Understanding derivation of Wigner function for the Harmonic oscillator In the document https://www.hep.anl.gov/czachos/aaa.pdf, they derive the Wigner functions, $f_n$ for the harmonic oscillator. However, I have some problems understanding some of the steps. On page 37 they write the equation:
$$\tag{1}
\left(\left(x+\frac{i \hbar}{2} \partial_{p}\right)^{2}+\left(p-\frac{i \hbar}{2} \partial_{x}\right)^{2}-2 E\right) f(x, p)=0
$$
Defining $z\equiv \frac{4}{\hbar}H= \frac{2}{\hbar}\left(x^{2}+p^{2}\right)$ they say that the real part of eq. $(1)$ can be written:
$$\tag{2}
\left(\frac{z}{4}-z \partial_{z}^{2}-\partial_{z}-\frac{E}{\hbar}\right) f(z)=0$$
and by setting $f(z)=\exp (-z / 2) L(z)$ we get Laguerre's equation:
$$\tag{3}
\left(z \partial_{z}^{2}+(1-z) \partial_{z}+\frac{E}{\hbar}-\frac{1}{2}\right) L(z)=0$$
Solved by Laguerre polynomials:
$$\tag{4}
L_n=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(-z)^{k}}{k !}$$
and the Wigner functions are then:
$$\tag{5}
f_{n}=\frac{(-1)^{n}}{\pi \hbar} e^{-z/2} L_{n}\left(z\right)$$
There are 3 things that I do not get:

*

*We can write the real part of eq. $(1)$ as $\left(x^{2}-\frac{\hbar^{2}}{4} \partial_{p}^{2}+p^{2}-\frac{\hbar^{2}}{4} \partial_{x}^{2}-2 E\right) f=0$. How is eq. $(2)$ obtained from this? (I don't get the $z \partial_{z}^{2}-\partial_{z}$ term)

*How is eq. $(3)$ obtained from eq. $(2)$?

*where does the $\frac{(-1)^{n}}{\pi \hbar}$ pre-factor in eq. $(5)$ come from?

 A: Product support.

*

*Recall, crucially,  f was shown to be a function of z only, f(z), so, acting on it,
$$\partial_x = \frac{\partial z}{\partial x } \partial_z  \leadsto \\
\partial_x^2 = \left (\partial_x \frac{\partial z}{\partial  x } \right )\partial_z  + \left ( \frac{\partial z}{\partial x }\right )^2  \partial_z^2 ~ ,
$$
and similarly for y, so that $\partial_x^2 + \partial_y^2 = 8(\partial_z + z\partial_z^2)/\hbar$.


*You've been there before with the integrating factor of the oscillator equation to Hermite's in Hilbert space. Analogously,
$$
\left(\frac{z}{4}-z \partial_{z}^{2}-\partial_{z}-\frac{E}{\hbar}\right) ( e^{-z / 2}~L(z) )=0, ~~~~~\leadsto\\
 e^{-z / 2}
 \left(z \partial_{z}^{2}+(1-z) \partial_{z}+\frac{E}{\hbar}-\frac{1}{2}\right) L(z)=0.$$
But the exponential can never be zero, and can thus be dropped.


*Any multiple of these polynomials will solve this linear equation.
However, it is practical/convenient to simplify their Rodrigues formula
$$L_n(z)=\frac{e^z}{n!}\partial_z^n \left(e^{-z} z^n\right) =\frac{1}{n!} \left( \partial_z -1 \right)^n z^n,$$
and Sheffer sequence recursive formula,
$$
\tag{7} \partial_z L_n = \left (  \partial_z - 1 \right ) L_{n-1},$$ generating function, etc, as you probably learned  from your Hydrogen atom. So they are all unity at the origin. Recall, from the text,  these are all ingredients of Wigner functions f normalized to 1, whence the common normalization; trivially checkable for n=0,
$$
1=\int\!\! dxdp ~f(z)= \frac{\pi \hbar }{2}\int_0^\infty\!\! dz ~e^{-z/2} L_n (z)\frac{(-)^n}{\pi \hbar} ~.
$$
But from n=0 and the Sheffer sequence recursion (7), you may readily check the normalization for n=1, through integration by parts to be a mere change of sign. So, recursively, for all n, you show the alternating sign normalizations.
