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HandWiki. Gordon Decomposition. Encyclopedia. Available online: https://encyclopedia.pub/entry/31407 (accessed on 15 April 2024).
HandWiki. Gordon Decomposition. Encyclopedia. Available at: https://encyclopedia.pub/entry/31407. Accessed April 15, 2024.
HandWiki. "Gordon Decomposition" Encyclopedia, https://encyclopedia.pub/entry/31407 (accessed April 15, 2024).
HandWiki. (2022, October 26). Gordon Decomposition. In Encyclopedia. https://encyclopedia.pub/entry/31407
HandWiki. "Gordon Decomposition." Encyclopedia. Web. 26 October, 2022.
Gordon Decomposition

In mathematical physics, the Gordon-decomposition (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.

gordon-decomposition klein-gordon particle-number

## 1. 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 $\displaystyle{ j^\mu=\bar\psi \gamma^\mu\psi }$ can 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

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

### 1.1. Proof

You can see that from Dirac equation,

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

and from conjugation of Dirac 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, 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,

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

which is just Gordon decomposition after some algebra.

## 2. Massless Generalization

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

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

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

$\displaystyle{ \hat {\mathbf S}=\frac 12 \left[\begin{matrix}{\boldsymbol \sigma}&0 \\ 0 &{\boldsymbol \sigma}\end{matrix}\right], }$

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, we can 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

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

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

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

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[1]

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

## 3. Angular Momentum Density

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

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

Using the Dirac equation we can evaluate $\displaystyle{ T^{0\mu}_{\rm BR}=({\mathcal E},{\mathbf P}) }$ to find the energy density to be $\displaystyle{ {\mathcal E}=E\psi^\dagger \psi }$, and the momentum density to be given by

$\displaystyle{ {\mathbf P}= \frac 1{2i}\left (\psi^\dagger (\nabla \psi)- (\nabla \psi^\dagger)\psi\right) +\frac 12 \nabla\times {\mathbf S}. }$

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

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

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

$\displaystyle{ \int {\mathbf r}\times\left(\frac 12 \nabla\times {\mathbf S}\right)\,d^3x = \int {\mathbf S}\, d^3x. }$

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 $\displaystyle{ g=2 }$ 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.

## 4. Spin in Maxwell's Equations

Motivated by the Riemann-Silberstein vector form of Maxwell's equations, Michael Berry[2] 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

$\displaystyle{ \lt \mathbf P\gt =\frac 1{4c^2} [{\mathbf E}^*\times {\mathbf H}+ {\mathbf E}\times {\mathbf H}^*] }$
$\displaystyle{ = \frac{\epsilon_0}{4i\omega }[{\mathbf E}^*\cdot(\nabla {\mathbf E})- (\nabla {\mathbf E}^*)\cdot{\mathbf E} +\nabla\times({\mathbf E}^*\times {\mathbf E})] }$
$\displaystyle{ = \frac{\mu_0}{4i \omega }[{\mathbf H}^*\cdot(\nabla {\mathbf H})- (\nabla {\mathbf H}^*)\cdot{\mathbf H} + \nabla\times({\mathbf H}^*\times {\mathbf H})]. }$

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 instrinsic angular momentum density $\displaystyle{ {\mathbf S} }$ we have

$\displaystyle{ {\mathbf P}_{\rm bound}= \frac 12 \nabla\times {\mathbf S}, }$

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. D.T.Son, N.Yamamoto (2013). "Kinetic theory with Berry curvature from quantum field theories". Physical Review D 87: 085016. doi:10.1103/PhysRevD.87.085016. Bibcode: 2013PhRvD..87h5016S.  https://dx.doi.org/10.1103%2FPhysRevD.87.085016
2. M.V.Berry (2009). "Optical currents". J. Opt. A 11: 094001 (12 pages). doi:10.1088/1464-4258/11/9/094001. Bibcode: 2009JOptA..11i4001B.  https://dx.doi.org/10.1088%2F1464-4258%2F11%2F9%2F094001
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