Electric dipole moment (EDM) and CP violation

Question

It's well known that a non-zero value for the electric dipole moment (EDM) would imply CP violation. If we consider the interaction Hamiltonian of an EDM $d$ with an electric field $\vec{E}$,

$$ H = -\frac{d}{S}\vec{S}\cdot \vec{E}, \quad \mbox{$\vec{S}$ is the spin} $$

we see it is P- and T-odd, but is it C-even? If CPT theorem holds, then it should be C-even but because $\vec{E}$ is C-odd it should imply that the spin changes sign under C. Is this true? My question comes from the answer to this post Does Charge conjugation change the spin momentum?, where it is said that it should be clear the spin does not change under C.

Therefore, there is a contradiction, isn't it?

Answer 1

No, there's no contradiction, the issue is that you're mixing up ideas from nonrelativistic quantum mechanics and relativistic quantum field theory.

You've written down the Hamiltonian for a single electron in an external field. This automatically puts you in the realm of nonrelativistic quantum mechanics. In the restricted setting you are considering, the operation $\hat{C}$ doesn't even make any sense, because it should map the electron to a positron, but your Hilbert space explicitly includes only the states of a single electron. (That's why we never talk about $\hat{C}$ in undergraduate QM courses.) So in such a setting, you can't even state the CPT theorem, much less apply it!

If you were working in relativistic QFT, you would instead have a term like $$\mathcal{L} \supset \bar{\psi} \sigma^{\mu\nu} \gamma^5 \psi F_{\mu\nu}$$ which in the low energy limit reduces to your term. This term is invariant under CPT. Speaking loosely, the "extra" sign flip that seems to be missing in the nonrelativistic analysis comes from the fact that C flips the sign of the electron/positron charge.