# Are these bra-ket manipulations correct?

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.

• 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. Jan 29 at 13:26
• Oh, it looks like you are correct; that matrix element is something like $\psi(q)\psi^*(q'')$, right? Thanks, Cosmas. Jan 29 at 14:11
• Right. Do you still need to see lack of reality? It is straightforward, and you never need to go down your path. Jan 29 at 14:49
• Yes — although I did follow through with it anyway, that result alone is sufficient for me to get it intuitively. Thanks again. Jan 29 at 15:01

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.