Why Does the Upper Integration Limit Change in Quantum Mechanics Problems?

[Woozie]

By the way, this isn't a homework problem, it's self study.

Homework Statement

We're given that we have a wave function in an infinite potential well, with the well having 0 PE from 0 to a. The wave function would be described by

{\Phi}_n={\sqrt{\frac{2}{a}}Sin({\frac{n{\pi}x}{a})}. It was given that the function is originally in it's second exicited state, meaning it's original state can be described by {\Phi}_3={\sqrt{\frac{2}{a}}Sin({\frac{3{\pi}x}{a})}

The box is suddenly expanded, so that it's rightmost boundary is now 2a. This means the new basis vectors are of the form
|{\Theta}_n>={\sqrt{\frac{1}{a}}} Sin{\frac{n{\pi}x}{2a}})

What the book asks is for me to find the probability of this particle being found in states n=1 and 2. I figured the probability of finding it in any given state would be <{\Theta}_n|{\Phi_3}>
={\int}_0^{2a}({\sqrt{\frac{1}{a}}} Sin({\frac{n{\pi}x}{2a}}))({\sqrt{\frac{2}{a}}Sin({\frac{3{\pi}x}{a})})dx

I had no problems up to here. It's the next step in their analysis that threw me off.

={\frac{{\sqrt{2}}}{a}{\int}_0^aSin({\frac{n{\pi}x}{2a}})Sin({\frac{3{\pi}x}{a})}dx

Looks like they simplified the equation in the exact same way that I did. The problem is that the upper limit of integration has changed from 2a to a. I've been looking at this problem for a while now and I cannot figure out why the limit of integration changes. Why is it that the upper limit of integration changes from 2a to a when all they did was simplify a few terms? Shouldn't the limit of integration still be 2a? When I do use 2a, all of the probabilities vanish, so obviously I'm wrong and the book is right. I just cannot figure out why that is the case. Why does the upper limit become a instead of 2a?

 

[dwintz02]

Shouldn't it be 0 to a from the beginning since that is the area the particle is confined to immediately after the walls have been expanded?

 

[Galileo]

The integration is from 0 to 2a, but your initial wavefunction is zero for |x|>a

 

[Woozie]

The integration is from 0 to 2a, but your initial wavefunction is zero for |x|>a

Oh, thanks, I didn't forgot about that. Thank you.

 

[Woozie]

Wait, in that case, shouldn't the equations be:

={\int}_0^{a}({\sqrt{\frac{1}{a}}} Sin({\frac{n{\pi}x}{2a}}))({\sqrt{\frac{2}{a}}Sin({\frac{3{\pi}x}{a})})dx ?

Because the book does in fact write this integral out with 2a as the upper limit of integration.

 

[nrqed]

Wait, in that case, shouldn't the equations be:

={\int}_0^{a}({\sqrt{\frac{1}{a}}} Sin({\frac{n{\pi}x}{2a}}))({\sqrt{\frac{2}{a}}Sin({\frac{3{\pi}x}{a})})dx ?

Because the book does in fact write this integral out with 2a as the upper limit of integration.

then that's a typo in the book.

 

[Woozie]

Oh okay, thanks.