Find Result <x> for Particle in n=3 Excited State of Rigid Box

[SpaceTrekkie]

Homework Statement

find the result <x> (the expectation value) found when the position of a particle in the second (n=3) excited state of a rigid box.

Homework Equations

<x> = (2/a)\intxsin²(xpi/a)dx

evaluated from 0 to a

The Attempt at a Solution

well when that integral is evaulated using the ground state the answer is a/2. I am not sure how they even got that. But my real problem is, is where does the n = 3 come in?

and where do I go from there?

thanks...

 

[buffordboy23]

"n" appears in the numerator of the argument of the sine-squared term. "n" denotes the state of the particle.

a/2 for the ground state (n = 1) is correct only if the boundaries of the box occur at 0 and a. To get the answer, all you need to do is integrate. In QM, the integrals may get a little complicated to do by hand, so a good integral-solver software application (e.g. Maple) will likely benefit you.

 

[SpaceTrekkie]

ooo ok. Thanks, I figured it out. The use of the boundary conditions are what was confusing me.