Wightman functions in quantum field theory (Source of the 'theorem': click) Given a field $\Phi(x)$ with spin $s$ and its adjoint $\Phi^*(x)$, define the expectation values
\begin{align}
    f(x-y)&:=\langle v,\Phi(x)\Phi^*(y)v\rangle, \\ g(x-y)&:=\langle v,\Phi^*(x)\Phi(y)v\rangle.
\end{align}
where $v$ is the vacuum state. The condition
\begin{equation}\tag{*}
    \Phi(x)\Phi^*(y)=-(-1)^{2s}\Phi^*(y)\Phi(x)
\end{equation}
should then imply $$f(x)+(-1)^{2s}g(-x)=0. \tag{**}$$ But how are $f(x)$ and $g(x)$ defined explicitly? Taking the expectation value of (*) one gets
$$ \langle v,\Phi(x)\Phi^*(y)v\rangle+(-1)^{2s}\langle v,\Phi^*(y)\Phi(x)v\rangle=0$$ so working backwards it should be $f(x)=\langle v,\Phi(x)\Phi^*(y)\rangle$ and $g(x)=\langle v,\Phi^*(y)\Phi(x)v\rangle$, but I don't see how these definitions are compatible with $f(x-y)$ and $g(x-y)$ respectively. In fact, $f(x-y)$ looks exactly like $f(x)$, and $g(x)$ like $g(y-x)$.
Where does the dependence on $y$ go?
 A: The issue I believe OP is having here is actually entirely independent of the Wightman functions. Note that the functions $f$ and $g$ are functions of only a single argument, not two. That is, they are not functions of $x$ and $y$ as the RHS of their definitions would seem to imply, the are a function only of the difference. Almost certainly the author of whatever reference OP is looking at is assuming translation invariance of the correlators, which is why $f$ and $g$ have this property.
In any case, let's look at the RHS a little more and, in particular, note that
$$
\langle v, \Psi(x + z)\Psi^*(y+z)v\rangle = f(x-y) = \langle v, \Psi(x)\Psi^*(y)v\rangle
$$
for any vector $z$ we like since it will always just cancel in the difference of the evaluation points of the operators $\Psi,\Psi^*$.
This is also why I said that the fact $f$ depends only upon the difference really follows from translation invariance. Here we are using the fact that such an $f$ exists (as asserted by OP) to show that the correlator is translation invariant.
Anyway, the point is that there's not much meaning to asking "where did the $y$-dependence go" then the function you're thinking about doesn't depend on $y$ except through its difference with some other vector.
