Where is the velocity term in Dirac current hiding? 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 $\psi$.


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*(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).

*(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)
 A: 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
