I am reading Gauge Theory of Elementary Particle Physics by Tapei Cheng and Lingfong Li. Proceeding equation 5.54, there is a statement which says

Then from Lorentz covariance, we can include the other components of the currents.

One would then have

$$ \left[ Q^a \left( t \right) , J_{\mu}^b \left( \vec{x} , t \right) \right] = \mathrm{i} C^{a b c} J_{\mu}^c \left( \vec{x} , t \right). $$

I am having a hard time trying to figure out how would one be able to derive the equation above from Lorentz covariance and the following

$$ \left[ Q^a \left( t \right) , J_0^b \left( \vec{x} , t \right) \right] = \mathrm{i} C^{a b c} J_0^c \left( \vec{x} , t \right). $$

Can someone give me a hint on this?


If your current is a Lorentz tensor, $$ \Lambda^{0}_{\mu} J_{0}(\vec{x},t) = J'_{\mu}(\vec{x}',t'). $$ and the charge $Q(t)$ is a Lorentz scalar: $$ Q^{'}_a(t') = Q_a (t), $$ we have that $$ [Q^{'a} (t') , J_{\mu}^{'b} ( \vec{x}',t')] = i C^{abc} J^{'c}_{\mu}(\vec{x},t) $$ can be written by a Lorentz transformation as $$ [Q^{a} (t), \Lambda^{0}_{\mu} J_0^{b} (\vec{x} ,t) ] = \Lambda^{0}_{\mu} [Q^{a} (t), J_0^{b} (\vec{x} ,t) ] = i C^{abc} \Lambda^{0}_{\mu} J_0^{c}(\vec{x},t). $$ Therefore, it follows that if $$ [Q^{a} (t), J_0^{b} (\vec{x} ,t) ] = i C^{abc} J_0^{c}(\vec{x},t). $$ is true, then $ [Q^{'a} (t') , J_{\mu}^b ( \vec{x}',t')] = i C^{abc} J^{'c}_{\mu}(\vec{x},t) $ must also be true being derived by linear transformation from the previous one.

Some intuition. The notion of covariance stablishes that there is no prefered direction. In absence of interactions every dimension should appear in equal footing in every equation.


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