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Most sources say that Wigner distribution acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula

$$\int_{\mathbb{R}^6}w(x,p)a(x,p)dxdp=<\psi|\hat{a}\psi>$$ as this allows us to compute the expectation value for the operator $\hat{a}$ corresponding to the physical quantity $a$. However, without the knowledge of this formula, how did Wigner come up with this definition? $$w(x,p)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\psi(x-\frac{v}{2})\psi^*(x+\frac{v}{2})e^{iv.p}dv.$$ I would be greatly indebted for any mathematical motivation that anyone could provide. Similarly, I also wonder at the mathematical motivation for the Weyl quantization.

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If you like this question you may also enjoy reading this post. –  Qmechanic May 11 '13 at 12:47
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up vote 2 down vote accepted

Let us for simplicity work in 1D with $\hbar=1$. (The generalization to higher dimensions is straightforward.) Moreover, let us for simplicity take an operator $\hat{f}(\hat{X},\hat{P})$ without any ordering ambiguities, i.e., each monomial term in the symbol $f(x,p)$ depends only on either $x$ or $p$, but not on both. Then one possible motivation of Wigner's phase space distribution $$ \tag{1} w(x,p)~:=~ \int_{\mathbb{R}}\! {dy\over2\pi}e^{ipy}\psi^{*}(x+\frac{y}{2}) \psi(x-\frac{y}{2}). $$

goes as follows. The expectation value of the operator $\hat{f}(\hat{X},\hat{P})$ in the Schrödinger position representation

$$\tag{2} \hat{X} ~\longrightarrow~x, \qquad \hat{P} ~\longrightarrow~ -i \frac{\partial}{\partial x}, $$


$$\langle\psi| \hat{f}(\hat{X},\hat{P})|\psi \rangle ~\stackrel{(2)}{=}~ \int_{\mathbb{R}} \!dx~ \psi^{*}(x) f\left(x, -i \frac{\partial}{\partial x}\right) \psi(x) \qquad\qquad $$ $$ ~\stackrel{\begin{matrix}\text{substitute}\\ x=x^{\pm}\end{matrix}}{=}~ \int_{\mathbb{R}^{2}}\! dx^{+}dx^{-}~ \delta(x^{+}-x^{-})\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, -i \frac{\partial}{\partial x^{-}}\right) \psi(x^{-}) $$ $$ ~\stackrel{\delta\text{-fct}}{=}~ \int_{\mathbb{R}^{3}}\! {dx^{+}dx^{-}dp\over2\pi}~ e^{ip(x^{+}-x^{-})}\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, -i \frac{\partial}{\partial x^{-}}\right) \psi(x^{-}) $$ $$ ~\stackrel{\text{int. by part}}{=}~\int_{\mathbb{R}^{3}}\! {dx^{+}dx^{-}dp\over2\pi}~ e^{ip(x^{+}-x^{-})}\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, p\right) \psi(x^{-}) $$ $$ ~\stackrel{\begin{matrix}\text{substitute}\\ x^{\pm} = x\pm \frac{y}{2}\end{matrix}}{=}~\int_{\mathbb{R}^{3}}\! {dxdydp\over2\pi}~ e^{ipy}\psi^{*}(x+\frac{y}{2}) f(x, p) \psi(x-\frac{y}{2}) $$ $$ \tag{3} ~\stackrel{(1)}{=}~\int_{\mathbb{R}^{2}}\! dxdp~w(x,p) f(x, p) .$$

That's the motivation! The above manipulations make sense for a sufficiently well-behaved function $f(x, p)$.

For more general operators $\hat{f}(\hat{X},\hat{P})$, we leave it for OP to show that if $f(x,p)$ is interpreted as the Weyl-symbol of the operator $\hat{f}(\hat{X},\hat{P})$, then the equation

$$\tag{3'} \langle\psi| \hat{f}(\hat{X},\hat{P}) |\psi\rangle ~=~\int_{\mathbb{R}^{2}}\! dxdp~w(x,p) f(x, p)$$

continues to hold [at least for a sufficiently well-behaved function $f(x, p)$].

One important virtue from a physics perspective of Weyl-ordering (as opposed to other ordering prescriptions) is that the operator $\hat{f}(\hat{X},\hat{P})$ formally becomes Hermitian for real functions $f:\mathbb{R}^2 \to\mathbb{R}$ and two Hermitian operators $\hat{X}$ and $\hat{P}$. Recall that Hermitian operators correspond to physical observables in quantum mechanics. OP might also find this Phys.SE post useful.

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So Wigner was aware of the Weyl Quantization rule? –  Vivek May 11 '13 at 22:15
Weyl quantization (1927) predates Wigner's phase space distribution (1932) by five years, so it seems likely. –  Qmechanic May 11 '13 at 22:35
I am not a physics student so I was wondering whether physics students at the graduate level are familiar with Weyl Quantization? –  Vivek May 13 '13 at 14:17
That seems to be a good question for our chat room. My impression is that students generally have heard that one is supposed to symmetrize the $\hat{X}$s and $\hat{P}$s. Whether they have pondered how to do this systematically depends on their specialization. –  Qmechanic May 13 '13 at 20:31
I updated the answer. –  Qmechanic May 13 '13 at 21:30
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