In QFT for gifted amateur pg. 13, the functional derivative for the functional

$$J[f]=\int [f(y)]^p \phi(y)dy$$

is given by

$$\frac{\delta J[f]}{\delta f(x)}= \lim_{\epsilon\rightarrow0} \frac{1}{\epsilon} (\int [f(y)+\epsilon\delta(y-x)]^p \phi(y) dy - \int[f(y)]^p \phi(y)) =p[f(x)]^{p-1} \phi(x). $$

To check that this is true, I expanded the first integral on the RHS,

$$\lim_{\epsilon\rightarrow0} \frac{1}{\epsilon} \int [f(y)+\epsilon\delta(y-x)]^p \phi(y) dy=\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon}\int [f(y)^p+pf(y)^{p-1}\epsilon\delta(y-x) + \binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2+...]dy.$$

I have trouble showing that the integral $$\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int\binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2dy=0.$$

Carrying out the integration using one of the Dirac delta function, we get

$$\lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\int\binom{p}{2}f(y)^{p-2}\epsilon^2\delta(y-x)^2dy= \epsilon\binom{p}{2}f(x)^{p-2} \delta(x-x).$$

Since $\delta(x-x)$ is infinity, how can we say that this expression is zero? It seems to me that it is undefined since $\epsilon$ is also a very small number. Why can we say $\epsilon \delta(x-x) = 0$?


1 Answer 1


Since $\delta(x)$ is not an operational function, it can only be defined by a limiting process. Your variation is in fact including two limiting processes. For multi-limiting processes, the order of taking limitation is relevant to the final result.

In your formulation, you may replace the $\epsilon \delta(x-y)$ by a Gaussian function $$ \epsilon \delta(x-y) = \lim_{\sigma\to 0}\frac{\epsilon}{\sigma\sqrt{\pi}} e^{-\left(\frac{x}{\sigma}\right)^2} $$ It means that you impose small a Gaussian deviation from the original function at near $x$ with a strength $\epsilon$. Then, the limit of $\sigma$ is performed after the limit of $\epsilon.$

  • $\begingroup$ Is there anywhere I can read more about the order of limiting process? $\endgroup$
    – TaeNyFan
    Feb 2, 2022 at 4:03
  • $\begingroup$ Also why must we take the limit of $\epsilon$ first instead of the limit of $\sigma$ first? $\endgroup$
    – TaeNyFan
    Feb 2, 2022 at 4:04
  • $\begingroup$ Some good practices about the limiting process appear in the area of complex analysis. Where you did integral for a singular point on the integral path, you have to move the singular up or down, then restore after the integration, These processes generating the advanced and retarded green functions. $\endgroup$
    – ytlu
    Feb 3, 2022 at 1:56
  • $\begingroup$ As for this problem, consider that you are doing a numerical computation for the functional variation. You put a small deviation (strength $\epsilon$) about a point $x_0$, perform the integration, find out $\Delta J (\epsilon)$ as function of the strength, take the limit $\epsilon \to 0$. After all this had done, you then may narrow the deviation to a $\delta$ function. If you limit you deviation into a $\delta$ function prior to the process mentioned above. The numerical process will not able to proceed using a $\delta$ function. $\endgroup$
    – ytlu
    Feb 3, 2022 at 2:04

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