The interpretation of the Dirac equation states depend on what representation you choose for your $\gamma^\mu$-matrices or your $\alpha_i$ and $\beta$-matrices depending on what you prefer. Both are linked via $\gamma^\mu=(\beta,\beta\vec{\alpha})$. Choosing your representation will (more or less) fix your basis in which you consider the solutions to your equation (choosing another representation will rotate your entire solution).
The representation that I will choose is the Dirac-Pauli representation, given by:
$$\beta=\left(\begin{array}{c c}\mathbb{I}_{2\times2}&0\\0&-\mathbb{I}_{2\times2}\end{array}\right) \quad\text{and}\quad \alpha^i=\left(\begin{array}{c c}0&\sigma^i\\\sigma^i&0\end{array}\right),$$
where $\sigma^i$ are the Pauli-matrices.
If you would solve the Dirac-equation in this representation, you will find 4 independent solutions:
$$
\psi_1(x)=N_1\left(\begin{array}{c}1\\0\\\frac{p_z}{E+m}\\\frac{p_x+ip_y}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu)
$$
$$
\psi_2(x)=N_2\left(\begin{array}{c}0\\1\\\frac{p_x-ip_y}{E+m}\\\frac{-p_z}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu)
$$
$$
\psi_3(x)=N_3\left(\begin{array}{c}\frac{p_z}{E-m}\\\frac{p_x+ip_y}{E-m}\\1\\0\end{array}\right)\exp(ip_\mu x^\mu)
$$
$$
\psi_4(x)=N_4\left(\begin{array}{c}\frac{p_x-ip_y}{E-m}\\\frac{-p_z}{E-m}\\0\\1\end{array}\right)\exp(ip_\mu x^\mu)
$$
The way to interpret these states is to look at them in the rest-frame, so the frame in which they stand still $p^\mu=(E,0,0,0)$, the states will become simply the following:
$$\psi_1=N_1\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-iEt}, \psi_2=N_2\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)e^{-iEt}, \psi_3=N_3\left(\begin{array}{c}0\\0\\1\\0\end{array}\right)e^{iEt}\text{ and } \psi_4=N_4\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)e^{iEt},$$
by inspection of the time-evolution of the phase factor we can already see that $\psi_1$ and $\psi_2$ represent positive energy states (particles) and the $\psi_3$ and $\psi_4$ represent negative energy states (so anti-particles).
In order to know the spin you should use the helicity-operator, given by:
$$\sigma_p=\frac{\hat{\vec{p}}\cdot \hat{S}}{|\vec{p}|},$$
In the case of the Dirac-equation the spin operator is given by the double Pauli-matrix:
$$\hat{S}=\frac{1}{2}\left(\begin{array}{cc}\vec{\sigma}&0\\0&\vec{\sigma}\end{array}\right),$$
if we let this one work on the spinors $\psi_1$, $\psi_2$, $\psi_3$ and $\psi_4$, we find that their spin is respectively up, down, up, down. So looking at electrons the Dirac-spinor can be interpreted in the Pauli-Dirac representation as (for example for the electron):
$$
\psi=\left(\begin{array}{c}e^-\uparrow\\e^-\downarrow\\e^+\uparrow\\e^+\downarrow\end{array}\right).
$$
When the momentum is NOT equal to zero these different states mix up and you can't make such a simple identification. Usually one says that the electron becomes a mixture of an electron with positrons when it starts moving.
