Solution of Dirac equation-Positive and Negative energy For particles defined with positive energy, we use $$\phi=
\begin{pmatrix}
       1  \\
        0  \\
        \end{pmatrix}
 $$ or $$
\begin{pmatrix}
       0  \\
        1  \\
        \end{pmatrix}
$$
while for particles defined with negative energy, we use this instead
$$\chi=
\begin{pmatrix}
       1  \\
        0  \\
        \end{pmatrix}
 $$ or $$
\begin{pmatrix}
       0  \\
        1  \\
        \end{pmatrix}
$$
where ∅ is the upper component while $\chi$ is the lower component of the bispinor in Dirac equation.
Can we do it the other way round or
$$\phi_p=
\begin{pmatrix}
       1  \\
        0  \\
        0  \\
        0  \\
        \end{pmatrix}
 ....$$ instead? I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.
 A: In the Dirac equation:
$$(\beta mc^2+\sum_i \alpha_i p_i)\psi = i\hbar \frac{\partial\psi}{\partial t}$$
there are four 4x4 matrix operators, which can be defined as follows in terms of 2x2 blocks:
$$\alpha_i=\begin{bmatrix} 0 & \sigma_i \\ \sigma_i & 0\end{bmatrix}$$
$$\beta=\begin{bmatrix} I & 0 \\ 0 & -I\end{bmatrix}$$
where the $\sigma_i$ are the standard Pauli matrices, $I$ is the 2x2 identity matrix, and 0 is the 2x2 zero matrix. The fact that the operators in the Dirac equation naturally split into 2x2 blocks should suggest that splitting the 4x1 bispinor $\psi$ into two 2x1 spinors would also behave nicely. Indeed, as developed in your source, it does.
A: I don't understand is why is the 4-spinor split into two components, one for the positive energy and the other for the negative energy. It did not say anything here.
You are misinterpreting the large and small components of the Dirac 4-spinor for the positive and negative energy solutions (a common mistake).  Each Dirac spinor (whether a positive or negative energy solution) has four components.  For a positive energy solution the upper 2-spinor is called the large component because the lower 2-spinor is much smaller (of order $\frac {v}{c}$ in comparison, $v$ is the average velocity of the particle described). The same applies to the negative energy solutions except that the relative size of the upper and lower components is reversed.
Your link describes the situation correctly as does the answer by @probably_someone.  
