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Wikipedia describes an accidental symmetry as

a symmetry which is present in a renormalizable theory only because the terms which break it have too high a dimension to appear in the Lagrangian

but I don't understand what this means: are non-accidental symmetries always present, no matter the dimension of terms that I allow in the renormalized Lagrangian? Also, I know that some examples of accidental vs non-accidental symmetries are the lepton number vs the electric charge, but I fail to see why the first one would be accidental and the second one wouldn't.

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    $\begingroup$ Accidental is a synonym for "small explicit" breaking. A gauge symmetry (such as the one associated with the electric charge) is unforgiving, however: it has to be exact. $\endgroup$
    – Cosmas Zachos
    Commented Oct 20, 2021 at 13:07

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Typically, when you want to require a symmetry you choose to restrict the allowed terms in your Lagrangian to a smaller subset, this subset being the terms that are singlets under the symmetry group. For example, consider the Lagrangian of two fields $\phi_1$ and $\phi_2$. If it depends only on the combinations $$ \phi_1^2+\phi_2^2\,,\quad(\partial\phi_1)^2+(\partial\phi_2)^2\,, $$ then there is an $\mathrm{SO}(2)$ symmetry. This statement amounts to saying that terms such as $\phi^2_1$ or $\phi_1\phi_2$ are absent.

Now suppose that you write down all terms that involve the particles you are considering and which are relevant. Relevant here is meant in an RG sense, namely terms that survive in the infra-red. If all these terms happen to be singlets of a certain group, then so be it, your Lagrangian is going to have that symmetry. But it was not a deliberate choice, it's just what is forced upon you when you go to low energies.

Because of this fact such symmetries are called "accidental."

An example of this is QED which is $\mathsf{P}$ invariant but just because you can't write any $\mathsf{P}$-breaking term that has dimension below four which is made only of electrons and photons.

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  • $\begingroup$ Another example would be provided by the Dirac Lagrangian, right? If one tries to produce a renormalizable, Lorentz invariant Lagrangian describing (at most) locally interacting fermions, one is forced to end up with $\mathcal{L} = \overline{\psi}(i{\partial\!\!\!\!/} - m)\psi$ which, not only is interaction-free (which by itself is quite interesting to me) but possesses a global U(1) symmetry that was not put by hand. Is this a valid example of an accidental symmetry? $\endgroup$
    – Albert
    Commented Jan 26 at 16:16
  • $\begingroup$ I guess that's another valid example, yes $\endgroup$
    – MannyC
    Commented Jan 28 at 16:50
  • $\begingroup$ What does it mean explicitly that "a term survives in the infrared"? $\endgroup$
    – Koschi
    Commented Jul 12 at 9:27
  • $\begingroup$ That it doesn't get suppressed by powers of the energy. You can think of it in terms of amplitudes. If your Lagrangian has terms $a_1 \mathcal{O}_1 + a_2 \mathcal{O}_2 + ...$ the amplitude can be expanded as $a_1 \# E^{p_1} + a_2 \# E^{p_2} + ...$ The "#" are order one coefficients and the powers depend on the dimension of the operator. If the power is positive, those terms will not survive in the infrared. If you want to be more concrete $E$ can be identified as $\sqrt{s}$ the Mandelstam variable. $\endgroup$
    – MannyC
    Commented Jul 12 at 19:06

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