In paragraph 19.5 Peskin and Schroeder discuss the difference between the canonical energy-momentum tensor $T^{\mu\nu}$ and the symmetric and gauge invariant energy-momentum tensor $\Theta^{\mu\nu}$. Ultimately they state that (I am taking the gauge fields zero here) for fermions the latter reads $$ \Theta^{\mu\nu} = \frac{i}{2}\bar{\psi}\big(\gamma^{\mu}\partial^{\nu}+\gamma^{\nu}\partial^{\mu}\big)\psi - \eta^{\mu\nu}\bar{\psi}\big(i\gamma^{\rho}\partial_{\rho}-m\big)\psi. \tag{19.150} $$ Taking the divergence yields $\partial_{\mu}\Theta^{\mu\nu}\neq0$. The last term is zero because it is simply the Lagrangian, which vanishes when imposing the equations of motion. The first term ($\propto \gamma^{\mu}$) also vanishes by simply employing the equations of motion. The second term ($\propto \gamma^{\nu}$) however, does not vanish. Instead, one can show that the tensor $$ \tilde{\Theta}^{\mu\nu} = \frac{i}{4}\Big[\bar{\psi}\gamma^{\mu}\partial^{\nu}\psi+\bar{\psi}\gamma^{\nu}\partial^{\mu}\psi - (\partial^\nu\bar{\psi})\gamma^{\mu}\psi -(\partial^{\mu}\bar{\psi})\gamma^{\nu}\psi \Big]- \eta^{\mu\nu}\bar{\psi}\big(i\gamma^{\rho}\partial_{\rho}-m\big)\psi, $$ is conserved by using the equations of motion.

Hence, my question is: did Peskin and Schroeder make a mistake or am I missing something here?

[I know that similar questions have been asked, for instance here and here, but they don't help me solve this problem]

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    $\begingroup$ What are the "three terms" you are talking about (I only see two), and why would the Lagrangian vanish when imposing the equations of motion? $\delta L = 0$ on-shell, not $L=0$! $\endgroup$
    – ACuriousMind
    Oct 31, 2017 at 13:39
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    $\begingroup$ @ACuriousMind The Lagrangian is first order so it vanishes upon enforcing the equations of motion. Or am I mistaken? $\endgroup$
    – Funzies
    Oct 31, 2017 at 13:52
  • $\begingroup$ Ah, you're right. physics.stackexchange.com/q/348085/50583 might be relevant. $\endgroup$
    – ACuriousMind
    Oct 31, 2017 at 14:00

1 Answer 1


The symmetric SEM tensor for a Dirac fermion is indeed not P&S's guess (19.150) but OP's last formula. The latter agrees e.g. with Ref. 1 and formula (10) in my Phys.SE answer here.


  1. A. Bandyopadhyay, Improvement of the Stress-Energy Tensor using Spacetime symmetries, PhD thesis (2001); eq. (3.20).
  • $\begingroup$ Yes, textbooks discuss the symmetrization of the SEM tensor by the Belinfante procedure only for the em field (equivalently would be to derive it from the em-gravity coupling, see Dirac's booklet on GR), not for the spinorial ones. $\endgroup$
    – DanielC
    Oct 31, 2017 at 16:11
  • $\begingroup$ @Qmechanic Thanks for the reference. I am happy to hear that indeed there was something wrong with the SEM tensor as presented in Peskin and Schroeder. However, in research articles I very often encounter it in the form of Peskin and Schroeder (that's why I was asking this question in the first place). For instance, here: arxiv.org/abs/1303.6266 (equation 7). So I guess in some calculations it does not matter which one to take? Do you know perhaps why a lot of authors use this "wrong" SEM tensor? $\endgroup$
    – Funzies
    Nov 1, 2017 at 8:51

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