# 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.

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

• Formatting hints: Use #Heading instead of $\textbf{Heading}$ for section heading, use double $$ instead of single  for equations that shall have their own line, and use \mathrm{i} instead of i for the imaginary unit. It looks better that way :) – ACuriousMind Feb 27 '15 at 2:32 • @AliMoh How do you get the addional \frac{\vec\sigma\cdot\vec\Pi}{2m} factor in the first equation of the section '2. Dirac to Schrodinger-Pauli'? – Matthias Mar 2 '15 at 14:59 • @AliMoh The QED Lagrangian I know looks like:$$\mathcal{L}_\text{QED}=\bar\psi(i\gamma\cdot\partial-m)\psi-\frac14 (F_{\mu\nu})^2-e\bar\psi\gamma^\mu\psi A_\mu From where did you get your Ansatz? – Matthias Mar 2 '15 at 15:01
• in the relativistic limit, energy almost equals rest energy (mass) + electrostatic so$\mathcal{E} + e - e\phi\approx 2m$, this gives $\psi_- = -(-\vec{\sigma}\cdot\vec{\Pi} + \vec{\sigma}\cdot{\vec{B}}/2m$ which I substitute back in the first equation. (Note that I dropped $\vec{E}$ in comparison to the rest energy as well). But then you're right I forgot about the $\sigma\cdot B$ when substituting $\psi_-$ from eq2 back into eq1.. to be corrected now – Ali Moh Mar 2 '15 at 15:12
• the first term in your lagrangian is the same as the fist in mine. I dropped the $F^2$ because it's irrelevant to my discussion. Finally, I added the an new gauge invariant parity violating interaction term, because the asker wanted to see how such term corresponds to a dimple moment having electro – Ali Moh Mar 2 '15 at 15:15