# Time dependence of expectation value $\hat O$ if $\frac{\partial \hat O}{\partial t} = 0$

I am given the following derivation in my lectures:

$$\frac{\partial}{\partial t} \langle \hat O \rangle = \frac{\partial}{\partial t}\int_{-\infty}^{\infty} \psi^* \hat O \ \psi \ dx$$

$$\implies \frac{\partial}{\partial t} \langle \hat O \rangle = \int_{-\infty}^{\infty} \frac{\partial}{\partial t}\left(\psi^* \hat O \ \psi \right) \ dx$$

$$\implies \frac{\partial}{\partial t} \langle \hat O \rangle = \int_{-\infty}^{\infty} \frac{\partial \psi^*}{\partial t} \ \hat O \ \psi \ dx + \int_{-\infty}^{\infty} \psi^* \ \hat O \ \frac{\partial \psi}{\partial t} \ dx$$

Which doesn't make sense to me. $$\hat O$$ and $$\frac{\partial}{\partial t}$$ are both linear operators, and for the farthest-right integral in the last line above, the implication is..

$$\frac{\partial}{\partial t} \left( \hat O \ \psi \right) = \hat O \ \frac{\partial \psi}{\partial t}$$

Which implies, generally, for two linear maps $$S$$ and $$T$$ respectively, applying to vector $$\psi$$ in vector space $$H$$ that

$$(S \ \circ\ T)(\psi) = (T \ \circ \ S)(\psi)$$ which is untrue, as far as I'm aware. The only situation where it might perhaps be true is if they are inverse maps of eachother, but this exception cannot prove the derivation generally which it aims to do.

What's going on here? Is it because $$\frac{\partial}{\partial t} \left( \hat O \ \psi \right) = \frac{\partial \hat O}{\partial t} \psi + \hat O \frac{\partial \psi}{\partial t}$$

If so, I suppose this is true, although in general, I find the concept of $$\frac{\partial \hat O}{\partial t}$$ generally weird if $$\hat O$$ is not applied to $$\psi$$ or anything.

• I think they have assumed that the operator has no explicit time dependence in position representation (in which you seem to work). Hence the commutation. Also, in the Schrodinger picture, the operators are not changed with time. – Naman Agarwal Dec 10 '18 at 16:00

What's going on here? Is it because $$\frac{\partial}{\partial t} \left( \hat O \psi\right) = \frac{\partial \hat O}{\partial t} \psi + \hat O\frac{\partial \psi}{\partial t}$$

Yes, that's right.

If I'm understanding you correctly, then you're concerned about the notion of the derivative of an operator with respect to some parameter (in this case, time). It's actually pretty much what you'd expect; if an operator $$\hat O$$ depends on some parameter $$\sigma$$, then we can formally define its derivative (acting on some state $$\psi$$) to be $$\left(\frac{\partial \hat O}{\partial \sigma}\right)\psi := \lim_{\epsilon \rightarrow 0} \frac{\hat O(\sigma +\epsilon)\psi - \hat O(\sigma) \psi}{\epsilon}$$ This is reasonable because $$\hat O$$ can be thought of as a linear map $$\hat O : \mathbb R \rightarrow \mathcal L(\mathcal H)$$ (from the real numbers to the linear operators on the Hilbert space $$\mathcal H$$), so differentiation takes place pretty much exactly like it does for your standard functions of a real variable.

The domain of the derivative is obviously some subset of the domain of $$\hat O$$ for which the limit exists, which must be decided on an operator-by-operator basis.

If you're wondering what kinds of operators might have such an explicit time dependence, then you may want to reference Stone's theorem on one-parameter unitary groups. As another example, you can consider the adiabatic theorem which concerns slowly-varying potentials.