# Problem with the proof of Ehrenfest theorem

My quantum mechanics book has this proof of Ehrenfest theorem: let $$A$$ be an observable and $$\hat{A}$$ the operator that represents it. Then we have $$\frac{d}{dt}\langle \hat{A}\rangle=\frac{i}{\hbar}\langle [\hat{H},\hat{A}]\rangle+\left\langle \frac{\partial \hat{A}}{\partial t}\right\rangle.$$ The proof is: \begin{align}\frac{d}{dt}\langle \hat{A}\rangle=&\frac{d}{dt}\langle\Psi| \hat{A}\Psi\rangle=\left\langle \frac{\partial \Psi}{\partial t}|\hat{A}\Psi\right\rangle+\left\langle \Psi|\frac{\partial \hat{A}\Psi}{\partial t}\right\rangle \\ \overset{!}{=}&\left\langle \frac{\partial \Psi}{\partial t}|\hat{A}\Psi\right\rangle+\left\langle \Psi|\hat{A}\frac{\partial \Psi}{\partial t}\right\rangle+ \left\langle \Psi|\frac{\partial \hat{A}}{\partial t}\hat{\Psi}\right\rangle.\end{align} I can't understand how to evaluate the term $$\frac{\partial \hat{A}\psi}{\partial t}$$. $$\hat{A}$$ is an operator, if we have a multiplicative operator like the potential then the result is trivial and follow from applying the product rule.

How to extent this product rule like result to a generic operator $$\hat{A}$$ though?

Consider a harmonic potential with a constant $$K$$ which explicitly depends on time: $$U(t,x) = \frac{K(t)}{2} x^2\:.$$ Next define $$A(t) := -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + U(t,x)\:.$$ Here the derivative you pointed out gives a contribution. Another possibility is $$m=m(t)$$, giving rise to a temporal dependence in the kinetical energy operator which is not multiplicative as you wanted.

• Ok, this is one of the cases in which I have problems What does it mean to take the derivative of the operator $\frac{\partial \hat {A}}{\partial t}$? Jan 26 '20 at 18:47
• Imagine the operator being a function of time $A: \mathbb{R} \rightarrow \mathbb{L}, t \rightarrow A(t)$, it should be clear what is meant by $A'(t)$? @Rhino
– user224659
Jan 26 '20 at 18:53
• Allright, it's a function that gives an operator for any value ov $t$, but why $\frac{\partial \hat{A}\Psi}{\partial t}=\frac{\partial \Psi}{\partial t}\hat{A}+\frac{\partial \hat{A}}{\partial t}\Psi$? Jan 26 '20 at 19:02
• To compute the derivative of an operator parametrically depending on $t$ you can use the strong or the weak operator topology. Here, the weak one is sufficient. Jan 26 '20 at 19:08
• $A_n \to A$ strongly if $A_nf \to Af$ for all $f$. Similarly $A_n \to A$ weakly if $\langle g, A_nf \rangle \to\langle g, Af\rangle$ for all $g, f$. Derivatives are defined accordingly. Jan 26 '20 at 20:48

Let's go through the proof for the sake of completion.

$$\frac{d\langle A \rangle}{dt} = \frac{d}{dt}\langle\psi|A|\psi\rangle$$

Applying the product rule gives

$$\frac{d}{dt}\langle\psi|A|\psi\rangle =\frac{d\langle \psi|}{dt}A|\psi \rangle + \langle \psi |\frac{dA}{dt} |\psi \rangle + \langle \psi| A \frac{d|\psi \rangle}{dt}$$

Now to reveal some more information we look at the time dependent Schrödinger equation,

$$H|\psi \rangle=i\frac{d|\psi\rangle}{dt}$$

We can write the first two terms as:

$$\frac{d\langle \psi|}{dt}A|\psi \rangle + \langle \psi |\frac{dA}{dt} |\psi \rangle = i\langle\psi|HA|\psi\rangle + -i\langle\psi|AH|\psi\rangle$$ $$=i[H,A]$$

Thus we have Erehnfests theorem $$\frac{d}{dt} \langle A \rangle = i[H,A] + \langle \psi|\frac{dA}{dt}|\psi \rangle$$

Within the Schrodinger picture we assume $$A \neq A(t)$$ i.e. all the time dependence is in the state $$|\psi\rangle = |\psi (t) \rangle$$. Meaning the expressions simplifies to just the commutator of the Hamiltonian.