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Consider the simplest possible case in which the time reversal operator $\hat{\mathrm{T}}$ is given by the operation of complex conjugation $\hat{\mathrm{K}}$.

We can view $\mathrm{T}$ is an anti-untary operator on the Hilbert space of our quantum system. In particular, we can determine its action on the momentum operator $\hat{\mathbf{p}}$ as

$$ \hat{\mathrm{T}} \hat{\mathbf{p}} \hat{\mathrm{T}}^{-1} = - \hat{\mathbf{p}}. \hspace{2cm}(1) $$

Question: How can we describe this operation within the phase space formulation of quantum mechanics? Can the form of this operation be derived from the Wigner transformation?

Some ideas: In this case, all non-commutativity of quantum mechanics is absorbed into the Moyal product, denoted by the binary $\star$ operation. The relation above will be transformed to its phase-space counterpart

$$ \mathrm{T} \star \mathbf{p} \star \mathrm{T}^{-1} = - \mathbf{p}. \hspace{2cm}(2) $$

It seems to me that the phase-space representation $\mathrm{T}$ of the operator $\hat{\mathrm{T}}$ should be rather non-trivial. My suspicion is that it will still contain the operation of complex conjugation (because the relations above should also hold when we consider the canonical momentum in electromagnetic fields).

The difficulty arises because $\mathbf{p}$ is now a real $c$-number and not a complex operator. Thus, $\mathrm{T}$ must have a dependence on position and momentum.

Conjecture: Based on these observations, I would conjecture that

$$ \mathrm{T} = \mathcal{U}(\mathbf{x},\mathbf{p})\mathrm{K}, $$

where $\mathcal{U}$ is a yet to be determined $\star$-unitary function on phase space. One thing which can easily be done is a translation $t$ by $\mathbf{p}'$, i.e.,

$$ t(\mathbf{p'})\star \mathbf{p} \star t(\mathbf{p'})^{-1} = \mathbf{p} +\mathbf{p}'. $$

This operation is easily defined. After we take the limit $\mathbf{p'}\to -2 \mathbf{p}$ we get the desired result. Can we incorporate this into the conjectured form?

EDIT 1: Following the discussion in the comment section we could interpret the statement (2) from above as an integral operator, let's call it $\tilde{\mathrm{T}}$ which acts via the functional

$$ \tilde{\mathrm{T}}[f(\mathbf{x},\mathbf{p})] = \int \text{d}\mathbf{x}'\text{d}\mathbf{p}' \delta(\mathbf{x}-\mathbf{x}')\delta(\mathbf{p}+\mathbf{p}') f(\mathbf{x}',\mathbf{p}') $$

Because of its operational definition, it is clear that this coincides with the Wigner transformation $\mathcal{W}$ of Eq. (1). It remains to show, that we can write this in terms of a $\star$-product, i.e.

$$ \tilde{\mathrm{T}}[f(\mathbf{p})] = \mathcal{W} [ \hat{\mathrm{T}} f(\hat{\mathbf{p}}) \hat{\mathrm{T}}^{-1} ] \overset{\text{to show}}{=} \mathrm{T} \star f( \mathbf{p} ) \star \mathrm{T}^{-1}. $$

EDIT 2: I might have found some additional constraints. If we assume that $\mathrm{T} \star \mathrm{T}^{-1} = \mathrm{T}^{-1} \star \mathrm{T} = 1 $, Eq. (2) can be used to derive

$$ \left\lbrace \mathrm{T} ~\overset{\star}{,} ~\mathbf{p} \right\rbrace= 0, $$

which is only fulfilled for $\mathrm{T} = 0$ which is not a solution to the original problem. This means $\mathrm{T}^{-1}$ cannot coincide with the $\star$-inverse of $\mathrm{T}$.

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  • $\begingroup$ My first impression is that time reversal has its classical expression here: $W(x,p) \mapsto W(x,-p)$; ie you just change the sign of all momenta. Applying this, then the momentum operator $p\times -\frac{1}{2}\mathrm{i}\hbar\partial_x$, then again time reversal we get $W(x,p)\mapsto W(x,-p)\mapsto pW(x,-p)-\frac{1}{2}\mathrm{i}\hbar\partial_xW(x,-p) \mapsto -pW(x,p)-\frac{1}{2}\mathrm{i}\hbar\partial_xW(x,p)$, so the action on the momentum operator seems to be $(p\times -\frac{1}{2}\mathrm{i}\hbar\partial_x) \mapsto -p\times -\frac{1}{2}\mathrm{i}\hbar\partial_x$. Not sure about this, though. $\endgroup$ – pglpm Apr 23 '18 at 13:19
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    $\begingroup$ Your translation stunt won't work, but its cousin, a dilatation / squeezing, will, for p. However , it will reverse x as well. The only way out is to reverse p but also complex conjugate to preserve the heisenberg commutation rule... or else flip the sign of $\hbar$ everywhere.... $\endgroup$ – Cosmas Zachos Apr 23 '18 at 16:01
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    $\begingroup$ Sorry I still don't understand what you mean by "explicit representation". Your states are functions, operators are functional maps, and these come in all guises. If you mean an integral-kernel representation like $W(x,p)\mapsto W'(x',p')=\iint K(x',p';x,p)\,W(x,p)\,\mathrm{d}x\,\mathrm{d}p$, then the kernel in this case should be something like $\delta(x'-x)\,\delta(p'+p)$. $\endgroup$ – pglpm Apr 23 '18 at 17:28
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    $\begingroup$ Alright. I suggest that you modify your question "How can we describe this operation within the phase space formulation of quantum mechanics?" and make this requirement explicit. Representations like $W(x,p)\mapsto W(x,-p)$, or like $\delta(x'-x)\,\delta(p'+p)$ in terms of a kernel, are legitimate phase-space representations (I personally find the kernel one the most useful). $\endgroup$ – pglpm Apr 23 '18 at 18:42
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    $\begingroup$ I think the problem is, that $K$ interferes with the $\star$-product. This makes this really tricky $\endgroup$ – sagittarius_a Apr 24 '18 at 7:40
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I might have found a possible route. Consider the parity operator

$$ \hat{\Pi} = \int\text{d}\mathbf{x}\text{d}\mathbf{x}'~\delta(\mathbf{x} + \mathbf{x}')~ |\mathbf{x}\rangle\langle \mathbf{x}'|. $$

Its Wigner transform is given by

$$ \Pi = (\pi \hbar)^d~ \delta(\mathbf{x})\delta(\mathbf{p}). $$

This means we have to interpret $\Pi$ in a distributional sense. We can show that

$$ \Pi \star \mathbf{p} \star \Pi = - \mathbf{p}. $$

proof.

For simplicity, we consider the one-dimensional case. Here we have

$$ \delta(p)\delta(x) \star p = p~\delta(p)\delta(x) + \frac{i \hbar}{2} \delta(p)\delta'(x) $$

As we interpret this in a distributional sense we view $p~\delta(p)$ as equivalent to $0$ (this is the sloppy part, I guess one could improve here). Thus

$$ \delta(p)\delta(x) \star p \star \delta(p)\delta(x) = \frac{i\hbar}{2} \delta(p)\delta'(x) \star \delta(p)\delta(x) $$

This term can be simplified as

$$ \delta(p)\delta'(x) \star \delta(p)\delta(x) = \Big(\delta(p) \star \delta(x) \Big) \partial_x \Big( \delta(x) \star \delta(p) \Big) $$

In one dimension, one has the identity

$$ \delta(x) \star \delta(p) = \frac{2}{h} \exp \left\lbrace \frac{2i xp}{\hbar}\right\rbrace, $$

which is why we find

$$ \delta(p)\delta'(x) \star \delta(p)\delta(x) = \frac{8 i}{h^2 \hbar} p $$

Multiplying by $(\pi \hbar)^2$ from the definition of $\Pi$ and the factor $i \hbar /2 $ we finally obtain $$ \Pi \star \mathbf{p} \star \Pi = - \mathbf{p}. $$

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I think this is a promising way to a proper definition of $\mathrm{T}$. The missing piece is how complex conjugation looks like in the Wigner transformation. I'll extend this answer later on.

Let me continue... Let us consider a scalar phase space function $A$ and its quantum counterpart, the observable $\hat{A}$. They are related by the Wigner transformation

$$ A(\mathbf{x},\mathbf{p}) = \int \text{d}x'~e^{-i \mathbf{x}'\cdot\mathbf{p}/\hbar} ~ \Big\langle \mathbf{x} + \frac{\mathbf{x}'}{2} \Big|\hat{A}\Big|\mathbf{x} - \frac{\mathbf{x}'}{2} \Big\rangle . $$

So we see that the action of complex conjugation on phase space has the result

$$ K A(\mathbf{x},\mathbf{p} ) K = A^\ast(\mathbf{x},-\mathbf{p}), $$

which coincides with the Wigner transformation of $\hat{K}\hat{A}\hat{K}$. From this we might argue that $K$ is the Wigner transformation of $\hat{K}$, but its action is independent $\star$-product and we have

$$K\star A \star K = K~A~K$$

At the moment this is the only consistent way I know of. As Cosmas Zachos points out, $K$ is a really different object compared to parity $\Pi$ introduced above. But defined in the outlined way it keeps the symplectic structure intact, we have

$$ K \star [ x \overset{\star}{,} p ] \star K = - [ x \overset{\star}{,} p ]= - i \hbar =K \star (i\hbar) \star K $$

For spinless systems we then identify $K$ as the operator of time reversal.

Let me know what you think of this

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  • $\begingroup$ But... that was the point I made above. Your parity operator reverses both x and p, and the misplaced assignment of the question requires only p to reverse, but not x, so the symplectic structure and the commutation relation is smashed. It might not look it, but the Royer operator you are looking at is squeezing with squeezing parameter -1, so to speak... $\endgroup$ – Cosmas Zachos Apr 24 '18 at 21:06
  • $\begingroup$ As a lark, it is virtually trivial to check that $e_\star^{\pi x\star p /2} \star (x,p)\star e_\star^{-\pi x\star p /2}=-(x,p)$ does the job, and condenses to the equivalent image, as per Exercise 5 of our book, but I don't want to pretend this resolves any of your quandary. Parity is easy. Loss of symplectic structure isn't. $\endgroup$ – Cosmas Zachos Apr 24 '18 at 21:17
  • $\begingroup$ Thank you very much, Cosmas Zachos. I will look into it! $\endgroup$ – sagittarius_a Apr 25 '18 at 5:10
  • $\begingroup$ @CosmasZachos could you elaborate on the Royer operator? I am not familiar with this concept $\endgroup$ – sagittarius_a Apr 25 '18 at 19:55
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    $\begingroup$ Ugh... but both arguments in their (2.15) are space-like. You probably mean integrand function $\langle x+x'/2|\hat{A}|x-x'/2\rangle$, instead. But then you must write exactly what you mean... $\endgroup$ – Cosmas Zachos Apr 25 '18 at 21:49

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