# Confusion about functional derivative in path integral

If we act a functional derivative $$\frac{\delta}{\delta J(z)}$$On the expression$$\int\int d^4x d^4y \space J(x)\Delta(x-y)J(y)$$ where $$\Delta(x-y)$$ is Feynman propagator. What one should get is $$2\int d^4y\Delta(z-y)J(y).$$

But what I get is $$\int d^4xd^4y\space\delta(x-z)\Delta(x-y)J(y)+ \int d^4xd^4yJ(x)\Delta(x-y)\delta(y-z).$$ After calculating delta function and change of variable, this is equivalent as: $$\int d^4y\Delta(z-y)J(y)+\int d^4y\Delta(y-z)J(y).$$

Is Feynman propagator "even"? How do i get the correct result?

If it's even, how do we prove it?

In the second equation you quote, only the symmetric $$x\leftrightarrow y$$ part contributes, so you may as well assume that $$\Delta(x-y)=\Delta(y-x)$$.
In other words $$\Delta(x-y) =\frac 12 (\Delta(x-y)+\Delta(y-x))+ \frac 1 2(\Delta(x-y)-\Delta(y-x))$$ and $$\frac 12 \int d^nx d^ny J(x) \{\Delta(x-y)-\Delta(y-x)\}J(y)=0.$$
• Im not saying that $\Delta(x-y)= \Delta(y-x)$. What I am saying is that only the symmetric part contributes. I will edit my answer say this more clearly. – mike stone Feb 15 at 0:26
The Feynman propagator of a scalar particle is even. Proof: $$\Delta(x-y) = \int \frac{d^dp}{(2\pi)^d} \frac{e^{-i(x-y)\cdot p}}{p^2-m^2+i \varepsilon}\,.$$ Changing variable to $$p^\mu = - q^\mu$$ (the Jacobian is $$1$$) is equivalent to send $$x-y \to y-x$$ .