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  Chiral effective lagrangian and higher order terms

+ 4 like - 0 dislike
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At scales below $\Lambda_{QCD}$ global QCD symmetry $SU_{L}(3)\times SU_{R}(3)$ is spontaneously broken down to $SU_{f}(3)$, and pseudogoldstone degrees of freedom called mesons arise. It is possible to obtain explicit for of lagrangian for them: we substract goldstone degrees of freedom from quarks field, $q = U \tilde{q}$, where $U = e^{i\gamma_{5}t^{a}\epsilon_{a}}$ and $\epsilon_{a}$ parametrise goldstone degrees of freedom; then we replace $\bar{\tilde{q}}\tilde{q}, \bar{\tilde{q}}\gamma_{5}\tilde{q}$ by their VEVs $v, 0$. In the result chiral effective theory arises.

The question: I understand how (by using method described above) to derive term $\text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U]$, but I don't understand how terms like
$$
\tag 1 \text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U\partial_{\nu}U^{\dagger}\partial^{\nu}U]
$$ arise, since in QCD lagrangian there are only bilinear combinations of quark fields (if I understand correctly, term $(1)$ arise without taking into account gluon field), so product of only two $U$ matrix can arise.

I would be grateful for explanation how terms like $(1)$ arise.

asked Oct 9, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Oct 9, 2015 by NAME_XXX

1 Answer

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The method described at the beginning of the question seems to be a classical one: assume that we have a non-trivial expectation value for some bilinear combination of quark, expand the classical Lagrangian of QCD around this expectation value to derive a Lagrangian for the Goldstone bosons. This approach clearly does not work in the full quantum mechanical QCD: in this setting, the spontaneous symmetry breaking is a complicated dynamical non-perturbative effect and similarly the effective theory describing the Goldstone bosons should emerge from the QCD Lagrangian via a complicated path integral calculation and in general it is not known how to do it. But the whole point of the chiral effective theory is that it is possible to say something non-trivial on the low energy behavior of the Goldstone bosons only by symmetry considerations, without having to know the complicated underlying microscopic dynamics it comes from. One simply writes the most general Lagrangian compatible with the symmetries of the problem. At low energy, the most relevant terms are those with the smaller number of derivatives. Up to two derivatives, there is only one such a term: $\text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U]$. But then one has next to leading order with four derivatives and so on.

answered Oct 9, 2015 by 40227 (5,140 points) [ revision history ]

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