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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:

  1. We can write the real part of eq. $(1)$ as $\left(x^{2}-\frac{\hbar^{2}}{4} \partial_{p}^{2}+p^{2}-\frac{\hbar^{2}}{4} \partial_{x}^{2}-2 E\right) f=0$. 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?
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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 $\partial_x^2 + \partial_y^2 = 8(\partial_z + z\partial_z^2)/\hbar$.

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

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  • $\begingroup$ I get point 2 and 3 now. But I am still a bit lost on point 1. Is the following argument correct: $\partial f=\partial z \frac{\partial f}{\partial z} \Rightarrow \frac{\partial f}{\partial x}=\frac{\partial z}{\partial x} \frac{\partial f}{\partial z}$? And I do not see how you get the expression for $\partial_x^2$. As far as I know, the square of the product of two operators is the product of the squares, so I would assume $\partial_{x}=\frac{\partial z}{\partial x} \partial_{z} \Rightarrow \partial_{x}^{2}=\left(\frac{\partial z}{\partial x}\right)^{2} \partial_{z}^{2}$ $\endgroup$ – Physics101 Apr 24 at 18:00
  • $\begingroup$ Notice in my first equation in the comment I use notation $\frac{\partial}{\partial x}=\partial_{x}$, so my denominator is not the same as the one you wrote above. $\endgroup$ – Physics101 Apr 24 at 18:04
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    $\begingroup$ @Physics101 You have to apply the product rule, check this math.stackexchange.com/questions/1313764/… and apply it here. $$\frac{\partial}{\partial x}\frac{\partial}{\partial x} f= \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\frac{\partial f}{\partial z}\right)\\ =\left(\frac{\partial^2 z}{\partial x^2}\frac{\partial f}{\partial z}\right) +\left(\frac{\partial z}{\partial x}\right)^2 \left(\frac{\partial^2 f}{\partial z^2}\right)\\ $$ $\endgroup$ – Hans Wurst Apr 24 at 18:37
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    $\begingroup$ @Physics101 $$\frac{\partial^2 }{\partial x^2}\rightarrow \frac{\partial^2 z }{\partial x^2}\frac{\partial}{\partial z} + \left(\frac{\partial z}{\partial x}\right)^2 \frac{\partial^2 }{\partial z^2} $$ $\endgroup$ – Hans Wurst Apr 24 at 18:38
  • $\begingroup$ Thanks @Hans , I couldn't have said it better. I corrected my TeX-typos $\endgroup$ – Cosmas Zachos Apr 24 at 18:39

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