Suppose I have an Lagrangian $$\mathcal{L} = \frac{1}{2}g_{ab} \bar{\psi}^a \Gamma^k \partial_k \psi^b $$ and I want to show it's invariance under the infinitesimal Lorentz transformations $$\delta \psi^a = -\Lambda_{mn}x^m \partial^n \psi^a + \frac{1}{2} \Lambda_{mn} \Sigma^{mn}\psi^a ,$$ $$\delta \bar{\psi}^a = -\Lambda_{mn}x^m \partial^n \bar{\psi}^a - \frac{1}{2} \Lambda_{mn} \bar{\psi}^a \Sigma^{mn},$$ where $\Lambda_{mn}$ are the components of an infinitesimal Lorentz transformation and hence antisymmetric, and $\Sigma^{mn}$ are the generators of the spinor representation of the Lorentz group. I proceed the usual way and after some algebra get that $$\delta \mathcal{L} = -\frac{1}{2}g_{ab}\Lambda_{mn}[x^m \partial^n \bar{\psi}^a \Gamma^k \partial_k \psi^b + x^m \bar{\psi}^a \Gamma^k \partial_k(\partial^n \psi^b) + \bar{\psi}^a \Gamma^m \partial^n \psi^b ].$$ This needs to be written as a total derivative, but I can't seem to achieve this. For example if I try $$\partial^m (\Lambda_{mn} x^n \mathcal{L}),$$ I get the first two, but not the third term. Can anyone tell me how to proceed?

  • $\begingroup$ In your algebra did you include the transformation of the covariant derivative? $\partial_\mu \rightarrow \Lambda_\mu^\nu \partial_\nu$ $\endgroup$ Commented May 8, 2015 at 14:21
  • $\begingroup$ Hi SCFT. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented May 8, 2015 at 15:39

1 Answer 1


It was pointed out by @Peter Anderson in the comment that you forgot the transformation of the derivative, which in infinitesimal form should read $$\delta \partial_n = - g^{lm} \Lambda_{mn}\partial_l$$ which comes from the Lorentz transformation $$\partial_n \to g^{lm}(L^{-1})_{mn} \partial_l$$ (the metric is there to keep the indices in agreement with OP's choice) which expands to $$ g^{lm}(L^{-1})_{mn} = \delta^l_n - g^{lm} \Lambda_{mn} + ... $$ where I'm inferring, from your other transformation laws, that you are applying active Lorentz transformations, i.e. under the symbolic perspective $$x \to L x$$ with a field transforming as $$\phi(x) \to M \phi(L^{-1} x)$$ being $M$ a representation of the Lorentz group.

If you use this you will get a new term in the variation of your Lagrangian $$ \delta \mathcal{L} -\frac{1}{2} g_{ab} \Lambda_{mn} \overline \psi^a \Gamma^n g^{lm}\partial_l \psi^b$$ and this new term with the last term of your variation hold $$-\frac{1}{2} g_{ab} \Lambda_{mn}\left[ \overline \psi^a \Gamma^m \partial^n \psi^b+\overline \psi^a \Gamma^n \partial^m \psi^b \right]$$ and so you have a contraction between an anti-symmetric tensor $\Lambda_{mn}$ with a symmetrised quantity $\Gamma^m \partial^n+\Gamma^n \partial^m$ and hence these two terms vanish. You are then left with the first two, which you already said you can rewrite them in a total derivative form.

  • $\begingroup$ Thanks a lot for pointing that out. I was taking for granted that $\delta$ commutes with $\partial_{m}$, but it doesn't in this case. $\endgroup$
    – SCFT
    Commented May 11, 2015 at 11:55
  • $\begingroup$ Glad it helped :) and yes, in a nutshell in this case it does not commute as the derivative is not a Lorentz scalar. $\endgroup$
    – romanovzky
    Commented May 11, 2015 at 18:33

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