finish TeXing the transmission

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Valentin Boettcher 2023-06-22 14:54:11 -04:00
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@ -951,6 +951,8 @@ as can be ascertained from \cref{fig:delta_choice}.
\end{figure}
\subsection{Effects of Asymmetric Damping}
\label{sec:effects-asymm-damp}
@ -982,7 +984,7 @@ The diagonalizing transformation is then
\eu^{\iu ϕ} & \eu^{\iu ϕ}
\end{pmatrix}
& T^{-1}_{\mp;i,0} &\triangleq
\frac{1}{\sqrt{2}\sqrt{1-Δ^{2}}}
\frac{1}{\sqrt{2}\cos(ψ)}
\begin{pmatrix}
-1 & \eu^{-\iu (ϕ-ψ)}\\
1 & \eu^{-\iu (ϕ+ψ)}
@ -1007,7 +1009,7 @@ With this, we have for \(m=\pm,-N,-N+1,\ldots,-1,1,2,\ldots,N\)
\begin{equation}
\label{eq:94}
\begin{aligned}
ω^{0}_{m\neq \pm} &= ω_{n}^{B} & ω^{0}_{\pm} &= \pm δ \sqrt{1-Δ^{2}}\\
ω^{0}_{m\neq \pm} &= ω_{n}^{B} & ω^{0}_{\pm} &= \pm δ \cos(ψ)\\
η^{0}_{m\neq\pm} &= η_{B} = η_S+2δΔ & η^{0}_{\pm} &= n_{S} + δΔ =
\frac{η_{S}+ η_{B}}{2}.
\end{aligned}
@ -1017,16 +1019,90 @@ Next, we compute the transformation of the interaction
\begin{equation}
\label{eq:96}
\begin{aligned}
V_{\pm,n\not\in \{+,-\}} &= \frac{J_{0,n}}{\sqrt{2}\sqrt{1-Δ^{2}}}\eu^{-\iu\pm ψ)} &
V_{\pm,n\not\in \{+,-\}} &= \frac{J_{0,n}}{\sqrt{2}\cos(ψ)}\eu^{-\iu\pm ψ)} &
V_{n\not\in \{+,-\},\pm}&= \frac{J_{n,0}}{\sqrt{2}}\eu^{\iu ϕ} & V_{n\neq \pm,
m\neq \pm} &=
J_{nm} \\
V_{+,-}&=\frac{J_{00}}{2\sqrt{1-Δ^{2}}}\eu^{-\iu ψ} & V_{-,+} &=\frac{J_{00}}{2\sqrt{1-Δ^{2}}}\eu^{\iu ψ}.
V_{+,-}&=\frac{J_{00}}{2\cos(ψ)}\eu^{-\iu ψ} & V_{-,+} &=\frac{J_{00}}{2\cos(ψ)}\eu^{\iu ψ}.
\end{aligned}
\end{equation}
Evidently, the matrix \(T^{-1}VT\) is not hermitian except in the
limit \(Δ\to 0\).
\subsection{Rotating-Wave Interaction}
\label{sec:rotat-wave-inter}
The modulation term in \cref{eq:89} can be written as
\begin{equation}
\label{eq:103}
V_{nm}(t) = \hat{V}_{nm} f(t),
\end{equation}
where \(f(t)\) is proportional to the voltage applied to the
EOM. Note that \(V_{nm}\) is already expressed in the basis of the
\(c_{m}\) (see \cref{eq:96}).
Let us now assume that the voltage modulation takes the form of
\begin{equation}
\label{eq:104}
f(t)=∑_{j}\frac{\hat{f}_{j}}{2} \sin\qty[\pqty{\hat{ω}_{j}+\hat{δ}_{j}}t
+ \varphi_{j}] = -\iu_{j}\hat{f}_{j}\bqty{\eu^{\iu
\pqty{\hat{ω}_{j}+\hat{δ}_{j}}t}\eu^{\iu \varphi} - \eu^{-\iu
\pqty{\hat{ω}_{j}+\hat{δ}_{j}}t}\eu^{-\iu \varphi_{j}}}.
\end{equation}
Transforming into the rotating frame of the \(\tilde{c}_{m}\) (see
\cref{eq:66}), we have
\begin{equation}
\label{eq:105}
\tilde{V}_{mn}=-\iu_{j}\hat{V}_{mn}\hat{f}_{i}\bqty{\eu^{i \pqty{\hat{ω_{j}} +\hat{δ}_{j} -
\pqty{ω^{0}_{n}^{0}_{m}}}t} \eu^{\iu \varphi_{j}}
- \eu^{-i \pqty{\hat{ω_{j}} +
\hat{δ}_{j}+\pqty{ω^{0}_{n}^{0}_{m}}}t} \eu^{-\iu \varphi_{j}}}.
\end{equation}
As discussed in \cref{sec:choice-hybridization}, we want to couple the
\(+\) and \(m\neq \pm\). Comparing with \cref{eq:12}, we set
\begin{align}
\label{eq:107}
\hat{ω_{j}}&^{0}_{j} - ω^{0}_{+} = - \pqty{ω^{0}_{-j} - ω^{0}_{-}}& \hat{δ}_{j} = ε_{+}_{j}
\end{align}
where \(0<j\leq N\) and obtain from \cref{eq:105} by neglecting all
counter-rotating terms
\begin{equation}
\label{eq:108}
\begin{aligned}
\tilde{V}_{+,j} &= -\iu \hat{f}_{j} \hat{V}_{+,j} \eu^{i\varphi_{j}}
\eu^{\iu \pqty{ε_{+}_{j}}t}
& \tilde{V}_{j,+} &= \iu \hat{f}_{j} \hat{V}_{j,+}
\eu^{-\iu\varphi_{j}} \eu^{\iu
\pqty{ε_{j}_{+}}t}\\
\tilde{V}_{-j,-} &= -\iu \hat{f}_{j} \hat{V}_{-j,-} \eu^{i\varphi_{j}}
\eu^{\iu \pqty{ε_{+}_{j}}t}
& \tilde{V}_{-,-j} &= \iu \hat{f}_{j} \hat{V}_{-,-j}
\eu^{-\iu\varphi_{j}} \eu^{\iu
\pqty{ε_{j}_{+}}t}.
\end{aligned}
\end{equation}
Note, that \cref{eq:108} implies \(ε_{-} = -ε_{+}\) and
\(ε_{-j} = -ε_{j}\).
Comparing \cref{eq:108} with \cref{eq:96} and identifying
\(J_{mn}=f(t)\hat{J}_{mn}\) we get
\begin{align}
\label{eq:113}
\begin{aligned}
V^{0}_{+,j} &= -\iu \hat{f}_{j}
\frac{\hat{J}_{0,j}}{\sqrt{2}\cos(ψ)}
\eu^{-i\pqty{ϕ+ ψ-\varphi_{j}}}
& V^{0}_{j,+} &= \iu \hat{f}_{j} \frac{\hat{J}_{0,j}}{\sqrt{2}}
\eu^{i\pqty{ϕ-\varphi_{j}}}\\
V^{0}_{-j,-} &= -\iu \hat{f}_{j} \frac{\hat{J}_{-j,0}}{\sqrt{2}}
\eu^{i\pqty{ϕ+\varphi_{j}}}
& V^{0}_{-,-j} &= \iu \hat{f}_{j} \frac{\hat{J}_{0,-j}}{\sqrt{2}\cos(ψ)}
\eu^{-i\pqty{ϕ- ψ+\varphi_{j}}}.
\end{aligned}
\end{align}
\subsection{Steady-state Transmission on the Small Ring}
\label{sec:steadyst-transm}
@ -1048,17 +1124,17 @@ Using \cref{sec:effects-asymm-damp}, we identify
\eu^{\iu m
ψ}\pqty{δ_{m,+}_{m,-}}\\
&& U^{-1}_{m,β}&\to U^{-1}_{m,0} =
T^{-1}_{m;S,0}=\frac{1}{\sqrt{2}\sqrt{1-Δ^{2}}}\pqty{δ_{m,+}_{m,-}}
T^{-1}_{m;S,0}=\frac{1}{\sqrt{2}\cos(ψ)}\pqty{δ_{m,+}_{m,-}}
\end{aligned}
\end{equation}
which yields using \cref{eq:88}
\begin{equation}
\label{eq:98}
η_{m} = \frac{\abs{κ} πn_{B}}{2c\sqrt{1-Δ^{2}}} \pqty{δ_{m,+}
η_{m} = \frac{\abs{κ} πn_{B}}{2c\cos(ψ)} \pqty{δ_{m,+}
\eu^{\iu ψ} + δ_{m,-}\eu^{-\iu ψ}}.
\end{equation}
To obtain the \(ω_{γ}\) and \(O_{m,γ}\) we have to diagonalize
To obtain the \(ω_{γ}\), \(λ_{γ}\) and \(O_{m,γ}\) we have to diagonalize
\begin{equation}
\label{eq:99}
V^{0}_{n,m} + \pqty{ε_{m} -\iu \pqty{η^{0}_{m} + η_{m}} δ_{nm}}.
@ -1066,7 +1142,7 @@ To obtain the \(ω_{γ}\) and \(O_{m,γ}\) we have to diagonalize
As
\begin{equation}
\label{eq:100}
\Im η_{\pm}= \pm \frac{\abs{κ} πn_{B}}{2c\sqrt{1-Δ^{2}}}Δ,
\Im η_{\pm}= \pm \frac{\abs{κ} πn_{B}}{2c\cos(ψ)}Δ,
\end{equation}
we obtain a correction to the on-site energies \(ε_{\pm}=\pm ε_{A}\),
whereas
@ -1074,7 +1150,115 @@ whereas
\label{eq:101}
\Re η_{\pm} = \frac{\abs{κ} πn_{B}}{2c}
\end{equation}
is independent of \(Δ\) which is consistent with the symmetry of \(η^{0}_{\pm}\).
is independent of \(Δ\) which is consistent with the symmetry of
\(η^{0}_{\pm}\). Note however that \cref{eq:100} is of very small
magnitude as it is the product of two quantities that are small
compared to \(Ω_{B}\) and \(1\) respectively.
In light of \cref{eq:113} we may choose\footnote{The right hand sides
denote quantities from the previous set of notes.}
\begin{gather}
\label{eq:114}
ε_{+} = ε_{A}- \Im η_{+}= ε_{A} - \frac{\abs{κ}
πn_{B}}{2c\cos(ψ)}Δ\\
\begin{aligned}
ε_{j} &= ω_{j}, & \varphi_{j} &= 0, & \hat{f}_{i}&= \frac{\sqrt{2}}{\hat{J}_{0,i}}η_{j}
\end{aligned}
\end{gather}
so that \(V^{0}_{n,m}\) most closely resembles the target
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.
To calculate the susceptibility (see \cref{eq:87}), we evaluate
\begin{align}
\label{eq:116}
U_{γ} &= ∑_{σ=\pm} \eu^{\iu \tilde{ω}_{σ} t}O^{-1}_{γσ}
T^{-1}_{σ;S0} = ∑_{σ=\pm} \frac{σ
}{\sqrt{2}\cos(ψ)} \eu^{\iu \pqty{ω_{σ}^{0}_{σ}} t}O^{-1}_{γσ}\\
\bar{U}_{γ}&= ∑_{σ=\pm} \eu^{-\iu \pqty{ω_{σ}^{0}_{σ}}
t}O_{σγ}T_{S0;σ} = ∑_{σ=\pm} \frac{σ \eu^{\iu σ ψ}}{\sqrt{2}}\eu^{-\iu \pqty{ω_{σ}^{0}_{σ}}t}O_{σγ}
\end{align}
with
\begin{equation}
\label{eq:117}
\tilde{ω}_{\pm} \equiv ω_{σ}^{0}_{σ} = ω_{S} \pm δ\cos(ψ) - ε_{A}
+ \frac{\abs{κ}
πn_{B}}{2c\cos(ψ)}Δ.
\end{equation}
Finally we arrive at
\begin{equation}
\label{eq:118}
\begin{aligned}
χ(t,s) &= χ_{0}Θ(s)
_{γ,σ,σ'}\eu^{{-\iu\bqty{\tilde{ω}_{σ}t -
\tilde{ω}_{σ'}s + ω_{γ}(t-s)} -
λ_{γ}(t-s)}}O_{σ,{γ}}O^{-1}_{γ,σ'}\frac{σ σ' \eu^{\iu σ
ψ}}{\cos(ψ)}\\
&= θ(s) χ_{1}(t-s) + χ_{2}(t,s).
\end{aligned}
\end{equation}
where
\begin{equation}
\label{eq:119}
χ_{0}=\frac{\abs{κ}π n_{B}}{2c}.
\end{equation}
The stationary and non-stationary susceptibilities work out to be
\begin{align}
\label{eq:121}
χ_{1}(t) &= χ_{0}
_{γ,σ}\eu^{{-\bqty{\iu\pqty{\tilde{ω}_{σ} + ω_{γ}} +
λ_{γ}}t}}O_{σ,{γ}}O^{-1}_{γ,σ'}\frac{\eu^{\iu σ
ψ}}{\cos(ψ)}\\
χ_{2}(t,s) &= -χ_{0}_{γσ} \eu^{-\bqty{\iu\pqty{\tilde{ω}_{σ} +
ω_{γ}} + λ_{γ}}t} \eu^{\bqty{\iu\pqty{\tilde{ω}_{\bar{σ}} +
ω_{γ}} + λ_{γ}}s} O_{σ,{γ}}O^{-1}_{γ,\bar{σ}},
\end{align}
where \(\bar{σ}=-σ\). As \(V^{0}_{mn}\) decomposes into two blocks,
where modes with \(n=-,-1,-2,\dots,-N\) don't couple to modes with
\(n=+,1,2,\dots,N\) so that we obtain two non-overlapping sets of
eigenstates, we have
\begin{equation}
\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.
We can now proceed to calculate the response of the system to a
coherent input beam with frequency \(ω\) in the limit of
\(t\gg λ_{γ}\)
\begin{equation}
\label{eq:123}
_{0}^{t}\eu^{-\iu ω s} χ(t-s)\dd{s} = \eu^{\iu ω t} χ_{0}_{σγ}\frac{O_{σ,{γ}}O^{-1}_{γ,σ'}}{\iu\pqty{\tilde{ω}_{σ} + ω_{γ}} +
λ_{γ}}\frac{\eu^{\iu σψ}}{\cos(ψ)}\equiv \eu^{-\iu ω t}T_{S}(ω).
\end{equation}
Both the magnetic and the electric field are proportional to
\(\Im b_{\outputf}\) so that the absolute value of the pointing
vector, e.g. the intensity, averaged over the oscillation period of
the input light becomes
\begin{equation}
\label{eq:128}
\bar{I} = I_{0}\pqty{1-2\Re{T_{S}(ω)} + \abs{T_{S}(ω)}^{2}} \approx
I_{0}\pqty{1-2\Re{T_{S}(ω)}} ,
\end{equation}
where we have used \(b_{\inputf}=b_{0}\eu^{-\iu ωt}\).
For \(Δ=0\) we have
\begin{equation}
\label{eq:130}
\Re{T_{S}(ω)} = χ_{0}_{σγ}\frac{λ_{γ}\abs{O_{σ,γ}}^{2}}{\pqty{\tilde{ω}_{σ} + ω_{γ}}^{2} +
λ_{γ}^{2}},
\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.
\newpage
\printbibliography{}