My question is related to the following post: Extracting Electric Dipole Moment from Matrix Element via Form Factor

There, it is said that the electric dipole moment (EDM) is giving by a term that goes with $\gamma^5$ in the vertex correction. So 2 things:

1) Where did this $\gamma^5$ term come from? From electroweak SM's sector? It couldn't come from QED since this is a vector theory.

2) When you renormalize QED you obtain just the terms in $F_1, F_2$ and if you take $A = (A_0 = 0, \vec{A})$ and recall that $$\vec{E} = -\frac{\partial \vec{A}}{\partial t}, $$ then you can write $$\vec{A} = -\vec{E}·t$$ assuming $\vec{E}$ constant (copying Srednicki's method to get the magnetic dipole moment). Therefore, you can write a Hamiltonian that goes with the electric field and that gives you the electric dipole moment contribution.

So, finally, why do I need $\gamma^5$ term to get the EDM and where did it come from?

  • $\begingroup$ You don't. see the comments of the link you posted. $\endgroup$ – Dwagg Apr 4 at 13:57
  • $\begingroup$ There it is not explained anything I ask $\endgroup$ – Vicky Apr 4 at 14:42

The dipole moment of the Neutron $\vec{d}$ couples to the electric field $\vec{E}$ in the standard way $ \mathcal{E} = \vec{d} \cdot \vec{E}\,, $ where $\mathcal{E}$ is the term that is added to the Hamiltonian. The neutron dipole moment can be defined as $$ \vec{d}_{n,s} = \int \mathrm{d}^3x\,\vec{x}\,\langle n,s | J^0_{\mathrm{em}} | n,s \rangle\,, $$ where $J^0_{\mathrm{em}}$ is the electromagnetic current and $| n, s\rangle$ the state of the neutron with spin $s$. Now the neutron is a neutral particle and its charge distribution is uniform, the only vector object that I can build out of it must be proportional to the spin $$ \vec{d}_{n,s} = d_n\,\vec{s}\,. $$ This implies that $\vec{d}_{n,s}$ is a pseudo-vector (one that does not change sign under parity $\vec{x}\to-\vec{x}$) and therefore, since $\vec{E}$ is a vector, $\mathcal{E}$ is a pseudo-scalar. That is, the Hamiltonian is not parity invariant.

This implies that the dipole moment is in the form factor that breaks parity, which naturally contains a $\gamma^5$. For definiteness check out Chapter 8 of 1611.00048. $$ \langle n,\vec{p}',s'| J^\mu_{\mathrm{em}} | n,\vec{p},s\rangle = \bar{u}_{s'}(p')\,\Gamma^\mu(p'-p) \, u_s(p)\,, $$ with $$ \Gamma^\mu(k) = (\mbox{parity even terms}) + F_A(k^2) \left(\gamma^\mu\gamma^5 k^2 - 2 m_n \gamma^5k^\mu\right) + \frac{1}{m_n}F_3(k^2) \,\sigma^{\mu\nu}\gamma_5 k_\nu\,. $$ And the dipole moment is given by $$ d_n = \frac{F_3(0)}{2m_n}\,. $$ Now these form factors are based purely on Lorentz symmetry considerations. As you correctly ask: where does the $\gamma^5$ come from when I actually perform the computation? In electroweak theory one might expect it to show up and it does: it is a result of the Cabibbo-Kobayashi-Maskawa matrix. However just by dimensional analysis we know that this is not going to be the leading contribution to the dipole moment, because it is ridiculously small.

In QCD naively one does not expect any parity breaking terms, but there is a renormalizable interaction that can be introduced. It is a total derivative so it vanishes order by order in perturbative calulations (see e.g. Sec. 5.2) and we usually ignore it. But it can have non-perturbative CP-breaking effects $$ \mathcal{L}_\theta = \frac{\theta}{32\pi^2} \varepsilon^{\mu\nu\rho\lambda} \,\mathrm{Tr}\,F_{\mu\nu}F_{\rho\lambda}\,. $$ The $\varepsilon$ transforms as $\varepsilon^{\mu\nu\rho\lambda} \to \varepsilon^{\mu\nu\rho\lambda} \det \Lambda$ under a Lorentz transformation $\Lambda^\mu_{\phantom{\mu}\nu}$ so it's parity breaking. The actual $\gamma^5$ may come from an effective theory that you have to build in order to compute the matrix element that leads to the form factors. The terms with the $\gamma^5$ will be proportional to $\theta$ as that is the only source of CP-breaking in QCD.

Useful link if you are interested: Villadoro's GGI lectures.


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