Shape of a free electron wave

[jaketodd]

What is the shape of a free electron wave? Is it a sinusoidal plane wave like an EM wave? Does it expand laterally like an EM wave? I'm thinking the answers to these questions is yes because the double slit experiment works for electrons. But I'm not totally sure.

Thanks!

Jake

 

[Bob_for_short]

Any linear combination of plane waves with different frequencies and wave vectors will be also a solution to the "free" Schroedinger equation. Actual solution is determined with the electron source (preparation device).

 

[jaketodd]

Any linear combination of plane waves with different frequencies and wave vectors will be also a solution to the "free" Schroedinger equation. Actual solution is determined with the electron source (preparation device).

So what are the possible shapes and electron sources? In particular, I'm interested in the shape of an electron wave from a standard electron emitter/gun (if there is a standard one) and from a particle accelerator.

Thank you!

 

[JK423]

Im interested in this question as well!

 

[Bob_for_short]

So what are the possible shapes and electron sources? In particular, I'm interested in the shape of an electron wave from a standard electron emitter/gun (if there is a standard one) and from a particle accelerator.
Thank you!

I am not an experimentalist. Think of electron wave as of a usual wave from some source. To obtain a "plane" wave you have to be far from the source, maybe limit the flux with some diaphragm, etc. I want to say that all the wave properties are applicable to the electron case.

A particle accelerator uses diaphragms so behind a diaphragm you have some plane wave with small diffraction due to the ratio (electron de Broglie wave-length)/(diaphragm width).

 

[Civilized]

In particular, I'm interested in the shape of an electron wave from a standard electron emitter/gun (if there is a standard one) and from a particle accelerator.

The electrons in an accelerator have a well-defined momentum (and energy) , and so they are in momentum eigenstates to a good approximation, which means that they are plane waves $\psi = e^{i \vec{k} \cdot \vec{x}}$, especially on the length scales relevant for scattering in QM. The corrections which would arise by not treating the electrons as a plane wave are too small to be conceivably measured.