# Is the free electron wavefunction stable?

The wavefunction of a free electrons is variously described as a plane wave or a wave packet. I am fairly happy with the wave packet, as it is localised.

But if we change to the electron's rest frame, then there is no direction of motion, and thus the wavefunction must be isotropic, presumably as a Gaussian wavepacket.

Yet the wavepacket supposedly expands over time in real space, permitting it to drop in energy in momentum space.

Does this mean that the free-electron wavefunction is not stable? In its own frame, will it simply expand ad infinitum until it interacts with another particle?

If the electron can continue to expand arbitrarily, then the probability that it retains its original momentum grows vanishingly small, which would deny conservation of momentum for any free particle with a significant path length.

## migrated from theoreticalphysics.stackexchange.comMar 13 '12 at 8:58

• If you like this question, you may also enjoy reading this question. – Qmechanic Mar 13 '12 at 19:51

The magnitudes of the spatial $~$Fourier spectrum (the momentum space) of a spreading localized free field do not change over time. The momentum components stay the same over time except for their phase.

The time evolution of each momentum component is of course simply: $e^{-i\sqrt{p^2+m^2}}t$

This may seem to run counter to the Fourier transform scaling rules for a Gaussian but the spreading wave packet isn't a Gaussian. The spreading edges show a wave front indicating the momentum by which they spread. It is the magnitude $\varphi^* \varphi$ of the field which is Gaussian like, the field $\varphi$ itself however is not.

See for instance figure 9.8 here (In the chapter of my book which handles the Klein Gordon equation) http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf The image shows the real and imaginary components of the spreading field as well as the envelope.

Hans