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.

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

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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

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Is the electron wave function stable? Nope, never. In free space it always spreads.

And as Hans de Vries noted, the momentum components do not change. That is, simply making the wave function larger in xyz space does not automatically imply higher precision of its Fourier momentum components in momentum space. It just creates a different set of phase relationships between them. (I used to ponder this same point.) Schrodinger once hoped stable packets could be created, but he got firmly slapped down by one of his more classical mentors (I forget which one).

To me, this simple incomplete symmetry between the spatial and momentum wave functions is absolutely fascinating. It is so simple and so "almost" symmetric, except for one especially critical difference. It is the same difference that keeps wavefunctions compact in momentum space while letting the Fourier equivalent of the wave function in ordinary xyz space spread without limit: Expanding length in momentum space requires energy, while expanding length in xyz space does not. Therein lies a very interesting tale that as it grows ends up encompassing most of quantum mechanics.

[1] Hans: Alas, I'm still waiting for your complete print version of your quantum textbook. I'd snap up a copy of it from Amazon in about 10 minutes, for just about any price!

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Hi Terry, thank you for your nice words. –  Hans de Vries Mar 14 '12 at 11:45