How is the charge carried by a field determined from its Lagrangian? I don't understand how we can determine the charge carried by some field from a Lagrangian.
In my lectures, I was told that I could determine the charges carried by $\phi$ from a lagrangian:
$$\mathcal{L}= \partial_\mu \phi^\dagger \partial^\mu \phi \tag{1}$$
and if $\phi \to \phi ' = e^{i\theta}\phi$
I have been told I must consider the charge under the symmetry group, but I am not sure how this will influence the charge carried by the field or how to determine said charge.

From what I have read, the conserved charge is given by:
$$ Q = \int d^3 x j^0 \tag{2} $$
But what parameter is $j^0$ ? How can I derive it? Will this allow me determine the carried charge?
 A: If you have a Lagrangian of the form $$\mathcal{L}=\partial_\mu\phi^\dagger\partial^\mu\phi=\eta^{\mu\nu}\partial_\mu\phi^\dagger\partial_\nu\phi$$the corresponding Noether's current is given by $$j^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi^\dagger)}\delta\phi^\dagger=\eta^{\mu\nu}\partial_\nu(\phi^\dagger)\delta\phi+\eta^{\mu\nu}\partial_\nu(\phi)\delta\phi^\dagger.$$ To work out $\delta\phi$ and $\delta\phi^\dagger$, consider an infinitesimal transformation of the form $\phi\rightarrow e^{-i\alpha}\phi$. We can expand this in first order to see how both $\phi$ and $\phi^\dagger$ transform under an infinitesimal transformation $$\phi\rightarrow\phi-i\alpha\phi\quad\text{and}\quad\phi^\dagger\rightarrow\phi^\dagger+i\alpha\phi^\dagger.$$Thus we can read off the values for $\delta\phi$ and $\delta\phi^\dagger$, given by $$\delta\phi=-i\alpha\phi\quad\text{and}\quad\delta\phi^\dagger=i\alpha\phi^\dagger.$$Substituting this into the expression of the current, we end up with $$j^\mu=i\left(\frac{\partial\phi}{\partial x^\nu}\phi^\dagger-\frac{\partial\phi^\dagger}{\partial x^\nu}\phi\right)\eta^{\nu\mu}.$$ This is the Noether's current associated with the $U(1)$ symmetry. To compute the corresponding charge, we just plug $j^\mu$ into the corresponding formula $$Q=\int d^3x\left(i\dot{\phi}^\dagger\phi-i\dot{\phi}\phi^\dagger\right).$$Note that the $\alpha$ factors drop out since the variations $\delta\phi$ and $\delta\phi^\dagger$ are with respect to the symmetry factor $\alpha$. If you want a more explicit form for $Q$, then you should get explicit constructions for $\phi$ and $\phi^\dagger$. In QED, the charge will be associated to the electric charge.
