The Lagrangian (density) for a free Dirac field is given as $${\mathcal L}_\mathrm{Dirac} = \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi,$$ but given that $\psi$ obeys the Dirac equation, $$\left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi = 0$$ Doesn't this mean the Lagrangian (density) is zero?


1 Answer 1


What you have found out is that the lagrangian evaluated at the solution of the equation of motion is constant (and equal to zero).

However the lagrangian density is defined for a generic field configuration, not just the solution of the equation of motion.

Since eoms are found by stabilising the action $S = \int \mathcal{L}$, considering only the solution of the eoms in place of a generic field configuration is like considering a function $f(x)$ only at its stationary point; in general it is not sufficient because you need to know how the funciton behaves in a whole neighbourhood of such points.

  • $\begingroup$ The Lagrangian (density) for the KG field, ${\mathcal L}_\mathrm{KG} = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) - \frac{1}{2} m^2 \phi^2$, doesn't vanish? $\endgroup$
    – jim
    Jan 1, 2017 at 13:34
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    $\begingroup$ It does, in the sense that your $\mathcal{L}_\text{KG}$ is equivalent to $\mathcal{L}' = -\frac{1}{2} \phi (\Box + m^2)\phi$ up to a 4-divergence. But even if it did not vanish, it would have been just a constant (with respect to the fields), and you can redefine the lagrangian subtracting it so that it vanishes. $\endgroup$
    – user139175
    Jan 1, 2017 at 14:14
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    $\begingroup$ Just to specify, what vanishes is the Lagrangian evaluated in the solution of equation of motions, not for a generic field configuration. But applying the variational principle we are interested in the behaviour of the Lagrangian in a neighbourhood of the solution of the eoms, so evaluating a Lagrangian for that specific point is often not what you want to do. $\endgroup$
    – user139175
    Jan 1, 2017 at 14:23
  • $\begingroup$ @yoric your answer is reeeeally helpful. I don't understand though how, calculating the energy momentum tensor $T_{\mu \nu}$ for a dirac spinor field I can remove the term with the lagrangian (based on the considerations above), while when calculating the one for a scalar field i need to consider it in order to get the right tetramomentum vector $\endgroup$
    – user129511
    Jun 4, 2017 at 15:07

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