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

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.

• To the close voters: there seems to be $1+\epsilon$ questions combined here, but they both stem from trying to understand what the displacement operator does. As for the homework style, it is tough because the underlying physics concept is the computation itself. I will try an edit Aug 15 at 15:39

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.