# Understanding this phrasing of a conserved current in QFT

I'm trying to derive a version of Noether's theorem. I found this online and I'm really confused as to what's going on on page 202 (the second slide in the link).

I'll start off by saying I am familiar with Noether's theorem, but the way I usually see it is: You have a symmetry of $\mathcal{L}$ such that $\delta \mathcal{L} = \partial_{\mu} F^{\mu}$. Then the conserved current is $$J^{\mu} = \frac{\partial \mathcal{L} }{\partial (\partial_{\mu}\phi)} \delta \phi - F^{\mu},$$ such that $\partial_{\mu} J^{\mu}=0.$

On this page, it says to consider a theory with a continuous symmetry.

Firstly, it says we can consider transformations that result in $\delta \mathcal{L} = 0$. Why on earth is this the case? I can see that sometimes this may be true, but most generally, shouldn't you have to write $\delta \mathcal{L} = \partial_{\mu} F^{\mu}$ (for some $F$) as I mentioned above?

Also, it says that $$\partial^{\mu} J_{\mu}(x) = \delta \mathcal{L}(x) - \frac{\delta S}{\delta \phi(x)} \delta \phi(x).$$ Why is this true? Shouldn't this be equal to 0? And how would the actual $J$ be defined?

Can someone clarify this? Or send me a reference that explains Noether's theorem in such a way?

Q1) $\delta \mathcal L$ is zero for a global symmetry transformation, for example: assume that $\phi$ is a field that transforms under the fundamental representation of $SU(N)$, $\phi \mapsto U \phi$, then the Lagrangian for this field can look like: $$\partial_m \phi^\dagger \partial^m \phi + m^2 \phi^\dagger \phi + \lambda (\phi^\dagger \phi)^2 + \cdots\,.$$ The important thing here is that all these terms are invariant under a $\mathrm{SU(N})$ transformation, $\phi^\dagger \phi \mapsto \phi^\dagger U^\dagger U \phi = \phi^\dagger \phi$. This means that the Lagrangian is invariant under a $\mathrm{SU(N})$ transformation, i.e., $\delta \mathcal{L} = 0$.
This would not be true for space-time transformations. If $x^m \mapsto x^m + v^m$ is an infinitesimal space-time transformation then any space-time scalar, including the Lagrangian, will transform like $f(x) \mapsto f(x) - v^m \partial_m f(x)$.
Q2) Variation of the Lagrangian $\mathcal L(\phi, \partial_m \phi)$ can always be written as: $$\delta \mathcal L = \left(\frac{\partial \mathcal L}{\partial \phi} - \partial_m \frac{\partial \mathcal L}{\partial(\partial_m \phi)} \right) \delta \phi + \partial_m \left(\frac{\partial \mathcal L}{\partial(\partial_m\phi)} \delta\phi \right) \,. \tag{*}$$ Note that, the first pair of braces contain the equation of motion. And the equation of motion is, by definition, the functional derivative of the action with respect to the field (so that, extremizing action $\Rightarrow$ functional derivative with respect to field vanishes $\Rightarrow$ equation of motion satisfied). Therefore, you can write the variation of the Lagrangian as: $$\delta \mathcal L = \frac{\delta S}{\delta \phi} \delta \phi + \partial_m \left(\frac{\partial \mathcal L}{\partial(\partial_m\phi)} \delta\phi \right) \,.$$ This is what is written on page 202 of the pdf you linked, with the identification: $$j^m = \frac{\partial \mathcal L}{\partial(\partial_m\phi)} \delta\phi \,.$$ Now, it is important that this $j^m$ is required to be divergenceless only on shell. For example, in the first question we saw that under a transformation generated by a global symmetry group $\delta\mathcal L$ is zero. The first term of the equation (*) vanishes only when the equation of motion is satisfied, so it is only when the equation of motion is satisfied that we require $\partial_m j^m = 0$. For more general symmetry transformations, the definition of symmetry is that the variation of the Lagrangian be a total derivative, $\delta \mathcal L = \partial_m F^m$ on shell. Since the first term of (*) vanishes on shell, we have to require for symmetry that, on shell, the following must be true: $$\partial_m F^m = \partial_m j^m \Rightarrow \partial_m (j^m - F^m) = 0\,.$$ So the general definition of current is indeed what you wrote: $$J^m := j^m - F^m\,.$$ But this is divergenceless only when the equation of motion is satisfied.