# Taking functional derivatives of generating functional

I'm learning how to compute functional derivatives of generating funtionals. Suppose I have the generating functional $$Z[J] = \exp\left\{\int{dy_1 \; dz_1\; J(y_1) \Delta(y_1 - z_1) J(z_1)}\right\}.$$

Now suppose I want to compute the functional derivative $$\frac{\delta}{\delta J(y)}\frac{\delta}{\delta J(x)} Z[J].$$ I'm not sure how to compute it. My main confusion is about how to take derivatives of a function like $$J(y_1)$$ with respect to $$J(x)$$. For example, what is the value of $$\frac{\delta J(y_1)}{\delta J(x)}$$? Is that equal to some kind of delta function? How does one compute the derivative above? Do I have to use the product rule? Could you explain what are the rules for functional differentiation?

Following the first comment, my guess would be that, for example, \begin{align} \frac{\delta}{\delta J(x)} [ J(y_1) \Delta(y_1 - z_1) J(z_1) ] &= (\delta(x-y_1) \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \delta(x-z_1)), \\ &= \Delta(x-z_1)J(z_1) + J(y_1) \Delta(y_1 - x), \end{align} where I have used the ordinary product rule. Is that correct?

• In this answer physics.stackexchange.com/a/590696/93729 I introduce functional derivatives in an intuitive way (I think). The take away is that as a rule you can use $\frac{\delta J(y)}{\delta J(x)}=\delta(x-y)$ with $\delta(x-y)$ the Dirac delta function along with the usual rules of differentiation. Commented Apr 30, 2021 at 21:36
• Your edit seems correct, note that a second functional derivative is going to give you a second delta function, which, when evaluated in the integral will give you the Feynman propagator between $x$ and $y$. Also also note that you will also have to take another derivative of the exponential which is going to pull down another factor, however when you evaluate this at $J=0$ (as you do) any terms containing $J$ (other than in the exponential) will be equal to zero. Commented Apr 30, 2021 at 21:46

## 3 Answers

My main confusion is about how to take derivatives of function like $$J(y_1)$$ wrt to $$J(x)$$. For example what is the value of $$\frac{\delta J(y_1)}{\delta J(x)}$$. Is that equal some kind of delta function?

Yes, it is equal to a delta function. To see this, write $$J(y_1)$$ as a functional of $$J$$: $$J(y_1) = \int dx J(x)\delta(x - y_1)\;.\tag{A}$$

From Eq. (A) it should be clear that $$\frac{\delta J(y_1)}{\delta J(x)} = \delta(x-y_1)$$

$$\frac{\delta}{\delta J(x)} [ J(y_1) \Delta(y_1 - z_1) J(z_1) ]$$ $$= (\delta(x-y_1) \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \delta(x-z_1))$$ $$= \Delta(x-z_1)J(z_1) + J(y_1) \Delta(y_1 - x)$$

where I have used the ordinary product rule. Is that correct?

The first equality is correct, the second one is not, since you seem to have just removed the delta functions by hand for no reason following the second equality.

A functional derivative is a natural continuation of the concept of a multivariable derivative to an infinite-dimensional setting. Recall that in multivariable calculus, the gradient operator is defined so that \begin{align*} f(\vec x+\vec h)\approx f(\vec x) + \vec h\cdot \nabla f(\vec x)+\mathcal O(\|\vec h\|^2). \end{align*} By analogy, the functional derivative $$\delta$$ is such that $$F[g+h]\approx F[g] + \langle \delta F[g],h\rangle$$, or \begin{align*} F[g+h]\approx F[g] + \int\delta F[g](x)h(x)d\mu(x) \end{align*} (where $$\mu$$ is the underlying measure.)

Calculations can be performed on the basis of this definition. For example, consider the 'evaluation functional', $$F_x[\cdot]$$ say, that evaluates its argument at $$x$$. Then $$F_x[J]=J(x)$$, and its functional derivative $$\delta F_x[J](y)\equiv K(y)$$ satisfies \begin{align*} \int K(y) h(y) d\mu(y) &\approx F_x[J+h]-F_x[J]\\ &=J(x) + h(x) - J(x)\\ &=h(x). \end{align*} As this holds for all values of $$x$$, we can recognize that $$\delta F_x[J](y)$$ behaves exactly like $$\delta(x-y)$$.

[In the second example, the expression $$J(y_1)\Delta(y_1-z_1)J(z_1)$$ is evaluated at two points, and so the functional derivative would need to either be restricted to one argument, either $$y_1$$ or $$z_1$$, producing something like $$\delta(x-y_1)\Delta(y_1-z_1)J(z_1)$$, or apply to both simultaneously, in which case the result would have the form $$\Delta(y_1-z_1)\delta(x-y_1)\delta(x'-z_1)$$. Note that the result is only tangentially related to what you would obtain by taking the functional derivative with respect to $$J(x)$$ of $$F_\Delta[J]\equiv\int d\mu(y_1)d\mu(z_1)\big[J(y_1)\Delta(y_1-z_1)J(z_1)\big]$$ which yields $$2\int d\mu(y_1)J(y_1)\Delta(y_1-x)$$, or perhaps more accurately $$\int d\mu(y_1)J(y_1)(\Delta(y_1-x)+\Delta(x-y_1))$$, as can be checked by comparing with the linear term in the evaluation of $$F_\Delta[J+\eta]$$. Taking a second derivative yields $$2\Delta(x-x')$$, or $$\Delta(x-x') + \Delta(x'-x)$$, depending on the symmetries of $$\Delta$$.]

The examples that we encounter in QFT are somewhat more complicated, but nonetheless can be approached using the standard technique of renormalized perturbation theory with Feynman diagrams. This is often performed with respect to a finite (discretized, finite volume) 'regularized' theory whose calculations should approach a well-defined limit as the lattice spacing approaches zero and the volume approaches infinity (keeping in mind that the lattice spacing is itself typically measured in terms of the values of the fields themselves.) In cases relevant to the vast majority of graduate students, the most important punchline of the finite regularized limiting definition is that all the standard tricks of calculus can be used at a formal level when introducing and evaluating perturbative expansions. For example, it can otherwise be somewhat difficult to approach even "seemingly innocuous" expressions such as $$\exp(-\frac{\lambda}{4!}\int\phi^4)$$ which may appear under a path integral, given that the properties of typical $$\phi$$ instances, which may be somewhat erratic, are not known a priori, and may even call into question the meaning of expressions such as $$\phi^4$$ (which as it turns out is worth worrying about.) For an uncomfortably long time, a non-trivial danger to the perturbative QFT enterprise was the possibility that non-trivial field theories simply didn't exist. This concern has been allayed to some extent through agreement between the standard model and experimental results at CERN, as well as a somewhat heroic accomplishment of Arthur Jaffe and others in proving the existence of non-trivial quantum field theories.

Be careful. Your first line is correct, but the statement in the second line is not. You can only use it inside the integration to get rid of the delta function. As I will show you how to compute the first functional derivative $$\frac{\delta}{\delta J(x)} Z[J]$$: $$\frac{\delta Z[J]}{\delta J(x)} = Z[J] \cdot \frac{\delta}{\delta J(x)} \left\{ \int d y_1 d z_1 J(y_1) \Delta(y_1 - z_1) J(z_1) \right\}.$$

Using the chain rule for functional derivatives: $$\frac{\delta}{\delta J(x)} \left\{ \int d y_1 d z_1 J(y_1) \Delta(y_1 - z_1) J(z_1) \right\} = \int d y_1 d z_1 \left[ \frac{\delta J(y_1)}{\delta J(x)} \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \frac{\delta J(z_1)}{\delta J(x)} \right].$$

Since $$\frac{\delta J(y_1)}{\delta J(x)} = \delta(y_1 - x)$$ and $$\frac{\delta J(z_1)}{\delta J(x)} = \delta(z_1 - x)$$, we get: $$\int d y_1 d z_1 \left[ \delta(y_1 - x) \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \delta(z_1 - x) \right] = \int d y_1 d z_1 \left[ \delta(y_1 - x) \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \delta(z_1 - x) \right].$$

NOW, simplifying using the properties of the delta function: $$\int d y_1 d z_1 \left[ \delta(y_1 - x) \Delta(y_1 - z_1) J(z_1) + J(y_1) \Delta(y_1 - z_1) \delta(z_1 - x) \right] = \int d z_1 \Delta(x - z_1) J(z_1) + \int d y_1 J(y_1) \Delta(y_1 - x).$$

Since $$\Delta$$ inside the integrand is symmetric, $$\Delta(x - z_1) = \Delta(z_1 - x)$$: $$\int d z_1 \Delta(x - z_1) J(z_1) + \int d y_1 J(y_1) \Delta(y_1 - x) = 2 \int d z_1 \Delta(x - z_1) J(z_1).$$

Thus, $$\frac{\delta Z[J]}{\delta J(x)} = 2 \int d z_1 \Delta(x - z_1) J(z_1) Z[J].$$

Now using the same idea, you can compute the second derivative.