Gordon decomposition of Dirac current for massless electron?

Question

We know Gordon decomposition of Dirac current is applicable only for massive (nonzero mass) Dirac particles. Is there an analog for massless Dirac particles? (I have made an attempt to answer arxiv:physics-1107.4536)

Answer 1

For solutions of the massive Dirac equation the standard Gordon decomposition is $$ \bar\psi \gamma^\mu\psi =\frac{i}{2m} (\bar \psi \partial^\mu \psi -(\partial^\mu \bar \psi) \psi)+\frac{1}{2m} \partial_\nu(\bar\psi \Sigma^{\mu\nu}\psi). $$ What happens when $m=0$? For the both the massive and massless case assume that $\psi({\bf r},t)=\psi({\bf r})\exp\{-iEt\}$ and use the Dirac equation in the form $$ \partial_t\psi = -{\alpha}\cdot \nabla\psi -im\beta \psi, \nonumber\\ $$ $$ \partial_t \bar\psi = +\nabla\bar\psi\cdot {\alpha} +i m\bar\psi \beta,\nonumber $$ with $\beta=\gamma^0$, $\alpha^i=\gamma^0\gamma^i$ to show that $$ {\bf j}= \bar \psi {\gamma} \psi = \frac{1}{2iE} \left(\psi^\dagger \nabla \psi - (\nabla \psi^\dagger)\psi\right) +\frac{1}{E} (\nabla \times{\bf S}). $$ Here $\gamma\equiv (\gamma^1,\gamma^2,\gamma^3)$, and $$ {\bf S} =\psi^\dagger \hat {\bf S}\psi $$ with $$ (\hat S_x,\hat S_y,\hat S_z)= (\Sigma_{23},\Sigma_{31},\Sigma_{12}) $$ and $$ \Sigma^{ij} = \frac{i}{4} [\gamma^i,\gamma^j] , $$ so $$ \hat {\bf S}=\frac 12 \left[\matrix{{\sigma}&0 \cr 0 &{\sigma}}\right]. $$

For the both massive and massless case we also have the Belinfante-Rosenfeld symmetric momentum density $$ T^{\mu\nu}_{\rm BR}= \frac{i}{4}(\bar \psi \gamma^\mu \partial^\nu \psi - (\partial^\nu \bar\psi) \gamma^\mu\psi +\bar \psi \gamma^\nu \partial^\mu \psi-(\partial^\mu \bar\psi) \gamma^\nu\psi). $$ Using the Dirac equation we evaluate $T^{0\mu}_{\rm BR}=({\mathcal E},{\bf P})$ to find ${\mathcal E}=E\psi^\dagger \psi$, and $$ {\bf P}= \frac 1{2i}\left (\psi^\dagger (\nabla \psi)- (\nabla \psi^\dagger)\psi\right) +\frac 12 \nabla\times {\bf S}. $$ If we use the non-symmetric canonical energy-momentum tensor $$ T^{\mu\nu}_{\rm canonical}= \frac{i}{2}(\bar \psi \gamma^\mu \partial^\nu \psi - (\partial^\nu \bar\psi) \gamma^\mu\psi), $$ we do not get the bound spin-momentum.

Note that $$ \frac 12 \int {\bf r}\times(\nabla\times {\bf S})\,d^3x = \int {\bf S}\, d^3x, $$ so the division by 2 in the spin contribution to the momentum density is correct. The lack of such a division in the formula for the current reflects the $g=2$ gyromagnetic ratio of the electron. (I.e. a spin-density gradient is twice as effective at making a current as it is at contributing momentum). We confirm this by computing the spin contribution to the magnetic moment: $$ {\mu}\stackrel{\rm def}{=} \frac{e}{2}\int {\bf r}\times {\bf j}_{\rm spin}\,d^3x =\frac{e}{2E}\int {\bf r}\times (\nabla \times {\bf S})\,d^3 x = \frac{e}{E}\int {\bf S}\,d^3 x . $$

Michael Berry uses the Gordon strategy for solutions to Maxwell's equations: Assume ${\bf E}={\bf E}({\bf r})e^{-i\omega t}$, ${\bf H}={\bf H}({\bf r})e^{-i\omega t}$ so that the time average momentum desity is given by $$ {\bf P}=\frac 1{4c^2} [{\bf E}^*\times {\bf H}+ {\bf E}\times {\bf H}^*] $$ $$ = \frac{\epsilon_0}{4i\omega }[{\bf E}^*\cdot(\nabla) {\bf E}- (\nabla ){\bf E}^*\cdot{\bf E} +\nabla\times({\bf E}^*\times {\bf E})] \nonumber\\ $$ $$ = \frac{\mu_0}{4i \omega }[{\bf H}^*\cdot(\nabla) {\bf H}- (\nabla ){\bf H}^*\cdot{\bf H} +\nabla\times({\bf H}^*\times {\bf H})] \nonumber $$ As $$ {\bf P}_{\rm tot}= {\bf P}_{\rm free}+ {\bf P}_{\rm bound} $$ and for a spinful fluid $$ {\bf P}_{\rm bound}= \frac 12 \nabla\times {\bf S} $$ These identities suggest that the photon spin density is either $$ {\bf S}= \frac{\mu_0}{2i \omega }{\bf H}^*\times {\bf H} $$ or $$ {\bf S}= \frac{\epsilon_0}{2i \omega }{\bf E}^*\times {\bf E} $$ The two decompositions coincide the field is pure helicity state with ${\bf E}=i\sigma c {\bf B}$, $\sigma=\pm 1$.

In other cases we should take the average so as to preserve the electric-magnetic duality.