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This is probably a very basic question, but I'm not sure I understood correctly if a current, like \begin{equation} \bar{\psi}\gamma^{\mu\nu}\psi \end{equation} with $\gamma^{\mu\nu}=\frac{1}{2}[\gamma^{\mu},\gamma^{\nu}]$, or even something simpler like QED's \begin{equation} \bar{\psi}\gamma^{\mu}\psi \end{equation} is a Minkowski 4-vector or if it is a completly different object (because, after all, it deals with Dirac spinors which transform in a different way than 4-vectors).

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    $\begingroup$ It is a vector. Honestly, you should check this yourself. You know how Dirac spinors transform, right? $\endgroup$ – Prahar Apr 1 at 13:59
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This is indeed a Lorentz vector. We are used to seeing a 4-vector which is a representation of group $SO(3,1)$(i.e. the Lorentz group). But we can write the same group locally as $SO(3,1)\sim SU(2)\times SU(2)$. Now the spinors $\psi$ are spin-$1/2$ representations of $SU(2)$ group. Thus we can say the left and right spinor basicallt are the $(1/2,0)$ and $(0,1/2)$ representations of the total symmetry group. Thus their combination $\bar{\psi}\gamma^{\mu}\psi$ can be shown to be a $(1/2,1/2)$ representation of $SU(2)\times SU(2)$.

The next bit of the proof is already done here.

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This are what are called bilinear covariants. Any object of the form $$\bar\psi\Gamma\psi$$ where $\Gamma = \{1, \gamma^\mu, \sigma^{\mu\nu}, \gamma_5, \gamma^\mu\gamma_5\}$, transform in a different way. Note that these are $16$ linearly independent $4\times 4$ matrices, this means that the furnish a basis for all the Clifford algebra. Now you can see easily that this quantities transform, under a Lorentz transformation, in the following way $$ \begin{align} &\bar\psi 1\psi \to \text{ Scalar}\\ &\bar\psi\gamma^\mu\psi\to \text{ Vector}\\ &\bar\psi\sigma^{\mu\nu}\psi \to \text{ Rank } 2\text{ antisymmetric tensor}\\ &\bar\psi\gamma_5\psi\to\text{ Pseudoscalar}\\ &\bar\psi\gamma^\mu\gamma_5\psi\to\text{ Axial vector} \end{align} $$ Let's see how this can be proven: take first the scalar $$\bar\psi^\prime\psi^\prime = \bar\psi S^{-1}S\psi = \bar\psi\psi$$ then the vector $$\bar\psi^\prime\gamma^\mu\psi^\prime = \bar\psi S^{-1}\gamma^\mu S\psi = \Lambda^\mu_{\;\nu}\bar\psi\gamma^\nu\psi$$ the tensor, just by taking one of the two products in the commutator, $$\bar\psi^\prime\gamma^\mu\gamma^\nu\psi^\prime = \bar\psi S^{-1}\gamma^\mu\underbrace{SS^{-1}}_{1}S\psi = \Lambda^{\mu}_{\;\sigma}\Lambda^{\nu}_{\;\rho}\bar\psi\gamma^\rho\gamma^\sigma\psi$$ and for the others i encourage you to try and find yourself the transformation rule to see if they transform as I said, it's a little bit more involved. You should use the following formulas $$\gamma_5 = \frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma$$ and $$\epsilon_{\mu\nu\rho\sigma}\Lambda^\mu_{\;\alpha}\Lambda^\nu_{\;\beta}\Lambda^{\rho}_{\;\eta}\Lambda^\sigma_{\;\delta} = \det(\Lambda)\epsilon_{\alpha\beta\eta\delta}$$

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