# Why does a plane wave leave the position of the particle unspecified?

I'm covering a book on QM, and just started recently and I'm stuck at understanding something. It says that we can describe the state of motion of a particle with an infinite plane wave equation:

$\psi(r,t)=Ae^{i(\mathbf{k}.\mathbf{r}-wt)}$

It says that it corresponds to a motion with precisely defined momentum.

"but having amplitudes $|\psi|=const. \forall r,t$, the infinite harmonic plane waves leave the position of the particle entirely unspecified."

1. What does the word "infinite" aim to imply here? Does it mean we assume the imaginary "waves" extend infinitely in both directions? Also, if so, does that matter?
2. Of course any planar wave has constant amplitude! It's not that they are imposing some mathematical restriction on the wave. Please explain the statement.

So as you say, a planar wave has a constant amplitude. This then implies that they do extend all the way to infinity in both directions--there is no edge to it. Of course, this is never physically realized, but it's a good first (or zeroth) approximation in a lot of cases--just like it is in electromagnetism, if you've dealt with that before.

The reason this leaves the position unspecified is the way that you would extract information on where the particle is is to integrate the absolute square $\psi \psi^* = \left| \psi \right|^2$ over a volume. This quantity represents the probability that the particle will be measured in that region. If $\left| \psi \right|$ is constant everywhere in space, this probability is also constant everywhere in space--so we're equally likely to find the particle in any place we look. To do this mathematically, it's usually best to bound it somewhat so you can normalize the wavefunction--say that you have a volume $V$ that the particle can be found in. Still, though, a pure momentum plane-wave state will then just be entirely indeterminate in position in the volume.

This is why people build wave packets by combining many plane waves; then their distribution in space will be nontrivial.

1) Yes, that's pretty much it - a perfect/ideal wave will extend to infinity. It matters in the sense that a sine wave, $\sin (kx)$ (and similarly a complex plane wave) is only a momentum eigenfunction if it is taken to be infinite in extent, as the domain of sine is all real numbers.

2) If the amplitude was not constant, it would have to vary. If it varied, the envelope function for the amplitude would have to decay at infinity (or at least be square integrable). If it decayed, it could not be infinite in extent.

The result of the infinite extent of the wave-function means that the particle could be (except at the zeroes of $\psi^2$) anywhere, and so it's position is not determinable.

• sorry for unmarking you answer, but I think it was very helpful in making my understand. However the answer by @sirelderberry pointed out the reason for using the amplitude, which I think was essential. But again, both answers were equally useful. Mar 11, 2015 at 15:27