We all know that the canonical commutation relation give you
$$[x^i,p_j]=i\hbar~\delta^i_j,\qquad i,j=1,2,3.$$
Is there a relativistic version such as
$$[x^a,p_b]=i\hbar~\delta_b^a,\qquad a,b=0,1,2,3~?$$
If so what is the time operator $x^0$?
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$\begingroup$ Related: physics.stackexchange.com/q/34243/2451 , physics.stackexchange.com/q/6584/2451 , physics.stackexchange.com/q/56081/2451 , physics.stackexchange.com/q/83701/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 3, 2015 at 23:22
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$\begingroup$ No. The relativistic version of QM is QFT, where position and momentum are no longer operators on a Hilbert space, rather variables that fields depend on. $\endgroup$– gentedCommented Jul 24, 2017 at 22:04
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Actually, if the energy of particles is high enough to take relativity into consideration, the concept of particles in quantum mechanics is no longer as valid. For example, the uncertainty relationship $\Delta E \cdot \Delta t \approx \hbar$ and energy-mass relation $E=mc^2$ suggest that there will be new particles created and annihilated in those cases. So physicists developed quantum field theory to describe the quantum theory in relativistic case. For an introduction course for QFT, maybe you can click here to view the lecture of Professor David Tong.
In QFT, we treat both position and time as indexes rather than treat both as operators.So generally we do not have time operator and position operator in QFT. And we have the canonical commutation relation between field operators.
