How do quantum probabilities transform under Lorentz transformations? I think I get how scattering probabilities transform under Lorentz transforms. Once the interaction phase is over, the final probabilities become time independent. Hence, every observer could describe the final state using the same probabilities.
But I don't understand how time-dependent probabilities would transform under a change of frame. Suppose there's a quantum system in a box whose probabilistic state at time $t$ is described by some wavefunction/wavefunctional $\psi (t)$. How would a moving observer describe the probabilistic state of the same system? I think the concept of "probability at a time" gets screwed up because of different planes of simultaneties for the two observers.
 A: There is no universal answer here. Transformation formulas depend on the way you describe (enumerate) system states: it can be done in invariant and non-invariant way, consistent with system symmetry or not. So the only answer to your question is: they transform somehow, as some representation of Lorentz group.
ADDENDUM
In general case we have some Hilbert space $\mathcal H$. We can imagine a time dependent state as a moving point in the $\mathcal T \times \mathcal H$ fiber space, where $\mathcal T$ is the time axe. To put this theory into special relativity context, some Lorentz group representation $L: \mathcal T \times \mathcal H \to \mathcal T \times \mathcal H$ should be defined.
A: Wavefunctions are not compatible with Special Relativity where the number of particles can be changed over the course of an experiment, perhaps what you mean then is something like the electron field? If so the problem is just not there since all formulations of QFT are manifestly Lorentz invariant (just look at their lagrangians), meaning that observables like squared scattering amplitudes (which are the only thing you can observe about a system) are automatically Lorentz Invariant and thus all observers agree on their value.
