# Gordon decomposition

In mathematical physics, the Gordon decomposition[1] (named after Walter Gordon one of the discoverers of the Klein–Gordon equation) 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 ${\displaystyle \psi }$ of the massive Dirac equation,

${\displaystyle (i\gamma ^{\mu }\nabla _{\mu }-m)\psi =0,}$

the Lorentz covariant number-current ${\displaystyle j^{\mu }={\bar {\psi }}\gamma ^{\mu }\psi }$ may be expressed as

${\displaystyle {\bar {\psi }}\gamma ^{\mu }\psi ={\frac {i}{2m}}({\bar {\psi }}\nabla ^{\mu }\psi -(\nabla ^{\mu }{\bar {\psi }})\psi )+{\frac {1}{m}}\partial _{\nu }({\bar {\psi }}\Sigma ^{\mu \nu }\psi ),}$

where

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

is the spinor generator of Lorentz transformations.

The corresponding momentum-space version for plane wave solutions ${\displaystyle u(p)}$ and ${\displaystyle {\bar {u}}(p')}$ obeying

${\displaystyle (\gamma ^{\mu }p_{\mu }-m)u(p)=0}$
${\displaystyle {\bar {u}}(p')(\gamma ^{\mu }p'_{\mu }-m)=0,}$

is

${\displaystyle {\bar {u}}(p')\gamma ^{\mu }u(p)={\bar {u}}(p')\left[{\frac {(p+p')^{\mu }}{2m}}+i\sigma ^{\mu \nu }{\frac {(p'-p)_{\nu }}{2m}}\right]u(p)~,}$

where

${\displaystyle \sigma ^{\mu \nu }=2\Sigma ^{\mu \nu }.}$

### Proof

One sees that from Dirac's equation that

${\displaystyle {\bar {\psi }}\gamma ^{\mu }(m\psi )={\bar {\psi }}\gamma ^{\mu }(i\gamma ^{\nu }\nabla _{\nu }\psi )}$

and, from the conjugate of Dirac's equation,

${\displaystyle ({\bar {\psi }}m)\gamma ^{\mu }\psi =((\nabla _{\nu }{\bar {\psi }})(-i\gamma ^{\nu }))\gamma ^{\mu }\psi .}$

${\displaystyle {\bar {\psi }}\gamma ^{\mu }\psi ={\frac {i}{2m}}({\bar {\psi }}\gamma ^{\mu }\gamma ^{\nu }\nabla _{\nu }\psi -(\nabla _{\nu }{\bar {\psi }})\gamma ^{\nu }\gamma ^{\mu }\psi ).}$

From Dirac algebra, one may show that Dirac matrices satisfy

${\displaystyle \gamma ^{\mu }\gamma ^{\nu }=\eta ^{\mu \nu }-i\sigma ^{\mu \nu }=\eta ^{\nu \mu }+i\sigma ^{\nu \mu }.}$

Using this relation,

${\displaystyle {\bar {\psi }}\gamma ^{\mu }\psi ={\frac {i}{2m}}({\bar {\psi }}(\eta ^{\mu \nu }-i\sigma ^{\mu \nu })\nabla _{\nu }\psi -(\nabla _{\nu }{\bar {\psi }})(\eta ^{\mu \nu }+i\sigma ^{\mu \nu })\psi ),}$

which amounts to just the Gordon decomposition, after some algebra.

### Utility

The second, spin-dependent, part of the current coupled to the photon field, ${\displaystyle -A_{\mu }j^{\mu }}$ yields, up to an ignorable total divergence,

${\displaystyle -{\frac {e\hbar }{2mc}}\partial _{\nu }A_{\mu }{\bar {\psi }}\sigma ^{\nu \mu }\psi =-{\frac {e\hbar }{2mc}}{\tfrac {1}{2}}F_{\mu \nu }{\bar {\psi }}\sigma ^{\mu \nu }\psi ,}$

that is, an effective Pauli moment term, ${\displaystyle -(e\hbar /2mc){\vec {B}}\cdot \psi ^{\dagger }{\vec {\sigma }}\psi }$.

## Massless generalization

This decomposition of the current into a particle number-flux (first term) and bound spin contribution (second term) requires ${\displaystyle m\neq 0}$.

If one assumed that the given solution has energy ${\displaystyle E={\sqrt {|{\mathbf {k} }|^{2}+m^{2}}}}$ so that ${\displaystyle \psi ({\mathbf {r} },t)=\psi ({\mathbf {r} })\exp\{-iEt\}}$, one might obtain a decomposition that is valid for both massive and massless cases[2] .

Using the Dirac equation again, one finds that

${\displaystyle {\mathbf {j} }\equiv e{\bar {\psi }}{\boldsymbol {\gamma }}\psi ={\frac {e}{2iE}}\left(\psi ^{\dagger }\nabla \psi -(\nabla \psi ^{\dagger })\psi \right)+{\frac {e}{E}}(\nabla \times {\mathbf {S} }).}$

Here ${\displaystyle {\boldsymbol {\gamma }}=(\gamma ^{1},\gamma ^{2},\gamma ^{3})}$, and ${\displaystyle {\mathbf {S} }=\psi ^{\dagger }{\hat {\mathbf {S} }}\psi }$ with ${\displaystyle ({\hat {S}}_{x},{\hat {S}}_{y},{\hat {S}}_{z})=(\Sigma ^{23},\Sigma ^{31},\Sigma ^{12}),}$ so that

${\displaystyle {\hat {\mathbf {S} }}={\frac {1}{2}}\left[{\begin{matrix}{\boldsymbol {\sigma }}&0\\0&{\boldsymbol {\sigma }}\end{matrix}}\right],}$

where ${\displaystyle {\boldsymbol {\sigma }}=(\sigma _{x},\sigma _{y},\sigma _{z})}$ is the vector of Pauli matrices.

With the particle-number density identified with ${\displaystyle \rho =\psi ^{\dagger }\psi }$, and for a near plane-wave solution of finite extent, one may interpret the first term in the decomposition as the current ${\displaystyle {\mathbf {j} }_{\rm {free}}=e\rho {\mathbf {k} }/E=e\rho {\mathbf {v} }}$, due to particles moving at speed ${\displaystyle {\mathbf {v} }={\mathbf {k} }/E}$.

The second term, ${\displaystyle {\mathbf {j} }_{\rm {bound}}=(e/E)\nabla \times {\mathbf {S} }}$ 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

${\displaystyle {\boldsymbol {\mu }}{\stackrel {\rm {}}{=}}{\frac {1}{2}}\int {\mathbf {r} }\times {\mathbf {j} }_{\rm {bound}}\,d^{3}x={\frac {1}{2}}\int {\mathbf {r} }\times \left({\frac {e}{E}}\nabla \times {\mathbf {S} }\right)\,d^{3}x={\frac {e}{E}}\int {\mathbf {S} }\,d^{3}x~.}$

For a single massive particle in its rest frame, where ${\displaystyle E=m}$, the magnetic moment reduces to

${\displaystyle {\boldsymbol {\mu }}_{\rm {Dirac}}=\left({\frac {e}{m}}\right){\mathbf {S} }=\left({\frac {eg}{2m}}\right){\mathbf {S} }.}$

where ${\displaystyle |{\mathbf {S} }|=\hbar /2}$ and ${\displaystyle g=2}$ 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 ${\displaystyle {\hat {\mathbf {k} }}}$ of its kinetic momentum and the magnetic moment becomes[3]

${\displaystyle {\boldsymbol {\mu }}_{\rm {Weyl}}=\left({\frac {e}{E}}\right){\frac {\hbar {\hat {\mathbf {k} }}}{2}}.}$

## Angular momentum density

For the both massive and massless cases, one also has an expression for the momentum density as part of the symmetric Belinfante–Rosenfeld stress–energy tensor

${\displaystyle T_{\rm {BR}}^{\mu \nu }={\frac {i}{4}}({\bar {\psi }}\gamma ^{\mu }\nabla ^{\nu }\psi -(\nabla ^{\nu }{\bar {\psi }})\gamma ^{\mu }\psi +{\bar {\psi }}\gamma ^{\nu }\nabla ^{\mu }\psi -(\nabla ^{\mu }{\bar {\psi }})\gamma ^{\nu }\psi ).}$

Using the Dirac equation one may evaluate ${\displaystyle T_{\rm {BR}}^{0\mu }=({\mathcal {E}},{\mathbf {P} })}$ to find the energy density to be ${\displaystyle {\mathcal {E}}=E\psi ^{\dagger }\psi }$, and the momentum density,

${\displaystyle {\mathbf {P} }={\frac {1}{2i}}\left(\psi ^{\dagger }(\nabla \psi )-(\nabla \psi ^{\dagger })\psi \right)+{\frac {1}{2}}\nabla \times {\mathbf {S} }.}$

If one used the non-symmetric canonical energy-momentum tensor

${\displaystyle T_{\rm {canonical}}^{\mu \nu }={\frac {i}{2}}({\bar {\psi }}\gamma ^{\mu }\nabla ^{\nu }\psi -(\nabla ^{\nu }{\bar {\psi }})\gamma ^{\mu }\psi ),}$

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

${\displaystyle \int {\mathbf {r} }\times \left({\frac {1}{2}}\nabla \times {\mathbf {S} }\right)\,d^{3}x=\int {\mathbf {S} }\,d^{3}x.}$

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 ${\displaystyle g=2}$ 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[4] 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 ${\displaystyle {\mathbf {E} }={\mathbf {E} }({\mathbf {r} })e^{-i\omega t}}$, ${\displaystyle {\mathbf {H} }={\mathbf {H} }({\mathbf {r} })e^{-i\omega t}}$. The time average of the Poynting vector momentum density is then given by

${\displaystyle \langle \mathbf {P} \rangle ={\frac {1}{4c^{2}}}[{\mathbf {E} }^{*}\times {\mathbf {H} }+{\mathbf {E} }\times {\mathbf {H} }^{*}]}$
${\displaystyle ={\frac {\epsilon _{0}}{4i\omega }}[{\mathbf {E} }^{*}\cdot (\nabla {\mathbf {E} })-(\nabla {\mathbf {E} }^{*})\cdot {\mathbf {E} }+\nabla \times ({\mathbf {E} }^{*}\times {\mathbf {E} })]}$
${\displaystyle ={\frac {\mu _{0}}{4i\omega }}[{\mathbf {H} }^{*}\cdot (\nabla {\mathbf {H} })-(\nabla {\mathbf {H} }^{*})\cdot {\mathbf {H} }+\nabla \times ({\mathbf {H} }^{*}\times {\mathbf {H} })].}$

We have used Maxwell's equations in passing from the first to the second and third lines, and in expression such as ${\displaystyle {\mathbf {H} }^{*}\cdot (\nabla {\mathbf {H} })}$ the scalar product is between the fields so that the vector character is determined by the ${\displaystyle \nabla }$.

As

${\displaystyle {\mathbf {P} }_{\rm {tot}}={\mathbf {P} }_{\rm {free}}+{\mathbf {P} }_{\rm {bound}},}$

and for a fluid with intrinsic angular momentum density ${\displaystyle {\mathbf {S} }}$ we have

${\displaystyle {\mathbf {P} }_{\rm {bound}}={\frac {1}{2}}\nabla \times {\mathbf {S} },}$

these identities suggest that the spin density can be identified as either

${\displaystyle {\mathbf {S} }={\frac {\mu _{0}}{2i\omega }}{\mathbf {H} }^{*}\times {\mathbf {H} }}$

or

${\displaystyle {\mathbf {S} }={\frac {\epsilon _{0}}{2i\omega }}{\mathbf {E} }^{*}\times {\mathbf {E} }.}$

The two decompositions coincide when the field is paraxial. They also coincide when the field is a pure helicity state – i.e. when ${\displaystyle {\mathbf {E} }=i\sigma c{\mathbf {B} }}$ where the helicity ${\displaystyle \sigma }$ takes the values ${\displaystyle \pm 1}$ for light that is right or left circularly polarized respectively. In other cases they may differ.

## References

1. ^ W. Gordon (1928). "Der Strom der Diracschen Elektronentheorie". Z. Phys. 50: 630–632. Bibcode:1928ZPhy...50..630G. doi:10.1007/BF01327881.
2. ^ M.Stone (2015). "Berry phase and anomalous velocity of Weyl fermions and Maxwell photons". International Journal of Modern Physics B. 30: 1550249. arXiv:1507.01807. doi:10.1142/S0217979215502495.
3. ^ D.T.Son, N.Yamamoto (2013). "Kinetic theory with Berry curvature from quantum field theories". Physical Review D. 87: 085016. arXiv:1210.8158. Bibcode:2013PhRvD..87h5016S. doi:10.1103/PhysRevD.87.085016.
4. ^ M.V.Berry (2009). "Optical currents". J. Opt. A. 11: 094001 (12 pages). Bibcode:2009JOptA..11i4001B. doi:10.1088/1464-4258/11/9/094001.