# Solving For the Time Dependence of an Operator Question

I am trying to find $$\frac{d}{dt}\left<X\right>$$ So I expanded this in the X basis and got $$<\dot{\Psi}|X|\Psi> + <\Psi|\dot{X}|\Psi> + <\Psi|X|\dot{\Psi}>$$ My book by Shankar says that we assume that X has no explicit time dependence so the middle term vanishes, but I don't understand how he can do that when we are literally solving for the time dependence of our operator: $$\left<\dot{X}\right>$$ what am I missing here? If you want to look it's Shankar's Principles of QM pages 179-180. Is it just bad notation that Shankar is using? I guess my question stems on the fact that $$\left<\dot{X}\right> \ \neq \ \frac{d}{dt}\left<X\right>$$ but Shankar uses them like they are interchangeable.

• I don't see how Shankar is interchanging these two things. There's no mistake in the textbook. – knzhou Aug 4 '16 at 3:46
• Line 6.4 Says, "We consider $X$ for $\Omega$ and he starts that line with $$\left<\dot{X}\right>=...$$ Instead of $$\frac{d}{dt}\left<X\right>=...$$ – Shrodinger 2016 Aug 4 '16 at 3:49
• I agree, that's an ambiguity of dot notation. The dot is supposed to be over the entire thing, not over just the $X$. – knzhou Aug 4 '16 at 3:50
• Oh my, that is some confusing notation. Thank you. – Shrodinger 2016 Aug 4 '16 at 3:51

You are presumably working in the Schrodinger picture where states carry the time dependence due to evolution. You are right. $\frac{d\langle\psi_S| X_S| \psi_S\rangle}{dt}$ is not to be taken as $\langle\psi_S| \dot{X}|\psi_S \rangle$.
But suppose you work in the Heisenberg picture where operators carry the time dependence. There indeed $$\frac{d\langle \psi_H| X_H|\psi_H \rangle}{dt}=\langle\psi_H |\dot{X_H}|\psi_H\rangle$$ where the expectation is taken with respect to the Heisenberg picture state which is time independent.
The way to relate these is, as you might know, $\langle\psi_S| \dot{X}|\psi_S \rangle=\langle\psi_H |\dot{X_H}|\psi_H\rangle$ where $|\psi_H\rangle=|\psi_S(t=t_0)\rangle$ for some $t_0$