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So I was reading Henk Stoof's Ultracold Quantum Physics and there was this simple example of a particle in a finite space (not infinite well) with no potential whatsoever. It is solved in the position basis and then Fourier transformed to get the solution in the momentum basis. The answer is:

$$\chi_{n}(p)=\frac{2\hslash}{\sqrt{2\pi\hslash L(p-p_n)}}\sin\left((p-p_n)\frac{L}{2\hslash}\right)$$

edit: I should note that $p_n$ is equal to $2\pi\hslash n/L$ where $n$ is an integer, $L$ is the system size and the energy is (of course) calculated by $p_n^2/2m$.

I'm pretty sure this answer is correct. but now here comes my question:

What if we start in the momentum basis? Shouldn't make a difference right? So we have the basis-independent time independent Schrödinger equation and multiply it with a momentum bra from the left while we use a completeness relation to get: $$\frac{p^2}{2m}\chi_n(p)=\epsilon_n \chi_n(p)$$ where orthonormality of the p-vectors got rid of the integral in the completeness relation. when we simply insert the above $\chi_n(p)$ in here I find that it does not really equate the same wave function multiplied by some factor. What am I missing here?

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  • $\begingroup$ What exactly are the boundary conditions on this finite space that is not an infinite well? $\endgroup$ – By Symmetry Sep 11 at 12:26
  • $\begingroup$ Periodic, but that's not so important because I'm sure that the first expression is the right solution. If you're still curious for the derivation, you can find it in chapter 2 of the book I mentioned (which probably can be found easily on the web). $\endgroup$ – Antaios Sep 12 at 10:06

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