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In the $U(1)$-Peccei-Quinn-Symmetry an axion field $a$ is introduced in order to solve the strong CP-Problem. It is said, that below a certain scale $f_{a}$ this symmetry is broken and you are able to write the field as $a = \langle a\rangle +\,a_\mathrm{phy}$ where $\langle a\rangle$ is the VEV and $a_\mathrm{phy}$ denotes the physical axion field, that is associated with the axion particle.

The VEV is $\langle a\rangle = -\theta f_{a}$ and so the CP-Violating term vanishes, since: $\mathcal{L} \supset (a/f_{a} + \theta) G \tilde{G}$. This is what I've understood so far. However, what I am wondering about now is: Why doesn't the interaction term $\mathcal{L} \supset a_\mathrm{phy} \,G\,\tilde{G}$ violate CP-Symmetry? The only possibility I see is if $a_\mathrm{phy}$ also transforms under CP-Symmetry but I wasn't able to find anything about that.

What I'm also not able to understand is: Why is the symmetry breaking scale parameter $f_a$ also called the decay constant of the axion? I'm not sure why these two are equivalent in this context.

AccidentalFourierTransform
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F.ert
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1 Answers1

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The Lagrangian is as I understand an effective Lagrangian. The Lagrangian for the $a$ field is $$ \frac{1}{2}\partial_\mu a\partial^\mu a + F\left(\frac{a}{f}\right)Tr(E_{QCD}\cdot B_{QCD}) $$ This function is to lowest order $F(x) = \frac{1}{2}x^2$. The $F\tilde F = Tr(E_{QCD}\cdot B_{QCD})$ is then a mass source term, with the trace over the color indices. This is a mass term. This compensates for the $\theta$ angle for $CP$ violations and so $$ F\left(\frac{a}{f}\right) \rightarrow F\left(\frac{a}{f}\right) + \theta $$ We now have the $CP$ violating angle $\theta$ and the mass of this scalar field, called the axion, cancelling each other. In this way the effective Lagrangian remains $CP$.

Lawrence B. Crowell
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