# Generating Noether charges for Dirac Lagrangian

1. Jan 19, 2012

### JustMeDK

I have been calculating the currents and associated Noether charges for Lorentz transformations of the Dirac Lagrangian. Up to some spacetime signature dependent overall signs I get for the currents expressions in agreement with Eq. (5.74) in http://staff.science.uva.nl/~jsmit/qft07.pdf [Broken].

What confuses me is the 'inner' term, the anticommutator term. The associated charges vanish for boost generators, simply because the anticommutator itself vanishes for boosts, so how can these Noether charges generate all Lorentz transformations? Have I misunderstood something fundamental?

PS: The charges resulting from Eq. (5.74) are hermitian. This by itself is, of course, inconsistent with the fact that boost generators are antihermitian, generating as they do the non-compact part of the Lorentz group.

Last edited by a moderator: May 5, 2017
2. Jan 19, 2012

### samalkhaiat

You need to be careful when you use the Hermitian Dirac Lagrangian; it is always a good idea to throw away a total divergence and work instead with
$$\mathcal{L} = \bar{\psi}( i \gamma^{a}\partial_{a} - m) \psi$$

Only the spin part vanishes; there is a non-vanishing contribution from the orbital (boost) part of the angular momentum tensor $J^{0k}$.

There is no inconsistency; these charges generate infinite dimensional unitary representation of the Lorentz group. Non-compact groups do admit such representations.

sam

Last edited by a moderator: May 5, 2017
3. Jan 20, 2012

### JustMeDK

Since yesterday I have myself realized that it is best to start from ${\cal{L}}_{D} = \bar{\psi}(i\gamma^{\rho}\partial_{\rho} - m)\psi$. I'am aware of the fact that for boosts only the spin part of Eq. (5.74) vanishes. I apologize if that was not apparent from my formulation.
I'am also aware of the fact that unitary representations of a non-compact group can only be infinite-dimensional. Although I first realize it now, this tallies nicely with $L_{\mu\nu} = -i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})$ being an infinite-dimensional representation of the Lorentz group, while $S_{\mu\nu} = \frac{i}{4}[\gamma_{\mu},\gamma_{\nu}]$ is finite-dimensional.
I have never before explicitly carried out such a Noether current/charge calculation and what surprises me the most is, if not misconceived, that ${\cal{L}}_{D}$ and its hermitized version, which differ only by a four-divergence, can end up with Noether charges which either generates internal boosts or do not.