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.

For any solution ${\displaystyle \psi }$ of the massive Dirac equation

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

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

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

where

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

is the spinor generator of Lorentz transformations.

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

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

${\displaystyle \overline{u}(p^{\prime })({\gamma }^{\mu }p{\text{'}}_{\mu }-m)=0,}$

is

${\displaystyle \overline{u}(p^{\prime }){\gamma }^{\mu }u(p)=\overline{u}(p^{\prime })[\frac{(p+p^{\prime }{)}^{\mu }}{2m}+i{\sigma }^{\mu \nu }\frac{(p^{\prime }-p{)}_{\nu }}{2m}]u(p)}$

where

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

You can see that from Dirac equation,

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

and from conjugation of Dirac equation

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

Adding two equations yields

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

From Dirac algebra, you can 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,

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

which is just Gordon decomposition after some algebra.

This decomposition of the current into a particle number-flux (first term) and bound spin contribution (second term) requires ${\displaystyle m\ne 0}$. If we assume that the given solution has energy ${\displaystyle E=\sqrt{|\mathbf{𝐤}{|}^2+m^2}}$ so that ${\displaystyle \psi (\mathbf{𝐫},t)=\psi (\mathbf{𝐫})\exp \{-iEt\}}$, we can obtain a decomposition that is valid for both massive and massless cases. Using the Dirac equation again we find that

${\displaystyle \mathbf{𝐣}}\equiv e\overline{\psi }\mathit{\gamma }\psi =\frac{e}{2iE}({\psi }^{†}\mathrm{\nabla }\psi -(\mathrm{\nabla }{\psi }^{†})\psi )+\frac{e}{E}(\mathrm{\nabla }\times \mathbf{𝐒}).$

Here ${\displaystyle \mathit{\gamma }}=({\gamma }^1,{\gamma }^2,{\gamma }^3)$, and ${\displaystyle \mathbf{𝐒}}={\psi }^{†}\widehat{\mathbf{𝐒}}\psi$ with ${\displaystyle ({\hat{S}}_x,{\hat{S}}_y,{\hat{S}}_z)=({\mathrm{\Sigma }}^{23},{\mathrm{\Sigma }}^{31},{\mathrm{\Sigma }}^{12})}$ so that

${\displaystyle \widehat{\mathbf{𝐒}}=\frac{1}{2}\left[\begin{array}{cc}\mathit{\sigma }& 0\\ 0& \mathit{\sigma }\end{array}\right],}$

where ${\displaystyle \mathit{\sigma }}=({\sigma }_x,{\sigma }_y,{\sigma }_z)$ is the vector of Pauli matrices.

With the particle-number density identified with ${\displaystyle \rho ={\psi }^{†}\psi }$, and for a near plane-wave solution of finite extent, we can interpret the first term in the decomposition as the current ${\displaystyle {\mathbf{𝐣}}_{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}=e\rho \mathbf{𝐤}/E=e\rho \mathbf{𝐯}}$ due to particles moving at speed ${\displaystyle \mathbf{𝐯}}=\mathbf{𝐤}/E$. The second term, ${\displaystyle {\mathbf{𝐣}}_{\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}=(e/E)\mathrm{\nabla }\times \mathbf{𝐒}}$ 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

${\displaystyle \mathit{\mu }}\stackrel{}{=}\frac{1}{2}\int \mathbf{𝐫}\times {\mathbf{𝐣}}_{\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}{d}^{3}x=\frac{1}{2}\int \mathbf{𝐫}\times (\frac{e}{E}\mathrm{\nabla }\times \mathbf{𝐒})d^{3}x=\frac{e}{E}\int \mathbf{𝐒}{d}^{3}x.$

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

${\displaystyle {\mathit{\mu }}_{\mathrm{D}\mathrm{i}\mathrm{r}\mathrm{a}\mathrm{c}}=\left(\frac{e}{m}\right)\mathbf{𝐒}=\left(\frac{eg}{2m}\right)\mathbf{𝐒}.}$

where ${\displaystyle |\mathbf{𝐒}|=\hslash /2}$ 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 ${\displaystyle \widehat{\mathbf{𝐤}}}$ of its kinetic momentum and the magnetic moment becomes^[2]

${\displaystyle {\mathit{\mu }}_{\mathrm{W}\mathrm{e}\mathrm{y}\mathrm{l}}=\left(\frac{e}{E}\right)\frac{\hslash \widehat{\mathbf{𝐤}}}{2}.}$

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

${\displaystyle T_{\mathrm{B}\mathrm{R}}^{\mu \nu }=\frac{i}{4}(\overline{\psi }{\gamma }^{\mu }{\mathrm{\nabla }}^{\nu }\psi -({\mathrm{\nabla }}^{\nu }\overline{\psi }){\gamma }^{\mu }\psi +\overline{\psi }{\gamma }^{\nu }{\mathrm{\nabla }}^{\mu }\psi -({\mathrm{\nabla }}^{\mu }\overline{\psi }){\gamma }^{\nu }\psi ).}$

Using the Dirac equation we can evaluate ${\displaystyle T_{\mathrm{B}\mathrm{R}}^{0\mu }=({ℰ},\mathbf{𝐏})}$ to find the energy density to be ${\displaystyle {ℰ}}=E{\psi }^{†}\psi$, and the momentum density to be given by

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

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

${\displaystyle T_{\mathrm{c}\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{n}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{l}}^{\mu \nu }=\frac{i}{2}(\overline{\psi }{\gamma }^{\mu }{\mathrm{\nabla }}^{\nu }\psi -({\mathrm{\nabla }}^{\nu }\overline{\psi }){\gamma }^{\mu }\psi ),}$

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

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

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^[3] 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

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

${\displaystyle =\frac{{ϵ}_0}{4i\omega }[{\mathbf{𝐄}}^{*}\cdot (\mathrm{\nabla }\mathbf{𝐄})-(\mathrm{\nabla }{\mathbf{𝐄}}^{*})\cdot \mathbf{𝐄}+\mathrm{\nabla }\times ({\mathbf{𝐄}}^{*}\times \mathbf{𝐄})]}$

${\displaystyle =\frac{{\mu }_0}{4i\omega }[{\mathbf{𝐇}}^{*}\cdot (\mathrm{\nabla }\mathbf{𝐇})-(\mathrm{\nabla }{\mathbf{𝐇}}^{*})\cdot \mathbf{𝐇}+\mathrm{\nabla }\times ({\mathbf{𝐇}}^{*}\times \mathbf{𝐇})].}$

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

As

${\displaystyle {\mathbf{𝐏}}_{\mathrm{t}\mathrm{o}\mathrm{t}}={\mathbf{𝐏}}_{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}}+{\mathbf{𝐏}}_{\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}},}$

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

${\displaystyle {\mathbf{𝐏}}_{\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}=\frac{1}{2}\mathrm{\nabla }\times \mathbf{𝐒},}$

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

${\displaystyle \mathbf{𝐒}}=\frac{{ϵ}_0}{2i\omega }{\mathbf{𝐄}}^{*}\times \mathbf{𝐄}.$

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.

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