I've found the following Lagrangian $$\mathcal{L}=i\bar{\psi}\gamma^\mu\left(\partial_\mu-ieA_\mu -ieA'_\mu\right)\psi-m\bar\psi\psi -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{4}F'_{\mu\nu}F'^{\mu\nu}.$$ Could I think of it as concealing the "standard" QED Lagrangian, i.e. $$\mathcal{L}=i\bar{\psi}\gamma^\mu\left(\partial_\mu-ieA_\mu\right)\psi-m\bar\psi\psi -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}?$$

I tried to make a trivial field redefinition, like $$A_\mu(x)\rightarrow A_\mu(x)+A'_\mu(x),$$ but I've found that the $F_{\mu\nu}F^{\mu\nu}\equiv F^2$ term transforms as $$F^2\rightarrow F^2+F'^2+F_{\mu\nu}F'^{\mu\nu}+F'_{\mu\nu}F^{\mu\nu}$$ under this field redefinition. Am I wrong or is it possible to trace this theory back to QED, at least for the $\psi$ field?


1 Answer 1


The theory is equivalent to QED with an extra copy of a photon field – under which the Dirac field is neutral so it's decoupled. Under the gauge fields $A,A'$, the Dirac field has charges $+1,+1$. If you introduce fields $$B^\pm_\mu = \frac{A_\mu \pm A'_\mu}{\sqrt{2}} $$ you will be able to see that $\psi$ has charges $+1,0$ under $B^+,B^-$, well, perhaps with some permutation and the $\sqrt{2}$ factor must be used to redefine the coupling $e$. So these $B^\pm_\mu$ fields may be referred to as the visible and dark photon, respectively.

If you do this redefinition, you will get a usual QED with one of the fields, either $B^+$ or $B^-$, but you will still get the kinetic term $F^2$ for the other field among the gauge fields $B^\pm_\mu$. This field will allow transverse physical polarizations of the "dark photons". But these dark photons will be decoupled from the regular visible photon as well as the Dirac fermions that carry a charge under the visible photon.

  • $\begingroup$ Won't the redefinition $B^\pm_\mu = \frac{A_\mu +A'_\mu}{\sqrt{2}}$ mess up the kinetic terms for the gauge fields? Do they stay diagonal? $\endgroup$ Oct 12, 2018 at 21:03
  • $\begingroup$ Did you mean to place a $\pm$ sign between the $A$-fields in defining $B_\pm$? $\endgroup$
    – J.G.
    Oct 12, 2018 at 21:56
  • 1
    $\begingroup$ Dear @AccidentalFourierTransform, yes, the switch from $A,A'$ is an orthogonal transformation, so the free terms $F^2$ keep their quadratic form. J.G., yes, thanks. $\endgroup$ Oct 13, 2018 at 5:37
  • $\begingroup$ Thank you, dear @Luboš Motl, very clear. I asked myself if there were differences (with respect to QED) calculating, for example, the mass shift for the $\psi$-field. The Lagrangian should be $$\mathcal{L}=-\frac{1}{4}F^2_+-\frac{1}{4}F_-^2 + i \bar{\psi}\left(\partial\!\!\!/-m\right)\psi +e\bar\psi\gamma^\mu B^+_\mu\psi,$$ indicating that there are two photonic kinetic terms, but just one interaction of the same form as in simple QED. So, should my calculation for electron self-energy, $\Sigma_{(2)}(p\!\!\!/)$, be the same as in QED? $\endgroup$
    – Stig
    Oct 13, 2018 at 16:25
  • 1
    $\begingroup$ Yes, the self-energy will be unchanged - assuming you identify the value of $e$ correctly - because the field you called $B^-_\mu$ doesn't interact with $\psi$ and $B^+_\mu$ at all. The $B^-_\mu$ basis photons are just flying through space and through the rocks of Earth and are invisible for everyone else, like if they were in a parallel world. Braneworlds make this "parallel" character literally true. $\endgroup$ Oct 13, 2018 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.