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Obviously I know that the $U(1)$ symmetry is associated with the (number of particles - number of antiparticles) conservation.

However I thought, by Nother's theorem, that every conserved quantity should have an associated symmetry. So what's the symmetry for conservation of (number of particles + number of antiparticles) which is also conserved?

(This question doesn't seem to answer it: Free Complex scalar field and separate conservation of particle and antiparticle number)

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  • $\begingroup$ The notation is messy, but the idea is that you get a continuous symmetry for each individual Fourier mode of the field, which you can rotate by a phase, because the number of excitations in every mode is conserved. Rotating them all by the same phase gives the usual $U(1)$ symmetry. Rotating them by arbitrary phases gives you all of the other symmetries, including the one you're talking about, though the expressions are messy and generally not local. Maybe somebody else can come along and work out the notation. $\endgroup$
    – knzhou
    Commented Mar 29, 2022 at 15:37
  • $\begingroup$ Minor comment to the post (v1): That is strictly speaking the inverse Noether theorem. $\endgroup$
    – Qmechanic ♦
    Commented Mar 29, 2022 at 16:28

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The infinitesimal symmetry $\delta=\frac{\epsilon}{i\hbar}[\cdot,Q]$ for particle (anti-particle) conservation is as usually generated by the corresponding Noether charge $Q$, i.e. the total number operator for particles (anti-particles), respectively.

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  • $\begingroup$ I'm not quite sure how to use this symmetry to get the Noether current. Are you saying that the symmetry is $$\phi(x) \rightarrow \epsilon \int dp \hat a^\dagger_p \hat a_p \phi(x) - \epsilon \phi(x) \int dp \hat a^\dagger_p \hat a_p$$ or something? $\endgroup$
    – Alex Gower
    Commented Mar 29, 2022 at 16:57
  • $\begingroup$ Yes, or its classical counterpart. $\endgroup$
    – Qmechanic ♦
    Commented Mar 29, 2022 at 17:02
  • $\begingroup$ I'm probably being stupid, but wouldn't the commutator evaluate to 0 for classical fields? $\endgroup$
    – Alex Gower
    Commented Mar 29, 2022 at 17:26
  • $\begingroup$ Well, it becomes a Poisson bracket. $\endgroup$
    – Qmechanic ♦
    Commented Mar 29, 2022 at 17:36
  • $\begingroup$ Oh I see, but ultimately this problem is very messy to show easily? $\endgroup$
    – Alex Gower
    Commented Mar 29, 2022 at 17:38

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