$-k^2f''(x)=(E^2-V(x))f(x)$ 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)?
Time Independent Schrodinger equation
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ordinary-differential-equations
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0orthogonal means int z(k,x)*z(b,x) dx vanishes when b not equal to k – 2012-12-28
1 Answers
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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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0In 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