Dirac Current

The Dirac current is a positive-density probability current in a first-quantized description and, with a specified charge normalization, an electric current. Its spatial flow contains more than the phase gradient of a scalar amplitude. The Gordon decomposition separates convective and spin contributions, and the nonrelativistic limit retains a divergence-free magnetization current that the continuity equation alone cannot determine.

Required background. The Covariant Dirac Equation derives current conservation; Free Dirac Spinors fixes on-shell normalization; Dirac to Pauli derives the leading small component. Use $\hbar=c=1$ and $m>0$ below.

Density, flux, and plane-wave velocity

With the established adjoint, the number-normalized current is

$$j^\mu=\bar\psi\gamma^\mu\psi,\qquad j^0=\psi^\dagger\psi,\qquad \mathbf j=\psi^\dagger\boldsymbol\alpha\psi.$$

Its conservation and the bound $|\mathbf j|\leq j^0$ are derived in the covariant equation and geometric current pages. For a normalized single-particle amplitude of charge $q$, the electric current is $qj^\mu$. For quantized fields the charge operator and particle versus antiparticle contributions require the field normalization and ordering.

For a single $u_s(p)$ mode in the $2m$ convention,

$$\bar u_s(p)\gamma^\mu u_s(p)=2p^\mu,\qquad \frac{\mathbf j}{j^0}=\frac{\mathbf p}{E_{\mathbf p}}.$$

The same diagonal bilinear holds for $v_s(p)$, with the future-directed label $p$ of the spinor convention. Its mode’s canonical energy and momentum are both reversed, so its branch group velocity is again $\mathbf p/E_{\mathbf p}$. A positive time component is compatible with either frequency sign. It does not assign the antiparticle’s electric charge before quantization.

Gordon decomposition

For a minimally coupled spinor let $D_\mu\psi=(\partial_\mu+iqA_\mu)\psi$ and $\bar\psi\overleftarrow D_\mu=\partial_\mu\bar\psi-iqA_\mu\bar\psi$. The original and adjoint equations imply

$$\begin{aligned} 2m\bar\psi\gamma^\mu\psi ={}&i\bar\psi\gamma^\mu\gamma^\nu D_\nu\psi\\ &-i(\bar\psi\overleftarrow D_\nu)\gamma^\nu\gamma^\mu\psi. \end{aligned}$$

Use $\gamma^\mu\gamma^\nu=\eta^{\mu\nu}-i\sigma^{\mu\nu}$. The metric terms give the antisymmetric derivative bilinear; the sigma terms combine into an ordinary derivative of a gauge-neutral bilinear. Therefore

$$j^\mu= \frac{i}{2m}\bar\psi\overleftrightarrow D^\mu\psi +\frac1{2m}\partial_\nu (\bar\psi\sigma^{\mu\nu}\psi),$$

where $\bar\psi\overleftrightarrow D^\mu\psi =\bar\psi D^\mu\psi-(\bar\psi\overleftarrow D^\mu)\psi$. The first contribution is often called convective and the second spin or magnetization current. The divergence of the second vanishes identically for smooth fields because $\sigma^{\mu\nu}$ is antisymmetric and the derivatives commute.

This decomposition does not divide probability into two separately positive distributions. Surface terms also matter when integrating the spin contribution over a finite region. It is the total Dirac current that has the positive density and causal bound. At $m=0$ the current itself remains defined, but this decomposition with $1/m$ is not an available identity.

On-shell transition current

Between two free positive-frequency spinors with the same mass, the bilinear identity reads

$$\bar u_{s'}(p')\gamma^\mu u_s(p) =\frac1{2m}\bar u_{s'}(p') \left[(p'+p)^\mu+i\sigma^{\mu\nu}(p'-p)_\nu\right]u_s(p).$$

The transfer is $p'-p$; reversing its definition reverses the displayed sigma term. One derivation uses $\bar u(p')p'\!\!\!/=m\bar u(p')$ on the left and $p\!\!\!/u(p)=mu(p)$ on the right, then splits the gamma products into symmetric and antisymmetric parts. The identity is on shell and assumes equal masses. It is useful for organizing electromagnetic matrix elements; it is not an identity for arbitrary unconstrained four-spinors.

The Pauli current includes a curl

Remove the rest phase and write $\psi=(\varphi,\chi)^T$. The leading low-energy relation is $\chi=(\boldsymbol\sigma\cdot\boldsymbol\pi)\varphi/(2m)$, with $\boldsymbol\pi=-i\nabla-q\mathbf A$. Substituting into $j_i=\varphi^\dagger\sigma_i\chi+\chi^\dagger\sigma_i\varphi$ and using $\sigma_i\sigma_j=\delta_{ij}+i\epsilon_{ijk}\sigma_k$ gives

$$\mathbf j= \frac1m\operatorname{Im}(\varphi^\dagger\nabla\varphi) -\frac qm\mathbf A\,\varphi^\dagger\varphi +\frac1{2m}\nabla\times (\varphi^\dagger\boldsymbol\sigma\varphi)$$

to leading order in the nonrelativistic reduction. The first two terms form the usual gauge-covariant convective current. The curl has identically zero divergence, so adding it does not change the Pauli continuity equation. Its coefficient is fixed by matching the Dirac current, not by conservation alone. Restoring units puts $\hbar$ in the phase-gradient and spin-curl numerators.

For a localized snapshot with $\mathbf A=0$, $\varphi=f(\mathbf x)(1,0)^T$ and real $f$, the convective term vanishes, but

$$\mathbf j_{\rm spin} =\frac1{2m}(\partial_y f^2,-\partial_x f^2,0).$$

If $f^2$ is proportional to $e^{-r^2/\sigma^2}$, this becomes $f^2(-y,x,0)/(m\sigma^2)$: a circulating current, not an outward loss of probability. This is a slowly varying packet snapshot; it need not be a stationary free eigenstate. For an unbounded Gaussian the local expansion is not uniform in its extreme tails: replacing the density by $f^2$ there can give a ratio $|\mathbf j|/f^2>1$. The full Dirac density includes the small component and retains the exact causal bound.

For decay sufficient to discard surface terms, $\boldsymbol\mu=(q/2)\int\mathbf x\times\mathbf j_{\rm spin}\,d^3x =q\langle\mathbf S\rangle/m$, where $\langle\mathbf S\rangle=\tfrac12\int\varphi^\dagger\boldsymbol\sigma\varphi\,d^3x$. This agrees with the signed $g=2$ convention in the Pauli Equation.

Exercises

1. Why can continuity alone not fix the Pauli spin-curl term?

Solution

$\nabla\cdot(\nabla\times\mathbf M)=0$ for a smooth $\mathbf M$. Thus currents differing by such a term have the same local continuity equation. Matching to the relativistic current or to the electromagnetic coupling supplies information beyond that equation.

2. Derive the magnetic-moment factor from the spin curl.

Solution

Integration by parts gives $\int\mathbf x\times(\nabla\times\mathbf M)\,d^3x =2\int\mathbf M\,d^3x$ when the surface term vanishes. With $\mathbf M=\varphi^\dagger\boldsymbol\sigma\varphi/(2m)$, $\boldsymbol\mu=q\int\mathbf M\,d^3x=q\langle\mathbf S\rangle/m$. For $q=-e$ the magnetic moment points opposite to the spin.

3. Put $p'=p$ in the on-shell Gordon identity and recover the diagonal vector bilinear.

Solution

The sigma term vanishes and $\bar u_s\gamma^\mu u_s=(p^\mu/m)\bar u_su_s=2p^\mu$. The derivation uses both the on-shell identity and $\bar uu=2m$; changing normalization changes the final factor.

References

• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw–Hill, 1964 — Gordon decomposition and spin currents.
• C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw–Hill, 1980 — on-shell bilinear identities.
• B. Thaller, The Dirac Equation, Springer, 1992 — currents, position, and the nonrelativistic limit.