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Consider a free electron in space. Let us suppose we measure its position to be at point A with a high degree of accuracy at time 0. If I recall my QM correctly, as time passes the wave function spreads out, and there is a small but finite chance of finding it pretty much anywhere in the universe. Suppose it's measured one second later by a different observer more than one light second away and, although extremely unlikely, this observer discovers that electron. I.e. the electron appears to have traversed the intervening distance faster than light speed. What's going on here?

I can think of several, not necessarily contradictory, possibilities:

  1. I'm misremembering how wave functions work, and in particular the wave function has zero (not just very small) amplitude beyond the light speed cone.
  2. Since we can't control this travel, no information is transmitted and therefore special relativity is preserved (similar to how non-local correlations from EPR type experiments don't transmit information)
  3. Although the difference between positions is greater than could have been traversed by the electron traveling at c, had we measured the momentum instead, we would have always found it to be less than $m_e c$ and it's really the instantaneous momentum that special relativity restricts; not the distance divided by time.
  4. My question is ill-posed, and somehow meaningless.

Would anyone care to explain how this issue is resolved?

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This is one reason why we need quantum field theory. –  leongz Dec 31 '12 at 19:58
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Welcome Elliotte, good question. I don't know the answer to it, I hope someone with better knowledge in QM will be able to help you. I have a small correction for you about the momentum. In special relativity, the momentum is $p=\gamma m v$, where m is the rest mass, $\gamma = \frac{1}{\sqrt{1-(\frac{v}{c})^2}}$, and v is the velocity. As v tends to c, $\gamma$ tends to infinity, so the momentum can actually be much larger than $mc$. –  Andrey B Dec 31 '12 at 20:07
    
Related: physics.stackexchange.com/q/15282/2451 and links therein. –  Qmechanic Jan 27 '13 at 18:17

3 Answers 3

Excellent question. You are correct about wavepacket spreading, and in fact you do get superluminal propagation in non-relativistic QM - which is rubbish. You need a relativistic theory.

You should read the first part of Sidney Coleman's lecture notes on quantum field theory where he discusses this exact problem: http://arxiv.org/abs/1110.5013

The short answer is that you need antiparticles. There is no way to tell difference between an electron propagating from A to B, with A to B spacelike separated, and a positron propagating from B to A. When you add in the amplitude for the latter process the effects of superluminal transmission cancel out.

The way to gaurantee that it all works properly is to go to a relativistic quantum field theory. These theories are explicitly constructed so that all observables at spacelike separation commute with each other, so no measurement at A could affect things at B if A and B are spacelike. This causality condition severely constrains the type of objects that can appear in the theory. It is the reason why every particle needs an antiparticle with the same mass, spin and opposite charge, and is partially responsible for the spin-statistics theorem (integer spin particles are bosons and half-integer spin particles are fermions) and the CPT theorem (the combined operation of charge reversal, mirror reflection and time reversal is an exact symmetry of nature).

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So, is it correct to say that if one releases an electron at r=0 at t=0, and waits, then the probability of measuring the electron outside the light cone will be zero, but this is due to the field of a positron which cancels out the propagation of electron outside the light cone? Can one then measure a positron anywhere inside the light cone? –  Alexey Bobrick Jan 1 '13 at 22:30
    
Another comment is that one does not need relativistic quantum field theory for this problem. Dirac theory describes a propagating particle-electron (non field) well enough. –  Alexey Bobrick Jan 1 '13 at 22:31
    
Thanks for the question, Elliotte! In my QFT class, we briefly touched on how the antiparticle field cancels out the superluminal effects of the particle field. But what I don't understand is that the particle still can travel faster than the speed of light. Is there no way one can observe only that? I'm sorry if this is a silly question, I've taken just one semester of QFT..Thanks! –  user34801 Jan 2 '13 at 7:19
    
Good questions, and not at all silly! @Alexey Bobrick: You are correct, there is no amplitude to measure an electron outside the light cone. There is also no amplitude to measure a positron inside the lightcone (if you start with an electron state rather than a positron state!). –  Michael Brown Jan 2 '13 at 10:59
    
@user34801: Your question and Alexey's other question are answered by the same discussion: The "electron field" $\psi$ really is the sum of two terms: a term that annihilates an electron (the convention is backwards - blame Heisenberg) and a term that creates a positron. The conjugate field $\bar{\psi}$ does the reverse. (The opposite action on electron vs. positron states makes it an operator with definite electric charge.) Any operator that acts on electron or positron states must be built up out of these combinations to preserve causality. This is the restriction I mentioned before. –  Michael Brown Jan 2 '13 at 11:00

Comment on the answer of @Michael:

The short answer is that you need antiparticles

is false. In Quantum Field Theory you have perfectly working solutions also without antiparticles, i. e. for real fields. Even if you do want to consider antiparticles, always have in mind that despite the misleading name they are in fact different particles from the original ones and saying that an electron propagating from A to B is equivalent to a positron propagating from B to A is also wrong: there is indeed a way to distinguish between the two, namely the former is represented by the field and the latter by its hermitian conjugate and they transform differently under representation of the Poincaré group. Moreover, adding the two contributions does not cancel out the possible superluminar factors.

To answer the original question: QM is indeed not a relativistic theory, end of the story. The correct relativistic extension is QFT by the fact that the cancellations happen if you take into account the degrees of freedom carried by the field themselves on top of the ones of the particles (no need to have antiparticles at all).

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In a real field, a particle is its own antiparticle. You still have a zero conmutator for spacelike separated measurements, because of the particle propagating forward and backward cancel out. And I'd love to see why the two contributions don't cancel out superluminal factors (remember that the only observable superluminal effect would be a nonzero conmutator of spacelike separated fields) –  Bosoneando Jun 26 at 7:07
    
The interpretation that a real field is its own antiparticle is still at least misleading (if not wrong). Likewise, (anti)-particles never travel backwards: they always do so forward and the backward terminology is just to (wrongly) justify the minus sign. About the commutator: the two contributions only cancel out for free fields; if you try to calculate the same commutator for any type of interaction you will see that the contributions do not generally cancel each other (unless you postulate so and derive the fields accordingly, but that's a different story). –  Gennaro Tedesco Jun 26 at 17:47

The very useful solutions of the Shrodinger equation that are usually taught in beginning quantum mechanics are not Lorenz invariant and therefore paradoxes with respect to special relativity may be constructed.

The relativistic equations of Dirac:

the Dirac equation is a relativistic wave equation derived by British physicist Paul Dirac in 1928. In its free form, or including electromagnetic interactions, it describes all spin-½ massive particles, for which parity is a symmetry, such as electrons and quarks, and is consistent with both the principles of quantum mechanics and the theory of special relativity,

The Klein Gordon equation :

( sometimes Klein–Gordon–Fock equation) is a relativistic version of the Schrödinger equation.

Thus there is no problem with the simple solutions of the underlying wave functions that are needed to build up the Quantum Field Theories discussed in the other answers. Those are a meta level using the solutions of the relativistic equations as a basis on which the QFT creation and annihilation operators operate.

As far as Lorenz invariance is concerned it is enough that the Hilbert space on which the QFT operators operate is Lorenz invariant in order not to have any light cone problem with any modeling .

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