Why and how does the term $\frac{\theta}{32\pi^2}F_{\mu\nu a}\tilde{F}^{\mu\nu a}$ induce electric dipole moment of the neutron?

Question

It is well-known that the operator $$\delta \mathcal{L}_{QCD}=\frac{\theta}{32\pi^2}F_{\mu\nu a}\tilde{F}^{\mu\nu a}$$ violates CP, it can contribute to the neutron electric dipole moment, $d_n$. For example, see the opening statement here.

The $\theta$-term interaction is an interaction between gluons, and it's reasonable that it affects the property of nucleons. But I don't understand how can gluonic interactions induce electric dipole moment of the neutron. What does strong interaction have to do with electric dipole moment?

Answer 1

The term $F_{\mu\nu}\tilde{F}^{\mu\nu}$ can be written as $\vec{E}\cdot\vec{B}$, with $\vec{E}$ and $\vec{B}$ the so-called chromoelectric/magnetic fields (which you can think of as the strong force versions of their classical counterparts).

A derivation of the nEDM can be found here (beware: it is quite long and tedious).

As you mention, $\theta F\tilde{F}$ violates $\textsf{CP}$ and allows for processes such as those depicted below, which in turn mean non-zero nEDM.

Answer 2

The actual reason is that the operator $F_{\mu\nu}^{a}\tilde{F}^{\mu\nu,a}$ in general has non-zero matrix element $$ \tag 1 \lim_{\mathbf p' \to \mathbf p}\langle N(\mathbf p')| F^{a}_{\mu\nu}\tilde{F}^{\mu\nu,a}|N(\mathbf p)\rangle = i\mu \bar{N}(\mathbf p)\gamma_{5}N(\mathbf p), $$ where $N$ is the nucleon state and $\mu$ is dimensional constant. The same argument applies for the matrix element $$ \tag 2 \lim_{\mathbf p' \to \mathbf p}\langle N(\mathbf p')|F^{a}_{\mu\nu}F^{\mu\nu,a}|N(\mathbf p) \rangle \simeq m_{N}\bar{N}(\mathbf p)N(\mathbf p), $$ making actually one of main contributions in the nucleon mass.

Although $(1)$ looks reasonable, it is of course hard to evaluate it for getting the value of $\mu$, because it is non-perturbative. However, here can be used the chiral anomaly: any chiral transformation $$ q_{f} \to e^{i\alpha_{f}\gamma_{5}}q_{f}, \quad f = u,d,s $$ generates the shift $\theta \to \theta + 2\alpha_{f}$. Therefore it is possible to eliminate the $\theta$-term with appearing the phase in the quark mass matrix: $$ M_{q} \to M_{q}\cdot \text{diag}\big(\text{exp}\left[2i\alpha_{u}\gamma_{5}\right],\text{exp}\left[2i\alpha_{d}\gamma_{5}\right],\text{exp}\left[2i\alpha_{s}\gamma_{5}\right]\big) $$ It's then possible to realize that for small $\theta$ the phases matrix generates the shift $$ \bar{q}M_{q}q \to \bar{q}M_{q}q + i\theta \frac{m_{u}m_{d}m_{s}}{m_{u}m_{d}+m_{d}m_{s}+m_{u}m_{s}}\bar{q}\gamma_{5}q, $$ which effectively defines $\mu$ (up to numerical factor).

Once $(1)$ is derived, it's simple to obtain the dipole term. Classically, the dipole moment $d$ is defined through the hamiltonian $$ H_{\text{EDM}} = d(\mathbf E \cdot \mathbf S), $$ where $S$ is the spin operator and $\mathbf E$ is electric field. By using $(1)$, one realizes that one just needs to generate the operator $$ L_{\text{EDM}} = d\cdot \bar{n}i\sigma_{\mu\nu}F^{\mu\nu}n $$ This can be done only through the proton and pion fields, since the neutron doesn't have the tree-level coupling to photons. The latter is purely technical thing.