Why doesn't the entropy portion of thermodynamics have anything to say about time reversal symmetry? Or does it say that but we're not smart enough to see it yet?

Θ⁻¹HΘ is a way to write a new Hamiltonian which is the “reflection” with respect to time of the original Hamiltonian H, see https://en.wikipedia.org/wiki/Hamiltonian_mechanics for an explanation of what the Hamiltonian is.

Θ⁻¹HΘ = H means that when you reverse time, your physics looks the same as it did before. Such a thing (time-reversal invariance) is being called “time-even”. In the same way a function f(x) would be “even” with respect to its argument if f(x) = f(–x).

The author of that comment is pointing out that if we assume the Hamiltonian is time-even, then we can’t have a particle with an electric dipole moment. This is because the contribution to the Hamiltonian from the dipole interacting with an external electric field would be time-even. That interaction can be expressed as a kind of dot product of two terms, an electric dipole operator part and an electric field part, and the action of our time-reversal operator Θ can be applied separately to each part. Since the electric field is time-even, that means the dipole itself must also be time-even.

But this can’t really work, because a solitary particle has spin but no other symmetries which could generate the electric dipole moment. The spin gets reversed if we run time backwards, a “time-odd” symmetry. So if we find a particle with a static electric dipole moment in nature, then there must be some contradiction in our assumption that the Hamiltonian was time-even.

The rest of his comment is a somewhat analogous kind of argument related to a slightly more complicated situation and slightly more complicated kind of symmetry.

See https://en.wikipedia.org/wiki/Electric_dipole_moment#Dipole_...