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When we solve Schrodinger equation with potential goes to 0 at large distance, if $E>0 $, the wave function dies away to zero (as this shows).

My idea to prove this fact is using curvature, since normalizable it must vanish as the curvatual show.But there is a possibility that curve of wave function can look like a bowl,since normalizable we may claim , it can't happen.

The question here is for free particle,we allow unnormalizable solution,why here unnormalizable solution is not allowed.(the intuition is quite natural,big potential barrier the wave function,but what's the difference compared with free particle case?)

The similar question for harmonic oscillator. Why unnormalizable solution is allowed for free particle but not harmonic oscillator solution?

Qmechanic
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1 Answers1

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You are correct that a plane wave does not represent a normalizable wave function. Therefore, it is not a valid wave function.

However, from a mathematical point of view, a plane wave can still be used to construct properly normalizable wave functions, and this is why they are discussed in many books, although you are correct in pointing out that many books do not always make it clear why they discuss them.

To give you an example, in the case of a free particle or of a simple potential like a step or a barrier, the most common wave function you will be looking at will be a wave packet: $$ \psi(x,t)=\int_{-\infty}^{\infty}g(k)e^{i(kx-\omega t)}dk, $$
which is made of a superposition of plane waves and, for a good choice of $g(k)$, will be properly normalizable.

ProfM
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