The free particle is the particle in no potential:(adsbygoogle = window.adsbygoogle || []).push({});

Hψ = Eψ

with V=0

you get waves travelling in + and - direction but they are not normalizeable. So for some reason my book comes up with the idea of constructing:

ψ(x,t) = ∫β(k)exp(i(kx-hk^{2}/2m t)) dk

I have no idea where this came from, so far all the complete solutions to the SE I have seen have been discrete. What makes us suddenly go to continuous solutions?

I can see how the above resembles a fourier transform, but I am unsure what to make of it.

Can someone guide me through a more intuitive way of seeing the above. I have watched extra lectures in QM and know how the momentum space wave function is the fourier transform of the position space. But how does it exactly relate to the above? I mean all I can see is that for a free particle the solutions must be eigenstates to the momentum operator. But why is it that they are not normalizeable etc. etc.

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# The free particle

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