# What is the physical meaning of $\overline{\Psi} \Psi$ in the Dirac current's "Gordon Decomposition"?

When writing the Dirac (charge) current out in a way that resembles the (charge) current in the Pauli/Schrödinger theories, one obtains the following:

$j^\mu = -\frac{\mathrm{i} e\hbar}{2m} \left[ \overline{\Psi} (\partial^\mu \Psi) - (\partial^\mu \overline{\Psi}) \Psi \right] - \frac{e^2}{m} A^\mu \overline{\Psi} \Psi + \frac{\mathrm{i} e \hbar}{2m} \partial_\nu (\overline{\Psi} \sigma^{\mu \nu} \Psi).$

This resembles the Pauli current quite closely, and indeed reduces to it in the non-relativistic limit. However, within the Dirac theory, the term containing $e A^\mu$ contains the object $e \overline{\Psi} \Psi$, which as far as I know, has no physical interpretation. In the non-relativistic limit, it of course reduces to $e \Psi^\dagger \Psi$, i.e. the (charge) density (and gives a term in the current that is equal to the vector potential times the density), but what does the missing $\gamma^0$ imply in the relativistic Dirac case?

• The $A^\mu \bar\psi\psi$ term is just there to ensure gauge invariance. It can be combined with the gradient terms so that the current is expressed in the gauge covariant derivative. May 20 '15 at 12:07

You can find all the terms of the Gordon Decomposition back in the covariant treatment of classical electro-magnetism. The Dirac equation correctly reproduces them all!

The term $\overline{\Psi}\Psi$ is just the scalar density of the field. It is the same in all reference frames while $\overline{\Psi}\gamma^0\Psi$ transforms as the time component of a four-vector. The current given by the Gordon Decomposition is.

$$j^\mu = -\frac{\mathrm{i} e\hbar}{2m} \left[ \overline{\Psi} (\partial^\mu \Psi) - (\partial^\mu \overline{\Psi}) \Psi \right] - \frac{e^2}{m} A^\mu \overline{\Psi} \Psi + \frac{\mathrm{i} e \hbar}{2m} \partial_\nu (\overline{\Psi} \sigma^{\mu \nu} \Psi)$$

First note that $j^\mu$, the charge-current density, is proportional to the kinetic momentum $p^\mu=\frac{m}{e}j^\mu$ defined by the operator $p^\mu=-i\hbar\partial^\mu-eA^\mu$, where $-i\hbar\partial^\mu$ is the operator for the canonical momentum which is proportional to the spatial/temporal frequencies and $-eA^\mu$ is the term that tells us that the momentum of a charged particle changes proportional to the electro-magnetic vector potential field. So we have.

$$j^\mu=-\frac{ie\hbar}{m}\partial^\mu-\frac{e^2}{m}A^\mu$$

These terms correspond with those from the Gordon decomposition. The subtraction of the two terms between square brackets makes the canonical part of the current real valued.

The last term is the bound current from the magnetization-polarization tensor. An electron at rest is has a magnetic moment. The magnetization field gives rise a non zero polarization field when the electron is boosted.

$${J^{\mu}}_{\text{bound}}=-\partial_{\nu} \mathcal{M}^{\mu \nu} ~~~~~~~~\mbox{with}~~~~~~ \mathcal{M}^{\mu \nu} = \begin{pmatrix} 0 & P_xc & P_yc & P_zc \\ - P_xc & 0 & - M_z & M_y \\ - P_yc & M_z & 0 & - M_x \\ - P_zc & - M_y & M_x & 0 \end{pmatrix},$$