4 QUANTIZATION OF GAUGE THEORIES

Transcription

1 4 QUANTIZATION OF GAUGE THEORIES 4.1 Vector fields in the Coulomb gauge We have learned that to develop a perturbative approach to QFT for gauge theories one needs to fix a gauge lest the propagator not exist. Among the several choices of gauge, it is instructive to consider the Coulomb gauge as it allows for an explicit removal of the redundant degrees of freedom. In the Coulomb gauge one extracts out the longitudinal modes of the field from the very start. The gauge condition is A = 0. (354) Then if A = A+ Λ ( A+ Λ) = 0 (355) and Λ = 1 2 A = not very different from the situation in the Lorenz gauge altough the last expression needs a more careful handling. Let us remember the canonical conjugate momenta (the lagrangian does not depend on Ȧ0) and d 3 x 1 4π x x A(x ), (356) Λ = 1 µa µ, (357) π 0 = 0 (358) π i = δl M δ A i = 0 A i + i A 0 = Ȧi i A 0 = F 0i = E i. (359) The lagrangian L M = 1 4 F2 = 1 2 F2 0i 1 4 F2 ij (360) can be written, introducing E i as an independent field as L M = 1 2 E2 i E i F 0i 1 4 F2 ij. (361) By using the equation of motion E i = F 0i (362) one recovers the previous lagrangian. 60

2 Note the following conventions: ( i ), E (E i ). Using (361) and replacing F 0i by its expression in terms of the four-potential, the lagrangian of QED can be written after integration by parts as L M = E A A0 E E F2 ij + ψ(i D m)ψ (363) = E A+ 1 2 ( E 2 B 2 )+L(ψ, A) A 0 ( E e ψγ 0 ψ), (364) showing that A 0 is a Lagrange multiplier that provides the constraint E = eψγ 0 ψ = ρ (365) i.e. Gauss law. Note that the previous reasoning is actually independent of the gauge chosen. However, in Lorenz gauge A 0 does appear in the gauge condition itself and so it cannot disappear so easily from the theory. Therefore if we count degrees of freedom, we see that in the Coulomb gauge A 0 drops out as a dynamical variable. In addition, the longitudinal mode can be eliminated by a gauge transformation as we saw at the beginning of this section thus leaving only 2 degrees of freedom. We can separate E = E T + E L where E T = 0 (this defines E T ) and E L = ρ. Solving for E L EL = 1 2ρ, (366) we see that it is fully determined and the following term appears in the lagrangian 1 2 E 2 L = e2 8π d 3 xd 3 y ( ψγ 0 ψ)(x)( ψγ 0 ψ)(y). (367) x y This is an instantaneous four-fermion Coulom interaction (but the whole theory is of course consistent with special relativity). Next we impose the canonical commutation relations. Naively we require the following ETC [A i ( x,t),π j ( y,t)] = iδ j i δ(3) ( x y) (368) but this is obviously impossible because if we take the divergence on the l.h.s. we do not get zero on the r.h.s. This problem is fixed by replacing the previous ETC with [A i ( x,t),π j ( y,t)] = iδ j i δ (3) ( x y), (369) where δ ij δ(3) ( x y) = d 3 k k( x y) (2π) 3ei (δ ij k ik j ). (370) k 2 61

3 The mode decomposition of the field will be A(x) = d 3 k (2π) 3 [ ǫ( 2E k,λ)a( k,λ)e ikx + ǫ ( k,λ)a ( k,λ)e ikx ]. (371) k λ To ensure the gauge condition the polarization vectors must verify k ǫ( k,λ) = 0 A = 0. (372) The creation and annihilation operators are given by a( k,λ) = i d 3 xe ikx ǫ( k,λ) 0 A(x) (373) and an analogous expression for a ( k,λ). Then the following commutation relation holds [a( k,λ),a ( k,λ )] = (2π) 3 2k 0 δ (3) ( k k ). (374) All states are positive norm states. The issue of negative norm states does not exist in the Coulomb gauge beacuse A 0 can be fully eliminated from the theory. It is quite instructive to compare this gauge fixing procedure with the one in the Lorenz gauge. 4.2 Covariant quantization and the Faddeev-Popov procedure QED The Coulomb gauge is useful in understanding the decoupling of unphysical degrees of freedom, but it is clearly difficult to handle because of the lack of Lorentz invariance. Also, it is more easy to implement in the operator formalism than in the path-integral formulation of QFT. In path integrals it becomes obvious that in order include quantum fluctuations of the theory one must select a gauge. The bilinear part in the A µ field is in Yang-Mills 1 2 Aa µ (k2 g µν k µ k ν )A a ν 1 2 Aa µ Mµν A a ν ; (375) M µν cannot be inverted to find the propagator. The way out is to add the piece Then M µν = k 2 g µν (1 1 a )kµ k ν 1 2a ( µ A a µ) 2. (376) (M 1 ) µν = gµν (1 a) kµ k ν k 2 k 2 +iǫ. (377) We can now write Feynman diagrams. The added term(376) breaks the local gauge symmetry which, generally speaking, is only recovered for S-matrix elements. 62

4 How can be sure, though, that by adding the gauge breaking term we are not unduly forcing the theory? We shall present here a proof for the somewhat simpler case of an abelian gauge theory and then comment the differences in the non-abelian case. The generating functional is Z[J] = N [da µ ]exp S[J] S[J] = d 4 x( 1 4 F µνf µν +A µ J µ ) (378) If we stick to an abelian theory (e.g. electromagnetism), F µν F µν = ( µ A ν ν A µ ) 2. We decompose A T µ = PT µν A ν, A L µ = PL µν A ν (379) with (in momentum space) P T µν = g µν k µk ν k 2 P L µν = k µk ν k 2. (380) Then S = d 4 x 1 2 ( µa T ν) 2 (381) and therefore the integral over A L µ diverges. Note that a gauge transformation A µ A µ = A µ µ θ changes A L µ but not A T µ To avoid this the functional integral must only include one element of each gauge orbit. We want (for instance) to select those representants of the gauge equivalence class that fulfill some condition. For instance F(x) = µ A µ (x) f(x) = 0. (382) We can write da T = da T dfδ(f), (383) or [da T ] = [da T ][dθ]det δf δ(f), (384) δθ where θ is a gauge transformation function. Under such a transformation F F θ, and the previous integral equals [da µ ]det( )δ( µ A µ f), (385) where obviously dθ = da L µ. We can write Z then as Z = N [df][da µ ]det( )δ( µ A µ f)g[f]exp S, (386) 63

5 where G[f] is an arbitrary functional of f. For instance G[f] = exp 1 d 4 xf 2. (387) 2a Then Z = N [da µ ]det( ) exp( S 1 2a Note that det( ) is field-independent. We shall denote d 4 x( µ A µ ) 2 ). (388) FP = det( ). (389) Non-abelian gauge theories Now we can attemp to reformulate the same procedure for non-abelian theories where the identification of the gauge degrees of freedom is not so directly straightforward. We write 1 = FP (A) dωδ(f Ω (A)) (390) with F Ω (A) F(A ), where A is the gauge transformed of A by Ω. The factor FP, which we shall determine later, ensures the correct measure. Here dω is the invariant group measure (we have already seen how to define one for SU(2) and similar constructions exist for other groups). Note that FP is gauge invariant: a gauge transformation of A can becompensated by a change in Ω. FP (A) = FP (A Ω ). (391) Inserting 1 into the functional we have da µ FP dωδ(f Ω (A))e S. (392) Now let us make a gauge transformation in the path integral over A µ : A µ Aµ Ω 1. The only factor that is not gauge invariant is δ(f Ω (A)). The previous integral becomes ( dω) da µ FP δ(f(a))e S, (393) and the piece dω (394) can be factored out and the overcounting is removed. The problem is thus reduced to finding an explicit expression for FP. This determinant can be found from the relation δ(f Ω (A)) = δ(ω Ω 0 )det δfω δω 1, (395) 64

6 where Ω 0 solves the Dirac delta. i.e. In order to compute this determinant in general let us expand FP = det δfω δω F(A)=0. (396) F θ (A) = F(A)+ d 4 ym(x,y)θ(y)+... (397) and FP = detm. (398) We already know from the QED case that the determinant is actually independent of A and it thus can be dropped from calculations; it does not couple to anything. The situation is different in non-abelian theories. Let us take for instance the Lorenz gauge µ A a µ = 0. Then M ab (x,y) = δ δθ a (x) µ A b µ θ (y) = δ 1 δθ a (x) g µ ( µ θ b +gf bcd A c µ θd )(y) (399) = 1 g ( µ ( µ δ ab gf abc A c µ))δ (4) (x y) (400) = 1 g ( µd ab µ )δ(4) (x y). (401) We can rescale away the 1/g overall factor and use a pair of anticonmuting variables c,c to write detm = d cdce ca µ D ab µ c b. (402) according to the rules of fermion integrations. c, c are ghosts. They have a boson-looking lagrangian, but they anticommute. Finally, we can raise the gauge condition to the exponent in the same way as we did for QED, thus adding a piece to the action. 1 2a ( µ A a µ )2 (403) 4.3 Ghosts in Yang-Mills, Feynman rules and unitarity As we have seen ghosts are unnecessary in QED and other abelian theories because the Faddeev-Popov determinant is independent of the dynamical fields. However this is not the case and ghosts need to be included as any other field in the calculation. 65

7 We shall write the Feynman rules in Minkowski space conventions to which we stick in this section. In order to extract the Feynman rules we decompose the action into a free and an interacting part L 0 = 1 4 ( µa a ν νa a µ )2 1 2a ( µ A a µ )+ ca c a, (404) L I = 1 2 g( µa a ν νa a µ )fabc A bµ A cν + g2 4 fabc f ade A b µ Ac ν Adµ A eν +ig µ c a f abc A c µ cb. (405) From this expression we immediately observe the presence of a vertex involving ghosts and vector fields. The associated Feynman rules are: Ghost propagator: Ghost-vector boson: i ab (k) = iδ ab 1 k 2 +iǫ (406) iγ abc µ = gf abc k µ, (407) where k µ is the momentum of the outgoing ghosts. Note that due to their anticommuting nature ghost loops acquire a (-1) factor. As we have seen, ghosts are produced as we quantize the theory, they are a quantum phenomenon and as such they need not appear in tree diagrams at all; they only make their presence in loop diagrams through internal loops. However, this statement needs some qualification. In QED there is exact cancellation between the longitudinal and the temporal degrees of freedom. Therefore it makes no difference for the external photon states to sum over all polarizations or only over the two physical ones. Let us see this. On top of fixing the gauge, one has to define what is meant by a physical state as there are states of negative norm, namely those created by temporal photons as Lorentz covariance forces upon us the canonical commutation relations or ETC (Equal Time Commutators) that eventually leads to δ(x 0 y 0 )[A µ (x),π ν (y)] = iη µν δ (4) (x y) (408) [a( k,λ),a ( k,λ )] = η λλ 2E k δ (3) ( k k ), (409) showing clearly the presence of states of negative norm in the Hilbert space. To eliminate these states one usually forces upon states the Gupta-Bleuler quantization condition. Physical states satisfy ( µ A µ (k)) (+) Ψ = 0 (410) 66

8 where the symbol (+) indicates that only annihilation operators are kept. Then taking e.g. k ẑ one gets (ai a( k,λ i )) k(a 0 a 3 ) Ψ = 0 (411) or a 0 Ψ = a 3 Ψ. Then because the QED hamiltonian can be written as H = 1 2 d 3 k (2π) 3( i=1,2,3 a a a 0 a 0) (412) it is clear that in the evolution of Ψ the two polarizations cancels exactly each other. However the situation is different in non-abelian Yang-Mills theories and there is no such exact cancellation. The sum has to be done only over physical polarizations. If we insist on including all four polarizations, then external states with ghost lines have to be included. This is clearly sen recalling that the total cross-section of a production is related to the imaginary par of the forward amplitude. Let us separate the S matrix as S = I +it. (413) The identity corresponds to the rather trivial case where there is no interaction at all. Unitarity implies S S = I = I +i(t T )+T T. (414) This implies i(t T ) = T T. (415) Recalling the definition of the reduced S-matrix element f T i = (2π) 4 δ (4) (p f p i )M i f, (416) f T i = (2π) 4 δ (4) (p f p i )Mf i, (417) we can write this as M i f M f i = i(2π) 4 n δ (4) (p n p i )M f nm i n. (418) The above relation thus provides non trivial constrains between different orders in perturbation theory. Another consequence of this result is the following. Take f = i, then ImM i i = 1 2 (2π)4 n δ (4) (p n pi) M i n 2. (419) 67

9 The quantity in the r.h.s. is related to the total cross-section. This is the contents of the socalled optical theorem: the imaginary part of the forward scattering amplitude is proportional to the total cross-section. Then it is clear that a second order in the coupling constant, O(g 2 ), ghost loops need to be included in the r.h.s. 4.4 BRST symmetry and BRST quantization Let us rewrite the abelian QED lagrangian in a slightly different way (Minkowski notations used) L M = 1 4 F µνf µν b( µ A µ )+ a 2 b2 µ c µ c. (420) Using the equations of motion for the auxiliary field b(x) one gets and recovers the usual expression for the lagrangian. b = 1 a µ A µ (421) It is easy to see that the above (gauge-fixed) lagrangian is invariant under the following set of transformations δa µ = ǫ µ c, δ c = ǫb, δc = 0, δb = 0. (422) Note that ǫ must be a constant Grassmann variable for consistency. This transformation is thus a global symmetry. Likewise for non abelian theories one can write the lagrangian as L M = 1 4 Fa µν Faµν b a ( µ A a µ )+ a 2 b2 µ c a D µab c b. (423) Again, using the equation of motion for b a (x) one gets the already familiar gauge-fixed Yang- Mills lagrangian. This lagrangian is invariant under δa a µ = ǫd ab µ c b, δ c a = ǫb a, δc a = ǫ 2 gfabc c b c c, δb = 0, (424) which is the non-abelian version of the previous one. This invariance is called BRST symmetry. Sometimes (very rarely) is referred to as gauge invariance of the second kind. The auxiliary field b(x) is called the Nakanishi-Laudrup field. It is necessary for a linear realization of the BRST symmetry. Note that BRST symmetry is the remnant of the gauge symmetry after the lagrangian has been gauge fixed and hence gauge invariance, as such, has been lost. In this sense it is related to a residual gauge symmetry. 68

10 Since this is a symmetry there is a conserved current J µ BRST = D νf νµ c g 2 µ c (c c)+b D µ c (425) and a conserved charge Q BRST = d 3 x[d ν F ν0 c g 2 0 c (c c)+b D 0 c] (426) that generates the previous transformations [ǫq BRST,φ] = iδ BRST φ. (427) In addition there is another global symmetry δc = θc, δ c = θ c, (428) with θ a constant bosonic quantity. This leads to the conserved current J µ ghost = µ c c+ c D µ c (429) and charge Q ghost = d 3 x( c D 0 c 0 c c). (430) This leads to ghost number conservation. These two charges fulfill the following set of relations (on-shell) Q BRST = Q BRST, Q 2 BRST = 0, (431) [Q ghost,q BRST ] = iq BRST, [Q ghost,c] = ic, [Q ghost, c] = i c. (432) It is particularly important the nilpotency property of Q BRST as it will be evident in the following. This property follows from the definition of the BRST symmetry implying that δ 2 BRST 0. It holds only on-shell. We shall assume that the vacuum respects BRST symmetry, namely that this global symmetry is not spontaneously broken. This is in fact tantamount to saying that gauge symmetry is never broken spontanously, something that can be proven explicitly: the expectation value of any gauge non-invariant operator is always zero, as it is fairly intuitive. Let O(A) be a gauge-invariant operator 0 δ BRST O(A) 0 = 0. (433) On the other hand 0 δ BRST O(A) 0 0 [Q BRST,O(A)] 0, (434) 69

11 and from this by simply choosing O = 0 n, with n 0, it follows that n Q BRST 0 = 0 (435) guaranteeing that there is no spontaneous symmetry breaking of the BRST symmetry. On one-particle states BRST acts in the following way Q BRST A µ (k) k µ c(k), Q BRST c(k) = 0, Q BRST c(k) k µ A µ (k), (436) where the result is projected onto 1-particle states too. By the previous notation we mean A µ (k) = A µ (k) 0 = λ ǫ µ (λ)a ( k,λ) 0, (437) and so on. If we take the third expression in (436) and apply Q BRST again using the first one we see that Q 2 BRST = 0 on one particle states if and only if k2 = 0. Nilpotency of the BRST charge (which can be proven only on shell) requires that vector bosons are massless (also on shell) which is a necessary condition for gauge invariance, and viceversa. In general the physical space generated by creation and annihilation vector meson operators acting on the vacuum is not a positive definite Hilbert space. This is of course a well known fact in covariant gauges A a µ (k) Ab ν (k ) = η µν δ ab 2k 0 (2π) 3 δ (3) ( k k ). (438) We shall postulate that physical states Ψ are those satisfying Q BRST Ψ = 0. (439) In particular this implies that the vacuum is physical according to this definition. The general solution of this equation has the form (always restricted to 1-particle states) Ψ(k) = ξ µ A µ (k) +λ c(k), ξ k = 0, k 2 = 0, k 0 > 0. (440) One can prove that this implies that Ψ Ψ 0!. In the above expression λ is a Grassmann parameter. We should think of ξ µ as a polarization vector; when contracted with the λ ǫ µ(λ)a (λ) in the definition of A µ (k) it projects on a given polarization, characterizing a physical state. Note that the condition on ξ reflects what we naturally expect in the Lorenz gauge. While the space of physical states is of semidefinite positive norm, there are many states of zero norm. Indeed any state of the form Φ = Q BRST Φ 0 (441) 70

12 is automatically of zero norm due to the nilpotency of the BRST charge. In fact it can be proven that all zero-norm states are necessarily of this form. Thus the Hilbert space of positive definite norm H will be the physical space H phys modded out by the spaces of zero norm H 0. In other words, the genuine states are equivalence classes where Ψ and Ψ +Q BRST Φ are identified. Identifying Q BRST with a differental operator, the elements of H would be cohomology classes. Continuing with this analogy, physical states would be closed forms, while zero norm states would correspond to exact forms. It is instructive to investigate further what this identification of states modulo zero norm states means in slightly more familiar terms. If Ψ = ξ µ A µ +λ c(k). (442) To get the compatibility of ghost numbers we write Φ = τ µ A µ +σ c(k), (443) with τ µ of fermionic character and ghost number -1. Then a new physical state would be Ψ = Ψ +Q BRST Φ (444) and Q BRST Φ = ασk µ A µ (k) +βτ µ k µ c(k), (445) with α and β being some constants. This amounts to redefining ξ µ ξ µ +ασk µ, (446) which corresponds to the usual residual gauge invariance that allows to eliminate one of the degrees of freedom, provided that k 2 = 0. In the Gupta-Bleuler formalism, physical states are defined by the condition ( µ A µ ) (+) Ψ = 0, (447) where the (+) indicates the positive frequencies (annihilation operators). It is not difficult to see that this is equivalent to the Q BRST Ψ = 0 condition. This follows from (440) ( µ A µ ) (+) [ξ ν A ν +λ c(k) ] = ξ ν ( µ A µ ) (+) A ν k ξ = 0, (448) where the commutation relations between creation and annihilation operators have been used. Due to the fact that [H,Q BRST ] = 0, if we assume that only physical states are considered as in states, the positive definite character of the Hilbert space is preserved at all times. Note that we have considered only one-particle states. This is justified for initial states where assumed to factorize, valid if interactions decay sufficiently fast or the initial individual states are sufficiently separated. 71

13 4.5 Ward identities Gauge symmetry transform a (gauge-noninvariant) Green function into another one (or several ones). Therefore gauge invariance provides a number of relations amongst Green functions. These relations play a pivotal role in e.g. proving the renormalizability of the theory. To begin with we shall omit the matter fields (we will see later how the treatment has to be modified when we add scalar fields that mix with the longitudinal polarization of vector bosons in a broken theory). We include source terms for the ghosts as well as for the gauge field in the generating functional Z[J,σ, σ] = N [da][d c][dc] exp S Let us start, as usual, with an abelian theory S = d 4 x[j µ A µ + σc+ cσ] (449) d 4 x[ 1 4 F2 1 2a ( A)2 µ c µ c]. (450) As we already know this gauge-fixed action has a residual global symmetry under the BRST transformations δa µ = ǫ µ c, δ c = ǫ 1 A, δc = 0. (451) a The e-o-m for the b field has been used. We now make this change to the integration variables in Z[J,...]. Obviously the functional integral itself does not change. In addition the measure of integration does not change either because the jacobian matrix is [da ][d c ][dc ] = [da][d c][dc] (452) 1 ǫ 1 a µ (453) ǫ µ 0 1 whose determinant is 1. The action does not change either, so all the changes must be in the source terms 0 = δz = N [da][d c][dc](j µ µ c+ 1 a µ A µ σ)exp S d 4 x[j µ A µ + σc+ cσ] (454) = J µ µ c+ 1 a µ A µ σ J, σ,σ. (455) The vanishing of this full Green function obviously extends too to its connected part, as a similar argument holds for W[J,...]. 72

14 Let us now recall the definition of the effective action (the superindex C stands for the classical fields Γ[A C µ, c C,c C ] = W[J, σ,σ] The vanishing of the previous connected Green function implies d 4 x(j µ A C µ + σc C + c C σ). (456) (J µ δ µ δ σ + 1 a σ δ µ δj µ)w = δ (Jµ µ δ σ + 1 a σ δ µ δj µ)[γ+ d 4 x(j µ A C µ, σcc + c C σ)] = 0, (457) or J µ µ c C + 1 a σ µ A C µ = 0. (458) Finally recalling and analogous expressions we get J µ = δγ δa C µ (459) δγ δa C µ µ c C + 1 a µ A C µ δγ = 0. (460) δ c C This is the general form of the so called Ward identities in an abelian theory, such as QED. Note that these are expressions valid for non-zero sources. When expanded in sources they imply an infinite number of relations amongst the different terms in the Taylor expansion of the effective action Γ. Let us see how this works in a particular case. We have to remember that Γ is the generating funcional of the 1PI Green functions. Γ = d 4 xd 4 y[ c C (x) 1 (x,y)c C (y)+ 1 2 ACµ (x)d 1 µν (x,y)acν (y)]+... (461) The dots stand for terms with 3 or more fields. Applying the Ward identity we get d 4 xd 4 y[ µ c C (x)d 1 µν ACν (y)+ 1 a µ A C µ (x) 1 (x,y)c C (y)]+... = 0 (462) Again the dots refer to terms with two or more classical fields. Integrating by parts and using that (x y) = (y x) we immediatey get the relation µ D 1 µν (x y)+ 1 a ν 1 (x y) = 0. (463) Since we are in an abelian theory and the ghost is a free field we know that 1 (x y) = δ (4) (x y). (464) 73

15 Also, in momentum space D 1 µν (k) = Π µν(k) = (k 2 η µν k µ k ν )Π T (k 2 )+k µ k ν Π L (k 2 ). (465) It follows immediately that k 2 k ν Π L 1 a k νk 2 = 0 (466) The longitudinal part of the propagator does not get any radiative corrections. Ward identities in a non-abelian theory The lagrangian at the quantum level will be invariant under L 1 2a ( λ A a λ) 2 c a λ (D λ c) a, (467) A λ A λ +ǫ(d λ c), c c 1 a ǫ λ A λ, c c 1 ǫgc c. (468) 2 In order to ease the notation we shall attempt not to write group indices as far as possible. We shall add to the gauge fixed or quantum lagrangian the following source terms d 4 x[j λ A λ +c σ + cσ +u λ (D λ c) 1 vgc c]. (469) 2 Note that we have added two additional sources with respect to the abelian case. Notice also the ordering of source and field in the second term. This is in order to preserve the relation σ = δγ δcc. (470) As before, the jacobian of the BRST transformation has unit determinant. In addition, we shall use the fact that (see exercises) δ(dc) = 0, δ(c c) = 0, (471) It follows that J λ δa λ +δc σ +δ cσ = 0, (472) where the expectation value is of course with non-zero sources (it vanishes identically otherwise). This can be written at the level of connected Green functions as J λδw δu λ + δw δv σ 1 a ( λδw δjλ)σ = 0. (473) 74

16 Just like in the abelian case we replace the sources by derivatives of the effective action Γ Γ[A C,c C, c C,u,v] = W[J, σ,σ,u,v] d 4 x[j λ A C λ +cc σ + c C σ] (474) w.r.t. the classical fields A C,c c and c C (note that the sources u,v are not Legendre transformed), getting δγ δa C λ δγ δu λ + δγ δv δγ δc C 1 a ( λ A C λ) δγ = 0. (475) δ c C Let us now use the equations of motion for c (this can be done because this field, after integration by parts appears only linearly and with no derivatives sort of an auxiliary field) or, in terms of expectation values, 0 = λ D λ c σ (476) Replacing this into (475) we get δγ δγ = λ δ c C δuλ. (477) Defining the Ward identity reads δγ δγ δuλ( δa C λ 1 a λ ν A C ν )+ δγ δγ = 0. (478) δv δcc Γ = Γ+ 1 2a δγ δu λ δγ δa C λ + δγ δv d 4 x( λ A C λ )2 (479) δγ = 0. (480) δcc This is the simplest form of the generating functional of the Ward identities in a non-abelian theory. The meaning of Γ is the following: the gauge-fixing term appears unchanged in Γ. Once we remove that piece, the remaining part satisfies an invariance under where the variations are given by δγ A C µ A C µ +δa C µ, c C c C +δc C, (481) δu and µ δγ δv, respectively. These variations correspond to the vacuum expectation values of the variations at the lagragian level, that is δγ δu µ = D µc, δγ δv = g c c, (482) 2 which of course do not factorize trivially in terms of classical fields, i.e. for instance c c = c C c C. 75

17 Itisalso possibletoexpresstheward identities for theabelian theory intermsof themodified generating functional Γ, that has the (unmodified by radiative corrections) gauge-fixing term removed. From (460) we have which can be expressed as ( δγ δa C µ + 1 a µ ν A C ν ) µ c = 0 (483) δγ δa C µ µ c C = 0. (484) This equation expresses the conservation of the vector gauge current in all green functions. Everytime we have an insertion of A µ and act on this Green function with µ we get zero. 4.6 Spontaneous symmetry breaking and renormalizability When studying the abelian Higgs model we have seen that it is always possible to make a gauge transformation to remove completely one of the two degrees of freedom of the scalar field. If we examine the gauge boson propagator, this takes the following form ( i) ηµν kµ k ν M 2 k 2 M 2 +iǫ (485) which is the naive propagator of a massive vector field. This however gives a terrible UV behaviour since, at least naively, goes as 1/M 2 as k. The theory looks non-renormalizable in this gauge. On the other hand the distribution of the degrees of freedom looks fairly normal: one massive scalar particle (one physical degree of freedom) and three degrees of freedom corresponding to a massive spin one particle. For this reason this particular choice is called the unitary gauge. While the particle content of the theory is manifest in this gauge, it may be more convenient to work in another gauge where the symmetry does not look so manifestly broken. This can be achieved by the use of the so-called t Hooft-Feynman or R ξ gauges. Let us add to the lagrangian the gauge fixing term L gf = 1 2ξ F2, (486) where F = µ A µ +Mξφ 2. (487) In this class of gauges the vector boson propagator has the form i k µ k ν k 2 M 2 +iǫ [ηµν (1 ξ) k 2 ξm 2 ]. (488) +iǫ 76

18 This propagator has some notable properties. First of all, it reproduces in the ξ limit the result in the unitary gauge as naively ξ forces φ 2 = 0 in the path integral due to the gauge condition. On the other hand it has for k the same UV behaviour as the propagator of the unbroken theory. Finally, we note that it exhibits a dual pole structure with single poles at k 2 = M 2 and k 2 = ξm 2. The second one reflects the presence of the unremoved Goldstone boson. However the fact that this pole is manifestly gauge dependent indicates that it may not show in any physical S-matrix element. Indeed, at least very naively, this is the case as it multiplies the longitudinal structure k µ k ν that carries all the gauge dependence at tree level in QED. Itisworthreflectingforonesecondontheissueofthe normal UVbehaviourofpropagatorsin R ξ gauges. ThisshowsthattheUV singularitiesof agiven theorydoesnotdependonwhether the symmetry is spontaneously broken or not. t Hooft showed that spontaneously broken non-abelian theories are renormalizable. The presence of the underlying symmetry implies secret relations among the renomalization constants, even if the symmetry is spontaneously broken and not directly realized in the physical states. For future use we provide below the decomposition of the propagators in pure longitudinal and pure transverse part with D µν = D T µν +DL µν, (489) Dµν T = ( i) η µν kµkν k 2 k 2 M 2, DL µν = ξk µ k ν k 2 (k 2 ξm 2 ). (490) Finally for completeness we provide the t Hooft-Feynman gauge fixing term for the electroweak theory L gf = 1 2ξ (F2 γ +F 2 Z +2F + F ), (491) with F ± = µ W ± µ im Wξφ ±, F Z = µ Z µ +M Z ξχ, F γ = µ A µ. (492) 4.7 R ξ gauges and modified Ward identities Apart from allowing us to interpolate smoothly between covariant and unitary gauges, there is another nice feature of the R ξ gauges. Let us look at the bilinear piece of the scalar lagrangian of the abelian Higgs model, We see that there is a piece of the form ea µ φ 1 µ φ 2. (493) 77

19 when the field φ 1 picks up a vacuum expectation value, φ 1 = v +φ 1 this gives the annoying piece MA µ µ φ 2. (494) It is annoying because it would make the matrix of two point functions non-diagonal and complicate perturbative calculations enormously. However a similar piece is produced by the gauge fixing term which that upon integration by parts cancels the previous one. Mφ 2 µ A µ, (495) Letusnow turntotheissueof howward identities havetobemodifiedwhenthephenomenon of spontaneous symmetry breaking is present. First we consider some spontaneous symmetry breaking G H pattern leading to a number of Goldstone bosons φ a. The corresponding gauge fixing term is F a = µ A a µ +Mξφa, (496) where the φ a is a subset of the φ i, the scalars present in the theory, the ones corresponding to the broken directions. The gauge-fixed action is invariant under the following set of BRST transformations δa a µ = ǫ(d µ c) a, δφ i = iǫc a T a ijφ j, (497) δ c a = ǫ 1 ξ Fa, δc a = ǫ 1 2 gfabc c b c c. (498) We add the following set of source terms to W, the generating functional J a λa aλ +c a σ a + c a σ a +J i φ i u a λ(d λ c) a +u i (ic a T a ijφ j )+v a ( 1 2 gfabc c b c c ). (499) Repeating the same steps as we did for a non-abelian theory in the Lorenz gauge and ignoring the fermion fields we get the following Ward identity (obvious group indices and integration over x are omitted, in what follows v should not be confused with the v.e.v.). Γ is the effective action with the gauge fixing term removed δγ δγ δu λ δa C λ + δγ δu δγ δφ C + δγ δv δγ = 0. (500) δcc Note that this can be understood as constituting a statement of invariance of the modified effective actionγ [A C,c C ;u i,u,v] undertheabovebrsttransformations, takingintoaccount that δγ δu λ = D λc, etc. (501) 78

20 Let us consider the implications of this Ward identity on two-point funcions. We differentiate with respect to A C and c C and we consider the sector of Green functions with zero ghost number (i.e., the physical sector) and set sources to zero. Then δ 2 Γ δa C λ δac µ δ 2 Γ δc C δu λ + δ2 Γ δφ C δa C µ δ 2 Γ δc C = 0. (502) δu Introducing all variables explicitly d 4 x[π ab δ λµ(x, y) δc Cd (z) (Dλ c) d (x) +Π ib δ µ(x,y) δc Cd (z) (ica Tijφ a j )(x) ] = 0. (503) We note that the last term vanishes for any theory with invariance φ φ unless there is spontaneous symmetry breaking. If there is no such breaking, we are left with the first term that is just the direct generalization to the non-abelian case of the Ward identity that we derived in the abelian case. We then see that the form of the Ward identity depends on the form the symmetry is realized. To see the implications of this ward identity let us work in momentum space δ 2 Γ δa C A(p 2 )(η µλ p µp λ λ δac µ p 2 )+B(p2 ) p µp λ p 2. (504) Note that the transverse part due to the gauge fixing term is not included here because we are using the modified effective action Γ = Γ+ 1 2ξ F2. δ 2 Γ δφ C δa C ip µ C(p 2 ), µ Then the Ward identity reads δ 2 Γ δc C δu λ pλ J(p 2 ), δ 2 Γ δc C δu I(p2 ). (505) p µ B(p 2 )J(p 2 )+ip µ C(p 2 )I(p 2 ) = 0. (506) Note that the longitudinal part of the gauge field two-point function A(p 2 ) is not constrained at all by the Ward identity. This Ward identity represents the mixing between the longitudinal part of this Green function and the mixed gauge-scalar two point function. Likewise we could have derived the master Ward identity w.r.t. φ C and c C. We then get d 4 δ 2 Γ δ 2 Γ x[ δa C λ (x)δφc (y) δc C (z)δu λ (x) + δ 2 Γ δ 2 Γ δφ C (x)δφ C (y) δc C ] = 0. (507) (z)δu(x) Introducing the Fourier transform of the scalar 1PI Green functions we get δ 2 Γ δφ C δφ C p2 F(p 2 ), (508) ip 2 C(p 2 )J(p 2 )+p 2 F(p 2 )I(p 2 ) = 0. (509) None of this holds if there is no spontaneous symmetry breaking. 79