# How does the wave function of free particle $\psi(x,t)=A\exp \{ i(kx-\omega t)\}$ satisfy normalisation condition? [duplicate]

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?

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.
• The introduction of the $\delta$ function allows you to introduce some kind of "orthogonality condition", but no normalization. – valerio Jun 5 '17 at 9:45