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I am currently reading Zee's book on quantum field theory, and I am in the chapter where he is introducing Grassmann integrals.

He re-introduces the path integral evaluated for the vacuum, i.e. no sources: \begin{equation} Z=Ce^{-\frac{1}{2}Tr(log(\partial^2+m^2))}. \end{equation} We note that the trace of an operator can be written as $$ Tr(O)=\int d^4x\int \frac{d^4k}{(2\pi)^4}\frac{d^4q}{(2\pi)^4}\langle x|k\rangle\langle k|O|q\rangle\langle q|x\rangle. $$

Then we note that $Z=\langle 0|e^{-iHt}|0\rangle=e^{-iET}$ for the vacuum, therefore $$ iET=\frac{1}{2}Tr(log(\partial^2+m^2)). $$ Now what I don't understand is when he said that this is evaluated as $$ iET=\frac{1}{2}VT\int\frac{d^4 k}{(2\pi)^4}log(k^2+m^2+i\epsilon)+A. $$ Where $A$ is for divergent terms in the constant "$C$" earlier (this I understand).

So, my question is how did he end up from the second to the last equation to the last equation using the trace identity?

He didn't give any information at all and he just straight up gave it as is.

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  • $\begingroup$ I am a bit confused— have you tried applying the equation you wrote for $Tr(O)$ in order to calculate $Tr(log(\partial^2 + m^2))$ yet? $\endgroup$ – Diffycue Oct 14 '18 at 23:52
  • $\begingroup$ Yes, but I am getting nowhere. I don't understand how that simplifies to the final one.... I am just as confused as you. $\endgroup$ – Gradient137 Oct 15 '18 at 0:38
  • $\begingroup$ Should the last $k$ in the integral expression for $Tr(O)$ be an $x$? And the $V$ in the final formula, is it just the volume? $\endgroup$ – Sunfoil Oct 15 '18 at 1:05
  • $\begingroup$ Oh ya I edited it. And ya it is the volume...It's after integrating in 3 space. I still don't get how he ended up with that. I don't get how he ended up with anything. $\endgroup$ – Gradient137 Oct 15 '18 at 3:52
  • $\begingroup$ "He didn't give any information at all and he just straight gave it as is." That's Zee for you. This step isn't so bad, but there are many like it. His textbook is fine if you have already seen everything he is doing somewhere else first. $\endgroup$ – octonion Oct 15 '18 at 5:26
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Use the following representations and integrals:

$$ \langle x | p \rangle = \int \mathrm d^4 x\; e^{-ipx}$$

$$ \int \mathrm d^4 x\;e^{-ix(k-q)} = (2 \pi)^4 \delta(k-q) $$

Explicitly substituting the expression into the trace representation of an operator:

$$ \mathrm{Tr}\left[\log(\partial^2 + m ^2)\right] =\int \mathrm d^4x\int \frac{\mathrm d^4k}{(2\pi)^4}\frac{\mathrm d^4q}{(2\pi)^4}\langle x|k\rangle\langle k|\log(\partial^2 + m^2)|q\rangle\langle q|x\rangle $$

$$ = \int \mathrm d^4x\int \frac{\mathrm d^4k}{(2\pi)^4}\frac{\mathrm d^4q}{(2\pi)^4}e^{-ix(k-q)}\langle k|\log(\partial^2 + m^2)|q\rangle $$ $$ = \int \frac{\mathrm d^4k \; \mathrm d^4q}{(2\pi)^4}\langle k|\log(\partial^2 + m^2)|q\rangle \delta(k-q) = \int \frac{\mathrm d^4k}{(2\pi)^4}\langle k|\log(\partial^2 + m^2)|k\rangle $$

$$ = \int \frac{\mathrm d^4k}{(2\pi)^4}\log(-k^2 + m^2)$$

because the Fourier transform of $\partial_{\mu}$ is $ik_{\mu}$, and it's diagonal in the momentum representation. Another way to determine this is to note that $\mathrm{Tr}(\log M) = \log\det M$. This is equivalent to the final expression, with some damping factor (the $i\epsilon$ prescription) introduced for convergence.

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  • $\begingroup$ Oh wow I'm so stupid. In the days of me doing Fourier transforms to calculate amplitudes. This time I forgot to do Fourier transforms! Thanks a lot. $\endgroup$ – Gradient137 Oct 15 '18 at 18:16
  • $\begingroup$ Do you know where the 4-volume $VT$ comes from? It should come from the integral for $d^4x$ which should give a divergent delta, but your derivation looks right yet it doesn't have that. $\endgroup$ – Gradient137 Oct 16 '18 at 17:20
  • $\begingroup$ Nvm, I edited the answer, there should be an extra integral over spacetime giving an additional factor of 4-volume. $\endgroup$ – Gradient137 Oct 16 '18 at 20:48

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