Answer to Question #161009 in Quantum Mechanics for sunny

Question #161009

. 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 + (βˆ’1) π‘˜ π‘›πœ‹ , where k = 0, 1, 2, 3... and n = 2k + 1. 


1
Expert's answer
2021-02-03T16:11:35-0500

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

P=∫∣Ψ∣2dxP=\int |\Psi|^2 dx

=2L∫L43L4∣sin(nΟ€xL)∣2dx=\frac{2}{L}\int_ \frac{L}{4}^\frac{3L}{4} |sin( \frac{n\pi x}{L})|^2 dx

=1L∫L43L4∣1βˆ’cos(2nΟ€xL)∣dx=\frac{1}{L}\int_ \frac{L}{4}^\frac{3L}{4} |1-cos( \frac{2n\pi x}{L})| dx

=1L[xβˆ’1(2nΟ€L)sin(2nΟ€xL)]=\frac{1}{L}[x- \frac{1}{( \frac{2n\pi }{L})}sin (\frac{2n\pi x}{L})] between limits L/4 and 3L/4

=1L[3L4βˆ’1(2nΟ€L)sin(2nΟ€(3L4)L)βˆ’([L4βˆ’1(2nΟ€L)sin(2nΟ€(L4)L))]=\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}))]

=[12βˆ’12nΟ€(βˆ’1)3n+12nΟ€(βˆ’1)n]=[\frac{1}{2}- \frac{1 }{2n\pi}(-1)^{3n}+\frac{1 }{2n\pi}(-1)^{n}]


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