In the quantum version of the symmetric infinite well, the energy eigenvalues are, in principle, well-determined. Why would the momentum then have a spread or distribution(adsbygoogle = window.adsbygoogle || []).push({}); for a given energy eigenvaluei.e.

[tex] \phi(p) = 1/(2\pi\hbar) \int_{-a}^{a}dx u_n (x) e^{-ipx/\hbar}[/tex]

where u_n is the solution with wavenumber n = 1,3,5... or n = 2,4,6...

when there is the relationship E = p^2/2m?

So transforming the solutions into momentum space gives a distribution, while transforming the solution into energy space give the Dirac delta function I think is another way to put it?

Is it because of the time-dependence of momentum?

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# Homework Help: Energy eigenvalues and momentum distributions

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