[Following the derivation presented here (pp. 498-504)]

Each plane wave spinor is a solution of the Dirac equation, \begin{align} \left(\gamma_\mu \frac{\partial}{\partial x_\mu}+m\right)\psi=0 \quad : \quad \gamma_0 = \begin{pmatrix} 1 &0 \\ 0 & -1 \end{pmatrix}, \ \ \gamma_k = \begin{pmatrix}0 & -i\sigma_k \\ i\sigma_k & 0 \end{pmatrix}, \end{align} and takes the form \begin{align} \psi(x)=u(p)e^{ip\cdot x}, \end{align} where $u(p)$ is a bispinor with components \begin{align} u^{(1)} = \sqrt{\frac{\text{E} + m}{2m }} \begin{pmatrix} 1\\ 0 \\ \frac{\text{p}_z}{\text{E} + m}\\ \frac{\text{p}_x+i\text{p}_y}{\text{E} + m}\end{pmatrix},& \quad \quad u^{(2)} = \sqrt{\frac{\text{E} + m}{2m }} \begin{pmatrix} 0\\ 1 \\ \frac{\text{p}_x-i\text{p}_y}{\text{E} + m}\\ \frac{-\text{p}_z}{\text{E} + m}\end{pmatrix}, \\ u^{(3)} = \sqrt{\frac{-\text{E} + m}{2m }} \begin{pmatrix} \frac{-\text{p}_z}{-\text{E} + m}\\ \frac{-(\text{p}_x+i\text{p}_y)}{-\text{E} + m} \\ 1\\ 0\end{pmatrix},& \quad \quad u^{(4)} = \sqrt{\frac{-\text{E} + m}{2m }} \begin{pmatrix} \frac{-(\text{p}_x-i\text{p}_y)}{-\text{E} + m}\\ \frac{\text{p}_z}{-\text{E} + m} \\ 0\\ 1 \end{pmatrix}. \end{align} The solutions can be further understood in terms of the conserved probability current, \begin{align} j_\mu = i \bar \psi \gamma_\mu \psi, \end{align} where $\bar \psi = \psi^\dagger \gamma_0$. Written in terms of $\psi = (A_1 \ A_2 \ B_1 \ B_2)^T$, we have \begin{align} j_\mu &= i\begin{pmatrix} A_1^* & A_2^* & -\!B_1^* & -\!B_2^*\end{pmatrix}\gamma_\mu \begin{pmatrix} A_1 \\ A_2 \\ B_1 \\ B_2 \end{pmatrix}\\ &= \begin{pmatrix} i[A_1^*A_1 + A_2A_2^* + B_1^*B_1 + B_2B_2^*]\\ A_1^*B_2 + A_1B_2^* + A_2^*B_1 + A_2B_1^* \\ -i[A_1^*B_2 - A_1B_2^* - A_2^*B_1 + A_2B_1^*] \\ A_1^*B_1 + A_1B_1^* - A_2^*B_2 - A_2B_2^* \end{pmatrix}. \end{align} If we compare the probability currents of the particle and anti-particle plane waves, we find that \begin{align} \begin{cases} j_\mu^{(1)} = j_\mu^{(2)} = \frac{1}{|E|}\begin{pmatrix} i\text{E},& \text{p}_x, & \text{p}_y, & \text{p}_z \end{pmatrix}, \\ j_\mu^{(3)} = j_\mu^{(4)} = \frac{1}{|E|}\begin{pmatrix} i\text{E},& -\text{p}_x, & -\text{p}_y, & -\text{p}_z \end{pmatrix}, \end{cases} \end{align} where the $1/|E|$ is a normalization factor.

Can anyone provide an explanation for, or reveal any deeper significance to, the temporal component of $j$ having an imaginary value? As this is a four-vector, the components can mix under Lorentz transformations. This appears to imply that the flow of probability is complex-valued in general. Or am I missing something?


1 Answer 1


All the gamma matrices you quote obey $(\gamma^\mu)^2=1$, so the author of the notes seems to be using the obsolete $x^4=ict$ convention. In this metric convention the Lorentz group is obscured by making it look like 0(4) and the time components of physical quantities are complex. A modern version would have no factor of $i$ in the current and no $i$'s multiplying the $\sigma$'s in the other gamma matrices. Whose notes are these? I'd get a better set.

Edit: I just checked, and yes the author of the notes states on page 46 that he/she uses the $(x,y,z,ict)$ convention that gives and illusion of mathematical simplicity at the expense of obscuring the physics. This convention has not been used in field theory since the 1950's.

  • $\begingroup$ Doesn't the factor of $i$ in the gamma components and the $i$ in the current simply change the sign of the resulting solutions as they are multiplied together in the end, no? $\endgroup$ Oct 14, 2022 at 12:54
  • $\begingroup$ I understand now from your edit. Thank you. $\endgroup$ Oct 14, 2022 at 12:57
  • $\begingroup$ No -- because there is a factor of $i$ between the space and time components. Dropping the overallk factor of $i$ in the current would make the $0$ component real, but then the space componts would be complex. You need to drop all the explicit "i"'s to get the usual "west coast" Bjorken and Drell metric. $\endgroup$
    – mike stone
    Oct 14, 2022 at 12:57
  • 1
    $\begingroup$ @MasterDrifter To reinforce the answer, these sorts of subpar resources often show up high in search results just because they've been on the internet for a long time (20 years in this case). They should almost always be avoided -- get a solid set of modern notes or a real book, and you'll find everything much easier. $\endgroup$
    – knzhou
    Oct 14, 2022 at 17:46

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