- #1
ehrenfest
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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 for a given energy eigenvalue i.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?
[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?