Easy proof of Noether's theorem? I' looking for an easy proof of Noether's theorem? I mean I know that the variation must be
$ 0=\delta S = (EULER-LAGRANGE)+ (CONSERVED\, \, \, CURRENT) $
for the case of a particle $q(t)$. I know how to obtain it by I have doubts for the case of fields $ \phi (x) $, any hint? I have checked several books but I cannot find any easy proof of Noether's theorem anywhere; they use too complicated methods.
 A: We consider infinitesimal transformations of a field in the form,
$$\phi\to\phi' = \phi(x)+\alpha \Delta \phi(x)$$
for an infinitesimal parameter $\alpha$. The system is said to be invariant under such a transformation if it changes up to a total derivative or surface term, i.e.
$$\mathcal{L}\to\mathcal{L}'=\mathcal{L}(x)+\alpha \,\partial_\mu F^{\mu}(x)$$
By varying with respect to the fields,
$$\alpha \Delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}\alpha \Delta \phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\mu (\alpha\Delta \phi)$$
$$= \alpha \underbrace{\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\Delta \phi\right)}_{F^\mu(x)} + \alpha \left[ \frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right)\right]\Delta \phi$$
where in the last line we employed the equations of motion which arise by demanding $\delta S=0$. Notice the second term is zero for that reason, and hence we can declare,
$$j^{\mu}(x)=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\Delta \phi - F^{\mu}(x)$$
which satisfies the continuity equation, $\partial_\mu j^{\mu}=0$, or in vector calculus language,
$$\frac{\partial j^{0}}{\partial t}+\nabla \cdot \vec{j}=0$$
The corresponding Noether charge is given by,
$$Q=\int \mathrm{d}^{d-1} x \, j^{0}$$
which one can verify via the continuity equation and Stokes' theorem that $Q$ is conserved locally.

Useful Resources: Peskin and Schroeder's Introduction to Quantum Field Theory
