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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?

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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 $$ \frac 12 \int d^nx d^ny J(x) \{\Delta(x-y)-\Delta(y-x)\}J(y)=0. $$

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  • $\begingroup$ It seems like one cannot switch x and y in Feynman propagator with out changing the value of it. Can you prove it? $\endgroup$ – Universe Maintainer Feb 15 at 0:19
  • $\begingroup$ 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. $\endgroup$ – mike stone Feb 15 at 0:26
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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 $x-y \to y-x$ .

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