Related papers: Residual detuning in the laboratory-frame anti-Jaynes--Cummings model with a squeezed vacuum

Residual detuning in the laboratory-frame anti-Jaynes--Cummings model with a squeezed vacuum

URL: http://arxiv.org/abs/2609.04580v1
Date: Fri, 04 Sep 2026 00:20:23 GMT
Title: Residual detuning in the laboratory-frame anti-Jaynes--Cummings model with a squeezed vacuum
Abstract summary: AJC interaction retains a residual detuning $2f$ that is absent from the rotating-frame model.<n>All quoted contrasts are the amplitude of the first turning point of the atomic ground-state population.<n>The calculation is a numerical control landscape, not a new solvable limit.
Score: 0.0
License: http://arxiv.org/licenses/nonexclusive-distrib/1.0/
Abstract: The laboratory-frame anti-Jaynes--Cummings (AJC) interaction retains a residual detuning $2fλ$ that is absent from the rotating-frame model. We map two proposed remedies---a Kerr shift $χ(\hat a^\dagger\hat a)^2$ and collective Dicke coupling of $N$ two-level emitters---for a squeezed vacuum on that ladder ($r=1$, $\langle n\rangle=\sinh^2 r\simeq 1.38$). All quoted contrasts are the amplitude of the first turning point of the atomic ground-state population, which coincides with the two-level formula $\mathcal{C}=1/[1+(2f+χ)^2]$ at $r=0$ to $10^{-9}$. Three results follow. (i)~A single Kerr strength never restores unit contrast at $r=1$; the $r=0$ $n$-dependent shift $χ(2n+1)$ cannot cancel $2fλ$ on every occupied Fock component. (ii)~The exact $r=1$ contrast at $χ=0$ is not reproduced by an incoherent sum $\sum_n P_n(r)\,\mathcal{C}_n$ built from the $r=0$ two-level formula; pointwise deviations are several tenths. (iii)~Collective coupling raises the contrast systematically. At $f=5$ one finds $\mathcal{C}=0.140$ ($N=1$) and $\mathcal{C}=0.575$ ($N=8$), above the unsqueezed value $8/[8+(2f)^2]=0.074$. The $N=16$ point at this $f$ remains truncation-limited and is not quoted to three digits. The same $N\sim(2f)^2$ estimate for $\mathcal{C}=1/2$ places trapped-ion values $f\sim 10^{2}$--$10^{3}$ outside the present construction. The relevant platform is ultrastrong circuit QED with $f\sim 1$--$10$. The calculation is a numerical control landscape, not a new solvable limit.

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