Gordon decomposition

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

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

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]

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

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.

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.