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)?


1 Answer 1


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.

  • $\begingroup$ How should a state of $\psi_1+\psi_3$ then be interpreted? Will a measurement find sometimes electron, other times positron? $\endgroup$
    – Ruslan
    Commented May 15, 2014 at 10:55
  • 4
    $\begingroup$ I wouldn't say that the Dirac representation is "the most common". The Chiral (or Weyl) representation $$ \gamma^{\mu} = \left( \begin{array}{c c} 0_2 & \sigma^{\mu} \\ \bar{\sigma}^{\mu} & 0_2 \end{array} \right) $$ where $$ \sigma^{\mu} = (I_2, \sigma^i) \quad \mbox{ and } \quad \bar{\sigma}^{\mu} = (I_2, - \sigma^i) $$ with the $\sigma^i$ the usual Pauli matrices, is also very common. As with the metric convention, it is an unfortunate case that everyone won't agree on. $\endgroup$
    – Flint72
    Commented May 15, 2014 at 14:40
  • 1
    $\begingroup$ @Ruslan, as Flint72 pointed out, I mad a typo, it's indeed the states $\psi_1$ and $\psi_3$ I was talking about. $\endgroup$
    – Nick
    Commented May 15, 2014 at 15:40
  • 1
    $\begingroup$ @Ruslan 1) Bear in mind that $\frac{1}{\sqrt{2}}(\Psi_1+\Psi_3)$ isn't an eigenstate of the Dirac Hamiltonian, since one has energy>0 while the other <0. Their superposition is possible but undefined in the Hamiltonian basis. 2) Sure, $\Psi_{3,4}$ are electrons with energy<0. You may then interpret them as belonging to the Dirac sea, as Dirac did, and say that positrons are holes; or you may look at the conserved current and reinterpret them as positively charged states.In either way, you realize that the Dirac equation does admit energy<0 states for electrons. That was Pauli's concern,indeed. $\endgroup$
    – Wizzerad
    Commented May 17, 2014 at 9:50
  • 2
    $\begingroup$ @Ruslan, indeed in semi-conductors you also have electrons and holes, and if you want you can draw that kind of anology since a hole also acts like a positive charged particle, see also Wizzerad's comment! Because $\psi_1$, $\psi_2$, $\psi_3$ and $\psi_4$ all follow Dirac's equation we know it are particles with mass $m$ (so all the same mass), if you look at the charge you see that the charge of $\psi_1$ and $\psi_2$ is opposite to that of $\psi_3$ and $\psi_4$ (you can do this by symmetry, or by calculating the probability density and interpreting this as a charge distribution as Pauli did). $\endgroup$
    – Nick
    Commented May 17, 2014 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.