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I have a (possibly) fundamental question, which is driving me crazy.


When considering the Dirac action (say reading Peskin's book), one have

$\int dV\;\bar{\psi}\left(\imath\not\partial-m\right)\psi,$

or (following Weinberg)

$\int dV\;\bar{\psi}\left(-\not\partial-m\right)\psi.$

Where Dirac conjugation is defined according to the metric signature.


Is there a reason why the kinetic term in those actions have that particular sign choice?

Are these actions: $\int dV\;\bar{\psi}\left(-\imath\not\partial-m\right)\psi$ and $\int dV\;\bar{\psi}\left(\not\partial-m\right)\psi$ ill-possed?

Personal Thoughts

  • Elements of Clifford algebra are well defined up to a sign, therefore, I would say the second set of actions is Ok.

  • I'm afraid that the different sign would spoil the positivity of energy. However, I'm not sure how to prove that statement.

Any help or thoughts are welcome! Thank you.

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This is the most annoying part of modern QFT--- the lack of good standardized Dirac notation and results. This can't be solved by committee, because it was standardized in the 1950s, but in the 1970s, when we needed higher dimensions, it was realized the standardized thing was suboptimal, and the dust hasn't settled yet. I would go with Weinberg, but I don't have a good answer for the optimal conventions, and there are about half a dozen reasonable ones floating around (Euclidean/Lorentzian Weyl/Dirac Majorana/C-matrix are each discrete choices, also complex-coordinates/real-coordinates). – Ron Maimon Oct 24 '12 at 17:15
What do you mean @Dox by Elements of Clifford algebra are well defined up to a sign? that $\gamma^\mu$ and $-\gamma^\mu$ yield the same? you are actually changing by $i$. Didn't you write $\partial\!\!\! /$ instead of $\mathrm{i}\partial\!\!\! /$? – c.p. Oct 24 '12 at 19:08
@JorgeCampos: You are right, but I mean one a convention is picked (say signature of metric), changing $\gamma^\mu\to(\gamma^\mu)'= -\gamma^\mu$ span the same Clifford algebra. – Dox Oct 24 '12 at 19:53

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