# Does $[P_j,B_k]=i(Mc^2)\delta_{jk}$ imply particle number conservation?

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.

• Only the latter relation (M =0) is part of the Galilean group. Nonvanishing M is part of the central extension (Bargmann) algebra. – Cosmas Zachos Feb 11 '18 at 20:52
• $[P_j,B_k] = iH\delta_{jk}$ 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. – Luke Pritchett Feb 12 '18 at 15:19
• 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. – SRS Feb 12 '18 at 18:08
• 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. – Luke Pritchett Feb 12 '18 at 18:56
• 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. – Tomáš Brauner Feb 25 '18 at 19:55