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DEvens

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The usual thing is to have the wave function go to zero as radius goes to infinity. Is that enough?

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Quantum Defect

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Integral of Psi*Psi needs to be finite (square integrable). You can have a function go to zero as r->inf. that would not be square integrable.The usual thing is to have the wave function go to zero as radius goes to infinity. Is that enough?

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This may be excessive, as the requirement that the final wave function be square-integrable permits wave functions that are periodic with finite integrability.

The other boundary conditions also imposed by the method are that the components of the wave function in polar coordinates are orthogonal and normal.

The polar coordinates themselves are part of this, in that in other coordinate systems (rectilinear, cylindrical) the wave functions are NOTsquare-integrable.

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DEvens

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That is normalization, not boundary condition.Integral of Psi*Psi needs to be finite (square integrable). You can have a function go to zero as r->inf. that would not be square integrable.

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And you can have square integrable functions which don't go to 0 when their argument goes to infinity.Integral of Psi*Psi needs to be finite (square integrable). You can have a function go to zero as r->inf. that would not be square integrable.

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