using D 2 D 2 D 2 = 16p 2 D 2

Transcription

1 PHY 396 T: SUSY Solutions for problem set #4. Problem (a): Let me start with the simplest case of n = 0, i.e., no good photons at all and one bad photon V = or V =. At the tree level, the S tree 0 is just a propagator while the S tree comprises two propagators with one vertex between them, cf. eqs. (5). When V happens to be chiral, the operator V = commutes with the D α operators (because [D α,] = (D α ) = 0), hence 2qS tree () = id2 D 2 6p 2 (2qi) id2 D 2 6p 2, 2 S tree () = id2 D 2 6p 2 D2 D 2 6p 2 2 = id2 6p 2 D 2 D2 D 2 6p 2 2 commuting D 2 through = id2 6p 2 D 2 using D 2 D 2 D 2 = 6p 2 D 2 = id2 D 2 6p commuting D 2 back through 2 S0 tree. (S.) Likewise, when V happens to be antichiral, the operator V = commutes with the D α operators, hence 2qS tree () = id2 D 2 6p 2 (2qi) id2 D 2 6p 2, 2 S tree () = D2 D 2 6p 2 id2 D 2 2 6p 2 2 = D2 D 2 6p 2 2 = D 2 id2 6p 2 2 = id2 D 2 6p 2 2 S0 tree. D 2 id2 6p 2 2 commuting D 2 through using D 2 D 2 D 2 = 6p 2 D 2 commuting D 2 back through (S.2)

2 Graphically, eqs. (S.) and (S.2) may be summarized as = = (S.3) Now consider the amplitudes with several good photons besides the bad one. At the tree level, diagrams contributing to the S tree n (V,...,V n ) have k n internal vertices and k + propagators. Note that k could be less then n (as long as k ) because several photons can connect to the same vertex, for example (S.4) Lets focus on one such diagram and consider all the ways we may insert one more photon V n+ into it. Clearly, there are 2k + possibilities: we may attach the V n+ to any one of the k existing internal vertices, or we may splice a new vertex (with the V n+ attached) into any one of the k + propagators, i.e. or (S.5) According to eq. (S.3), when the new photon is chiral, V n+ =, splicing it into a propagator is equivalent up to the overall sign to attaching to the vertex at the arrowhead side 2

3 of the propagator. Thanks to the sign, this means that out of 2k + possible attachments of the new photon, 2k attachments cancel each other, + = 0. (S.6) The only un-canceled insertion is the splicing on the right-most propagator whose arrowhead end is the external vertex rather than any of the k internal vertices. Consequently, V V n V V n = 0 + insertions of Vn+ = into V V n = (S.7) where the stands for any particular tree diagram for,, and n vectors V,...,V n. Summing both sides over all such diagrams contributing to the S tree n half) at the tree level, i.e. we arrive at eq. (7) (top S tree n+ (V,...,V n ;V n+ = ) = S tree n (V,...,V n ). (S.8) 3

4 The anti-chiral photons behave in a similar way. The only difference is that when we splice a new vertex with a V n+ = attached to it into an existing propagator, this is equivalent to attaching to the tail end rather than head end of the propagator. So instead of eq. (S.6) we now have + = 0. (S.9) Again, 2k out of 2k+ alternative insertions of the V n+ = cancel each other. But this time, the one un-cancelled insertion is the splicing into the left-most propagator, which is equivalent to attaching the to the external vertex rather than, V V n V V n = 0 + insertions of Vn+ = into V V n = Summing both sides of this formula over all tree diagrams contributing to the S tree n (S.0) gives the 4

5 bottom half of eq. (7) (at the tree level), i.e. S tree n+(v,...,v n ;V n+ = ) = S tree n (V,...,V n ). (S.) Quod erat demonstrandum. Problem (b): Consider a general one-loop n external-vectors diagram with k n vertices, for example (S.2) There are 2k ways to add another photon V n+ to this diagram: we may attach it to any one of the k existing vertices, or we may splice it into any one of k existing propagators. If the new photon happens to be chiral, V n+ =, then according to eq. (S.3), splicing in the middle of a propagator cancels attaching it to the arrowhead end of that propagator, for example + = 0. (S.3) Likewise, for an antichiral V n+ =, splicing it in the middle of a propagator cancels attaching 5

6 it to the tail end of that propagator, e.g. + = 0. (S.4) In either case, each propagator splicing pairs up with an attachment to an existing vertex and vice verse, so the cancellation is complete and the amplitude vanishes V V n = 0. insertions of Vn+ = or into (S.5) Herethe standsforanyparticularone-loopdiagramwithnexternal vectorlegsv,...,v n. Summing over all such diagrams gives us V V n all loop = d 4 θiv loop n+ (V,...,V n,v n+ ) 0 when V n+ = or V n+ =. V n+ (S.6) 6

7 Moreover, once we sum over all the one-loop diagrams, the amplitude is symmetric under permutations of the vector superfields V,...,V n+, so if any one of them happens to be chiral or antichiral, the amplitude must vanish. Quod erat demonstrandum. Warning: We have worked our the cancellations (S.3), (S.4), and (S.5) at the graphical level only, without paying any attention to the integrals over the loop momenta. In terms of the actual UV-regulated amplitudes V[diagram#i] = reg d 4 k i (2π) 4 F i(k i ), (S.7) our argument shows that diagrams i F i (k + i ) = 0 (S.8) before the integration, at least for some suitable identification of all the loop momenta, k i = k+ i. However, cancellation of the un-regulated integrands before the momentum integration does not guarantee that the regulated integrals cancel out, F i (k + i ) = 0 i i reg d 4 k i (2π) 4 F i(k i ) = 0. (S.9) In general, shifting the integration variables or interchanging the order of i and d 4 k are OK for only for diagrams that converge or diverge no worse than logarithmically. Unfortunately, the individual SQED diagrams with vector external legs only suffer from a quadratic UV divergence cf. homework set#3, problem#4 so we must regulate all the diagrams before shifting any momenta or combining the integrands under a common integral. This raises a big question: Do the cancellations (S.5), etc., survive the UV regularization? That is, can we find a UV regulator that would render all the diagrams finite (for finite UV ) and at the same time preserve the cancellations between the regulated integrands of related diagrams? If yes, then the Ward identities (7) and () indeed hold true in the quantum theory. Otherwise, some of those identities may be broken by the anomalies. As I have explained in class, there is a regulator preserving the U() vector symmetry of the SQED, namely a combination of supersymmetric higher-derivative term for the vector 7

8 superfield, plus a Pauli Villars compensators for the loops of chiral superfields. Thanks to the unbroken U() vector, the regulated diagrams continue to cancel each other just as suggested by the graphic argument, so the Ward identities work as advertised. Ontheotherhand, thesameregulatorbreakstheu() axial symmetry ofthemassless SQED, so the Ward identities for the axial symmetry fail at the quantum level. In terms of the axial current, this leads to the Konishi anomaly. Problem (c): Let s start by adding photonic propagators to the tree-level Ward identities of part(a). Consider a generic diagram D with a single continuous line of charged propagators, starting at the external vertex and ending at the external vertex. In general, this line is connected to n external and m internal photonic lines, and since there are no other charged lines, the diagram has L = m loops. Let s compare D to a tree diagram D which obtains from D by cutting all the photonic propagators in halves. D has n+2m external photons, but they are attached to exactly the same vertices on the charged line as in D; indeed, the charged lines of the two diagrams are identical. Consequently, both diagrams offer exactly the same places where one more external photon can be added to the diagram. As we saw in part (a), if that new photon happens to be chiral or antichiral, then most ways of adding it to the D cancel each other and we get S[D +γ ] = S[D ], S[D +γ ] = S[D ]. (S.20) The cancellations leading to this formula depend only on the charged line of the D, which is exactly the same as the charged line of the original diagram D. Hence, if we total up the ways of adding a new external photon before we integrate over the internal photons momenta, we would obtain ( ) Integrand S[D +γ ] ( ) Integrand S[D +γ ] ( ) = Integrand S[D], ( ) = Integrand S[D] (S.2) in exactly the same way as in part (). And since the momentum integrals for these diagrams have only a logarithmic UV divergence (cf. homework set#3, problem #4), we may total the 8

9 integrands before the momentum integration, hence S[D +γ ] = S[D], S[D +γ ] = S[D] (S.22) as in eqs. (7). For example, consider the one-loop amplitude for n = 0 external photons which comes from two diagrams S loop 0 = + (S.23) and all the ways one can add an external photon to these diagrams, S loop (V) = + V V + + V V 9

10 + + V V + + (S.24) V V When the vector field V happens to be chiral, V =, the pairs of diagrams on the last three lines of (S.24) cancel each other exactly as in eq. (S.6). Also, the two diagrams on the top line of (S.24) survive but lose their right-most propagators, thus S loop (V = ) = (S.25) = S loop 0. Similar arguments apply to eqs. (0) for amplitudes without external or legs. Indeed, consider an L+ loop diagram with n external and L internal photonic lines and a single closed loop of charged lines. If we cut all the internal photon lines in halves, we would get a one-loop diagram with n + 2L external photons. Again, both diagrams offer exactly the same places where you can attach another external photon, and if that new photon happens to be chiral or antichiral, alternative insertions would cancel each other in exactly the same way. In part (b) we saw that for the one-loop diagram, the cancellation was complete, so exactly the same proof applied to the L+ loop diagram would show that it also satisfies eq. (0). 0

11 Finally, consider a multi-loop diagram with several charged loops connected to each other by photonic propagators. If we want to add another external photon to this diagram, we may attach it to any one of those charged loops. Also, each loop offers several places for attachment of the new photon, and we must sum over all the alternatives. This summation works similarly to the one-loopdiagrams in part (b), so if the new photon happens to be chiral or antichiral, the alternative attachments to any particular charged loop cancel each other and we get a big fat zero. This completes the proof of the Ward Takahashi identities (0) for the general multi-loop amplitudes. Similarly, consider a general multi-loop diagram with two charged external legs and. In general, the charged lines of such a diagram forms several distinct continuous lines: one open line from to and several closed loops, all connected to each other by photonic propagators. If want to add a new external photon to the diagram, we may attach it to any charged line, and then we have a choice of vertices and propagators on that line: The open line works as in part (a), and the closed loops as in part (b). So when the new photon happens to be chiral or antichiral, we get a massive cancellation when we sum over alternative attachments within any particular line. For any closed charged loop this cancellation is complete, but for the open charged line there is one left-over un-canceled alternative: splicing the new photon into the external line attached to or (depending on whether V new = or.) And as in part (a), this is equivalent to attaching the directly to the or the directly to the, thus S n+ (V,...,V n ;V n+ = ) = S n (V,...,V n ), S n+ (V,...,V n ;V n+ = ) = S n (V,...,V n ). (7) Quod erat demonstrandum. Problem 2(a): Let s start with eqs. (3). In a diagram without external charged lines or, all charged propagators form closed loops, so cutting any one such propagator cannot severe the loop. This means that the one-particle-irreducibility of a diagram depends only on the internal photons connecting the charged loops to each other, but not on the charged loops themselves or on the external photons. So if we add a new external photon to an PI diagram, we would always get a PI diagram regardless of where we attach the new photon, OOH, no such attachment would change a non-pi diagram into a PI diagram.

12 In parts (b) and (c) we have proved the Ward identities () not just for the net n-vector amplitudes but also for for partial sums over much smaller sets of diagrams, namely over all possible attachments of a bad photon to any particular n diagram. Since all such sets are either all-pi or none-pi, summing only over the PI sets immediately gives us eqs. (3) for the net PI n-vector amplitudes V PI n (V,...,V N ). Next, let s work out eqs. (5) and (6). Beyond the tree level, the un-amputated S 0 amplitude is the dressed chiral propagator is 0 (p) = + PI + PI PI + = iπ a + iπ a iγ 0 iπ a + iπ a iγ 0 iπ a iγ 0 iπ a + = iπ a + Π a Γ 0 (p)π a + ˆΓ 0 (p) id2 D 6p 2 +i0 where ˆΓ 0 = Π a Γ 0 Π a is indistinguishable from the PI 2 point bubble Γ 0 itself when sandwiched between the chiral on the left and the antichiral on the right. The un-amputated vector amplitude is (V) comprises one PI dressed vertex iγ (V) and two dressed charged propagators, 2 (7) = PI (S.26) i.e., is (p,p 2 ;V) = is 0 (p ) iγ (p,p 2 ;V) is 0 (p 2 ). (S.27) When the vector field V happens to be chiral, eq. (4) tells us that S (p,p 2 ;V = ) = S 0 (p ). (S.28) 2

13 Combining this formula with eq. (S.27) and eq. (7) for the S 0, we have and consequently Π a + ˆΓ 0 (p ) Γ Π a (p,p 2 ;V = ) + ˆΓ 0 (p ) = + Π a + ˆΓ (S.29) 0 (p 2 ) Π a Γ (p,p 2 ;V = ) Π a = (+ ˆΓ 0 (p 2 )). (S.30) In the context of d 4 θγ (V), this formula is equivalent to Γ (p,p 2 ;V = ) = (+Γ 0 (p 2 )). (5a) Similarly, when the vector field V happens to be antichiral, we have S (p,p 2 ;V = ) = S 0 (p 2 ) (S.3) and consequently Π a + ˆΓ 0 (p ) Γ Π a (p,p 2 ;V = ) + ˆΓ 0 (p 2 ) = + Π a + ˆΓ 0 (p 2 ) (S.32) Π a Γ (p,p 2 ;V = ) Π a = (+ ˆΓ 0 (p )), (S.33) which is equivalent to Γ (p,p 2 ;V = ) = +Γ 0 (p ). (5b) Finally, let s prove eqs. (6) for multi-vector amplitudes. The two-vector un-amputated amplitude comprises two kinds of diagrams, = PI + PI PI (S.34) 3

14 while for n > 2 vectors there are even more possibilities: n n k n k = PI + PI PI k k 2 n k k 2 + PI PI PI + n k n k = PI + PI (S.35) The last line in this graphic summary gives a recursive formula for the S n in terms of the PI amplitudes and the S k with k < n. Specifically, S n (V,...,V n ) = S 0 Γ n (V 0,...,V n ) S 0 n k=partitions S k (kvectors) Γ n k (n kvectors) S 0 (S.36) where the second sum is over which k vectors out of V,...,V n appear in the S k while the rest appear in the Γ n k ; the minus sign on the RHS follow from the i factors in each amplitude, dressed propagators, or dressed vertex. Eqs. (S.36) give rise to an induction-in-n proof that eqs. (7) imply eqs. (6) for the PI amplitudes. To prove the induction base for n = 2, consider what happens to the two-vector amplitude S 2 (V,V 2 ) = S 0 Γ 2 (V,V 2 ) S 0 S 0 Γ (V ) S (V 2 ) S 0 Γ (V 2 ) S (V ) (S.37) when one of the vectors happens to be chiral, say V 2 =. In this case, in the second term on the RHS we have S (V 2 = ) = S 0 while for the third term eqs. (5a) gives us 4

15 Γ (V 2 = ) = + (+Γ 0 ). At the same time, on the LHS we have S(V,V 2 = ) = S (V ) = +S 0 Γ (V ) S 0. (S.38) Plugging all these formulae into eq. (S.37), we obtain +S 0 Γ (V ) S 0 = S 0 Γ 2 (V,V 2 = ) S 0 + S 0 Γ (V ) S 0 (S.39) S 0 (+Γ 0 ) S (V ) where the second term on the RHS cancels against the LHS. The remaining terms give us S 0 Γ 2 (V,V 2 = ) S 0 = S 0 (+Γ 0 ) S (V ) = +S 0 (+Γ 0 ) S 0 Γ (V )S 0 (S.40) where in the context of d 4 θγ, Γ 2 (V,V 2 = ) = (+Γ0 )S 0 Γ (V ) = (+ ˆΓ0 )S 0 Γ (V ) since is chiral just like = Π a Γ (V ) in light of eq. (7) (S.4) = Γ (V ) in accordance with eq. (6). That was the induction base for n = 2. Now, we need to prove that if eqs. (6) holds true for all n < n then they also hold true for n. Our starting point is the recursion formula (S.36) for the n-vector amplitude S(V,...,V n ). Suppose V n =, and let s split the sum over the partitions of n vectors (into k arguments of the S k and n k arguments of the Γ n k ) into two sums according to whether the V n = is an argument of the S k or the Γ n k, thus S n (V,...,V n ;) = S 0 Γ n (V 0,...,V n ;) S 0 n k=selections n k=selections S k (V,...,V) Γ n k (V,...,V;) S 0 S k (V,...,V;) Γ n k (V,...,V) S 0. (S.42) 5

16 By the induction assumption, on the second line here S k (V,...,V) Γ n k (V,...,V;) S 0 = +S k (V,...,V) Γ n k (V,...,V) S 0, (S.43) while on the third line of (S.42) we may use eq. (7) to rewrite S k (V,...,V;) Γ n k (V,...,V) S 0 = S k (V,...,V) Γ n k (V,...,V) S 0. (S.44) Due to opposite signs in eqs. (S.43) and (S.44), these terms cancel each other for appropriate partitions of the (V,...,V n ;) into arguments of the S k and Γ n k. In the context of the big sums in eq. (S.42), the only terms that survive this cancellation are the k = n term on the second line and the k = term on the third line, thus S n (V,...,V n ;) = S 0 Γ n (V 0,...,V n ;) S 0 S n (V,...,V n ) Γ () S 0 (S.45) S () Γ n (V,...,V n ) S 0. Applying eqs. (7) to the LHS here and to the S () in the third term on the RHS, and also applying eq. (5) to the second term on the RHS, we rewrite eq. (S.45) as S n (V,...,V n ) = S 0 Γ n (V 0,...,V n ;) S 0 S n (V,...,V n ) (+Γ 0 )S 0 + S 0 Γ n (V,...,V n ) S 0 (S.46) where the LHS and the second term on the RHS cancel each other in the context of an operator between and, S n (V,...,V n ) = S n (V,...,V n ) (+Γ 0 )S 0. (S.47) Consequently, S 0 Γ n (V 0,...,V n ;) S 0 = +S0 Γ n (V,...,V n ) S 0 (S.48) and hence in the context of d 4 θγ, Γ n (V 0,...,V n ;) = Γ n (V,...,V n ). (6a) This completes the proof-by-induction of eqs. (6a) for the V n =. The antichiral case V n = is left out as an exercise for the students. 6

17 PS: Strictly speaking, the gray bubbles labeled PI in eq. (S.35) are only one-charged particle irreducible, but they might be severed by cutting a vector propagator, for example PI PI PI (S.49) However, when one adds a bad photon V n+ = or to a diagram like this, the bad photon has to be attached to the lower bubble; thanks to eqs. (3), attachments to the upper bubbles cancel out. Consequently, as far as eqs. (6) for the diagrams like (S.49) are concerned, we may disregard the upper bubbles with all the vectors attached to them and focus on the lower bubble only, thus PI = PI (S.50) In other words, once we have proved the Ward identities (6) for the one-charged particle irreducible amplitudes, the completely-pi amplitudes must also satisfy the same identities(6). 7

18 Preamble to problems 2(b) and 2(d): Consider a divergent PI amplitude with superficial degree of divergence 0. In the ordinary perturbation theory, the infinite part of the amplitude must be a polynomial function of the external moments, of degree. The reason for this is simple: For each power of an external momentum we extract from the integral over the loop momenta, we have one less power of an internal momentum in the numerator inside the integral, which reduces the degree of divergence. Consequently, all the kext n terms for k > have convergent coefficients. For the superfield Feynman amplitudes there is a similar rule: the infinite part of a divergent amplitude is an d 4 θ of some local combination of external fields and no more then 2 spinor derivatives D α and D α acting on them. Also, the divergence of a term involving 2k spinor derivatives is no worse than (UV cutoff) n k (or log(cutoff) for k = n). The reason is similar to the ordinary perturbation theory: For every pair of D α and D α acting on the external lines of the diagram, there are fewer spinor derivatives available to generate the momenta in the numerator via {D α,d α } = 2p α α. Problem 2(b): In the previous homework (set#3, problem#3) we saw that all the PI amplitudes (9) have superficial degree of divergence = 0. Consequently, the infinite parts of those amplitudes may not involve any spinor derivatives acting on the external charged or vector fields, so all we have inside the d 4 θ is the product of the fields themselves, V V 2 V n PI = O(log(cutoff)) i(2q) n d 4 θv V 2 V n + finite. (S.5) We may cancel all such divergences with a series of local counterterms, L counter terms ( d 4 θ δ 2 + n= ) (2qV)n, δ 2,δ (n) = O(log(cutoff)) constants. n! δ (n) (S.52) 8

19 In terms of the Γ n, eqs. (S.5) and (S.52) mean Γ loops 0 = δ 2 + finite, Γ loops n (V,...,V n ) = δ (n) V V 2 V n + finite(v,...,v n ). (S.53) Note that the infinite parts here do not care whether the photons V,...,V n are good or bad, so when the V n happens to be chiral or antichiral we also have Γ loops n (V,...,V n ;V n = or ) = δ (n) V V n ( or ) + finite. (S.54) On the other hand, according to eq. () Γ loops n (V,...,V n ;V n = ) = Γ loops n (V,...,V n ) = δ (n ) V V n + finite, (S.55) and the only way both formulae can be true at the same time is to have δ (n) = δ (n ) and hence all δ (n) = same δ n. (S.56) Also, for n = while at the same time Γ loops () = δ + finite (S.57) Γ loops () = ( +Γ loops 0 ) = ( δ 2 +finite) = δ 2 + finite, hence δ = δ 2 and all δ (n) = δ 2. (S.58) In terms of the amplitudes Γ n, these Ward identities may be summarized as (2q) n Γ loop n (V,...,V) = δ 2 exp(2qv) + finite. (9) n! n= In terms of the counterterms (S.52), the Ward identities (S.58) mean that we may cancel 9

20 an infinite number of divergent amplitudes with a single gauge-invariant counterterm L counter terms δ 2 d 4 θe 2qV. (S.59) In general, there is a separate δ 2 for every charged chiral superfield. However, the SQED theory () is symmetric with respect to the charge conjugation A B, V V, which assures δ (A) 2 = δ (B) 2. Consequently, the entire renormalized Lagrangian for the charged fields can be summarized as L ren = L classical + L counter terms (Z = +δ 2 ) ( ) d 4 θ Ae +2V A + BE 2V B. (20) Problem 2(c): Eqs. (3) tell us that the all-vector PI amplitudes are invariant under SUSY gauge transformations V V + i i. Consequently, these amplitudes may be written in terms of the gauge-invariant tensor superfields W α = 4 D2 D α V and W α = 4 D2 D α V. For example, we saw in class that the two-photon amplitude has form PI = (2q)2 i 2 d 4 θf(k) W α (k)w 2α ( k), (S.60) or in terms of eq. (3) V2 PI (V,V 2 ) = F(k) V D α D 2 D α V 2. (S.6) 8 Note four spinor derivatives acting on the external vector fields in this formula. Similarly, the four-photon amplitude is required by gauge invariance to have form PI = d 4 θg(momenta) (WW)(WW) 4 + d 2 θh(momenta) (WW)2 8 + H.c., (S.62) 20

21 or in terms of eq. (3), V PI 4 (V,...,V 4 ) = G 4 Wα W 2βW 3 βw β 4 H 32 (Dα V )W 2α W β 3 W 4β H 32 (D αv )W α 2 W 3 + permutations of photons,2,3,4. β W β 4 (S.63) On the top line here we have 2 spinor derivatives acting on the V superfields 3 derivatives for each W α or W α. On the second line each term has only only 0 spinor derivatives, but that s because an overall D 2 or D 2 was lost while converting a d 2 θ or d 2 θ integral into the d 4 θ form. But we cannot have any fewer spinor derivatives without losing the gauge invariance. Similarly, an n photon gauge-invariant amplitude must involve at least 3n 2 spinor derivatives acting on the vector superfields V,...,V n. Problem 2(d): In the preamble (back on page 8) we saw that the infinite part of an amplitude with superficial degree of divergence may have no more then 2 spinor derivatives acting on the external fields; allterms withmorespinor derivatives must befinite. Fortheall-vector amplitudes = 2 regardless of the number n of the vectors (cf. homework set#3, problem#4), but in part (c) we saw that such amplitudes always involve at least 3n 2 spinor derivatives. For n > 2 this is more than 2 = 4 derivatives, hence all the multi-vector PI amplitudes are UV-finite and do not need any counterterms. As to the two-vector amplitude, it has only 4 spinor derivatives so its effective degree of divergence is 4/2 = 0. In other words, the two-vector amplitude has a logarithmic UV divergence. As explained in the preamble, the infinite part of a log-divergent amplitude is independent of the external momenta. In terms of the function F(k) in eq. (S.6), this means F(k) = ( ) δ 3 = O(log(cutoff)) constant + a finite function(k). (S.64) Clearly, we may cancel this divergence using a local gauge-invariant counterterm L counter term δ 3 d 4 θv Dα D 2 D α V 8 (S.65) which looks precisely like the tree Lagrangian for the vector field. Consequently, the renormal- 2

22 ized Lagrangian becomes L ren V = L tree V + L CT V = ( ) g 2 + δ 3 d 4 θv Dα D 2 D α V. (7) 8 Quod erat demonstrandum. 22