Detrimentally symmetrical cases: $ 0<G\leq g$

Two equal QDs create a symmetric system of levels in the ``horizontal'' sense: $ \gamma_{E1}=\gamma_{E2}=\gamma$ and $ P_{E1}=P_{E2}=P$ ( $ \Gamma_1=\Gamma_2$ but $ \gamma_\mathrm{TOT}\neq
P_\mathrm{TOT}$). Effectively, the populations behave as if the two dots were uncoupled

$\displaystyle n_1=n_2=\frac{P}{\gamma+P}\,,\quad n_B= n_1 n_2\,,\quad n_{12}=0\,,$ (4.40)

although the system is always in strong coupling (FSC) with

$\displaystyle R=\frac{\vert\gamma-P\vert}{\gamma+P}g\,,$ (4.41)

and $ D^s=2\sqrt{\gamma P}/(\gamma+P)$. If $ P\gg\gamma$, or the other way around, the symmetry in the ``vertical'' sense is completely broken and $ D^s=0$.

The second possibility is the case $ \gamma_{E1}=P_{E1}$ and $ \gamma_{E2}=P_{E2}$ ( $ \Gamma_1\neq\Gamma_2$ but $ \gamma_\mathrm{TOT}
= P_\mathrm{TOT}$), which is symmetrical in the vertical sense. This is equivalent to both 2LS in a thermal bath of infinite temperature, so that pump and decay become equivalent. Now $ D^s=2\sqrt{\gamma_{E1}\gamma_{E2}}/(\gamma_{E1}+\gamma_{E2})$ and the populations are also effectively uncoupled, $ n_1=n_2=1/2$, $ n_B=n_1n_2$, $ n_{12}=0$. The Rabi reads

$\displaystyle R=\frac{\vert\Gamma_-\vert}{\Gamma_+}\sqrt{g^2-\Gamma_+^2}\,,$ (4.42)

and gives the conditions for SC/WC regions in terms of $ \Gamma_+$. If $ \gamma_{E1}\gg\gamma_{E2}$ or the other way around, the symmetry in the ``horizontal'' sense is broken and $ D^s=0$.

In the particular case where all the parameters are equal to $ \gamma$, the symmetry is perfect in all senses ($ D^s=1$). Again, all the levels are equally populated ( $ n_1=n_2=1/2$, $ n_B=1/4$) and uncoupled ($ n_{12}=0$) but the Rabi vanishes. The SC/WC regimes become conventional with $ z_{1,2}=\sqrt{g^2-\gamma^2}$. Some asymmetry in the parameters is needed to reach the SSC and MC regimes.

Elena del Valle ©2009-2010-2011-2012.