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I´m wondering how can one get to the definition of Functional Derivative found on most Quantum Field Theory books:

$$\frac{\delta F[f(x)]}{\delta f(y) } = \lim_{\epsilon \rightarrow 0} \frac{F[f(x)+\epsilon \delta(x-y)]-F[f(x)]}{\epsilon}$$

from the definitions of Functional Derivatives used by mathematicians (I´ve seen many claims that it is, in effect, the Fréchet derivative, but no proofs). The Wikipedia article says it´s just a matter of using the delta function as “test function” but then goes on to say that it is nonsense.

Where does this $\delta(x-y)$ comes from?

Qmechanic
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Forever_a_Newcomer
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4 Answers4

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Whenever I have troubles with functional derivative things, I just do the replacement of a continuous variable $x$ into a discrete index $i$. If I'm not mistaken this is what they call a "DeWitt notation".

The hand waiving idea is that you can think of a functional $F[f(x)]$ as of a "ordinary function" of many variables $F(f_{-N},\cdots,f_0,f_1,f_2,\cdots,f_N) = F(\vec{f})$ with $N$ going to "continuous infinity".

In that language your functional derivative transforms into partial derivative over one of the variables: $$\frac{\delta F}{\delta f(x)} \to \frac{\partial F}{\partial f_i}$$ And the delta-function is just an ordinary Kronecker delta: $$\delta(x-y) \to \delta_{ij}$$

So, gathering this up we for your expression: $$\frac{\delta F}{\delta f(x)} = \lim_{\epsilon\to\infty}\frac{F[f(x)+\epsilon\delta(x-y)]-F[f(x)]}{\epsilon} \to$$ $$\frac{\partial F}{\partial f_j} = \lim_{\epsilon\to\infty}\frac{F[f_i+\epsilon\delta_{ij}]-F[f_i]}{\epsilon} $$ Which is, to my taste, a bit redundant. But true.

Kostya
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7

This is a formal notation for the following general thing:

$$F(f+\delta f) = F(f) + \int A(x) \delta f(x) $$

Where $\delta f$ is the infinitesimal change in f, and it is a smooth test function, and then on the right hand side, $A(x)$ is just a linear operator on the space of functions. The notation for the $A(x)$ is then

$$ A(x) = {\delta F\over \delta f(x)}$$

Because if you formally substitude $\delta f(x) = \delta(x-y)$, you find $A(y)$ as the value of the integral. This is just a notational trick--- $\delta f$ is an everywhere small variation, which is impossible if it is infinite at one point. Another way of saying this is that the point-delta-function limit has to be taken after the small epsilon limit in the definition you give, so that the variation becomes small before it becomes infinitely concentrated.

Ron Maimon
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5

The physicist's derivative notation denotes the components of a Frechet derivative in the direction of the delta-function supported at $y$.

This is one of those places where the habit of denoting the function $f$ by its value $f(x)$ gets confusing. It's somewhat clearer if you write $\delta_y$ for the delta function at $y$, and

$$ \frac{\delta F}{\delta (\delta_y)}[f] = \lim_{\epsilon \to 0} \frac{1}{\epsilon} ( F[f + \epsilon \delta_y] - F[f]). $$

Obviously, the delta function isn't actually a function. But this use of it makes exactly as much sense as the position basis (and the latter can be made perfectly rigorous using rigged Hilbert spaces).

David Z
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user1504
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3

Let the definition of the functional derivative be

$$ \int \frac{\delta F}{\delta \rho(x)} \phi(x) dx = \lim_{\epsilon \rightarrow 0} \frac{F[\rho + \epsilon \phi] - F[\rho]}{\epsilon} $$

Choose

$$ \phi(x) = \delta(x-y) $$

And complete the integration on the left-hand side

$$ \int \frac{\delta F}{\delta \rho(x)}\delta(x-y) dx = \frac{\delta F}{\delta \rho(y)} = \lim_{\epsilon \rightarrow 0} \frac{F[\rho + \epsilon \delta(x-y)] - F[\rho]}{\epsilon} $$

Let a physicist choose a function in an integrand, and he'll choose a delta function every time.

Justin Ratner
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