13.2 Abelian anomalies

Henceforth, we work in an even-dimensional manifold $M$ (dim $M=m=2l$ ) with a Euclidean signature. Four-dimensional results will readily be obtained by putting $m=4$ . We assume our system is non-chiral, namely, the gauge field couples to the right and the left components in the same way. Our convention is

$$\begin{array}{rl}& {\gamma }^{\mu †}={\gamma }^{\mu }\left\{{\gamma }^{\mu },{\gamma }^{\nu }\right\}=2{\delta }^{\mu \nu }{\gamma }^{m+1}=(\mathrm{i}{)}^l{\gamma }^1\dots {\gamma }^m\\ & {\gamma }^{m+1†}={\gamma }^{m+1}({\gamma }^{m+1}{)}^2=+I.\end{array}$$

The Lie group generators $\{T_{\alpha }\}$ satisfy

$$T_{\alpha }^{†}=-T_{\alpha }[T_{\alpha },T_{\beta }]=f_{\alpha \beta }^{\gamma }{T}_{\gamma }\mathrm{tr}(T^{\alpha }{T}^{\beta })=-\frac{1}{2}{\delta }^{\alpha \beta }.$$

13.2.1 Fujikawa’s method

Among several methods of deriving anomalies, Fujikawa’s way (Fujikawa 1979, 1980, 1986) reveals the topological and geometrical nature of the problem most directly. This method is equivalent to the heat kernel proof of the relevant index theorem.

Let $\psi$ be a massless Dirac field interacting with an external non-Abelian gauge field ${\mathcal{A}}_{\mu }$ . The effective action $W[\mathcal{A}]$ is given by

$${\mathrm{e}}^{-W[\mathcal{A}]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi }{\mathrm{e}}^{-\int \mathrm{d}x\bar{\psi }\mathrm{i}\bar{\nabla }\psi }\text{\hspace{1em}(13.10)}$$

where $\mathrm{i}\overline{\nabla }=\mathrm{i}{\gamma }^{\mu }{\nabla }_{\mu }=\mathrm{i}{\gamma }^{\mu }({\partial }_{\mu }+{\omega }_{\mu }+{\mathcal{A}}_{\mu })$ , with ${\omega }_{\mu }=\frac{1}{2}{\omega }_{\mu \alpha \beta }{\Sigma }^{\alpha \beta }$ being the spin connection of the background space. We compactify the space in such a way that the geometry (the spin connection) plays no role. For example, this can be achieved by compactifying ${\mathbb{R}}^4$ to $S^4={\mathbb{R}}^4\cup \{\infty \}$ , for which the Dirac genus $\hat{A}(TM)$ is trivial; see example 12.5. If this is the case, the spin connection is irrelevant and may be dropped from $\mathrm{i}\overline{\nabla }$ . The classical action $\int \mathrm{d}x\bar{\psi }\mathrm{i}\overline{\nabla }\psi$ is invariant with respect to the chiral rotation,

$$\psi \to {\mathrm{e}}^{\mathrm{i}{\gamma }^{m+1}\alpha }\psi \bar{\psi }\to \bar{\psi }{\mathrm{e}}^{\mathrm{i}{\gamma }^{m+1}\alpha }.\text{\hspace{1em}(13.11)}$$

We expand $\psi$ and $\bar{\psi }$ as

$$\psi =\sum _i{a}_i{\psi }_i\bar{\psi }=\sum _i{\bar{b}}_i{\psi }_i^{†}\text{\hspace{1em}(13.12)}$$

where $a_i$ and ${\bar{b}}_i$ are anti-commuting Grassmann variables,

$$\left\{a_i,a_j\right\}=0\left\{{\bar{b}}_i,{\bar{b}}_j\right\}=0\left\{a_i,{\bar{b}}_j\right\}=0$$

and ${\psi }_i$ is an eigenvector of the Dirac operator

$$\mathrm{i}\nabla {\psi }_i={\lambda }_i{\psi }_i.\text{\hspace{1em}(13.13)}$$

Since $\mathrm{i}\overline{\nabla }$ is Hermitian, ${\lambda }_i$ is real. Since $M$ is compact, ${\psi }_i$ can be normalized as

$$⟨{\psi }_i|{\psi }_j⟩=\int \mathrm{d}x{\psi }_i^{†}(x){\psi }_j(x)={\delta }_{ij}.$$

Now the path integrals over $\psi$ and $\bar{\psi }$ are replaced by those over $a_i$ and ${\bar{b}}_i$ .

Consider an infinitesimal chiral transformation,

$$\psi (x)\to \psi (x)+\mathrm{i}\alpha (x){\gamma }^{m+1}\psi (x)\text{\hspace{1em}(13.14a)}$$

$$\bar{\psi }(x)\to \bar{\psi }(x)+\mathrm{i}\bar{\psi }(x)\alpha (x){\gamma }^{m+1}.\text{\hspace{1em}(13.14b)}$$

As usual, we take $\alpha =\alpha (x)$ to be $x$ -dependent. Under this change, the classical action transforms as

$$\begin{array}{rl}& \int \mathrm{d}x\bar{\psi }\mathrm{i}\nabla \psi \to \int \mathrm{d}x(\bar{\psi }+\mathrm{i}\bar{\psi }\alpha {\gamma }^{m+1})\mathrm{i}\nabla (\psi +\mathrm{i}\alpha {\gamma }^{m+1}\psi )\\ & =\int \mathrm{d}x\bar{\psi }\mathrm{i}\nabla \psi +\mathrm{i}\int \mathrm{d}x[\alpha \bar{\psi }{\gamma }^{m+1}\mathrm{i}\nabla \psi +\bar{\psi }\mathrm{i}\nabla (\alpha {\gamma }^{m+1}\psi )]\\ & =\int dx\bar{\psi }i\bar{\nabla }\psi -\int dx[\alpha \bar{\psi }{\gamma }^{m+1}{\gamma }^{\mu }({\partial }_{\mu }+{\mathcal{A}}_{\mu })\psi \\ & +\bar{\psi }{\gamma }^{\mu }\left({\partial }_{\mu }+{\mathcal{A}}_{\mu }\right)\left(\alpha {\gamma }^{m+1}\psi \right)]\\ & =\int \mathrm{d}x\bar{\psi }\mathrm{i}\bar{\nabla }\psi +\int \mathrm{d}x\alpha (x){\partial }_{\mu }{j}_{m+1}^{\mu }(x)\end{array}\text{\hspace{1em}(13.15)}$$

where we have used the anti-commutation relations $\{{\gamma }^{\mu },{\gamma }^{m+1}\}=0$ and

$$j_{m+1}^{\mu }(x)\equiv \bar{\psi }(x){\gamma }^{\mu }{\gamma }^{m+1}\psi (x)\text{\hspace{1em}(13.16)}$$

is the chiral current. This is the higher-dimensional analogue of $j_5^{\mu }$ defined previously. If (13.15) were the only change caused by (13.14), naive application of the Ward-Takahashi relation would imply the conservation of the axial current ${\partial }_{\mu }{j}_{m+1}^{\mu }=0$ . In quantum theory, however, we have an additional change, namely the change of the path integral measure. Define the chiral-rotated fields by

$${\psi }^{\prime }=\psi +\mathrm{i}\alpha {\gamma }^{m+1}\psi =\sum a_i^{\prime }{\psi }_i\text{\hspace{1em}(13.17a)}$$

$${\bar{\psi }}^{\prime }=\bar{\psi }+\mathrm{i}\bar{\psi }\alpha {\gamma }^{m+1}=\sum {\bar{b}}_i^{\prime }{\psi }_i^{†}.\text{\hspace{1em}(13.17b)}$$

Now the measure changes as

$$\int \prod _i\mathrm{d}{a}_i\mathrm{d}{\bar{b}}_i\to \int \prod _i\mathrm{d}{a}_i^{\prime }\mathrm{d}{\bar{b}}_i^{\prime }.\text{\hspace{1em}(13.18)}$$

From the orthonormality of $\{{\psi }_i\}$ , we find that

$$\begin{array}{rl}{a}_i^{\prime }& =⟨{\psi }_i|{\psi }^{\prime }⟩=⟨{\psi }_i|(1+\mathrm{i}\alpha {\gamma }^{m+1})\psi ⟩\\ & =\sum _j⟨{\psi }_i|\left(1+\mathrm{i}\alpha {\gamma }^{m+1}\right){\psi }_j⟩a_j\equiv \sum _j{C}_{ij}{a}_j\end{array}\text{\hspace{1em}(13.19a)}$$

where

$$C_{ij}=⟨{\psi }_i|\left(1+\mathrm{i}\alpha {\gamma }^{m+1}\right){\psi }_j⟩={\delta }_{ij}+\mathrm{i}\alpha ⟨{\psi }_i\mid {\gamma }^{m+1}{\psi }_j⟩.\text{\hspace{1em}(13.20)}$$

The measure in terms of the new variables is

$$\begin{array}{rl}\prod \mathrm{d}{a}_j^{\prime }& =[\det C_{ij}{]}^{-1}\prod \mathrm{d}{a}_i=\exp (-\mathrm{tr}\ln C_{ij})\prod \mathrm{d}{a}_i\\ & =\exp \left[-\mathrm{tr}\ln \left(I+\mathrm{i}\alpha ⟨{\psi }_i|{\gamma }^{m+1}{\psi }_j⟩\right)\right]\prod \mathrm{d}{a}_i\\ & \approx \exp (-\mathrm{tr}\mathrm{i}\alpha ⟨{\psi }_i|{\gamma }^{m+1}{\psi }_j⟩)\prod \mathrm{d}{a}_i\\ & =\exp \left(-\mathrm{i}\alpha \sum _i⟨{\psi }_i|{\gamma }^{m+1}{\psi }_i⟩\right)\prod \mathrm{d}{a}_i\end{array}\text{\hspace{1em}(13.21)}$$

where the inverse of the determinant appears since $a_i$ and $a_i^{\prime }$ are Grassmann variables, see Berezin (1966).¹ As for ${\bar{b}}_i\to {\bar{b}}_i^{\prime }$ , we have

$${\bar{b}}_i^{\prime }=\sum _j{\bar{b}}_j⟨{\psi }_j|\left(1+\mathrm{i}\alpha {\gamma }^{m+1}\right)|{\psi }_i⟩=\sum _j{C}_{ji}{\bar{b}}_j.\text{\hspace{1em}(13.19b)}$$

The Jacobian for the change ${\bar{b}}_i\to {\bar{b}}_i^{\prime }$ agrees with (13.21). Thus, the measure transforms under the chiral rotation (13.17) as

$$\prod _i\mathrm{d}{a}_i\mathrm{d}{\bar{b}}_i\to \prod _i\mathrm{d}{a}_i^{\prime }\mathrm{d}{\bar{b}}_i^{\prime }\exp \left(-2\mathrm{i}\int \mathrm{d}x\alpha (x)\sum {\psi }_n^{†}(x){\gamma }^{m+1}{\psi }_n(x)\right).\text{\hspace{1em}(13.22)}$$

Now the effective action has two expressions:

$$\begin{array}{rl}& {\mathrm{e}}^{-W[\mathcal{A}]}=\int \prod _i\mathrm{d}{a}_i\mathrm{d}{\bar{b}}_i\exp \left(-\int \mathrm{d}x\bar{\psi }\mathrm{i}\nabla \psi \right)\\ & =\int \prod _i\mathrm{d}{a}_i^{\prime }\mathrm{d}{\bar{b}}_i^{\prime }\exp (-\int \mathrm{d}x\bar{\psi }\mathrm{i}\bar{\mathbb{V}}\psi -\int \mathrm{d}x\alpha (x){\partial }_{\mu }{j}_{m+1}^{\mu }(x)\\ & -2\mathrm{i}\int \mathrm{d}x\alpha (x)A(x))\end{array}\text{\hspace{1em}(13.23)}$$

where

$$\mathbf{A}(x)\equiv \sum _i{\psi }_i^{†}(x){\gamma }^{m+1}{\psi }_i(x).\text{\hspace{1em}(13.24)}$$

Since $\alpha (x)$ is arbitrary, we have

$${\partial }_{\mu }{j}_{m+1}^{\mu }(x)=-2\mathrm{i}A(x).\text{\hspace{1em}(13.25)}$$

Thus, naive conservation of an axial current does not hold in quantum theory. This non-conservation of the current $j_{m+1}^{\mu }$ is called the Abelian anomaly (or chiral anomaly or axial anomaly).

How is this related to the topology? Let us look at the Jacobian (13.22) and assume that $\alpha (x)$ is independent of $x$ . The integral in (13.22) is not well defined and must be regularized. We introduce the Gaussian cut-off (heat kernel regularization) as

$$\begin{array}{rl}\int \mathrm{d}xA(x)& =\int \mathrm{d}x\sum _i{\psi }_i^{†}(x){\gamma }^{m+1}{\psi }_i(x)\exp [-({\lambda }_i/M{)}^2]{|}_{M\to \infty }\\ & =\sum ⟨{\psi }_i\mid {\gamma }^{m+1}\exp \left[-(\mathrm{i}\bar{\mathcal{V}}/M{)}^2\right]\mid {\psi }_i⟩{|}_{M\to \infty }.\end{array}\text{\hspace{1em}(13.26)}$$

In (13.26), $1/M^2$ corresponds to the ‘time’ parameter $t$ in the previous chapter and $M\to \infty$ implies $t\to \epsilon$ . Let $|{\psi }_i⟩$ be an eigenstate of $\mathrm{i}\overline{\nabla }$ with non-vanishing eigenvalue ${\lambda }_i$ . Among the eigenstates, there exists a state $|{\psi }_i{⟩}^{\chi }\equiv {\gamma }^{m+1}|\psi ⟩$ with eigenvalue $-{\lambda }_i$ :

$$\begin{array}{rl}\mathrm{i}\bar{\nabla }|{\psi }_i{⟩}^{\chi }& =\mathrm{i}\bar{\nabla }{\gamma }^{m+1}|{\psi }_i⟩=-{\gamma }^{m+1}\mathrm{i}\bar{\nabla }|{\psi }_i⟩\\ & =-{\lambda }_i{\gamma }^{m+1}|{\psi }_i⟩=-{\lambda }_i|{\psi }_i{⟩}^{\chi }\end{array}$$

where use has been made of the anti-commutation relation $\{{\gamma }^{m+1},\mathrm{i}\overline{\mathcal{V}}\}=0$ . Since $\mathrm{i}\overline{\mathcal{V}}$ is a Hermitian operator, eigenvectors which belong to different eigenvalues are orthogonal, hence $⟨{\psi }_i|{\psi }_i{⟩}^{\chi }=⟨{\psi }_i|{\gamma }^{m+1}|{\psi }_i⟩=0$ . This shows that

$$⟨{\psi }_i|{\gamma }^{m+1}\exp [-(\mathrm{i}/\forall /M{)}^2]|{\psi }_i⟩=⟨{\psi }_i|{\gamma }^{m+1}|{\psi }_i⟩\exp [-({\lambda }_i/M{)}^2]=0.$$

Thus, the contribution to the RHS of (13.26) comes only from the zero-energy modes. Let $|0,i⟩$ be the zero-energy modes of $\mathrm{i}\vec{\nabla }$ , $(1\le i\le n_0)$ . They are not in an irreducible representation of the spin algebra and should be classified according to the eigenvalue of ${\gamma }^{m+1}$ . We write

$${\gamma }^{m+1}|0,i{⟩}_{\pm }=\pm |0,i{⟩}_{\pm }.\text{\hspace{1em}(13.27)}$$

Then, (13.26) becomes

$$\begin{array}{rl}\int \mathrm{d}x\mathbf{A}(x)& =\sum ⟨{\psi }_i|{\gamma }^{m+1}\exp [-(\mathrm{i}\nabla /M{)}^2]|{\psi }_i⟩{|}_{M\to \infty }\\ & =\sum _{i+}⟨0,i|0,i{⟩}_{+}-\sum _{i-}⟨0,i|0,i{⟩}_{-}\\ & =v_{+}-v_{-}=\mathrm{indi}{\nabla }_{+}\end{array}\text{\hspace{1em}(13.28)}$$

where ${\nu }_{+}({\nu }_{-})$ is the number of zero-energy modes with positive (negative) chirality $({\nu }_{+}+{\nu }_{-}=n_0)$ and $\mathrm{i}{\vec{\mathbb{V}}}_{+}$ is defined by

$$\mathrm{i}\vec{\mathbb{V}}=\left(\begin{array}{cc}0& \mathrm{i}{\vec{\mathbb{V}}}_{-}\\ \mathrm{i}{\vec{\mathbb{V}}}_{+}& 0\end{array}\right)\mathrm{i}{\vec{\mathbb{V}}}_{-}=(\mathrm{i}{\vec{\mathbb{V}}}_{+}{)}^{†}.$$

2 We are looking at the zero-momentum Ward-Takahashi relation.

The Atiyah-Singer index theorem now comes into the problem.

To show that (13.28), indeed, represents an integral of the relevant Chern character, we first note that

$$\begin{array}{rl}(\mathrm{i}\nabla {)}^2& =-{\gamma }^{\mu }{\gamma }^{\nu }{\nabla }_{\mu }{\nabla }_{\nu }=-\left\{{\delta }^{\mu \nu }+\frac{1}{2}\left[{\gamma }^{\mu },{\gamma }^{\nu }\right]\right\}\frac{1}{2}\left[\left\{{\nabla }_{\mu },{\nabla }_{\nu }\right\}+{\mathcal{F}}_{\mu \nu }\right]\\ & =-{\nabla }_{\mu }{\nabla }^{\mu }-\frac{1}{4}\left[{\gamma }^{\mu },{\gamma }^v\right]{\mathcal{F}}_{\mu v}\end{array}\text{\hspace{1em}(13.29)}$$

where use has been made of the relation $[{\nabla }_{\mu },{\nabla }_{\nu }]={\mathcal{F}}_{\mu \nu }$ . Then

$$\mathbf{A}(x)=\sum _i⟨{\psi }_i|x⟩⟨x\mid {\gamma }^{m+1}\exp \left[\left({\nabla }^2+\frac{1}{4}\left[{\gamma }^{\mu },{\gamma }^{\nu }\right]{\mathcal{F}}_{\mu \nu }\right)/M^2\right]\mid {\psi }_i⟩{|}_{M\to \infty }.\text{\hspace{1em}(13.30)}$$

Let us take $m=4$ for definiteness. We introduce the plane wave basis as

$$⟨x|{\psi }_i⟩=\int \frac{{\mathrm{d}}^{4}k}{(2\pi {)}^4}⟨x|k⟩⟨k|{\psi }_i⟩.$$

Then (13.30) becomes

$$\begin{array}{rl}& A(x)=\int \frac{\mathrm{d}k}{(2\pi {)}^4}\int \frac{\mathrm{d}{k}^{\prime }}{(2\pi {)}^4}\sum _i⟨{\psi }_i|k^{\prime }⟩⟨k^{\prime }|x⟩\\ & \times {\gamma }^{m+1}\exp [({\nabla }^2+\frac{1}{4}[{\gamma }^{\mu },{\gamma }^v]{\mathfrak{F}}_{\mu \nu })/M^2]⟨x|k⟩⟨k|{\psi }_i⟩{|}_{\begin{array}{c}M\to \infty \\ y\to x\end{array}}\\ & =\int \frac{\mathrm{d}k}{(2\pi {)}^4}\mathrm{tr}{\gamma }^{m+1}\exp {\left[\left(-k^2+\frac{1}{4}\left[{\gamma }^{\mu },{\gamma }^{\nu }\right]{\mathcal{F}}_{\mu \nu }\right)/M^2\right]}_{M\to \infty }\end{array}\text{\hspace{1em}(13.31)}$$

where use has been made of the completeness property

$$\sum _i⟨k|{\psi }_i⟩⟨{\psi }_i|k^{\prime }⟩=(2\pi {)}^4{\delta }^4(k-k^{\prime }).$$

In (13.31), we have replaced ${\nabla }^2$ by the symbol $-k^2$ since the residual terms containing $\mathcal{A}$ do not survive in the limit $M\to \infty$ . If we put ${\tilde{k}}^{\mu }\equiv k^{\mu }/M$ , (13.31) becomes

$$\mathbf{A}(x)=\mathrm{tr}[{\gamma }^5\exp (\frac{1}{4}[{\gamma }^{\mu },{\gamma }^{\nu }]{\mathfrak{F}}_{\mu \nu }/M^2)]M^4\int \frac{\mathrm{d}\tilde{k}}{(2\pi {)}^4}\exp (-{\tilde{k}}^2).$$

We expand the first exponential and use

$$\begin{array}{rl}& \mathrm{tr}{\gamma }^5=\mathrm{tr}{\gamma }^5{\gamma }^{\mu }{\gamma }^{\nu }=0\mathrm{tr}{\gamma }^5{\gamma }^{\kappa }{\gamma }^{\lambda }{\gamma }^{\mu }{\gamma }^{\nu }=-4{ϵ}^{\kappa \lambda \mu \nu }\\ & \int \mathrm{d}\tilde{k}\exp (-{\tilde{k}}^2)={\pi }^2\end{array}$$

to obtain

$$\begin{array}{rl}A(x)& =\frac{1}{2}\mathrm{tr}\left[{\gamma }^5\frac{1}{4^2}{\left\{\left[{\gamma }^{\mu },{\gamma }^v\right]{\mathcal{F}}_{\mu v}\right\}}^2\right]\frac{1}{16{\pi }^2}\\ & =\frac{-1}{32{\pi }^2}\mathrm{tr}{ϵ}^{\kappa \lambda \mu \nu }{\mathcal{F}}_{\kappa \lambda }(x){\mathcal{F}}_{\mu \nu }(x).\end{array}\text{\hspace{1em}(13.32)}$$

Note that the higher-order terms in the expansion of the exponential vanish in the limit $M\to \infty$ . The anomalous conservation law (13.25) now becomes

$$\begin{array}{rl}{\partial }_{\mu }{j}_5^{\mu }& =\frac{1}{16{\pi }^2}\mathrm{tr}{ϵ}^{\kappa \lambda \mu \nu }{\mathcal{F}}_{\kappa \lambda }{\mathcal{F}}_{\mu \nu }\\ & =\frac{1}{4{\pi }^2}\mathrm{tr}\left[{ϵ}^{\kappa \lambda \mu v}{\partial }_{\kappa }\left({\mathcal{A}}_{\lambda }{\partial }_{\mu }{\mathcal{A}}_v+\frac{2}{3}{\mathcal{A}}_{\lambda }{\mathcal{A}}_{\mu }{\mathcal{A}}_v\right)\right].\end{array}\text{\hspace{1em}(13.33)}$$

This is regarded as a local version of the AS index theorem. Let us write (13.33) in terms of the field strength $\mathcal{F}=\frac{1}{2}{\mathcal{F}}_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }$ . We easily verify that

$${\nu }_{+}-{\nu }_{-}={\int }_{M}dx{\partial }_{\mu }{j}_{m+1}^{\mu }={\int }_M{\mathrm{ch}}_2(\mathcal{F}).\text{\hspace{1em}(13.34)}$$

This is the index theorem for a twisted spinor complex with trivial background geometry $(\hat{A}(TM)=1)$ .

For $\dim M=m=2l$ , we have the following identity:

$$v_{+}-v_{-}={\int }_{M}dx{\partial }_{\mu }{j}_{m+1}^{\mu }={\int }_M{\mathrm{ch}}_l(\mathcal{F})={\int }_M\frac{1}{l!}\mathrm{tr}{\left(\frac{\mathrm{i}\mathcal{F}}{2\pi }\right)}^l.\text{\hspace{1em}(13.35)}$$

Endnotes

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正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

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Footnotes

1. See section 1.5. For example, we have $\int a\mathrm{d}a=\int ca\mathrm{d}(ca)=1$ , $c\in \mathbb{R}$ and $a$ being a real Grassmann number. This shows that $\mathrm{d}(ca)=\mathrm{d}a/c$ . ↩