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.

## Contents

* 1 Original Statement
* 1.1 Proof
* 2 Massless Generalization
* 3 Angular Momentum Density
* 4 Spin in Maxwell's equations
* 5 References

## Original Statement

For any solution [math]\displaystyle{ \psi }[/math] of the massive Dirac equation

[math]\displaystyle{ (i\gamma^\mu \nabla_\mu-m)\psi=0, }[/math]

the Lorentz covariant number-current [math]\displaystyle{ j^\mu=\bar\psi \gamma^\mu\psi }[/math] can be expressed as

[math]\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), }[/math]

where

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

is the spinor generator of Lorentz transformations.

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

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

[math]\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) }[/math]

where

[math]\displaystyle{ \sigma^{\mu\nu} = 2\Sigma^{\mu\nu}. }[/math]

### Proof

You can see that from Dirac equation,

[math]\displaystyle{ \bar\psi \gamma^\mu (m \psi) = \bar\psi \gamma^\mu (i\gamma^\nu \nabla_\nu\psi) }[/math]

and from conjugation of Dirac equation

[math]\displaystyle{ (\bar\psi m )\gamma^\mu \psi = ((\nabla_\nu \bar\psi)(-i\gamma^\nu))\gamma^\mu\psi }[/math]

Adding two equations yields

[math]\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) }[/math]

From Dirac algebra, you can show that Dirac matrices satisfy

[math]\displaystyle{ \gamma^\mu \gamma^\nu = \eta^{\mu\nu} - i\sigma^{\mu\nu}= \eta^{\nu\mu} + i\sigma^{\nu\mu} }[/math]

Using this relation,

[math]\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) }[/math]

which is just Gordon decomposition after some algebra.

## Massless Generalization

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

[math]\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}). }[/math]

Here [math]\displaystyle{ {\boldsymbol \gamma}= (\gamma^1,\gamma^2,\gamma^3) }[/math], and [math]\displaystyle{ {\mathbf S} =\psi^\dagger \hat {\mathbf S}\psi }[/math] with [math]\displaystyle{ (\hat S_x,\hat S_y,\hat S_z)= (\Sigma^{23},\Sigma^{31},\Sigma^{12}) }[/math] so that

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

where [math]\displaystyle{ {\boldsymbol \sigma}=(\sigma_x,\sigma_y,\sigma_z) }[/math] is the vector of Pauli matrices.

With the particle-number density identified with [math]\displaystyle{ \rho= \psi^\dagger\psi }[/math], and for a near plane-wave solution of finite extent, we can interpret the first term in the decomposition as the current [math]\displaystyle{ {\mathbf j}_{\rm free}= e\rho {\mathbf k}/E= e\rho {\mathbf v} }[/math] due to particles moving at speed [math]\displaystyle{ {\mathbf v}={\mathbf k}/E }[/math]. The second term, [math]\displaystyle{ {\mathbf j}_{\rm bound}= (e/E)\nabla\times {\mathbf S} }[/math] 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

[math]\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. }[/math]

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

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

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

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

## 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

[math]\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). }[/math]

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

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

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

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

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

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

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

[math]\displaystyle{ \lt \mathbf P\gt =\frac 1{4c^2} [{\mathbf E}^*\times {\mathbf H}+ {\mathbf E}\times {\mathbf H}^*] }[/math]
[math]\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})] }[/math]
[math]\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})]. }[/math]

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

[math]\displaystyle{ {\mathbf P}_{\rm tot}= {\mathbf P}_{\rm free}+ {\mathbf P}_{\rm bound}, }[/math]

and for a fluid with instrinsic angular momentum density [math]\displaystyle{ {\mathbf S} }[/math] we have

[math]\displaystyle{ {\mathbf P}_{\rm bound}= \frac 12 \nabla\times {\mathbf S}, }[/math]

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

[math]\displaystyle{ {\mathbf S}= \frac{\mu_0}{2i \omega }{\mathbf H}^*\times {\mathbf H} }[/math]

[math]\displaystyle{ {\mathbf S}= \frac{\epsilon_0}{2i \omega }{\mathbf E}^*\times {\mathbf E}. }[/math]

The two decompositions coincide when the field is paraxial. They also coincide when the field is a pure helicity state 
 i.e. when [math]\displaystyle{ {\mathbf E}=i\sigma c {\mathbf B} }[/math] where the helicity [math]\displaystyle{ \sigma }[/math] takes the values [math]\displaystyle{ \pm 1 }[/math] 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. doi:10.1007/BF01327881. Bibcode: 1928ZPhy...50..630G.
2. ↑ 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.
3. ↑ M.V.Berry (2009). "Optical currents". J. Opt. A 11: 094001 (12 pages). doi:10.1088/1464-4258/11/9/094001. Bibcode: 2009JOptA..11i4001B.

0.00

(0 votes)

|

Retrieved from "https://handwiki.org/wiki/index.php?title=Physics:Gordon_Decomposition&oldid=289375"