Wigner transform & convolution I'm trying to understand the gradient expansion within the Keldysh formalism. In particular, I am reading "Quantum Field Theory of Non-equilibrium States" by J. Rammer, section 7.2, regarding Wigner or mixed coordinates. There, in Eq. (7.23), the author presents the Wigner transform of a convolution,
$$(A\otimes B)(x,k)\equiv \int d^4r\ e^{ik\cdot r}\int d^4w\ A(u,w)B(w,v)=e^{(i/2)(\partial_x^A\partial_k^B-\partial_k^A\partial_x^B)}A(x,k)B(x,k),\tag{1}$$
where $x=(u+v)/2$ and $r=u-v$. A similar result can be found in this reference, Eq. (2.52).
The proof provided by Rammer considers the convolution
$$C(u,v)=\int d^4w\ A(u,w)B(w,v)=\int d^4w\ A(x+r/2,w)B(w,x-r/2)\equiv C(x,r).\tag{2}$$
Rewriting Eq. (2) in mixed coordinates,
$$C(x,r)=\int d^4w\ A\left(\frac{x+r/2+w}{2},x+r/2-w\right)B\left(\frac{w+x-r/2}{2},w-x+r/2\right),\tag{3}$$
Shifting the $w-$integral in Eq. (3) gives
$$C(x,r)=\int d^4w\ A(x+w/2,r-w)B(x-r/2+w/2,w).\tag{4}$$
After Wigner transforming the last expression, one ends up with
$$\begin{align}C(x,k)=&\int d^4r\ e^{ik\cdot r}\int d^4w\ A(x+w/2,r-w)B(x-r/2+w/2,w),\tag{5}\\
=&\int d^4r\ e^{ik\cdot r}\int d^4w\ \int\frac{d^4k'}{(2\pi)^4}e^{-ik'\cdot(r-w)}A(x+w/2,k')\\
&\int\frac{d^4k''}{(2\pi)^4}e^{-ik''\cdot w}B(x-r/2+w/2,k'').\tag{6}\end{align}$$
Apparently, the argument to proceed from Eq. (6) to get Eq. (1) relies on a Taylor expansion and partial integrations, but I have not been able to complete the proof.
Suggestions to complete the missing steps are very welcomed.
 A: I'll start you from the r.h.side of (1) as per my comment, and get you to the standard convolution similar to Wikipedia's, which you'll then have to take to your expressions, assuming they hold, by changes of variables. I'll stick to one-dimensional integrals, as your four dimensions are merelymultiplexing the structure by four.
Recall the Fourier transform of the inverse Fourier transform is
$$
B(x,k)= \frac{1}{2\pi}\! \int\!\! dk'dx'~ e^{ixk'+ikx'}\!\! \int\!\! dx''dk''~ e^{-ix'k''-ik'x''} \!B(x'',k''),
$$
which we insert in the rightmost side of (1),
$$
e^{(i/2)(\partial_x^A\partial_k^B-\partial_k^A\partial_x^B)}\!A(x,k)B(x,k)= A\!\left(x+\tfrac{i}{2}\partial_k^B,k - \tfrac{i}{2}\partial_x^B\right ) B(x,k)\\ =\!
\frac{1}{2\pi}\!\! \int\!\! dk'dx' dx''dk''~ e^{ixk'+ikx'}\! A(x-\tfrac{1}{2} x',k + \tfrac{1}{2} k')   ~ e^{-ix'k''-ik'x''}\! B(x'',k'').
$$
Four integrals. As usual in this game, you should be able to easily check this by taking "test functions" $A=e^{i(ax+bk)}$ and $B=e^{i(cx+dk)}$. Can you say the same for all of your expressions?
The original Groenewold derivation might be useful.
A: Hints:

*

*Eq. (1) is the Moyal/Groenewold star product $f\star g$ of Weyl-symbols $f$ and $g$.


*If $\hat{f}$ and $\hat{g}$ denote the corresponding Weyl-ordered operators (cf. e.g. this Phys.SE post), the task is to show that
$$ \widehat{f\star g}~=~\hat{f}\circ\hat{g},$$
where $\circ$ denotes composition of operators.
