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# Answer to Question #2251 in Quantum Mechanics for raj

Question #2251
A quantum particle confined to one dimensional box of width 'a' is in its first exited state. What is the probability of finding the particle over an interval of (a/2) marked symmetrically at the centre of the box.
1
2011-04-04T10:21:09-0400
At the first exited state the probability density is equal to
<img src="/cgi-bin/mimetex.cgi?%5CPsi%5E2%20=%20%5Cfrac%7B2%7D%7Ba%7D%5Csin%5E2%7B%5Cfrac%7B2%5Cpi%20x%7D%7Ba%7D%7D" title="\Psi^2 = \frac{2}{a}\sin^2{\frac{2\pi x}{a}}">,
so
<img src="https://latex.codecogs.com/gif.latex?P%280%3Cx%5Cleq%20a/2%29%20=%20%5Cint_0%5E%7Ba/2%7D%5Cleft%20%28%5Cfrac%7B2%7D%7Ba%7D%5Csin%5E2%7B%5Cfrac%7B2%5Cpi%20x%7D%7Ba%7D%7D%20%5Cright%20%29dx%20=%20%5Cfrac%7B2%7D%7Ba%7D%20%5Cleft%20[%20%5Cfrac%7Bx%7D%7B2%7D%20-%20%5Cfrac%7Ba%20%5Csin%5E2%7B%5Cfrac%7B4%5Cpi%20x%7D%7Ba%7D%7D%7D%7B8%20%5Cpi%7D%20%5Cright%20]_0%5E%7Ba/2%7D%20=%20%5Cfrac%7B1%7D%7B2%7D" title="P(0<x\leq a/2) = \int_0^{a/2}\left (\frac{2}{a}\sin^2{\frac{2\pi x}{a}} \right )dx = \frac{2}{a} \left [ \frac{x}{2} - \frac{a \sin^2{\frac{4\pi x}{a}}}{8 \pi} \right ]_0^{a/2} = \frac{1}{2}">;
or simplier: if the full probability is 1 and the probability function is symmetrical relatively the center of& the box, the probability of finding the particle over the half of the box is equal to the 1/2.

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