What does this equation mean?

The obvious next question is why ordinary relativistic quantum mechanics gets along without any of this. Run the identical calculation on the Poincaré group. The relativistic analogue of the boost–momentum commutator is $[K_i,P_j]=\frac{i\hbar}{c^2}\,\delta_{ij}\,H$. where H is the total energy operator [ 2 ] . The right-hand side is not proportional to the identity. It is proportional to H , an operator with a spectrum, different on every energy eigenstate. Repeating the loop calculation of the previous section on a state with sharp energy $\langle$ H$\rangle$ gives a phase $\langle$ H$\rangle$\,$\mathbf b$$\cdot$$\mathbf v$/($\hbar$ $c^2$) that varies from state to state. A central extension, by definition, must give the same phase to every vector in the representation; a state-dependent phase is not a central extension, it is an ordinary consequence of ordinary dynamics, and it can be removed by working with true, non-projective unitary representations of the Poincaré group throughout. This is the content of Bargmann’s own cohomology theorem: semisimple factors such as the Lorentz group admit no continuous central charge, so relativistic quantum mechanics carries no analogue of the mass superselection rule that follows from the Galilei case [ 1 , 4 ] . Mass in special relativity shows up instead as the ordinary Casimir invariant $P^\mu$ $P_\mu$=$m^2$$c^2$ : a quantum number you diagonalize, not a phase a loop leaves behind.