(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]P = \int_a^b \, \left| \psi(x) \right|^2 \, dx[/tex]

If the particle in the box is in the second excited state (i.e., n=3), what is the probability P that it is between x=L/3 and x=L? To find this probability, you will need to evaluate the integral:

[tex]\int_{L/3}^L \left(\sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) \right)^2 \, dx = \frac{2}{L} \int_{L/3}^L \sin^2 \left( \frac{n \pi x}{L} \right) \, dx .[/tex]

2. Relevant equations

This one I think.

[tex]

* \int \sin^2(kx) dx = \frac{x}{2} - \frac{1}{4k} \sin(2 k x) +C, and

[/tex]

3. The attempt at a solution

Set k = n(pi)/L or np/L

[2/L][(x/2 - 1/4k)sin(2kx)]|L and L/3

put k back in as it is.

[2/L]{[L/2 - L/4np]sin(2*np/L*L)]-[L/3*2 - L/4np]sin(2 * np/L * L/3)]}

cancel out the L's and plug-in n = 3

2{[(1/2 - 1/12p)sin(6p)] - [(1/6) - (1/12p)]sin(2p)]}

well the answer isn't 0. but sin6p and sin2p both equal 0

2[0 - 0] = 0

Where did I go wrong?

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# Probablities from wave function

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