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I know for a (free) complex scalar field $\psi$ the Lagrangian is: $$ \mathcal{L} = \partial^\mu \psi^\ast\partial_\mu \psi$$ and that Noether's theorem from the $U(1)$ symmetry of the system gives a conserved current $j_\mu \propto iq(\psi\partial_\mu\psi^\ast-\psi^\ast\partial_\mu\psi)$, which can be interpreted as the difference of the number of particles and anti-particles and hence as the conservation of electrical charge.

For real scalar field, though, I would have: $$ \mathcal{L} = \partial^\mu \phi\partial_\mu \phi$$ so there is not $U(1)$ symmtry... but I still expect particle number to be conserved? Shouldn't particle number be the conserved charge?

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  • $\begingroup$ Hint: note that $\mathcal{L} = \frac{1}{2} \partial^\mu \phi\partial_\mu \phi$ is invariant under spacetime symmetries i.e. the Poincaré symmetry-group; compute the noether's conserved current/charges for that group $\endgroup$
    – user164843
    Apr 14, 2019 at 0:08
  • $\begingroup$ yeah but that's just energy and momentum. I meant does particle number never come up as a conserved charge? $\endgroup$
    – SuperCiocia ♦
    Apr 14, 2019 at 0:14
  • $\begingroup$ $\mathcal{L} = \partial^\mu \phi\partial_\mu \phi$ defines an free classical field theory, and therefore the number of particles is arbitrary, however after the procedure of quantization, $\hat{N}=\hat{a}^{\dagger}\hat{a}$ (particle number operator) commutes with $\hat{H}$ and implies that the number of particles is conserved; also, i find an related question here physics.stackexchange.com/questions/332189/… $\endgroup$
    – user164843
    Apr 14, 2019 at 1:07

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Conservation of charge does not mean conservation of particle number. It means conservation of total charge during interactions. Conservation of particle number is automatic if there are no other fields that your complex scalar could interact with, in which case conservation of energy-momentum leads to conservation of particle number. Similarly, in the real scalar field case with no other fields, particle number is conserved because energy-momentum is conserved.

By the way, there is a shift-symmetry in your Lagrangian of the real scalar. Although you can break it by adding a mass term.

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  • $\begingroup$ What about O(N) symmetry? What charge does that give? $\endgroup$
    – SuperCiocia ♦
    Apr 14, 2019 at 18:56
  • $\begingroup$ @SuperCiocia O(N) charge of course. Or whatever you want to call it. And it must be specified by the representation of O(N) in which the scalars transform. See e.g. QCD and color conservation physics.stackexchange.com/questions/56886/… $\endgroup$
    – Kosm
    Apr 14, 2019 at 19:23

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