Multiple functional derivatives

[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

 

[jvicens]

Hello Dan,

Thank you for your post and for bringing up this interesting problem. I can definitely understand your confusion with double functional derivatives. Let me try to explain this concept to you in a simple and clear manner.

First, let's define what a functional derivative is. A functional derivative is a generalization of the concept of a derivative to functions of multiple variables. In simple terms, it is the rate of change of a functional with respect to its argument. In your example, \Psi[\phi] is the functional and \phi(x) is the argument.

Now, when we take the first functional derivative of \Psi[\phi], we are essentially treating \phi(x) as a variable and taking the derivative with respect to it. This gives us \frac{\delta \Psi}{\delta \phi(x)} = - \phi(x) \Psi. This means that for every value of \phi(x), we have a corresponding value for the functional \Psi[\phi].

But when we take the second functional derivative, we are essentially taking the derivative of \frac{\delta \Psi}{\delta \phi(x)} with respect to \phi(x) again. This is where your confusion arises. The key thing to remember here is that \frac{\delta \Psi}{\delta \phi(x)} is not just a function, but a functional. So when we take the derivative of a functional with respect to its argument, we get another functional. This is why the first term in your naive approach is \delta(0), as you correctly pointed out.

To avoid this confusion, we can treat the functional derivative as a partial derivative when there is a function of \phi(x) sitting in front of the functional. This is because in this case, we are essentially taking the derivative of a function with respect to one of its variables. In your example, the function in front of the functional is just \phi(x), so we can treat the functional derivative as a partial derivative and get the correct result.

I hope this explanation helps clear up your confusion. If you have any further questions, please feel free to ask. Keep up the good work in your studies!