Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading Arthur Jaffe's Introduction to Quantum Field Theory. (You can find it here.) There is an interesting question posed in Exercise 2.5.1:

Solutions to the Klein-Gordon equation propagate with finite speed. But $f^t$ instantly spreads from its localization at the origin [...] Does this fact not contradict the laws of special relativity?

$f^t$ is defined in equation (2.30) and it is explicitly a solution to the KG equation, see (2.32).

Is the correct answer that $f^t$ is not a propagating solution? I.e., the group velocity isn't well-defined so nothing can be said to propagate/violate finite speed information flow? Or is there something more subtle here?

share|improve this question
    
I don't think you're really answering Arthur's question. Would you agree that according to relativity, if the signals/perturbations are confined to a small region at time $t$, they should be confined to a region at most $dt\cdot c$ further at time $t+dt$? Right or wrong? Is it supposed to be true for $f$ being nonzero in a region? Hasn't Arthur found a way to violate this condition of relativity? If he has, does it contradict the relativistic invariance of the full KG theory? The first-quantized one? The second-quantized one? Can you use it to instantly affect remote places in QFT? Class. FT? –  Luboš Motl Apr 11 '13 at 13:18
    
Your "solution" just tries to overlook Arthur's way of violating relativity by claiming that it doesn't agree with your way how relativity could be violated. But that doesn't matter. Your way isn't necessarily the only way how relativity could be violated. In particular, it's wrong if you claim that the group velocity must be well-defined in any situation that violates relativity. Arthur has presented you with something that could be interpreted as (ultimately wrong) proof that in KG equations, one may send signals to spacelike separated regions, after all. What's wrong with this proof? –  Luboš Motl Apr 11 '13 at 13:24
    
And no, the fact that Arthur's construction doesn't agree with your ad hoc excuse that "disproofs of relativity should have well-defined group velocities" clearly can't be a valid reason why Arthur's proof of relativity is wrong. You must know very well that your solution isn't a solution, it's just an unjustifiable excuse to "eliminate" Arthur's proof although you don't see anything wrong with it. Why are you talking about more subtle things if you clearly don't understand even the non-subtle ones? –  Luboš Motl Apr 11 '13 at 13:26

1 Answer 1

up vote 2 down vote accepted

Please, if this is a homework from Arthur to you and if you will use any information in the answer below, indicate it clearly in your answer (and send my best regards to Arthur).

Your comments about the group velocity have nothing to do with the right resolution to Arthur's "would-be violation of relativity". The right solution is that ${\mathfrak f}(\vec x,t)$ solves the first-order (in time) Schrödinger equation (2.5). As sketched in (2.34) etc., the solution immediately becomes nonzero at time $t+dt$ everywhere in space although ${\mathfrak f}=0$ were true everywhere except for a small region at time $t$.

However, because (2.5) is a first-order equation that is nonlocal in space, it follows that even $\partial {\mathfrak f}/\partial t$ is nonzero (almost) everywhere in space at time $t+dt$. This first time derivative is proportional to the action of the nonlocal Hamiltonian $H$ on the Gothic $f$.

This fact prevents us from saying that the original "impulse" was confined to the small region. To claim so in the context of solutions to the actual second-order (in time) Klein-Gordon equation, we need all the initial conditions i.e. both ${\mathfrak f}$ and $\partial {\mathfrak f}/\partial t$ (canonical coordinates and canonical momenta) to vanish everywhere outside the small region. The latter doesn't vanish in the initial time $t$ so it's not true that the initial "impulse" was confined to the small region, and it's therefore allowed for the responses to appear everywhere in space at $t+dt$, too.

If one could construct any solutions that vanish and whose first time derivative vanish everywhere outside a small region at time $t$, but that are nonzero arbitrarily far at the time $t+dt$, it would prove that the theory violates relativity, regardless of any additional excuses about ill-defined group velocities or anything else.

share|improve this answer
    
Thanks for this answer. I'm just a non-specialist trying to understand these notes. From your answer I understand that positive frequency solutions, f^t, can't be prepared locally. Any local disturbance in the field must then but a superposition of positive and negative energy solutions. Do you know if the same true for field detection? If I measure a field locally, must I be measuring some mixed +/- superposition solution? I'm confused because in quantum optics (the Glauber theory) the positive frequency part of the field operator is used to annihilate a particle at the spacetime point (x,t). –  Jase Uknow Apr 11 '13 at 14:13
    
Hi! Good if it's no homework or exam, but the downside is that I can't send greetings to Arthur in this way. ;-) Indeed, not only in quantum optics but also in QFT, the negative/positive frequency modes are really split by producing creation/annihilation operators from the two groups, respectively. However, this doesn't imply that you can't measure certain observables. You may measure whatever (Hermitian) operator you want, anywhere. These statements about what you can prepare and what you may measure aren't analogous in any sense. –  Luboš Motl Apr 11 '13 at 16:52
    
The first claim, here, says that a localized packet inevitably contains both positive and negative energy modes as contributions. But that's a statement about which states may exist and what are their "localization" properties in space. This has nothing to do with the question which obserables may be measured. –  Luboš Motl Apr 11 '13 at 16:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.