# What is the next step in solving the time derivative of expectation value of the momentum of a particle [duplicate]

So this question is a direct follow on from my last question

I am trying to get an expression for $$\frac{d\left}{dt}$$.

I have my working out below, which was thanks to the Anonymous Physicist's help, (not to mention all the patient people in the mathematics chatroom) I have got to this expression: $$i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right).$$

And now I'm stuck. What can I do about $$\int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx$$

In particular how to deal with $$\frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x}?$$ With the $$\partial t$$ in play I don't know how solve the integral...

As before, I really would appreciate hints to get me moving in the right direction more than an outright solution.

## My working so far

We are given $$\left$$ in the text, which is determined by computing $$\frac{d\left}{dt}$$ and multiplying by $$m$$.

Lets just start with what we know: $$\left = \int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx.$$

So taking the time derivative, $$\frac{d\left}{dt} = \frac{d}{dt}\int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx.$$

We probably should at this point pause, and listen to Ted, (and RobJohn). So take out the constants before going any further. We get, $$-i\hbar \frac{d}{dt}\int\limits_{-\infty}^{+\infty}{ \left( \Psi^* \left[ \frac{\partial}{\partial x} \right] \Psi \right) }dx.$$

(1) Since the variables t and x are independent, you can bring the time derivative into the integral. You can also use the derivative product rule to get two terms in the integral.

Bringing the time derivative into the integral, gives us $$-i\hbar \int\limits_{-\infty}^{+\infty}{\frac{\partial}{\partial t}\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi}dx.$$

Then differentiating with the product rule yields, $$-i\hbar \int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x}+ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx.$$

Now we have two terms in our integral, let's split em up. Let's consider the left handside first, we have $$\int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} }dx.$$

By Schrödinger this can become, $$\int\limits_{-\infty}^{+\infty}{ \left(\frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar}V \Psi^* \right) \frac{\partial \Psi}{\partial x} }dx.$$

Expand to get, $$\int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x}+ \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx.$$

So, just to review, our full problem can be written as, $$i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right).$$