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.

  • 2
    $\begingroup$ 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). $\endgroup$
    – Ron Maimon
    Oct 24, 2012 at 17:15
  • $\begingroup$ 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 $\mathrm{i}\partial\!\!\! /$? $\endgroup$
    – c.p.
    Oct 24, 2012 at 19:08
  • $\begingroup$ @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. $\endgroup$
    – Dox
    Oct 24, 2012 at 19:53
  • $\begingroup$ The Dirac action leads to the correct equation of motion but is otherwise unacceptable. It leads to an energy expression of indefinite sign and charge of definite sign. It should be the other way around. It also requires the unacceptable narrative of zitter bewegung. $\endgroup$
    – my2cts
    Jul 6, 2021 at 17:12

2 Answers 2


The severity of this problem is exaggerated far too often.

These actions (which have nothing wrong with them) look different because the first defines the Dirac bar by $\bar{\psi} = \psi^\dagger i \gamma^0$ while the second defines it by $\bar{\psi} = \psi^\dagger \gamma^0$. This is enough to show that both of them are \begin{equation} S = \int dV \, \psi^\dagger \gamma^0 (-\gamma^\mu \partial_\mu - m) \psi. \end{equation} We can tell that the kinetic term doesn't need an extra $i$ because it is Hermitian. This follows from Hermiticity of \begin{equation} \gamma^0 \gamma^\mu = \begin{cases} I \\ \gamma^0 \gamma^i \end{cases} \end{equation} which is true in mostly plus and mostly minus because $\gamma^0 \gamma^i$ has one Hermitian guy and one anti-Hermitian guy either way.

Does the kinetic term need a minus sign then? Only if you also change the conventions for how you quantize the theory as explained in this answer. You are right that positivity of the energy is the key requirement but do not be thrown off by the fact that $\pm \gamma^\mu$ both satisfy the Clifford algebra. The kinetic term has two gamma matrices in Minkowski space and Minkowski space is where unitarity matters.


Dirac field is (classically) Grassmanian (anticommuting objects) valued son there is no concept of positivity of its energy. You need to second quantize the field to define value of the energy associated to quantum state.

  • $\begingroup$ The OP wouldn't cite Peskin and Weinberg if he didn't want to second quantize the field. $\endgroup$ Jul 6, 2021 at 1:06

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