What is the quantum mechanical explanation for the speed limit of light? Does it arise from more fundamental principles?
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There is the approach to explain it in terms of vacuum fluctuations of charged particles allowed because of Heisenberg's uncertainty principle: Mainland & Mulligan, Foundations of Physics 50(5), 457–480, "Polarization of Vacuum Fluctuations: Source of the Vacuum Permittivity and Speed of Light" (2020). See also my question on it: Why is the speed of light in vacuum frequency-independent? – A. P. Jan 09 '21 at 19:51
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And Does the speed of light in our definitions take vacuum energy into consideration? – A. P. Jan 09 '21 at 20:02
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9The speed of light isn’t explained by QM. You can have SR without QM and QM without SR. – G. Smith Jan 09 '21 at 20:15
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Related/possible duplicate: https://physics.stackexchange.com/q/230703/50583 and its linked questions – ACuriousMind Jan 09 '21 at 20:34
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So far as we know, there is no quantum mechanical reason for the speed of light $c$ to be an absolutely limit. The limiting speed is a consequence of special relativity, and the validity of special relativity has nothing to do with the validity of quantum mechanics. You can have a consistent non-quantum yet relativistic theory; or you can have a consistent nonrelativistic quantum theory. As it happens, the actual universe is both quantum and relativistic, but quantum mechanics does not necessitate relativity.
Buzz
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Standard quantum mechanics is not even compatible with relativity. If you try to calculate the probability of a particle being found at the time $t$ at the position $\mathbf{x}$, given that it was measured at $\mathbf{x}_0$ at the time $t_0$, you find that $$P(t,x)=\left(\frac{m}{2 \pi i (t-t_0)}\right)^{\frac{3}{2}} \exp \left(i m \frac{\left|\mathbf{x}-\mathbf{x}_{0}\right|^{2}}{2 (t-t_0)}\right)$$ which is non-zero everywhere, implying that you can find the particle anywhere in the universe.
As you can see, there is no limiting speed.
JGBM
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