How to calculate functional derivative correctly?

Question

Let $phi$  be a real scalar field and $J$ an arbitrary source function. Consider
$$S_{E}[phi, J]=int d^{4} xleft[frac{1}{2}(partial_{mu} phi)(partial^{mu}phi)+frac{1}{2} m^{2} phi^{2}+V(phi)-J(x) phi(x)right].$$
I would like to calculate the functional derivative of $S_E$ with respect to $phi(x)$. My attempt (the bar over the derivatives is just to indicate that we take them with respect to $y$)
$$
begin{align*}
frac{delta S_E[phi,J]}{delta phi(x)} 
&= frac{delta}{delta phi(x)} int d^{4} yleft[frac{1}{2}(barpartial_{mu} phi)(barpartial^{mu}phi)+frac{1}{2} m^{2} phi^{2}+V(phi)-J(y) phi(y)right]
&= m^{2} phi(x)+V'(phi)-J(x) + frac{1}{2}frac{delta}{delta phi(x)} int d^{4} y(barpartial_{mu} phi)(barpartial^{mu}phi)
&overset{P.I}{=}m^{2} phi(x)+V'(phi)-J(x) - frac{1}{2}frac{delta}{delta phi(x)} int d^{4} y,(barpartial^2 phi)phi(y)
end{align*}
$$
and here is where I'm a bit stuck. I thought that
$$frac{delta}{delta phi(x)} int d^{4} y,(barpartial^2 phi)phi(y) = partial^2phi(x),$$
i.e. we treat $barpartial^2phi(y)$ and $phi(y)$ as "independent variables" when we take the derivative with respect to them. But then the final result would be
$$
begin{align*}
frac{delta S_E[phi,J]}{delta phi(x)} 
&= m^{2} phi(x)+V'(phi)-J(x) - frac{1}{2}partial^2 phi(x),
end{align*}
$$
which is wrong. So, could someone explain to me how one takes the functional derivative in this expression properly?

J. Murray · Accepted Answer

Once nice way to calculate functional derivatives is to use the concept of the Gateaux derivative as follows:
$$frac{d}{depsilon}S[phi+epsilon eta]bigg|_{epsilon=0} = int d^4xfrac{delta S}{delta phi} eta$$
In your case,
$$S[phi+epsilon eta]= int d^4x  bigg{frac{1}{2}big((partial phi)^2 + 2epsilon (partial_muphi)(partial^mueta) + epsilon^2(partial eta)^2big)$$
$$ + frac{1}{2}m^2big(phi^2+2epsilon etaphi+epsilon^2eta^2big)+ V(phi+epsilon eta)- J(phi+epsiloneta)bigg}$$
Differentiating and setting $epsilon$ to zero yields
$$ frac{d}{depsilon}S[phi+epsiloneta]bigg|_{epsilon=0} = int d^4x bigg{(partial_muphi)(partial^mueta) + m^2phi eta + V'(phi)eta - Jetabigg}$$
We can cast this into the desired form by integrating by parts, yielding
$$frac{d}{depsilon}S[phi+epsiloneta]bigg|_{epsilon=0} = int d^4xbigg{-partial^2phi + m^2phi + V'(phi) - Jbigg} eta$$
We can therefore read off
$$frac{delta S}{delta phi} = -partial^2phi + m^2 phi+ V'(phi) - J$$

Andrew · Answer

The safest way to compute the functional derivative is to use the following prescription:
begin{equation}
S[phi + delta phi] = S[phi] + int {rm d}^4 x frac{delta S}{delta phi}delta phi + O(delta phi^2)
end{equation}
In other words, add a small perturbation to the field, and manipulate the action so it has the form of an integral times the variation (ignoring terms higher than linear order in the variation). Then the part of the integrand multiplying the variation is the functional derivative.
Here's how to apply this in your example.
We start with the action (I am going to absorb the mass term into the potential since it doesn't really make any difference for this calculation)
begin{equation}
S[phi] = int {rm d}^4 x left(frac{1}{2}(partial phi)^2 + V(phi) + phi J right)
end{equation}
Then we add a perturbation to the field and only keep terms to first order
begin{equation}
S[phi+delta phi] = S[phi] + int {rm d}^4 x left(partial_mu phi partial^mu delta phi + frac{partial V}{partial phi}delta phi + delta phi J right) + O(delta phi^2)
end{equation}
Then we do an integration by parts on the kinetic term so that we remove the derivative from the variation. This leads to
begin{equation}
S[phi+delta phi] = S[phi] + int {rm d}^4 x left[left(-square phi + frac{partial V}{partial phi}+  J right) delta phi right] + O(delta phi^2)
end{equation}
Comparing with the definition above, we see that
begin{equation}
frac{delta S}{delta phi} = -square phi + frac{partial V}{partial phi} +  J
end{equation}

kaylimekay · Answer

Here is a second way to see the correct result for taking the functional derivative of the spacetime derivative of the field, which I hope will be helpful.
Recall that the definition of the functional derivative is
$$ frac{deltaphi(y)}{deltaphi(x)}=delta(y-x) .$$
You know that Dirac deltas are distributions. That is, you should always think of them living under an integral with some test function. So the above definition should really be thought of as
$$ frac{delta}{deltaphi(x)} int phi(y) f(y), dy = int delta(y-x) f(y), dy=f(x)$$ for some arbitrary function $f(y)$.
Now suppose instead you have the spacetime derivative of $phi$.
$$ frac{delta}{deltaphi(x)} int partialphi(y) f(y), dy $$
To understand what this means, just integrate by parts.
$$ -frac{delta}{deltaphi(x)} int phi(y), partial f(y), dy=
-int delta(y-x), partial f(y), dy =-partial f(x) $$
But this is exactly the definition of how the derivative of the Dirac delta is supposed to act on an arbitrary test function. Informally, you can just integrate by parts back to get
$$ -int delta(y-x), partial f(y), dy = int partialdelta(y-x),  f(y), dy .$$
Pulling this result out of its nice safe integral home, we can  write the definition
$$frac{delta}{deltaphi(x)}partialphi(y) = delta'(y-x).$$
Applying this definition to your problem gives the desired result. In shorthand, we can say that the functional derivative just "moves past" the spacetime derivative on the field ($deltapartialphi=partialdeltaphi$), so that it acts "as you'd expect" on the two factors of $partialphi$ and gives you the factor of 2 that you need.

How to calculate functional derivative correctly?

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