# Proper time for a relativistic particle in quantum mechanics?

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:

1. 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?
2. 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)?

• In reality, the muon is in a state which is close to the coherent state — the uncertainty principle is saturated. Try calculating the uncertainty in time dilation and compare it to the absolute value. – Prof. Legolasov Jun 13 at 20:36
• I realise well that in practice in cosmic rays the effect is tiny. But tiny isn't the same as non-existent. "I can't measure it" is one thing, but if I could, what would I measure? – Okarin Jun 13 at 20:41
• well in reality there isn’t such a thing as “one muon”, it’s all quantum field theory, which is also likely incomplete. – Prof. Legolasov Jun 13 at 20:47
• That still feels like an unsatisfactory answer to me though. Sure, if you go down enough, you get QFT, but then does that mean that a simple Dirac equation can not describe something as the proper time experienced by a particle? And if not, if I did the QFT calculation, what would it predict? – Okarin Jun 13 at 20:59
• Ok, what you said in those last two comments is fair enough. – PM 2Ring Jun 15 at 16:49