mairzydoats said:
And M is by definition a macroscopic value. It is the magnetic dipole moment per unit volume. Undefined on the quantum level, and hence so too is H.
That's not correct. M and P can be defined microscopically. Only in the simplest cases, it is the dipole moment density.
However, the definition is model dependent, namely, it depends on whether one splits the electric charges into internal and external ones or into bound and free + external. Any of these groups fulfills a continuity equation
##\nabla \cdot j =\partial \rho /\partial t ##.
Introducing the four vector ##j^\mu## with ##j^0=\rho## and ##j^\mu=j_i## for ##\mu \in \{1,2,3\}## and ##i \in \{x,y,z\}##, and ##\partial_\mu## as ##\partial_0=\partial_t##, ##\partial_\mu =-\partial_i##,
we can write this as
##\partial_\mu j^\mu=0##.
This equation will be fulfilled iff ##j^\mu = \partial_\nu \Pi^{\mu\nu}## where ##\Pi## is an antisymmetric tensor, i.e. ##\Pi^{\mu\nu}=-\Pi^{\nu\mu}##.
We call it the polarisation tensor.
It may be parametrized as
##\Pi = \begin{pmatrix}
0 &P_x &P_y &P_z\\
-P_x& 0 &M_z& -M_y\\
-P_y &-M_z &0 & M_x\\
-P_z & M_y & -M_x& 0
\end{pmatrix} .
##
This tensor is not unique, as the solution of ##j^\mu = \partial_\nu \Pi^{\mu\nu}## is defined only up to a solution of the homogeneous problem ##\partial_\nu \Pi^{\mu\nu}=0 ##. I suppose this is what Hendrik meant with the possibility to shuffle terms.