The dirac current is $$J^\mu = \bar{\psi}\gamma^\mu \psi $$

It looks weird at first because there is no derivative in the expression. So the velocity must be hidden somewhere in either $\gamma$ or $\psi$.

  1. (argument for $\gamma$) From Gordon decomposition, we get $$\bar{u}(p)\gamma^\mu u(p) = \bar{u}(p)\frac{p^\mu}m u(p)$$ Which is reassuring because it is roughly in the form of rho * velocity. It is tempting to treat $\gamma^\mu$ as the "operator" for velocity from this context. More 'justification' of this: an operator which mixes between the left & right handed spinor component can generate translation because one component is the derivative of the other (yeah this is very sloppy).

  2. (argument for $\psi$) Now, if I examine the usual amplitude in e.g. unpolarized elastic electron-electron scattering $$i\mathcal{M(ee\rightarrow ee)}= \frac{ie^2}{q^2}\bar{u}(p')\gamma^\mu u(p)\bar{u}(p)\gamma^\nu u(p') \propto \frac1{q^2}J^\mu J^\nu$$ It turns out that all the momentum terms in the final expression originate from the spin sum (u) $$\sum_s u^s(p)\bar{u}^s(p)= \ \not \!\!\!\!\! p + m$$ while all the momentum term coming from Gordon decomposition on the $\gamma$'s will be contracted out into m's after taking the trace of $\mathcal{M}$.

Back to my question: which one should I call velocity? What about the other?

I just want to keep track the meaning of each term (because recently I feel guilty of blindly computing the traces of my matrices). My question is messy because I'm confused.

hidden question: what is the physical role of $\gamma^\mu$? (nvm, just ignore this one)


1 Answer 1


The velocity gets into the spinor via the boost operator. At rest $\psi_L$ and $\psi_R$ are equal. After a boost they are multiplied by.

$\psi_L ~\rightarrow~ \Lambda\psi_L ~~=~~ \exp\big\{-\eta\cdot\frac{\sigma}{2}\big\} $

$\psi_R ~\rightarrow~ \Lambda\psi_R ~~=~~ \exp\big\{+\eta\cdot\frac{\sigma}{2}\big\} $

So the momentum is indeed doubly "encoded" in the Dirac field, via the spatial derivatives as well as via the spinor values.

The physical role of the $\gamma^\mu$ and why they can be used to extract the momentum is understood by the eigenfunctions of the Pauli matrices $\sigma^i$. For instance $\sigma^x$ has as eigenvectors the spinors pointing in the positive x-direction and the negative x-direction. The first has an eigenvalue +1 and the second has an eigenvalue of -1.

In the rest frame we have:

$\left(x^\uparrow\right)^* \sigma^x \left(x^\uparrow\right) ~~=~~ +1$

$\left(x^\downarrow\right)^* \sigma^x \left(x^\downarrow\right) ~~=~~ -1$

After a boost in the x-direction and combining the two chiral components you will get.

$\exp\big\{+\eta^x\big\}-\exp\big\{-\eta^x\big\} ~~=~~ 2\sinh\big\{\eta^x\big\}$

Which is (proportional to) the momentum.

Regards, Hans

  • $\begingroup$ thanks. Your answer suggests that the velocity term comes from the field itself (something like $\sqrt{\sigma\cdot p}$ ). I need to confirm this one: so is it true that $\gamma^\mu$ is used mainly to denote the spin change on the vertex, and that the extra momentum factor ($\sinh \eta$) is just a side effect? $\endgroup$
    – pcr
    Nov 5, 2011 at 4:34
  • $\begingroup$ I probably need to be more specific here. "Side effect" in a sense that the $\sinh \eta$ averages roughly to $m$ when we average over all the direction of the light polarization $\endgroup$
    – pcr
    Nov 5, 2011 at 4:51
  • $\begingroup$ Via the vertex $\bar{u}_1\,\gamma^\mu\,u_3$ is the interference (transition) current part of $\overline{(\psi_1+\psi_3)}\gamma^\mu(\psi_1+\psi_3)$ This current is the source field of the emitted/absorbed photon. $\endgroup$ Nov 5, 2011 at 4:53
  • $\begingroup$ To see how you get at the $\sqrt{p\cdot\sigma}$ in P&S from the boost operator you can have a look at section 16.16 from my book: physics-quest.org/Book_Chapter_Dirac.pdf $\endgroup$ Nov 5, 2011 at 5:02
  • $\begingroup$ Thanks for the book =), okay I'm actually fine with $\sqrt{p\cdot \sigma}$. Yeah I'm more concerned with the overall scattering process now: if I dissect $\bar{u} \gamma^\mu u$ carefully (I hope), the $\; \not \! \! \! \! p +m$ from the u's will contribute to the dynamical (motion of electron) portion, while the $(p^\mu + p'^\mu) +i\sigma^{\mu\nu}q_{\nu}$ from the $\gamma$'s will contribute to the spin change portion. So we also use $p^\mu$ to describe spin change in this case. $\endgroup$
    – pcr
    Nov 5, 2011 at 6:28

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.