# Ehrenfest's theorem derivation

I'm stuck at a question from Griffiths which ask to prove that:

$$\dfrac{d \left\langle p \right\rangle}{dt}=\left\langle -\dfrac{\partial V}{\partial x}\right\rangle.$$

What I did is the following:

$$\dfrac{d \left\langle p \right\rangle}{dt}=\dfrac{d}{dt}\int\psi^*\dfrac{h}{i}\dfrac{d\psi}{dx}=\dfrac{h}{i}\int\left(\dfrac{d\psi^*}{dt}\dfrac{d\psi}{dx}+\psi^*\dfrac{d}{dx}\dfrac{d\psi}{dt}\right)$$

And after inserting the time derivative of $\psi^*$ and $\psi$ and taking the integration I found the following equation:

$$-\dfrac{d\psi^*}{dx}\dfrac{d\psi}{dt}-\int \psi^*\dfrac{\partial V}{\partial x}\psi$$ First term is evaluated at infinity and minus inifnity.

My question is, can I say that derivative of wave function at inifinity always goes to zero ? Or am I making a mistake somewhere?

Lets start by deriving Ehrenfest's theorem. The expectation value is given as:

$$\left< A \right> = \left< \psi \left| \hat{A} \right| \psi \right>$$

We can now take the time derivative of the expectation value:

$$\frac{d}{dt}\left< A \right> = \frac{d}{dt}\left< \psi \left| \hat{A} \right| \psi \right>$$

Now by expanding the right hand side:

$$\frac{d}{dt}\left< A \right> =\left< \frac{d}{dt}\psi \left| \hat{A} \right| \psi \right> + \left< \psi \left| \frac{\partial}{\partial t}\hat{A} \right| \psi \right>+ \left< \psi \left| \hat{A} \right| \frac{d}{dt}\psi \right>$$

Now we can see that the first and last term can be replaced directly by considering the time-depend Schrödinger equation:

$$i\hbar\frac{\partial}{\partial t}\left| \psi \right>=\hat{H}\left| \psi \right>$$

Now giving:

$$\frac{d}{dt}\left< A \right> =-\frac{1}{i\hbar}\left< \hat{H} \psi \left| \hat{A} \right| \psi \right> + \left< \psi \left| \frac{\partial}{\partial t}\hat{A} \right| \psi \right>+ \frac{1}{i\hbar}\left< \psi \left| \hat{A} \right| \hat{H}\psi \right>$$

By the definition of commutators this can be seen to reduce to:

$$\frac{d}{dt}\left< A \right> =\frac{i}{\hbar}\left< \psi \left| \left[ \hat{H} , \hat{A} \right] \right| \psi \right> + \left< \psi \left| \frac{\partial}{\partial t}\hat{A} \right| \psi \right>$$

Note here that I have made no assumptions about the positional derivative of the wave function. Lets now inspect the expectation value of the momentum:

$$\frac{d}{dt}\left< p \right> =\frac{i}{\hbar}\left< \psi \left| \left[ \hat{H} , \hat{p} \right] \right| \psi \right> + \left< \psi \left| \frac{\partial}{\partial t}\hat{p} \right| \psi \right>$$

Since $\hat{p}$ is time-independent, the last term vanishes:

$$\frac{d}{dt}\left< p \right> =\frac{i}{\hbar}\left< \psi \left| \left[ \hat{H} , \hat{p} \right] \right| \psi \right>\tag{1}$$

Now we define our Hamiltonian operator as:

$$\hat{H}=\frac{1}{2m}\hat{p}^2 - \hat{V}$$

We there have that:

$$\left[ \hat{H} , \hat{p} \right] = \left[ \frac{1}{2m}\hat{p}^2 , \hat{p} \right] + \left[ \hat{V} , \hat{p} \right]$$

Clearly the first commutator is equal to zero. Now lets inspect the second commutator. Here we will let the commutator work on a function:

$$\left[ \hat{V} , \hat{p} \right]f = V(-i\hbar)\frac{\partial f}{\partial x} - (-i\hbar)\frac{\partial \left(V f\right)}{\partial x}$$

Now expanding the second term:

$$\left[ \hat{V} , \hat{p} \right]f = -i\hbar V\frac{\partial f}{\partial x} +i\hbar f\frac{\partial V}{\partial x} +i\hbar V\frac{df}{dx} = i\hbar f\frac{\partial V}{\partial x}$$

Thus:

$$\left[ \hat{V} , \hat{p} \right] = i\hbar \frac{\partial V}{\partial x}$$

Now by inserting in equation (1):

$$\frac{d}{dt}\left< p \right> =\frac{i}{\hbar}\left< \psi \left| i\hbar \frac{\partial V}{\partial x} \right| \psi \right> = -\left< \frac{\partial V}{\partial x} \right>$$

Which is the result we were looking for. I know that I didn't directly answer you what you were asking for. But I think that you can find use in this indirect answer also.