# 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?

## marked as duplicate by John Rennie quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 5 '17 at 15:42

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