# A Confusion about Derivation of Vacuum Rabi Rate

1. Sep 22, 2016

### Twigg

Hi all,

I am learning about the vacuum Rabi frequency for the first time in Metcalf's book Laser Cooling and Trapping (Section 2.2). I am trying to make sense of the following two equations Metcalf gives:
$$\hbar \Omega_{S} = -\vec{\mu} \cdot \vec{E_{\omega}}$$ and $$\vec{E_{\omega}} = \sqrt{\frac{\hbar\omega}{2\epsilon_{0}V}}\vec{\epsilon}$$ where $\Omega_{S}$ is the vacuum Rabi frequency for the mode $S = | \vec{k}, \vec{\epsilon} \rangle$, $\mu$ is the dipole moment of a two-level atom, and $\vec{E_{\omega}}$ is the "electric field per mode".

I see that the expression for $\vec{E_{\omega}}$ is obtained from setting the energy of the field equal to the zero-point energy of the radiation field: $$\epsilon_{0} E_{\omega}^{2} V = \frac{1}{2} \hbar \omega$$
However, I am confused why $E_{\omega}$ is the vacuum-state field in this case. If I'm not mistaken, this field is the expected value of the electric field per mode for the vacuum state: $\langle 0 | \vec{E_{\omega}} | 0 \rangle$, since it has the same energy per mode as the vacuum state. On the other hand, the Rabi rate is typically an off-diagonal element of the interaction Hamiltonian. So why isn't the first equation $\hbar \Omega_{S} = -\vec{\mu} \cdot \langle 1_{S} | \vec{E} | 0 \rangle$?

2. Sep 23, 2016

### Twigg

I came up with a possible solution. I believe the key lies in where the electric field is being evaluated. The field of a photon emitted spontaneously will propagate away from the atom, so while there may be a photon in the radiation field, the field near the atom is still the expected field of the ground state. Am I on the right track?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted