I was reviewing past exams and I found a question where I could give no satisfactory answer. The core of the question is the following: If there are boundary conditions, does this necessarily imply, that the solutions to the time-independent Schrödinger equation are quantized, i.e. there are only countably many solutions? By the expression "has a boundary conditions" I want to specifically exclude the free particle i.e. $V=0$.
Is there a rigid mathematical proof for this?

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    $\begingroup$ Please note that you must always have boundary (and initial) conditions of some kind when you're solving a partial differential equation, such as Schrödinger's. Without them there's no specific solution.Which conditions do you have in mind? $\endgroup$ – pglpm Jun 6 at 8:59
  • $\begingroup$ What are the boundary conditions for a free particle? $\endgroup$ – Makkabi Jun 6 at 9:16
  • $\begingroup$ They are expressed in the form of limits, for example that $\psi(x,t) \to 0$ in a specific way as $x \to \infty$. That's a boundary condition. $\endgroup$ – pglpm Jun 6 at 9:17
  • $\begingroup$ That actually makes my question a lot easier: Is this the only boundary condition for which there are uncountably many possible states ? $\endgroup$ – Makkabi Jun 6 at 9:29
  • $\begingroup$ Note that "free particle" is ambiguous though: do you have a bounded or unbounded domain? $\endgroup$ – pglpm Jun 6 at 9:31