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Understanding derivation of Wigner function for the Harmonic oscillator

Physics Asked by Physics101 on July 17, 2021

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:

  1. 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)
  2. How is eq. $(3)$ obtained from eq. $(2)$?
  3. where does the $frac{(-1)^{n}}{pi hbar}$ pre-factor in eq. $(5)$ come from?

One Answer

Product support.

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

  2. 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.

  3. 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.

Correct answer by Cosmas Zachos on July 17, 2021

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