Understanding derivation of Wigner function for the Harmonic oscillator

Question

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:
$$tag{1}
left(left(x+frac{i hbar}{2} partial_{p}right)^{2}+left(p-frac{i hbar}{2} partial_{x}right)^{2}-2 Eright) f(x, p)=0
$$
Defining $zequiv frac{4}{hbar}H= frac{2}{hbar}left(x^{2}+p^{2}right)$ 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:
$$tag{5}
f_{n}=frac{(-1)^{n}}{pi hbar} e^{-z/2} L_{n}left(zright)$$
There are 3 things that I do not get:

We can write the real part of eq. $(1)$ as $left(x^{2}-frac{hbar^{2}}{4} partial_{p}^{2}+p^{2}-frac{hbar^{2}}{4} partial_{x}^{2}-2 Eright) f=0$. 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?

Cosmas Zachos · Accepted Answer

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 $partial_x^2 + partial_y^2 = 8(partial_z + zpartial_z^2)/hbar$.

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
$$L_n(z)=frac{e^z}{n!}partial_z^n left(e^{-z} z^nright) =frac{1}{n!} left( partial_z -1 right)^n z^n,$$
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.

Understanding derivation of Wigner function for the Harmonic oscillator

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