This question is regarding Noether's Theorem in general, but also in the application to an example. The example is:
Find the conserved current for the Lagrangian
$$L=\bar{\psi}(\frac{i}{2}\gamma^{\mu}\partial_{\mu}-m)\psi.$$
Is my first step to find a transformation that leaves the action invariant? But isn't there more than one symmetry meaning this question has several answers?
From Peskin and Schroeder, it says we need to find
$$\partial_{\mu}j^{\mu}(x)=0 \; \text{ for } \; j^{\mu}(x)=\frac{\partial L}{\partial (\partial_{\mu}\psi)}\Delta\psi-J^{\mu}.$$
So I think I'm right in saying this $J^{\mu}$ is dependent on the transformation we're making? e.g. if it's just $\psi \rightarrow \psi +a$ then $J^{\mu}=0$.
The Lagrangian changes under the transformation of the field as
$$L \rightarrow L+\alpha \partial_{\mu}J^{\mu}.$$
But this doesn't help me because plugging $\psi \rightarrow \psi +\alpha \Delta \psi$ (an arbitrary transformation of the field $\psi$) into the Lagrangian at the top gives me (after no more than one line)
$$L \rightarrow L+\alpha \bar{\psi}(\frac{i}{2}\gamma^{\mu}\partial_{\mu}-m)\Delta \psi$$
and how do we use this to find $J^{\mu}$?
(Something I've assumed throughout all of this is that $\psi$ and $\bar{\psi}$ are treated completely separate and here we're only considering $\psi$. I hope I am right in doing this.)
I haven't really pin pointed a question here as my understanding breaks down at many points and when I think I finally understand it, I'm given a new Lagrangian and become stuck again. There must be some general procedure?
Best Answer
Firstly, you ask
Yes! In general, a given theory can have all sorts of symmetries, and each of these symmetries leads to its own conserved quantity via Noether's theorem.
As for what's going on with Noether's theorem and applying it in general, I'd like to strongly encourage you to read my answer here:
Noether's current expression in Peskin and Schroeder
If you still have questions after that, then let me know in the comments to this answer, and I'll gladly add an addendum to address your remaining confusions/questions.