Question
A "spinless" electron can interact with $A^\mu$ only via its charge; the coupling involves $\left(p_f+p_t\right)^\mu$. Show that$$\bar{u}_f \gamma^\mu u_i=\frac{1}{2 m} \bar{u}_f\left(\left(p_f+p_i\right)^\mu+i \sigma^{\mu \nu}\left(p_f-p_i\right)_v\right) u_i,$$from which it is possible to establish that the physical spin $-\frac{1}{2}$ electron interacts via both its charge and its magnetic moment; see also Exercise 6.2. Equation (6.7) is known as the Gordon decomposition of the current.
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$$ In terms of the spinor solutions $u(p)$ and $\bar{u}(p)$, this translates to: $$ (\not p - m) u(p) = 0 \quad \text{and} \quad \bar{u}(p) (\not p - m) = 0, $$ where $\not p = \gamma^\mu p_\mu$. Show more…
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