Functional Quantization

Transcription

1 Functonal Quantzaton In quantum mechancs of one or several partcles, we may use path ntegrals to calculate transton matrx elements as out Ût out t n n = D[allx t] exps[allx t] Ψ out Ψ n n 1 where the paths x t are ndependent of each other and are not subject to any boundary condtons, and the acton depends on all partcles paths as S[all x t] = t out t n Ldt, L = m ẋ V allx. Let s generalze formula 1 to the quantum feld theory. In classcal feld theory, a fnte set of dynamcal varables x becomes a feld Φx where x s now a contnuous label rather than a dynamcal varable of ts own. Thus, a set of all the partcles paths x t becomes a feld confguraton Φx,t, wth acton functonal where the Lagrangan densty s t out S[Φx,t] = dt d 3 x L 3 t n L = 1 µφ µ Φ m Φ λ 4 Φ4 4 for a scalar feld, or somethng more complcated for other knds of felds. In quantum feld theory, the path ntegral D[allx t] D[x t] 5 generalzes to the functonal ntegral over the feld confguratons D[Φx,t] 6 A careful defnton of ths functonal ntegral nvolves dscretzaton of both tme and space to a 4D lattce of N = TV/a 4 spacetme ponts x n and then takng a zero-lattce-spacng 1

2 lmt D[Φx, t] = lm a 0 Normalzaton factor n dφx n. 7 I shall address ths ssue later ths week. For now, let s focus on usng the functonal ntegral. The QFT analogue of eq. 1 for the transton matrx element s out Ût out t n n = D[Φx,t] exps[φx,t] Ψ out ] Ψ n n ]. 8 Note that n the feld theory s Hlbert space, the wave-functons Ψ n and Ψ out depend on the entre 3-space feld confguraton Φx for a fxed tme t = t n or t = t out. In terms of such wave-functons, the out n bracket s gven by a 3-space analogue of the functon ntegrals over Φx at a fxed tme, out n = D[Φx] Ψ out[φx] Ψ n [Φx]. 9 Now, let s go to the Hesenberg pcture and consder the matrx element of the quantum feld ˆΦ H x 1,t 1 at some pont x µ 1 = x 1,t 1. The states and the operators n Hesenberg and Schrödnger pctures are related to each other va the evoluton operator Ût t 1, hence out ˆΦ H x 1,t 1 n = out Ût out t 1 ˆΦ S x 1 Ût 1 t n n = D[Φ 1 x] out Ût out t 1 Φ 1 Φ 1 x 1 Φ 1 Ût 1 t n n = D[Φx,t for the whole tme] Ψ out ] exp S[Φ] from t 1 tot out = Φx 1,t 1 exp S[Φ] from t n tot 1 Ψn n ] D[Φx,t for the whole tme] Ψ out ] Ψ n n ] exp S[Φ] for the whole tme Φx 1,t 1. Lkewse, matrx elements of tme-ordered products of several felds ˆΦ H x n,t n ˆΦ H x 1,t 1 10

3 for t n > > t 1 are gven by functonal ntegrals out ˆΦ H x n,t n ˆΦ H x 1,t 1 n = = D[Φx,t for the whole tme] Ψ out ] Ψ n n ] exp S[Φ] for the whole tme Φx n,t n Φx 1,t Now, let s take the tme nterval between the n and out states to nfnty n both drectons, or rather let s take the lmt t out + 1 ǫ, t n 1 ǫ. 1 As we saw back n November cf. class notes, n ths lmt out e Ĥtout Ω c-number factor, e +Ĥtn n Ω c-number factor, 13 where Ω s the vacuum state of the quantum feld theory. Therefore, we may obtan the correlaton functon of n quantum felds G n x 1,x,...,x n = Ω TˆΦ H x 1 ˆΦ H x n Ω 14 from the functonal ntegrals G n x 1,x,...,x n = const D[Φx] exp S[Φ] = d 4 x L Φx 1 Φx n. Note that the acton ntegral S = d 4 x L for each feld confguraton Φx s taken over the whole Mnkowsk space tme, so t s Lorentz nvarant, and the correlatons functons 15 are also Lorentz nvarant. 15 Strctly speakng, the lmt 1 mples that all the tme coordnates x 0 of all felds n 15 lve on the slghtly tlted tmes axs that runs from 1 ǫ to + 1 ǫ. Consequently, the acton has a complexfed verson of the Lorentz symmetry, but after analytc contnuaton of all the correlaton functons back to the real axs, we recover the usual Lorentz symmetry SO3, 1. Alternatvely, we may analytcally contnue to the magnary tmes; ths gves us SO4 symmetry n the Eucldean spacetme. 3

4 We do not know the constant factor n front of the functonal ntegral 15 and we do not know the overall normalzaton of the functonal ntegral. But we can get rd of both of these unknown factors by takng ratos of correlaton functons. In partcular, usng G 0 = Ω Ω = 1 16 we can wrte all the other correlaton functons as normalzaton-ndependent ratos G n x 1,x,...,x n = Ω TˆΦ H x 1 ˆΦ H x n Ω Ω Ω = D[Φx] exp D[Φx] exp d 4 x L Φx 1 Φx n. d 4 x L 17 In a few pages, we shall learn how to calculate such ratos n perturbaton theory and rederve the Feynman rules. But before we do perturbaton theory, let s see how functonal ntegrals work out for the free scalar feld. Functonal Integrals for the Free Scalar Feld The free scalar feld has a quadratc acton functonal S[Φx] = d 4 x L = 1 µφ µ Φ 1 m Φ = 1 d 4 x Φx + m Φx. 18 Hence, the functonal ntegral Z = D[Φx] exp S[Φx] 19 s a knd of a Gaussan ntegral, generalzaton of the ordnary Gaussan ntegral ZA = dξ 1 dξ dξ N exp 1 to a contnuous famly of varables, {ξ j } D[Φx]. N j,k=1 A jk ξ j ξ k 0 4

5 Lemma 1: The ordnary Gaussan ntegral 0 evaluates to ZA = πn/ deta. 1 Note: A jk s a symmetrc N N matrx, real or complex. To make the ntegral 0 converge, the real part ReA jk of the A matrx should be postve defnte. Proof: Any symmetrc matrx can be dagonalzed, thus A jk ξ j ξ k = B k ηk jk k where η 1,...,η N are some ndependent lnear combnatons of the ξ 1,...,ξ N. Changng ntegraton varables from the ξ k to η k carres a Jacoban J = det ξj ; 3 η k by lnearty of the ξ η transform, ths Jacoban s constant. Therefore, Z = J d N η exp 1 B k ηk = J = J N k=1 N π k=1 k dη k exp B k η k B k = πn/ J detb 4 where B s the dagonal matrx B jk = jk B k. Ths matrx s related to A accordng to B = ξ ξ A, 5 η η hence and therefore detb = deta det ξ η Z = π N/ J detb = = deta J 6 πn/ deta. 1 5

6 Lemma concerns the ratons of Gaussan ntegrals for the same matrx A: d N ξ exp 1 j A jξ ξ j ξ k ξ l = A 1 d N ξ exp 1 j A kl. 7 jξ ξ j Proof: d N ξ exp 1 A j ξ ξ j ξ k ξ l = A kl j d N ξ exp 1 A j ξ ξ j, 8 j hence d N ξ exp 1 d N ξ exp 1 j A jξ ξ j ξ k ξ l = log j A jξ ξ A j kl [ ] d N ξ exp 1 j A jξ ξ j = log πn/ A kl deta = + log deta A kl = A 1 kl. 9 Both lemmas apply to the Gaussan functon ntegrals such as 19. For example, consder the free scalar propagator G x,y. Eq. 17 gves us a formula for ths propagator n terms of a rato of two Gaussan functonal ntegrals G x,y = D[Φx] exp d 4 x Φ + m Φ ΦxΦy D[Φx] exp d 4 x Φ + m Φ. 30 The role of the matrx A here s played by the dfferental operator +m. Ths operator s purely ant-hermtan, whch corresponds to a purely magnary matrx A. To make the Gaussan ntegrals converge, we add an nfntesmal real part, A j A j + ǫ j, or + m + m + ǫ. Consequently, lemma tells us that G x,y = x 1 d 4 + m + ǫ y = p π 4 e px y p m + ǫ. 31 Note that the rule A A + ǫ for makng the Gaussan ntegrals wth magnary A matrces converge automagcally leads to the Feynman propagator 31 rather than some other Green s functon. 6

7 To obtan correlaton functons G n for n > we need analogues of Lemma for Gaussan ntegrals nvolvng more than two ξs outsde the exponental. For example, for four ξs we have d N ξ exp 1 j A jξ ξ j ξ k ξ l ξ m ξ n = 3 d N ξ exp 1 j A jξ ξ j = A 1 kl A 1 mn + A 1 km A 1 ln + A 1 kn A 1 lm. Applyng ths formula to the Gaussan functonal ntegral for the free scalar felds, we obtan for the 4 pont functon G 4 x,y,z,w = G x y G z w + G x z G y w + G x w G y z 33 = + +. Lkewse, for hgher even numbers of ξs, d N ξ exp 1 j A jξ ξ j ξ 1 ξ N = d N ξ exp 1 j A jξ ξ j A 1 34 ndex par pars parngs and smlarly G free N x 1,...,x N = D[Φx] exp d 4 x Φ + m Φ Φx 1 Φx N D[Φx] exp d 4 x Φ + m Φ = parngs pars. 35 7

8 Feynman Rules for the Interactng Feld A self-nteractng scalar feld has acton S[Φx] = S free [Φx] λ d 4 x Φ 4 x. 36 4! Expandng the exponental e S n powers of the couplng λ, we have exp S[Φx] = exp S free [Φx] n=0 n 1 λ d 4 x Φ x n! 4! Ths expanson gves us the perturbaton theory for the functonal ntegral of the nteractng feld, D[Φx] exp S[Φx] Φx 1 Φx k = = D[Φx] exp S free [Φx] = d 4 x Φ + m Φ = n=0 λ n n! 4! n d 4 z 1 d 4 z n Φx 1 Φx k n=0 n 1 λ d 4 x Φ 4 x n! 4! D[Φx] exp d 4 x Φ + m Φ Φx 1 Φx k Φ 4 z 1 Φ 4 z n usng eq. 35 for the free-feld functonal ntegrals [ ] = D[Φx] exp d 4 x Φ + m Φ λ n n! 4! n d 4 z 1 d 4 z n products of k + 4n n=0 parngs [ ] = D[Φx] exp d 4 x Φ + m Φ n=0 free propagators Feynman dagrams wth k external and n nternal vertces Ths s how the Feynman rules emerge from the functonal ntegral formalsm

9 At ths stage, we are summng over all the Feynman dagrams, connected or dsconnected, and even the vacuum bubbles are allowed. Also, the functonal ntegral 38 carres an overall factor Z 0 = D[Φx] exp d 4 x Φ + m Φ whch multples the whole perturbaton seres. Fortunately, ths pesky factor as well as all the vacuum bubbles cancel out when we dvde eq. 38 by a smlar functonal ntegral for k = 0 to get the correlaton functon G k. Indeed, 39 G k x 1,...,x k = D[Φx] exp S[Φx] Φx1 Φx k D[Φx] exp S[Φx] = Z 0 all Feynman dagrams wth k external vertces Z 0 all Feynman dagrams wthout external vertces = Feynman dagrams wth k external vertces and no vacuum bubbles. 40 Generatng Functonal In the functonal ntegral formalsm there s a compact way of summarzng all the connected correlaton functons n terms of a sngle generatng functonal. Let s add to the scalar s feld Lagrangan L a lnear source term, LΦ, Φ LΦ, Φ + Jx Φ 41 where Jx s an arbtrary functon of x. Now let s calculate the partton functon for Φx as a functonal of the source Jx, Z[Jx] = D[Φx] exp d 4 xl + JΦ = D[Φx] exp S[Φx] exp d 4 xjxφx. 4 To make use of ths partton functon, we need functonal.e., varatonal dervatves /Jx. Here s how they work: JyΦyd 4 y = Φx, 43 Jx 9

10 µ Jy A µ yd 4 y = µ A µ x, 44 Jx Jx exp JyΦyd 4 y = Φx exp JyΦyd 4 y, 45 etc., etc. Applyng these dervatves to the partton functon 4, we have and smlarly Jx 1 Jx Jx Z[J] = D[Φy] exp d 4 yl + JΦ Φx 46 Jx k Z[J] = D[Φy] exp d 4 yl + JΦ k Φx 1 Φx Φx k. Thus, the partton functon 4 generates all the functonal ntegrals of the scalar theory: k, D[Φy] exp d 4 yl Φx 1 Φx k = k Jx 1 47 Jx k Z[J]. Jy 0 Consequently, all the correlaton functons 40 follows from the dervatves of the Z[J] accordng to G k x 1,...,x k = k Z[J] Jx 1 48 Jx k Z[J]. 49 Jy 0 In terms of the Feynman dagrams, the correlaton functons 49 contan both connected and dsconnected graphs; only the vacuum bubbles are excluded. To obtan the connected correlaton functons G conn k x 1,...,x k = connected Feynman dagrams only 50 we should take the /Jx dervatves of log Z[J] rather than Z[J] tself, G conn k x 1,...,x k = k Jx 1 Jx k log Z[Jy]. 51 Jy 0 In other words, log Z[Jx] s the generatng functonal for for all the connected correlaton functons of the quantum feld theory. 10

11 Proof: For the sake of generalty, let s allow for G 1 x = Φx 0. For k = 1, G conn 1 G 1 and eq. 51 s equvalent to eq. 49. For k = we use log Z[J] JxJy = 1 Z Z JxJy 1 Z Z Jx 1 Z Z Jy. 5 Let s multply ths formula by, set J 0 after takng the dervatves, and apply eqs. 49 to the rght hand sde; then log Z[J] Jx Jy = G x,y G 1 x G 1 y = G conn x,y. 53 J 0 The k = 1 and k = cases serve as a base of a proof by nducton n k. Now we need to prove that s eqs. 51 hold true for all l < k that they also hold true for k tself. A k-pont correlaton functon has a cluster expanson n terms of connected correlaton functons, G k x 1,...,x k = G conn k x 1,...,x k + cluster decompostons cluster G conn cluster where each cluster s proper subset of x 1,...,x n and we sum over decompostons of the whole set x 1,...,x n nto a unon of dsjont clusters. For example, for k = 3 54 G 3 x,y,z = G conn 3 x,y,z + G 1 x G 1 y G 1 z + G conn x,y G 1 z + G conn x,z G 1 y + G conn y,z G 1 x = two smlar. There s a smlar expanson of dervatves of Z n terms of dervatves of log Z, 1 k Z Z Jx 1 Jx k = k log Z Jx 1 Jx k + clusters l log Z, Jx 1 Jx l x 1,...,x l cluster decompostons

12 Now let s set Jy 0 after takng all the dervatves n ths expanson. On the left hand sde, ths gves us k G k x 1,...,x k by eq. 49. On the rght hand sde, for each cluster l log Z Jx 1 Jx l = l G conn l x 1,...,x l 57 by the nducton assumpton. Altogether, we have G n x 1,...,x n = k k log Z Jx 1 Jx k + J 0 cluster decompostons cluster G conn cluster, 58 and comparng ths formula to eq. 54 we mmedately see that G conn n x 1,...,x n = k k log Z Jx 1 Jx k, 51 J 0 quod erat demonstrandum. We may summarze all the equatons 51 n a sngle formula whch makes manfest the role of log Z as a generatng functonal for the connected correlaton functons, log Z[Jx] = log Z[0] + k=1 k d 4 x 1 d 4 x k G conn k x 1,...,x k Jx 1 Jx k. 59 k! To see how ths works, consder the free theory as an example. A free scalar feld has only one connected correlaton functon, namely G conn x y = G F x y; all the other G conn k requre nteractons and vansh for the free feld. Hence, the partton functon of the free feld should have form log Z free [Jx] = log Z free [0] + d 4 d 4 y G F x y JxJy + nothng else. 60 To check ths formula, let s calculate the partton functon. For the free feld Φ, L free + JΦ = 1 Φ + m Φ + JΦ = 1 Φ + m Φ + 1 J + m 1 J 61 1

13 up to a total dervatve where Φ = Φ + m 1 J. In other words, S free [Φx,Jx] def = L+JΦd 4 x = S free [Φ x,j ] + d 4 d 4 y JxG F x yjy 6 where Φ x = Φx d 4 y G F x yjy. 63 In the path ntegral, D[Φx] = D[Φ x], hence Z free [Jx] = = = exp = exp D[Φx] exp S free [Φx,Jx] D[Φ x] exp d 4 x S free [Φ x,j ] exp 1 d 4 y JxG F x yjy Z free [0]. 1 1 d 4 x d 4 y JxG F x yjy d 4 x d 4 y JxG F x yjy D[Φ x] exp S free [Φ x,j ] Takng the log of both sdes of ths formula gves us eq. 60, whch confrms that log Z free [J] s ndeed the generatng functonal for the correlaton functons of the free feld