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  Fermion field redefinition in anomalous gauge theory

+ 2 like - 0 dislike

Consider the model

\mathcal{L} = \bar{\psi}\big[i\gamma^{\mu}(\partial_{\mu}-i\gamma_{5}A_{\mu} - iV_{\mu}) -me^{2i\gamma_{5}\phi}\big] \psi+\frac{m_{Z}^{2}}{2}(A_{\mu}+\partial_{\mu}\phi)^{2} \quad (1)

Here $A_{\mu}$ is an axial-vector gauge boson, $V_{\mu}$ is a vector-like gauge boson, $\phi$ is the Goldstone boson associated with the Higgs mechanism which generates the masses $m_{A}$ and $\psi$. The Lagrangian is classically invariant under the axial transformations

\psi \to e^{i\gamma_{5}\alpha(x)}\psi, \quad A_{\mu} \to A_{\mu}+\partial_{\mu}\alpha, \quad \phi \to \phi - \alpha​$$

Since $m_{A}$ and $m$ are non-zero, the physical  particle content is $\{\psi, Z_{\mu}\equiv A_{\mu}+\partial_{\mu}\phi, V_{\mu}\}$. If I perform the redefinition

\psi \equiv e^{-i\gamma_{5} \phi }\kappa, \quad (2)

where $\kappa$ is an axial gauge invariant fermion, then classically, the Lagrangian $(1)$ will transform to the form

\mathcal{L} = \bar{\kappa}\big[ i\gamma^{\mu}(\partial_{\mu} - i\gamma_{5}Z_{\mu}-iV_{\mu}) - m\big]\kappa + \frac{m_{Z}^{2}}{2}Z_{\mu}^{2} \quad (3)

Naively, now the axial symmetry is completely rendered out, since all the quantities in (3) are gauge invariant by definition. However, actually the model is anomalous, and on the quantum level an additional gauge variant term will arise.

My question is the following. Am I right that the redefinition $(2)$ induce the anomalous term to the Lagrangian as well as if we consider it as the axial transformation, inducing the well known anomalous term $\frac{\phi}{4 \pi}V_{\mu\nu}\tilde{V}^{\mu\nu}$?

asked Dec 22, 2017 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Dec 22, 2017 by NAME_XXX

1 Answer

+ 3 like - 0 dislike

You can either think about (2) as an axial rotation, the anomaly giving rise to the axion term or as a field redefinition, during which one must consider how the path integral measure transforms, leading again to the axion term as the Jacobian determinant.

answered Dec 26, 2017 by Ryan Thorngren (1,925 points) [ no revision ]

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