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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!

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    $\begingroup$ Already your 3rd equation has a superfluous pathological normalization. Recall $1\!\!1= \int dp ~~|p\rangle\langle p|$... Remember the delta function? Further, $\langle x|p\rangle= \exp(ixp/\hbar)~~/\sqrt{2\pi \hbar}$. $\endgroup$ – Cosmas Zachos May 7 at 0:41
  • $\begingroup$ Can't I choose it either way though? Or is it necessary to normalize in this fashion? I understand that this will rid me of my problem, but I am wondering what the imperative for this normalization is. $\endgroup$ – user132849 May 7 at 0:53
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    $\begingroup$ No, you can't; or, if you do, your must trail the consequences in delta functions, etc... You may do anything you like, provided you do it consistently: changes of definition cannot change answers. However, you saw the pain you inflict on yourself with spotty redefinitions.... $\endgroup$ – Cosmas Zachos May 7 at 0:56
  • $\begingroup$ But if I define $\textbf{1} = \frac{1}{2\pi \hbar} \int \vert p \rangle \langle p \vert dp$ and $\int \exp{\frac{-ipx}{\hbar}} dx = 2 \pi \hbar \delta(p) $, and $\langle x \vert p \rangle = \exp{\frac{ipx}{\hbar}}$, then what is inconsistent about this normalization convention? $\endgroup$ – user132849 May 7 at 3:11
  • $\begingroup$ I instead used: $\langle p \vert p' \rangle = 2 \pi \hbar \delta(p-p')$, which I believe should be consistent, right? So then: $\vert p \rangle = \frac{1}{2 \pi \hbar} \int \vert p' \rangle \langle p' \vert p \rangle dp' = \int \vert p' \rangle \delta(p-p') dp'$. Isn't that consistent? $\endgroup$ – user132849 May 7 at 15:08
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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...

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  • $\begingroup$ Ok this makes sense. Thanks again for your help! Although, as to the origins of the convention, this is the way I learned it in my first quantum course, and Richard Mackenzie also uses the same convention as me in his lectures on path integrals ( arxiv.org/pdf/quant-ph/0004090v1.pdf (page 7) ). But, nonetheless, I thoroughly appreciate the clarity you provided! $\endgroup$ – user132849 May 7 at 17:20
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    $\begingroup$ Oh... OK, he intercalates xs and ps so there is a premium of the simplest $\langle x|p\rangle$ possible... But it's really eccentric... Most applications try to be even-handed between x and p . $\endgroup$ – Cosmas Zachos May 7 at 17:47
  • $\begingroup$ Yeah I guess either has an advantage, but it seems to me that the mainstream convention is a little simpler given that the identity operator in the momentum space doesn't come with any baggage. $\endgroup$ – user132849 May 7 at 17:54

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