# Do non-vanishing boundaries conditions necessarily force a discrete/quantized set of solutions for the Schrödinger Equation? [duplicate]

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?

• 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? – pglpm Jun 6 at 8:59
• What are the boundary conditions for a free particle? – Makkabi Jun 6 at 9:16
• 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. – pglpm Jun 6 at 9:17
• That actually makes my question a lot easier: Is this the only boundary condition for which there are uncountably many possible states ? – Makkabi Jun 6 at 9:29
• Note that "free particle" is ambiguous though: do you have a bounded or unbounded domain? – pglpm Jun 6 at 9:31