Quantum Mechanics: Expected momentum of a real wavefunction

[galaxyrocker]

Homework Statement

Given the wavefunction, $\psi(x) = Cx$ for 0 < x < 10 and $\psi(x) = 0$ for all other values.

What is the normalization constant of C? I got $\sqrt{3/1000}$.
What is <x>? I got 30/4.
What is <p>? Here is where I'm confused.

Homework Equations

$$\langle p \rangle = C^2 \frac{\hbar}{i} \int_0^{10} \psi^* \frac{d}{dx} \psi \, dx$$

The Attempt at a Solution

I worked out the integral, and got $(3/20)(\hbar/i)$.

It's here that I can't figure out whether to leave it as that, or say it's 0, since the momentum is an observable, and must be real. I realize that the function is odd, but I'm not integrating it from -a to a, which means I can't just say it's 0 for that reason.

[strike]PS: Sorry for my horrible usage of LaTeX.[/strike]
Mod note: Fixed your LaTeX for you.[/color]

 

[PhysicsGente]

Try integrating by parts. Also, your notation is not good. You already pulled the C's out of the integral so it should be $\langle p \rangle = C^2 \frac{\hbar}{i} \int_0^{10} x \frac{d}{dx} x \, dx$NVM the above ;(. Think about it. You know that <p> is real and you get a purely imaginary result. What can you conclude? ;)

 

[galaxyrocker]

I conclude that I am either wrong, or that <p> = 0. I'm inclined to the former, but the latter makes sense, because if I was integrating around the origin (say, from -5 to 5 instead), it would cancel.

 

[Marioqwe]

I conclude that I am either wrong, or that <p> = 0. I'm inclined to the former, but the latter makes sense, because if I was integrating around the origin (say, from -5 to 5 instead), it would cancel.

But your integral goes from 0 to 10, so <p> = 0.

 

[PhysicsGente]

But your integral goes from 0 to 10, so <p> = 0.

Yes.