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A Confusion about Derivation of Vacuum Rabi Rate

  1. Sep 22, 2016 #1

    Twigg

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    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##?
     
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  3. Sep 23, 2016 #2

    Twigg

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    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?
     
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