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In following an extension course on quantum field theory, a problem popped up that my TAs couldn't quite explain to my satisfaction. I suspect the answer is really simple, so I hope somebody with a mathematical background can elucidate.

What is given is as follows:

  • A field variable $\phi$ can be rewritten into another field variable, $\phi'$ arbitrarily.
  • The boundary conditions do not contribute, when integrating by parts.

Given the above and some arbitrary expression $F$, the following should allegedly hold: \begin{align} \int [d\phi]\frac{\delta F}{\delta\phi}=0 \end{align} where the square around the measure absorbs any constants from the measure that may appear (so we normalise any expression to have no constants and only the relevant parts).

The argument the TAs made is that by the above, also \begin{align} \int [d\phi']\frac{\delta F}{\delta\phi'}=0. \end{align}

Expanding $\phi'$ around $\phi$, we must argue that the leading order coefficient in the series expansion must vanish, the other coefficients are zero since a second order variation is "doubly infinitesimal" and thus vanishing.

But I don't consider this a satisfactory explanation. I've been looking at Zee, Quantum Field Theory in a Nutshell, which does gloss over it, but I'm still confused:

  • How does the measure change?

  • Locality of $\phi$ appears to dictate whether the 'derivative chain rule' applies. But these locality properties are not declared initially, and I'm not sure what the considerations are to determine it. What I'm getting at is whether \begin{align} \frac{\delta}{\delta\phi(x)} = \int d^d y \frac{\delta\phi(y)}{\delta \phi(x)} \frac{\delta}{\delta \phi(y)} \end{align} for example, with/without the integral over some $y$.

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  • $\begingroup$ Possible duplicate: Why is the functional integral of a functional derivative zero? $\endgroup$ – Qmechanic Feb 22 at 17:04
  • $\begingroup$ @Qmechanic I believe this answers the question in part. Perhaps you could explain the ideas behind picking the functional derivative chain rule in some way in an answer, or you could mark it as duplicate and answer this in chat (?). Just thinking out loud. Thanks! $\endgroup$ – 1010011010 Feb 24 at 9:47
  • $\begingroup$ Consider to reword/stress more clearly the parts of your question that are not answered by the linked post. $\endgroup$ – Qmechanic Feb 24 at 14:03

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