Would it be meaningful to think of it as the 0 component of the 4-vectors x and p? In which case the regular uncertainties between position and momenta are the 1,2,3 components of the 4 vectors and the ΔEΔt is just the 0 component?
I can't speak to the QFT part of your response, because I don't know QFT.No, that can't be correct. The rigorous derivation of the uncertainty relation for position and momentum considers them to be operators. But there is no time operator in QM, at least not in the standard formulation. And when you go to QFT, even position is no longer an operator, let alone time.
Some say that the uncertainty relation between energy and time is only a myth. And to some extent its true because it can't be thought of in the same way you think about the uncertainty relation for position and momentum. But there are some situations where it is true in a different meaning. You can check "Modern Quantum Mechanics" by Sakurai for such different meanings of this relation but I don't remember which section.
And at the end, there is the famous saying of Griffiths:
"In general, when you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."
In non-Relativistic QM, Minkowski spacetime is not used, but a three dimensional space with a Newtonian absolute time.I see no reason not to have a time operator which would just be the projection into the t (or 0) coordinate in Minkowski space, it's even hermitian. The issue is in considering E its fourier transform.
I understand that, the coincidence I was referring to was that each component of the 4 momentum has an uncertainty relation (of sorts) with its corresponding component of 4 position. Notwithstanding their differences, notwithstanding that for three components we have quantum operators and for the fourth we don't. Notwithstanding all that, we can still write them all asThe two uncertainty relations are not on the equal footing. So there is nothing for you to consider as a coincidence or not!
This is not meaningful in standard formulation of QM, but makes sense in a non-standard one:Would it be meaningful to think of it as the 0 component of the 4-vectors x and p? In which case the regular uncertainties between position and momenta are the 1,2,3 components of the 4 vectors and the ΔEΔt is just the 0 component?
How true that is.Although Schrodinger's equation can be made relativistic by using e.g. the Dirac Hamiltonian, it is not manifestly relativistic