# How to get the time derivative of an expectation value in quantum mechanics?

The textbook computes the time derivative of an expectation value as follows: $$\frac{d}{dt}\langle Q\rangle=\frac{d}{dt}\langle \Psi|\hat Q\Psi\rangle=\langle \frac{\partial\Psi}{\partial t}|\hat Q\Psi\rangle+\langle\Psi|\frac{\partial\hat Q}{\partial t}\Psi\rangle+\langle\Psi|\hat Q\frac{\partial\Psi}{\partial t}\rangle$$ I can't see how this could be done. The text seems to treat $\hat Q\Psi$ as a multiplication of two functions of $t$ and use the product rule of differentiation to get the result. But $\hat Q$ is a functional, its parameter is an element from the Hilbert space, not time. And $\hat Q\Psi$ means $\hat Q(\Psi)$, not $\hat Q$ times $\Psi$. So isn't $\frac{\partial\hat Q}{\partial t}$ a meaningless expression?

I guess the chain rule should be used, but the result should be the product of two derivatives instead of the sum.

$Q$ is not a functional, but a linear operator. Since it is linear, there are no problems in using the chain rule. I will pretend there are no domain problems here, and that you can exchange limits and integrals as you wish (e.g. I'm supposing you can use the dominated convergence theorem), and obviously that each quantity is differentiable. $$\frac{d}{dt}\langle\psi(t),Q(t)\psi(t)\rangle= \lim_{h\to 0}\frac{1}{h}\Bigl( \langle\psi(t+h),Q(t+h)\psi(t+h)\rangle - \langle\psi(t),Q(t)\psi(t)\rangle\Bigr)\\=\lim_{h\to 0} \langle\frac{1}{h}(\psi(t+h)-\psi(t)+\psi(t)),Q(t+h)\psi(t+h)\rangle - \frac{1}{h}\langle\psi(t),Q(t)\psi(t)\rangle\\= \langle\partial_t\psi(t),Q(t)\psi(t)\rangle +\lim_{h\to 0}\frac{1}{h}\langle\psi(t),Q(t+h)\psi(t+h)\rangle - \frac{1}{h}\langle\psi(t),Q(t)\psi(t)\rangle\\=\langle\partial_t\psi(t),Q(t)\psi(t)\rangle +\lim_{h\to 0}\langle\psi(t),\frac{1}{h}(Q(t+h)-Q(t)+Q(t))\psi(t+h)\rangle \\-\frac{1}{h}\langle\psi(t),Q(t)\psi(t)\rangle\\=\langle\partial_t\psi(t),Q(t)\psi(t)\rangle+\langle\psi(t),(\partial_tQ(t))\psi(t)\rangle+\lim_{h\to 0}\langle\psi(t),Q(t)\frac{1}{h}(\psi(t+h)-\psi(t))\rangle\\=\langle\partial_t\psi(t),Q(t)\psi(t)\rangle+\langle\psi(t),(\partial_tQ(t))\psi(t)\rangle+\langle\psi(t),Q(t)\partial_t\psi(t)\rangle\; .$$ I remark that $\frac{1}{h}\bigl(Q(t)\psi(t+h)-Q(t)\psi\bigr)=Q(t)\frac{1}{h}(\psi(t+h)-\psi(t))$ because by definition of linearity, given $\psi,\phi\in\mathscr{H}$, and $\lambda\in\mathbb{C}$: $$A(\lambda(\psi+\phi))=\lambda(A(\psi)+A(\phi))$$ for any linear operator $A$.