# In chiral perturbation theory with complex $\phi$-s, would the next lo leading order renormalization $\gamma$-s change?

+ 2 like - 0 dislike
303 views

The Lagrangian of chiral perturbation theory (with two quark flavors) is written using the following matrix $U$
$$U=e^{i\sigma^i\phi_i/f}$$
where $\sigma^i$ are the Pauli matrices, $\phi_i$ are three scalar fields and $f$ is a constant with mass dimension. $U$ is unitary, which makes the $\phi_i$ fields real.

The Lagrangian at $O(p^2)$ order is
$$\mathcal{L}_2=\frac{f^2}{4}tr(D_{\mu}U^{\dagger}D^{\mu}U)$$
the Lagrangian at next to leading order $O(p^4)$ is
$$\mathcal{L}_4=\frac{l_1}{4}tr{}(D_{\mu}U^{\dagger}D^{\mu}U)tr(D_{\nu}U^{\dagger}D^{\nu}U)+\frac{l_2}{4}tr(D_{\mu}U^{\dagger}D_{\nu}U)tr(D^{\mu}U^{\dagger}D^{\nu}U)\\ +\frac{l_3+l_4}{16}\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\frac{l_4}{8}tr(D_{\mu}UD^{\mu}U^{\dagger})tr(\chi{}U^{\dagger}+U\chi^{\dagger})\\ +l_5tr(U^{\dagger}F_R^{\mu\nu}UF_{L\mu\nu})+i\frac{l_6}{2}tr(F_R^{\mu\nu}D_{\mu}UD_{\nu}U^{\dagger}+F_L^{\mu\nu}D_{\mu}U^{\dagger}D_{\nu}U)\\ -\frac{l_7}{16}\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2+\frac{h_1+h_3-l_4}{4}tr(\chi^{\dagger}\chi)\\ +\frac{h_1-h_3-l_4}{16}\bigg(\big[tr(\chi{}U^{\dagger}+U\chi^{\dagger})\big]^2+\big[tr(\chi{}U^{\dagger}-U\chi^{\dagger})\big]^2\\ -2tr(\chi{}U^{\dagger}\chi{}U^{\dagger}+U\chi^{\dagger}U\chi^{\dagger})\bigg)-\frac{4h_2+l_5}{2}tr(F_{R\mu\nu}F_R^{\mu\nu}+F_{L\mu\nu}F_L^{\mu\nu})$$

at next to leading order the coefficients $l_i$ and renormalize with loops with vertices coming only from the $O(p^2)$ Lagrangian like

$$l_i=l_i^r(\mu)+\frac{\gamma_i}{32\pi^2}\big(-\frac{1}{\epsilon}-\ln4\pi+\gamma_e-1\big)$$

where the gammas are

$$\gamma_1=1/3\qquad\gamma_2=2/3\qquad\gamma_3=-1/2\qquad\gamma_4=2\qquad\gamma_5=-1/6\qquad\gamma_6=-1/3\qquad\gamma_7=0$$

Now comes the question. Let's consider the same theory with the same Lagrangian but allowing the $\phi$ fields in $U$ to be complex.

My question is, would the $\gamma_i-s$ change?

asked Aug 28, 2015
edited Aug 30, 2015

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.