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The wavefunction $\Psi(x,t)$ for a free particle is given by

$$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$

This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?

Why then do we use free particles $\psi(\vec{x}) = e^{ikz}$, for example, in scattering theory?

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The wavefunction:

$$ \Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)} $$

is an infinite plane wave. So it describes a particle that has an infinite extent in both time and space. That is, it exists for $-\infty \le x \le \infty$ and for $-\infty \le t \le \infty$. Unsurprisingly, if the particle has an infinite extent then it's amplitude is everywhere zero and normalisation requires multiplying $\infty$ by zero which is meaningless. This is a mathematical idealisation, not an attempt to describe a real particle, and you are quite correct that the particle described by the equation does not exist in nature.

In reality the particle has a finite lifetime and during that lifetime can travel a finite distance. However in many experiments, for example scattering, we are not concerned where the particle originally came from or where it is eventually going to, and it is a convenient approximation to describe it as an infinite plane wave.

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  • $\begingroup$ $\psi$ function describes systems of particles in it's configuration space, not in physical space. If $x$ represents three coordinates, $\psi$ describes point particle, not "particle with an infinite extent in space". $\psi$ function being infinite does not imply particle is infinite; particle is not $\psi$ function anymore dust particle is its probability distribution $\rho(x)$ one may use to describe it. $\endgroup$ – Ján Lalinský Nov 28 '15 at 11:41
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This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?

No, it does not; it means that it is not a valid $\psi$ function to use in a theory based on the Born interpretation of $|\psi(x)|^2|$ as density of probability for configuration $x$.

You're right there are no free particles in nature, but the reason is in nature things interact with their neighbours and we do not know any way to isolate part of the world so as to prevent that entirely. What we can do is minimize interactions, but there is always some left.

Why then do we use free particles $\psi(\vec{x}) = e^{ikz}$, for example, in scattering theory?

It is much easier to do calculations with such functions, and some useful results can be obtained with it.

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