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In studying QFT and General Relativity, I came across two different definitions of Functional Derivative, and I'd like to know if they are equivalent.

  1. Firstly, in Wald's book General Relativity, as well as in other GR references (Baez), the variation of the action is given in terms of a "one-parameter family of field configurations $\psi_\epsilon$" and this family can be defined as $\psi_\epsilon = \psi_0+\epsilon\phi$, where $\phi$ is an arbitrary field. Then, the following definitions are made: $$d\psi_\epsilon/d\epsilon|_{\epsilon=0}:=\delta\psi \ \ \ \ \ \ \ \ \ \ \ \delta S:=\frac{d}{d\epsilon}S[\psi+\epsilon\phi]\Big|_{\epsilon=0}$$ where $S[\psi]$ is the functional of interest. Finally, the Functional Derivative $\delta S/\delta\psi(x)$ is defined as follows:

$$\delta S= \int \mathrm{d}^4x\frac{\delta S}{\delta\psi(x)}\phi(x) = \frac{d}{d\epsilon}S[\psi+\epsilon\phi]\Big|_{\epsilon=0}$$

  1. When I turn to a QFT reference, like Greiner's Field Quantization, the definition is practically the same, but the arbitrary field $\phi$ is now specified as $\delta^4(x-x')$. I understand that this specification can be interpreted as a variation at the position $x'$ alone, so that the integral can be seen as an analogous of $dS = \sum \frac{\partial S}{\partial x_i}dx_i$. It also seems important when dealing with Generating Functionals, but I haven't studied them yet, so I might be wrong.

I would like to know why is this specification ($\phi=\delta^4(x-x')$) made and if both definitions are equivalent.

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In most practical cases the two definitions are equivalent, but the latter definition is not mathematically sane. The functional $S$ is defined for smooth functions, and $\delta^4$ is not even a function.

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Aside from the latter definition not being "mathematically sane" as md2perpe pointed out, the equivalence can be established easily as follows:

Let $\delta_{x_0}$ denote the Dirac delta distribution centered on $x_0$, eg. $\delta_{x_0}(x)=\delta(x-x_0)$.

Assume that $S$ is functionally differentiable at $\psi$. then for any variation $\phi$ we have $$ \delta S[\psi]=\int d^4x \frac{\delta S[\psi]}{\delta\psi(x)}\phi(x). $$ Since this is true for any variation (or at least those that vanish at the boundary), we can replace $\phi$ with $\delta_{x_0}$. Then we get $$ \delta S[\psi]_{\text{specific}}=\int d^4x\frac{\delta S[\psi]}{\delta\psi(x)}\delta_{x_0}(x)=\frac{\delta S[\psi]}{\delta\psi(x_0)}, $$ so using the Dirac delta distribution as a "test function" will get you the functional derivative directly, rather than indirectly.

But of course, as it has been pointed out, this is a formal manipulation and is not actually sane mathematically.

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