13.3 Non-Abelian anomalies

In the last section we considered the chiral current which is a gauge singlet (no gauge indices). Now we turn to the study of the gauge current $j_{\alpha }^{\mu }$ where $\alpha$ is the gauge index. Here we consider a chiral theory in which the gauge field $\mathcal{A}$ couples only to the left-handed Weyl fermion $\psi$ . Suppose $\psi$ transforms in a complex representation $r$ of the gauge group $G$ . For example, suppose $\psi$ belongs to a 3 of SU(3). The effective action $W_r[\mathcal{A}]$ is given by

$${\mathrm{e}}^{-W_r[\mathcal{A}]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi }\exp \left(-\int \mathrm{d}x\bar{\psi }\mathrm{i}{\bar{\mathcal{V}}}_{+}\psi \right)\text{\hspace{1em}(13.36)}$$

where

$$\mathrm{i}{\vec{\mathcal{V}}}_{+}=\mathrm{i}{\gamma }^{\mu }\left({\partial }_{\mu }+{\mathcal{A}}_{\mu }\right){\mathcal{P}}_{+}{\mathcal{P}}_{\pm }=\frac{1}{2}\left(1\pm {\gamma }^{m+1}\right).\text{\hspace{1em}(13.37)}$$

The gauge current is

$$j_{\alpha }^{\mu }=\mathrm{i}\bar{\psi }{\gamma }^{\mu }{T}_{\alpha }{\mathcal{P}}_{+}\psi .\text{\hspace{1em}(13.38)}$$

Let $v=v^{\alpha }{T}_{\alpha }$ be an infinitesimal gauge transformation parameter, $g=1-v$ under which we have

$${\mathcal{A}}_{\mu }\to (1+v)({\mathcal{A}}_{\mu }+\mathrm{d})(1-v)={\mathcal{A}}_{\mu }-{\mathcal{D}}_{\mu }v\text{\hspace{1em}(13.39)}$$

where ${\mathcal{D}}_{\mu }v\equiv {\partial }_{\mu }v+[{\mathcal{A}}_{\mu },v]$ is the covariant derivative for a field in the adjoint representation. The effective action transforms as

$$\begin{array}{rl}& W_r[\mathcal{A}]\to W_r[\mathcal{A}-\mathcal{D}v]\\ & =W_r[\mathcal{A}]-\int \mathrm{d}x\mathrm{tr}\left(\mathcal{D}v\frac{\delta }{\delta \mathcal{A}}{W}_r[\mathcal{A}]\right)\\ & =W_r[\mathcal{A}]-\int \mathrm{d}x\mathrm{tr}\left({\partial }_{\mu }{v}^{\alpha }+f_{\alpha \beta \gamma }{A}_{\mu }^{\beta }{v}^{\gamma }\right){\frac{\delta }{\delta A_{\mu }}}^{\alpha }{W}_r[\mathcal{A}]\\ & =W_r[\mathcal{A}]+\int dx\mathrm{tr}\left(v^{\alpha }\mathcal{D}\frac{\delta }{\delta \mathcal{A}}{W}_r[\mathcal{A}{]}_{\alpha }\right).\end{array}\text{\hspace{1em}(13.40)}$$

Since

$$\frac{\delta }{\delta A_{\mu }^{\alpha }}{W}_r[\mathcal{A}]=⟨\mathrm{i}\bar{\psi }{\gamma }^{\mu }{T}_{\alpha }\frac{1}{2}(1+{\gamma }^{m+1})\psi {⟩}_{\mathcal{A}}=⟨j_{\alpha }^{\mu }⟩$$

we obtain

$$W_r[\mathcal{A}-\mathcal{D}v]-W_r[\mathcal{A}]=\int dx\mathrm{tr}\left(v^{\alpha }{\mathcal{D}}_{\mu }⟨j^{\mu }{⟩}_{\alpha }\right).\text{\hspace{1em}(13.41)}$$

We are naively tempted to regard (13.36) as $\det (\mathrm{i}\vec{\mathbb{V}})=\prod {\lambda }_i^{\prime }$ , ${\lambda }_i$ being the ‘eigenvalue’ of $\mathrm{i}\vec{\mathbb{V}}$ . A subtlety arises here: $\mathrm{i}{\vec{\mathbb{V}}}_{+}$ maps sections of $S_{+}\otimes E$ to those of $S_{-}\otimes E$ , where $E$ is the vector bundle associated with the $G$ bundle and $S_{\pm }$ are spin bundles with chirality $\pm$ . Accordingly, the equation $\mathrm{i}{\vec{\mathbb{V}}}_{+}\psi =\lambda \psi$ is meaningless. To avoid this difficulty, we formally introduce a Dirac spinor $\psi$ and define

$${\mathrm{e}}^{-W_r[\mathcal{A}]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi }\exp \left(-\int \mathrm{d}x\bar{\psi }\mathrm{i}\hat{D}\psi \right)\text{\hspace{1em}(13.42)}$$

where $\mathrm{i}\hat{D}$ is defined by

$$\mathrm{i}\hat{D}\equiv \mathrm{i}{\gamma }^{\mu }\left({\partial }_{\mu }+\mathrm{i}{\mathcal{A}}_{\mu }{\mathcal{P}}_{+}\right)=\left(\begin{array}{cc}0& \mathrm{i}/{\partial }_{-}\\ \mathrm{i}/{\overline{\mathbb{V}}}_{+}& 0\end{array}\right)\text{\hspace{1em}(13.43)}$$

where we have diagonalized ${\gamma }^{m+1}$ . In (13.43), the gauge field $\mathcal{A}$ couples only to the positive chirality field. Now the eigenvalue problem $\mathrm{i}\hat{D}{\psi }_i={\lambda }_i{\psi }_i$ is well defined. Note that $\mathrm{i}\hat{D}$ is not Hermitian and ${\lambda }_i$ is a complex number in general. Moreover, we need to introduce right and left eigenfunctions separately by

$$\mathrm{i}\hat{D}{\psi }_i={\lambda }_i{\psi }_i\text{\hspace{1em}(13.44a)}$$

$${\chi }_i^{†}(\mathrm{i}\overleftarrow{\hat{D}})={\lambda }_i{\chi }_i^{†}(\mathrm{i}\hat{D}{)}^{†}{\chi }_i={\bar{\lambda }}_i{\chi }_i.\text{\hspace{1em}(13.44b)}$$

Since $\int {\chi }_i^{†}{\psi }_j\mathrm{d}x=0$ for $i\ne j$ , we may choose an orthonormal basis,

$$\int {\chi }_i^{†}{\psi }_j\mathrm{d}x={\delta }_{ij}.\text{\hspace{1em}(13.45)}$$

It should be noted that the eigenvalue ${\lambda }_i$ is not gauge invariant. This follows from the observation that

$$\begin{array}{rl}g(\mathrm{i}\hat{D}({\mathcal{A}}^g))g^{-1}& =g\mathrm{i}{\gamma }^{\mu }[{\partial }_{\mu }+g^{-1}({\mathcal{A}}_{\mu }+{\partial }_{\mu })g{\mathcal{P}}_{+}]g^{-1}\\ & =\mathrm{i}\hat{D}(\mathcal{A})-\mathrm{i}\partial /g{g}^{-1}+\mathrm{i}\partial /g{g}^{-1}{\mathcal{P}}_{+}\ne \mathrm{i}\hat{D}(\mathcal{A}).\end{array}\text{\hspace{1em}(13.46)}$$

If the equality were to hold in (13.46), $g^{-1}{\psi }_i$ would satisfy $\mathrm{i}\hat{D}({\mathcal{A}}^g)g^{-1}{\psi }_i={\lambda }_i{g}^{-1}{\psi }_i$ when $\mathrm{i}\hat{D}(\mathcal{A}){\psi }_i={\lambda }_i{\psi }_i$ . Then $\mathrm{Spec}\mathrm{i}\hat{D}(\mathcal{A})$ would be gauge invariant. Although individual eigenvalues are not gauge invariant, the absolute value of the product of eigenvalues of $\mathrm{i}\hat{D}$ is gauge invariant. In fact,

$$\begin{array}{rl}\det (\mathrm{i}\hat{D})\det ((\mathrm{i}\hat{D}{)}^{†})& =\det (\mathrm{i}\hat{D}(\mathrm{i}\hat{D}{)}^{†})\\ & =\det \left(\begin{array}{cc}(\mathrm{i}/j_{-})(\mathrm{i}/j_{+})& 0\\ 0& (\mathrm{i}/{\nabla }_{+})(\mathrm{i}/{\nabla }_{-})\end{array}\right)\\ & =\det (\mathrm{i}/{\partial }_{-}\mathrm{i}/{\partial }_{+})\det (\mathrm{i}{\nabla }_{+}\mathrm{i}{\nabla }_{-})\end{array}\text{\hspace{1em}(13.47)}$$

where $\mathrm{i}/{\partial }_{+}=(\mathrm{i}/{\partial }_{-}{)}^{†}$ and $\mathrm{i}{\nabla }_{-}=(\mathrm{i}{\nabla }_{+}{)}^{†}$ . This is simply the Dirac determinant (up to an irrelevant factor $\det (\mathrm{i}/{\partial }_{-}\mathrm{i}/{\partial }_{+}))$

$$[\det (\mathrm{i}\vec{\forall }){]}^2=\det \left(\begin{array}{cc}\mathrm{i}{\vec{\forall }}_{-}\mathrm{i}{\vec{\forall }}_{+}& 0\\ 0& \mathrm{i}{\vec{\forall }}_{+}\mathrm{i}{\vec{\forall }}_{-}\end{array}\right)=[\det (\mathrm{i}{\vec{\forall }}_{+}\mathrm{i}{\vec{\forall }}_{-}){]}^2\text{\hspace{1em}(13.48)}$$

where $\mathbf{i}\nabla$ is given by

$$\mathrm{i}\vec{\mathbb{V}}=\left(\begin{array}{cc}0& \mathrm{i}{\vec{\mathbb{V}}}_{-}\\ \mathrm{i}{\vec{\mathbb{V}}}_{+}& 0\end{array}\right).\text{\hspace{1em}(13.49)}$$

The Dirac determinant is gauge invariant, hence so is $|\det (\mathrm{i}\hat{D})|$ . It then follows that $\mathrm{Re}{W}_r[\mathcal{A}]$ is gauge invariant since

$$\exp (-W_r[\mathcal{A}])\exp (-\overline{W_r[\mathcal{A}]})=\det (\mathrm{i}\hat{D})\det ((\mathrm{i}\hat{D}{)}^{†})\propto \det (\mathrm{i}{\hat{\nabla }}_{+}\mathrm{i}{\hat{\nabla }}_{-})$$

is gauge invariant. Therefore, only the imaginary part of $W_r[\mathcal{A}]$ , that is the phase of $\det (\mathrm{i}\hat{D})$ , may gain an anomalous variation under gauge transformations.

The anomaly may be computed by evaluating the Jacobian as before. The functional measure is taken to be ${\prod }_i\mathrm{d}{a}_i\mathrm{d}{\bar{b}}_i$ . We consider an infinitesimal gauge transformation,

$$\mathcal{A}\to \mathcal{A}-\mathcal{D}v\psi \to \psi +v{\psi }_{+}\bar{\psi }\to \bar{\psi }-{\bar{\psi }}_{-}v\text{\hspace{1em}(13.50)}$$

where the gauge transformation rotates the positive chirality parts only. The Jacobian factor is

$$\int \mathrm{d}x\mathrm{tr}v(x)\sum _n(n|x⟩{\gamma }^{m+1}⟨x|n⟩\text{\hspace{1em}(13.51)}$$

where $⟨x|n⟩={\psi }_n(x)$ and $(n|x⟩={\chi }_n^{†}(x)$ (note that $(n|$ is not the Hermitian conjugate of $|n⟩$ ). This integral is ill defined and must be regularized. As before, we employ the Gaussian regulator,

$$\begin{array}{rl}& \int \mathrm{d}x\underset{\begin{array}{c}M\to \infty \\ x\to y\end{array}}{\lim }\mathrm{tr}v(x)\sum _n(n|y⟩{\gamma }^{m+1}⟨x|{\mathrm{e}}^{-(\mathrm{i}\hat{D}{)}^2/M^2}|n⟩\\ & =\int \mathrm{d}x\underset{\begin{array}{c}M\to \infty \\ x\to y\end{array}}{\lim }\mathrm{tr}v(x){\gamma }^{m+1}{\mathrm{e}}^{-(\mathrm{i}{\hat{D}}_x{)}^2/M^2}\delta (x-y)\end{array}\text{\hspace{1em}(13.52)}$$

where use has been made of the completeness relation

$$\sum _n|n⟩(n|=I.\text{\hspace{1em}(13.53)}$$

It follows from (13.41) and (13.52) that

$$\int \mathrm{d}x{v}^{\alpha }{\mathcal{D}}_{\mu }\left(\frac{\delta }{\delta A_{\mu }^{\alpha }}{W}_r[\mathcal{A}]\right)=\int \mathrm{d}x\underset{\begin{array}{c}M\to \infty \\ x\to y\end{array}}{\lim }\mathrm{tr}\left[v{\gamma }^{m+1}{\mathrm{e}}^{-(\mathrm{i}{\hat{D}}_x{)}^2/M^2}\delta (x-y)\right].\text{\hspace{1em}(13.54)}$$

In the present case $W_r$ really changes under (13.50). The trace may be written as

$$\begin{array}{rl}\mathrm{tr}\left[v{\gamma }^{m+1}{\mathrm{e}}^{-(\mathrm{i}{\hat{D}}_x{)}^2/M^2}\right]& =\mathrm{tr}\left[v\left({\mathcal{P}}_{+}-{\mathcal{P}}_{-}\right){\mathrm{e}}^{-(\mathrm{i}{\mathcal{Y}}_{-}\mathrm{i}{\overline{\mathcal{Y}}}_{+})-(\mathrm{i}{\overline{\mathcal{Y}}}_{-}\mathrm{i}{\mathcal{Y}}_{+})/M^2}\right]\\ & =\mathrm{tr}[v{P}_{+}{\mathrm{e}}^{(\mathrm{i}\overline{\partial }\mathrm{i}\overline{\partial })/M^2}]-\mathrm{tr}[v{P}_{-}{\mathrm{e}}^{(\mathrm{i}\overline{\partial }\mathrm{i}\partial )/M^2}].\end{array}\text{\hspace{1em}(13.55)}$$

(13.55) can be evaluated in the plane wave basis, which is straightforward but tedious (see Gross and Jackiw (1972), for example). We derive the non-Abelian anomaly from a topological viewpoint in the next section. For $m=4$ , the anomalous variation is

$$\begin{array}{rl}{W}_r[\mathcal{A}-\mathcal{D}v]-W_r[\mathcal{A}]& =\int dx{v}^{\alpha }{\mathcal{D}}_{\mu }⟨j^{\mu }{⟩}_{\alpha }\\ & =\frac{1}{24{\pi }^2}\int dx\mathrm{tr}\left\{v^{\alpha }{T}_{\alpha }{ϵ}^{\kappa \lambda \mu v}{\partial }_{\kappa }\left[{\mathcal{A}}_{\lambda }{\partial }_{\mu }{\mathcal{A}}_v+\frac{1}{2}{\mathcal{A}}_{\lambda }{\mathcal{A}}_{\mu }{\mathcal{A}}_v\right]\right\}\\ & =\frac{1}{24{\pi }^2}\int \mathrm{tr}\left\{v\mathrm{d}\left[\mathcal{A}\mathrm{d}\mathcal{A}+\frac{1}{2}{\mathcal{A}}^3\right]\right\}.\end{array}\text{\hspace{1em}(13.56)}$$

The anomalous divergence of the gauge current is

$${\mathcal{D}}_{\mu }⟨j^{\mu }{⟩}_{\alpha }=\frac{1}{24{\pi }^2}\mathrm{tr}\left\{T_{\alpha }{ϵ}^{\kappa \lambda \mu \nu }{\partial }_{\kappa }\left[{\mathcal{A}}_{\lambda }{\partial }_{\mu }{\mathcal{A}}_{\nu }+\frac{1}{2}{\mathcal{A}}_{\lambda }{\mathcal{A}}_{\mu }{\mathcal{A}}_{\nu }\right]\right\}.\text{\hspace{1em}(13.57)}$$

This should be compared with (13.33). There are two differences between these results: the two-thirds in front of ${\mathcal{A}}^3$ is replaced by a half and the overall factor is different.

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