# 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?

-
4 components means two spinors $\phi$ and $\chi$. You can obtain one from the other. See en.wikipedia.org/wiki/Dirac_spinor – jinawee Dec 29 '13 at 9:27

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.
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 $\xi^{A}$.