I am confused about the wave function of a free particle
$$\psi(x,t)=A\exp \{ i(kx-\omega t)\}$$
How does this satisfy the normalization condition? Since this corresponds to a plane wave, what meaning does the probability have?
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Qmechanic
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john melon
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It does not satisfy it, and therefore it does not represent a physical state. – valerio Jun 05 '17 at 09:26
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Then why do we use it? – john melon Jun 05 '17 at 09:27
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2Like for many other things in physics, it is a useful idealization. – valerio Jun 05 '17 at 09:28
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Also, a superposition of plane waves can represent a physical state. – valerio Jun 05 '17 at 09:33
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Hi John. Also see ACuriousMind's characteristically excellent answer to Is the existence of a sole particle in an hypothetical infinite empty space explicitly forbidden by QM? – John Rennie Jun 05 '17 at 15:43
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It does NOT satisfy the usual normalization condition. The plane wave has the same probability density everywhere. The normalization of plane waves requires the introduction of $\delta$-functions.
ZeroTheHero
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2The introduction of the $\delta$ function allows you to introduce some kind of "orthogonality condition", but no normalization. – valerio Jun 05 '17 at 09:45