- #1

JosephButler

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1. In Ramond's text (Field Theory: A Modern Primer), he explains that the Lagrangian for fermions should have the derivative operator antisymmetrized in order for the kinetic term to be real: [itex]\psi^\dag \sigma^\mu \overleftrightarrow\partial_\mu \psi[/itex] .This is equation (1.7.3), with a discussion above equation (1.4.40). I don't quite understand why the antisymmetrization should do this since the derivative really only acts on a factor of [itex]e^{ip\cdot x}[/itex] in the Fourier transform of the spinors?

[I understand that at the end of the day one can always integrate by parts to get the "only-right-acting" derivative... but this becomes more subtle in cases where one dimension is compact or when the space has curvature since then integration by parts picks up surface terms and/or derivatives of [itex]\sqrt{g}[/itex]]

2. This, I'm almost sure, is a stupid question: I'm confused about the factors of i in Ramond's fermion Lagrangian. He seems to use the same conventions as particle physicists, e.g. (+---) metric and chiral basis of gamma matrices, but his kinetic term appears to be missing a factor of i and the mass term appears to have gained a factor of i. Doesn't this mean that the Lagrangian is no longer real? Where is this factor of i coming from?

Thanks,

Joe