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

  • $\begingroup$ Formatting hints: Use #Heading instead of $\textbf{Heading}$ for section heading, use double $$ instead of single $ for equations that shall have their own line, and use \mathrm{i} instead of i for the imaginary unit. It looks better that way :) $\endgroup$ – ACuriousMind Feb 27 '15 at 2:32
  • $\begingroup$ @AliMoh How do you get the addional $\frac{\vec\sigma\cdot\vec\Pi}{2m}$ factor in the first equation of the section '2. Dirac to Schrodinger-Pauli'? $\endgroup$ – Matthias Mar 2 '15 at 14:59
  • $\begingroup$ @AliMoh The QED Lagrangian I know looks like:$$\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? $\endgroup$ – Matthias Mar 2 '15 at 15:01
  • $\begingroup$ 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 $\endgroup$ – Ali Moh Mar 2 '15 at 15:12
  • $\begingroup$ 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 $\endgroup$ – Ali Moh Mar 2 '15 at 15:15

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