Local symmetry restoration via a gauge field In the book Quantum Field Theory for the Gifted Amateur, the author stated that, having a field that transforms locally via $\psi(x) \rightarrow \psi(x)e^{i \alpha(x)}$ will destroy local symmetry -and he is right- but he said we can fix that right up by adding a field $A_\mu(x)$ and replacing derivatives in the Lagrangian by covariant ones of the form: $D_\mu = \partial_\mu + iq A_\mu(x)$, I don't really get how he got from here to there, why covariant derivative with some field would fix that up, and why exactly this form of a derivative? If anyone can help, many thanks!
 A: 1)The situation is similar to general relativity, where you cannot compare vectors in different tangent planes, and must  supply a connection to 'transport' vectors between different planes. Here, once you allow gauge transforms, you cannot compare fields at 2 different points before and after the transform, since they transformed differently(depending on $x$). This naturally gives rise to a covariant derivative. The analogy is(schematically)-$$D_{GR}(v)=\partial v+\Gamma v \leftrightarrow D_{QFT}(\psi)=\partial\psi+iA\psi$$
2) In either case, you need a connection-the christoffel symbols $\Gamma$ in GR and $A_\mu$ here. The name can also be motivated as follows-just as in GR, $\partial_\mu v^\alpha$ doesn't transform like a tensor under coordinate transformations(you get extra terms), and the non-tensorial transformation of $\Gamma$ is required to exactly cancel these terms to get a tensorial object-one that transforms 'nicely'. Hence, 'covariant'.
3) Similarly, it is easy to see that that under $$\psi\to  e^{i\alpha(x)}\psi, \bar{\psi}\to e^{-i\alpha(x)}\bar{\psi}, A_\mu\to A_\mu+\partial_\mu\alpha$$
$$\implies D_{\mu}\psi\to e^{i\alpha(x)}D_\mu\psi,\hspace{5mm} D_\mu\bar{\psi}\to e^{-i\alpha(x)}D_\mu\bar{\psi}$$ 
And together these leave the Lagrangian gauge invariant. A simple $\partial_\mu\psi$ wouldn't have transformed this elegantly; you'd have extra terms and the Lagrangian would NOT be gauge invariant. So you add the connection $A$ to cancel the 'bad' terms.
A: I want to give you an initial argument and a point from where you can start to find the result yourself.
To have a symmetry in the lagrangian given by a local $U(1)$ transformation you should have that
$$\bar\psi e^{-i\alpha(x)}(i\not \partial - m) e^{i\alpha(x)} \psi= \bar\psi(i\not \partial-m)\psi$$
now, it's evident that the mass term is invariant since it's just a constant, but the derivative term is not. To keep invariance you should find a new derivative such that $$(D\psi)^\prime = e^{i\alpha(x)}D\psi\tag{1}$$ so that the exponential cancels with the one from the transformation of $\bar\psi$ and you get the first equality. Broadly speaking a covariant derivative is given by the normal derivative plus a term which is called connection, so that a covariant derivative is of the form $$D_\mu =\partial_\mu +i\Gamma_\mu \tag{2}$$ where $\Gamma_\mu$ is the connection. Now using $(1)$ and $(2)$ try and find that the covariant derivative is given by $$D_\mu = \partial_\mu+iA_\mu$$ the $e$ in your covariant derivative is just a matter of definition.
