How does the Stratonovich-Weyl operator kernel, used to find the Wigner function, work?

Question

Recently during my studies, I came across an alternative construction of the Wigner function. This construction starts from the notion of the Stratonovich-Weyl operator kernel. I saw this construction in a paper called 'Quantum phase-space representation for curved configuration spaces'. And the development is as follows:

The author defines the Stratonovich-Weyl operator kernel as:

$$\hat{\Delta}(x,p) = \hat{D}(x,p)\hat{\Delta}(0,0)\hat{D}^{\dagger}(x,p)$$

where $\hat{D}$ is the already known displacement operator, given by:

$$\hat{D}(x,p) = e^{\frac{-ix\hat{p}}{\hslash}}e^{\frac{ip\hat{x}}{\hslash}}$$

with the undisplaced kernel:

$$\hat{\Delta}(0,0) =\frac{1}{2\pi\hslash}\int dp' \int dx'\hat{D}(x',p')e^{\frac{ix'p'}{2\hslash}} = \int dx' |x'/2\rangle\langle -x'/2|$$

I couldn't understand how he got to the last equality. I think he inserted the completeness relation to get the last equality. However, I performed the calculations but could not get the correct result. What does it mean to take the Fourier transform of a displacement operator? Is there a standard way to manipulate displacement operators, such as to express them in bra-ket notation?

Answering these questions will hopefully help me understand how to perform other calculations that I cannot currently do involving the Stratonovich-Weyl operator kernel, such as proving that $$\hat{\Delta}(x,p) = \int dx'e^{\frac{px'}{\hslash}}|x+x'/2\rangle\langle x-x'/2|.$$

I tried to get to that relationship, but I couldn't either.

Answer 1

The non-standard definition of displacement operators got me confused initially, but that just changes the phase factors.

All you need to do is rewrite an $x$-displacement as $$\exp(-i x\hat{p}/\hbar)=\int dx^\prime |x^\prime+x\rangle\langle x^\prime|$$ and verify that it acts appropriately on all position eigenstates. Then we can rewrite the displacement as $$\hat{D}(x,p)=\int dx^\prime |x^\prime+x\rangle\langle x^\prime| e^{i p\hat{x}/\hbar}=\int dx^\prime |x^\prime+x\rangle\langle x^\prime| e^{i px^\prime/\hbar}.$$ For the undisplaced kernel, we observe that the integral over $p^\prime$ now gives us a Dirac delta function, which simplifies everything as necessary: \begin{aligned} \frac{1}{2\pi\hbar}\int dx^\prime dp^\prime D(x^\prime ,p^\prime)e^{i p^\prime x^\prime/2\hbar}&=\frac{1}{2\pi\hbar}\int dx^\prime \int dp^\prime \int dx^{\prime\prime} |x^{\prime\prime}+x^\prime\rangle\langle x^{\prime\prime}| e^{i p^\prime x^{\prime\prime}/\hbar}e^{i p^\prime x^\prime/2\hbar}\\ &=\frac{1}{\hbar}\int dx^\prime \int dx^{\prime\prime} |x^{\prime\prime}+x^\prime\rangle\langle x^{\prime\prime}| \delta\left(\frac{x^{\prime\prime}}{\hbar}+\frac{x^{\prime}}{2\hbar}\right)\\ &=\int dx^\prime \int dx^{\prime\prime} |x^{\prime\prime}+x^\prime\rangle\langle x^{\prime\prime}| \delta\left(x^{\prime\prime}+\frac{x^{\prime}}{2}\right)\\ &= \int dx^\prime |-\frac{x^{\prime}}{2}+x^\prime\rangle\langle -\frac{x^{\prime}}{2}| . \end{aligned}

As for the second question, you should now be able to do this. You have the operator in the position basis, so you know how $\exp(i p\hat{x}/\hbar)$ acts in its eigenbasis and you know that $\exp(-i x\hat{p}/\hbar)$ shifts the eigenstates.