# Proving that free-particle Lagrangian is not invariant under $SU(3)$ local gauge invariance?

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..

• Hi MathIsFun. Res. recom. qs are restricted under Phys.SE. Why not just ask your physics question directly? Feb 7, 2021 at 20:48
• Hi Qmechanics, alright, thanks for the edit. Maybe the tag for specific resources should be edited? :)
– user248824
Feb 7, 2021 at 21:02
• 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.
– Gold
Feb 8, 2021 at 13:24

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

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

Therefore,

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

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.

• 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..
– user248824
Feb 8, 2021 at 13:40