# Non-linear sigma model and QCD

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In the exponential representation after spontaneous symmetry breaking of the $SU(2) \times SU(2)$ by a VEV, $v$, the non-linear sigma model can be written,

\begin{align} {\cal L} & = \frac{1}{2} \left[ ( \partial ^\mu S ) ^2 - 2 \mu ^2 S ^2 \right] + \frac{ ( v + S ) ^2 }{ 4} \mbox{Tr} \left( \partial _\mu \Sigma \partial ^\mu \Sigma ^\dagger \right) - \lambda v S ^3 - \frac{ \lambda }{ 4} S ^4 \\  & + \bar{\psi} i  \partial_\mu \gamma^\mu \psi - g ( v + S ) \left( \bar{\psi} _L \Sigma \psi _R + \bar{\psi} _R \Sigma \psi _L \right) \end{align}

where $S$ and $\Sigma$ are related to the linear representation of $\sigma$ and ${\vec \pi}$ by,

\begin{equation}
\pi = \sigma + i {\vec \tau} \cdot {\vec \pi} = ( v + S ) \Sigma \quad , \quad \Sigma = \exp \left( \frac{ i {\vec \tau} \cdot {\vec \pi} ' }{ v } \right)
\end{equation}

Spontaneous symmetry breaking lead to three Goldstone bosons, $\vec \pi$. These are typically identified as the pions. But where did this identification come from? We started with 4 unknown scalar fields and an $SU(2)\times SU(2)$ flavor symmetry. Why would we expect the Goldstone bosons of this broken symmetry give rise to composite particles made of the fermions?

edited Apr 14, 2014

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I don't know a full answer myself, and look forward to other replies.

However, in the S-matrix (which defines the physical  fields), all bound states appear on equal footing (nuclear democracy).

Thus compositeness is nothing intrinsic to a (relativistic) quantum field but a property of the (bare) Lagrangian formulation.

answered Apr 14, 2014 by (15,787 points)
edited Apr 18, 2014

Thanks, I guess if we are to identify the mesons as the Goldstones I would have expected to have have a state "$\bar{q}q$" come out in the spontaneously broken Lagrangian, but I admit I'm not sure how that would work.

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