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Physics
Quantum Physics
Ballentine on construction of the (Galilean) symmetry generators
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[QUOTE="vanhees71, post: 6848842, member: 260864"] I think the logic is the other way, i.e., in the approach to construct non-relativistic QM from the symmetries of the Newtonian spacetime model, you have the generators ##\hat{H}##, ##\hat{\vec{P}}##, ##\hat{\vec{J}}##, and ##\hat{\vec{K}}## for the Galilei algebra. As it turns out this algebra has a non-trivial central charge, ##\hat{M}##, which is crucial to get a physically useful ray representation of the Galilei group. Physically, ##\hat{M}## represents the total mass of the system. After eliminating all "trivial central charges" the relevant part of the "quantum Galilei algebra" (also known as "Bargmann algebra") reads $$[\hat{H},\hat{P}_j]=0, \quad [\hat{H},\hat{K}_j]=-\mathrm{i} \hat{P}_j, \quad [\hat{K}_j,\hat{P}_k]=\mathrm{i} \hat{M} \delta_{kl} \hat{1}, \quad [\hat{K}_j,\hat{K}_k]=0, \quad [\hat{P}_j,\hat{P}_k]=0.$$ From Noether's theorem you get that ##\hat{P}_k## and ##\hat{H}## are all not explicitly time-dependent, but that $$\mathring{\hat{K}}_j=0=\frac{1}{\mathrm{i}} [\hat{K}_j,\hat{H}]+\partial_t \hat{K}_j = \hat{P}_j+\partial_t\hat{K}_j.$$ From this you get $$\hat{K}_j=\hat{M} \hat{X}_j-\hat{P}_j t,$$ where ##\hat{X}_j## are explicitly time-independent operators, which necessarily fulfill the commutation relations for the usual position operators with all the generators of the Galilei group, which shows that there are always position operators (for the "center of mass position" of the considered quantum system). [/QUOTE]
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Ballentine on construction of the (Galilean) symmetry generators
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