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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
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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 [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]

is

[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]

1.1. 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.

2. 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[1]

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

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

[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.

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 [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].

As

[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]

or

[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. 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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