What is the meaning of the field deformation (delta phi) for continuous symmetries?

Question

I'm taught in class that for a symmetry $\phi \rightarrow \phi + \delta\phi$ (and leaving the spacetime coordinates alone), the Noether current is $$ J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \delta\phi $$ but the expression $\delta\phi$ is very unclear to me exactly what it means. Sometimes, $J^\mu$ is written as $$ J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \frac{\delta\phi}{\delta\alpha}$$ if the symmetry depends on a parameter $\alpha$. This makes slightly more sense, but it still makes me uncomfortable because of the following reason. For the U(1) symmetry $\phi \rightarrow e^{i\alpha}\phi$, if I blindly take a derivative with respect to $\alpha$, then I get $i e^{i\alpha}\phi$. However my textbook clearly states that the expression $\frac{\delta\phi}{\delta\alpha} = i\phi$ (without the extra phase).

So my question is: is it at least correct (though less concise) to say that a symmetry $\phi'(\phi,\alpha)$ is a function of the field and the parameter $\alpha$ such that $\phi'(\alpha=0) = \phi$, and that $$ J^\mu = \frac{\partial L}{\partial (\partial_\mu\phi_n)} \frac{\partial \phi'}{\partial \alpha}\Big|_{\alpha=0} $$ is the Noether current? My concern with this is, what about symmetries that are not parameterized by some parameter $\alpha$? What would be a more clear definition of $\delta\phi$ in that case?

Answer 1

The transformation $\phi\rightarrow\phi+\delta \phi$ is assumed to be small (infinitessimal). You can write $\delta\phi$ as \begin{align}\delta\phi&=\phi'-\phi\\ &=e^{i\delta \alpha}\phi-\phi\\ &\approx(1+i\,\delta\alpha)\phi-\phi\\ &=i\,\delta\alpha\,\phi \end{align} So this means $\frac{\delta\phi}{\delta\alpha}=i\phi$. Why does this coincide with $\left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}$? This is exactly because $\left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}$ gives the first order in a Taylor expansion of $\phi'$ in $\alpha$ around zero: $$\phi'(\alpha)=\phi'(0)+\alpha \left.\frac{\partial\phi'}{\partial\alpha}\right|_{\alpha=0}+\mathcal O(\alpha^2)$$ Note that $\phi'(0)=\phi$. Where does this Taylor expansion appear in the derivation of $\delta\phi$?

So to answer your question, yes it would be correct to write $\phi'$ like you did. What about symmetries that can't be parametrised by some parameter $\alpha$? We are talking about continuous symmetries and I'm not 100% sure but I think all continuous symmetries can be parametrised that way.