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It seems that the Lagrangian of QED describing electrons and muons cannot include terms like that: $$\overline{\psi_{(e)}}i\not\!\partial\psi_{(\mu)}$$ where $\psi_{(e)}$ and $\psi_{(\mu)} $ are 4-component dirac spinors describing the electron and the muon respectively.

I don't understand why this is illicite, as it seems to me that this term is Lorentz-invariant (both spinors transform in the same way), and we could make it gauge invariant. I'm missing something.

Thank you very much,

Anthony.

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    $\begingroup$ Near duplicate. $\endgroup$ – Cosmas Zachos Sep 27 at 18:48
  • $\begingroup$ Thank you Cosmas Zachos, I wouldn't have found it. So it's basically due to the conservation of lepton number. $\endgroup$ – Anthony Sep 27 at 19:19
  • $\begingroup$ (and it is very consistent with the exercise I was dealing with, since by making this term gauge-invariant by the gauge prescription, we make the decay $\mu^- \rightarrow e^- \gamma$ possible because of the consecutive new interacting term in the Lagrangian, hence the lepton number violation) $\endgroup$ – Anthony Sep 27 at 19:28
  • $\begingroup$ Careful, as always (below), e and μ are defined by the respective mass terms, so the $\mu \to e\gamma$ amp at the tree level is gone, as part of the minimal coupling prescription. (It is thought to possibly exist at the loop level with enormous suppression.) $\endgroup$ – Cosmas Zachos Sep 27 at 20:02
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You are not missing anything: no nature law enforcement agency would stop you from writing terms such as this. It would be Lorentz and made gauge invariant. You are presumably asking what observable consequences such a mixing term would entail. In the absence of other recondite couplings, none.

One assumes you also have the conventional diagonal kinetic and unequal-mass terms: the definition of $\psi_e$ and $\psi_\mu$ follows from the propagating mass eigenstates. If the masses were the same, you'd diagonalize the kinetic term which includes your off-diagonal proposal and its h.c., consistently with the (identity) mass matrix in the space of these two states.

If the masses are not the same, as in our world, diagonalizing the kinetic term would then throw the mass matrix off-kilter, and would introduce mixing terms at the mass level, $\epsilon ~\overline \psi_e \psi_\mu$ + h.c. Moreover, you'd then have to adjust your normalizations of the canonical fields in the diagonal kinetic term, to ensure the term is the identity matrix! This adjustment would further alter the newly off-diagonal, but still symmetric, mass term.

This mass term may now be diagonalized by a different orthogonal transformation than the previous one, by dint of the normalization adjustment indicated, and the e and μ masses will be shifted as a result of this adjustment. But the effect of this second rotation on the identity kinetic term will not be visible, of course. You will then have a Lagrangian with diagonal kinetic and mass terms, pretty much as in the non-mixed case baseline you tweaked, except with modified lepton masses. Can you estimate the ε-order of their tweak?

I gather you don't need explicit formulas implementing the above description, a useful exercise if you have never done such, and you pursued particle physics, where you wake up and go to sleep diagonalizing mass matrices.

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  • $\begingroup$ Thank you so much for this explanation! Indeed, I haven't quite understood with the first link you provided me with. Your very clear post leads me to this one that I'm going to read through. $\endgroup$ – Anthony Sep 27 at 21:23
  • $\begingroup$ Good find! Indeed, @JeffDror 's answer provides the general formulas. In your case, these are 2x2 orthogonal matrices! $\endgroup$ – Cosmas Zachos Sep 27 at 21:32

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