# Why is the functional integral of a functional derivative zero?

I'm reading Quantum Field Theory and Critical Phenomena, 4th ed., by Zinn-Justin and on page 154 I came across the statement that the functional integral of a functional derivative is zero, i.e. $$\int [d\phi ]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} = 0$$ for any functional $F[\phi ]$.

I would be most thankful if you could provide a mathematical proof for this identity.

• Try discretizing spacetime. – Michael Brown Aug 17 '13 at 8:06
• The functional integral of a total derivative vanishes, which results from the variation of fixed boundary conditions. – soliton Aug 17 '13 at 8:27
• @soliton Could you please write some equations to better understand your statement? I tried a simple case, $F[\phi] = \int_{-\infty}^{\infty} C(x)\phi (x)dx$. Then $\int [d\phi]\frac{\delta F}{\delta\phi (x)} = C(x)\int [d\phi] \neq 0$ – user22208 Aug 17 '13 at 8:54
• Sorry for late reply. As Trimok says, $F(-\infty) - F(\infty)$ makes sense. – soliton Aug 18 '13 at 11:29

If the functional derivative

$$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}$$

exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally

$$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x).$$

OP's functional integral formula

$$\tag{3} \int [d\phi ]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} ~=~ 0$$

is really a shorthand for infinitely many integrations

$$\tag{4} \left[\prod_{y\in M,\beta\in J} \int d\phi^{\beta}(y) \right]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} ~=~ 0.$$

Before we can proceed, the functional integral measure in (4) must be given a mathematical definition. The precise definition depends on context and method. Needless to say that a general mathematically rigorous definition of functional integrals is a well-known open problem in mathematics. For instance, one may try to construct the functional integral as an appropriate continuum limit of a discretized space-time $M$, as Michael Brown suggests in a comment.

Let us use DeWitt's condensed notation, where all indices (both continuous and discrete indices) are lumped together as

$$\tag{5} i~=~(\alpha,x)~\in~ I~:=~ J\times M,$$

and fields are written as

$$\tag{6} \phi^i ~:=~ \phi^{\alpha}(x)~,\qquad i~\in~ I.$$

We now discretize space-time $M$. The discretization means that we think of $I$ as a finite index-set. In other words, we now only have finitely many variables $\phi^i$, $i\in I$, in the theory. The functional derivative (1) [times$^1$ the volume $\Delta x$ of a single cell of the discretization] is replaced by a partial derivative

$$\tag{1'} \frac{\partial F[\phi]}{\partial\phi^{i}}.$$

An infinitesimally variation is given by the standard formula from calculus in several variables

$$\tag{2'}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\sum_{i\in I} \frac{\partial F[\phi]}{\partial\phi^{i}}\delta\phi^{i}.$$

Finally, OP's functional integral formula (4) becomes

$$\tag{4'} \left[\prod_{j\in I} \int d\phi^{j} \right] \frac{\partial F[\phi]}{\partial\phi^{i}}~=~ 0.$$

Equation (4') follows from the fact that an integral of a total derivative vanishes if the boundary contributions are zero.

--

$^1$ Concerning dimensions of functional derivatives versus partial derivatives, see also this Phys.SE post.

• +1. Yes, this is exactly what I meant. Never knew that was called DeWitt notation though. – Michael Brown Aug 17 '13 at 13:39

The expression : $[d\phi(x)] \frac{\delta F}{\delta \phi(x)}$ could be interpreted as a formal $dF(\phi)$ : $$\int [d\phi(x)] \frac{\delta F}{\delta \phi(x)} \sim \int \frac {\partial F}{\partial \phi_i} d\phi_i \sim \int dF(\phi) =F(+\infty) - F(-\infty)$$

So the left hand side of the expression is zero only for identical boudary conditions, for instance $F(-\infty) = F(+\infty)$

For instance, the function $F(\phi) = e^{- \frac{1}{2}\int dx~\phi^2(x)} =\Pi_x ~(e^{- \frac{1}{2} \phi^2(x)}$), is a valid function, because, at positive and negative infinite $\phi$, we have $F(\phi) = 0$

Your function $F = \int dx ~ C(x) \phi(x)$ is not valid because it takes different values at negative and positive infinite values of $\phi$. Moreover, the values of $F$ are infinite, for infinite $\phi$, so it is difficult to give a sense to $F(-\infty) - F(+\infty)$

I might be misinterpreting what "functional derivative" and "functional integral" here particularly mean but if I recall correctly,

Given a functional

$$\int_{x_0}^{x_1} L(x,y(x), y'(x), y''(x) ... y^{[n]}(x))$$

The functional derivative

$$\frac{\delta L}{\delta y}$$

Is defined such that if $$\frac{\delta L}{\delta y} = 0$$

For some y, then there exists a local optima to the aforementioned functional by fitting appropriate boundary conditions on that y.

Derivation of the functional derivative itself requires substitution of

$$y = U + e k(x)$$

where U is the optimal solution, e is a variable (that we will manipulate and should be thought of as infinitesmally small) and k(x) is an arbitrary test function such that $k(x_0) = k(x_1) = 0$

then:

$$\frac{d}{de} [\int_{x_0}^{x_1} L(x,u(x) + ek(x), ... u^{[n]}(x) + ek^{[n]}(x)) ]$$

Yields (after fidgeting with integration by parts)

$$\int_{x_0}^{x_1} [k(x) \sum_{i=0}^{n} {(-1)^{i} \frac{d^{i}}{dx^{i}}[\frac{\partial L}{\partial y^{[i]}} ]}] = 0$$

Which implies

$$\int_{x_0}^{x_1} [\sum_{i=0}^{n} {(-1)^{i} \frac{d^{i}}{dx^{i}}[\frac{\partial L}{\partial y^{[i]}} ]}] = 0$$

Whereas:

$$\sum_{i=0}^{n} {(-1)^{i} \frac{d^{i}}{dx^{i}} \frac{\partial L}{\partial y^{[i]}} } = \frac{\delta L}{\delta y}$$

The inverse of the functional derivative is a problem in partial differential equations.

Consider for example the one variable, single order derivative case: aka inverting the euler lagrange equations of

$$\frac{\partial L}{\partial y} - \frac{d}{dx}[\frac{\partial L}{\partial y'}] = H$$

for some functional H(x,y,y'). We can expand the total derivative with respect to x (Assuming our arguments are only x, y,y' to find:

$$\frac{\partial L}{\partial y} - \frac{\partial^2 L}{\partial x y'} - y'\frac{\partial^2 L}{\partial y y'} - y''\frac{\partial^2 L}{\partial (y')^2} = H$$

These can be quite challenging for general H (i'm not even sure if solutions exist for general H). But if $$\frac{\partial^2 H}{\partial (y'')^2} = 0$$ which WILL ALWAYS BE THE CASE if H = $\frac{\delta L}{\delta y}$ for any well defined $L(x,y,y')$. In this case follow the procedure below:

$$H = u(x,y,y') + y''w(x,y,y')$$

such that

$$\frac{\partial u}{\partial y''} = 0$$

and

$$\frac{\partial w}{\partial y''} = 0$$

Then it is trivial to showL

$$L = \int \int [w] \partial y'' \partial y'' + y'a_1(x,y) + a_2(x,y)$$

Now take the truncated euler lagrange equation and substitute $\int \int [w] \partial y'' \partial y'' + y'a_1(x,y) + a_2(x,y)$ for $L$ in:

$$\frac{\partial L}{\partial y} - \frac{\partial^2 L}{\partial x y'} - y'\frac{\partial^2 L}{\partial y y'} = u$$

To ultimately derive an expression for $a_2$ in terms of $a_1$, substituting this back into your original definition we conclude that in general the inverse of the functional derivative for a functional H (ie the functional integral of H(x,y,y')) is:

$$\int H \delta y = y' a_1(x,y) + \int{[u + \int \int [\frac{\partial w}{\partial y}] \partial (y')^2 + \frac{\partial a_1}{\partial x} - \int [\frac{\partial w}{\partial x} + \frac{\partial w}{\partial y}] \partial y'} \partial y + (1 - y')a_1(x,y) + g(x) - \int \int [w] \partial (y')^2$$

for arbitrary functions of their arguments: $a_1(x,y)$ and $g(x)$

This is certainly not equal to 0. It is also definitely not simplified either. But I suppose if that was soo important you could manage it.