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In page 666 (it couldn't be other way - bad joke), chapter 19, the Eq. (19.73) claims (see properties of the $\phi_n(x)$ functions in this post: Change of variables in path integral measure):

$$ \sum_n \phi^\dagger_n(x)\gamma^5 \phi_n(x) = \lim_{\substack{M \rightarrow \infty}}\sum_n \phi^\dagger_n(x)\gamma^5 e^{(i\not{D})^2/M^2}\phi_n(x) = \lim_{\substack{M \rightarrow \infty}}\langle x| tr\{\gamma^5 e^{(i\not{D})^2/M^2}\} |x\rangle $$

Can anyone explain me the appearance of $tr$, i.e., trace inside the bracket and the bracket itself? In principle, the trace is given by the sum over $n$ so I don't understand why it is inside the bracket

EDIT I've seen another ways to compute this and the last step is changed by

$$ \lim_{\substack{M \rightarrow \infty}}\sum_n \phi^\dagger_n(x)\gamma^5 e^{(i\not{D})^2/M^2}\phi_n(x) = \lim_{\substack{M \rightarrow \infty \\ y \rightarrow x}}tr\{\int \frac{d^4p}{(2\pi)^4}e^{-ipy}\gamma^5 e^{(i\not{D})^2/M^2}e^{ipx}\} $$

But how can you go from $\phi_n(x)$ that are supposed to be an spinor basis to just an exponential and also there is a trace that I don't undertand if we are changing $\sum_n$ by $\int \frac{d^4p}{(2\pi)^4}$ and this is already representing the trace.

Source: http://www.int.washington.edu/users/dbkaplan/572_16/PhysRevD.21.2848.pdf Eq. (2.15)

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  • $\begingroup$ Just integrate over $x$ across the second equal sign. I believe he was using the notation $Tr ... \equiv \int dx \langle x|tr ...| x\rangle$ $\endgroup$
    – pathintegral
    Commented Apr 1, 2019 at 23:18
  • $\begingroup$ @pathintegral I'm almost sure that the original trace, what you call $Tr...$, is given by the sum in $n$-type indices and not over the $x$ since after the calculus of this trace you have to integrate in this variable $\endgroup$
    – Vicky
    Commented Apr 1, 2019 at 23:27
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    $\begingroup$ Define $G=\gamma^5\exp((i\gamma{D})^2/M^2)$ and think of $G$ as a "matrix" with components $G_{n,m}(x,y)$, with both discrete and continuous "indices". Let $\phi_n(x)$ be an orthonormal basis with respect to the discrete index so that $G_{n,m}(x,y)=\langle x|\phi_n^\dagger G\phi_m|y\rangle$, and define $tr\{G(x,y)\} = \sum_n G_{n,n}(x,y)$ to be the partial trace over the discrete index. The trace is just a sum, so it can be moved in or out of the bracket: $\sum_n\langle x|f_n|y\rangle=\langle x|\sum_n f_n|y\rangle$. Does that help? $\endgroup$
    – Chiral Anomaly
    Commented Apr 2, 2019 at 0:01
  • $\begingroup$ @ChiralAnomaly More or less, but there is still something that I don't like: you are not calculating the trace of $\langle x|\phi^\dagger_n(x)G\phi_m(y)|y \rangle$ but the trace of $\phi^\dagger_n(x)G\phi_m(y)$, i.e., $tr\{\phi^\dagger(x)·G·\phi(y)\} = \sum_n \phi^\dagger_n(x)G\phi_n(y)$, where did the bra and the ket come up? Recall that $\phi_n$ is not the $n$-component of a vector but a whole vector labelled by a mode index $n$ $\endgroup$
    – Vicky
    Commented Apr 2, 2019 at 0:29
  • $\begingroup$ @ChiralAnomaly I'm not very sure about that since I've read papers where to go from sum over n to the part on brackets, they don't even use them but change $\phi_n(x)$ by $e^{ipx}$ as 'plane waves'. See: int.washington.edu/users/dbkaplan/572_16/PhysRevD.21.2848.pdf Eq. (2.15) Also I don't see how you can go from $\phi_n(x)$ that are supposed to be an spinor basis to just an exponential $\endgroup$
    – Vicky
    Commented Apr 2, 2019 at 0:34

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