Wavefunction of an electron Electron is a spin $\frac{1}{2}$ particle, so needs 2-component wave function but the Dirac theory of electron is based on 4-component wave function, are all components of it independent or only two of them are independent? 
 A: You have discovered the fact that the Dirac spinors form a reducible representation of Spin(3,1) $\simeq$ SL(2,C), the covering group of SO(3,1)$^+$. The left and right Weyl spinors, which have two components, are irreducible representations.
A: The 4-component Dirac spinor $\psi$ is formed by stacking two two-component Lorentz spinors $\xi^{A},\eta^{\dot{B}}$. The Dirac equation is the pair of coupled equations,
$$
\hat{p}^{A}_{\dot{B}}\eta^{\dot{B}}=m\xi^{A}
$$
$$
\hat{p}^{\dot{A}}_{B}\xi^{B}=m\eta^{\dot{A}}
$$
where $p^{A}_{\dot{B}}$ is equivalent to the relativistic momentum operator $\hat{p}_{\mu}=i\partial/\partial x^{\mu}$. The spinor form of the momentum operator can be written,
$$
\hat{p}^{A}_{\dot{B}}=2i\frac{\partial}{\partial X^{\dot{B}}_{A}}
$$
and the Hermitian tensor $X^{\dot{A}}_{B}$ is equivalent to the spacetime point $x^{\mu}$. Now try the plane wave ansatz,
$$
\xi^{A}=u^{A}\exp(iK^{C}_{\dot{D}}X^{\dot{D}}_{C}/2)
$$
$$
\eta^{\dot{A}}=v^{\dot{A}}\exp(iK^{C}_{\dot{D}}X^{\dot{D}}_{C}/2)
$$
where $u$ and $v$ are constant spinors. Substituting in the Dirac equation,
$$
mv^{\dot{A}}=-u^{B}K^{\dot{A}}_{B}
$$
shows that (in this case) the $v$ spinor is fixed by the choice of the $u$ spinor: only two of the four components of the Dirac spinor are free. The wave number is found by putting the last equation into the Dirac equation,
$$
m^{2}v^{\dot{A}}=v^{\dot{C}}K^{\dot{A}}_{B}K^{B}_{\dot{C}}=v^{\dot{C}}k_{\mu}k^{\mu}\delta^{\dot{A}}_{\dot{C}}=v^{\dot{A}}k_{\mu}k^{\mu} \ .
$$
So, the two components $\eta^{\dot{A}}$ are fixed by the choice of the $\xi^{A}$.
