Now lets

We have all the elements to finish the calculation of the Hamiltonian.
Before pulling this all together in a brute force way,
its good to realize that **almost all the terms will give zero**.
We see that the derivative of
is proportional to a 4-vector, say
and to a polarization vector,
say
.
The dot products of the 4-vectors, either
with itself or
with
are zero.
Going back to our expression for the Hamiltonian density, we can eliminate some terms.

The remaining term has a dot product between polarization vectors which will be nonzero if the polarization vectors are the same. (Note that this simplification is possible because we have assumed no sources in the region.)

The total Hamiltonian we are aiming at, is the integral of the Hamiltonian density.

When we integrate over the volume only products like will give a nonzero result. So when we multiply one sum over by another, only the terms with the same will contribute to the integral, basically because the waves with different wave number are orthogonal.

This is the

It should not be a surprise that the terms that made up the Lagrangian gave a zero contribution because and we know that E and B have the same magnitude in a radiation field. (There is one wrinkle we have glossed over; terms with .)

Jim Branson 2013-04-22