The Rabi frequency is an expression of the light-matter interaction (Hamiltonian) and gives a relation between field polarization and electric dipole.

Reading the book of Cohen-Tannoudji (Complement $E_X$), I have calculated the dipole transition elements of electronic states $\langle a|D_1^m|b \rangle$ (using Clebsch-Gordan coefficients). Then I tried to calculate from that the Rabi frequency $$\Omega_{a\to b}=\frac{1}{2\hbar}E_m \langle a|D_1^m|b \rangle.$$

But I got stuck with the E-field vector $E_m$. What is its direction for the $\pi$, $\sigma^+$, $\sigma^-$ transitions?


The transitions $\pi$, $\sigma^+$, $\sigma^-$ correspond to the linear, and the two directions of circular polarization. You can look at it using the Jones-vector but then you need a trick to deal with the z-direction since there should be none.

  • $\begingroup$ Could you please rephrase this question? Direction of electric field acting on a molecule is determined by the electric field vector. This can be related to position and motion of its charged sources, but has nothing to do with various "transitions" in the theory of molecules, because these are merely pairs of Hamiltonian eigenfunctions that have fixed orientation with respect to molecular body (rigid body made of nuclei) but have no fixed orientation with respect to electric field or coordinate system. $\endgroup$ May 10, 2016 at 20:16
  • $\begingroup$ Yes, I meant the direction of the E-field with respect to the dipole. $\endgroup$
    – strpeter
    Jun 6, 2016 at 18:54

1 Answer 1


The directions are:

$$\vec{e}_{\sigma^+}=\frac{1}{\sqrt{2}}(\vec{e}_x+i\vec{e}_y)$$ $$\vec{e}_{\pi} =\vec{e}_z$$ $$\vec{e}_{\sigma^-}=-\frac{1}{\sqrt{2}}(\vec{e}_x-i\vec{e}_y)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.