Question on quantum numbers

The Schrodinger equation: [itex]H \psi = E \psi[/itex] can be solved for arbitrary values of [itex]E[/itex], but when [itex]E[/itex] is not an eigenvalue of the hamiltonian, then [itex]\psi[/itex] will be unnormalizable--it will blow up at infinity, or at the origin, or somewhere. For example, a solution to the free particle Schrodinger equation with [itex]E < 0[/itex] is: [itex]\psi = e^{\hbar K x}[/itex], which corresponds to an energy of [itex]-\hbar^2 K^2/(2 m)[/itex].

The differential equations for the wave function are forced to have discrete values by boundary conditions: to make sure that the wave function is single-valued in the whole domain, and to make sure that the integral of [itex]|\psi|^2[/itex] is finite.