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I'm confused about the evaluation of the functional derivative of Equation 1.12, $$J[f] = \int [f(y)]^p \phi(y) dy$$ on page 13 of Quantum Field Theory for the Gifted Amateur in Chapter 1.

Here are the steps shown in the text: $$\begin{align} \frac{\delta J[f]}{\delta f(x)} &= \lim \limits_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int [f(y) + \epsilon \delta(y - x)]^p\phi(y) dy - \int [f(y)]^p \phi(y) dy \right] \\ &= p[f(x)]^{p-1}\phi(x) \end{align}\tag{1.12}$$

How do they get from the first line to the second?

Qmechanic
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aadithyaa
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1 Answers1

2

Working at $\mathcal{O}(\epsilon)$ and the binomial expansion you have $$[f(y)+\epsilon\delta(x-y)]^p\approx[f(y)]^p+\epsilon{p\choose 1}[f(y)]^{p-1}\delta(x-y)+\mathcal{O}(\epsilon^2).\tag{1}\label{1}$$ The $\mathcal{O}(\varepsilon^2)$ terms divided by $\epsilon$ yield $\mathcal{O}(\epsilon)$ terms, which vanish in the $\varepsilon\to 0$ limit. Using \eqref{1} and the last statement, you have \begin{align} \frac{\delta J[f]}{\delta f(x)}&=\\ &=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int [f(y) + \epsilon \delta(y - x)]^p\phi(y) dy - \int [f(y)]^p \phi(y) dy \right]=\\ &\overset{\eqref{1}}{=}\lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int \left\{[f(y)]^p\phi(y)+ \epsilon p[f(y)]^{p-1}\delta(y - x)]^p\phi(y)\right\} dy - \int [f(y)]^p \phi(y) dy +\mathcal{O}(\epsilon^2)\right]=\\ &=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int\epsilon p[f(y)]^{p-1}\delta(y - x)]^p\phi(y) dy\right]=\\ &=p[f(x)]^{p-1}\phi(x). \end{align}

Does the book do this using the limit definition? After defining the functional derivative that way you may see that it behaves "well" under composition and use what you know from ordinary calculus (cf. point 2 in this answer by Qmechanic), in OP'case case the power rule.

Feynmate
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