# Transformation in non-linear sigma model

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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 $\pi$:

\pi = \sigma + i  {\vec \tau} \cdot {\vec \pi}

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:

\psi _L \rightarrow L \psi _L  \quad , \quad \psi _R \rightarrow R \psi _R \quad , \quad \pi \rightarrow L \pi R ^\dagger

But what is the justification for allowing this transformation for $\pi$? Normally matrices transform under

U A U ^\dagger

so shouldn't this be

\pi \rightarrow R L \pi L ^\dagger R ^\dagger

asked Apr 4, 2014
edited Apr 4, 2014

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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 (775 points)

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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