- #1
EoinBrennan
- 12
- 0
Homework Statement
I understand the premise of Noether's theorem, and I've read over it in as many online lectures as I can find as well as in An Introduction to Quantum Field Theory; Peskin, Schroeder but I can't seem to figure out how to actually calculate it. I feel like I'm missing a step that all of the explanations are glossing over.
Homework Equations
I want to find the Noether current [itex]J^{\mu}_{a b}[/itex] with Lagragian:
[itex]L = \frac{1}{2} \partial_{\mu} \Phi^{a} \partial^{\mu} \Phi^{a} - \frac{1}{2} m^{2} \Phi^{a} \Phi^{a} - \frac{1}{4} \lambda ( \Phi^{a} \Phi^{a} )^{2}[/itex]
Under the SO(N) symmetry:
[itex]\Phi^{a} \rightarrow \Lambda^{a b} \Phi^{b}[/itex]
Which is infinitesimally:
[itex]\Phi^{a} \rightarrow \Phi^{a} + \epsilon^{a b} \Phi^{b}[/itex]
And given the following relations:
[itex]\Lambda^{T} \Lambda = I ; \Lambda = I + \epsilon ; \epsilon^{a b} = - \epsilon^{b a}[/itex]
The Attempt at a Solution
So I know that [itex]J^{\mu} = \frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a} - F^{\mu}[/itex]
with
[itex]F^{\mu} = \frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a}[/itex]
This is one of things I'm confused by, as it seems to be a meaningless statement.
Regardless, I calculated
[itex]\frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a} = ∂^{\mu} \Phi^{a} Δ \Phi^{a}[/itex]
And I'm fairly certain that [itex]Δ \Phi^{a} = \epsilon^{a b} \Phi^{b}[/itex]
This would give me that:
[itex]J^{\mu} = ∂^{\mu} \Phi^{a} \epsilon^{a b} \Phi^{b} - F^{\mu}[/itex]
And a few of the similar examples I found seem to take F as zero here. Giving me:
[itex]J^{\mu}_{a b} = ∂^{\mu} \Phi^{a} \epsilon^{a b} \Phi^{b}[/itex].
Am I vastly wrong? Where am I going astray if so?
Any help would be really appreciated, especially any help understanding the process step by step.
Cheers,
-Eoin