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I'm reading an old paper ("Wigner's Function and Other Distributions in Mock Phase Spaces," Balazs and Jennings, Phys. Rep. 104(6), 1984), and came across the following statement (in which $\hat{q}$ and $\hat{p}$ are a pair of generalized canonical operators and $\hat{\rho}$ is some density operator):

We may also evaluate $\text{Tr}[\delta(\hat{q}-q')\delta(\hat{p}-p')\hat{\rho}]$ or $\text{Tr}[\delta(\hat{p}-p')\delta(\hat{q}-q')\hat{\rho}]$ ... These quantities, however, are not the same, are not symmetrical in $\hat{p}$ and $\hat{q}$ and are not positive everywhere. Furthermore, they will be complex.

It seemed intuitive to me that these traces could be different in general (since $\hat{q}$ and $\hat{p}$ don't commute), but it wasn't so clear that this quantity could be negative or imaginary even though clearly the wavefunction in position or momentum basis may be.

So I tried seeing this by evaluating a bit. Let $\hat{\rho} = \left|\psi\right\rangle \left\langle \psi \right|$ and $\left|\psi\right\rangle = \int dq \, \psi(q)\left|q\right\rangle$ in the position basis. Then I thought to do the trace in the position basis, i.e.

$$ \begin{aligned} \text{Tr}[\delta(\hat{q}-q')\delta(\hat{p}-p')\hat{\rho}] &= \int dq'' \left\langle q'' | \delta(\hat{q}-q')\delta(\hat{p}-p')\hat{\rho} | q'' \right\rangle \\ &= \int dq'' \int dq \left\langle q''|\delta(\hat{q}-q')\delta(\hat{p}-p')|q \right\rangle \langle q|q'' \rangle |\psi(q)|^2 \\ &= \int dq'' \int dq \left\langle q''|\delta(\hat{q}-q')\delta(\hat{p}-p')|q'' \right\rangle |\psi(q)|^2 \\ &= \langle q' | \delta(\hat{p}-p') | q' \rangle |\psi(q)|^2 \\ &= \int dp'' \langle q'|\delta(\hat{p}-p')|p'' \rangle \langle p''|q' \rangle |\psi(q)|^2 \\ &= \langle q'|p' \rangle \langle p'|q' \rangle |\psi(q)|^2 \\ &= |\langle q'|p' \rangle|^2 |\psi(q)|^2 \end{aligned} $$

which is always real and positive, contrary to the statement of the paper. So I have to believe I've done something wrong. Any help in pointing out my error would be greatly appreciated.

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    $\begingroup$ Your damning mistake is on the 2nd line: $\langle q|\hat \rho|q''\rangle \neq \langle q|q''\rangle |\psi(q)|^2$ ...!! Eliminate all operators first, by insertion of complete states, and only then insert wave functions into the highly non-symmetric matrix element of the density matrix. $\endgroup$ Jan 29 at 13:26
  • $\begingroup$ Oh, it looks like you are correct; that matrix element is something like $\psi(q)\psi^*(q'')$, right? Thanks, Cosmas. $\endgroup$ Jan 29 at 14:11
  • $\begingroup$ Right. Do you still need to see lack of reality? It is straightforward, and you never need to go down your path. $\endgroup$ Jan 29 at 14:49
  • $\begingroup$ Yes — although I did follow through with it anyway, that result alone is sufficient for me to get it intuitively. Thanks again. $\endgroup$ Jan 29 at 15:01
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It's straightforward to see your expression is not always real. Writing your second line correctly, you have $$ \text{Tr}[\delta(\hat{q}-q')\delta(\hat{p}-p')\hat{\rho}] = \int dq'' \left\langle q'' | \delta(\hat{q}-q')\delta(\hat{p}-p')\hat{\rho} | q'' \right\rangle \\ = \int dq'' \delta(q''-q') \langle q''|\delta(\hat{p}-p')\hat \rho|q'' \rangle = \langle q'|\delta(\hat{p}-p')\hat \rho|q' \rangle \\ =\int dp \langle q'|p\rangle \langle p|\delta(\hat{p}-p')\hat \rho|q' \rangle\propto e^{iq'p'/\hbar} \langle p'|\hat \rho|q'\rangle = e^{iq'p'/\hbar} \phi(p') \psi^*(q')~, $$ so for a real Gaussian wave function, which is also a real Gaussian in momentum space, you manifestly see the phase multiplying a real quantity.

Some of the exercises here cover these.

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