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.

  • $\begingroup$ I don't see how Shankar is interchanging these two things. There's no mistake in the textbook. $\endgroup$ – knzhou Aug 4 '16 at 3:46
  • $\begingroup$ Line 6.4 Says, "We consider $X$ for $\Omega$ and he starts that line with $$\left<\dot{X}\right>=...$$ Instead of $$\frac{d}{dt}\left<X\right>=...$$ $\endgroup$ – Shrodinger 2016 Aug 4 '16 at 3:49
  • 1
    $\begingroup$ I agree, that's an ambiguity of dot notation. The dot is supposed to be over the entire thing, not over just the $X$. $\endgroup$ – knzhou Aug 4 '16 at 3:50
  • $\begingroup$ Oh my, that is some confusing notation. Thank you. $\endgroup$ – 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 $t_0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.