In the Clifford algebra framework, the transformation law of spinor $\psi$ (a rotation along $z$ axis as an example)
$$
\psi \rightarrow
e^{\frac{\theta}{2}\gamma_1\gamma_2}
\psi
$$
is fundamental, while all other transformations are derived.
Accordingly, a vector $V^\mu$ such as fermion current
$$
V^\mu = \bar{\psi}\gamma^\mu\psi =\psi^\dagger\gamma^0\gamma^\mu\psi,
$$
transforms as
\begin{align}
V^\mu &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(\Lambda^\nu_\mu(\theta)\gamma^\mu) \psi \\
&=\Lambda^\nu_\mu(\theta)\psi^\dagger \gamma^0\gamma^\mu \psi \\
&=\Lambda^\nu_\mu(\theta)V^\mu.
\end{align}
Note that
- The fermion current $V^\mu$ is defined as $\bar{\psi}\gamma^\mu\psi= \psi^\dagger\gamma^0\gamma^\mu\psi$ not as $\psi^\dagger\gamma^\mu\psi$. Bear in mind that for Lorentz boost $(e^{\frac{\theta}{2}\gamma_0\gamma_i})^\dagger\gamma^0 = e^{+\frac{\theta}{2}\gamma_0\gamma_i}\gamma^0 = \gamma^0e^{-\frac{\theta}{2}\gamma_0\gamma_i}$. The additional $\gamma^0$ in $\bar{\psi} = \psi^\dagger\gamma^0$ is essential in guaranteeing the Lorentz covariance of the vector $V^\mu$.
- $\Lambda^\nu_\mu(\theta)$ are just real numbers, so that they can be
moved freely around Clifford algebra elements.
And you can also derive that according to the spinor transformation rule, an antisymmetric tensor $T^{\mu\nu}$ such as
$$
T^{\mu\nu} = \bar{\psi}\gamma^\mu\gamma^\nu\psi =\psi^\dagger\gamma^0\gamma^\mu\gamma^\nu\psi,
$$
transforms as
\begin{align}
T^{\mu\nu} &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2} )(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\
&=\psi^\dagger \gamma^0(\Lambda^{\mu'}_\mu(\theta)\gamma^\mu)(\Lambda^{\nu'}_\nu(\theta)\gamma^\nu) \psi \\
&=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)\psi^\dagger \gamma^0\gamma^\mu\gamma^\nu \psi \\
&=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)T^{\mu\nu} .
\end{align}
For related discussions,
- See here for how to embed vectors and tensors in a spinor.
- See here for how to correctly define covariant derivatives
with respect to single-sided vs double-sided gauge transformations.