.
However, it is possible to construct a classical model that has non-trivial
dynamics which depends on underlying variables not yet stated in equation
8.1.
It turns out that the equations of motion of this classical model have
the same form as the quantum dynamical equations in the Heisenberg representation.
This is demonstrated next for a system of rotating ``corner-charged squares''.
The example given appears to be the simplest classical system involving
discrete charged particles that reproduces in form the equations of motion
of the angular momentum of the quantum coupled spin system. Another example,
perhaps simpler, but not involving discrete charges, would be a system
of charged rotating circles or current loops. Note that the exchange interaction
approximation that the magnetic field at any square is given in direction
by the sum of the spins of the other rotating squares is unphysical. For
the magnetic interaction we need to consider a dipole field, but this yields
a different equation of motion, as is shown. In effect, the true spin-spin
coupling arises not from a fictitious magnetic field as is assumed here,
but from the fermion exchange asymmetry. See for example [48].
Consider a classical system of charges consisting of quadruples of charges,
with the four charges of the ith quadruple each having the same
charge value
and mass value 
,
and with the charges in each quadruple constrained to lie at the corners
of a rigid rotating massless square. Let the positions of the corners of
the ith square relative to its center be given by
in counterclockwise order (the direction of the rotation), with 
.
Each square rotates about the axis thru its center and perpendicular to
itself, with the corners of the square moving with unit magnitude velocity 
.
Thus the mass, length, and charge values in this example can be ignored
for now; we may reconstitute them at the end of the calculation. Let there
be a uniform magnetic field 
in the vicinity of the square. We easily compute that the net force on
any square is zero
(Note that this is true for any charge configuration which is any number
of charges equally spaced on a circle.) The torque on the square is nonzero.
We find the torque for two interaction Hamiltonians: the exchange energy 
and the dipole energy
.
These Hamiltonians are given by the expressions
where 
is the vector separation of the squares. Using Hamilton's equation 
with 
, 
,
noting that the spin 
we have
The torque is 
.
In summing over the charges on the square note also that only terms with
an even number of 
's
appearing may be nonzero (this is true for any charge configuration which
is an even number of charges equally spaced on a circle). The identity 
is useful when computing the torque, and note that 
always (this is true for any charge configuration which is an even number
of charges equally spaced on a circle). Note also that for the square 
,
where the corner subscript is taken mod(4). Finally, the torque
for square i due to 
is
In three places so far the generality of the various even charge symmetry
simplifications has been noted, so we might ask if the result above holds
for all even number charge configurations. The answer is no! The single
symmetry used to arrive at the equations of motion above that is not general
is ,
which holds for the square, octagon (in the form 
),
4k-gons, etc. For the circular charged loop there is square symmetry
too, and the above result also holds. However, for the rod (two charges),
the hexagon, 4k+2-gons, etc., the result does not hold. For simple
derivations of the results above for current loops, see for example [40].
The quantum equations of motion of the operators 
,
for the Hamiltonian of equation 8.1,
in the Heisenberg representation are, from equation 7.18
and assuming that 
,
and that 
(diagonal and spin direction isotropic)
The identity 
is useful in showing this result. Compare equation 8.9
with equation 8.7
to see that the spin-spin coupling terms have the same form in the classical
description and the quantum description, though the objects in the two
equations are different. The spin-field terms also have the same form.