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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?

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  • $\begingroup$ 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

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$\pi^0$ 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.

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  • $\begingroup$ 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

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