lagrangian formalism - 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.

Answer

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.

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