Probability of Momentum in Infinite Well

[Poop-Loops]

Homework Statement

Need to find the probability density of a ground state in an infinite square well.

Homework Equations

Ground State psi(x) = Asin(Bx), don't want to look up the constants, don't think they are relevant anyway.

The Attempt at a Solution

Took the derivative of psi(x) with respect to x to get momentum, times h/(2*pi*i).

Then tried to integrate psi*psi, but before I did that I stopped myself, because taking the limits of integration to be infinity and negative infinity like for the position probability, I'd end up with 0. Since I get cos(x)^2 = 1 - sin(x)^2

Sin(x)^2 integrated from -inf. to inf. = 1, so this would be 0. That doesn't make sense, so it's clearly not the way to go about it.

Should I go a step deeper and just try to solve Schrodinger's equation for momentum? Would that mean I have to take d/dx for the whole thing and then solve it? This problem is harder than I thought it would be.

 

[HungryChemist]

probability density function is the integrand! since the problem is asking you to find this integrand, you don't need to integrate!

Also, when you do integrate the probability density function of momentum, you find that <p>= 0 (as you claimed) and this is perfectly fine. Since, in infinite square well potential, you would expect <x> to be constant! (to wit: <p> = m d<x>/dt)

 

[Poop-Loops]

I thought it was only constant in the classical case? Since the ground state is an arc (half a sine wave), there's more probability in the middle, so wouldn't the momentum be smaller in the middle?

 

[Poop-Loops]

Wait, you're just using the expectation value, which isn't what I am trying to find at all.