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In a set of lectures I'm watching on Effective Field Theory the professor introduces a spurion vector field, $\ell_\mu$. He then says that we take it to transform as a "left handed gauge field" and writes, \begin{equation} \ell ^\mu \rightarrow L (x) \ell ^\mu L ^\dagger (x) + i \left(\partial _\mu L (x) \right) L ^\dagger (x) \end{equation} where $L (x)$ is the transformation matrix. I've never encountered a left-handed gauge field and I would have naively only included the first part, $L \ell L^\dagger$. How did he get this transformation law?

Though the question is self contained, for more context feel free to take a look at my lecture notes under Effective Field Theory (pg 53).

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Are you sure that the second term is correct ? Don't you have $L(x)L^\dagger(x)=1$? I think it should be $(\partial_\mu L(x))L^\dagger(x)$. See for example eqs. 12-10 and 12-11 here: – Adam May 19 '14 at 19:21
Good point, thanks! I switched it. – JeffDror May 19 '14 at 22:13
Then it seems to me that the pdf file I linked answer your question: you need the second term to have a gauge invariant interaction. You can also show that the field strength is gauge invariant only if you include the second term. – Adam May 19 '14 at 22:45
Great, thanks @Adam. That helped a lot. I posted an answer based on your link. – JeffDror May 20 '14 at 1:44

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The gauge covariant derivative is given by \begin{equation} D _\mu \psi = \partial _\mu \psi - i g W _\mu ^a t _a \psi \end{equation} rearranging gives, \begin{equation} i g W _\mu = \partial _\mu \psi - D _\mu \psi \end{equation} where we write $W_\mu \equiv W_\mu^a t _a $. Under a gauge transformation we have, \begin{align} i g W _\mu \psi & \rightarrow \partial _\mu \left( L (x) \psi \right) - L (x) D _\mu \psi \\ i g W ' _\mu L\psi & = ( \partial _\mu L ) \psi + L \partial _\mu \psi - L D _\mu \psi \\ i g W ' _\mu \psi '& = ( \partial _\mu L ) L ^\dagger L \psi + L W_\mu L^\dagger L \psi \\ i g W ' _\mu \psi ' & = \left((\partial_\mu L ) L ^\dagger + L W _\mu L ^\dagger \right) \psi' \end{align} which implies that a gauge field transforms as \begin{equation} i g W _\mu \rightarrow (\partial_\mu L ) L ^\dagger + L W _\mu L ^\dagger \end{equation}

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