- #1
Dixanadu
- 254
- 2
Homework Statement
Hey guys,
Consider the U(1) transformations
[itex] \psi'=e^{i\alpha\gamma^{5}}\psi [/itex] and [itex] \bar{\psi}'=\bar{\psi}e^{i\alpha\gamma^{5}} [/itex] of the Lagrangian [itex]\mathcal{L}=\bar{\psi}(i\partial_{\mu}\gamma^{\mu}-m)\psi[/itex].
I am meant to find the expression for [itex]\partial_{\mu}J^{\mu}[/itex].
Homework Equations
Gamma matrices anticommute
Noether current is [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\delta x \mathcal{L} [/itex]
not sure of anything else...
The Attempt at a Solution
So here's what I've done so far. Since its a U(1) transformation, the coordinates arent changing, so the Noether current is given by [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi [/itex]. I found that [itex]\delta\psi=i\alpha\gamma^{5}\psi[/itex], so that
[itex]\delta J^{\mu}=-\bar{\psi}\alpha\gamma^{\mu}\gamma^{5}\psi[/itex], then I drop the infinitesimal parameter to get
[itex]J^{\mu}=-\bar{\psi}\gamma^{\mu}\gamma^{5}\psi[/itex].
So the next step is to calculate the derivative of this. Doing so, I get
[itex]\partial_{\mu}J^{\mu}=-(\partial_{\mu}\bar{\psi}\gamma^{\mu}\gamma^{5}\psi+\bar{\psi}\gamma^{\mu}\gamma^{5}\partial_{\mu}\psi)[/itex]
And at this point I am stuck...im not sure if this is right and/or if I can simplify this or do something neat with it? because I think I'm meant to be using the transformed Lagrangian
[itex]\mathcal{L}'=i\bar{\psi}\partial_{\mu}\gamma^{\mu}\psi-m\bar{\psi}e^{2i\alpha\gamma^{5}}\psi[/itex]
for something but I don't really know.
Thanks guys..