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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?

Hunter
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Phibert
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1 Answers1

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Firstly, you ask

isn't there more than one symmetry meaning this question has several answers?

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.

Community
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joshphysics
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    Ok that's the best explanation of Noether's theorem I've seen before and it's helped me a lot. A couple of questions that I'm developing as I go through it using problems: First and foremost, step (1) says "Let a particular flow ϕ^(α,x) be given". In the problem above, I'm not given any flow (transformation of the field) and just asked to find the conserved current. So immediately I'm a bit lost. Do I just take an arbitrary flow and work from there? Could you also tell me what I should consider $\bar{\psi}$ as if we're varying $\psi$? – Phibert Feb 23 '14 at 00:11
  • Incidentally, are you not missing a $\partial_{\mu}$ in the $\delta \mathcal{L}$ term in Lemma 1. of the answer you linked me to? – Phibert Feb 23 '14 at 00:54
  • @user13223423 Thanks for the careful read; I was indeed missing a $\partial_\mu$! Recall that $\bar\psi = \psi^\dagger\gamma^0$, so any flow on $\psi$ induces a flow on $\bar\psi$ in the natural way; I can write more details about this if "natural way" isn't descriptive enough. Lastly, your problem statement is a bit unclear, but I would guess that it wants you to find the most general flow that transforms the field by both a spacetime transformation, and a target space transformation that is a symmetry. To find this, you can put in a general flow ansatz, and see what symmetry demands. – joshphysics Feb 23 '14 at 01:54
  • I've only recently come back to Noether's Theorem and I understand it quite well now thanks to you. One more thing though, for the transformation $\phi \rightarrow e^{i\alpha} \phi$, before we even start using Noether's theorem, to show the Lagrangian is invariant under this transformation, do we plug in $\Delta \phi = i \phi$ or $\Delta \phi = i \alpha \phi$ (because we only consider the $\textit{infinitesimal}$ transformation)? The former leaves my Lagrangian invariant, the latter gives $L \rightarrow \alpha^2 L$. (This is for the complex scalar field by the way). – Phibert Apr 16 '14 at 18:12
  • @user13223423 Yeah so in this case, (using my notation from the linked post), we have $\hat\phi(\alpha, x) = e^{i\alpha}\phi(x) = \phi(x) + i\alpha\phi(x) + O(\alpha^2)$. Now, plug this right hand side into the Lagrangian and expand the result in $\alpha$. Recall that we only want to exhibit infinitesimal invariance which means that to first order in $\alpha$, the Lagrangian doesn't change. So you should find that $L \to L + O(\alpha^2)$ which is what I think you probably mean by $L\to \alpha^2L$. I'm guessing you erroneously only plugged in $i\alpha\phi$ instead of $\phi+i\alpha\phi$? – joshphysics Apr 16 '14 at 18:27
  • Ah so I should still be plugging in $\phi+i \alpha \phi$? The issue was Peskin and Schroeder write $\alpha \Delta \phi = i \alpha \phi$ so I thought perhaps the $\alpha$'s cancelled. I guess for this transformation it's obvious that we're going to get $L \rightarrow L + O(\alpha^2)$ just from looking at it. But sometimes there's going to be a $J^{\mu}$ which pops up in the form $L \rightarrow L+\partial_{\mu} J^{\mu}$? (By the way, I'm not sure if it's just me, but this final point I made was that largest obstacle in coming to understand Noether's Theorem). – Phibert Apr 16 '14 at 18:33
  • @user13223423 Yeah you should plugin the whole "flowed" field to first first in $\alpha$, not just the change. You can just use the formula for $\delta \mathcal L$ that I gave in the linked post formulated in terms of a derivative with respect to $\alpha$. Yes, sometimes you are only going to have inifinitesimal invariance up to a total derivative term $\partial_\mu J^\mu$. I agree that's one of the really confusing parts which is why I tried really hard in the proof sketch in the link to highlight the logic behind it. Maybe I'll write a more detailed blog post... – joshphysics Apr 16 '14 at 21:20