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}| $

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

2 Answers 2


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.


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