# Wigner-Weyl transform for a function of coordinates only

I am reading this paper by Tatarskii, which serves as an introduction to the Wigner representation of quantum mechanics.

There is a step in the paper involving the Weyl transform that does not seem valid to me. Define the ordered operator function assigned to the function $$f(p,q)$$ as \begin{align} \left\{ f(\hat{p}, \hat{q}) \right\} &\triangleq f\left(\frac{1}{i} \frac{\partial}{\partial \lambda}, \frac{1}{i} \frac{\partial}{\partial \mu} \right) \hat{F}(\lambda, \mu) \vert_{\lambda, \mu=0} \\ &= \frac{1}{4\pi^2} \iiiint\!\! d\lambda d\mu dp dq ~~f(p,q)\exp\left(-i(\lambda p + \mu q)\right) \hat{F}(\lambda, \mu) . \end{align} \tag{2.1} This is equation 2.1 in the above paper for reference. The author goes on to explain that our choice of function $$\hat{F}$$ will determine the actual ordering of the resultant operator equation, but that, no matter what, it should satisfy some listed properties. One of these is that we should have $$\hat{F}(0, \mu) = e^{i \mu \hat{q}}$$. He then claims that

If $$\hat{F}$$ satisfies these conditions, then in constructing by means of Eq 2.1 the functions $$\{f(\hat{p}, \hat{q}) \}$$, functions of the form $$f(\hat{p})$$ or $$f(\hat{q})$$, which depend only on one variable, are obtained from the corresponding functions $$f(p)$$, $$f(q)$$ by the simple replacement $$q \rightarrow \hat{q}$$ and $$p \rightarrow \hat{p}$$.

I am willing to believe this is the case, but I fail to see how it works operationally. If we assume we just have a function of one variable, and that $$\hat{F}$$ satisfies the above condition, then we have \begin{align} \left\{f(\hat{q}) \right\} &= \frac{1}{4\pi^2}\int \int \int \int d\lambda d\mu dp dq f(q)\exp\left(-i\mu q)\right) \exp(i \mu \hat{q}) \\ &= \frac{1}{4\pi^2} \int \int d\mu dq f(q) \exp(-i\mu(q-\hat{q})) \\ &= \frac{1}{2\pi} \int dq f(q) \delta(q-\hat{q})~, \end{align} where I have used the fact that $$(1/2\pi) \int \exp(ik(x-y)) dk = \delta(x-y)$$.

This final equation seems reasonable, in the sense that it is telling me to replace my $$q$$s with $$\hat{q}$$s and be done with it. But I do not understand how to interpret $$\delta(q-\hat{q})$$, since we usually have that $$\delta(0) = \infty$$, but this implies $$q=\hat{q}$$ which doesn't even make sense since one is a c-number and the other is an operator. Is there a way to interpret a delta-function like this, or have I made a mistake somewhere in my derivation? Any pointers are greatly appreciated.

Imagine you've diagonalized your operator $$\hat q= \operatorname {diag} (q_1,q_2,q_3,...)$$; you may undiagonalize it later by the inverse unitary transformation, a similarity transformation, so all its equations will revert to what you had in the general case.
Then the δ function, like the exponential it came from, is also a diagonal matrix, where q means q multiplying the identity matrix, which most physicists skip, $$\int\!\! dq~~f(q)~ \delta(q 1\!\! 1-\hat{q})\\ =\int\!\! dq~~f(q) ~\delta(\operatorname {diag} (q-q_1,q-q_2,q-q_3,...)) \\ =\int\!\! dq~~f(q) ~\operatorname {diag} (\delta(q-q_1),\delta (q-q_2),\delta (q-q_3),...) \\ =\operatorname {diag} (f(q_1),f(q_2),f(q_3),...) = f(\hat q).$$