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The quantum virial thoerem is derived by arguing that the left-hand-side of the following expression is zero for stationary/bound states: $$ \frac{d}{dt}\langle{\bf{r} \cdot \bf{p}}\rangle = \bigg\langle\frac{\bf{p}^2}{m}\bigg\rangle - \langle \bf{r} \cdot \nabla{V} \rangle $$ In the classical case this follows from a consideration of the time average of this expression, which tends to $0$. Why is the same true, intuitively, in the quantum case?

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  • $\begingroup$ WP. Hint: go to the Heisenberg picture. $\endgroup$ – Cosmas Zachos Sep 29 '19 at 23:58
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For any operator $O$, the expectation value in a stationary (eigen-) state $|\psi(t) \rangle $ is constant, $$ \langle O \rangle = \langle \psi(t) | O | \psi(t) \rangle = \langle \psi(0) | e^{i E t} O e^{-i E t} | \psi(0) \rangle = e^{(i E - i E)t} \langle \psi(0) | O | \psi(0) \rangle = \langle \psi(0) | O | \psi(0) \rangle $$ where $E$ is the energy of that state. Since this expectation value doesn't depend on time, the time derivative is identically 0.

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