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This question already has an answer here:

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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marked as duplicate by John Rennie quantum-mechanics Jun 5 '17 at 15:42

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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.

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  • $\begingroup$ Then why are we using it? $\endgroup$ – john melon Jun 5 '17 at 9:27
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    $\begingroup$ The introduction of the $\delta$ function allows you to introduce some kind of "orthogonality condition", but no normalization. $\endgroup$ – valerio Jun 5 '17 at 9:45

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