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.

  • $\begingroup$ Ok, this is one of the cases in which I have problems What does it mean to take the derivative of the operator $\frac{\partial \hat {A}}{\partial t}$? $\endgroup$
    – Rhino
    Jan 26 '20 at 18:47
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    $\begingroup$ 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 $\endgroup$
    – user224659
    Jan 26 '20 at 18:53
  • $\begingroup$ Allright, it's a function that gives an operator for any value ov $t$, but why $\frac{\partial \hat{A}\Psi}{\partial t}=\frac{\partial \Psi}{\partial t}\hat{A}+\frac{\partial \hat{A}}{\partial t}\Psi$? $\endgroup$
    – Rhino
    Jan 26 '20 at 19:02
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    $\begingroup$ 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. $\endgroup$ Jan 26 '20 at 19:08
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    $\begingroup$ $A_n \to A$ 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. $\endgroup$ 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 $$\frac{d}{dt} \langle A \rangle = i[H,A] + \langle \psi|\frac{dA}{dt}|\psi \rangle$$

Some notes on this which address your concerns:

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.

But if we shift to the Heisenberg picture then there is an explicit time dependence on the operator. Meaning that the derivative is well defined and thus has to be considered.


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