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Average result <x>

1. 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.


2. Homework Equations
<x> = (2/a)[tex]\int[/tex]xsin2(xpi/a)dx

evaluated from 0 to a



3. 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...
 
"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.
 
ooo ok. Thanks, I figured it out. The use of the boundary conditions are what was confusing me.
 

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