How could the effective electric dipole interaction be derived In some papers (e.g. Bernreuther equation (1.4), The electric dipole moment of the electron) you can find the electric dipole interaction defined as
$$L_I=-\frac i2 d_f\bar\psi\sigma_{\mu\nu}\gamma_5\psi F^{\mu\nu}$$
of a fermion $\psi$, where $d_f$ is the dipole moment, $\sigma^{\mu\nu} :=\frac i2\left[\gamma^\mu,\gamma^\nu\right]$ and $F^{\mu\nu}=\partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$ the Electromagnetic tensor.
In the classical electrodynamic the dipole is defined as 
$$d_f = -\int \rho(\vec x) \vec x d^3x$$
where $\rho(\vec x)$ is the charge distribution.
Can someone explain qualitatively or give a useful hint, how the interaction Lagrangian above can be derived or motivated?
What I know so far is: The electric dipole moment of a charged particle is generated out of its spin (or more generally: its angular momentum). The classical picture of the dipole moment is $CP$ odd and $\gamma_5$ and the spin are $CP$ odd too.
 A: Well you want to go from QFT to Classical mechanics.
Let's do this in three steps
1. QED to Dirac Equation
QED lagrangian with electric dipole is
$\mathcal{L} = \bar{\psi}\left(\gamma\cdot\Pi - \frac{\mathrm{i}d}{2}\sigma^{\mu\nu}\gamma^5 F_{\mu\nu}\right)\psi\\$
Where $\Pi\equiv \partial - \mathrm{i}eA$. This implies the hamiltonian $$\mathcal{H} = \gamma^0 m + \gamma^0 \vec{\gamma}\cdot \vec{\Pi} + e\phi -\mathrm{i}d\left(-\vec{\gamma}\gamma^0\cdot \vec{B} + \mathrm{i}\vec{\gamma}\gamma^0\gamma^5\cdot\vec{E}\right)$$
Now using the Dirac representation of gamma matrices, the eigenvalue equation $\mathcal{H}\Psi = \mathcal{E}\Psi$ becomes $$\left( \begin{array}{cc} e\phi + m - \mathcal{E} + \mathrm{i}d\vec{\sigma}\cdot\vec{E} & \vec{\sigma}\cdot\vec{\Pi} + d \vec{\sigma}\cdot\vec{B} \\ -\vec{\sigma}\cdot\vec{\Pi} + d \vec{\sigma}\cdot\vec{B} & -e\phi + m + \mathcal{E} + \mathrm{i}d\vec{\sigma}\cdot\vec{E}   \end{array}  \right) \left( \begin{array}{c} \psi_+ \\ \psi_-\end{array} \right) = 0$$
in units of $c=1$.
2. Dirac to Schrodinger-Pauli
In the weak field non relativistic limit eq2 shows that $\psi_- \ll\psi_+$, so that solving for the latter we get an equation for the former
$$\left( \mathcal{E} - m\right)\psi_+ = \left(\frac{1}{2m}\left(\vec{\sigma}\cdot\vec{\Pi}\right)^2  - \frac{d}{2m}\left[ \vec{\sigma}\cdot\vec{\Pi}, \vec{\sigma}\cdot\vec{B}\right] + d \vec{E}\cdot\vec{\sigma} + e\phi\right)\psi_+$$
Now using the pauli matrix identities the commutator can be rewritten as $\frac{\hbar d}{m}\vec{\sigma}\cdot\left(\vec{\nabla}\times\vec{B}\right)+ -d^2B^2$, now we used the weak field limit (only first order in $\vec{E}$ and $\vec{B}$) so that we can also approximate $\vec{\Pi}\approx \vec{p} \rightarrow -i\hbar\vec{\nabla} $.
Using maxwell's equations for stationary fields we finally get that the hamiltonian in a stationary EM field is $$H = \frac{\left(\vec{p}- e\vec{A}\right)^2}{2m} + \frac{e\hbar}{2m}\vec{\sigma}\cdot\vec{B} + e\phi +   d\vec{\sigma}\cdot \vec{E}$$
3. to Classical Electrodynamics
we deduce from he last term that the electron has a dipole moment of $\vec{d} \propto \vec{S}d$, where $\vec{S}$ stands for spin. Because as you know Electrostatic energy is$$ E= \int dx^3\phi\rho = \phi_0\int dx^3\rho + \nabla \phi_0\cdot\int dx^3 \vec{x}\rho + \ldots\rightarrow q\phi_0 + \vec{d}\cdot\vec{E}_0+\ldots$$
Note: no guarantee there are no signs or i's mistakes, but this is irrelevant to demonstrating the point
