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The following is from page 363 of Weinberg volume II.

We wish to evaluate the RHS of

\begin{align}\label{EQbbvbv} [d \psi][d \bar{\psi}] \rightarrow(\operatorname{Det} \mathscr{U} \operatorname{Det} \overline{\mathscr{U}})^{-1}[d \psi][d \bar{\psi}], \end{align}

in order to find the effect on the measure of a change of path integration variables corresponding to a local matrix transformation $\begin{align} \psi(x) \rightarrow U(x) \psi(x) \end{align}$. We define

\begin{align}\notag \mathscr{U}_{x n, y m} &:=U(x)_{n m} \delta^{4}(x-y), \text{ and}\\\label{EQdwdf22vfr} %% %% \bar{\mathscr{U}}_{x n, y m} &:=\left[\gamma_{4} U(x)^{\dagger} \gamma_{4}\right]_{n m} \delta^{4}(x-y), \end{align} where $\gamma_{4} := i \gamma^{0}$ is used in defining $\bar{\psi}=\psi^{\dagger} \gamma_{4}$.

Also note that the indices $n,m$ run over flavour labels and Dirac spin indices.

Let us consider $\alpha(x)$ to be an infinitesimal scalar function in the transformation \begin{align}\label{EQmnmmb4} U(x)=\exp \left[i \gamma_{5} \alpha(x) t\right] \end{align} where $t$ is a general hermitian matrix.

Note that Weinberg omits the working out from here, so the following 3 equations are my own work.

In this case, since the Taylor expansion of the exponential has negligible contributions from terms of order greater the one in $\alpha$, one obtains that \begin{align} [\mathscr{U}-1]_{n x, m y}=i \alpha(x)[\gamma _5 t]_{n m} \delta^{4}(x-y). \end{align} Therefore, \begin{align} \operatorname{Det} \mathscr{U} &= \exp{\text{Tr} \ln \{1+i \alpha(x)\left[\gamma_{5} t\right]_{n m} \delta^{4}(x-y)\}}\\ %% %% &= \exp{\, i \alpha(x)\text{Tr}\{\gamma_{5} t\} \delta^{4}(x-y)}, \end{align} where we have used the identity for the determinant of a matrix $M$, $\operatorname{Det} M=\exp \operatorname{Tr} \ln M$, and that $\ln (1+x)\rightarrow x$ as $x\rightarrow 0.$ But since $\mathscr U$ is pseudo-Hermitian, $(\begin{align} \overline{\mathscr{U}}=\mathscr{U} \end{align})$ we have

\begin{align} [d \psi][d \bar{\psi}] \rightarrow(\operatorname{Det} \mathscr{U})^{-2}[d \psi][d \bar{\psi}]. \end{align}

Weinberg now claims that the measure changes under this transformation as \begin{align}\label{EQnmnnghr3} [d \psi][d \bar{\psi}] \rightarrow \exp \left\{i \int d^{4} x \alpha(x) \mathscr{A}(x)\right\}[d \psi][d \bar{\psi}], \end{align} where we define the anomaly function \begin{align}\label{EQvccvezz33r43} \mathscr{A}(x)=-2 \operatorname{Tr}\left\{\gamma_{5} t\right\} \delta^{4}(x-x). \end{align} We use `Tr' to denote a trace to be taken over Dirac and species indices.

Question 1: Where does the $\delta^{4}(x-x)$ comes from? I can't see any reason why the argument of the delta function might change in the calculations I went through.

Question 2: Where does the integral over $d^4 x$ comes from in the second last equation? If we're working with the Jacobian as on the RHS of the first equation, I don't see how an integral could pop up.

Question 3: This is definitely a rather trivial question, but I've never encountered $\gamma^4$ before... I'm used to the definition $\bar{\psi}=\psi^{\dagger} \gamma_{0}$. This is probably just a different representation of the spinors. If so, could I get a name for it please? I can't seem to find it. Final trivial question: what is a species index?

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Questions 1. and 2. are intimately related.

To try to make sense out of the $\det = \exp \rm Tr \ln$ formula, the meaning of the trace has to be elaborated, I quote QFT and the Standard Model by Schwartz, line 30.58:

$$\det U = \exp \rm Tr \log U = \exp \int d^4x \langle{x}|\rm tr \ln U |x \rangle{}$$

Meaning that the $\rm Tr$ written with uppercase letters here has a double meaning - Tracing over the internal indices (spelled out $n,m$) in your question, and represented by the lowercase $\rm tr$ in the above formula, and "tracing" over the "spacetime indices", represented by the spacetime integral, and by using a one particle Hilbert space here only as a mathematical trick to be able to perform the tracing out.

So in the example above you need the trace of $(\gamma_{5} t) \mathbf{I} $ where the first part $(\gamma_{5} t)$ carries fermionic/color indices and $\mathbf{I}$ is an identity operator with "spacetime" indices, which is above represented by $\delta^{(4)} (x-y)$. Thus

$$\rm Tr\{\gamma_{5} t \mathbf{I} \} = \int d^4 x \langle{x}|\rm tr (\gamma_{5} t) \mathbf{I} |x \rangle{} \\ = \rm tr (\gamma_{5} t) \int d^4x \langle{x}|x \rangle{} \\ = \int d^4x \rm tr (\gamma_{5} t) \delta^{(4)}(x-x)$$

For Question 3, I think this amounts only to a specific notation shown at the beggining of Weinberg's textbook.

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  • $\begingroup$ Exactly what I needed, thank you! $\endgroup$ – user4580791 Oct 8 '19 at 10:34
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    $\begingroup$ glad to help :) $\endgroup$ – Mateo Oct 8 '19 at 10:57
  • $\begingroup$ In the final equation in my question, I assume Weinberg is just using bad notation for 'Tr'? Clearly there are no spacetime indices to be traced in that expression, so I'm wondering why he didn't use 'tr' like you did in your final expression. Or should it be clear from context? I'm finding it hard to imagine why Weinberg wouldn't elaborate on his Tr notation involving the spacetime trace. I wouldn't have needed to ask this question if he just explained what he means by 'Tr' like he does every other time. $\endgroup$ – user4580791 Oct 8 '19 at 11:19
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    $\begingroup$ It is actually explained in the volume 1, on page 392. The textbook by Schwartz was nearer to me, so I quoted from there first. Note that uppercase/lowercase notation might mean reverse things for different sources, but the idea is to discern between tracing only group indices and tracing over spacetime. $\endgroup$ – Mateo Oct 8 '19 at 12:19

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