# Multiple functional derivatives

1. Jul 17, 2010

### twoform

Hi all,

Long time stalker, first time poster. I've finally got stumped by something not already answered (as far as I can tell) around here. I'm trying to make sense of double functional derivatives: specifically, I would like to understand expressions like

$$\int dx \frac{\delta^2}{\delta \phi(x) \delta \phi(x)} \Psi[\phi]$$.

What can happen is that taking one derivative gives me an expression with a $$\phi(x)$$ sitting in front of $$\Psi$$, and then I'm not sure how to act with the second functional, since I now have something like a function times a functional; the naive approach gives me a bunch of delta functions. For example, for a Gaussian

$$\Psi[\phi] = exp \left[ -\frac{1}{2} \int dx' \phi(x') \phi(x') \right]$$

the first derivative is

$$\frac{\delta \Psi}{\delta \phi(x)} = \left[ -\int dx' \delta(x-x') \phi(x') \right] \Psi = - \phi(x) \Psi$$.

Now naively

$$\frac{\delta^2 \Psi}{\delta \phi(x) \delta \phi(x)} = \frac{\delta \phi(x)}{\delta \phi(x)} \Psi + \phi(x) \frac{\delta \Psi}{\delta \phi(x)}$$

but the first term is just $$\delta(0)$$! Which is even worse when I then try to integrate this over dx.

So, my guess is that I'm supposed to instead treat the functional derivative as a partial derivative when there's some function of $$\phi(x)$$ sitting in front of the functional. But I

a. don't know if this is true
b. don't know why it should be true.

Any help appreciated!

Thanks,
Dan