From b5c8cab2ed8505b15db943b577162501f5476bfb Mon Sep 17 00:00:00 2001 From: Valentin Boettcher Date: Fri, 23 Jun 2023 18:21:28 -0400 Subject: [PATCH] add transmission on big loop --- .../index.tex | 121 +++++++++++++++++- 1 file changed, 115 insertions(+), 6 deletions(-) diff --git a/roam/data/d6/e44fbc-975b-463d-87f9-878d50758b55/index.tex b/roam/data/d6/e44fbc-975b-463d-87f9-878d50758b55/index.tex index debea6d..30dbe3c 100644 --- a/roam/data/d6/e44fbc-975b-463d-87f9-878d50758b55/index.tex +++ b/roam/data/d6/e44fbc-975b-463d-87f9-878d50758b55/index.tex @@ -698,7 +698,7 @@ where we have set \(g_{β}^{0}=\sqrt{κ}\) and defined This also simplifies \cref{eq:64} to \begin{equation} \label{eq:88} - η_{m}=\abs{κ}\frac{πn_{B}}{c}∑_{σ=\pm,β,β'}U_{βm}U^{-1}_{m'β}δ_{\sgn(β),σ} δ_{\sgn(β'),σ}. + η_{m}=\abs{κ}\frac{πn_{B}}{c}∑_{σ=\pm,β,β'}U_{βm}U^{-1}_{mβ}δ_{\sgn(β),σ} δ_{\sgn(β'),σ}. \end{equation} Further defining @@ -1103,7 +1103,7 @@ Comparing \cref{eq:108} with \cref{eq:96} and identifying \end{aligned} \end{align} -\subsection{Steady-state Transmission on the Small Ring} +\subsection{Steady-state Transmission on the Small Loop} \label{sec:steadyst-transm} We can now proceed to calculate the steady state transmission for the @@ -1139,6 +1139,10 @@ To obtain the \(ω_{γ}\), \(λ_{γ}\) and \(O_{m,γ}\) we have to diagonalize \label{eq:99} V^{0}_{n,m} + \pqty{ε_{m} -\iu \pqty{η^{0}_{m} + η_{m}} δ_{nm}}. \end{equation} +Both apriori loss \(η_{m}^{0}\) and the loss induced by the coupling +to the transmission line \(η_{m}\), as well as the detunings \(ε_{m}\) +enter the final interaction hamiltonian. + As \begin{equation} \label{eq:100} @@ -1171,7 +1175,10 @@ Hamiltonian. The drive phases \(\varphi_{j}\) can also be set to \(\varphi_{j}=ϕ + \frac{π}{2}\), to remove the phase in the interaction if it is known. The phase in the interaction does not influence the observable \(ρ_{A}\). However it does influence the -interference with a reference light beam. +interference with a reference light beam. Also, the magnitude of +\cref{eq:100} is likely negligible. If it is required, we can +determine it by choosing \(N=1\) and measuring the eigenenergies with +the result obtained below. To calculate the susceptibility (see \cref{eq:87}), we evaluate \begin{align} @@ -1226,7 +1233,9 @@ eigenstates, we have \label{eq:122} O_{σ,γ}=0 \wedge O^{-1}_{\bar{σ},γ} =0\; \forall σ=\pm\implies χ_{2}(t,s) = 0 \end{equation} -and the non-stationary contribution to the susceptibility vanishes. +and the non-stationary contribution to the susceptibility +vanishes. Persistent oscillations in the output intensity therefore +would likely signal a breakdown of the RWA. We can now proceed to calculate the response of the system to a coherent input beam with frequency \(ω\) in the limit of @@ -1257,8 +1266,108 @@ For \(Δ=0\) we have \end{equation} whereas \(Δ\neq 0\) will very slightly shift the peaks and influence the peak heights. We also see, that we only have a good signal on the -states that have some overlap with the small loop. We therefore have -to attach an additional transmission line to the big loop. +states that have some overlap with the small loop. + +\subsection{Steady-state Transmission on the Big Loop} +\label{sec:steady-state-transm} + +To probe the structure of the states in bath, we have to probe the big +loop. + +Analogous to \cref{sec:steadyst-transm} we can also obtain the +transmission for a transmission line (inclusive laser) attached to the +big loop. Note, that the phase \(ϕ\) is now \(ϕ=k_{0}L_{B}/2\), +whereas all the other model parameters retain their meaning. The main +difference to the calculations for the small loop is the number of +modes that interact with the transmission line, leading to +non-vanishing stationary (low-frequency) oscillations in the output +intensity. + +Just as in \cref{sec:steadyst-transm}, we begin by identifying the +relation between the bare modes in the big loop and the eigenmodes of +the unmodulated system. +\begin{equation} + \label{eq:55} + \begin{aligned} + T_{i_{0},β;m} &\to T_{B,β;\pm} &\implies U_{β,m} &= T_{S,0;m} = δ_{β0} + \frac{\eu^{\iu ϕ}}{\sqrt{2}} + ∑_{σ=\pm} δ_{m,σ} + + ∑_{j\neq 0}δ_{βj}δ_{mj}\\ + && U^{-1}_{m,β} &= + T^{-1}_{m;S,0}=δ_{β,0}\frac{\eu^{-\iu ϕ}}{\sqrt{2}\cos(ψ)} + ∑_{σ=\pm} δ_{m,σ}\eu^{-\iu σ ψ} +∑_{j\neq 0}δ_{βj}δ_{mj} + \end{aligned} +\end{equation} +where \(j\in [-N,N] \setminus 0\) and \(β\in [-N,N]\). + +The decay rate introduced by the coupling to the transmission line +works out to be +\begin{equation} + \label{eq:71} + \begin{aligned} + η_{\pm} = \frac{\abs{κ} πn_{B}}{2c\cos(ψ)} \eu^{\mp \iu ψ} + \end{aligned} +\end{equation} +where the sign in the exponent is \emph{inverted} compared to +\cref{eq:98}. Therefore the effective on-site energy for the \(\pm\) +states will be shifted in the inverse direction. Note however, that it +is not guaranteed that \(κ\), \(η_{S}\), and \(Δ\) will be the same as +in \cref{sec:steadyst-transm}. + +Using the fact that either \(O_{σ,γ}=0\) or \(O^{-1}_{\bar{σ},γ} =0\) +for any value of \(σ\) we find for the transmission +\begin{equation} + \label{eq:79} + T_{B}(ω,t) = θ(s) χ_{0}' ∑_{σ=\pm}\bqty{T^{B}_{σ,σ}(ω) + ∑_{n\neq\pm}\pqty{T^{B}_{σ,n}(ω,t) + T^{B}_{n,σ}(ω,t)} + ∑_{n,m\neq\pm} T^{B}_{σ,m,n}(ω,t)}, +\end{equation} +with +\begin{equation} + \label{eq:91} + \begin{aligned} + χ_{0}' &= \abs{κ}\frac{π n_{T}}{c} & \tilde{ω}_{m} = ω^{0}_{m} - ε_{m} + \end{aligned} +\end{equation} +and +\begin{align} + \label{eq:85} + T^{B}_{σ,σ}(ω) &= ∑_{γ}\frac{O_{σ,{γ}}O^{-1}_{γ,σ}\eu^{\iu σψ}}{2\cos(ψ)}\frac{1}{\iu\pqty{\tilde{ω}_{σ} + ω_{γ}-ω} + + λ_{γ}}\\ + \label{eq:92} + T^{B}_{σ,n}(ω,t) &=δ_{σ,\sgn(n)}\frac{O_{σ,{γ}}O^{-1}_{γ,n}\eu^{\iu ϕ}}{\sqrt{2}} + \frac{\eu^{-\iu(\tilde{ω}_{σ}-\tilde{ω}_{n})t}}{\iu\pqty{\tilde{ω}_{n}+ ω_{γ}-ω} +λ_{γ}}\\ + \label{eq:102} + T^{B}_{n,σ}(ω,t) &= δ_{σ,\sgn(n)}\frac{O_{n,{γ}}O^{-1}_{γ,σ}\eu^{-\iu (ϕ+ψ)}}{\sqrt{2}} + \frac{\eu^{-\iu(\tilde{ω}_{n}-\tilde{ω}_{σ})t}}{\iu\pqty{\tilde{ω}_{σ}+ ω_{γ}-ω} + λ_{γ}}\\ + \label{eq:106} + T^{B}_{σ,m,n}(ω,t) &= δ_{σ,\sgn(n)} δ_{σ,\sgn(m)}O_{nγ}O^{-1}_{γm} \frac{\eu^{\iu(\tilde{ω}_{m}-\tilde{ω}_{n})t}}{\iu\pqty{\tilde{ω}_{m}+ ω_{γ}-ω} + λ_{γ}}. +\end{align} + +The stationary transmission peaks around \(\tilde{ω}_{\pm}\) and has +subpeaks shifted by \(ω_{γ}\) just as in \cref{sec:steadyst-transm}, +where the peak height is roughly proportional to the overlap of the +\(γ\) and \(\pm\) states. In the regime we're interested in, there +will only be one state with substantial overlap with the \(\pm\) +states (a.k.a. the \(A\) site). In the same frequency region, +\cref{eq:102} will also have peaks. Those peaks however will be +suppressed, as their height is proportional to the overlap of the +\(\pm\) states and the \(n\neq \pm\), i.e. the \(A\)-site, and bath +sates with the eigenstate \(γ\). The frequency of the steady state +oscillations of \cref{eq:102} allows to tune the relative energies of +the \(A\) sites and the bath site. The same signal may be retrieved +more cleanly from \cref{eq:92}, where the peaks are situated around +\(\tilde{ω}_{n}\). The transmission component \cref{eq:106} will only +be significant if \(m=n\), as the \(γ\) states that don't overlap with +the \(\pm\) states are almost identical to the \(n\neq\pm\), +i.e. bath, states. In this case the transmission does not exhibit +oscillations making the signal from \cref{eq:92} even clearer. +Comparing \cref{eq:92} and \cref{eq:102}, we can extract the damping +asymmetry \(Δ = \sin(ψ)\). + +Time-averaging \cref{eq:79} leaves us with the stationary transmission +\begin{equation} + \label{eq:110} + T_{B}(ω) = θ(s) χ_{0}' ∑_{σ=\pm}\bqty{T^{B}_{σ,σ}(ω) + ∑_{n}T^{B}_{σ,n,n}(ω)}. +\end{equation} \newpage \printbibliography{}