Wigner Transform in momentum space coefficient I am currently reading some about the Wigner transform, and I ran into a problem. The Wigner transform (in the literature I am reading) is defined as:
$$\tilde{A} = \int \exp{\big[\frac{-ipy}{\hbar}\big]} \langle x + \frac{y}{2} \vert \hat{A} \vert x -\frac{y}{2} \rangle dy $$
It is then stated that the same representation can be derived using a momentum representation as:
$$\tilde{A} = \int \exp{\big[\frac{ixu}{\hbar}\big]} \langle p + \frac{u}{2} \vert \hat{A} \vert p -\frac{u}{2} \rangle du$$
When I attempt to derive the second equation from the first, I get the following:
$$\tilde{A} = \frac{1}{(2\pi \hbar)^2} \int \exp{\big[\frac{-ipy}{\hbar}\big]} \langle x + \frac{y}{2} \vert p' \rangle \langle p' \vert \hat{A} \vert p'' \rangle \langle p'' \vert x -\frac{y}{2} \rangle dy  dp' dp'' \rightarrow  $$
$$\tilde{A} = \frac{1}{(2\pi \hbar)^2} \int \exp{\big[\frac{i(-py + p'(x+y/2) - p''(x-y/2))}{\hbar} \big]}  \langle p' \vert \hat{A} \vert p'' \rangle dy  dp' dp'' $$
Gathering the y terms and integrating over y gives:
$$\tilde{A} = \frac{1}{2\pi \hbar} \int \exp{\big[\frac{ix(p' - p'')}{\hbar} \big]} \delta(p - \frac{p'}{2} - \frac{p''}{2}) \langle p' \vert \hat{A} \vert p'' \rangle  dp' dp'' \rightarrow $$
$$\tilde{A} = \frac{1}{\pi \hbar} \int \exp{\big[\frac{2ix(p' - p)}{\hbar} \big]}  \langle p' \vert \hat{A} \vert 2p - p' \rangle  dp' $$
Using $p' = p + \frac{u}{2}$ ($u$ being the variable here), we can write this as:
$$\tilde{A} = \frac{1}{h} \int \exp{\big[\frac{ixu}{\hbar} \big]}  \langle p + \frac{u}{2} \vert \hat{A} \vert p - \frac{u}{2} \rangle  du $$
Which is almost what they have written (the second equation I wrote), but the factor of $h$ is frutrating my efforts. Does anyone have any insight as to why my results doesn't match theirs? Thanks in advance!
 A: I will first review the mainstream conventions that all good texts on the subject review when asserting the identity of your first with your second equation (ours does). If the expectation values between x eigenstates are normalized differently than those between p eigenstates, you get the mismatch factors you observed. The differences of the Wigner transform you observe are a red herring, and I'll pare down the problem to simple traces. The mismatch you observed is but a feature of your weird imbalanced conventions, addressed in the end of this answer.
The conventions of QM in phase space are normally dimension-balanced units treating both x and p on the same footing, so they both have units of $\sqrt{\hbar}$. You may thence convince yourself from the normalizations of wavefunctions that  the units of both kets $|x\rangle$ and  $|p\rangle$ must be
$\hbar^{-1/4}$ and
$$
1\!\!1=\int\!\!dx~~ |x\rangle\langle x| = \int\!\!dp~~ |p\rangle\langle p| ~,\\
\langle x|x'\rangle=\delta(x-x'), ~~~ \langle p|p'\rangle=\delta(p-p'),\\
\langle x|p\rangle= \frac{e^{ixp/\hbar}}{\sqrt{2\pi\hbar}}~,~~~~~~ 2\pi \hbar ~\delta(x)=\int\!\!dp~~ e^{ixp/\hbar}, ... 
$$
Now, more simply than the Wigner transform, look at traces,
$$
\int\!\!dx~~\langle x|\hat A| x\rangle= \int\!\!dx~dp~dp'~\langle x|p\rangle\langle p| \hat A| p'\rangle\langle p'| x\rangle\\
= \int\!\!dx~\frac{e^{ix(p-p')/\hbar}}{ 2\pi\hbar} dp~dp'~ \langle p| \hat A| p'\rangle = \int\!\!  dp ~ \langle p| \hat A| p\rangle . 
$$
Now using overbars for your tilted conventions, whose origin is puzzling,
$$
1\!\!1= \frac{1}{2\pi \hbar} \int\!\!d\bar p~~ |\bar p\rangle\langle \bar p| ~,~~\langle \bar x|\bar p\rangle=  e^{ixp/\hbar} ~~\leadsto \\
 \langle\bar p|\bar p'\rangle     =2\pi \hbar ~\delta(\bar p-\bar p'), \qquad \langle\bar x|\bar x'\rangle = \delta(\bar x-\bar x'), ~~\leadsto \\
1\!\!1=   \int\!\!d\bar x~~ |\bar x\rangle\langle \bar x| ,
$$
etc.
Repeating the above trace in your conventions, you find, instead, for the trace,
$$
\int\!\!d\bar x~~\langle\bar x|\hat A|\bar x\rangle=\frac{1}{(2\pi \hbar)^2} \int\!\!d\bar x~d\bar p~d\bar p'~\langle \bar x|\bar p\rangle\langle\bar  p| \hat A|\bar  p'\rangle\langle\bar  p'| \bar x\rangle\\
=\frac{1}{2\pi \hbar} \int\!\! d\bar p~d\bar p'~ \delta(\bar p-\bar p') ~ \langle \bar p| \hat A|\bar p'\rangle = \frac{1}{2\pi \hbar}\int\!\!  d\bar p ~ \langle \bar p| \hat A| \bar p\rangle . 
$$
So, indeed, you may track down the effect of these conventions all over, and monitor the changes they bring about, including the first two relations you wish to compare, but the whole point of the construction is to treat x and p equitably. In fact, unless one were interested in the classical limit, one simply absorbs the universal unit $\sqrt \hbar$ in the variables, by setting it equal to 1.
NB I suspect you might get your results off the mainstream treatment above, through the substitution $| p\rangle = \sqrt{2\pi \hbar} |\bar p\rangle$, leaving everything else, including p as a variable, alone, but I can't be sure...
