Relativistic contraction for a wave packet and uncertainty on momentum

Consider an electron described by a wave packet of extension $\Delta x$ for experimentalist A in the lab. Now assume experimentalist B is flying at a very high speed with regard to A and observes the same electron. The extension of the wave packet will appear contracted, and the uncertainty on momentum will increase. What happens when the later become larger than the electron's rest mass?

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What happens? Nothing special. The momentum and the energy increase as well. Maybe this answer is too naive, could you say what you have in mind?

Answer to comments: Physical particles and antiparticles always have positive energy. Particles also have positive frequency in a free field, while antiparticles have negative frequency. One can prove that the sign of the frequency (positive or negative) is invariant under Poincare transformations. You can ask this as a separate question. (Edit: Finally, I have added that at the end of the answer.)

Let's say that for observer A the particle is at rest on average, so energy-momentum expectation is $(c=1)$: $$(m,0)$$ And the momentum has an indetermination $\Delta p$, that should be lower than $m$ if A really knows that there is one particle.

Then one can see that for B (it is an exercise on Lorentz transformations):$$\frac{\Delta E'}{E'}=\frac{\Delta p}{m}v < \frac{\Delta p}{m}$$ where $v$ is the relative speed (norm of velocity) between A and B. So, if $\frac{\Delta p}{m}\ll1$, then $\frac{\Delta E'}{E'}\ll1$

Absoluteness of particle/antiparticle concept under Lorentz transformations.

The positive frequency solution (connected with particles) is defined by:

$$i\partial _t \, f_+ =\omega \, f_+ \, , \; \omega >0$$

Taking $f_+ \propto e^{-i(\omega \, t - p\cdot x)}$ ($f_+$ must also verify the Klein-Gordon equation), with $\omega \equiv +\sqrt {m^2+p^2}$

The boosted observer (with rapidity $\theta$) uses his time $t'$:

$$i\partial _{t'}\, f_+=(\cosh \theta \, i\partial _t - \sinh \theta \, i\partial _x )\, f_+=(\omega \, \cosh\theta + p \sinh\theta )f_+ \equiv \omega ' f_+, \; \omega'>0$$ He thus obtains that $f_+$ is also a positive frequency solution with the boosted eigenvalue. Note that this does not happen for a general transformation and hence the distinction between particles and antiparticles (negative eigenvalue of $i\partial _t$) is absolute for observers connected by Lorentz transformations (inertial observers), but accelerating observers disagree about what is a particle, an antiparticle or vacuum.

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He means when the uncertainty is larger than rest mass, it would mean non-zero probability amplitude of negative energy. –  Siyuan Ren Sep 14 '12 at 5:57
Yes, this is what I meant. In this situation, would B observe positrons? How would that fit with charge conservation? –  Whelp Sep 14 '12 at 9:53
Just because the uncertainty is larger than the mass doesn't mean that there's a non-zero probability of negative energy--the uncertainty doesn't tell you anything about the distribution of the possible momentum states, just the minimum spread in them. –  Jerry Schirmer Sep 15 '12 at 5:33
@JerrySchirmer "anything"? Could you develop a little more your comment perhaps in a new answer? I'm not sure if I understand it. Thanks. –  drake Sep 15 '12 at 20:55
@drake: the uncertainty relation is a statement about the standard deviation of two distributions. But it doesn't necessarily tell you what the two distributions are--your assumption is that they're some sort of square wave, which is going to have a fixed width, which, in some cases, will go over an edge. But they could be, say, a Poisson distribution, which is constrained to be greater than zero. The uncertainty principle can't tell you anything about this, just that the spread in the distribution must be bigger than a number. –  Jerry Schirmer Sep 15 '12 at 23:39