The chiral perturbation theory Lagrangian is written $$\mathcal{L}_2=\frac{f_{\pi}^2}{4}Tr(D_{\mu}U^{\dagger}D^{\mu}U)$$ where $$U=e^{i\sqrt{2}\Phi/f}$$ and $$\Phi= \begin{pmatrix} \frac{1}{\sqrt{2}}\pi^0+\frac{1}{\sqrt{6}}\eta&\pi^+&K^+\\ \pi^-&-\frac{1}{\sqrt{2}}\pi^0+\frac{1}{\sqrt{6}}\eta&K^0\\ K^-&\bar{K}^0&-\frac{2}{\sqrt{6}}\eta \end{pmatrix}$$ expanding the exponential in $U$ and keeping only the first nontrivial term gives $$\mathcal{L}_2=\frac{1}{2}D_{\mu}\pi^{0*}D^{\mu}\pi^0+\frac{1}{2}D_{\mu}\pi^{-*}D^{\mu}\pi^-+\frac{1}{2}D_{\mu}\pi^{+*}D^{\mu}\pi^+\frac{1}{2}D_{\mu}K^{0*}D^{\mu}K^0+\frac{1}{2}D_{\mu}K^{-*}D^{\mu}K^-+\frac{1}{2}D_{\mu}K^{+*}D^{\mu}K^++\frac{1}{2}D_{\mu}\bar{K}^{0*}D^{\mu}\bar{K}^0+\frac{1}{2}D_{\mu}\eta^{*}D^{\mu}\eta+\ldots$$ now, I have written the complex conjugates of the fields but I am not sure if I should take the fields as complex. What makes me think this i the $1/2$ in front of all the kinetic terms which i characteristic of real kinetic terms.

But then I have another problem. Assume we get electromagnetic interaction in the covariant derivative. What sense does it have then to couple a real scalar to a $U(1)$ gauge field? I mean, a gauge transformation would transform a pion field in a complex field since it would involve a complex phase, shouldn't it?

So, summarizing, 1)are the pion fields real or complex? if they are complex why do I get $1/2$ with the kinetic term? and 3) if they are real, what sense does it have to make a $U(1)$ gauge theory with them?

  • $\begingroup$ I think the $\dagger$ is meant to act on $(D_\mu U)$ as a whole. See Schwartz's book Page 569. $\endgroup$
    – Nahc
    Jun 27, 2015 at 20:15

1 Answer 1


$\pi^0$ is a real field, and uncharged.

$\pi^{\pm}$ are both complex fields, and satisfy $\pi^- = (\pi^+)^*$.

So we can rewrite the pion kinetic terms (focusing only on the electromagnetic interaction, ignoring completely the weak interactions) as \begin{equation} \frac{1}{2} (D \pi^0)^2 + \frac{1}{2}(D_\mu\pi^-)^\dagger (D^\mu \pi^-) + \frac{1}{2} (D_\mu \pi^+)^\dagger(D^\mu \pi^+) = \frac{1}{2}(\partial\pi^0)^2 + D_\mu \pi^- D^\mu \pi^+ \end{equation} which satisfies all of your criteria.

  • $\begingroup$ and what if we considered weak interaction as well? $\endgroup$
    – Yossarian
    Jul 9, 2015 at 18:48
  • $\begingroup$ @Scardenalli That starts to get a bit complicated (beyond the scope of this simple answer), the short answer is that the W bosons couple to the pion (and K) fields in a way that depends on the CKM matrix. If you are interested in the details I recommend the review by Scherer (hep-ph/0210398), especially section 4.6.1 which gives a worked example of a decay process involving couplings between the $\pi^{\pm}$ and the $W^{\pm}$. $\endgroup$
    – Andrew
    Jul 9, 2015 at 23:07

Your Answer

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

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