Suppose I have a quantum operator $\hat{Q}(t)$ in the Heisenberg picture. Is the expression $ \frac{d}{dt}\langle \psi | \hat{Q}(t) | \psi \rangle$ (where the brakets denote a quantum expectation value) equal to $\langle \psi | \frac{d\hat{Q}(t)}{dt} | \psi\rangle$? i.e. do time dependence and quantum expectation value operations commute?
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2$\begingroup$ Yes. That is the rationale behind the Heisenberg picture. $\endgroup$– oliverMar 24 '21 at 21:31
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$\begingroup$ The vector $\left|\psi\right>$ doesn't depend on time in Heisenberg picture. $\endgroup$– heaven-of-intensityMar 24 '21 at 21:56
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Yes, that is indeed correct. The time-dependence in the Heisenberg picture is purely in the operators and not in the states. So, what evolves is the operator, and correspondingly, its eigenstates. However, the state of a system does not evolve. Thus, as long as $\vert\psi\rangle$ is supposed to be a state of the system (which it ought to be if one is talking about expectation values), then $\frac{d}{dt}\langle\psi\vert\hat{Q}\vert\psi\rangle=\langle\psi\vert\frac{d}{dt}\hat{Q}\vert\psi\rangle$ is correct.
However, it should be noticed that if one is talking about an eigenstate $\vert q\rangle$ of the operator $\hat{Q}$ then $\vert q(t)\rangle = e^{i\hat{H}t}\vert q(0)\rangle$. Thus,
\begin{align} \frac{d}{dt}\langle q\vert\hat{Q}\vert q\rangle &=\frac{d}{dt} \langle q(0)\vert e^{-i\hat{H}t}\vert e^{i\hat{H}t}\hat{Q}(0)e^{-i\hat{H}t}\vert e^{i\hat{H}t}\vert q(0)\rangle\\&=0\\&\neq\langle q\vert\frac{d}{dt}\hat{Q}\vert q\rangle \end{align}
