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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2Please 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 06 '20 at 08:59
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What are the boundary conditions for a free particle? – Makkabi Jun 06 '20 at 09:16
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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 06 '20 at 09:17
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That actually makes my question a lot easier: Is this the only boundary condition for which there are uncountably many possible states ? – Makkabi Jun 06 '20 at 09:29
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Note that "free particle" is ambiguous though: do you have a bounded or unbounded domain? – pglpm Jun 06 '20 at 09:31
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Possible duplicates: https://physics.stackexchange.com/q/65636/2451 and links therein. – Qmechanic Jun 06 '20 at 10:01