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In mathematical physics, the Gordon-decomposition of the Dirac current is a splitting of the charge or  particle-number current into a part that arises from the motion of the center of mass motion of the  particles and a part that arises from gradients  of the spin density. It makes explicit use of the of the Dirac equation and so it applies only to "on-shell" solutions of the Dirac equation.

Statement
For any solution $$\psi$$  of the  massive Dirac equation
 * $$ i\gamma^\mu (\nabla_\mu-m)\psi=0,$$

the four-current can be expressed as
 * $$\bar\psi \gamma^\mu\psi =\frac{i}{2m} (\bar \psi \nabla^\mu\psi -(\nabla^\mu\bar \psi) \psi)+\frac{1}{m} \partial_\nu(\bar\psi \Sigma^{\mu\nu}\psi),$$

where
 * $$\Sigma^{\mu\nu} = \frac {i}{4} [\gamma^\mu,\gamma^\nu]$$

is the spinor generator of Lorentz transformations.

This decomposition of the current into a particle number-flux and bound spin contribution requires $$m\ne 0$$. A version that is valid in both massive and massless cases assumes only that that the solution has energy $$E$$ so that $$\psi({\mathbf r},t)=\psi({\mathbf r})\exp\{-iEt\}$$. We find

{\mathbf j}\equiv  e\bar \psi  {\boldsymbol \gamma} \psi = \frac{e}{2iE} \left(\psi^\dagger \nabla  \psi - (\nabla \psi^\dagger)\psi\right) +\frac{e}{E} (\nabla \times{\mathbf  S}). $$ Here $${\boldsymbol \gamma}= (\gamma^1,\gamma^2,\gamma^3)$$, and $$ {\mathbf S} =\psi^\dagger \hat {\mathbf S}\psi $$ with $$ (\hat S_x,\hat S_y,\hat S_z)= (\Sigma^{23},\Sigma^{31},\Sigma^{12}) $$ so that

\hat {\mathbf S}=\frac 12 \left[\begin{matrix}{\boldsymbol \sigma}&0 \\ 0 &{\boldsymbol \sigma}\end{matrix}\right]. $$ With the particle-number density identified with $$\rho= \psi^\dagger\psi$$, and for a near plane-wave solution of finite extent, we can  interpret the first term in the decomposition as the  current $${\mathbf j}_{\rm free}= e\rho {\mathbf k}/E= e\rho {\mathbf v}$$ due to particles moving at speed $${\mathbf v}={\mathbf k}/E$$. The second term, $${\mathbf j}_{\rm bound}= (e/E)\nabla\times {\mathbf S}$$ is the current due to the gradients in the intrinsic magnetic moment density. The magnetic  moment itself is found by integrating by parts to show that

{\boldsymbol \mu}\stackrel{\rm }{=} \frac{1}{2}\int {\mathbf r}\times {\mathbf j}_{\rm bound}\,d^3x =\frac{1}{2}\int {\mathbf r}\times \left(\frac e E\nabla \times {\mathbf S}\right)\,d^3 x = \frac{e}{E}\int {\mathbf S}\,d^3 x. $$ For a single   massless particle obeying the right-handed Weyl equation the spin-1/2  is locked to the direction  $$\hat {\mathbf k}$$ of its  kinetic momentum. In this case

{\boldsymbol \mu}_{\rm Weyl}= \frac{e \hat {\mathbf k}}{2E}. $$ For the both massive and massless case we also have an expression for the momentum density as part of the symmetric  Belinfante-Rosenfeld stress-energy tensor

T^{\mu\nu}_{\rm BR}= \frac{i}{4}(\bar \psi \gamma^\mu \nabla^\nu \psi - (\nabla^\nu \bar\psi) \gamma^\mu\psi +\bar \psi \gamma^\nu \nabla^\mu \psi-(\nabla^\mu \bar\psi) \gamma^\nu\psi). $$ Using the Dirac equation we can evaluate $$T^{0\mu}_{\rm BR}=({\mathcal E},{\mathbf P})$$ to find the energy density to be $${\mathcal E}=E\psi^\dagger \psi$$, and the momentum density to be given by

{\mathbf P}= \frac 1{2i}\left (\psi^\dagger (\nabla \psi)- (\nabla \psi^\dagger)\psi\right) +\frac 12 \nabla\times {\mathbf S}. $$ If we used the non-symmetric canonical energy-momentum tensor

T^{\mu\nu}_{\rm canonical}= \frac{i}{2}(\bar \psi \gamma^\mu \nabla^\nu \psi - (\nabla^\nu \bar\psi) \gamma^\mu\psi), $$ we would not  find  the bound spin-momentum contribution. By an integration by parts we  find that  the spin contribution to the total angular momentum  is

\int {\mathbf r}\times\left(\frac 12 \nabla\times {\mathbf S}\right)\,d^3x = \int {\mathbf S}\, d^3x. $$ This is what is expected, so the  division by 2 in the spin contribution  to the momentum density is necessary. The absence of a division by 2 in the formula for the current  reflects the $$g=2$$ gyromagnetic ratio of the electron. In other words, a spin-density  gradient is twice as effective at making an electric current as it is at contributing to the  linear momentum.

Spin in Maxwells' equations
Motivated by the Riemann-Siberstein vector form  of Maxwell's equations, Michael Berry uses the  Gordon strategy  to determine the intrinsic spin angular momrntum for solutions to Maxwell's equations.

Assume $${\mathbf E}={\mathbf   E}({\mathbf  r})e^{-i\omega t}$$,  $${\mathbf  H}={\mathbf   H}({\mathbf  r})e^{-i\omega t}$$ so that the time average  momentum desity is given by

<\mathbf P>=\frac 1{4c^2} [{\mathbf  E}^*\times   {\mathbf   H}+ {\mathbf   E}\times   {\mathbf   H}^*] $$

= \frac{\epsilon_0}{4i\omega }[{\mathbf  E}^*\cdot(\nabla) {\mathbf E}- (\nabla ){\mathbf   E}^*\cdot{\mathbf   E} +\nabla\times({\mathbf   E}^*\times {\mathbf  E})] $$

= \frac{\mu_0}{4i \omega }[{\mathbf  H}^*\cdot(\nabla) {\mathbf   H}- (\nabla ){\mathbf   H}^*\cdot{\mathbf   H} + \nabla\times({\mathbf  H}^*\times {\mathbf   H})]. $$ As

{\mathbf  P}_{\rm tot}= {\mathbf   P}_{\rm free}+ {\mathbf   P}_{\rm bound}, $$ and for a fluid with instrinsic angular momentum density $$S$$ we have

{\mathbf  P}_{\rm bound}= \frac 12 \nabla\times {\mathbf   S}, $$ these identities suggest that the spin density can be idenfied as either

{\mathbf  S}= \frac{\mu_0}{2i \omega }{\mathbf   H}^*\times {\mathbf   H} $$ or

{\mathbf  S}= \frac{\epsilon_0}{2i \omega }{\mathbf   E}^*\times {\mathbf   E}. $$ The two decompositions coincide the field is pure helicity state in which  $${\mathbf   E}=i\sigma c {\mathbf   B}$$. Here the helicity $$\sigma$$  takes the values $$\pm 1$$ for light that is right or left circularly polarized light respectively.