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Why does a spurion give the correct symmetry breaking terms regardless of the high energy physics?

+ 2 like - 0 dislike
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In short, my question is why does a spurion analysis work to produce the correct symmetry breaking terms regardless of the high energy physics?

The context that this question arose is from an Effective Field Theory course (for more context, see here, Eq. 5.50). Consider the QCD Lagrangian,
\begin{equation} 
{\cal L} _{ QCD} = \bar{\psi} \left( i \gamma^\mu D_\mu - m \right) \psi 
\end{equation} 
The kinetic part is invariant under a chiral transformation:
\begin{equation} 
\psi \rightarrow \left( \begin{array}{cc} 
L & 0 \\  
0 & R 
\end{array} \right) \psi 
\end{equation} 
however, the mass term is not. Now the claim I don't understand is as follows. Suppose the mass transformed as,
\begin{equation} 
m \rightarrow L m R ^\dagger 
\end{equation} 
In that case the mass term would be invariant under such a transformation. To write down the correct chiral symmetry breaking terms in our Lagrangian we find the terms invariant given this transformation for $ m $ and then make $ m $ a constant again. 

The way I understand this physically is that the breaking arises from a high energy spurion field, $ X $, which gets a VEV, $ m $. When we write down all possible chiral symmetry preserving terms using the transforming $m$, we are writing down all the terms that the spurion couples to. The VEV is then inserted and is equal to $ m $. But this procedure assumes that the spurion obeys the chiral symmetry, $ SU(2) _L \times SU(2) _R $, and transforms as, $ X \rightarrow L X R ^\dagger $. How do we know this assumption is true? In fact it seems to fail for the case of QCD since the ``spurion field'' is really the Higgs field, which is a singlet under $ SU(2) _R $.

asked Apr 19, 2014 in Theoretical Physics by JeffDror (650 points) [ revision history ]
edited Apr 19, 2014 by JeffDror

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