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From reading Weinberg's Quantum Theory of Fields, Vol. 1, I learnt that for the Galilean group $[P_j,B_k]=i(Mc^2)\delta_{jk}$, and for the Poincare group $[P_j,B_k]=iH\delta_{jk}$ where $P_j$ and $B_k$ respectively denote the generators of translation and boost, and $H$ denotes the Hamiltonian.

I have been told by one of my friends that $[P_j,B_k]=i(Mc^2)\delta_{jk}$, conserves the particle number in non-relativistic physics. I just want to ensure whether that is correct and if yes, how? In a way, I'm also asking what is the physical content/implication of this commutation relation.

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    $\begingroup$ Only the latter relation (M =0) is part of the Galilean group. Nonvanishing M is part of the central extension (Bargmann) algebra. $\endgroup$ – Cosmas Zachos Feb 11 '18 at 20:52
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    $\begingroup$ $[P_j,B_k] = iH\delta_{jk}$ is what necessitates particle non-conservation in QFT, which I believe Weinberg shows. And clearly nonrelativistic QM (which uses the Bargmann algebra) doesn't require particle non-conservation, but I don't know if the Bargmann algebra forbids particle non-conservation. My guess is that it does not forbid it, but I'm not sure. $\endgroup$ – Luke Pritchett Feb 12 '18 at 15:19
  • $\begingroup$ Dear @LukePritchett Can you expand a little bit on how $[P_j,B_k]=iH\delta_{ij}$ necessitates particle non-conservation in QFT? Perhaps that will partially help me figure out the answer. I'm not familiar with Bargmann algebra. $\endgroup$ – SRS Feb 12 '18 at 18:08
  • $\begingroup$ Weinberg covers that, though I don't know the chapter off the top of my head. Basically, if you have a Hamiltonian with an interaction part that commutation relation constrains the possible interactions to the ones that can be written as products of fields, which necessarily have particle non-conserving terms. $\endgroup$ – Luke Pritchett Feb 12 '18 at 18:56
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    $\begingroup$ Once you promote NR quantum mechanics to quantum many-body theory, i.e. go to the Fock space allowing for states with different number of particles, it turns out that $[P_j,B_k]$ is proportional to the operator of particle number. You can check this by constructing all the symmetry generators explicitly for a free Schrödinger field. This corresponds to the fact that in the NR limit, the Hamiltonian of a many-body system reduces to the rest energy of a single particle times the number of particles. It does not imply that particle number is necessarily conserved in the NR theory though. $\endgroup$ – Tomáš Brauner Feb 25 '18 at 19:55
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I now this is a little late, but I might be able to answer. Look in this paper. As stated in the comments, non-zero mass is consequence of the Bargmann extension of the Galilean Lie algebra. This leads to the Bargmann superselection rule, which forbids the superposition of two particles with unequal masses. On page 4 of the paper I link, the authors state that

The mass appears as a central element of the extended Lie algebra... that is, it commutes with all the other elements, in particular with the Hamiltonian H. It is thus a conserved quantity. It is even a ''super-conserved" quantity, giving rise to a superselection rule ...This is due to the fact that the extension...is nontrivial so that a physically trivial transformation may result in a definite modification of the phase of the state vector, depending on the mass of the system. In order not to alter the physical properties of the system by such a transformation, we must forbid any superposition of states with different masses, thus obtaining Bargmanns superselection rule. It has the effect of breaking the Hilbert space into mutually incoherent eigenspaces of the mass operator M. The consequences of this superselection rule are rather far-reaching with respect to the possible types of particle reactions allowed by Galilean invariance. Suppose, for instance, that we treat a theory with only one kind of particle, with definite mass m. In this case, conservation of the mass implies conservation of the number of particles, thus forbidding any type of production process.

Hence, for one kind of particle with one kind of mass, the Bargmann superselection rule directly implies particle number conservation.

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