## Apr 26, 2013

### Gordon Identity

Gordon Identity (Gordon Decomposition) says the vector coupling can be decomposed as a current coupling and a dipole moment coupling.

$\bar{u}_{s'}(p') \gamma^\mu u_s(p) = \frac{1}{2 m}\bar{u}_{s'}(p') \left[ (p'+p)^\mu + i \sigma^{\mu\nu}(p'-p)_\nu \right] u_s(p),\qquad (1)$
where $\sigma^{\mu\nu} = \frac{i}{4}\left[ \gamma^\mu, \gamma^\nu \right]$. The spinor $u_s(p)$ is normalized as $\sum_{s=\pm} u_s(p)\bar{u}_s(p) = {p\kern-0.45em/} + m.$ Gordon Identity provides an interesting perspective of the electron spin. The lhs of (1) is the fermion current. When coupling with a vector gauge field, the first term on the rhs $\propto p \cdot A$ gives the canonical potential of a charged particle in $A$; the second term on the rhs $\propto p \wedge A$ gives a dipole interaction. Compare with the Pauli equation that describes a non-relativistic charge-field interaction plus a magnetic dipole-field interaction: $i\hbar\frac{\partial}{\partial t}\left.|\psi\right> = \left( \frac{1}{2m} (-i\hbar\mathbf{\nabla}-e\mathbf{A})^2 + e\phi - \frac{e\hbar}{2m}\mathbf{\sigma}\cdot\mathbf{B}\right) \left.|\psi\right>$ But in Dirac equation as well as in QED, the two interaction is united as a consequence of selection rule of Lorentz symmetry. (We can also have charged particles without magnetic dipole moment or spin. Therefore, the answer for why electron has a spin is that it just has!) The value of the electron magnetic dipole moment is obtained from the eigen-values of $\sigma^{\mu\nu}$ (say $\sigma^{12}$ which gives to $m_z\cdot B_z$) in the low energy limit. They turns out to be $\pm\frac{1}{2}$.