Expectation value and momentum for an infinite square well

Homework Statement

√[/B]
A particle in an infinite square well has the initial wave function:

$Ψ(x, 0) = A x ( a - x )$

a) Normalize Ψ(x, 0)

b) Compute <x>, <p>, and <H> at t = 0. (Note: you cannot get <p> by differentiating <x> because you only know <x> at one instance of time)

The Attempt at a Solution

For part a, I figured out A = $\sqrt{30 / a^5}$

I'm sort of confused for part b. For <x>, I set up the integral like this:

$\int_{0}^{a}x Ψ(x, 0)^2 dx$

And got $\frac{a^6 A^2}{60}$, but I'm not sure if I got it right.

For <p> and <H>, I don't know how to set up these integrals. How would I set up these integrals? I don't need you to solve it for me, I just wanted to know how to set them up.

Thank you for reading and helping.

blue_leaf77
Homework Helper
How does the momentum operator look like in position space, i.e. ##p_x\psi(x) =\ldots##?

vela
Staff Emeritus
Homework Helper
For <x>, I set up the integral like this:

$\int_{0}^{a}x Ψ(x, 0)^2 dx$

And got $\frac{a^6 A^2}{60}$, but I'm not sure if I got it right.
If you plug in your result for ##A##, you get ##\langle x \rangle = \frac a2##. Given the symmetry of ##\Psi(x,0)##, does the answer seem reasonable?

How does the momentum operator look like in position space, i.e. ##p_x\psi(x) =\ldots##?

I figured it out and got 0 for <p>. That seems reasonable right?

If you plug in your result for ##A##, you get ##\langle x \rangle = \frac a2##. Given the symmetry of ##\Psi(x,0)##, does the answer seem reasonable?

Ah I didn't think of that. It makes sense.

I also figured out <H> which turns out to be ##\frac{5 h^2}{m a^2}## which seems pretty reasonable too right?

blue_leaf77