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I'm having a conceptual problem. The time-independent Schrodinger equation for a continuum state $\psi_{k, l}(r)$ is given by

$$ 0 = \left( - \frac{\nabla^2}{2} + v(r) - \frac{k^2}{2} \right) \psi_{k,l}(r) , $$

where $\nabla^2$ is the Laplacian, $v(r)$ some potential, $k$ the electron momentum. Expressing this as a matrix, I find real-valued eigenstates. However, textbooks and papers say the solution is

$$ \psi_{k,l}(r) \propto R_{k,l}(r) Y_{l,m}(\theta,\phi) e^{i \sigma_{k}}, $$

where $\sigma_k$ is known as the scattering phase shift. I understand how to interpret the phase shift - it is the phase difference in the asymptotic region between the eigenstate and a plane wave of the same momentum. Where, however, does this comparison come in when solving the Schrodinger equation?

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  • $\begingroup$ How were you arriving at real eigenstates? If nothing else in the limit that $v=0$ the eigenstates are plane waves and clearly not real in general. $\endgroup$ Commented Aug 29, 2021 at 13:54

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One way to look at this is to imagine sending a wave into some potential, but you are only looking at the wave when it is way out at large radii, where the potential is approximately zero. As the wave goes in and interacts with the potential, its wavelength will change: as the particle it represents would gain kinetic energy falling into an attractive potential, the wavelength gets shorter, and so on. After interacting with the potential, the wave comes back out to large radii where you are able to observe it again. The phase shift mathematically represents the fact that the squishing and squeezing of the wave at small radii means that the crests and troughs of the wave after it comes back out away from the potential will not necessarily line up with the crests and troughs of the wave that you sent in. So "where it comes from in the Schrodinger equation" is the fact that the Schrodinger equation would properly describe the change in wavelength that was talked about pretty schematically here.

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