wavefunction - Interpretation of Dirac equation states

In Pauli theory the components of two-component wavefunction were interpreted as probability amplitudes of finding the particle in particular spin state. This seems easy to understand.

But when talking about Dirac equation, we have four-component wavefunction, two of which correspond to usual spin components of Pauli electron, and another two... How do I interpret positron-related components of Dirac electron? Are they probability amplitudes for the particle to appear to be positron? Or maybe to appear to not be positron (taking Dirac sea picture into account)?

Answer

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.