Fredrik said:
This book seems to be saying that the theorem is only an problem for Hamiltonian action-at-a-distance theories, and not for field theories.
Yes, usually two "explanations" are suggested of why CJS theorem can be ignored. One of them says that Hamiltonian theories are not good. Another one proposes to reject the idea of particles (and their worldlines) and consider (quantum) fields only. I think that both these "explanations" are not adequate.
First, each relativistic quantum theory must involve a Hilbert space and a representation of the Poincare group in this Hilbert space. Then, inevitably, we have 10 Hermitian generators (H - time translations, \mathbf{K} - boosts, \mathbf{P} -space translations, \mathbf{J} - rotations), which are characteristic for Hamiltonian theories. Quantum field theories also follow the same pattern. See S. Weinberg "The quantum theory of fields", vol. 1.
By the way, quantum field theories have a theorem analogous to the CJS theorem. It is the famous Haag's theorem, which establishes that "interacting" quantum fields cannot have simple Lorentz-like transformation laws.
Second, it is true that traditional QFT does not provide any description of interacting particles (their wave functions, time evolution, etc.). This theory focuses only on properties described by the S-matrix. For such properties, only the movement of particles in the asymptotic regime is relevant. So, QFT is not a full dynamical theory. For example, it is impossible to talk about worldlines of interacting particles even in the classical limit of QFT. These deficiencies can be overcome in the "dressed particle" approach, which is basically a Hamiltonian direct interaction theory. CJS theorem cannot be ignored in the "dressed particle" approach.
Fredrik said:
I can't really make sense of this. To prepare an unstable particle as 100% undecayed should mean to produce it in an interaction, but there isn't a well-defined moment when the particle was created. Whatever A intends to measure, he would have to add up amplitudes associated with different events that he can think of as possible events where the particle "might have been created". So there isn't a moment where he can say that the probability is 100%. The closest match for the scenario you're describing that I can think of, is a particle (let's say a muon) that makes a track in a bubble chamber. The formation of the first bubble is a fairly well-defined classical event, and I suppose that at least after the fact, we can say that the particle existed there with probability 1. The thing is, A, B and B' will all agree about which bubble was the first, and if "the particle exists there with probability 1" is a valid conclusion for A, then it's a valid conclusion for B and B' too.
You've probably misunderstood my definitions of observers A, B, and B'. We've agreed that all these observers are "instantaneous". They "open their eyes" for a short time interval only.
Observer A opens his eyes at the exact time instant when the unstable particle is prepared. So, he cannot see any track in the bubble chamber. He can see at most one bubble at time 0.
Observer B is shifted in time with respect to A. So, he opens his eyes some time t after the unstable particle is prepared. He will definitely see particle tracks. Some of these tracks will be straight (meaning that the particle remains undecayed). Other tracks will be branched-off (meaining that the particle has decayed). The ratio of branched-off tracks to the total number of tracks is the decay probability from the point of view of B.
The situation with observer B' is a bit more complicated. We need to decide how our experimental setup is split between the observed system and the measuring apparatus. We have agreed that our physical system is the unstable particle only. Therefore, the bubble chamber should be considered as measuring apparatus, i.e., a part of the laboratory or observer. Therefore, if the question is "what the moving observer sees?" then to answer this question we need to use a moving bubble chamber (while the device preparing unstable particles for us should remain the same as in the two other examples, i.e., at rest). Needless to say that using bubble chambers moving with high speeds is a very very difficult proposition.
Moreover, your use of a single bubble (at time zero) as an indicator of the undecayed particle is not reliable. Bubbles are created due to the presence of a charged particle. The charge is conserved in decays. Therefore, by looking at the bubble you cannot say whether this bubble was created by the original unstable particle or by its decay products.
Eugene.