states of nterest, namely, bound states that arse because of the nteracton. For example, we may want to study collsons of an electron wth an H atom due to electromagnetc nteractons. But t s the electromagnetc nteractons that bnds a p and an e nto an H atom. More pognant s the case of collsons of protons by the strong nteractons when t s these nteractons that keeps quarks bound n protons. The dea of collson theory s that oen can set up a muck theory of free partcles that happen to have the same mass (and other quantum numbers, e.g., spn) as the bound states. These are the n and out states. For very early (or late) tmes these states descrbe the evoluton of the partcles that later (earler) partcpate n the collson. And the S-matrx uses nformaton n the nteractng theory to connect the pre- and post-collson states. For theores wthout bound states we can use the smpler approxmaton of turnng on and o the nteracton va f(t)h 0. Remarkably the expresson for the S-matrx obtaned va ths smplfed treatment s the same as n a more complete and rgorous analyss that does not employ t. If we adabatcally turn on and o the nteractons then we can use our prevous approach: (x) = n (x)+ d 4 yg ret (x y)j(y) (5.) But now (@ 2 + m 2 ) = (e.g., f L 0 = g then =3g ). So for J(x) usef(t) Note that n the absence of f(t) ths would not work, the source would not be localzed n tme. Now, we had two ways of obtanng the S-matrx from ths. One was S n(x)s = n (x)+ d 4 y [ n(x), n(y)]j(y). But now J(y) depends on (x) so t does not commute wth n (x). Ths makes t harder to solve for S usng ths method. The second method used = U (t) n (x)u(t) and constructed U(t) nterms of H 0. Ths wll work. The result was, and stll s, apple S = T exp d 4 x L 0 n whch we can use, but now wth, say, L 0 n = g 3 n + 4 n. However, the above dscusson has to be modfed, as we wll see shortly, because n general one cannot take (x)! n (x) as t!.

3 78 CHAPTER 5. ELEMENTARY PROCESSES 5. Källen-Lehmann Spectral Representaton Here we wll see that we cannot take (x)! n (x) as t!. We study [ (x), (y)] and compare wth [ n(x), n(y)]. In partcular, h0 [ (x), (y)] 0 = X (h0 (x) nhn (y) 0 x $ y) n Now h0 (x) n = h0 e ˆP x (0)e ˆP x n = e pn x h0 (0) n where ˆP µ n = p µ n n, ˆP µ 0 = 0, and h0 [ (x), (y)] 0 = X e pn (x y) h0 (0) nhn (0) 0 x $ y n = X d 4 k (4) (p n k) e k (x y) e k (x y) h0 (0) n 2 n d 4 k = (2 ) 3 e k (x y) e k (x y) (k) where n gong from the frst to the second lne we ntroduce a factor of = R d 4 k (4) (p n k) and n the last lne we defned (k) X n (2 ) 3 (4) (p n k) h0 (0) n 2 = (k 2 ) (k 0 ). The last equalty follows from () Lorentz nvarance and () p 0 n > 0. Now compare ths wth the case of free felds. We have computed ths, but t s easy to derve from above: n s only the one partcle states ~p, PR n s R (dp), and h0 n(0) ~p = (dk)h0 ~k ~p 0 =. Then (k) = (dp)(2 ) 3 (4) (p k) = d 4 p (p 0 ) (p 2 m 2 ) (4) (p k) = (k 0 ) (k 2 m 2 ), and h0 [ n(x), n(y)] 0 = d 4 k (2 ) 3 (k0 ) (k 2 m 2 ) = e k (x y) k (x e y) d 4 k (2 ) 3 "(k0 ) (k 2 m 2 )e k (x y) (x y; m)

4 5.. KÄLLEN-LEHMANN SPECTRAL REPRESENTATION 79 Hence h0 [ (x), (y)] 0 = or = d 4 k (2 ) 3 h0 [ (x), d 4 k (2 ) 3 (k 2 )"(k 0 )e 0 = k (x y) d m 2 (k 2 m 2 ) (k 2 )"(k 0 )e 0 (y)] 0 = d m 2 ( m 2 ) 0 k (x y) d 4 k (2 ) 3 "(k0 ) (k 2 m 2 )e k (x y) d m 2 (m 2 ) (x y; m) (5.2) Ths s the Källen-Lehmann representaton. Now we separate the contrbutons to of -partcle states from 2-partcles states. We wll assume that p 0 M > m for 2-partcle states. For two free partcles p 0 2m. If nteractng we expect p 0 2m " where " s some nteracton energy; we are assumng "<m. If we had ">mthen the 2-partcle energy would be smaller than m and the -partcle state s not a state because t can decay nto a lower energy state. Then, f n fact we could demand (x)! n (x) as t! we should have h0 (0) ~p =? h0 n(0) ~p =, and therefore Ths s llustrated n the followng fgure: ( m 2 )? = (m 2 m 2 )+ ( m 2 ) ( m M). (5.3) m 2 M 2 m 2 Insertng (5.3) n (5.2) gves h0 [ (x), (y)] 0? = (x y; m)+ M 2 d m 2 (m 2 ) (x y; m). 0 of ths, and then the lmt y 0! x 0 we obtan on the left hand sde the equal tme commutator [ (x), (y)] = (3) (~x ~y ). Then note that on the rght hand we can do ths agan snce (x y; m) = h0 [ n(x), n(y)] 0. Ths gves =+? d m 2 ( m 2 ) M 2

5 80 CHAPTER 5. ELEMENTARY PROCESSES If ths equaton holds then ( m 2 ) = 0 for m>m. That s ( m 2 )= ( m 2 m 2 ), whch means h0 (0) n = 0 for any state n whch has 2 partcles. Ths then gves (x) = n (x) for all tmes whch makes (x) a free feld. We conclude that we cannot demand (x)! n (x) as t!. Assume nstead Then, repeatng the steps above, (x)! 2 n(x) as t! = + M 2 d m 2 ( m 2 ) so that >0requres0apple< (that 0 s from t beng ( 2 ) 2 ). Smlarly, we assume (x)! 2 out(x) as t! +. We conclude the secton wth some useful observatons. Unqueness of the vacuum state gves 0 out = 0 n = 0. (In prncple one can have a relatve phase, 0 out = e 0 n but we conventonally set = 0). Snce we are assumng the -partcle states are stable, they are egenstates of the Hamltonan, so they evolve smply, by a phase, e Et. Hence ~ k out = ~ k n (up to a constant phase that we conventonally set to zero). Now h0 (x) ~ k = h0 (0) ~ k e k x so that the prescrpton to evaluate at t! n order to compare wth the correspondng expectaton value of n(x) s superfluous, and smlarly for t!and expectaton values of out(x). So we have h0 (x) ~ k = 2 h0 n(x) ~ k = 2 h0 out(x) ~ k. We collect some basc results for the S-matrx: n(x) =S out (x)s, n = S out, outh n = out h S out = n h S n. For, the vacuum or -partcle states h0 S 0 = h0 0 =, h ~ k S ~ k 0 = h ~ k ~ k 0 =(2 ) 3 2E ~k (3) ( ~ k 0 ~ k ). Fnally, f U = U(a µ, ) s a Poncare transformaton, covarance means USU = S. It s often stated n textbooks that (x)! 2 n(x) cannot hold n the strong sense. That s, that t can only hold for separate matrx elements. Else we d

6 5.2. LS REDUCTION FORMULA: STATED 8 have, the argument goes, for example, [ (x), (y)]! [ n(x), n(y)] for nonequal, early tmes, and snce for n felds ths s a c-number, one would be able to argue that (x) s a free feld. I thnk ths s overkll. Obvously (x0 and n are d erent, one s an nteractng feld and one s not. One can produce multple partcle states out of the vacuum that s the statement that >0 the other cannot. The statement that (x)! 2 n(x) at t! s useful because t gves us the correct way of relatng wldly separated ntal state (sngle)-partcles created by to those created by n. To make sense of ths we need partcles that are truly separated, whch means we have to consder wave-packets rather than plane waves. We wll comment on ths when we dscuss the S-matrx for mult-partcle states. 5.2 LS reducton formula: stated LS stands for Lehmann, Symanzk and mmermann. The LS formula gves the probablty ampltude for scatterng any number of partcles nto any number of partcles: outh~p,...,~p l ~ k,..., ~ k n n =( ly (@ y 2 + m 2 ) = Comments: 2 ) n+l ly d 4 y = n Y j= d 4 x j e P l = p y P n j= k j x j ny (@ x 2 j + m 2 )h0 T ( (y ) (y l ) (x ) (x n )) 0. (5.4) j= () Computng S matrx elements reduced to computng Green s functons, G (n) (x,...,x n )=h0 T ( (x ) (x n )) 0. () One can do a more general treatment n term of -partcle wave-packets. Snce the LS formula s multlnear n the plane waves for the n and out states, the result amounts to replacng e k x )! f (x ) and e p y )! f (y ). () Integratng by parts (@ x 2 + m 2 )e ±p x) = (p 2 m 2 )e ±p x) = 0. The result has to be nterpreted wth care. Let Y n d 4 x e P n = k x G (n) (x,...,x n )=(2 ) 4 (4) ( X k ) G e(n) (k,...,k n ). = (5.5) That we always have a (4) ( P k ) follows form translaton nvarance. We can change varables to the d erences x + x together wth the center

7 82 CHAPTER 5. ELEMENTARY PROCESSES of mass X = P x. Then snce G (n) does not depend on X we wll have R d 4 Xe R P k tmes the rest. Now, G e(n) (k,...,k n ) s defned for arbtrary four vectors, k,...,k n, not necessarly satsfyng the on-shell condton k 2 = m 2 ; we say that k s o -shell f k 2 6= m2, or alternatvely, that the energes, k 0, are arbtrary, not gven by ±E ~ k. Incdentally, the on/o -shell language s smply short for the momentum beng on/o the mass-shell. Itmaybe that G e(n) (k,...,k n ) has smple poles as k 0!±E ~k. In fact, by Lorentz nvarance the poles must be pared, appearng as poles n k 2 m 2. These poles cancel the zeroes from Q (@ 2 + m 2 ) and the S-matrx element s just the resdue: outh~p,...,~p l ~ k,..., ~ k n n =( l+n Y k= d 4 q k e P l = q y + P l+n n Y j= q j x j =( 2 ) n+l (2 ) 4 (4) ( X j lm p 2!m2 kj 2!m2 ly (p 2 m 2 ) = k j 2 ) n+l ly d 4 y = n Y j= =( q 2 k +m2 )(2 ) 4 (4) (X X p ) d 4 x j e P l = p y P n j= k j x j k ) e G (n) (q,...,q n+l ) ny (kj 2 m 2 ) G e(n) (k,...,k n, p,..., p l ). j= and note that each factor of p 2 m 2 comes wth a /( 2 ). (v) It s therefore useful to summarze ths n terms of a scatterng ampltude, A = A(k,...,k n ; p,...,p l ) = lm p 2!m2 ly (p 2 m 2 ) ny (kj 2 m 2 ) eg (n) (k,...,k n, p,..., p l ) kj 2 =!m2 2 j= 2 so that outh~p,...,~p l ~ k,..., ~ k n n =(2 ) 4 (4) ( X j k j X p ) A. 5.3 S-matrx: perturbaton theory Contnung to present results wthout justfcaton (whch wll be gven later) we now gve the Green s functons (or correlators or n-pont functons n terms of n

8 5.3. S-MATRIX: PERTURBATION THEORY 83 felds: G (n) (x,...,x n )=h0 T ( (x ) (x n )) 0 = h0 T ( n (x ) n(x n )e R d 4 xh 0 n ) 0 (5.6) Expandng the exponental, wrtng Hn 0 for H0 n,wth =, just a countng devce, and retanng up to some power n, say N, we are approxmatng G (n) as a perturbatve expanson. If Hn 0 s wrtten explctly n terms of some parameters, and these can be consdered as small, we are then approxmatng G (n) as an expanson n powers of these small parameters. For example, wth L = L 0 + L 0, L 0 = 2 (@ µ ) 2 2 m2 2 and L 0 = H 0 = ( /4!) 4, then we are expandng G (n) n powers of,thecouplng constant. Let s see ths explctly n ths example. Compute G (4) (x,...,x 4 ): 0th order We take the n the expanson fo the exponental: G (4) 0 (x,...,x 4 )=h0 T ( n (x ) n(x 4 )) 0 To make the notaton more compact we drop the label n for now and use (x, etc. Usng Wck s theorem, for G (4) 0 (x,...,x 4 )=h0 T ( 4) 0 = h0 : 4:+: 2 3 4:+: 2 3 4:+: 2 3 4: +: 2 3 4:+: 2 3 4:+: 2 3 4: +: 2 3 4: +: 2 3 4: +: 2 3 4: 0 Only the last lne s non-vanshng, the prevous two have normal ordered operators actng on the vacuum. The last lne gves G (4) 0 (x,...,x 4 )= Now, we compute e G (4) and then A: 4Y n= d 4 x n e P k n x n G (4) 0 (x,...,x 4 )= d 4 x d 4 x 2 e k x k 2 x 2 2 d 4 x 3 d 4 x 4 e k 3 x 3 k 4 x permutatons.

9 84 CHAPTER 5. ELEMENTARY PROCESSES We need d 4 x d 4 x 2 e k x d 4 q = (2 ) 4 = q 2 d 4 q (2 ) 4 d 4 q (2 ) 4 e q (x x 2 ) q 2 m 2 + m 2 d 4 x e x (k +q) d 4 x 2 e x 2 (k 2 q) + k 2 x 2 q 2 m 2 + (2 )4 (4) (k + q)(2 ) 4 (4) (k 2 q) =(2 ) 4 (4) (k + k 2 ) k 2 m 2 There s no n the last step snce we have performed the ntegral over q 0. Usng ths above we have eg (4) (k,...,k 4 )= k 2 m 2 k3 2 m 2 (2 )4 (4) (k + k 2 )(2 ) 4 (4) (k 3 + k 4 )+perms apple =(2 ) 4 (4) (k + k 2 + k 3 + k 4 ) k 2 m 2 k3 2 m 2 (2 )4 (4) (k + k 2 )+perms Moreover, the scatterng ampltude s A(k,k 2 ; p,p 2 )= p 2 m 2 p 2 2 m 2 k 2 m 2 k 2 2 m 2 eg (4) (k,...,k 4 )=0, lm p 2!m2 k 2 j!m2 whch makes sense snce at ths order n the expanson there s no nteracton so there s no scatterng. st order We now expand the exponental to lnear order so that G (4) (x,...,x 4 )=h0 T( 4( ) d 4 x 4 4! (x)) 0 To calculate ths we use Wck s theorem. We know from experence ganed above that we need the terms wth all felds contracted. Let s dstngush terms lke ( ) d 4 x (x) (x) (x) (x) 4! or 2 ( ) d 4 x 4! 3 4 (x) (x) where at least two of the,..., 4 are contracted among themselves, from terms lke 4! d 4 x (x) (x) (x) (x) (5.7)

10 5.3. S-MATRIX: PERTURBATION THEORY 85 We call the frst knd dsconnected, the second connected. The reason for the termnology wll become clear when we ntroduce a graphcal representaton of these contractons. We say the felds,..., 4 are external whle the ones that appear from nsertng powers of the Hamltonan nto the tme ordered product are nternal. For dsconnected terms each contracton j of a par of external felds wll gve a sngle factor of (2 ) 4 (4) (k + k j )(k 2 m 2 ), and the rest of the factors n that term wll be ndependent of k and k j. Then, when computng the ampltude A we ll have lm (k k,j 2 2 m 2 )(kj 2 m 2 )!m2 k 2 m 2 (2 )4 (4) (k + k j ) (k,j -ndependent) = 0 To get a non-vanshng ampltude we look n connected terms. Note that the one n (5.7) s but one of 4! contractons of ths type, and they all gve the same result. So we have G (4),conn (x,...,x 4 )= = = from whch we read o d 4 x d 4 x 4Y n= (x ) (x) (x 2 ) (x) (x 3 ) (x) (x 4 ) (x) 4Y n= d 4 q n (2 ) 4 e qn (xn x) m 2 + q 2 n d 4 q n (2 ) 4 e qn (xn x) (2 ) 4 (4) ( 4X q n ) n= 4Y q 2 n= n m 2 + eg (4),conn (k,...,k 4 )= 4Y q 2 n= n m 2 + We now use the LS formula to compute the scatterng ampltude. Ths entals removng the four propagators and multplyng by 2. However, tself has a perturbatve expanson, wth =+O( ), so to the order we are workng we can set =. We obtan A(k,k 2 ; p,p 2 )=, (5.8) that s outh~p ~p 2 ~ k ~ k 2 n = (2 ) 4 (4) (k + k 2 p p 2 ) 5.3. Graphcal Representaton A graphcal representaton gves an e ectve way of communcatng and organzng these calculatons. Consder the Green s functon that we would need to compute

11 86 CHAPTER 5. ELEMENTARY PROCESSES the scatterng ampltude at p-th order n ( ) p h0 T (x ) (x 4 ) n the perturbatve expanson 4 (y ) 4! 4 (y p ) 4! 0 (5.9) In usng Wck s theorem to compute ths s expanded nto a sum of many terms. Each term n the sum s represented by a dagram, and the set of all dagrams, s constructed by drawng:. An endpont of a lne for each x,...,x 4. We call these lnes external. 2. A pont, or vertex, from whch four lnes orgnate for each y,...,y p. 3. Connectons between the loose ends of the lnes, so that all ponts x,...,y p are connected wth lnes (wth one lne emergng from the x s and four from the y s). We assocate wth each lne a factor of (z a ) (z b ), where z a and z b are the two ponts connected by the lne. To each vertex we assocate a. Fnally there s a combnatoral factor arsng from equvalent contractons, to compensate for less than 4! possble equvalent contractons; see example and fuller explanaton below. Let s recover the results of our 0-th and st order calculatons. At lowest order, p = 0, so there are not vertces, only four endponts of external lnes: x x 3 x 2 x 4 + x x 2 x 3 x 4 + x x 2 x 3 x 4 = These are dsconnected terms, and the dagrams are dsconnected dagrams. As such they gve A = 0. Now at frst order n perturbaton theory, the p =term, we have dsconnected dagrams, x x 2 x 3 y x 4 + x y x 3 x 2 x 4 + = 2 y 3 y y y and one connected dagram,

12 5.4. FEYNMAN GRAPHS 87 x y x 3 x 2 x 4 = y 2 y 3 y 4 y Note that the frst dsconnected dagram has a combnatoral factor of 2, as ndcated. Ths can be seen as follows. Startng from (5.9) wth p =, and nsstng that s contracted wth 2 and both 3 and 4 are contracted wth y s, we see that there s only one possble contracton of wth 2,thenwehave4ways of contractng 3 wth 4 y,whch 3 y un-contracted, and fnally we have 3 ways of contractng 4 wth 3 y. The last step leaves 2 y whch can gve a sngle contracton of y wth y. That s, there are 4 3 contractons. Ths tmes the pre-factor of 4! gves the symmetry factor of 2. Here s another example of a combnatoral factor, now for a connected dagram, from p = 2: x x2 y x 3 y 2 x4 = 2 ( )2 y 2 y 3 y 2 y y 2 2 The combnatoral factor s obtaned as follows. There are 4 ways of contractng wth 4 y, whch leaves 3 ways of contractng 2 wth 3 y. Smlarly, here are 4 ways of contractng 3 wth 4 y 2, whch leaves 3 ways of contractng 4 wth 3 y 2. Fnally we have to contract 2 y wth 2 y 2, and there are 2 ways of dong ths. We have = 4! 2 As an exercse you should verfy there are fve other connected dagrams for the p = 2 case and they all have the same combnatoral factor of Feynman Graphs The dagrammatc language above s very useful n computng Green s functons n perturbaton theory. But often we are nterested n scatterng ampltudes whch are obtaned from the Fourer transform e G (n) (k,...,k N ) by the LS reducton formula. So t s convenent to replace the rules for the dagrammatc analyss above so that one obtans drectly the Fourer transforms e G (n) or even the correspondng scatterng ampltude A. To ths e ect, n computng (5.9) we use a b = d 4 q (2 ) 4 eq (xa x b) q 2 m 2 +. (5.0)

13 88 CHAPTER 5. ELEMENTARY PROCESSES R For perturbaton theory the nteracton term, d 4 y 4 (y), s as n (5.9) but ntegrated over space-tme. Each nteracton term gves a factor of d 4 ye y P n qn = (2 ) 4 (4) ( X q n ), n where the q n are from the four contractons Q 4 = (y) (x ). To keep track of the sgns ±q x n the arguments of the exponental, t s convenent to thnk of q as an arrow: n (5.0) t s drected from x a to x b. So we have new rules:. Draw dagrams wth n external legs (all topologcally dstnct dagrams). 2. For each topology assgn momenta q to each lne, ncludng external legs. The assgnment s drectonal: draw an arrow to ndcate the drecton of q, arbtrarly. 3. Every external lne carres a factor d 4 q n (2 ) 4 e±qn xn qn 2 m 2 +. wth the plus sgn f the arrow for q n s drawn pontng nto the dagram, mnus f t ponts out. 4. Every nternal lne carres a factor d 4 q (2 ) 4 q 2 m Each vertex carres a factor (2 ) 4 (4) ( X n (±)q n ) where q n are the momenta of the lnes at the vertex, wth the sgn assgnment + f q n s drected nto the vertex and f drected out of the vertex. 6. Introduce a correcton symmetry factor, as before. For example, the connected dagram, x q q 3 x 3 x 2 q 2 q 4 x 4

14 5.4. FEYNMAN GRAPHS 89 gves 4Y = d 4 q n (2 ) 4 eqn xn 4X qn 2 m 2 + ( )(2 )4 (4) ( q n ) If the contracton nvolves an external leg, when takng the Fourer transform the correspondng coordnate x s ntegrated, R d 4 x e k x. Ths gves d 4 x e k x n= d 4 q,eq (x (2 ) 4 y) q 2 m 2 + = e k y m 2 + Now recall, Eq. (5.5), that e G (n) (k,...,k n ) s not really the Fourer transform of G (n) (x,...,x n ), but rather n Y = d 4 x e P n = k x G (n) (x,...,x n )=(2 ) 4 (4) ( X k 2 k ) e G (n) (k,...,k n ). The Feynman rules tell us how to compute for (2 ) 4 (4) ( P k ) e G (n) (k,...,k n )n perturbaton theory:. Draw dagrams wth n external legs (all topologcally dstnct dagrams). 2. For each topology fnd the nequvalent ways of assgnng momenta k to each external leg. The assgnment s drectonal: k goes nto the dagram, out to n f k 0 > 0 (draw an arrow to ndcate ths). 3. Assgn a momentum q n, n =,...,I to each nternal lne. Draw an arrow to ndcate ths momentum drecton, arbtrarly. 4. Every external lne carres a factor k 2 m Every nternal lne carres a factor d 4 q (2 ) 4 q 2 m Each vertex carres a factor (2 ) 4 (4) ( X n (±)p n ) where p n are the momenta of the lnes at the vertex, wth the sgn assgnment + f p n s drected nto the vertex and f drected out of the vertex.

15 90 CHAPTER 5. ELEMENTARY PROCESSES 7. Introduce a correcton symmetry factor, as before. So, for example, to order for the 4-pont functon s n the perturbatve expanson, the connected dagram k k 3 k 2 k 4 correspondng to eg (4) (k,...,k 4 )= 4Y n= k 2 n m 2 + ( )(2 )4 (4) ( Here s another example, a contrbuton to order 2 to e G (6) : k k 4 q k 2 k 5 4X k n ) n= correspondng to 6Y n= k 2 n leadng to d 4 q m 2 + (2 ) 4 " ( )(2 ) 4 (4) 3X =(2 ) 4 (4) eg (6) conn(k,...,k 6 )= k 3 k 6 q 2 n= k n " 6X k n n= 2 n= m 2 + #" # 6X q ( )(2 ) 4 (4) k n + q 2 n= n=3 6Y # kn 2 m 2 (k + k 2 + k 3 ) 2 m 2 6Y kn 2 m 2 (k + k 2 + k 3 ) 2 m 2 + where the ellpses stand for other connected graphs at order 2 (can you dsplay them?) plus terms of hgher order n. Wehaveremovedthe from the propagators n the last lne snce all ntegrals have been performed. One more example:

16 5.4. FEYNMAN GRAPHS 9 k q k 3 k 2 q 2 k 4 gves 2 4Y n= =(2 ) 4 (4) k 2 n m 2 2Y = d 4 q (2 ) 4 h ( )(2 ) 4 (4) (k + k 2 q q 2 ) 4X n= k n 2 ( )2 4Y n= q 2 k 2 n m 2 m 2 + h ( )(2 ) 4 (4) (k 3 + k 4 + q + q ) d 4 q (2 ) 4 q 2 m 2 + (k + k 2 q ) 2 m 2 + Notce that ths result nvolves a non-trval ntegraton. Ths occurs n any dagram for whch there s a closed crcut of nternal lnes: momentum conservaton at each vertex, enforced by -functons, does not completely fx the momentum of the nternal lnes. In ths example the momentum q appears n two propagators, tracng a closed trajectory, a loop. The stuaton s depcted n a new type of dagram n whch the delta functons of momentum conservaton have been explctly accounted for (except an sngle factor that gves momentum conservaton of the external momentum): k q k 3 k 2 q k k 2 k 4 There s no longer an ntegraton for each nternal lne. Instead there as n ntegral only over the undetermned momentum q. We call q a loop momentum and R d 4 q a loop ntegral. Ths generalzes. It s always the case that that the -functons at each vertex mpose momentum conservaton and therefore one can always recast the product of delta-functons as one (4) ( P n k n) of external momentum tmes the remanng delta functons. The number L of loop ntegrals we are left to do n any gven dagram s the number of nternal lnes I mnus the number of delta-functons, takng away the one for overall momentum conservaton. If there are V vertces, we then have V -functons and therefore L = I V +

17 92 CHAPTER 5. ELEMENTARY PROCESSES In our example above, I = 2, V = 2 and we had L =2 few more examples: 2 + = loops. Here are L =3 2+=2 L =7 4+=4 We can also have a theory wth 3-pont and 4-pont vertces, as n 3! g 3 + 4! here s an example: 4 ; L =5 4+2=2 Ths suggest a more compact set of Feynman rules to compute e G (n) (k,...,k n ):. Draw dagrams wth n external legs (all topologcally dstnct dagrams). 2. For each topology fnd the nequvalent ways of assgnng momenta k to each external leg. The assgnment s drectonal: k goes nto the dagram. Draw an arrow to ndcate ths. 3. Assgn q, =,...,L momenta to nternal lnes; draw arrow ndcatng drecton. Assgn momenta to remanng I L = V nternal lnes by enforcng momentum conservaton: at each vertex P p n = P p out. 4. For each lne p = p 2 m For each vertex, = 6. Integrate:. LY n= d 4 q n (2 ) 4 7. Symmetry factor /S as needed.

18 5.5. CROSS SECTION Cross Secton In a common expermental setup two beams of elementary partcles are accelerated n opposte drectons and brought nto face to face encounter. Some fracton of the partcles n the beams collde. The collson results n a spray of elementary partcles emanatng form the collson pont and an array of detectors surroundng the area regster these outgong partcles. Here as a computer generated mage of the tracks made by charged partcles that are sprayed out of the head-to-head collson of two protons, projected onto a plane transverse to the drecton of the protons: In another common setup a beam of partcle mpnges on a collecton of statonary targets. Ths second set-up s, of course, just the frst one as seen by an observer at rest wth the second beam. We say ths observer s n the lab frame. We are after a measure of how lkely are these collsons to occur. The cross secton,, for scatterng s defned through number of collsons unt tme =(flux). To calculate, rather than computng the number of collsons per unt tme from the actual flux, we use unt flux (that of one-on-one partcles) and therefore collson probablty unt tme =(untflux). We have an ntal state that conssts of two partcles, a fnal state f that conssts of n-partcles (n/ge2). The probablty that evolves nto f s P!f = hf out n 2 = hf n S n 2 = hf S 2 where n the last step we gnore the f = case (no collson, hence subtract from S) and suppressed the n label (we wll get tred of carryng t around).

19 94 CHAPTER 5. ELEMENTARY PROCESSES We have to be careful to ask the rght queston: snce we have contnuum normalzaton of states, f we are overly selectve n what we want for f, the probablty of fndng t n wll vansh. Recall f you drop a pn on a pece of paper the probablty of httng a gven pont on the paper, say, (x 0,y 0 ), s zero, snce a pont s a set of measure zero n the set of ponts that comprse the area of the paper. Lkewse, f we set f = ~ k,..., ~ k n we ll fnd P!f = 0. Instead we project out a subspace of F, rather than a sngle state. Instead of we take hf S 2 = h (S ) fhf S h (S ) X fhf (S ) f some states In partcular, for n partcles n the fnal state we have X fhf! (dk ) (dk n ) ~ k,..., ~ k n h ~ k,..., ~ k n where f we may not want to sum over all possble momenta, so the ntegrals can be restrcted must avod double countng from ndstngushable partcles Suppose partcles and 2 are ndstngushable (but the rest are not). Then to avod double countng one should wrte X fhf! (dk ) (dk n ) 2 ~ k,..., ~ k n h ~ k,..., ~ k n f If,2,3 are ndstngushable then the pre-factor becomes /3! snce the order of ~ k, ~ k 2 and ~ k 3 n the label of the state s mmateral. More generally, X fhf! S f ny (dk ) ~ k,..., ~ k n h ~ k,..., ~ k n = where S = m!m 2! where m s the number of dentcal partcles of type ( P m = n). We are ready to gve a probablty: P!f = h (S ) P fhf S h

20 5.5. CROSS SECTION 95 We have dvded by h because states must be normalzed for proper nterpretaton. Dd not dvde by normalzaton of f because t s ncluded properly n relatvstc measure n the sum over states. We next recast ths n terms of the scatterng ampltude, hf S =(2 ) 4 (4) (P f P ) A(! f). At ths pont we choose an ntal state of plane waves wth defnte momentum, = ~p,~p 2.Wehave h = h~p,~p 2 ~p,~p 2 = h~p ~p h~p 2 ~p 2 but snce n general h~p ~ k =2E ~k (2 ) 3 (3) (~p ~ k ), we have, h~p ~p =2E ~k (2 ) 3 (3) (0). Ths s embarrassng, but not dsastrous. There are two ways of dealng wth ths problem. It s not very hard to use wave-packets, whch can be properly normalzed, nstead of plane waves, but won t do here; check t out n some of the textbooks n our bblography. We have an alternatve means of dealng wth ths problem, whch s by puttng the system n a fnte box of volume V. We have already done so n computng phase space. We dscovered there that the correct nterpretaton of ths nfnty s (2 ) 3 (3) (0)! V. In fact, we wll also use, more generally, (2 ) 4 (4) (0)! VT,whereT = t fnal t ntal. So we can wrte h =4E E 2 V 2. Smlarly 2 hf S 2 = (2 ) 4 (4) (P f P ) A(! f) 2 = (2 ) 4 (4) (0) (2 ) 4 (4) (P f P ) A(! f) 2 Puttng t all together: where probablty tme = T VT S = VT(2 ) 4 (4) (P f P ) A(! f) 2 R Q (dk )(2 ) 4 (4) (P f P ) A(! f) 2 4E E 2 V 2 = A(! f) 2 d n V 4E E 2 d n = S (2 )4 (4) (P f P )(dk ) (dk n ) s the Lorentz-nvarant n-partcle phase space. Fnally, n order to determne the cross secton we need to dvde the above by the unt flux. Assume partcle s unformly dstrbuted n a box of volume V. The probablty of fndng t s a sub-volume v s v/v. We want v to be the nteracton volume, so project a volume forward of partcle 2, n the drecton of the relatve moton ~v 2 ~v, wth cross sectonal area perpendcular to that drecton, when partcle 2 moves over a tme t, as n the followng fgure:

21 96 CHAPTER 5. ELEMENTARY PROCESSES partcle 2 ~v 2 ~v Ths has volume V = ( ~v 2 ~v t) V ~v 2 ~v t ) probablty unt tme = ~v 2 ~v V Comparng wth the above probablty per unt tme we have d = 4E E 2 ~v 2 ~v A 2 d where we have wrtten d rather than to remnd us that d wll be ntegrated over: = R R d = 4E E 2 ~v 2 ~v A 2 d. The factor E E 2 ~v 2 ~v s nvarant under boosts along the drecton of ~v 2 ~v. Ths s most easly seen n a frame where ~v and ~v 2 are along the z-axs. Then E E 2 ~v 2 ~v = E E 2 p 2 E 2 p E = p 2 E p E 2 = 2µ p µ p 2 s nvarant to boosts n the 3-drecton. It s useful to compute ths factor n the two most common frames, once and for all. In the Lab frame: ~p 2 = 0, E 2 = m 2 so E E 2 ~v 2 ~v = p 2 E p E 2 = m 2 p. In the center of mass, or CM frame, ~p 2 + ~p = 0, so that ~p = ~p 2 ~p and E E 2 ~v 2 ~v = p 2 E p E 2 = ~p p (p + p 2 ) 2 where p,2 are 4-vectors. Let s (p + p 2 ) 2 so that s = m 2 + m2 2 +2E E 2 +2 ~p 2.SnceE = relatng ~p 2 to s, whch we solve: q ~p 2 + m 2 we have an equaton Ths gves ~p 2 = (s m2 m 2 2 )2 4m 2 2 m2 2 4s 4E E 2 ~v 2 ~v = ~p p q s =2 (s m 2 m 2 2 )2 4m 2 m2 2. The last expresson s vald n any frame boosted along ~v 2 ~v, and we can wrte d = 2 p A 2 d (s m 2 m 2 2 )2 4m 2 m2 2

22 5.5. CROSS SECTION 97 Example: 2! 2 scatterng, dentcal partcles ~ k ~p ~p 2 ~ k 2 We have for dentcal partcles m = m 2 = m Change varables: d 2 = d3 k (2 ) 3 2E d 3 k 2 (2 ) 3 2E 2 (2 ) 4 (4) (P k k 2 ) = d4 k d 4 k 2 (2 ) 2 (k 0 ) (k 0 2) (k 2 m 2 ) (k 2 2 m 2 ) (4) (P k k 2 ) p = k + k 2 q = 2 (k k 2 ) Note that the Jacoban of the transformaton,, k = 2 p + q k 2 = 2 2 q =. Then, d 2 = (2 ) 2 d4 p (4) (P p)d 4 q ( 2 p0 + q 0 ) ( 2 p0 q 0 ) (( 2 p + q)2 m 2 ) (( 2 p q)2 m 2 ) = (2 ) 2 d4 q ( 2 P 0 q 0 ) ( 4 P 2 + P q + q 2 m 2 ) (2P q) In the CM frame, ~ P = 0, ths s smple: d 2 = (2 ) 2 dq0 ~q 2 d ~q d cos d ( 2 P 0 q 0 ) ( 4 P 2 +(q 0 ) 2 ~q 2 m 2 ) (q 0 ) 2P 0 = q (2 ) 2 d cos d 2P (P 0 ) 2 m 2 = 8(2 ) 2 d cos d p 4m 2 /s where s = P 2 =(p + p 2 ) 2 as before. Therefore d d cos d = p 4m 2 8(2 ) 2 2 /s 2 p (s 2m 2 ) 2 4m 2 A 2 = 32 (2 ) 2 s A 2

23 98 CHAPTER 5. ELEMENTARY PROCESSES Often A s ndependent of,so 5.6 LS reducton, agan d d cos = 64 s A 2 We want to establsh the LS reducton formula. We don;t pretend to gve a complete proof. The objectve s to express the S-matrx n terms of Green s functons, vacuum expectaton values of tme ordered products, of n felds. Consder outh ~ k, ~ k 2,..., ~ k n n = out h ~ k n = out h ~ k n n Recall n(x) t n (x) = (dq) ~q n e q x + eq x ~q n (dq) E ~q ~q n e q x + E ~q ~q n eq x Invertng these, ~q n = d 3 xe q t n (x) So we have outh ~ k n = d 3 xe k t out h n(x) n = 2 d 3 xe k t out h (x) n as t!. Now we use the fundamental theorem of calculus, g(t 2 )=g(t )+ R t 2 t t 2!and t! so that lm t! d 3 xe k t out h (x) n = lm t! dt dg dt,wth d 3 xe k t out h (x) n d 4 x@ t e k t out h (x) n The frst term on the rght hand sde tmes 2 s out h ~ k out n, as can be seen by reversng the steps. If out = ~p,~p 2,...,~p n 0 out then ~ k out out = (2 ) 3 2E (3) ~k ( ~ k ~p ) ~p 2,...,~p n 0 out + +(2 ) 3 2E (3) ~k ( ~ k ~p n 0) ~p,...,~p n 0 out corresponds to a partcle not partcpatng n the scatterng. We have no use

24 5.6. LS REDUCTION, AGAIN 99 for ths. For the second term on the rght hand sde we t t g(t)) = f@ t 2 g (@ t 2 f)g so t s d 4 xe q x (@ t 2 + E~q 2 ) outh (x) n Usng E~q 2 e q x =( ~q 2 + m 2 )e q x =( r ~ 2 + m 2 )e q x and ntegratng by parts we have fnally outh ~ k n = out h ~ n + 2 d 4 xe k x (@ 2 + m 2 ) out h (x) n k out Smlarly outh ~p n = out h ~p n n + 2 d 4 xe p x (@ 2 + m 2 ) out h (x) n (5.) We woud lke to repeat the process untl we remove all partcles from out and n. To see how ths goes move a partcle o from n from what we already had: outh (0) ~ k 0 n = out h (0) ~ 0 n = 2 lm d 3 xe k k n x x 0 0 outh (0) (x)! = 2 d 4 x@ x 0 e k x 0 outh (0) (x) 0 n 2 lm d 3 xe k x x 0 0 outh (0) (x) 0 n (5.2)! In the last expresson we would lke to move (x) to the left of (0) so that we may turn (x) as x 0!nto ~ actng on hout. To ths end we rewrte the k out frst term n the last expresson n (5.2) usng 0 n (0) (x) =( ( x 0 )+ (x 0 )) (0) (x)+ (x 0 )( (x) (0) (x) (0)) = T ( (x) (0)) + (x 0 )[ (0), (x)] Then n d 4 x@ x 0 e k x 0 outh (x 0 )[ (0), (x)] 0 n (5.3) x 0 hts (x 0 ) we get the equal tme commutator [ (0), (x)] = 0. So we have (5.3) s = h d 4 x@ x 0 (x 0 ) e k x 0 outh [ (0), (x)] 0 n = lm d 3 xe k x x 0 0 outh [ (0),! (x)] 0 n

25 00 CHAPTER 5. ELEMENTARY PROCESSES Combnng ths wth the last term n (5.2) gves precsely what we want: t reverses the order of (0) (x) so that outh (0) ~ k 0 n = 2 d 4 x e k x (@ 2 + m 2 ) out h T ( (0) (x)) 0 n + out h ~ (0) k out 0 n We thus arrve at outh ~ k ~ k 0 2 n =( 2 ) 2 d 4 x d 4 x 2 e k x k 2 x 2 (@ 2 x + m 2 )(@ 2 x 2 + m 2 ) out h T ( (0) (x)) 0 n + dsconnected One can repeat the process untl all partcles n n are removed and we are left wth 0 n = 0. The argument above can be streamlned by replacng T ( (0) (x)) for (0) (x) n the lne above (5.2)), and ths ndeed becomes very convenent n completng the argument for arbtrary number of partcles. Smlarly we can remove -partcle states from out h ~ k usng (5.) repeatedly. 5.7 Perturbaton theory, agan We now gve a proof of (5.6) that gves us the bass for perturbaton theory. Consder G (n) (x,...,x n )=h0 T ( (x ) (x n )) 0. Take for defnteness x 0 x 0 2 x 0 n. Recall (x) =U(t) nu(t), ans use ths n each n the Green s functon: G (n) (x,...,x n )=h0 U (t ) (x )U(t )U (t 2 ) (x 2 )U(t 2 ) U (t n ) (x n )U(t n ) 0. Let U(t, t 0 )=U(t)U (t 0 ). Ths satsfes Moreover, U(, ) =S (5.4) U(t, ) =U(t) (5.5) U(t, t 0 )U(t 0,t 00 )=U(t, t 00 ) t0 ) = H nu(t, t 0 ), wth U(t 0,t 0 )=. Ths s the same equaton satsfed by U(t), but wth a d erent boundary condton. So the soluton s the same only wth d erent lmts of ntegraton, tf U(t f,t )=T exp dthn(t) 0. t

26 5.7. PERTURBATION THEORY, AGAIN 0 We now have G (n) = h0 U ()U(,t ) n (x )U(t,t 2 ) n (x 2 )U(t 2,t 3 ) U(t n,t n ) n (x n )U(t n, )U( ) 0 = h0 U ()T (U(,t ) n (x )U(t,t 2 ) n (x 2 )U(t 2,t 3 ) U(t n,t n ) n (x n )U(t n, )) 0 = h0 U ()T (U(,t )U(t,t 2 )U(t 2,t 3 ) U(t n,t n )U(t n, ) n (x ) n (x 2 ) n(x n )) 0 = h0 U ()T (U(, ) n (x ) n (x 2 ) n(x n )) 0 Unqueness of the vacuum means that 0, 0 n, 0 out, are equal up to a phase. Moreover, U() 0 = S 0 must be 0 up to a phase. To see ths note that S commutes wth Poncare transformatons, U(a µ, )SU (a µ, ) = S and 0 s the unque state (up to a phase) that s left nvarant by a Poncare transformaton. Then U(a µ, )(S 0) =U(a µ, )SU (a µ, )U(a µ, ) 0 = S 0 so S 0 s nvarant and hence equal to 0 up to a phase. So we have h0 U () =h0 U () 0h0. Usng ths and U() =U(, ) =T exp dt Hn(t) 0 we fnally have G (n) (x,...,x n )= n h0 T n(x ) n(x n )e R d 4 x L 0 n nh0 T e R d 4 x L 0 n 0 n 0 n Note that we replaced 0 n for 0 snce the phases n numerator and denomnator cancel. Ths s not what we set out to prove. It s better. The denomnator corresponds to a set of graphs wthout external legs. These vacuum graphs can also appear n the numerator, just multplyng any graph wth external legs. It s a smple exercse to check that the vacuum graphs n the numerator are cancelled by the graphs n the denomnator Generatng Functon for Green s Functons Let Then and [J] =h0 Te R d 4 xj(x) (x) 0. G (n) (x,...,x n )= J(x ) J(x n ) [J] J=0 [J] = n h0 Te R d 4 x(l 0 n +J(x) n(x)) 0 n nh0 T e R d 4 xl 0 n 0 n

27 02 CHAPTER 5. ELEMENTARY PROCESSES Ths s a convenent way of summarzng the results above for G (n), all n. Note also that [J] = X n d 4 x d 4 x n J(x ) J(x n )G (n) (x,...,x n ). n! n Generatng Functon for Connected Green s Functons Smlarly we defne W [J] = X n n n! d 4 x d 4 x n J(x ) J(x n )G (n) conn(x,...,x n ). We wll now show that [J] =e W [J]. We use a dagrammatc notaton to see how ths works: W [J] = + + where the heavy dots stand for J(x), the hatch crcles wth n lnes stand for G (n) and an ntegral Q R n! d 4 x n each term s understood. Let streamlne notaton for the purposes of ths proof: remove the heavy dots (the ends of lnes are understood as havng them) and reduce the hatch crcle to a pont, so that the above fgure s the same as W [J] = + + Wth ths notaton we consder the exponental of W [J]: exp(w [J]) = exp exp exp Now expand ts exponental, as n exp =+ + 2! + 3! + and reorganze by powers of J, that s, number fo external legs:

28 5.7. PERTURBATION THEORY, AGAIN 03 exp W [J] = ! + + where the ellpses stand for terms wth fve or more legs. Let s analyze n more detal the term n parenthess: we want to show that t gves G (4) (tmes sources, /4! and ntegrated). The cross stands for Now, 4 J(x ) J(x 4 ) 4! d 4 y d 4 y 4 J(y ) J(y 4 )G (4) conn(y,...,y 4 ) of ths gves G(4) conn(x,...,x 4 ). Note that the 4! s absent snce there are 4! terms from the ntegral (same as n d4 dx 4 x 4 = 4!). Turnng to the other term, the dsconnected graph, we have 4 J(x ) J(x 4 ) 2! 2! 2 d 4 yd 4 zj(y)j(z)g (2) (y, z) = G (2) (x,x 2 )G (2) (x 3,x 4 )+G (2) (x,x 3 )G (2) (x 2,x 4 )+G (2) (x,x 4 )G (2) (x 2,x 3 ) If we take the for J-functonal dervatves and set J = 0 these are the only terms we pck up n the expanson, so we have 4 J(x ) J(x 4 ) ew [J] = J=0 G(4) conn(x,...,x 4 ) + G (2) (x,x 2 )G (2) (x 3,x 4 )+G (2) (x,x 3 )G (2) (x 2,x 4 )+G (2) (x,x 4 )G (2) (x 2,x 3 ) = G (4) (x,...,x 4 )= 4 J(x ) J(x 4 ) [J] J=0 In the general case, the term wth J n n e W [J] s a sum of all possble contrbutons of the form n2 n3 J J 2 G (2) c J J 2 J 3 G (3) c n 2! 2! n 3! 3! such that 2n 2 +3n 3 + = n, n a hopefully obvous condensed notaton. For example, n = 4 has (n 2 =2,n 6=2 = 0) + (n 4 =,n 6=4 = 0), and n = 6 has (n 6 =,n 6=6 = 0) + (n 4 =,n 2 =,n 6=4,2 = 0) + (n 3 =2,n 6=3 = 0) + (n 2 =3,n 6=2 = 0). Consder the term wth G (k) c : take kn k J(x ) J(x knk ) n k! k! nk J J k G (k) c

29 04 CHAPTER 5. ELEMENTARY PROCESSES Ths s completely symmetrc under permutatons of x,...,x knk. To make ths explct we rewrte t as n k! k! kn k nk J J k G (k) c = n k! Y = (k!) n k d 4 y J(y ) J(y knk )G (k) c (y,...,y k ) G (k) c (y (nk )k+,...,y nk k) Takng kn k J-dervatves we obtan n k! (k!) n G(k) c k (x,...,x k ) G (k) c (x (nk )k+,...,x nk k)+permutatons of x,...,x nk k Ths contans many repeated terms. We need to count the number of nequvalent permutatons. For each G (k) c there are k! equvalent permutatons of the arguments; ths gves (k!) n k. Then we can permute the G (k) c among themselves; there are n k! such permutatons. So we obtan kn k J(x ) J(x knk ) n k! k! nk J J k G (k) c = G (k) c (x,...,x k ) G (k) c (x (nk )k+,...,x nk k)+nequv-perms Fnally combne all terms and symmetrze over x,...,x n. We obtan all possble combnatons of G c s that can make G (n). But that s precsely what we ntended to show.