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  Non-linear sigma model and QCD

+ 3 like - 0 dislike

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, 

\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) 

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?

asked Apr 14, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited Apr 14, 2014 by JeffDror

1 Answer

+ 1 like - 0 dislike

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 Arnold Neumaier (15,787 points) [ revision history ]
edited Apr 18, 2014 by dimension10

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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