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  Coupling constant to pseudoscalar and chiral transformation

+ 3 like - 0 dislike

Suppose we have lagrangian of interaction of pseudoscalar field $\varphi$ with EM field $F_{\mu \nu}$
L_{\varphi \gamma \gamma} = \frac{\varphi (x)}{f_{\gamma}}\epsilon^{\mu \nu \alpha \beta}F_{\mu \nu}F_{\alpha \beta} \qquad (1)

There are also fermions in theory with lagrangian

L_{q} = \bar{f}(i\gamma_{\mu}D^{\mu} - m_{f})f, \quad D_{\mu} \equiv \partial_{\mu} - ieA_{\mu}

Lagrangian $(2)$ generates effective $qq\varphi$ interactions

L_{ff\varphi} = f_{ff\varphi }m_{f}\frac{\varphi}{f_{\gamma}} \bar{f}\gamma_{5}f + f_{ff\varphi}{'} \frac{\varphi}{f_{\gamma}} \bar{f}\gamma^{\mu}\partial_{\mu}\gamma_{5}f + ...

I want to get explicit expression for coupling constant $f_{ff\varphi }$. For doing this, I need to compute triangle diagram $\varphi \to f\bar{f}$, where two photons and one fermion $f$ running in the loop. In the result, the coupling constant $f_{ff\varphi }$ has form
f_{ ff\varphi} \sim \alpha_{EM}
But I have got some problem when I've compared this result with formal result which is obtained from chiral rotation. Below I describe this problem.

I can perform the rotation

$f \to e^{-ic\gamma_{5}\frac{\varphi}{f_{\gamma}}}$. This rotation induces summand
c\frac{\alpha_{EM}}{8 \pi} \epsilon^{\alpha \beta \gamma \delta}F_{\alpha \beta}F_{\gamma \delta}
in lagrangian, and for $c = -\frac{8 \pi }{f_{\gamma}}$ I eliminate $\varphi \gamma \gamma$ interaction term. The resulting interaction of $\varphi$ with fermions is
L{'}_{\varphi f f} = m_{f}\bar{f}_{L}e^{i16 \pi\frac{\varphi}{f_{\gamma}}}f_{R} + h.c. + 8\pi\frac{\partial_{\mu}\varphi}{f_{\gamma}}\bar{f}\gamma^{\mu}\gamma_{5}f \qquad (2)

By comparing $(1)$ and $(2)$ I have completely different interactions between $\varphi$ and $ff$. For example, I don't have term $c_{\varphi ff}m_{f}\frac{\varphi}{f_{\gamma}}\bar{f}\gamma_{5}f$. Where is the problem?

An edit. The problem has simple solution: an interaction of form $\varphi \bar{f}\gamma_{5}f$ isn't induced by loops (at least of order $\frac{1}{f_{\gamma}}$) because of approximate shift symmetry of underlying lagrangian.

asked Nov 20, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Jan 4, 2016 by NAME_XXX

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