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?


1 Answer 1


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:

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.