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The Noether charge associated with global $SU(3)$ invariance is \begin{equation} J^{\mu}_a=-f_{abc}F_c^{\nu\mu}A_{b\mu}-i\frac{\delta \mathcal{L}_{Matter}}{\delta D_{\nu}\psi}t_a\psi \end{equation} and is conserved along the equations of motion $\partial_{\mu}J^{\mu}_a=0$ therefore giving rise to the conserved charges: \begin{equation} Q^a=\int d^3x\,\, J^0_a \end{equation} However the current is not gauge invariant (nor gauge covariant) so how can we say that Hadrons are colorless? In particular, can one say that the charge operator annihilates Hadrons? \begin{equation} \hat{Q}^a|Hadron\rangle=0 \end{equation}

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$Q^a=Q^a_0 \neq 0$ for fixed $Q^a_0$ is not a gauge-invariant statement $Q^a=0$ is gauge invariant.

In words, being "red" is not a gauge invariant statement, but being "colorless" is.

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