Equivalence theorem and dynamical symmetry breaking

The equivalence theorem, between the longitudinal gauge bosons and the states eaten up by them in the process of symmetry breaking, is shown to be valid in a class of models where the details of dynamical symmetry breaking makes it obscure. In a recent paper [1], Donoghue and Tandean have made an intriguing point regarding the validity of the equivalence theorem [2, 3, 4, 5, 6, 7] in a class of gauge models which exhibit dynamical symmetry breaking. As a paradigm, they considered the pedagogical model [8, 9] where the electroweak gauge symmetry is dynamically broken by quark condensates which are also responsible for the chiral symmetry breaking: 〈uLuR〉 = 〈dLdR〉 6 = 0 . (1) The pions, ππ, which would have been the Goldstone bosons for the chiral symmetry breaking SU(2)L×SU(2)R → SU(2)V , are eaten up by the electroweak gauge bosons in this model in absence of any fundamental Higgs bosons. Donoghue and Tandean [1] then calculate the amplitude of the process ee → Zγ at energies larger than MZ mediated by the triangle diagram. Both quarks and leptons can appear in the triangle, and from the condition of vanishing of gauge anomalies, they obtain that the amplitude vanishes. Since π constitutes the longitudinal part of the Z in this model, they then compare it with the process ee → πγ, mediated through the triangle diagram. However, since the pion consists of quarks but no leptons, only quarks circulate in the loop now, and therefore the amplitude does not vanish. This, they claim, is a violation of the equivalence theorem. The purpose of this article is to examine this claim. In Higgs models of symmetry breaking, one can calculate the couplings of the unphysical Higgs bosons (which are eaten up by the gauge bosons) from those of the gauge bosons without having to make any assumption about the Higgs content of the model. This can be done by demanding that, if one uses the Rξ gauges to calculate any amplitude, the ξ-dependent poles must cancel [10, 11]. 1 Thus, for two fermions a and b which are represented by their field operators ψa and ψb, if the coupling to any gauge boson V is given by ψaγ μ (Gab + G ′ abγ5)ψbVμ , (2) this requirement demands that the corresponding unphysical Higgs boson, S, will have the coupling 1 MV ψa [(ma −mb)Gab + (ma +mb)G ′ abγ5]ψbS , (3) where MV is the mass of the gauge boson V after symmetry breaking. From this requirement alone, one can verify the equivalence theorem for any amplitude. However, the equivalence theorem is more deep-rooted than the Higgs mechanism of symmetry breaking for the following reason. In any gauge symmetry breaking, some gauge bosons obtain masses MV whose exact values depend on some parameters of the theory. For any MV 6= 0, the longitudinal components of the massive gauge bosons are physical states. The “Nambu-Goldstone” modes, the states absorbed by the gauge bosons, are unphysical. On the other hand, for MV = 0, the Nambu-Goldstone modes are physical states but the longitudinal components are not, since the symmetry is not broken. The equivalence theorem then merely states that all observables are continuous in the limit MV → 0. In other words, in that limit, the amplitudes with any process with longitudinal component of a vector boson is the same (apart from a phase maybe) with the amplitudes for the corresponding processes where the longitudinal gauge bosons are replaced by the states that are eaten up by them in the process of symmetry breaking. Stated this way, the equivalence theorem seems to be a statement of continuity of certain parameters of the theory, and hence is expected to be valid for any model with symmetry breaking. In light of this, the claim of Donoghue and Tandean [1] is indeed surprising. To examine their claim, let us use their paradigm of QCD condensates breaking the electroweak gauge symmetry [8, 9], and for the sake of definiteness, let us talk about processes involving Z bosons only. Obviously, one can make similar arguments for processes involving W bosons. Since the Z-boson does not provide any flavor changing neutral current with the standard model fermions, the indices a and b in Eq. (2) have to be equal, and we denote the couplings G and G in this case with a single index. From Eq. (3) now, we see that proving the equivalence theorem is tantamount to proving that the coupling of the π to the fermion field ψa is given by 2ma MZ G a ψaγ5ψaπ 0 . (4) If the fermion a in question is the electron, for example, it might naively seem that such a coupling with the pion cannot exist since the pion wave function does not have any electron. However, this is not true, as can be seen from the diagram of Fig. 1. Two important points need to be made before we calculate this coupling. First, the intermediate line can only be Z. The diagram with intermediate photon line cannot contribute since the photon couplings are vectorial, whereas the pion couples only to the axial vector current through the relation