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Wigner's function: Observables

Let a (pure or mixed) quantum state of a one-dimensional particle be represented by a Wigner function \(W(x,p)\). Here, \(x\) and \(p\) are the particle's position and momentum, respectively.

Density operator

The density operator \(\rho\) representing the state can be expressed in terms of the Wigner function: \[\rho = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dx' \int_{-\infty}^{+\infty} dp \, | x' \rangle \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \langle x | \,.\] Proof: According to the definition of the Wigner function, \[W(x,p) = \frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | \rho | x - \tfrac{1}{2} x' \rangle \,.\] Hence, \[\begin{align} \int_{-\infty}^{+\infty} dp \, e^{i p \xi' / \hbar} W(\xi,p) &= \frac{1}{2 \pi \hbar} \int_{-\infty}^{+\infty} d\xi'' \, \langle \xi + \tfrac{1}{2} \xi'' |\rho | \xi - \tfrac{1}{2} \xi'' \rangle \underbrace{ \int_{-\infty}^{+\infty} dp \, e^{i p (\xi' - \xi'') / \hbar} }_{2 \pi \hbar \delta(\xi' - \xi'')} \\ &= \langle \xi + \tfrac{1}{2} \xi' |\rho | \xi - \tfrac{1}{2} \xi' \rangle \,. \end{align}\] Introducing \(x = \xi - \tfrac{1}{2} \xi'\) and \(x' = \xi + \tfrac{1}{2} \xi'\), so that \(\xi = \tfrac{1}{2} (x + x')\) and \(\xi' = x' - x\), one gets \[\langle x' | \rho | x \rangle = \int_{-\infty}^{+\infty} dp \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \,.\]

Observables

Consider an observable represented by an operator \(A\). The observable can be described by a phase-space function \(\mathcal{A}(x,p)\) that is defined by requiring that its averaging over the Winger function yields the correct expectation value of the operator, i.e. \[\operatorname{Tr} \rho A = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dp \, W(x,p) \mathcal{A}(x,p) \,. \] It follows from this definition that \(\mathcal{A}(x,p)\) is given by \[\mathcal{A}(x,p) = \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | A | x - \tfrac{1}{2} x' \rangle \,.\] Proof: The expression \[\rho = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dx' \int_{-\infty}^{+\infty} dp \, | x' \rangle \, e^{i p (x' - x) / \hbar} W \left( \frac{x + x'}{2}, p \right) \langle x |\] implies that \[\begin{align} \operatorname{Tr} \rho A &= \int_{-\infty}^{+\infty} d\xi'' \, \langle \xi'' | \rho A | \xi'' \rangle \\ &= \int_{-\infty}^{+\infty} d\xi \int_{-\infty}^{+\infty} d\xi' \int_{-\infty}^{+\infty} d\xi'' \int_{-\infty}^{+\infty} dp \, \langle \xi'' | \xi' \rangle \, e^{i p (\xi' - \xi) / \hbar} W \left( \frac{\xi + \xi'}{2}, p \right) \langle \xi | A | \xi'' \rangle \\ &= \int_{-\infty}^{+\infty} d\xi \int_{-\infty}^{+\infty} d\xi' \int_{-\infty}^{+\infty} dp \, e^{i p (\xi' - \xi) / \hbar} W \left( \frac{\xi + \xi'}{2}, p \right) \langle \xi | A | \xi' \rangle \end{align} \,.\] Introducing \(x = \tfrac{1}{2} (\xi + \xi')\) and \(x' = \xi - \xi'\), so that \(\xi = x + \tfrac{1}{2} x'\) and \(\xi' = x - \tfrac{1}{2} x'\), one gets \[\operatorname{Tr} \rho A = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{+\infty} dp \, W(x,p) \underbrace{ \int_{-\infty}^{+\infty} dx' \, e^{-i p x' / \hbar} \langle x + \tfrac{1}{2} x' | A | x - \tfrac{1}{2} x' \rangle }_{\mathcal{A}(x,p)} \,.\]