Integration help (expectation value)

[Perillux]

I'll skip the format because this isn't for a course, just a textbook I'm reading. Also because it shows the steps but I'm unsure about one of them. It might be a dumb question, but here goes:

It's for calculating \frac{d<p>}{dt} Using the momentum operator we have:

\frac{d}{dt}<p> = -i\hbar \int_{-\infty}^{+\infty} \frac{\partial}{\partial t} (\Psi^* \frac{\partial \Psi}{\partial x})dx

then I'm not entirely sure how they get the next step:

= -i\hbar \int_{-\infty}^{+\infty} [\frac{\partial}{\partial t}\Psi^* \frac{\partial \Psi}{\partial x} + \Psi^*\frac{\partial}{\partial x}(\frac{\partial\Psi}{\partial t})]dx

I know this is probably just some fundamental rule of integration. But I put the whole equations up anyway. Please just explain it to me. If it is just a rule I have to memorize would you possibly be able to point me somewhere that explains how they derive it?
Thank you.

 

[Hootenanny]

It's not a rule of integration, but rather a rule of differentiation, the product rule to be more precise.

 

[Perillux]

Oh right! I should have known this... so ashamed. lol

ok, so I guess that \frac{\partial}{\partial t} and \frac{\partial}{\partial x} are interchangeable. They do it that way to set it up for the next step.

I knew I was going to slap myself on the forehead after seeing the answer... oh well.
Thank you.

 

[Hootenanny]

Oh right! I should have known this... so ashamed. lol

ok, so I guess that \frac{\partial}{\partial t} and \frac{\partial}{\partial x} are interchangeable. They do it that way to set it up for the next step.

I knew I was going to slap myself on the forehead after seeing the answer... oh well.
Thank you.

Don't be ashamed! To answer your question yes, if the function is sufficiently well behave, more specifically if all the mixed second order derivatives are continuous then the order of differentiation may be changed. I.e.

\Psi_{xt} = \Psi_{tx}