I have calculated the expectation value of a particle in a box of width a to be a/2. The wavefunction of the particle is:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]N Sin(k_n x) Exp[-i \frac{E_n t}{\hbar}][/tex]

Now, in the first excited state with [tex]k_n[/tex] equal to [tex]2\pi / a[/tex] the position probability density peaks at a/4 and 3a/4 but is zero at a/2! But the expectation value is still given by a/2 because it is still the average value of the position. How can it be that the expectation value of the position is illegal (position probability equal to zero)? I thought the position expectation value was the position where the particle was most likely to be found?

Thanks

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# Homework Help: Position expectation value of a particle in a box

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