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I will simplify the argument.

Let's consider a Gauge Boson (like the gauged one of U(1), $A_\mu$). Then, consider the Higgs boson with exponential representation, then $$H = e^{i\pi(x)/v}\left(\begin{array}{c}0 \\ v+h(x) \end{array}\right)$$ where $\pi$ is the goldstone boson coming from SSB of U(1) and $v$ is the Higgs vev. Let's do a Gauge transformation to kill the $\pi$. Under U(1) gauge transformation of parameter $\alpha(x)$ the field transforms as $$ A_\mu(x) \rightarrow A'_\mu(x) = A_\mu(x) + \frac{1}{e}\partial_\mu\alpha(x) $$ $$ \pi(x) \rightarrow \pi'(x) = \pi(x) + \alpha(x) v $$ so I choose $\alpha(x) = -\pi(x)/v$.

The Higgs boson becomes $$H=\left(\begin{array}{c}0\\v+h(x)\end{array}\right)$$ and the $A_\mu$ becomes $$A'_\mu = A_\mu -\frac{1}{e}\partial_\mu\frac{\pi(x)}{v}.$$

Now, the question is: How can I perform the inverse gauge transformation? If I use the same rules give above and I choose now $\alpha(x) =+\pi(x)/v$ I get

$$ A'_\mu \rightarrow A_\mu +\frac{1}{e}\partial_\mu\frac{\pi(x)}{v} - \frac{1}{e}\partial_\mu\frac{\pi(x)}{v} -\frac{1}{e}\frac{\partial_\mu \pi(x)}{v} = A'_\mu $$

where I performed the transformation over $A_\mu$ and $\partial_\mu\pi(x)$. So, it seems I can't define the inverse transformation in the usual way. I should expect $A'_\mu \rightarrow A_\mu$. Since $\pi(x)$ transforms under gauge transformation, does its derivative transforms in the same way?$$\partial_\mu\pi(x) \rightarrow \partial_\mu\pi'(x) = \partial_\mu\pi(x) + \partial_\mu\alpha(x) v$$

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  • $\begingroup$ May be you need Stueckelberg trick. $\endgroup$ – Saksith Jaksri Feb 22 '17 at 9:27

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