Reference frame of non-localized particles?

In physics, a frame of reference (or reference frame) consists of an abstract coordinate system and the set of physical reference points that uniquely fix (locate and orient) the coordinate system and standardize measurements within that frame.

Take an example of muon decay: thickness of earth's atmosphere as seen from an observer at rest at earth's surface is $$\Delta x' = 10km$$ and muon's lifetime in lab frame is around $$\tau =2\cdot10^{-6}s$$, imagine muon has a speed of $$v=0.9999c$$ where classically they would mostly decay around $$0.6km = (0.9999c)(2\cdot 10^{-6})$$ so they shouldn't reach the ground but they do because of the relativistic effects.

In muon's frame length will be seen as

$$\Delta x = \gamma^{-1}\Delta x' = 0.14km$$

since $$0.14km<0.6km$$ they can reach the ground or in earth's frame

$$\Delta t = \gamma \Delta t' =\gamma \tau = 1.41\cdot 10^{-4}s$$

hence the distance $$0.9999c \cdot 1.41\cdot 10^{-4}s = 42km$$ so they will reach the ground. I don't have any problem understanding these special relativistic effects, I know experimentally these values correspond to the right results but I have a conceptual question:

$$\cdot$$ When we think about the distance in muon's frame why do we localize the muon?

I'm aware that special relativity is a classical theory so it doesn't take the quantum mechanical effects into account, this might be the reason.

$$\cdot$$ How is it even possible to set a reference frame in quantum realm (e.g. in muon's frame) and measure the distance considering that particles are not always localized since probability of measuring a particle at position $$x$$ is determined by the $$|\psi(x)|^2$$?

• You choose the frame in which the expectation value of the muon's momentum is zero. Commented Feb 18, 2021 at 17:28
• That's a great answer, it's always possible to choose a frame where $\langle p \rangle$ is zero right? Commented Feb 18, 2021 at 18:41
• Yes. Of course you still need to decide where to put the origin, but that doesn't really matter. Commented Feb 18, 2021 at 18:48

The "size" of a particle at rest such as the muon, is its Compton wavelength, $$1/m_\mu$$ in natural units ($$\hbar=1, c=1$$, easy to reinstate uniquely).
You may then check that for the muon this is $$\lambda_\mu \approx \frac{1}{m_\mu} \approx 2~fm= 2 \cdot 10^{-15}m,$$ a mere speck, close to nuclear size. You are contrasting this to tenths of kilometers. The muon is really localized.