Hilbert Space-Phase Space map

Given that a classical phase space quantity $$A_{cl}(x, p_{x})$$ is related to the quantum operator $$\hat{A}$$ in Hilbert space as

$$A_{cl} = e^{-ik_{x}x} \langle x|\hat{A}|k_{x}\rangle = \langle x|\hat{A}|k_{x}\rangle \langle k_{x}|x\rangle$$

By integrating over phase space one obtains the following expression: $$\int dx \int \frac{dp}{2\pi\hbar} A_{cl}(x,p_{x}) = \int dx \langle x|\hat{A}|x\rangle = Tr( \hat{A})$$

However, it is not clear to me how the Hilbert space operator can now be expressed via the following expression: $$\hat{A} = \int dx \int \frac{dp}{2\pi\hbar} |x\rangle \langle x|\hat{A}|k_{x}\rangle \langle k_{x}|$$

• Are you sure you didn't mistype in the last formula? I think the last $x$ should have been $k_x$. Commented Jun 19, 2020 at 8:44
• Thanks for spotting the mistake, I edited it. Commented Jun 19, 2020 at 8:49
• Might be helpful to point out this is the "standard ordering prescription" of Terletzky, Blokhintsev, and Yvon, late 30s, early 40s. Commented Jul 1, 2020 at 22:19

This can be shown by noting that $$\hat{A} = \operatorname{id} \hat A \operatorname{id} =\left(\int dx\, |x\rangle \langle x|\right) \hat{A}\left(\int \frac{dp}{2\pi\hbar} |k_{x}\rangle \langle k_{x}|\right)$$
We have the completeness relations \begin{align} \int \mathrm dx \lvert x\rangle\langle x\rvert &= \hat 1 & \int \frac{\mathrm dp}{2\pi\hbar}\lvert k_x\rangle\langle k_x\rvert &= \hat 1 \end{align} where $$\hat 1$$ is the identity operator.
Seeing that $$\lvert x\rangle\langle x\rvert \hat A$$ does not depend on $$p$$, one can move that vactor out of the $$p$$ integral: \begin{align} \int\mathrm dx \int \frac{\mathrm dp}{2\pi\hbar} \lvert x\rangle \langle x\rvert\hat{A}\lvert k_{x}\rangle \langle k_{x}\rvert &= \int\mathrm dx \lvert x\rangle \langle x\rvert\hat{A}\int \frac{\mathrm dp}{2\pi\hbar} \lvert k_{x}\rangle \langle k_{x}\rvert\\ &= \int\mathrm dx \lvert x\rangle \langle x\rvert\hat{A}\hat 1 = \int\mathrm dx \lvert x\rangle \langle x\rvert\hat{A} \end{align} Similarly, $$A$$ also does not depend on $$x$$, therefore on the left side one can use the completeness relation for $$x$$ in the same way.