Wigner Function and Spin in the Classical Limit?

Question

This is something I got curious about. Let's say I have the Wigner function for an $n$ particle system:

$$W \equiv W(x_1,\dots,x_n,;p_1,\dots,p_n) $$

Now, let's say this system obeys has spin. As far as I understand that restricts the allowed eigen-basis (upon interchange of $r_i \leftrightarrow r_j$for bosons its symmetric for fermions it's antisymmetric). Does this have any bearing on the wigner function (Since the Wigner function itself is constructed out of these wavefunctions)? If so, I'm looking for a calculation where the effects of this manifestation disappear under the classical limit $\hbar \to 0$?

Answer 1

You appear to have a vision I have not yet fathomed, so I would deny you the opportunity for confusion and consider just two bosons, nonrelativistically (which gets rid of the spin-statistics talk). You could also use fermionic wavefunctions which are antisymmetric, but the combinatorics is similar, and no clarification could ensue from throwing every possible pot on the fire.

For two bosons, the Schroedinger wave function is symmetric, $$\psi(x_1,x_2) = \psi(x_2,x_1),$$ so the corresponding Wigner function $$ W(x_1,x_2,p_1,p_2) \\ = { 1\over \pi \hbar} \int\!\!dy dw ~ e^{2i(yp_1+wp_2)/\hbar}~\psi^*(x_1+y,x_2+w)~ \psi(x_1-y,x_2-w) $$ is likewise symmetric in tandem, $$ W(x_1,x_2,p_1,p_2)= W(x_2,x_1,p_2,p_2). $$

It should be easiest for you to visualize oscillator states, which are additionally x-p symmetric, so you are really looking at $W(r_1,r_2)=W(r_2,r_1)$ objects on a plane, with no 4d visualization requirements. Note the $\hbar$-small negative value puddles, now distributed symmetrically across the 45-degree axis, which disappear in the classical limit--however you'd like to take that one, i.e. by low-pass filtering, or inspecting high-quantum number ensembles.