- #51

reilly

Science Advisor

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**How's that again?**

marlon said:No, once again: the Schrodinger equantion (SE) is independent of boundary conditions. The SE is a mere manifestation of E = p²/2m, translated into derivative operators working on a wavefunction.

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What? You might consider explaining this notion. Yours is quite an unusual approach.

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The solutions (ie the wavefunctions) are dependent on boundary conditions in order to assure their normalizability (so they are "physical" if you will) and their continuity at boundaries (like in the case of the potential-well)

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Whether boundary conditions are part of the SE is a somewhat of a problematical matter. If, for example, you use a Laplace Transform, then the boundary conditions become part of the transformed equation, as would be the case with a "one-sided Fourier transform" But, a regular FT does not explicitely contain boundary conditions. But, there are integrability conditions with the FT, which can be tantamount to Sommerfeld's Radiation Condition, a standard requirement for solutions as their spatial argument goes to infinity.

Regards,

Reilly Atkinson