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$-k^2f''(x)=(E^2-V(x))f(x)$ I have checked for many V(x) that the set of solutions to this differential equation are complete and have an orthogonal basis set, is this always true for all V(x)?

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    orthogonal means int z(k,x)*z(b,x) dx vanishes when b not equal to k2012-12-28

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I'm not sure what you mean by "complete and have an orthogonal basis set". If $V(x)$ is continuous, there is always a two-dimensional vector space of solutions. The solutions may or may not be in $L^2$ near $+\infty$ or near $-\infty$. You might look up http://en.wikipedia.org/wiki/Spectral_theory_of_ordinary_differential_equations#Limit_circle_and_limit_point_for_singular_equations

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    In order to use the Spectral Theorem, you want a self-adjoint operator. For your operator to be essentially self-adjoint on $C_0^\infty$ (the smooth functions of compact support) you need $V(x)$ to be in the limit point case at $+\infty$ and $-\infty$.2012-12-28