Ω/2π =
MHz
γ
22
/2π =
MHz
γ
12
/2π =
MHz
Span =
MHz
×
Info
This graph presents the solution of the optical Bloch equations in the steady state:
$$ \sigma_{12} = \dfrac{ (i\gamma_{12}+\delta)\Omega_0 }{ \gamma_{12}^2 + \delta^2 + 4\Omega_0^2(\gamma_{12}/\ gamma_{22}) } \\ \rho_{22} = \dfrac{ 2\Omega_0^2 }{ \gamma_{22}\gamma_{12} + \delta^2(\gamma_{12}/\gamma_{22}) + 4\Omega_0^2 } $$
Read more:
Marco P. M. de Souza,
Excitação coerente de um vapor atômico por trens de pulsos ultracurtos e lasers contínuos
(Tese de Doutorado,undefined UFPE, 2012)
L. Allen, J. H. Eberly,
Optical Resonance and Two-Level Atoms
(Dover Publications, 1987).
×
Abrir/Salvar/Deletar
Download
Home
Reset
Fullscreen
Info
Save