# Gordon decomposition

In mathematical physics, the Gordon decomposition (named after Walter Gordon) 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 of the particles and a part that arises from gradients of the spin density. It makes explicit use of the Dirac equation and so it applies only to "on-shell" solutions of the Dirac equation.

## Original statement

For any solution $\psi$ of the massive Dirac equation,
    $(i\gamma ^{\mu }\nabla _{\mu }-m)\psi =0,$
the Lorentz covariant number-current $j^{\mu }={\bar {\psi }}\gamma ^{\mu }\psi$ may 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, and
    ${\bar {\psi }}=\psi ^{\dagger }\gamma ^{0}$
is the Dirac adjoint.
The corresponding momentum-space version for plane wave solutions $u(p)$ and ${\bar {u}}(p')$ obeying
    $(\gamma ^{\mu }p_{\mu }-m)u(p)=0$
    ${\bar {u}}(p')(\gamma ^{\mu }p'_{\mu }-m)=0,$
is
    ${\bar {u}}(p')\gamma ^{\mu }u(p)={\bar {u}}(p')\left[{\frac {(p+p')^{\mu }}{2m}}+i\sigma ^{\mu \nu }{\frac {(p'-p)_{\nu }}{2m}}\right]u(p)~,$
where
    $\sigma ^{\mu \nu }=2\Sigma ^{\mu \nu }.$

### Proof

One sees that from Dirac's equation that
    ${\bar {\psi }}\gamma ^{\mu }(m\psi )={\bar {\psi }}\gamma ^{\mu }(i\gamma ^{\nu }\nabla _{\nu }\psi )$
and, from the adjoint of Dirac's equation,
    $({\bar {\psi }}m)\gamma ^{\mu }\psi =((\nabla _{\nu }{\bar {\psi }})(-i\gamma ^{\nu }))\gamma ^{\mu }\psi .$
Adding these two equations yields
    ${\bar {\psi }}\gamma ^{\mu }\psi ={\frac {i}{2m}}({\bar {\psi }}\gamma ^{\mu }\gamma ^{\nu }\nabla _{\nu }\psi -(\nabla _{\nu }{\bar {\psi }})\gamma ^{\nu }\gamma ^{\mu }\psi ).$
From Dirac algebra, one may show that Dirac matrices satisfy
    $\gamma ^{\mu }\gamma ^{\nu }=\eta ^{\mu \nu }-i\sigma ^{\mu \nu }=\eta ^{\nu \mu }+i\sigma ^{\nu \mu }.$
Using this relation,
    ${\bar {\psi }}\gamma ^{\mu }\psi ={\frac {i}{2m}}({\bar {\psi }}(\eta ^{\mu \nu }-i\sigma ^{\mu \nu })\nabla _{\nu }\psi -(\nabla _{\nu }{\bar {\psi }})(\eta ^{\mu \nu }+i\sigma ^{\mu \nu })\psi ),$
which amounts to just the Gordon decomposition, after some algebra.

### Utility

The second, spin-dependent, part of the current coupled to the photon field, $-A_{\mu }j^{\mu }$ yields, up to an ignorable total divergence,
    $-{\frac {e\hbar }{2mc}}\partial _{\nu }A_{\mu }{\bar {\psi }}\sigma ^{\nu \mu }\psi =-{\frac {e\hbar }{2mc}}{\tfrac {1}{2}}F_{\mu \nu }{\bar {\psi }}\sigma ^{\mu \nu }\psi ,$
that is, an effective Pauli moment term, $-(e\hbar /2mc){\vec {B}}\cdot \psi ^{\dagger }{\vec {\sigma }}\psi$.

## Massless generalization

This decomposition of the current into a particle number-flux (first term) and bound spin contribution (second term) requires $m\neq 0$.
If one assumed that the given solution has energy E=√(|k|$^2$+m$^2$){\textstyle E={\sqrt {|{\mathbf {k} }|^{2}+m^{2}}}} so that ψ(r, t)=ψ(r)e$^{−iEt}${\textstyle \psi (\mathbf {r} , t)=\psi ({\mathbf {r} })e^{-iEt}}, one might obtain a decomposition that is valid for both massive and massless cases.
Using the Dirac equation again, one finds that
    ${\mathbf {j} }\equiv e{\bar {\psi }}{\boldsymbol {\gamma }}\psi ={\frac {e}{2iE}}\left(\psi ^{\dagger }\nabla \psi -(\nabla \psi ^{\dagger })\psi \right)+{\frac {e}{E}}(\nabla \times {\mathbf {S} }).$
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
    ${\hat {\mathbf {S} }}={\frac {1}{2}}\left[{\begin{matrix}{\boldsymbol {\sigma }}&0\\0&{\boldsymbol {\sigma }}\end{matrix}}\right],$
where ${\boldsymbol {\sigma }}=(\sigma _{x},\sigma _{y},\sigma _{z})$ is the vector of Pauli matrices.
With the particle-number density identified with $\rho =\psi ^{\dagger }\psi$, and for a near plane-wave solution of finite extent, one may 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
    ${\boldsymbol {\mu }}{\stackrel {\rm}{=}}{\frac {1}{2}}\int {\mathbf {r} }\times {\mathbf {j} }_{\rm {bound}}\,d^{3}x={\frac {1}{2}}\int {\mathbf {r} }\times \left({\frac {e}{E}}\nabla \times {\mathbf {S} }\right)\,d^{3}x={\frac {e}{E}}\int {\mathbf {S} }\,d^{3}x~.$
For a single massive particle in its rest frame, where $E=m$, the magnetic moment reduces to
    ${\boldsymbol {\mu }}_{\rm {Dirac}}=\left({\frac {e}{m}}\right){\mathbf {S} }=\left({\frac {eg}{2m}}\right){\mathbf {S} }.$
where $|{\mathbf {S} }|=\hbar /2$ and $g=2$ is the Dirac value of the gyromagnetic ratio.
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 and the magnetic moment becomes
    ${\boldsymbol {\mu }}_{\rm {Weyl}}=\left({\frac {e}{E}}\right){\frac {\hbar {\hat {\mathbf {k} }}}{2}}.$

## Angular momentum density

For both the massive and massless cases, one also has an expression for the momentum density as part of the symmetric Belinfante–Rosenfeld stress–energy tensor
    $T_{\rm {BR}}^{\mu \nu }={\frac {i}{4}}({\bar {\psi }}\gamma ^{\mu }\nabla ^{\nu }\psi -(\nabla ^{\nu }{\bar {\psi }})\gamma ^{\mu }\psi +{\bar {\psi }}\gamma ^{\nu }\nabla ^{\mu }\psi -(\nabla ^{\mu }{\bar {\psi }})\gamma ^{\nu }\psi ).$
Using the Dirac equation one may evaluate $T_{\rm {BR}}^{0\mu }=({\mathcal {E}},{\mathbf {P} })$ to find the energy density to be ${\mathcal {E}}=E\psi ^{\dagger }\psi$, and the momentum density,
    ${\mathbf {P} }={\frac {1}{2i}}\left(\psi ^{\dagger }(\nabla \psi )-(\nabla \psi ^{\dagger })\psi \right)+{\frac {1}{2}}\nabla \times {\mathbf {S} }.$
If one used the non-symmetric canonical energy-momentum tensor
    $T_{\rm {canonical}}^{\mu \nu }={\frac {i}{2}}({\bar {\psi }}\gamma ^{\mu }\nabla ^{\nu }\psi -(\nabla ^{\nu }{\bar {\psi }})\gamma ^{\mu }\psi ),$
one would not find the bound spin-momentum contribution.
By an integration by parts one finds that the spin contribution to the total angular momentum is
    $\int {\mathbf {r} }\times \left({\frac {1}{2}}\nabla \times {\mathbf {S} }\right)\,d^{3}x=\int {\mathbf {S} }\,d^{3}x.$
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 Maxwell's equations

Motivated by the Riemann–Silberstein vector form of Maxwell's equations, Michael Berry uses the Gordon strategy to obtain gauge-invariant expressions for the intrinsic spin angular-momentum density for solutions to Maxwell's equations.
He assumes that the solutions are monochromatic and uses the phasor expressions $\mathbf {E} =\mathbf {E} (\mathbf {r} )e^{-i\omega t}$, $\mathbf {H} =\mathbf {H} (\mathbf {r} )e^{-i\omega t}$. The time average of the Poynting vector momentum density is then given by ${\begin{aligned}\langle \mathbf {P} \rangle &={\frac {1}{4c^{2}}}[{\mathbf {E} }^{*}\times {\mathbf {H} }+{\mathbf {E} }\times {\mathbf {H} }^{*}]\\&={\frac {\epsilon _{0}}{4i\omega }}[{\mathbf {E} }^{*}\cdot (\nabla {\mathbf {E} })-(\nabla {\mathbf {E} }^{*})\cdot {\mathbf {E} }+\nabla \times ({\mathbf {E} }^{*}\times {\mathbf {E} })]\\&={\frac {\mu _{0}}{4i\omega }}[{\mathbf {H} }^{*}\cdot (\nabla {\mathbf {H} })-(\nabla {\mathbf {H} }^{*})\cdot {\mathbf {H} }+\nabla \times ({\mathbf {H} }^{*}\times {\mathbf {H} })].\end{aligned}}$ We have used Maxwell's equations in passing from the first to the second and third lines, and in expression such as ${\mathbf {H} }^{*}\cdot (\nabla {\mathbf {H} })$ the scalar product is between the fields so that the vector character is determined by the $\nabla$.
As $\mathbf {P} _{\text{tot}}=\mathbf {P} _{\text{free}}+{\mathbf {P} }_{\text{bound}},$ and for a fluid with intrinsic angular momentum density $\mathbf {S}$ we have $\mathbf {P} _{\text{bound}}={\frac {1}{2}}\nabla \times \mathbf {S} ,$ these identities suggest that the spin density can be identified 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 when the field is paraxial. They also coincide when the field is a pure helicity state – i.e. when ${\mathbf {E} }=i\sigma c{\mathbf {B} }$ where the helicity $\sigma$ takes the values $\pm 1$ for light that is right or left circularly polarized respectively. In other cases they may differ.