Answer to Question #160997 in Quantum Mechanics for Vivek

Question

A particle is in the n

th state of an infinite one-dimensional box of length L. Show that 

the probability of finding the particle between L/4 and 3L/4 is 1

2

+

(−1)

𝑘

𝑛𝜋

, where k = 0, 

1, 2, 3... and n = 2k + 1.

Answer 1

The probability of of finding the particle between L/4 and 3L/4 is

$P=\int |\Psi|^2 dx$

$=\frac{2}{L}\int_ \frac{L}{4}^\frac{3L}{4} |sin( \frac{n\pi x}{L})|^2 dx$

$=\frac{1}{L}\int_ \frac{L}{4}^\frac{3L}{4} |1-cos( \frac{2n\pi x}{L})| dx$

$=\frac{1}{L}[x- \frac{1}{( \frac{2n\pi }{L})}sin (\frac{2n\pi x}{L})]$ between limits L/4 and 3L/4

$=\frac{1}{L}[\frac{3L}{4}- \frac{1}{ (\frac{2n\pi }{L})}sin (\frac{2n\pi(\frac{3L}{4}) }{L})-([\frac{L}{4}- \frac{1}{ (\frac{2n\pi }{L})}sin (\frac{2n\pi(\frac{L}{4}) }{L}))]$

$=[\frac{1}{2}- \frac{1 }{2n\pi}(-1)^{3n}+\frac{1 }{2n\pi}(-1)^{n}]$