In mathematical physics, the Gordon decomposition[1] (named after Walter Gordon) 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
of the massive Dirac equation,
![{\displaystyle (i\gamma ^{\mu }\nabla _{\mu }-m)\psi =0,}](//wikimedia.org/api/rest_v1/media/math/render/svg/0fab99f10a0a3372dd46edfa3a562386afcebcfb)
the Lorentz covariant number-current
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 ),}](//wikimedia.org/api/rest_v1/media/math/render/svg/44bd33df73dc84680a653d9b43940b954e8687e3)
where
![{\displaystyle \Sigma ^{\mu \nu }={\frac {i}{4}}[\gamma ^{\mu },\gamma ^{\nu }]}](//wikimedia.org/api/rest_v1/media/math/render/svg/63135dc1f1ce0f6b128874cfa5c0ef8c045aa92f)
is the spinor generator of Lorentz transformations,
and
![{\displaystyle {\bar {\psi }}=\psi ^{\dagger }\gamma ^{0}}](//wikimedia.org/api/rest_v1/media/math/render/svg/dafa5ba70b83e82cde041f28d40f2d367580457f)
is the Dirac adjoint.
The corresponding momentum-space version for plane wave solutions
and
obeying
![{\displaystyle (\gamma ^{\mu }p_{\mu }-m)u(p)=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/0a4fb32dc6de89f0f45383f2d2e058ca32854ec4)
![{\displaystyle {\bar {u}}(p')(\gamma ^{\mu }p'_{\mu }-m)=0,}](//wikimedia.org/api/rest_v1/media/math/render/svg/a3caba650e524e28b50d96e73592fce0f64724e8)
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)~,}](//wikimedia.org/api/rest_v1/media/math/render/svg/a628617f1ecfb33300eddb708f1e0b3ac894aec2)
where
![{\displaystyle \sigma ^{\mu \nu }=2\Sigma ^{\mu \nu }.}](//wikimedia.org/api/rest_v1/media/math/render/svg/6820f1578a6fda49490e0d5b950b4ec734cc9118)
Proof
One sees that from Dirac's equation that
![{\displaystyle {\bar {\psi }}\gamma ^{\mu }(m\psi )={\bar {\psi }}\gamma ^{\mu }(i\gamma ^{\nu }\nabla _{\nu }\psi )}](//wikimedia.org/api/rest_v1/media/math/render/svg/3c95f6b216b6282c892368f28206f862831ace42)
and, from the adjoint of Dirac's equation,
![{\displaystyle ({\bar {\psi }}m)\gamma ^{\mu }\psi =((\nabla _{\nu }{\bar {\psi }})(-i\gamma ^{\nu }))\gamma ^{\mu }\psi .}](//wikimedia.org/api/rest_v1/media/math/render/svg/f0ce5a6822ce6b07289756f96cc814a4ac76c618)
Adding these two equations yields
![{\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 ).}](//wikimedia.org/api/rest_v1/media/math/render/svg/c6b6ef9d023fe78ed5088bd7a95a2ca07b542445)
From Dirac algebra, one may show that Dirac matrices satisfy
![{\displaystyle \gamma ^{\mu }\gamma ^{\nu }=\eta ^{\mu \nu }-i\sigma ^{\mu \nu }=\eta ^{\nu \mu }+i\sigma ^{\nu \mu }.}](//wikimedia.org/api/rest_v1/media/math/render/svg/74afdebdb28ee34fac1d110b6d1ea20c7810e5e9)
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 ),}](//wikimedia.org/api/rest_v1/media/math/render/svg/27546a20409b743d429f426e64f7636cb156eb6d)
which amounts to just the Gordon decomposition, after some algebra.
Utility
The second, spin-dependent, part of the current coupled to the photon field,
yields, up to an ignorable total divergence,
![{\displaystyle -{\frac {e\hbar }{2mc}}\partial _{\nu }A_{\mu }{\bar {\psi }}\sigma ^{\nu \mu }\psi =-{\frac {e\hbar }{2mc}}{\tfrac {1}{2}}F_{\mu \nu }{\bar {\psi }}\sigma ^{\mu \nu }\psi ,}](//wikimedia.org/api/rest_v1/media/math/render/svg/2a037d59442882d2753f1db83fba03c68a3bae49)
that is, an effective Pauli moment term,
.
This decomposition of the current into a particle number-flux (first term) and bound spin contribution (second term) requires
.
If one assumed that the given solution has energy
so that
, 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} }).}](//wikimedia.org/api/rest_v1/media/math/render/svg/d19b07d0e367b825ee8156fbd311d37be7fb24ac)
Here
, and
with
so that
![{\displaystyle {\hat {\mathbf {S} }}={\frac {1}{2}}\left[{\begin{matrix}{\boldsymbol {\sigma }}&0\\0&{\boldsymbol {\sigma }}\end{matrix}}\right],}](//wikimedia.org/api/rest_v1/media/math/render/svg/58fcd582263a2f61e4e77bc0f883b9ec7af4a03e)
where
is the vector of Pauli matrices.
With the particle-number density identified with
, and for a near plane-wave
solution of finite extent, one may interpret the first term in the decomposition as the current
, due to particles moving at speed
.
The second term,
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~.}](//wikimedia.org/api/rest_v1/media/math/render/svg/d6df420c681728827df368c0b6676ee1a2c2c64a)
For a single massive particle in its rest frame, where
, the magnetic moment reduces to
![{\displaystyle {\boldsymbol {\mu }}_{\rm {Dirac}}=\left({\frac {e}{m}}\right){\mathbf {S} }=\left({\frac {eg}{2m}}\right){\mathbf {S} }.}](//wikimedia.org/api/rest_v1/media/math/render/svg/22b64c9d1f368b839eafa74e203ffc966b256b7c)
where
and
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
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}}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/352361c968af847ac7b2e59be655dc553005ddc3)
For both the 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 ).}](//wikimedia.org/api/rest_v1/media/math/render/svg/eb03cd967f1743c60f65f988e55fbdc83cb305ad)
Using the Dirac equation one may evaluate
to find the energy density to be
, 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} }.}](//wikimedia.org/api/rest_v1/media/math/render/svg/1da5bc0c406eb0f73c0d352bfdc648d2514296fb)
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 ),}](//wikimedia.org/api/rest_v1/media/math/render/svg/3d9739b6f53e5b7cff7f35afc721392127d5f39e)
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.}](//wikimedia.org/api/rest_v1/media/math/render/svg/35430b25eaa1c2e43598a6390aab89f4e7a2ce56)
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
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.