It is definitely Lorentz invariant, and it is certainly describing spin-1/2 particles by virtue of the Lorentz transformation of the field.
It also describes 4 spin-0 particles!
On the level of field equations, there is no problem. One can show that
<br />
\{i\gamma^{\mu}_{ij}\partial_{\mu}\psi_{j}=0\}\ \ \Leftrightarrow \ \ \{\partial_{\mu}\partial^{\mu}\psi_{i}=0\}<br />
This “equivalence” though does not extend to the corresponding Lagrangians:
1) all Noether currents derived from your Lagrangian don’t agree with those derived from the Dirac Lagrangian. For example, the U(1) Dirac current is
J^{\mu}=\bar{\psi}_{i}\gamma^{\mu}_{ij}\psi_{j}
while your Lagrangian, leads to
j^{\mu}=\bar{\psi}_{i}\partial^{\mu}\psi_{i}
This current, when coupled to potential A_{\mu}, describes SCALAR QED.
2) the dimension of the Dirac field is [3/2] in mass unit. The field in your Lagrangian is [1] mass unit which is (again) the dimension of a boson field.
On the representation level, we proceed as follow
1) We have the two fundamental rep.
\phi^{a}\sim (1/2,0), \ \ \psi_{\dot{c}}\sim (0,1/2)
2) we have the operator \partial_{\mu} and the matrices \sigma^{\mu}
3) for massive particles, we also have the mass of the representation m
4) we form the vector rep.
(1/2,0)\otimes (0,1/2) = (1/2,1/2) \sim \partial^{a\dot{c}}= (\sigma_{\mu}\partial^{\mu})^{a\dot{c}}
Clearly, this can be translated into a pair of coupled differential equations
\partial_{a\dot{c}}\phi^{a} = m \psi_{\dot{c}},
\partial^{a\dot{c}}\psi_{\dot{c}}=-m\phi^{a}
5) in order to form the bispinor of Dirac, we first extend SL(2,C) with parity operation then we put the above two equations together. This leads to the Dirac equation.
Notice again that each component satisfies the K-G equation
(\partial_{\mu}\partial^{\mu}+m^{2}) \left( \begin{array}{c} \phi^{a} \\ \psi_{\dot{c}} \\ \end{array}\right)=0
