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.


1 Answer 1


There is a well-established nifty theory of functions of matrices, which these discussions of Hermitian operators follow, but I should just make you comfortable visualizing such things. It is easiest to visualize functions of diagonal matrices, which are also diagonal matrices, the function acting on each diagonal argument in each site.

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, $$ \int\!\! dq~~f(q)~ \delta(q 1\!\! 1-\hat{q})\\ =\int\!\! dq~~f(q) ~\delta(\operatorname {diag} (q-q_1,q-q_2,q-q_3,...)) \\ =\int\!\! dq~~f(q) ~\operatorname {diag} (\delta(q-q_1),\delta (q-q_2),\delta (q-q_3),...) \\ =\operatorname {diag} (f(q_1),f(q_2),f(q_3),...) = f(\hat q). $$

Check it out with a 2 by 2 matrix and a concrete function.


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