Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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##?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A Confusion about Derivation of Vacuum Rabi Rate

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**