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?
Answer
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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