I would like to show that the free-particle Lagrangian $\mathcal L = \bar{\psi}\left( i\gamma^{\mu}\partial_{\mu} - m\right)\psi$ is not invariant under the $SU(3)$ local gauge invariance transformation $\hat U = \exp\left[ ig_{S}\mathbf{\alpha}(x) \mathbf{\cdot} \mathbf{\hat{T}}\right]$.

I tried it myself, but I am stuck on the way..

  • $\begingroup$ Hi MathIsFun. Res. recom. qs are restricted under Phys.SE. Why not just ask your physics question directly? $\endgroup$
    – Qmechanic
    Feb 7, 2021 at 20:48
  • $\begingroup$ Hi Qmechanics, alright, thanks for the edit. Maybe the tag for specific resources should be edited? :) $\endgroup$
    – user248824
    Feb 7, 2021 at 21:02
  • $\begingroup$ It's very simple really. The transformed $\psi$ will have an extra dependence in $x$ comming from $\hat{U}(x)$, so when the derivative $\partial_\mu$ hits it it produces two terms according to the product rule. The term in which $\hat{U}(x)$ is kept and $\psi$ is differentiated will give you the original Lagrangian, but the other term in which $\hat{U}(x)$ is differentiated will give a non-zero contribution which prevents the transformed and untransformed Lagrangians to be equal. This also hints what must be done if you want the local symmetry: you change the derivative. $\endgroup$
    – Gold
    Feb 8, 2021 at 13:24

1 Answer 1


It all comes down to understanding how each of these objects are defined.

$$\psi \mapsto U \psi$$ $$\overline{\psi} \mapsto \overline{\psi} U^\dagger$$

Note that $U^\dagger U = 1$. If $\alpha(x)$ is a constant, meaning $U$ is a constant matrix, then

$$(i \gamma^\mu \partial_\mu - m) U = U (i \gamma^\mu \partial_\mu - m). $$

This is because $\gamma^\mu$ acts on the spinor indicies while $U$ acts on the color indicies, so they simply commute, and because $U$ is a constant so it doesn't get affected by the derivative $\partial_\mu$.

This means that, if $U$ is constant, then

\begin{align} \overline{\psi} (i \gamma^\mu \partial_\mu - m) \psi &\mapsto \overline{\psi} U^\dagger (i \gamma^\mu \partial_\mu - m) U \psi \\ &=\overline{\psi} U^\dagger U (i \gamma^\mu \partial_\mu - m) \psi \\ &=\overline{\psi} (i \gamma^\mu \partial_\mu - m) \psi. \end{align}

Now, what happens when $\alpha(x)$ isn't a constant? Then

$$ \partial_\mu ( U \psi) = (\partial_\mu U) \psi + U (\partial_\mu \psi).$$


$$(i \gamma^\mu \partial_\mu - m) U = U (i \gamma^\mu \partial_\mu - m) + i \gamma^\mu (\partial_\mu U)$$

which is different from what we had before. This implies that the Lagrangian changes by

\begin{align} \overline{\psi} (i \gamma^\mu \partial_\mu - m) \psi &\mapsto \overline{\psi} U^\dagger (i \gamma^\mu \partial_\mu - m) U \psi \\ &=\overline{\psi} (i \gamma^\mu \partial_\mu - m) \psi + i \overline{\psi} \gamma^\mu (U^\dagger \partial_\mu U) \psi \end{align} and is not invariant due to the last term.

  • 1
    $\begingroup$ Good answer! I also noticed that one can write your last term as follows: $i\bar{\psi}\gamma^{\mu}\left(U^{\dagger}\partial_{\mu}U\right)\psi = i\bar{\psi}\gamma^{\mu}\left( ig_{S}\left( \partial_{\mu}\alpha^{a}(x)\right)T^a \psi\right)$, but it's the same at the end, I guess.. $\endgroup$
    – user248824
    Feb 8, 2021 at 13:40

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