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Why does the Time-Independent Schrodinger Equation, which is the second order ODE $ \left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right]\psi(x) = E\psi(x), $ can have an infinite number of eigenvalues, hence also corresponding to an infinite number of eigenfunctions?

I thought there could be at most two eigenfunctions and two eigenvalues for a second order ODE.

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    @anon: you should make your comment an answer, it is definitely the point where bugged is confused. An eigenvalue equation is not just an ordinary differential equation. There's something more as you note, and that is that you look for all possible values of the parameter $\lambda$ for which a solution exists.2011-10-17

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The key to understanding this issue is that $\lambda$ is not a fixed parameter, so -\phi''+\lambda\phi=0 for example isn't just a single equation: it's an entire family of them. There are initial or boundary conditions imposed, in which case there are infinitely many $\lambda$ for which this form of differential equation has a solution. For any such eigenvalue $\lambda$, the solution is unique (up to rescaling), so there is exactly one eigenvalue associated to each eigenfunction. In the context of quantum mechanics this means there are an infinite number of allowable energy levels. Furthermore, the fact that the spectrum, or set of eigenvalues, of the operator is discrete (countably infinite with no limit points IIRC) means that the energy levels are quantized, as we should expect in quantum mechanics.