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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)

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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

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  • $\begingroup$ 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
    Commented 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
    Commented 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. $\endgroup$
    – Hans de Vries
    Commented 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 $\endgroup$
    – Hans de Vries
    Commented 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
    Commented Nov 5, 2011 at 6:28

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