There is an obvious property of the Wigner function when one takes the limit $p\rightarrow\infty$. By the Riemann's lemma, *the Wigner function is expected to go to zero in this limit but it does driven by the largest eigenvalue*. Your question can be stated in the following way. The Wigner function for a generic operator $A$ can be defined as

$$ W_A(x,p)=\frac{1}{2\pi\hbar}\int_{-\infty}^{+\infty}dy e^{-\frac{i}{\hbar}py}\langle x-y|A|x+y\rangle. $$

Now, assume that you have diagonalized $A$ so that $A|a_n\rangle=a_n|a_n\rangle$, you can write immediately

$$W_A(x,p)=\frac{1}{2\pi\hbar}\sum_n a_n\int_{-\infty}^{+\infty}dy e^{-\frac{i}{\hbar}py}\phi_n^*(x-y)\phi_n(x+y)=\frac{1}{2\pi\hbar}\sum_n a_n W_n(x,p) $$

being $\phi_n(x)=\langle a_n|x\rangle$. Of course, if the spectrum of $A$ will not run to infinity, taking the limit of increasing $p$ should grant the work done. This can also be argued from the unbounded case of the Hamiltonian of the harmonic oscillator. In this case you will get

$$W_A(x,p)=\sum_n E_n W_n^{H.O.}(x,p)$$

being now

$$ W_n^{H.O.}(x,p)=\frac{(-1)^n}{\pi\hbar}e^{-\frac{p^2}{\hbar^2\kappa^2}-\kappa^2x^2}L_n\left(2\frac{p^2}{\hbar^2\kappa^2}+2\kappa^2x^2\right)$$

and $L_n(x)$ the Laguerre polynomials and $\kappa^2=m\omega/\hbar$. In the limit $p\rightarrow\infty$, being $x$ fixed, the Laguerre polynomial will give the driving contribution $p^n$. In this particular case, it is the "energy" of the oscillator to drive to zero with the power of $n$.

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