It is an often cited fact that a muon falling through the atmosphere at great speed has a decay time longer than the one we would observe in the same particle at rest due to relativistic effects, and for this reason can reach the surface. That much is clear to me, as it's a basic effect of time dilation. In the frame of reference of the muon, only the usual $\sim2\,\mu s$ have passed.
However what puzzles me is the interaction of this phenomenon with relativistic quantum mechanics. To lay out two different cases:
- consider a wavepacket describing a muon free falling through the atmosphere with an expectation value of its speed of around 0.95c. The muon will experience time dilation, but the wavepacket will not have one sharp value of momentum - rather, it will have an average expectation value, and a probability distribution over momenta. What is the lifetime observed? Is it the one determined by the average expectation value? Or is it rather a distribution of lifetimes, corresponding to the distribution of momenta?
- consider now a negative muon tightly bound to a heavy nucleus. I know in this case muons tend to have shorter lifetimes due to their interactions with the nuclei themselves, but if that wasn't the case, would the muon's lifetime be altered? After all, it possesses a great deal of kinetic energy to balance out the low potential energy of its state. In a classical description, it would be 'orbiting' really fast and be subject to very large centripetal accelerations, which ought to make it experience time dilation. However in the quantum description it is not really orbiting at all, and the atom is at rest with respect to me. So what happens?
I imagine QFT has answers to both these questions, so I'm curious to know what is it. Also, are these answers also attainable with a simple Dirac equation approach, or is that disregarding some key physics that make it impossible to use in this case (for example, since it does not describe the muon's decay)?