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Transformation in non-linear sigma model

+ 4 like - 1 dislike
202 views

I'm studying the linear sigma model with regards in the context of effective field theory and the professor is applying a strange transformation for one of the fields. Consider for example a linear $ \sigma $-model with 4 complex scalar fields, $ \sigma $ and $ {\vec \pi} = \left\{ \pi ^1 , \pi ^2 , \pi ^3 \right\}  $ as well as 2 left and 2 right Weyl fermionic fields. Furthermore, we define $ \pi $:
\begin{equation} 
\pi = \sigma + i  {\vec \tau} \cdot {\vec \pi} 
\end{equation} 
where $ {\vec \tau} $ are the Pauli matrices. We then consider the Lagrangian,
\begin{align}
{\cal L} _\sigma & = \frac{1}{4} \mbox{Tr} \left( \partial _\mu \pi \partial ^\mu \pi \right)  + \frac{ \mu ^2 }{ 4 } \mbox{Tr} \left( \pi ^\dagger \pi \right) - \frac{ \lambda }{ 4 !} \left( \mbox{Tr} \left( \pi ^\dagger \pi \right) \right)  ^2 \notag \\ 
  & \qquad +  \bar{\psi} _L i \sigma_\mu\partial^\mu \psi _L + \bar{\psi} _R i \sigma_\mu\partial^\mu \psi _R - g \left( \bar{\psi} _L \pi  \psi _R + \psi _R \pi ^\dagger \psi _L \right)  
\end{align} 

The claim is that this theory has an $ SU(2) _L \times SU(2) _R $ symmetry:
\begin{equation} 
\psi _L \rightarrow L \psi _L  \quad , \quad \psi _R \rightarrow R \psi _R \quad , \quad \pi \rightarrow L \pi R ^\dagger 
\end{equation} 
But what is the justification for allowing this transformation for $ \pi $? Normally matrices transform under 
\begin{equation} 
U A U ^\dagger 
\end{equation} 
so shouldn't this be 
\begin{equation} 
\pi \rightarrow R L \pi L ^\dagger R ^\dagger 
\end{equation} 
instead?

asked Apr 4, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]
edited Apr 4, 2014 by JeffDror

1 Answer

+ 4 like - 0 dislike

A good way to see this is to think of the Weyl fields as part of a Dirac field \(\begin{pmatrix}ψ_L \\ ψ_R \end{pmatrix}\). Then the matrix \(π\)is actually the off-diagonal component of a more general coupling:

\(\begin{pmatrix}\bar ψ_L & \bar ψ_R \end{pmatrix}\begin{pmatrix}0 & π \\ π^\dagger & 0 \end{pmatrix} \begin{pmatrix}ψ_L \\ ψ_R \end{pmatrix}\)

The symmetry transformations are part of a larger matrix

\(U = \begin{pmatrix}L & 0 \\ 0 & R \end{pmatrix}\)

and now the coupling transforms as \(UAU^\dagger\)as you would expect.

But even if the situation could not be explained as neatly as above: if the Lagrangian is symmetric under a fancy transformation, then this transformation is a "good" transformation, regardless of whether it fits a preconceived notion of what symmetry transformations should look like.

answered Apr 4, 2014 by Greg Graviton (665 points) [ no revision ]

Thanks, that works well. I agree that it would be a symmetry regardless, but I think if it couldn't be put in this reducible form into two $SU(2)$ transformations, I would at the very least be reluctant to call it an $SU(2)_L\times SU(2)_R$ symmetry. 

What I mean is that not every \(SU(2)\times SU(2)\) symmetry needs to be part of a \(SU(4)\)transformation.

The relevant concept is that of a group action: a mapping that assigns each group element g a transformation \(h_g\). In this case, \(h_{(L,R)}(π)=LπR^\dagger\). However, there is a conditions on the assignment, namely that it is compatible with the group laws, so that \(h_e(π)=π\) and \(h_{gk}(π)=h_g(h_k(π))\).

Note that the transformation you suggested, \(h_{L,R}(π)= RLπ L^\dagger R^\dagger \), is not a group action! It's not compatible with the multiplication law.

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