Vector potential of a Fermion

The vector potential can't be that complicated. There are 2 vectors making up the problem:

1: The position vector $bf r$

2: The spin axial vector ${bf S} = frac{hbar} 2{bf sigma}$

From those, you need to construct a vector:

$$ {bf A} = a(r){bf r} + b(r){bf S} + c(r)({bf r times} {bf S}) $$

where the coefficients can be functions of position through $r= ||{bf r}||$.

The vector potential is odd under time reversal, so $a(r)=0$, as position is even.

As a vector, it is parity odd, so $b(r)=0$.

The final term is the cross product of a vector and an axial vector, so it's parity odd. It is also time odd, since ${bf S}$ is.

Hence:

$$ {bf A} = c(r)({bf r} times {bf S}) $$

describes the structure.

The field cannot be too complicated, as it is a dipole field, so it must have the same symmetries as a dipole. It is:

$$ {bf A} = frac{mu_0}{4pi} frac{{bf m} times {bf r}}{r^3} $$

with:

$$ {bf m} = -gfrac{ehbar}{4m_ec}{bf sigma}$$