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Provided a mathematical description of the RWA with accompanying derivations, and removed the Stub status |
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<math>H_0=\hbar\omega_0|\text{e}\rangle\langle\text{e}|</math>
Suppose the atom is placed at <math>z=0</math> in an external (classical) [[electric field]] of frequency <math>\omega_L</math>, given by <math>\vec{E}(z,t)=\vec{E}_0(z)e^{-i\omega_Lt}+\vec{E}_0^*(z)
<math>H_I=-\vec{d}\cdot\vec{E}</math>
where <math>\vec{d}</math> is the [[transition dipole moment|dipole moment operator]] of the atom. The total Hamiltonian for the atom-light system is therefore <math>H=H_0+H_I.</math> The atom does not have a dipole moment when it is in an [[energy eigenstate]], so <math>\langle\text{e}|\vec{d}|\text{e}\rangle=\langle\text{g}|\vec{d}|\text{g}\rangle=0.</math>
<math>\vec{d}=\vec{d}_{\text{eg}}|\text{e}\rangle\langle\text{g}|+\text{H.c.}</math>
(with `H.c.' denoting the [[Hermitean conjugate]]). The interaction Hamiltonian can
<math>H_I=-\hbar\left(\Omega e^{-i\omega_Lt}+\tilde{\Omega}e^{i\omega_Lt}\right)|\text{e}\rangle\langle\text{g}|
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===Making the approximation===
This is the point at which the rotating wave approximation is made. The dipole approximation has been assumed, and for this to remain valid the
<math>\bar{H}_\text{RWA}=-\hbar\Omega e^{-i\Delta t}|\text{e}\rangle\langle\text{g}|
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as stated. The next stage is to find the Hamiltonian in the interaction picture, <math>\bar{H}.</math> The unitary operator required for the transformation is
<math>U=e^{iH_0t/\hbar},</math>
and an arbitrary state <math>|\psi\rangle</math> transforms to <math>|\bar{\psi}\rangle=U|\psi\rangle.</math> The [[Schrödinger equation]] must still hold in this new picture, so
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<math>\begin{align}
H_{I,\text{RWA}}&=U^\dagger\bar{H}_\text{RWA}U
\left(-\hbar\Omega e^{-i\Delta t}|\text{e}\rangle\langle\text{g}|
-\hbar\Omega^*e^{i\Delta t}|\text{g}\rangle\langle\text{e}|\right)
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-\hbar\Omega^*e^{i\omega_Lt}|\text{g}\rangle\langle\text{e}|.
</math>
[[Category:Atomic, molecular, and optical physics]]
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