- #1
iafatel
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Homework Statement
For a particle in the ground state of a rigid box, calculate the probability of finding it between x=0 and x=[tex]\frac{L}{3}[/tex]
Homework Equations
[tex]\left|\psi^{2}\right| = \frac{2}{L}Sin^{2}\left(\frac{nx\pi}{L}dx\right)[/tex]
The Attempt at a Solution
[tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} Sin^{2}\left(\frac{x\pi}{L}\right)[/tex]
[tex]= \frac{2}{L}\int^{\frac{L}{3}}_{0} \frac{1-Cos \left(\frac{2x\pi}{L}\right)}{2}[/tex]
[tex]\frac{1}{L} \int^{\frac{L}{3}}_{0} 1 - \int^{0}_{\frac{L}{3}} Cos \left(\frac{2x\pi}{L}\right)}[/tex]
[tex]\frac{1}{L} \left[ \left|x\right|^{\frac{L}{3}}_{0} - \left|\frac{L}{2\pi}Sin\left(\frac{2x\pi}{L}\right)\right|^{\frac{L}{3}}_{0} [/tex][tex]\frac{1}{L} \left[ \frac{L}{3}} - \frac{L}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]
factor out the L
[tex]\frac{1}{3}} - \frac{1}{2\pi}Sin\left(\frac{2\pi}{3}\right)\right] [/tex]
stuck here...when I work it out I get a negative number...am I mising something?
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