# Does a free electron, one that's not either in an atom or a wire, have an associated wave-function?

Would a free electron, one that's not either in an atom or moving through a wire, but moving through empty space on its own, have an associated wave-function? Or, is an electron described as a wave-function only when inside an atom?

And if the type of free electron I'm describing does have a wave-function, how can it when the electron wouldn't be in an orbital within an atom?

• Free particles do have wave functions. Common ways of defining a free particle wave function is either as a plane wave, in which case the position is completely unknown, but the momentum is precisely determined, or a gaussian wave packet, which is the evolution of a free particle which had an initially well known position that is becoming increasingly uncertain. – CuriousOne Jun 30 '15 at 0:47
• CuriousOne -- That definitely helps clarify things for me. Thank you once again. If you don't mind my asking, how about inside a wire when an electron is either in AC or DC current? In this case would the idea of an associated wave-function be eliminated? – adam3033 Jun 30 '15 at 1:06
• That's a lot more complicated. Electrons in metals are reasonably well described by electronic band structure: en.wikipedia.org/wiki/Electronic_band_structure. With DC current flowing one needs to have an additional model for scattering on phonons and impurities and with AC current there is also coupling to the electromagnetic field. – CuriousOne Jun 30 '15 at 1:18
• I just took a look at that page. I see what you mean..a lot more to it. Will definitely keep me busy later tonight.. – adam3033 Jun 30 '15 at 1:42
• Might I add that the free particle wave function as described by a plane wave cannot be normalised. – Kolandiolaka Jun 30 '15 at 6:25

Yes, a free particle has a wavefunction. If it has sharp momentum $p$, it is given by the plane wave $$\psi_p(x) = \mathrm{e}^{\mathrm{i}px}$$ which might look a bit strange, because it is non-normalizable/not square-integrable, but that should not be all too troubling - no momentum uncertainty at all is rather unphysical, after all. The general wavefunction of a free particle is a square-integrable superposition of these plane waves. As to why such non-normalizable states arise, read Rod Vance's answer on rigged Hilbert spaces and their connection to scattering states.
There is no need of a potential for the Schrödinger equation to have a solution, namely $$i\hbar \frac{\partial}{\partial t} |\psi\rangle = H |\psi\rangle$$ does possess solutions even when the Hamiltonian contains no potential. Questions of normalisability may arise, but that is another point (and can anyway be solved by expanding in Fourier terms).
You seem to have a little odd idea of what a wave function is, probably due to some odd-written books in chemistry which associate the wave function to some sort of orbital cloud. That is by no means what the wave function is. Rather it is the representation on the position basis $\langle x|\psi\rangle$ of the solution of the Schrödinger equation above.