text:
The purpose of this section is to prove Theorem \ref{thm:quantum lower bound} in the main text. For the sake of generality, the version of the theorem which we will derive, Theorem \ref{Thm:R lower quant bound}, is more general than Theorem \ref{thm:quantum lower bound}, which has not been stated in the main text since it requires some more technical definitions. After summarising the setup in the example section \ref{sec:example}, we will start by introducing the \wso~Clock and stating the necessary theorems about its dynamical properties from \cite{WSO16}. Then we will summarise the exact form of the potential to which Theorem \ref{Thm:R lower quant bound} applies. Then we will proceed by proving the necessary lemmas, and mathematical statements  before proving Theorem \ref{Thm:R lower quant bound} in Section \ref{Final Theorem}.
\subsection{Setup}\label{sec:setup}
In this section, we will briefly recall the important results from the example of a quantum clock in Section \ref{sec:example}, since they will form the starting point of the proof.

The probability of not getting a tick in time interval $[0,t]$, is $\tr[\rho(t)]$, where $\ket{\bar{\psi}_t}$ is given by Eq. \eqref{eq:psi bar t ex}, namely
\be
\rho(t):=\ketbra{\bar{\psi}_t}{\bar{\psi}_t} =\me^{-\mi t H }\rho_0  \me^{\mi t H^\dag},
\ee
where $\rho_0=\ketbra{\psi_0}{\psi_0}$ is the initial state, and
\be \label{H total clock plus V}
H=\hat H_\cl-\mi \hat V_\cl,
\ee
describes the evolution of the system with $\hat H_\cl$ being the Hamiltonian and $\hat V_\cl$ the potential.
The probability of getting the 1st tick in the infinitesimal time step $\delta>0$ is then
\be\label{eq:def P(t)}
P(t)= 2\, \tr[\hat V_C \rho(t)] = \frac{d}{dt}\left(1- \tr[\rho(t)]\right)=-\frac{d}{dt}\tr[\rho(t)], \quad t\geq 0,
\ee
where the last equality follows by taking the derivative inside the trace and noting that
\be
\frac{d}{dt} \rho(t)= -\mi  \hat H \rho(t) +\mi \rho(t) \hat H^\dag.
\ee
The mean and standard deviation of the tick distribution is
\be
\mu:= \int_0^{\infty} t P(t) dt, \quad  \underline{\sigma}:= \sqrt{\int_0^{\infty} (t-\mu)^2 P(t) dt}.
\ee
Since this example of a quantum clock is a reset clock, we are only interested in the accuracy of the 1st tick, ${R_1}$ since the accuracy of later ticks is determined solely from it; $R_j= j R_1$ (see Section \ref{Accuracy of Quantum Clocks}, or \app~\ref{app:delaysequence}, and Remark \ref{remark:delaysequence} for a detailed argument). Namely, we are interested in bounding the quantity
\be
{R_1}:= \frac{\mu^2}{\underline{\sigma}^2}.
\ee

\subsection{Overview of the \wso~Clock}\label{Overview of the quasi-ideal clock}
In this section, we will recall some of the definitions and the core theorem from \cite{WSO16} which we will need in this \app. This section will set some of the terminology and definitions needed for the proof of Theorem \ref{Thm:R lower quant bound}.
\subsubsection{The generator of dynamics and potential function $\bar V_0$}\label{The generator of dynamics and potential function bar V0}

We start by introducing a generator of dynamics for the \wso~Clock.
\be \label{eq:qusi ideal clock gen}
\hat H_{\cl}=\hat V_d +\hat H_\cl,
\ee
The clock's free Hamiltonian, $\hat H_\cl$ is a truncated Harmonic Oscillator Hamiltonian. Namely, $\hat H_\cl=\sum_{n=0}^{d-1}\omega n \ketbra{n}{n}$. The free evolution of any initial clock state under this Hamiltonian has a period of $T_0=2\pi/\omega$, specifically, $\me^{-\mi T_0 \hat H_\cl}\rho_\cl\me^{\mi T_0\hat H_\cl} = \rho_\cl$ for all $\rho_\cl$. The clock interaction term $\hat V_d$, takes the form,
\be
\hat{V}_d = \frac{d}{T_0} \sum_{k=0}^{d-1} {V}_d(k) \ketbra{\theta_k}{\theta_k},
\ee
where the basis $\{\ket{\theta_k}\}_{k=0}^{d-1}$ is the Fourier transform of the energy eigenbasis $\{\ket{n}\}_{n=0}^{d-1}$, specifically,
\begin{align}\label{finitetimestates_main}
\ket{\theta_k} &= \frac{1}{\sqrt{d}} \sum_{n=0}^{d-1} e^{-i2\pi n k/d} \ket{E_n}.
\end{align}
 In other words, we choose $\{\ket{\theta_k}\}$ to be the time basis. It will also be useful later to have the range of $k$ extended to $\zz$.\footnote{Note that $k$ will belong to a set of only $d$ consecutive integers so that $\{\ket{\theta_k}\}$ form a complete orthonormal basis without repetition.} Extending the range of $k$ in Eq.  \eqref{finitetimestates_main} it follows that $\ket{\theta_k}=\ket{\theta_{k \textup{ mod. } d}}$ for $k\in\zz$. It is this basis which we identify with the time basis, i.e. we set $\ket{\theta_k}=\ket{t_k}$, $k\in\nno$.
The function ${V}_d:\rr\mapsto\rr\cup\hh^{-}$ (where $\hh^{-}:= \{ a_0+\mi b_0 : a_0\in\rr, b_0 < 0 \}$ denotes the lower-half complex plane) is defined by
\be
V_d(x)= \frac{2\pi}{d} V_0\left( \frac{2\pi}{d}x \right),
\ee
where $V_0$ is any infinitely differentiable periodic function of period $2\pi$. We will only be interested in a specialised case, and from here on will choose $V_d$ to map $ \rr \mapsto -\mi\, \rr_{\geq 0}$. As one might imagine, it will be more convenient to work with a real function. We will also want to make explicit the normalisation. We will thus write the potential function in terms of an explicitly positive, normalised function by defining $\bar V_d:\rr\rightarrow \rrp$ and $\bar V_0:\rr\rightarrow \rrp$,
\ba
-\mi \delta \,\bar{V}_d(x)&:=V_d(x),\\
-\mi \delta \,\bar{V}_0(x)&:=V_0(x),\quad\quad \int_0^{2\pi} dx \bar{V}_0(x)=1, \quad\quad \bar V_0(x)>0\quad \forall x\in\rr,\label{eq:pod eqs def}
\ea
where we will specialise to the case $\delta\geq 1$. The last constraint ensures that $\bar V_0$ has full support. Secondly, it will be useful to let $\bar V_0$ have a unique global maximum in the interval $x\in[0,2\pi]$ at $x=x_0$. Let the parameters $\tilde\epsilon_V, x_{vl}, x_{xr}$ be such that
\be\label{eq:tilde ep V def}
1-\tilde\epsilon_V=\int_{x_{vl}}^{x_{vr}} dx\, \bar{V}_0(x+x_0)
\ee
for some $-\pi\leq x_{vl}< 0< x_{vr}\leq \pi$, where, for simplicity, we set $x_{vl}=- x_{vr}$ Furthmore, we will find that it is convenient to set $x_0=\pi+x_{vr}+\pi\gamma$. These are all the properties of $\bar V_0$ which we will need for the lemmas in Sections \ref{Approximating R1}, \ref{Upper and lower bounds for trace rho}, \ref{Calculating Delta(0) and Delta(1) and an explicit lower bound on R1}, \ref{R1 to leading order}. In Section \ref{Showing that the limit requirements}, we will find explicit parametrisations of $x_{vr}$ and $\gamma$ in terms of $d$.

\subsubsection{Explicit form on the potential $\bar V_0$}\label{Explicit form on the potential bar V0}
In Section \ref{Showing that the limit requirements} we will need to introduce an explicit function for the potential $\bar V_0$ satisfying the criteria from Section \ref{The generator of dynamics and potential function bar V0}. The function will be written in terms of parameters $\delta \geq 1$, $n\geq 1$, $N\in\nnp$. Specific values for these parameters will be chosen in Section \ref{Showing that the limit requirements} in order to satisfy specific conditions. Here we summarise the final form of the potential $\bar V_0$ to which Theorem \ref{Thm:R lower quant bound} applies.

For $d\geq 3$, let
\be \label{eq:def of rapped pot 0}
\bar{V}_0(x):=\frac{1}{\delta d^2}+n A_0 \sum_{p=-\infty}^{\infty} V_B\left (n(x-x_0-2\pi p) \right),
\ee
where
\be
A_0=\frac{1-2\pi/(\delta d^2)}{\int_{-\infty}^\infty dx V_B(x)},
\ee
is a normalization parameter chosen such that
\be \label{eq:norr V0 requirement 0}
\int_0^{2\pi} \bar{V}_0(x)\,dx =1,
\ee
and
\be
V_B(x):=\textup{sinc}^{2 N}(x):=\left( \frac{\sin(\pi x)}{\pi x}\right)^{2 N},
\ee
where $V_B(0)=1$ and $N\in \nnp$. $N$ is given by
\be
N=\left\lceil\frac{3-4\epsilon_5-\epsilon_9}{2(\epsilon_7-\epsilon_8-\epsilon_5)}\right\rceil,
\ee
with $\lceil\cdot\rceil$ the ceiling function, and where $\epsilon_7=\eta/4,$ $\epsilon_5=\epsilon_8=\eta/16,$ and $\epsilon_9=\eta/2,$ and the role of $\eta\in(0,1)$ becomes apparent in Theorem \ref{eq:lower quantum}. The parametre $\delta$ is given by
\be
\delta=d^{\epsilon_5}.
\ee
Let $C_0: \nnp\mapsto\rrp$ be any function which is only a function of $N$, (i.e. independent of $n$, $d$, and $k$), such that
\be \label{eq:main eq lemma dev bound 0}
\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k}  \bar{V}_0(x)\bigg{|} \leq n^{k+1} C_0^{k+1},\quad \forall\, k\in\nno, \forall n\geq 1,
\ee
holds. We will prove later by explicitly construction that such functions exist.
We now define $n$ to be
\be\label{eq:n in terms of d def 0}
n= \frac{\ln(\pi \alpha_0\sigma^2)}{2\pi C_0\alpha_0\kappa} \frac{d^{1-\epsilon_5}}{\delta \sigma},
\ee
where $\kappa=0.792$ and $\alpha_0\in(0,1]$ depends on the mean energy of the initial  \wso~Clock state and is defined in Section \ref{The wso clock states}. Finally, the location of the peak of the potential, $x_0$, is given by $x_0=\pi+x_{vr}+\pi\gamma$, where
\ba \label{eq:x vr limit contraint 0}
x_{vr}&=d^{\epsilon_7} \frac{\sigma}{\pi\,d},\\
\gamma&= \frac{m-2}{d},
\ea
where
\be\label{Eq: bar m def 1 0}
m=
\begin{cases}
	2\lfloor \bar m \rfloor &\mbox{ if } d=2,4,6,\ldots,\\ \vspace{-0.3cm}\\
	2\lfloor \bar m \rfloor+1 &\mbox{ if } d=3,5,7,\ldots,
\end{cases}
\quad\quad
\bar m =
\begin{cases}
	\frac{d^{\eta/2}}{2}\sigma +1  &\mbox{ if } d=2,4,6,\ldots,\\ \vspace{-0.3cm}\\
	\frac{d^{\eta/2}}{2}\sigma +\frac{1}{2}   &\mbox{ if } d=3,5,7,\ldots.
\end{cases}
\ee
With the above definitions, one can write the function $\bar V_0(x)$ solely in terms of parameters $d$, $\eta$, $\sigma$, and $\alpha_0$.

\subsubsection{The \wso~Clock states}\label{The wso clock states}

Recall that for the quasi-ideal clock, the initial state is pure $\rho_\cl=\ketbra{\Psi_\textup{nor}(k_0)}{\Psi_\textup{nor}(k_0)},$ where
\begin{align}\label{gaussianclock}
\ket{\Psi_\textup{nor}(k_0)} &= \sum_{\mathclap{\substack{k\in \mathcal{S}_d(k_0)}}}\psi_\textup{nor}(k_0;k) \ket{\theta_k},\\
\psi_\textup{nor}(k_0;x) &= A e^{-\frac{\pi}{\sigma^2}(x-k_0)^2} e^{i2\pi n_0(x-k_0)/d}, \quad x\in\rr.\label{eq:psi qudi ideal def}
\end{align}
with $\sigma \in (0,d)$, $n_0 \in (0,d-1)$, $k_0\in\rr$, $A \in \rr^+$, and $\mathcal{S}_d(k_0)$ is the set of $d$ integers closest to $k_0$, defined as
\begin{align}\label{eq: mathcal S def}
\mathcal{S}_d(k_0) = \left\{ k \; : \; k\in \mathbb{Z} \text{   and  }  -\frac{d}{2} \leq  k_0-k < \frac{d}{2} \right\}.
\end{align}
$A$ is defined so that the state is normalised, namely
\be\label{eq:A normalised}
A=A(\sigma;k_0)=\frac{1}{\sqrt{\sum_{k\in \mathcal{S}_d(k_0)} \me^{-\frac{2\pi}{\sigma^2}(k-k_0)^2}}},
\ee
which is of order $A=\bo\left( \left(\frac{2}{\sigma^2}\right)^{1/4} \right)$ as $d\rightarrow \infty$ for all $k_0\in\rr$.
$\omega n_0$ is approximately the mean energy of the initial clock state. We use $n_0$ to define  $\alpha_0\in(0,1]$. It is defined in Def. 1 in \cite{WSO16} as
\begin{align}\label{eq:alpha_0 def}
\alpha_0&=\left(\frac{2}{d-1}\right) \min\{n_0,(d-1)-n_0\}\\
&=1-\left|1-n_0\,\left(\frac{2}{d-1}\right)\right|\in(0,1].
\end{align}
Physically, it is a measure of the distance of the mean energy from the edge of the energy spectrum, and has its maximum value $\alpha =1$ when the the mean energy is in the midpoint of the spectrum, namely when $n_0=(d-1)/2$. We pick $\alpha_0$ to be a fixed constant thought this \doc. This means that $n_0$ takes on the value $n_0=\tilde n_0(d-1)$ with $\tilde n_0\in(0,1)$ fixed constant.

In the proof of Theorem \ref{thm:quantum lower bound}, we will need the core theorem in \cite{WSO16} (Theorem IX.1, page 35). For brevity, we will state it in a slightly reduced which is nevertheless adequate for our purposes. Intuitively, the theorem states that the evolution of \wso~Clock states under the generator $\hat H$ (Eq. \eqref{eq:qusi ideal clock gen}) mimics the evolution of the Idealised clock to a good approximation. In order to state it, we need to recall some definitions from \cite{WSO16}. Namely Definitions 8 and 9 on page 21 \cite{WSO16}:

Let $b$ be any real number satisfying
\be\label{eq:b def eq 0}
b\geq\; \sup_{k\in\nn^+}\left(2\max_{x\in[0,2\pi]} \left|  V_0^{(k-1)}(x) \right|\,\right)^{1/k},
\ee
where $ V_0^{(p)}(x)$ is the $p^\textup{th}$ derivative with respect to $x$ of $ V_0(x)$ and $V_0^{(0)}:=V_0$.
We can use $b$ to define $\mathcal{N}\in\nn^0$ as follows
\be \label{eq:mathcal N def}
\mathcal{N}=
	\left\lfloor \frac{\pi\alpha_0^2}{2\left(\bar \upsilon+\frac{d}{\sigma^2}\right)^2} \left(\frac{d}{\sigma}\right)^2 \right\rfloor,
\ee
where $\kappa=0.792$ and

\be \label{eq:upsilon def}
\bar\upsilon= \frac{\pi\alpha_0\kappa}{\ln\left(\pi\alpha_0\sigma^2  \right)}b.
\ee

The theorem states that for all $t\geq 0$, $k_0\in\rr$,
\ba
\me^{-\mi t \left(\hat V_d+\hat H_\cl\right)} \ket{\Psi_\textup{nor}(k_0)} &= \ket{\bar\Psi_\textup{nor}(k_0+td/T_0, t d/T_0)}+ \ket{\varepsilon_\nu} \\
&= \sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \me^{-\mi \int_{k-t d/T_0}^k dy  V_d(y)} \psi_\textup{nor}(k_0+td/T_0;\,k)\ket{\theta_k} + \ket{\varepsilon_\nu}, \label{eq:non free evollution of Quasi ideal}
\ea
where $\ket{\varepsilon_\nu}=\ket{\varepsilon_\nu}(t,d)$. The important question, is how small can the error term be, namely how does
\be\label{eq:def varepsiln nu}
\varepsilon_\nu(t,d):=\| \ket{\varepsilon_\nu}(t,d)\|_2
\ee
scale with the properties of the \wso~Clock and generator. The theorem puts bounds on this scaling. Namely if
\ba
\begin{split}\label{eq:mathcal N contraints}
	\bar \upsilon &\geq 0, \quad \quad	\mathcal{N} \geq 8,
\end{split}
\ea
are both satisfied,  the theorem tells us that in the limits $d\rightarrow \infty$, $(0,d)\ni \sigma\rightarrow \infty$,
\be \label{eq:epsilon t d}
 \varepsilon_\nu(t,d)=\,          |t| \frac{d}{T_0}\!\left(\bo\left( \frac{\sigma^3}{\bar\upsilon \sigma^2/d+1}\right)^{1/2}\!\!+\bo\left(\frac{d^2}{\sigma^2}\right)\right) \exp\left(-\frac{\pi}{4}\frac{\alpha_0^2}{\left(\frac{d}{\sigma^2}+\bar\upsilon\right)^2} \left(\frac{d}{\sigma}\right)^2 \right)+\bo\left(|t|\frac{d^2}{\sigma^2}+1\right)\me^{-\frac{\pi}{4}\frac{d^2}{\sigma^2}}+\bo\left( \me^{-\frac{\pi}{2}\sigma^2} \right).
\ee

\subsubsection{An expression for ${R_1}$ for the \wso~Clock}\label{sec:expressing R in terms of WSO}
In this section, we will show how to express $\tr[\rho(t)]$ from Section \ref{sec:setup} in terms of the \wso~Clock from \cite{WSO16} discussed in Sections \ref{The generator of dynamics and potential function bar V0}, \ref{The wso clock states}.
Recalling the initial state of the \wso~Clock, namely $\rho_0=\ketbra{\psi_0}{\psi_0}=\ketbra{\psi_\textup{nor}(k_0)}{\psi_\textup{nor}(k_0)}$. From Eq. \eqref{eq:non free evollution of Quasi ideal} if follows
\ba\label{eq:tr rho t from WSO paper}
\tr[\rho(t)]=&\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \big{|} \bra{\theta_k}\me^{-\mi t \hat H}\ket{\psi_\textup{nor}(k_0)} \big{|}^2\\
=& \sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \left|  \me^{-\mi \int_{k-t d/T_0}^k dy V_d(y)} \psi_\textup{nor}(k_0;k-t d/T_0) +\braket{\theta_k|\varepsilon_\nu}(t,d) \right|^2,
\ea
where we have used that $\psi_\textup{nor}(k_0+t d/T_0;k)=\psi_\textup{nor}(k_0;k-t d/T_0)$ which follows directly from Eq. \eqref{eq:psi qudi ideal def}. Due to the normalisation of $\ket{\psi_\textup{nor}(k_0)}$,  we have
\be \label{eq:normalisation}
\sum_{k\in\mathcal{S}_d (k_0+td/T_0)}\left| \psi_\textup{nor}(k_0;k-t d/T_0)\right|^2=1.
\ee

\subsection{Approximating ${R_1}$}\label{Approximating R1}
We will now proceed to lower bound ${R_1}$ in terms of dimension and mean clock energy for the \wso~Clock. Specifically in this section, up to additive error terms, the aim will be to express $R_1$ in terms of the 1st and 2nd moments of $\tr[\rho(t)]$ with integration range limited to one period of the free clock Hamiltonian, namely to the interval $[0,T_0]$.

\subsubsection{A more useful expression for ${R_1}$}
\begin{lemma}\label{lem:1st up bound on R}
	Let $\Delta(0)$ and $\Delta(1)$ satisfy
\ba
\Delta^2(0)&\leq \left(\int_0^{T_0} dt \,\tr [\rho(t)]\right)^2,\quad\quad
\Delta(1)\geq\int_0^{T_0}dt\,t\,\tr [\rho(t)]\geq 0.
\ea

then
\ba\label{eq:R lo bound 2}
{R_1}&\geq \frac{\Delta^2(0)+\epsilon_2}{2\Delta(1)-\Delta^2(0)+\epsilon_1},

\ea
where the epsilon terms $\epsilon_2$, $\epsilon_1$ are defined by
\be \label{eq:epsilon 2}
\epsilon_2:= 2\mu_0\mu_\epsilon+\mu_\epsilon^2,
\ee
with
\ba \label{eqdef:mu0 and mu epsilon}
\mu_0&:=\int_0^{T_0} dt\, \tr[\rho(t)],\\
\mu_\epsilon&:=-\lim_{t\rightarrow \infty} t\,\tr[\rho(t)]
+\int_{T_0}^\infty dt\, \tr[\rho(t)],
\ea
and
\be \label{eq:epsilon 1}
\epsilon_1:=-2\mu_0\mu_\epsilon-\mu_\epsilon^2 +2\int_{T_0}^\infty dt \, t\, \tr[\rho(t)] - \lim_{t\rightarrow \infty} \left( 2\mu t+(t-\mu)^2 \right)\tr[\rho(t)].
\ee
\end{lemma}

\begin{proof}
By integration by parts,
\ba
\mu&= \int_0^{\infty} t P(t) dt=-\left(t\, \tr[\rho(t)]\right)\Big{|}_{0}^{\infty} + \int_0^{\infty} dt\, \tr[\rho(t)] \\
&= \mu_0+\mu_\epsilon.\label{mu simple}
\ea
Similarly,
\ba
\underline{\sigma}^2 &= \int_0^\infty (t-\mu)^2 P(t)dt = -\left( (t-\mu)^2\, \tr[\rho(t)]\right)\Big{|}_{0}^{\infty} + 2\int_0^{\infty} dt\, (t-\mu)\tr[\rho(t)] \\
&= -\lim_{t\rightarrow \infty}(t-\mu)^2 \tr [\rho(t)]+\mu^2+2\int_0^{\infty} dt\,t\,\tr[\rho(t)] -2\mu\big(\mu+\lim_{t\rightarrow \infty} t\,\tr [\rho(t)]\big)\\
&=-\mu^2+ 2\int_0^{\infty}dt\,t\,\tr [\rho(t)]-\lim_{t\rightarrow \infty} \big(2\mu t+(t-\mu)^2\big)\tr[\rho(t)]\\
&=-\mu_0^2+2\int_0^{T_0} dt \,t\,\tr[\rho(t)] +\epsilon_1\leq -\mu_0^2+2\Delta(1) +\epsilon_1\\
& \leq -\Delta^2(0)+2\Delta(1) +\epsilon_1 \label{sigmaSqrt}
\ea
From Eqs. \eqref{sigmaSqrt}, \eqref{mu simple}, we find
\ba
{R_1}&= \frac{\mu^2}{\underline\sigma^2}\geq \frac{\mu_0^2+2\mu_0\mu_\epsilon+\mu_\epsilon^2}{- \Delta^2(0)+2\Delta(1) +\epsilon_1}\\
&\geq \frac{\Delta^2(0)+2\mu_0\mu_\epsilon+\mu_\epsilon^2}{ -\Delta^2(0)+2\Delta(1) +\epsilon_1}.\label{eq:R lo bound inter}
\ea

\end{proof}
We will now bound $\epsilon_1$ and $\epsilon_2$ appearing in Lemma \ref{lem:1st up bound on R}. These terms are negligible in the large $d$ limit and originate from the tails of the integrals in the definition of $\mu$ and $\underline{\sigma}$.
\begin{lemma}\label{lem: ep1 and ep2 bounds} $\epsilon_1$ and $\epsilon_2$, defined in Eqs. \eqref{eq:epsilon 1} and \eqref{eq:epsilon 2} respectively, are bounded by
	\ba
	|\epsilon_1|&\leq \left( a^2\left(\me^{-\delta}+\varepsilon_\nu(T_0,d)\right)^2  +2(2a T_0+a^2)\right)\left(\me^{-\delta}+\varepsilon_\nu(T_0,d)\right)^2,\\
	0\leq\epsilon_2&\leq \left(2T_0 a+a^2 \left(\me^{-\delta}+\varepsilon_\nu(T_0,d)\right)^2\right) \left(\me^{-\delta}+\varepsilon_\nu(T_0,d)\right)^2 ,
	\ea
	where $a$ is any parameter satisfying
	\be \label{eqdef: a}
	a\geq \frac{T_0}{4\pi\,\delta} \left(\min_{x\in[0,2\pi]} \bar{V}_0(x) \right)^{-1}.
	\ee
	and $\varepsilon_\nu(T_0,d)$ is defined by Eq. \eqref{eq:def varepsiln nu}. Note that since we have chosen $\bar{V}_0$ to have full support, $a$ is finite.
\end{lemma}
\begin{proof}
	We will start by bounding $\tr[\rho(t)]$. We find
	\ba
	\tr[\rho(t)]&=  \tr[\me^{-\mi t\hat H}\rho_0\me^{\mi t\hat H^\dag}] \leq \| \me^{-\mi t\hat H}\me^{\mi t\hat H^\dag}\|_\infty\leq \| \me^{-\mi t\hat H}\|_\infty\|\me^{\mi t\hat H^\dag}\|_\infty,
	\ea
	where $\|\cdot\|_\infty$ is the Schatten $p=\infty$ norm also called the operator norm. Recalling $\hat H=\hat H_C-\mi \hat V_C$, and applying the Golden-Thompson inequality, which holds for the operator norm \cite{Bhatia} it we find
	\ba
	\tr[\rho(t)]&\leq 2\| \me^{-\mi t\hat H_C}\|_\infty\|\me^{\mi t\hat H_C}\|_\infty \|\me^{-t \hat V_C}\|_\infty=2 \|\me^{-t \hat V_C}\|_\infty=2\exp\left[-\delta\, t \frac{2\pi}{T_0}\left( \min_{k\in \{0,1,2,\ldots,d-1\}} \bar{V}_0\left( \frac{2\pi}{d}k \right)\right)\right],
	\ea
	where we have used $ \hat V_C=\mi \hat V_d:=\delta \frac{d}{T_0}\sum_{k=0}^{d-1}\bar V_d(k)\ketbra{\theta_k}{\theta_k}$, where $\frac{d}{T_0} \bar V_d(k)= \frac{2\pi}{T_0} \bar{V}_0\left( \frac{2\pi}{d}k \right)$. Now, in this manuscript, $\bar{V}_0$ is positive (Eq. \eqref{eq:pod eqs def}). As such, for all $d\in\nnp$
	\be
	\min_{k\in \{0,1,2,\ldots,d-1\}} \bar{V}_0\left( \frac{2\pi}{d}k \right)>0
	\ee
	and thus
	\be \label{eq:lim powers rho}
	\lim_{t\rightarrow \infty} t^n \tr[\rho(t)]=0
	\ee
	for all $n\geq 0$ and $d\in\nnp$.
	Our next task will be to bound the two integrals $\int_{T_0}^\infty dt\, \tr[\rho(t)]$ and $\int_{T_0}^\infty dt\,t\, \tr[\rho(t)]$. We will start by finding a $t$ independent parameter $a>0$ such that $\tr[\rho(t)]\leq a P(t)$ for all $t\geq 0$, where $P(t)=-\frac{d}{dt}\tr[\rho(t) ]$ was defined in Eq. \eqref{eq:def P(t)}.
	\ba
	\frac{\tr[\rho(t)]}{P(t)}&=
	\frac{\tr[\rho(t)]}{\mi\tr[\hat H\rho(t)]-\mi\tr[\hat H^\dagger\rho(t)]}=\frac{\tr[\rho(t)]}{\mi\tr[(\hat H_C-\mi\hat V_C)\rho(t)]-\mi\tr[(\hat H_C+\mi\hat V_C)\rho(t)]}=\frac{1}{2\tr [\hat V_C \bar{\rho}(t)]}\\
	&\leq \frac{1}{2}\left( \inf_{\ket{\psi}\in\mathcal{S}_p} \tr [\hat V_C \ketbra{\psi}{\psi}] \right)^{-1},\quad \forall\, t\geq 0
	\ea
	where $\bar{\rho}(t):=\rho(t)/\tr[\rho(t)]$ is a normalised rank one density matrix for all $t\geq 0$ and $\mathcal{S}_p$ is the set of normalised pure quantum states (rank one density matrices). Crucially, note that $P(t)>0$ since $\tr [\hat V_C \bar{\rho}(t)]>0$ because $\hat V_C$ is positive-definite since $\bar{V}_0$ has full support. Hence taking the trace in the orthonormal basis $\{\ket{\theta_k}\}_{k=0}^{d-1}$,
	\ba
	\frac{\tr[\rho(t)]}{P(t)}&\leq \frac{1}{2}\left( \min_{k\in\{0,1,2,\ldots,d-1\}} \bra{\theta_k}\hat V_C \ket{\theta_k} \right)^{-1}= \frac{T_0}{4\pi\,\delta} \left(\min_{k\in\{0,1,2,\ldots,d-1\}} \bar{V}_0\left(2\pi k/d\right) \right)^{-1}\\
	&\leq \frac{T_0}{4\pi\,\delta} \left(\min_{x\in[0,2\pi]} \bar{V}_0(x) \right)^{-1} \leq a ,\quad \forall\, t\geq 0. \label{eq:bound  tr/P}
	\ea
	Using Eq. \eqref{eq:bound  tr/P} can now bound the first integral:
	\be \label{eq:int rho}
	\int_{T_0}^\infty dt\, \tr [\rho(t)]\leq a \int_{T_0}^\infty dt P(t)=-a \Big[ \tr [\rho(t)]\Big]_{T_0}^\infty=a\left( \tr [\rho(T_0)]-\lim_{t\rightarrow \infty}\tr [\rho(t)] \right)=a\,\tr [\rho(T_0)].
	\ee
	For the second integral, we will additionally have to integrate by parts and recall $T_0>0$\,:
	\ba \label{eq:int 1 rho}
	\int_{T_0}^\infty dt\,t\, \tr [\rho(t)]&\leq -a\int_{T_0}^\infty dt\,t \frac{d}{dt}\tr [\rho(t)]=-a \left( \Big[t\,\tr[\rho(t)]\Big]_{T_0}^\infty-\int_{T_0}^\infty dt\, \tr [\rho(t)] \right)=a T_0 \,\tr[\rho(T_0)]+ a\int_{T_0}^\infty dt\, \tr [\rho(t)]\\
	&\leq(a T_0+a^2)\tr[\rho(T_0)],
	\ea
	where to achieve the last line, we have used Eq. \eqref{eq:int rho}. In order for Eqs. \eqref{eq:int rho} and Eq. \eqref{eq:int 1 rho} to be useful, we need to bound $\tr [\rho(T_0)]$. This can be achieved using Eq. \eqref{eq:tr rho t from WSO paper}. One finds
	\be \label{def:ep1}
	\tr[\rho(T_0)]= |\me^{-\delta} + \braket{\theta_k|\varepsilon_\nu}(T_0,d) |^2\leq \left(\me ^{-\delta} +\varepsilon_\nu(T_0,d)\right)^2.
	\ee
	Finally, recalling that $-\frac{d}{dt}\tr[\rho(t) ]>0$, we have
	\be \label{eq:int mu}
	\mu_0=\int_0^{T_0} dt \tr [\rho (t)]\leq T_0 \max_{t\in[0,T_0]}  \tr [\rho (t)]=T_0 \tr [\rho(0)]=T_0.
	\ee
	Thus using Eqs. \eqref{eq:lim powers rho}, \eqref{eq:int rho}, \eqref{eq:int 1 rho}, \eqref{def:ep1} and \eqref{eq:int mu}, it follows that $\epsilon_1$ defined in Eq. \eqref{eq:epsilon 1}
	is bounded by
	\be
	|\epsilon_1|\leq 2T_0\mu_\epsilon+\mu_\epsilon^2 +2(aT_0+a^2) \left(\me ^{-\delta} +\varepsilon_\nu(T_0,d)\right)^2.
	\ee
	Similarly, we find that $\mu_\epsilon$ defined in Eq. \eqref{eqdef:mu0 and mu epsilon} is bounded by
	\be
	\mu_\epsilon\leq a \left(\me ^{-\delta} +\varepsilon_\nu(T_0,d)\right)^2.
	\ee
	Thus using the definition of $\epsilon_2$ in Eq. \eqref{eq:epsilon 2}, we complete the proof.
\end{proof}

\subsection{Upper and lower bounds for $\tr[\rho(t)]$}\label{Upper and lower bounds for trace rho}
\begin{lemma}\label{lem:1st up lowe on trace rho t}
	\ba \label{eq:trace rho bound}
	\left(\sum_{k\in\mathcal{S}_d(k_0+td/T_0)} \Delta_k\right) -\epsilon_0\leq \tr[\rho(t)] \leq  \left(\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \Delta_k\right) +\epsilon_0,
	\ea
	where
	\ba
	\Delta_k:=   \me^{-2\delta \int_{k-t d/T_0}^k dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k-t d/T_0)   \right|^2, \quad \epsilon_0=\epsilon_0(t,d):= \varepsilon_\nu(t,d)\big(\varepsilon_\nu(t,d) +2\big)d,
	\ea
	and $\varepsilon_\nu(t,d)$ is defined by Eq. \eqref{eq:def varepsiln nu}.

\end{lemma}
\begin{proof}
	Here we will use the core theorem in \cite{WSO16} discussed in Section \ref{sec:expressing R in terms of WSO}.
	Using Eq. \eqref{eq:tr rho t from WSO paper}, Eq. \eqref{eq:trace rho bound} follows from noting $|a+\epsilon|^2= |a|^2+|\epsilon|^2+2\,\mathfrak{Re}(a\,\epsilon)$, $\mathfrak{Re}(a)\leq |a|$ for $a,\epsilon\in\cc$, and $|\braket{\theta_k|\varepsilon_\nu}(t,d)|\leq \varepsilon_\nu(t,d)$ for all $k$.
\end{proof}
We will now calculate time dependent upper and lower bounds for $\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \Delta_k$, where $\Delta_k$ is defined in Lemma \ref{lem:1st up lowe on trace rho t}.
\begin{lemma}\label{lem: t bouns for DeltaC}
	Let $k_0=0$. There exists $\Delta_L\geq 0$, $\Delta_R\geq 0$, $\Delta_C\geq 0$ such that
	\ba
	\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \Delta_k= \Delta_L+\Delta_C+\Delta_R,
	\ea

	where
$\Delta_L$, $\Delta_R$ satisfy the bounds
	\ba
	\Delta_L,\Delta_R &\leq A^2 \frac{\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-\bar k(t))^2}}{1-\me^{-4\pi|\gamma d/2- \bar k(t)|/\sigma^2}},
	\ea
	where $	\bar{k}(t):=\lfloor -d/2 +td/T_0+1\rfloor +d/2 -td/T_0\in[0,1]$ and

		\ba\label{eq:gamma in term of m}
	\gamma &=\gamma(m):= \frac{m-2}{d}\in(0,1), \quad \quad m=
	\begin{cases}
		4,6,8\ldots,d+2 \mbox{ if } d=2,4,6,\ldots\\
		3,5,7,\ldots,d+2 \mbox{ if } d=3,5,7,\ldots
	\end{cases}

	\ea
 \\
	Furthermore, there exists $\Delta_C^{(-)}\geq 0,$ $\Delta_C^{(+)}\geq 0,$ such that $\Delta_C$ satisfies the bounds
	\ba \label{eq:Delta C def simplified}
\Delta_C^{(-)}-\epsilon_\textup{C}\leq	\Delta_C \leq \Delta_C^{(+)}
	\ea
	where
	\be\label{eq: ep C def}

	 \epsilon_\textup{C}=\epsilon_\textup{C}(t):=
	2  A^2 \frac{\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-\bar k(t))^2}}{1-\me^{-4\pi|\gamma d/2- \bar k(t)|/\sigma^2}},
	\ee
	and if $x_0$ (defined in Section \ref{The generator of dynamics and potential function bar V0}) satisfies $x_{vr} +\pi\gamma \leq x_0 \leq 2\pi-x_{vr}-\pi\gamma$, then
	\ba
	\begin{cases}
		\me^{-2\delta \tilde \epsilon_V} \leq \Delta_C ^{(-)}(t)\leq \Delta_C ^{(+)}(t) \leq 1 \quad&\textup{if}\quad 0 \leq t\frac{2\pi}{T_0} \leq x_0-x_{vr}-\pi\gamma,\\

	\quad\quad\;	0  \leq \Delta_C ^{(-)}(t)\leq \Delta_C ^{(+)}(t)  \leq 	\me^{-2\delta(1-\tilde\epsilon_V)}  \quad&\textup{if}\quad x_{vr}+\pi\gamma+x_0 \leq t\frac{2\pi}{T_0}.
	\end{cases}
	\ea
	When the value of $t$ is such that the above bounds do not hold, we can also use the bounds
	\ba
	0 \leq \Delta_C (t) &\leq 1, \quad \forall\, t\geq 0,\,x_0\in\rr.
	\ea
\end{lemma}
\begin{proof}
	\ba
	\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \Delta_k &=  	\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \me^{-2\delta \int_{k-t d/T_0}^k dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k-t d/T_0)   \right|^2 \\
	&=  	\sum_{k=\min \{\mathcal{S}_d(k_0+td/T_0)\}-td/T_0}^{\max \{\mathcal{S}_d(k_0+td/T_0)\}-td/T_0}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2\\
	&=  	\sum_{k=\lfloor -d/2+k_0+1+td/T_0\rfloor-td/T_0}^{\lfloor d/2+k_0+td/T_0\rfloor-td/T_0}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2\\
		&=  	\sum_{k= k_0 -d/2+\bar k(t)}^{ k_0 +d/2+\bar k(t)-1}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2,
	\ea
	where recall $\bar{k}(t)=\lfloor -d/2 +k_0+td/T_0+1\rfloor +d/2 -k_0-td/T_0\in[0,1]$. This follows from noting that $\bar{k}(t)$ is a solution to both equations $\lfloor -d/2+k_0+1+td/T_0\rfloor-td/T_0=k_0 -d/2+\bar k(t)$ and $\lfloor d/2+k_0+td/T_0\rfloor-td/T_0= k_0 +d/2+\bar k(t)-1$, since we can use the identity $1=\lfloor-x+y+1\rfloor - \lfloor x+y\rfloor +2x$, for $2x\in\zz$, $y\in \rr$ and set $x=d/2$, $y=k_0+td/T_0$. For simplicity, we will now take into account that $k_0=0$. We will now break the sum up into three contributions where the first two will correspond to the ``Gaussian tails" to the ``left" ($\Delta_L$) and ``right" ($\Delta_R$) of $k=k_0=0$, and a ``central term" ($\Delta_C$) corresponding to the region $k\approx k_0=0$. Namely,
	\ba
		\sum_{k\in\mathcal{S}_d (k_0+td/T_0)} \Delta_k= \Delta_L+\Delta_C+\Delta_R.
	\ea
	We introduce $\gamma\in(0,1)$ and start with bounding $\Delta_L$:
	\ba\label{eq:bound for Delta L}
	\Delta_L&:= \sum_{k= -d/2+\bar k(t)}^{\bar k(t)-\gamma d/2}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2\leq \sum_{k= -d/2+\bar k(t)}^{\bar k(t)-\gamma d/2}  \left| \psi_\textup{nor}(k_0;k)   \right|^2= \sum_{y=\gamma d/2-\bar k(t)}^{d/2-\bar k(t)}A^2 \me^{-\frac{2\pi}{\sigma^2}(-y)^2}\\
	&\leq \sum_{y=\gamma d/2-\bar k(t)}^{\infty}A^2 \me^{-\frac{2\pi}{\sigma^2}(-y)^2}\leq A^2 \frac{\me^{-\frac{2\pi}{\sigma^2}(\bar k(t)-\gamma d/2)^2}}{1-\me^{-4\pi| \bar k(t)-\gamma d/2|/\sigma^2}}.\label{eq:last line}
	\ea
	where the summations are defined as,
	\be
	\sum_{y=a}^{b} f(y)=f(a)+f(a+1)+\ldots +f(b),
	\ee
	where $a,b\in\rr$, $b-a\in\zz$ and the sum is defined to be zero if $b-a\leq 0$. We will use this convention throughout. We will constrain $\gamma$ appropriately  later in Eq. \eqref{eq:gamma in term of m in proof}. In the last line of Eq. \eqref{eq:last line} we have used Lemma J.0.1, page 59 from \cite{WSO16}. Similarly, for $\Delta_R$,
	\ba\label{eq:bound for Delta R}
		\Delta_R&:= \sum_{k=\bar k(t) +\gamma d/2}^{\bar k(t)+ d/2}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2\leq \sum_{k=\bar k(t) +\gamma d/2}^{\bar k(t)+ d/2}  \left| \psi_\textup{nor}(k_0;k)   \right|^2\leq \sum_{k=\bar k(t) +\gamma d/2}^{\infty}  \left| \psi_\textup{nor}(k_0;k)   \right|^2 \\
		&\leq \sum_{k=\bar k(t) -\gamma d/2}^{\infty}  \left| \psi_\textup{nor}(k_0;k)   \right|^2 \leq A^2 \frac{\me^{-\frac{2\pi}{\sigma^2}(\bar k(t)-\gamma d/2)^2}}{1-\me^{-4\pi| \bar k(t)-\gamma d/2|/\sigma^2}}= A^2 \frac{\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-\bar k(t))^2}}{1-\me^{-4\pi|\gamma d/2- \bar k(t)|/\sigma^2}}.
	\ea
	For $\Delta_C$, we have
	\ba \label{eq:Delta C def}
	\Delta_C&:= \sum_{k=\bar k(t) -\gamma d/2+1}^{\bar k(t)+ \gamma d/2-1}  \me^{-2\delta \int_{k}^{k+td/T_0} dy {\bar V_d}(y)}\left| \psi_\textup{nor}(k_0;k)   \right|^2\\
	&
	\begin{cases}
	 \leq \bigg(\max_{y\in \mathcal{I}_\gamma} \Big{\{} \me^{-2\delta \int_{y}^{y+td/T_0} dx {\bar V_d}(x)} \Big{\}}\bigg)\sum_{k=\bar k(t) -\gamma d/2+1}^{\bar k(t)+ \gamma d/2-1} \left| \psi_\textup{nor}(k_0;k)   \right|^2,\\ \vspace{-0.3cm}\\
	 \geq \bigg(\min_{y\in \mathcal{I}_\gamma} \Big{\{} \me^{-2\delta \int_{y}^{y+td/T_0} dx {\bar V_d}(x)} \Big{\}}\bigg)\sum_{k=\bar k(t) -\gamma d/2+1}^{\bar k(t)+ \gamma d/2-1} \left| \psi_\textup{nor}(k_0;k)   \right|^2,\label{eq: Delta C upper bottom bound split}
	\end{cases}
	\ea
	where $\mathcal{I}_\gamma:=\{ \bar k(t)-\gamma d/2+1, \bar k(t)-\gamma d/2+2,\ldots,\bar k(t)+\gamma d/2-1 \}.$ In order for the summation in the definitions of $\Delta_L, \Delta_C, \Delta_R$ to be well defined, we need the constraints, $-d\gamma
	/2+\bar k(t)-(-d/2+\bar k(t))=n_1$, $d\gamma
	/2+\bar k(t)-1-(-d\gamma/2+\bar k(t)+1)=n_2$, $d\gamma
	/2+\bar k(t)-(d/2+\bar k(t))=n_3$, for $n_1,n_2,n_3 \in \zz$. A solution for $\gamma$ is
	\be\label{eq:gamma in term of m in proof}
	\gamma=\gamma(m)= \frac{m-2}{d}\in(0,1), \quad \text{where } m=
	\begin{cases}
		4,6,8,\ldots,d+2 \mbox{ if } d=2,4,6,\ldots\\
		3,5,7,\ldots,d+2 \mbox{ if } d=3,5,7,\ldots
	\end{cases}
	\ee
	Before proceeding to bound the maximization and minimization in Eq. \eqref{eq: Delta C upper bottom bound split}, we will bound the common factor term which is approximately one. We find
	\ba \label{eq:gam gam sum equality}
	\sum_{k=\bar k(t) -\gamma d/2+1}^{\bar k(t)+ \gamma d/2-1} \left| \psi_\textup{nor}(k_0;k)   \right|^2&=\sum_{k\in\mathcal{S}_d(k_0+t d/T_0)} \left| \psi_\textup{nor}(k_0;k-t d/T_0)   \right|^2-\sum_{k=\bar k(t) -d/2}^{\bar k(t)-\gamma d/2} \left| \psi_\textup{nor}(k_0;k)   \right|^2- \sum_{k=\bar k(t) +\gamma d/2}^{\bar k(t)+ d/2} \left| \psi_\textup{nor}(k_0;k)   \right|^2\\
	&=1-\sum_{k=\bar k(t) -d/2}^{\bar k(t)-\gamma d/2} \left| \psi_\textup{nor}(k_0;k)   \right|^2- \sum_{k=\bar k(t) +\gamma d/2}^{\bar k(t)+ d/2} \left| \psi_\textup{nor}(k_0;k)   \right|^2.
	\ea
	where in the last line we have used Eq. \eqref{eq:normalisation}.

The remaining two terms in Eq. \eqref{eq:gam gam sum equality} have been bounded in Eqs. \eqref{eq:bound for Delta L}, \eqref{eq:bound for Delta R}. Noting that $0\leq  \me^{-2\delta \int_{y}^{y+td/T_0} dx {\bar V_d}(x)} \leq 1$, we can thus simplify Eq. \eqref{eq: Delta C upper bottom bound split}, to find
	\ba \label{eq:Delta C def simplified in  proof}
\Delta_C
\begin{cases}
	\leq \Delta_C^{(+)} \\%+|\epsilon_\textup{A}|\\
	 \vspace{-0.3cm}\\
	\geq \Delta_C^{(-)}-\epsilon_\textup{C},
\end{cases}
\ea
where
$\epsilon_\textup{C}$ is defined in the statement of the Lemma and
\be
\Delta_C^{(+)}:= \max_{y\in \mathcal{I}_\gamma} \Big{\{} \me^{-2\delta \int_{y}^{y+td/T_0} dx {\bar V_d}(x)} \Big{\}},\quad \Delta_C^{(-)}:= \min_{y\in \mathcal{I}_\gamma} \Big{\{} \me^{-2\delta \int_{y}^{y+td/T_0} dx {\bar V_d}(x)} \Big{\}}.
\ee \\

	Our next aim will be to find bounds on $t$ which determine whether $\Delta_C$ is approximately $1$ or $\me^{-2\delta}$. First note,
	\be \label{eq:re writing int to t cond}
	\int_{y}^{y+td/T_0} dx {\bar V_d}(x)=\frac{2\pi}{d} \int_{y}^{y+td/T_0} dx \bar{V}_0\left(\frac{2\pi}{d}x\right)=  \int_{2\pi y/d-x_0}^{2\pi y/d+t2\pi/T_0-x_0} dx \bar{V}_0\left(x+x_0\right).
	\ee
	\begin{figure}
		\includegraphics[angle=-1,scale=1.1]{Pot_for_ProofV5.png}
		\centering
		\caption{Depiction of the periodic potential function $\bar{V}_0(x+x_0)$ with the parameters $x_{vl}$, $x_{vr}$ and $\tilde{\epsilon}_V$ introduced in Eq. \eqref{eq:tilde ep V def}. Observe that the area under the potential $\bar{V}_0(x+x_0)$ (i.e. its integral) between $x=x_{vr}-2\pi$ and $x=x_{vl}$ is $\tilde\epsilon_V$ due to symmetry. The quantities $\frac{2\pi}{d} y-x_0$ and $\frac{2\pi}{d} y+t\frac{2\pi}{T_0}-x_0$ are the lower and upper integration limits in Eq. \eqref{eq:re writing int to t cond}.\label{fig:pot pic proof} }
	\end{figure}
	We can now use Eq. \eqref{eq:re writing int to t cond} to find conditions for the time "before a tick is measured", namely conditions for which
	\be\label{eq:before mtick condition}
	\int_{y}^{y+td/T_0} dx {\bar V_d}(x)\leq 1-(1-\tilde{\epsilon}_V)=\tilde{\epsilon}_V.
	\ee
	From Fig. \ref{fig:pot pic proof} and Eq. \eqref{eq:re writing int to t cond} we observe that Eq. \eqref{eq:before mtick condition} is satisfied for all $y\in \mathcal{I}_\gamma$ if
	\ba
	x_{vr}-2\pi&\leq \frac{2\pi}{d}y-x_0\\
	\frac{2\pi}{d}y+t\frac{2\pi}{T_0} -x_0&\leq x_{vl}=-x_{vr},
	\ea
	for all $y\in \mathcal{I}_\gamma$. Or equivalently, if
	\ba
	\frac{2\pi}{d} \min_{y\in\mathcal{I}_\gamma}\{y\} &\geq x_{vr}-2\pi+x_0\\
	\frac{2\pi}{d}\max_{y\in\mathcal{I}_\gamma}\{y\}&\leq x_0-\frac{2\pi}{T_0}t-x_{vr},
	\ea
	from which it follows,
	\ba
	x_0 &\leq 2\pi -x_{vr}-\pi\gamma+\frac{2\pi}{d}(\bar k(t)+1)\label{eq:1LowBoundOnt pre}\\
	0\leq t\frac{2\pi}{T_0} &\leq x_0-x_{vr}-\pi\gamma-\frac{2\pi}{d}(\bar k(t)-1).\label{eq:2LowBoundOnt pre}
	\ea
	Thus recalling that $\bar k(t)\in[0,1]$, sufficient conditions on $x_0$ and $t$ for Eq. \eqref{eq:before mtick condition} to be satisfied are
	\ba
	x_0 &\leq 2\pi -x_{vr}-\pi\gamma\label{eq:1LowBoundOnt}\\
	0\leq t\frac{2\pi}{T_0} &\leq x_0-x_{vr}-\pi\gamma.\label{eq:2LowBoundOnt}
	\ea
Similarly, we can work out conditions for the time ``after a tick has occurred", i.e.
	\be \label{eq:after tick has happened condition}
	\int_{y}^{y+td/T_0} dx {\bar V_d}(x)\geq 1-\tilde{\epsilon}_V,
	\ee
	for all $y\in \mathcal{I}_\gamma$ if
	\ba

	\frac{2\pi}{d}y-x_0 &\leq -x_{vr}\\
	x_{vr} &\leq  \frac{2\pi}{d}y+t\frac{2\pi}{T_0}-x_0
	\ea
	for all $y\in \mathcal{I}_\gamma$. Or equivalently, if
	\ba
	x_{vr}+\max_{y\in\mathcal{I}_\gamma} \{y\}\frac{2\pi}{d} &\leq x_0
	\\
		x_{vr}+\frac{2\pi}{d}\max_{y\in \mathcal{I}_\gamma}\{-y\}+x_0 &\leq t\frac{2\pi}{T_0},
	\ea
	from which it follows
	\ba
	x_{vr} +\frac{2\pi}{d}(\bar k(t)-1)+\pi\gamma &\leq x_0
	\label{eq:x0 contraint pre}\\
	x_{vr}-\frac{2\pi}{d}(\bar k(t)+1)+\pi\gamma+x_0 &\leq t\frac{2\pi}{T_0}
	\label{eq:afterclick pre}
	\ea
		Thus recalling that $\bar k(t)\in[0,1]$, sufficient conditions on $x_0$ and $t$ for Eq. \eqref{eq:after tick has happened condition} to be satisfied are
\ba
x_{vr} +\pi\gamma &\leq x_0,
\label{eq:x0 contraint}\\
x_{vr}+\pi\gamma+x_0 &\leq t\frac{2\pi}{T_0}.
 \label{eq:afterclick}
\ea
	Thus from Eqs. \eqref{eq:Delta C def}, \eqref{eq:1LowBoundOnt}, \eqref{eq:2LowBoundOnt},\eqref{eq:x0 contraint}, and recalling $\bar{V}_0\geq 0$, we conclude that if $x_0$ satisfies
	\be \label{eq:up low bound on x0}
	x_{vr} +\pi\gamma \leq x_0 \leq 2\pi-x_{vr}-\pi\gamma,
	\ee
	then

	\ba
	\me^{-2\delta \tilde \epsilon_V} \leq \Delta_C ^{(-)}(t)\leq \Delta_C ^{(+)}(t) &\leq 1 \quad\textup{if}\quad 0 \leq t\frac{2\pi}{T_0} \leq x_0-x_{vr}-\pi\gamma,\\

	0  \leq \Delta_C ^{(-)}(t)\leq \Delta_C ^{(+)}(t)  &\leq 	\me^{-2\delta(1-\tilde\epsilon_V)}  \quad\textup{if}\quad x_{vr}+\pi\gamma+x_0 \leq t\frac{2\pi}{T_0}.
	\ea

\end{proof}

\subsection{Calculating $\Delta(0)$ and $\Delta(1)$ and an explicit lower bound on ${R_1}$}\label{Calculating Delta(0) and Delta(1) and an explicit lower bound on R1}
Define
\be\label{eq:def delt t}
 t_1:=(x_0-x_{vr}-\pi\gamma)\frac{T_0}{2\pi},\quad t_2:=(x_0+x_{vr}+\pi\gamma)\frac{T_0}{2\pi},\quad \Delta t:= t_2-t_1=(x_{vr}+\pi\gamma)\frac{T_0}{\pi}.
\ee
Furthermore, we will want $t_1$ to be in the centre of the range $[0,T_0]$. As such, we set $t_1=T_0/2$ which taking into account the definition of $t_1$ in Eq. \eqref{eq:def delt t} implies
\be \label{eq:x0 def}
x_0=\pi+x_{vr}+\pi\gamma.
\ee
Physically, Eq. \eqref{eq:x0 def} means that the potential is peaked near the mid point $\pi$ and thus that the continuous measurements will occur approximately at a time $T_0/2$.
Note that Eq. \eqref{eq:x0 def} is consistent with Eq. \eqref{eq:up low bound on x0} as long as $x_{vr}+\pi\gamma\leq \pi/2$. As we will see later, both $\gamma$ and $x_{vr}$ will be parametrized such that $\gamma$, $x_{vr}$ $\rightarrow 0$ as $d\rightarrow \infty$.\footnote{See Eqs. \eqref{eq:up bound gamma even},\eqref{eq:bound for gamma d over sigma odd},\eqref{eq:x vr limit contraint} for explicit parametrization.} As such, this constraint will always be satisfied for sufficiently large $d$.\footnote{It is expected that one could actually weaken the constraint imposed on $x_{vr}$ and $\gamma$ if one choose a slightly different parametrization of $t_1$, say $t_1=T_0/2-(x_{vr}+\pi\gamma)T_0/\pi$. But for simplicity, we will not do this here.}

\begin{lemma}\label{lem:R lowe pre leading order}
$R_1$ satisfies the bound
	\be
{R_1}	\geq\frac{-\left( \left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt |\epsilon_\textup{C}(t)|\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)  +\Delta t^2\right)/T_0^2+\me^{-4\delta \tilde{\epsilon}_V}/4}{\epsilon_4+\epsilon_3+\frac{3}{4}\me^{-\delta (1-\tilde{\epsilon}_V)}+(1-\me^{-4\delta \tilde{\epsilon}_V})/4+\frac{2}{T_0^2} \left( t_2-t_1\me^{-2\delta\tilde \epsilon_V}\right) \Delta t +\left(\frac{\Delta t}{T_0}\right)^2},
	\ee
where we have defined
\ba \label{eq:ep 3 4 def}
\epsilon_3&:= \frac{2}{T_0^2}\int_0^{T_0} dt\,t (\Delta_L(t)+\Delta_R(t)),\\
\epsilon_4&:= \epsilon_0(T_0)  + \frac{1}{T_0^2}    \left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \,\epsilon_\textup{C}(t)\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)  +\epsilon_1/T_0^2.
\ea
\end{lemma}

\begin{proof}
Using Lemma \ref{lem:1st up lowe on trace rho t}, followed by Lemma \ref{lem: t bouns for DeltaC},
\ba
\int_0^{T_0} dt \,\tr [\rho(t)]&\geq \int_0^{T_0} dt \left(\Delta_L(t)+\Delta_C(t)+\Delta_R(t)-\epsilon_0\right)\\
&\geq -T_0\epsilon_0(T_0)-\int_0^{T_0}dt \epsilon_\textup{C}(t) + \int_0^{t_1}dt\, \Delta_C^{(-)}(t)+\int_{t_1}^{t_2}dt\, \Delta_C(t)+\int_{t_2}^{T_0}dt\,\Delta_C(t)\\
&\geq -T_0\epsilon_0(T_0)-\int_0^{T_0}dt \epsilon_\textup{C}(t)+ t_1 \me^{-2\delta\tilde{\epsilon}_V}+\int_{t_1}^{t_2}dt\, \Delta_C(t).
\ea
Thus,
\ba
\left( \int_0^{T_0} dt \,\tr [\rho(t)]\right)^2 \geq& \left( -T_0\epsilon_0(T_0)-\int_0^{T_0}dt \epsilon_\textup{C}(t)+ t_1 \me^{-2\delta\tilde{\epsilon}_V}+\int_{t_1}^{t_2}dt\, \Delta_C(t)\right)^2\\
\geq& \left( -T_0\epsilon_0(T_0)-\int_0^{T_0}dt \epsilon_\textup{C}(t)\right)^2 -\left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \epsilon_\textup{C}\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\int_{t_1}^{t_2}dt\, \Delta_C(t)\right)\\
&+  \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\int_{t_1}^{t_2}dt\, \Delta_C(t)\right)^2\\
\geq&  -\left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \epsilon_\textup{C}(t)\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)+  t_1^2\me^{-4\delta\tilde \epsilon_V} +2\me^{-2\delta\tilde \epsilon_V}\,t_1 \int_{t_1}^{t_2}dt\, \Delta_C(t) \\&+\left(\int_{t_1}^{t_2}dt\, \Delta_C(t)\right)^2\\
\geq&  -\left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \epsilon_\textup{C}(t)\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)+  t_1^2 \me^{-4\delta\tilde \epsilon_V} +2\me^{-2\delta\tilde \epsilon_V}\,t_1 \int_{t_1}^{t_2}dt\, \Delta_C(t) -\Delta t^2\\
 =&\Delta^2(0),
\ea
where we have used Definition \eqref{eq:def delt t} and Lemma \ref{lem: t bouns for DeltaC}.
Similarly,
\ba
\int_0^{T_0} dt \,t\,\tr [\rho(t)]& \leq \int_0^{T_0} dt \left(\Delta_L(t)+\Delta_C(t)+\Delta_R(t)+\epsilon_0\right)\\
&\leq \frac{T_0^2}{2}\left(\epsilon_0(T_0) +\epsilon_3\right)+ \int_0^{t_1}dt\, t\,\Delta_C(t)+\int_{t_1}^{t_2}dt\, t\,\Delta_C(t)+\int_{t_2}^{T_0}dt\,t\,\Delta_C(t)\\
&\leq \frac{T_0^2}{2}\left(\epsilon_0(T_0) +\epsilon_3\right)+ \frac{t_1^2}{2}+\int_{t_1}^{t_2}dt\, t\,\Delta_C(t)+\me^{-\delta (1-\tilde{\epsilon}_V)}\frac{T_0^2-t_2^2}{2}\\
&=\Delta(1),
\ea
where $\epsilon_3$ is defined in Eq. \eqref{eq:ep 3 4 def}. Setting $t_1=T_0/2$ and simplifying Eq. \eqref{eq:R lo bound 2}, we achieve
\ba
{R_1}&\geq \frac{\Delta^2(0)+\epsilon_2}{2\Delta(1)-\Delta^2(0)+\epsilon_1}\\
&=\frac{\left( -\left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \epsilon_\textup{C}(t)\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)+  T_0^2\me^{-4\delta \tilde{\epsilon}_V}/4-\Delta t^2\right)/T_0^2 +\left(2\me^{-2\delta\tilde \epsilon_V}\,t_1 \int_{t_1}^{t_2}dt\, \Delta_C(t) +\epsilon_2\right)/T_0^2}{\epsilon_4+\epsilon_3+\frac{3}{4}\me^{-\delta (1-\tilde{\epsilon}_V)}+(1-\me^{-4\delta \tilde{\epsilon}_V})/4+\frac{2}{T_0^2}\int_{t_1}^{t_2}dt\left( t-t_1\me^{-2\delta\tilde \epsilon_V}\right) \Delta_C(t) +\left(\frac{\Delta t}{T_0}\right)^2}\\
&\geq\frac{-\left( \left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt \epsilon_\textup{C}(t)\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)  +\Delta t^2\right)/T_0^2+\me^{-4\delta \tilde{\epsilon}_V}/4}{\epsilon_4+\epsilon_3+\frac{3}{4}\me^{-\delta (1-\tilde{\epsilon}_V)}+(1-\me^{-4\delta \tilde{\epsilon}_V})/4+\frac{2}{T_0^2} \left( t_2-t_1\me^{-2\delta\tilde \epsilon_V}\right)\int_{t_1}^{t_2}dt \Delta_C(t) +\left(\frac{\Delta t}{T_0}\right)^2}\\
&\geq\frac{-\left( \left(2T_0\epsilon_0(T_0)+\int_0^{T_0}dt |\epsilon_\textup{C}(t)|\right) \left(t_1 \me^{-2\delta\tilde{\epsilon}_V}+\Delta t\right)  +\Delta t^2\right)/T_0^2+\me^{-4\delta \tilde{\epsilon}_V}/4}{\epsilon_4+\epsilon_3+\frac{3}{4}\me^{-\delta (1-\tilde{\epsilon}_V)}+(1-\me^{-4\delta \tilde{\epsilon}_V})/4+\frac{2}{T_0^2} \left( t_2-t_1\me^{-2\delta\tilde \epsilon_V}\right) \Delta t +\left(\frac{\Delta t}{T_0}\right)^2},
\ea
where $\epsilon_4$ is defined in Eq. \eqref{eq:ep 3 4 def}.
\end{proof}

\subsection{${R_1}$ to leading order}\label{R1 to leading order}

\begin{lemma}\label{R to leading order}

Assume $\tilde{\epsilon}_V\rightarrow 0$, $\epsilon(T_0,d)\rightarrow 0$, $\delta\rightarrow \infty$, $a\rightarrow\infty$, and $\delta\,\tilde{\epsilon}_V\rightarrow 0$, in the limit $d\rightarrow \infty$, then using Big-O notation, $\bo$, to leading order in $\me^{-\delta},$ $\Delta t$, and $\delta\,\tilde{\epsilon}_V$; ${R_1}$ is lower bounded by
\be \label{eq:R with order terms}
{R_1} \geq\frac{1/4+\bo\left(\epsilon(T_0,d)\,d\right)+\bo(\Delta t^2)}{\bo\left( A^2 \left(\frac{\sigma}{d^{\eta/2}}+1\right)\me^{-\frac{\pi}{2}d^\eta}  \right)+\bo\left(a^2d\,\epsilon(T_0,d)\right)+\bo(a^2\,\me^{-\delta})+\bo(\delta\tilde{\epsilon}_V)+3\left(\frac{\Delta t}{T_0}\right)^2},
\ee
for all fixed constants $\eta>0$ and $\sigma, d$ satisfying $4/\sigma< d^{\eta/2}\leq d/\sigma$.
\end{lemma}
\begin{proof}
	We start with a technical definition which allows us to upper and lower bound $\gamma$  (defined in terms of the integer $m$ in Eq. \eqref{eq:gamma in term of m}) in terms of $\sigma$ and a power of $d^{\eta/2}$, which will be crucial for the rest of the proof.\\

	For $\eta>0$ with $4/\sigma< d^{\eta/2}\leq d/\sigma$, parametrize $m$ by
	\be\label{Eq: bar m def 1}
	m=
	\begin{cases}
		2\lfloor \bar m \rfloor &\mbox{ if } d=2,4,6,\ldots,\\ \vspace{-0.3cm}\\
		2\lfloor \bar m \rfloor+1 &\mbox{ if } d=3,5,7,\ldots,
	\end{cases}
	\ee
	where
	\be\label{Eq: bar m def 2}
	\bar m :=
	\begin{cases}
		\frac{d^{\eta/2}}{2}\sigma +1 \in\big(3,d/2+1\big] &\mbox{ if } d=2,4,6,\ldots,\\ \vspace{-0.3cm}\\
		\frac{d^{\eta/2}}{2}\sigma +\frac{1}{2} \in\big(5/2,(d+1)/2\big]  &\mbox{ if } d=3,5,7,\ldots.
	\end{cases}
	\ee
	Note the consistency of the domains of $\bar m$ and $m$: the domain of $\bar m$ (which follows from the constraint $4/\sigma< d^{\eta/2}\leq d/\sigma$) in Eq. \eqref{Eq: bar m def 2} implies (via Eq. \eqref{Eq: bar m def 1}) a range for $m$ of $m\in\{6,8,10,\ldots,d+2\}$ for $d=2,4,6,\ldots,$ and $m\in\{5,7,9,\ldots,d+2\}$ for $d=3,5,7\ldots,$ which is within the domain of $m$ defined in Eq. \eqref{eq:gamma in term of m}.\\

	From the bound $x-1\leq \lfloor x \rfloor \leq x$, $x\in\rr$, it follows $2\bar m-2\leq m\leq 2\bar m$ and thus from Eqs. \eqref{eq:gamma in term of m}, \eqref{Eq: bar m def 1}, \eqref{Eq: bar m def 2}  it follows
	\be \label{eq:up bound gamma even}
	\gamma=\frac{m-2}{d} \leq \frac{2\bar m-2}{d}=d^{\eta/2}\frac{\sigma}{d}\quad \textup{for } d=2,4,6,\ldots,
	\ee
	and
	\be \label{eq:bound for gamma d over sigma even}
	\gamma =\frac{m-2}{d}\geq \frac{2\bar m-4}{d}= d^{\eta/2}\frac{\sigma}{d}-\frac{2}{d} \quad \textup{for } d=2,4,6,\ldots
	\ee
	Similarly, we find
	\ba\label{eq:bound for gamma d over sigma odd}
	d^{\eta/2} \frac{\sigma}{d} -\frac{1}{d} \leq \gamma \leq d^{\eta/2}\frac{\sigma}{d} \quad \text{for } d=1,3,5,
	\ea
	We will now bound $\epsilon_3$ using Eqs. \eqref{eq:bound for gamma d over sigma even}, \eqref{eq:bound for gamma d over sigma odd},
	\ba
	\epsilon_3 &=  \frac{2}{T_0^2}\int_0^{T_0} dt\,t (\Delta_L(t)+\Delta_R(t))\leq \left(\frac{2 A}{T_0}\right)^2\int_0^{T_0} dt   \frac{t\,\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-\bar k(t))^2}}{1-\me^{-4\pi|\gamma d/2- \bar k(t)|/\sigma^2}}\leq \left(\frac{2 A}{T_0}\right)^2\int_0^{T_0} dt  \frac{t\,\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-1)^2}}{1-\me^{-4\pi(\gamma d/2-1)/\sigma^2}}\\
	&= {2 A^2}  \frac{\me^{-\frac{2\pi}{\sigma^2}(\gamma d/2-1)^2}}{1-\me^{-4\pi(\gamma d/2-1)/\sigma^2}}
	\leq 2 A^2 \frac{\me^{-\frac{\pi}{2}(d^{\eta/2}-4/\sigma)^2}}{1-\me^{-2\pi (d^{\eta/2}-4/\sigma)/\sigma}},\label{eq:line last line in ep 3 bound}
	\ea
	where the denominator of line \eqref{eq:line last line in ep 3 bound} is finite since $d^{\eta/2}-4/\sigma>0$ and we have used the observation that $1/(1-\me^{-x})$ is monotonic decreasing in $x$ for $x>0$.
	In the $d\rightarrow\infty$ limit, there are a number of possibilities, depending on whether $d^{\eta/2}/\sigma$ tends to infinity, zero or a positive constant. By calculating these cases separately.
	\begin{itemize}
	\item [1)] If $\lim_{d\rightarrow \infty} d^{\eta/2}/\sigma=\infty$ we find
		\be
		\epsilon_3= \bo\left( A^2 \frac{\sigma}{d^{\eta/2}}\me^{-\frac{\pi}{2}d^\eta}  \right).
		\ee
	\item [2)] 	If $\lim_{d\rightarrow \infty} d^{\eta/2}/\sigma$ converges to a constant,  we find
\be
\epsilon_3= \bo\left( A^2 \me^{-\frac{\pi}{2}d^\eta}  \right).
\ee
	\end{itemize}

	Thus recalling that Big-O notation captures the worse case scenario (i.e. it is an upper bound in the limit), combining the conclusions from 1) and 2) above we find for arbitrary $d^{\eta/2}/\sigma$,
	\be
	\epsilon_3= \bo\left( A^2 \left(\frac{\sigma}{d^{\eta/2}}+1\right)\me^{-\frac{\pi}{2}d^\eta}  \right).
	\ee
	Similarly, we find that $\epsilon_\textup{C}$ defined in Eq. \eqref{eq: ep C def} is of order
		\be
	\epsilon_\textup{C}= \bo\left( A^2 \left(\frac{\sigma}{d^{\eta/2}}+1\right)\me^{-\frac{\pi}{2}d^\eta}  \right).
	\ee

	Recalling the definition of $\epsilon_0$, $\epsilon_1$, $\epsilon_2$ and $\epsilon_4$, we find
	\ba
	&\epsilon_0 = \bo(d\,\epsilon(T_0,d)),&\quad
	&\epsilon_1 = \bo(a^2\,\me^{-\delta})+\bo(a^2\,\epsilon(T_0,d)),&\quad

	&\epsilon_4 = \bo(d\,\epsilon(T_0,d))+\bo(\me^{-\delta})+\bo(\epsilon_\textup{C}),&
	\ea
	Thus concluding the proof.
\end{proof}

\begin{corollary}\label{corrolary w lims}
	Recall that $\Delta t/T_0=x_{vr}/\pi+\gamma$. Let the conditions in Lemma \ref{R to leading order} be satisfied. In addition, if $x_{vr}$ is of higher order than $\gamma$, $(x_{vr}=\lo(\gamma))$; and the order terms in Eq. \eqref{eq:R with order terms} are of higher order than $\gamma^{2}$, $\big(\text{i.e.\,\,\,} A^2 (\sigma/d^{\eta/2}+1)e^{-\frac{\pi}{2}d^{\eta}}=\lo(\gamma)$, $a^2 d \,\epsilon(T_0,d)=\lo(\gamma)$, $a^2 \me^{-\delta}=\lo(\gamma)$, $\delta \tilde \epsilon_V=\lo(\gamma)\big)$ it follows from Eqs. \eqref{eq:up bound gamma even}, \eqref{eq:bound for gamma d over sigma odd}, that
	\be
	{R_1}\geq \frac{1}{12} \gamma^{-2}+\lo(\gamma^{-2})\geq \frac{1}{12} \frac{d^{2-\eta}}{\sigma^2}+\lo\left(\frac{d^{2-\eta}}{\sigma^2}\right),
	\ee
	where $\lo$ is little-o notation.
\end{corollary}
\begin{proof}
	Follows directly from Lemma \ref{R to leading order} by keeping leading order terms only.
\end{proof}

\subsection{Showing that the limit requirements of Corollary \ref{corrolary w lims} can be met}\label{Showing that the limit requirements}
The difficulty in proving the conditions given in Lemma \ref{R to leading order} can be satisfied, and that $x_{vr}$ can be of higher order than $\gamma$, consists in finding a potential $\bar{V}_0$ which can be parametrized in terms of $d$ adequately. In this section, we will demonstrate via a particular $\bar{V}_0$ that this can indeed be achieved. Recall that the potential $\bar V_0$ which we will introduce here, has been summarised in Section \ref{Explicit form on the potential bar V0} along with the particular parametrizations which we will pick in this section, as need be.
\subsubsection{Definition of $\bar{V}_0$}
We start by defining $\bar{V}_0:\rr\rightarrow\rr$ appropriately. Let\footnote{One could also exchange the term $1/(\delta d^2)$ with $1/(\delta d^K)$ for some constant $K>0$ and the main result, Theorem \ref{Thm:R lower quant bound} would still go through.}
\be \label{eq:def of rapped pot}
\bar{V}_0(x):=\frac{1}{\delta d^2}+n A_0 \sum_{p=-\infty}^{\infty} V_B\left (n(x-x_0-2\pi p) \right),
\ee
where $n\geq 1$ is a free parameter, $A_0$ a normalization constant such that
\be \label{eq:norr V0 requirement}
\int_0^{2\pi} \bar{V}_0(x)\,dx =1,
\ee
and
\be
V_B(x):=\textup{sinc}^{2 N}(x):=\left( \frac{\sin(\pi x)}{\pi x}\right)^{2 N},
\ee
where $V_B(0)=1$ (defined by continuity in $x$) and $N\in \nnp$ is another free parameter. In the following if statements about $\bar{V}_0$ or $V_B$ are made without mentioning either $n$ or $N$, it is to be understood that these properties hold for all $n\geq 1$ and $N\in \nnp$.\\
We will now show some useful properties of $\bar{V}_0$. Later we will parameterize $n$ in terms of $d$, while $N$ will be a constant dependent on the small ($d$ independent) parameter $\eta$ introduced in Lemma \ref{R to leading order}.\\

\subsubsection{Finding a value for $a$}
It will now become apparent why we included the term $1/(\delta d^2)$ in the definition of the potential $\bar V_0$.
\begin{corollary}\label{cor:a parametrisation for a}
	The constant $a$ defined in Eq. \eqref{eqdef: a} can be set equal to the following
	\be\label{eq:a set to}
	a= \frac{T_0}{4\pi} d^2.
	\ee
\end{corollary}
\begin{proof}
	Since all the terms in the definition of $\bar V_0$ are non-negative, it follows $\min_{x\in[0,2\pi]} \bar V_0(x)\geq 1/(\delta d^2)$. Hence
	\be \label{eq:theorem q clock}
	\frac{T_0}{4\pi\delta}\frac{1}{\min_{x\in[0,2\pi]}\bar V_0(x)}\leq \frac{T_0}{4\pi}d^2.
	\ee
	Eq. \eqref{eq:a set to} is thus a direct consequence of Eq. \eqref{eqdef: a}.
\end{proof}

\subsubsection{Properties of $\bar{V}_0$}
First note that $\bar{V}_0$ is periodic with period $2\pi$, and is infinitely differentiable. The first of these properties is clear from Eq. \eqref{eq:def of rapped pot} while the second will be demonstrated in Lemma \ref{lem: devs grow like powers}. These two properties are required by definition in \cite{WSO16}. We will now show some additional properties, specific to this particular choice of potential function $\bar{V}_0$.
\begin{lemma}[Normalization]\label{lem:Normalization A0}
$A_0$ is $n$ independent. In particular, it satisfies $A_0<\underline{A}_0$, where $\underline{A}_0$ is only a function of $N$.
\end{lemma}
\begin{proof}
	Since for all $x\in[0,2\pi]$,
	\be
	\sum_{p=-\infty}^\infty\big{|}V_B(n(x-x_0-2\pi p))\big{|}\leq 1+\sum_{p\in\zz\backslash\{0\}} \frac{1}{ p^{2N}}<\infty,
	\ee
	it follows from the Weierstrass M-test (see Theorem 7.10 in \cite{rudin1976principles}), that the sum in Eq. \eqref{eq:def of rapped pot} converges uniformly for all $x\in[0,2\pi]$. We will use this in the following to exchange summation and integration limits over a finite interval.
Using Eq. \eqref{eq:norr V0 requirement}, we have
\ba
1&=\int_0^{2\pi} dx \bar{V}_0(x)=\int_0^{2\pi}dx \bar{V}_0(x+x_0)=A_0 n  \int_0^{2\pi}dx \sum_{p=-\infty}^\infty V_B(n(x-2\pi p))+\int_0^{2\pi}\frac{dx}{\delta d^2}\\
&=A_0 n \sum_{p=-\infty}^\infty \int_0^{2\pi}dx V_B(n(x-2\pi p))+\frac{2\pi}{\delta d^2}= A_0 n \sum_{p=-\infty}^\infty \int_{-2\pi p}^{2\pi(1-p)}dx V_B(nx)+\frac{2\pi}{\delta d^2}= A_0 n \int_{-\infty}^\infty dx V_B(n x)+\frac{2\pi}{\delta d^2}\\
&= A_0  \int_{-\infty}^\infty dx V_B(x)+\frac{2\pi}{\delta d^2}.
\ea
Hence,
\be
A_0=\frac{1-2\pi/(\delta d^2)}{\int_{-\infty}^\infty dx V_B(x)}< \frac{1}{\int_{-\infty}^\infty dx V_B(x)}=:\underline{A}_0 ,
\ee
which is $n$ independent, since $V_B$ is, and well defined since $V_B\in L^1$ with a non-zero integral.
\end{proof}
\begin{lemma}[Technical Lemma needed for Lemma \ref{lem: devs grow like powers}]\label{tech lemm dev interchange}
\be
\frac{d^k}{dx^k} \sum_{p=-\infty}^\infty  V_B(x n -2\pi n p)= 	 \sum_{p=-\infty}^\infty \frac{d^k}{dx^k} V_B(x n -2\pi n p),\label{lem stament pre power}
\ee
	for all $k\in\nnp$ and $x\in[0,2 \pi]$.

\end{lemma}
\begin{proof}
	For $p=\pm 2,\pm 3,\pm 4,\ldots$, $x\in[0,2\pi]$ and for all $k\in\nnp$,
\ba \label{eq:M p k}
\bigg{|}\frac{d^k}{dx^k} V_B&(x n -2\pi n p)\bigg{|}\\
&\leq \Bigg{|} \frac{n^k}{(\pi n ( x-2\pi\ p))^{2N}} \sum_{q=0}^k \binom{k}{q} \frac{ (-{2N})(-{2N}-1)\ldots (-{2N}-q+1) }{n^q (x -2 \pi p)^q} \bigg{[}\frac{d^{k-q}}{dz^{k-q}} \sin^{2N}(\pi z)\bigg{]}_{z=n(x-2\pi p)}	\Bigg{|}\\
& \leq  \frac{n^k}{(\pi^2 n |p|)^{2N}} \sum_{q=0}^k \binom{k}{q} \frac{ \big{|}(-{2N})(-{2N}-1)\ldots (-{2N}-q+1) \big{|} }{n^q(2 \pi)^q} D_{k-q}=:M_p^k,
\ea
where we have defined $D_k:=\sup_{z\in\rr} \big{|}{d^k}/{dz^k} \,\sin^{2N}(\pi z)	\big{|}<\infty$ for $k\in\nno$. Hence, since \be \label{eq:uniform converg M test}
\sum_{p\in\zz,\, |p|\geq 2} M_p^k<\infty,
\ee
from the Weierstrass M-test (see Theorem 7.10 in \cite{rudin1976principles}), uniform convergence of $\sum_{p=-\infty}^\infty \, {d^k}/{dx^k} V_B(x n -2\pi n p)$ follows for all $k\in\nnp$.

If $f_n:\rr\rightarrow\rr$ is a sequence of differentiable functions with domain $[a,b]$, $a,b\in\rr$ with derivatives $f_n'$, and $\sum_{n=0}^\infty f_n'$ converges uniformly on $[a,b]$, then if we define $f:=\sum_{n=0}^\infty f_n$ and $|f|<\infty$, it follows that $f'=\sum_{n=0}^\infty f_n'$, where $f'$ is the derivative of $f$. This is a well known Theorem and can be found in e.g. Theorem 7.17 \cite{rudin1976principles}. Thus from Eq. \eqref{eq:uniform converg M test} for $k=1$, Eq. \eqref{lem stament pre power} follows immediate for $k=1$. Proceeding inductively, we prove Eq. \eqref{lem stament pre power} for all $k\in\nnp$.

\end{proof}

\begin{lemma}[Technical Lemma needed for Lemma \ref{lem: devs grow like powers}]\label{lem:finite set of discont}
	Let $f$ denote the \textit{Triangle function}, namely
	\be \label{eq:tringle func tech lem}
	f(x):=
	\begin{cases}
		0 &\mbox{ if } |x|\geq 1\\
		1-|x| &\mbox{ otherwise},
	\end{cases}
	\ee
	and let $\star$ denote Convolution. Then,
		\be \label{eq:lech lemm Conve}
	f^{\,\mathlarger{\star} N}:=\underbrace{	f\star  f \star \ldots \star	f}_{N \text{-times convolution product}}\in L^1,
	\ee
	is a continuous function which is infinitely differentiable on the intervals $(-N,-N+1)\cup(-N+1,-N+2)\cup\ldots\cup(N-1,N)$ and zero on the intervals $(-\infty,-N)\cup(N,+\infty)$ for all $N\in\nnp$.
\end{lemma}
\begin{proof}
	The proof is by induction. Let us start by defining the continuous functions $g_n:\rr\rightarrow\rr$ for $n\in \nnp$,
	\be \label{eq:gn def}
	g_n(x):= P_{n,m}(x)\quad \mbox{if } m\leq x \leq m+1 \text{ for } m\in\zz,
	\ee
	where
	\be \label{eq:gn def 2}
	P_{n,m}(x)=0 \quad\mbox{if  } m\in(-\infty, -n-1)]\cup[n,+\infty),
	\ee
	and $\{P_{n,m}(x)\}$ are a set of real polynomials in $x$. Continuity of $g$ implies
	\be\label{eq:continuity Pnm}
	P_{n,m}(m+1)=P_{n,m+1}(m+1)\quad \text{for all } m\in\zz.
	\ee
	We denote the set of such functions $g_n$ by $\mathcal{S}_n$. Note that $f$ in Eq. \eqref{eq:tringle func tech lem} belongs to $\mathcal{S}_1$. We will start by showing that $(g_1\star g_n)(y)\in\mathcal{S}_{n+1}$ for all $n\in\nnp$, $g_n\in\mathcal{S}_n$ and $g_1\in\mathcal{S}_1$. For this we will calculate $(g_1\star g_n)(y)$ for $y\in[q,q+1]$; for $q\in\zz$ and define $\delta_y:=y-q$, $0\leq \delta_y\leq 1$. We find
	\ba \label{eq:g1 star gn 1sr eq}
	(g_1\star g_n)(y)=& \int_{-\infty}^\infty dx g_1(x) g_n(x-y)=\sum_{m=-1,0} \int_m^{m+1}dx P_{1,m}(x) g_n(x-y)\\
	=& \sum_{m=-1,0}\left( \int_m^{m+\delta_y}dx P_{1,m}(x) g_n(x-q-\delta_y)+\int_{m+\delta_y}^{m+1}dx P_{1,m}(x) g_n(x-q-\delta_y) \right)\\
	=& \sum_{m=-1,0}\left( \int_m^{m-q+y}dx P_{1,m}(x) P_{n,m-q-1}(x-y)+\int_{m-q+y}^{m+1}dx P_{1,m}(x) P_{n,m-q}(x-y) \right)\label{line: g1 gn}\\
	=:&P'(n,q;y),
	\ea
	where $P'(n,q;y)$ is a polynomial in $y$ with coefficient depending on $n$ and $q$. This follows by noting: 1) the integral of $P_{1,m}(x) P_{n,m-q-1}(x-y)$ and $P_{1,m}(x) P_{n,m-q}(x-y)$ w.r.t. $x$ are polynomials in both $x$ and $y$ (this is trivial to see by a formal power-law expansion). 2) The $x$ variable is then evaluated at a linear function in $y$ which leaves us with a polynomial in $y$. 3) these polynomials are then summed over the coefficients in $m$, which again is another polynomial in $y$. $(g_1\star g_n)(y)$ is continuous in $y\in\rr$, since both $g_1$ and $g_n$ are continuous and the convolution of two continuous functions is continuous.

	All that is left to show, to prove that $(g_1\star g_n)(y)\in\mathcal{S}_{n+1}$, is to show that $P'(n,q;y)=P_{n+1,q}(y)$ for some $P_{n+1,q}(y)$ obeying Eqs. \eqref{eq:gn def}, \eqref{eq:gn def 2}. That $P'(n,q;y)$ obeys Eq. \eqref{eq:gn def} follows from the relationship between $y$ and $q$, namely $y\in[q,q+1]$. Eq. \eqref{eq:gn def 2} is true for $P'(n,q;y)$ if
	\be \label{eq:to be verified for P'}
	P'(n,q;y)=0 \quad \text{if } q\in(-\infty, -n-2] \cup [n+1,\infty).
	\ee
	We now verify that Eq. \eqref{eq:to be verified for P'} is satisfied. $q\leq -n-2$ implies $m-q-1\geq m+n+1\geq n$ for $m=-1,0$. Thus it follows using Eq. \eqref{eq:gn def 2} that the term $P_{n,m-q-1}$ in line \eqref{line: g1 gn} is zero for $q\leq -n-2$. Furthermore $q\leq -n-2$ implies $m-q\geq m+n+2\geq n+1$ for $m=-1,0$. Thus it follows using Eq. \eqref{eq:gn def 2} that the term $P_{n,m-q}$ in line \eqref{line: g1 gn} is zero for $q\leq -n-2$. Hence $P'(n,q;y)=0$ for $q\leq -n-2$. Similarly we can verify that $P'(n,q;y)=0$ for $q\geq n+1$. This concludes the proof that $(g_1\star g_n)(y)\in\mathcal{S}_{n+1}$ for all $n\in\nnp$, $g_n\in\mathcal{S}_n$ and $g_1\in\mathcal{S}_1$.

	Now work inductively to conclude that $g_1^{\,\mathlarger{\star} N}\in\mathcal{S}_{N}$. Thus recalling that $f\in\mathcal{S}_1$ and noting that all functions in $\mathcal{S}_N$ satisfy the conditions on $f^{\,\mathlarger{\star} N}$ of the Lemma, we conclude the proof.

\end{proof}

\begin{lemma}\label{lem: devs grow like powers}
	There exists $C_0=C_0(N)>0$ which is only a function of $N$, i.e. independent of $n$, $d$, and $k$, such that
\be \label{eq:main eq lemma dev bound}
\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k}  \bar{V}_0(x)\bigg{|} \leq n^{k+1} C_0^{k+1},\quad \forall\, k\in\nno,\,\, \forall\, n \geq 1.
\ee
\end{lemma}
\begin{proof} The proof will consist in writing the $k$th derivative in Fourier space and using properties of the Fourier Transform to calculate and upper bound the resultant expression.\\
	We will start by simplifying the expression for the derivative.

	We start by noting that the Fourier Transform of $V_B\in L^1$ can be computed via the Convolution theorem,
	\be \label{F trans of VB}
	\mathcal{F}(V_B) (y)=	\mathcal{F}((\textup{sinc}^2)^N) (y)= \underbrace{[	\mathcal{F}(\textup{sinc}^2)\star  	\mathcal{F}(\textup{sinc}^2)\star \ldots \star	\mathcal{F}(\textup{sinc}^2)]}_{N \text{-times convolution product}}(y),
	\ee
	which is well-defined, since the Convolution Theorem maps two $L^1$ functions to an $L^1$ function, and $\textup{sinc}^2\in L^1$.
	By direct calculation, we have that
	\be
	\mathcal{F}(\textup{sinc}^2)(y)=
	\begin{cases}
	0 &\mbox{ if } |y|\geq 1\\
	1-|y| &\mbox{ otherwise},
	\end{cases}
	\ee
	thus since the Convolution of two finite support functions, has finite support, we conclude by induction from Eq. \eqref{F trans of VB} that $\mathcal{F}(V_B) (y)$ has finite support. We denote the finite interval containing the support of $\mathcal{F}(V_B) (y)$ by $[y_\textup{min}, y_\textup{max}]$. Furthermore, note that $\mathcal{F}(\textup{sinc}^2)$ has a discontinuous derivative at three points on its domain $(0, y_\textup{min},y_\textup{max})$. When $	\mathcal{F}(\textup{sinc}^2)$ is convoluted with itself, the resultant function might also have a finite set of points in its support interval at which it is not differentiable (see Lemma \ref{lem:finite set of discont})\footnote{In fact, it is not differentiable at these points, but we will not need to prove this for our purposes.\label{refnote}}. Similarly due to Eq. \eqref{F trans of VB}, $\mathcal{F}(V_B) (y)$ may have a finite set of points contained in $[y_\textup{min}, y_\textup{max}]$ at which the function is not differentiable\footref{refnote}. From Lemma \ref{lem:finite set of discont}, we conclude that a set containing all such points is $\{y_l=-N-1+l\}_{l=1}^{2N+1}$, where $y_1=y_\textup{min}$, $y_{2N+1}=y_\textup{max}$. Finally, the last property of $\mathcal{F}(V_B)(y)$ which we will need; is that since $\mathcal{F}(\textup{sinc}^2)(y)$ has finite right and left 1st and 2nd derivatives in the interval $y\in[y_\textup{min},y_\textup{max}]$, it follows that $\mathcal{F}(V_B)(y)$ also has finite left and right 1st and 2nd derivatives in the interval $y\in[y_\textup{min},y_\textup{max}]$ (see Lemma \ref{lem:finite set of discont}).\\
	Using Lemma \ref{tech lemm dev interchange} and the change of variable $z=nx -2\pi p$, we have for $k\in\nno$
	\ba\label{eq:max over x kth up inter}
	\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k} \bar{V}_0(x)\bigg{|}&=\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k} \bar{V}_0(x+x_0)\bigg{|}&\\
	&=
	nA_0	\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k}\sum_{p=-\infty}^\infty  V_B(x n -2\pi n p) \bigg{|}\\
		&=
	nA_0	\max_{x\in[0,2\pi]} \bigg{|} \sum_{p=-\infty}^\infty \frac{d^k}{dx^k} V_B(x n -2\pi n p) \bigg{|}\\
	&= n^{k+1} 	A_0	\max_{x\in[0,2\pi]} \bigg{|}	\sum_{p=-\infty}^\infty \bigg{[} \frac{d^k}{dz^k} V_B(z) \bigg{]}_{z=nx -2\pi n p}\bigg{|}\\
	&=  n^{k+1} 	A_0	\max_{x\in[0,2\pi]} \bigg{|} 	\sum_{p=-\infty}^\infty \bigg{[} \mathcal{F}^{-1}\bigg( (2 \pi \mi y)^k \mathcal{F}(V_B) \bigg)(z) \bigg{]}_{z=nx -2\pi n p}\bigg{|}\label{line:kth dev intermediate}\\
	&=  n^{k+1} 	A_0	\max_{x\in[0,2\pi]} \bigg{|}	\sum_{p=-\infty}^\infty  \int_{-\infty}^\infty dy  (2 \pi \mi y)^k \mathcal{F}(V_B)(y) \,\me^{-2\pi \mi (nx -2\pi n p)y}\bigg{|} \\
	& \leq  2  n^{k+1} 	A_0	\max_{x\in[0,2\pi]} 	\sum_{p=0}^\infty \bigg{|} \int_{y_\textup{min}}^{y_\textup{max}} dy  (2 \pi \mi y)^k \cos(4\pi^2 n p y)\mathcal{F}(V_B)(y)\, \me^{-2\pi \mi nxy}\bigg{|} .
	\ea
	Note that the inverse Fourier Transform is well defined on the domain $L^2$. Thus  line \eqref{line:kth dev intermediate} is well justified since $(2 \pi \mi y)^k \mathcal{F}(V_B)\in L^2$ $\forall\,k\in\nno$, because $\mathcal{F}(V_B)$ has finite support.\\
	For the term $p=0$ we have
	\ba \label{eq:p is zero term}
	\bigg{|}\int_{y_\textup{min}}^{y_\textup{max}} dy  (2 \pi \mi y)^k \mathcal{F}(V_B)(y) \me^{-2\pi \mi nxy}	\bigg{|}&\leq (y_\textup{max}-y_\textup{min}) |2\pi y_\textup{max}|^k \max_{x\in [y_\textup{max},y_\textup{min}]} \big{|}\mathcal{F}(V_B)(x) \big{|}\\
	&\leq y_\textup{M} (2\pi  y_\textup{M})^k \mathcal{F}_\textup{max}^{(0)}(V_B),
	\ea
	where we have introduced the notation $\max\{|y_\textup{min}|,|y_\textup{max}|,1 \}=N=:y_\textup{M}$,  $d^k/dy^k \mathcal{F}(V_B)(y)=:\mathcal{F}^{(k)}(V_B)(y)$, and $\sup_{y \in(y_1,y_2)\cup (y_2,y_3)\cup\ldots\cup(y_{2N},y_{2N+1})} |\mathcal{F}^{(k)}(V_B)(y)|=: \mathcal{F}^{(k)}_\textup{max}(V_B)$. We will now bound the other terms $p\neq 0$. In the following we will denote, $\lim_{\varepsilon\rightarrow 0^+} \mathcal{F}^{(k)}(V_B)(y_0\pm \varepsilon)=: \mathcal{F}^{(k)}_{\pm}(V_B)(y_0)$, and use primes to denote 1st derivatives w.r.t. $y$. We will proceed by integration by parts twice, after splitting the integral up into sections in which $\mathcal{F}(V_B)(y)$ has continuous derivatives. We find
	\ba
	\bigg{|}\int_{y_\textup{min}}^{y_\textup{max}} &dy  (2 \pi \mi y)^k\cos(4\pi^2 n py) \mathcal{F}(V_B)(y) \me^{-2\pi \mi nxy}	\bigg{|}=	\bigg{|}\sum_{r=1,\ldots,2N}\int_{y_r}^{y_{r+1}} dy  (2 \pi \mi y)^k\cos(4\pi^2 n py) \mathcal{F}(V_B)(y)\, \me^{-2\pi \mi nxy}	\bigg{|}\\
	=&\bigg{|}\underbrace{\sum_{r=1,\ldots,2N}\bigg{[}  (2 \pi \mi y)^k\frac{\sin(4\pi^2 n py)}{4\pi^2 n p} \mathcal{F}(V_B) (y)\,\me^{-2\pi \mi nxy}\bigg{]}_{y_r}^{y_{r+1}}}_{=\,0 \text{ Since $\mathcal{F}(V_B)(y)$ is continuos}}\\ &-\sum_{r=1,\ldots,2N}\int_{y_r}^{y_{r+1}} dy \frac{\sin(4\pi^2 n py)}{4\pi^2 n p} (2 \pi \mi y)^{k-1} \me^{-2\pi \mi nxy}\,2\pi \mi\bigg{(}(k-2\pi \mi y nx)  \mathcal{F}(V_B)(y)+  \mathcal{F}(V_B)(y)+  y\mathcal{F}^{(1)}(V_B)(y)    \bigg{)}	\bigg{|}.\\
	=&	\bigg{|}\lim_{\varepsilon\rightarrow 0^+}\underbrace{\sum_{r=1,\ldots,2N} \bigg{[}\frac{-\cos(4\pi^2 n py)}{16\pi^4 n^2 p^2} (2 \pi \mi y)^{k-1} \me^{-2\pi \mi nxy}\,2\pi \mi\bigg{(}(k-2\pi \mi y nx)  \mathcal{F}(V_B)(y)+  \mathcal{F}(V_B)(y)+  y\mathcal{F}^{(1)}(V_B)(y)     \bigg{)}	\bigg{]}_{y_r+\varepsilon}^{y_{r+1}-\varepsilon}}_{\text{May not be zero since $\mathcal{F}^{(1)}(V_B)(y)$ may be discontinuos at points $y_1,y_2\ldots,y_{2N+1}$}}\\
	&-\sum_{r=1,\ldots,2N} \int_{y_r}^{y_{r+1}}dy \frac{-\cos(4\pi^2 n py)}{16\pi^4 n^2 p^2} \bigg{(} (2 \pi \mi y)^{k-1} \me^{-2\pi \mi nxy}\,2\pi \mi\big{(}(k-2\pi \mi y nx)  \mathcal{F}(V_B)(y)+  \mathcal{F}(V_B)(y)+  y\mathcal{F}^{(1)}(V_B)(y)\big{)} \bigg{)}'\,\bigg{|}\\
	\leq
&\sum_{r=1,\ldots,2N+1}\bigg{|}	\frac{-\cos(4\pi^2 n py_r)}{16\pi^4 n^2 p^2} (2 \pi \mi y_r)^{k-1} \me^{-2\pi \mi nxy_r}\,2\pi \mi  y_r\Big{(}\mathcal{F}^{(1)}_{-}(V_B)(y_r)-\mathcal{F}^{(1)}_{+}(V_B)(y_r)     \Big{)}\bigg{|}\\
&+ \sum_{r=1,\ldots,2N} \frac{y_{r+1}-y_{r}}{16\pi^4 n^2 p^2} \sup_{y \in(y_{r},y_{r+1}) } \bigg{|}\bigg{(} (2 \pi \mi y)^{k-1} \me^{-2\pi \mi nxy}\,2\pi \mi\big{(}(k-2\pi \mi y nx)  \mathcal{F}(V_B)(y)+  \mathcal{F}(V_B)(y)+  y\mathcal{F}^{(1)}(V_B)(y)\big{)} \bigg{)}'\,\bigg{|}\\
\leq&  \frac{1}{16\pi^4 n^2 p^2} \Bigg( (2\pi y_\textup{M})^{k-1} y_\textup{M} 4\pi (2N+1) \mathcal{F}^{(1)}_\textup{max}(V_B)   \label{inter step 106}\\
&\quad\quad\quad\quad\quad +(y_\textup{max}-y_\textup{min}) 2\pi (2\pi y_\textup{M} )^{k-1} \bigg{[} \big{(}(k-1)(2\pi y_\textup{M})^{-1} +n x\big{)} \big{(}(k+2\pi y_\textup{M} n x + 1) \mathcal{F}_\textup{max}(V_B) +y_\textup{M}\mathcal{F}^{(1)}_\textup{max}(V_B)\big{)}  2\pi \nonumber\\
&\quad\quad\quad\quad\quad+2\pi n x \mathcal{F}_\textup{max}(V_B) + (k+2\pi n x +2) \mathcal{F}_\textup{max}^{(1)}(V_B)+y_\textup{M} \mathcal{F}_\textup{max}^{(2)}(V_B)	\Bigg{]} \Bigg{)}.\nonumber
	\ea
	Note the dependency on $n$ of the expression on the R.H.S. of the inequality line \eqref{inter step 106}. It is of the form $(c_{00}+n c_{01}+n^2 c_{02})/n^2$, where $c_{00}\geq 0$, $c_{01}\geq 0$, $c_{02}\geq 0$ are $n$ independent. Thus since $n\in[1,\infty)$, the R.H.S. of line \eqref{inter step 106} is upper bounded uniformly in $n$ by setting $n=1$. Similarly, the coefficients $c_{01}\geq 0$, $c_{02}\geq 0$ are upper bounded in $x\in[0,2 \pi]$, by setting $x=2 \pi$. We thus have for $|p|\in\nnp$,
	\ba\label{eq:penultimate inter for multi dev bound}
	\bigg{|}\int_{y_\textup{min}}^{y_\textup{max}} &dy  (2 \pi \mi y)^k\cos(4\pi^2 n py) \mathcal{F}(V_B)(y) \me^{-2\pi \mi nxy}	\bigg{|}\\
	\leq&  \frac{(2\pi N)^{k-1}}{16\pi^4  p^2} \Bigg(  N 4\pi (2N+1) \mathcal{F}^{(1)}_\textup{max}(V_B)
	\\
	&\quad\quad\quad\quad\quad +N 2\pi  \bigg{[} \big{(}(k+1)(2\pi N)^{-1} + 2\pi \big{)} \big{(}(k+(2\pi)^2 N  + 1) \mathcal{F}_\textup{max}(V_B) +N\mathcal{F}^{(1)}_\textup{max}(V_B)\big{)}  2\pi \nonumber\\
	&\quad\quad\quad\quad\quad+(2\pi)^2 \mathcal{F}_\textup{max}(V_B) + (k+(2\pi)^2 +2) \mathcal{F}_\textup{max}^{(1)}(V_B)+N \mathcal{F}_\textup{max}^{(2)}(V_B)	\Bigg{]} \Bigg{)},
	\ea
	where we have used $y_\textup{M}=N$. The R.H.S. of the inequality Eq. \eqref{eq:penultimate inter for multi dev bound} can be written as
	\be
	\frac{(2\pi N)^{k}}{16\pi^4  p^2}\left( c_{03}+c_{04}k +c_{05}k^2  \right),
	\ee
	with the coefficients $c_{03}\geq 0$, $c_{04}\geq 0$, $c_{05}\geq 0$ are $k$, $x$ and $p$ independent. Thus from Eqs. \eqref{eq:max over x kth up inter}, \eqref{eq:p is zero term}, \eqref{eq:penultimate inter for multi dev bound},
	\ba
		\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k} \bar{V}_0(x)\bigg{|}&=\max_{x\in[0,2\pi]} \bigg{|} \frac{d^k}{dx^k} \bar{V}_0(x+x_0)\bigg{|}\\
		&\leq  2  n^{k+1} 	A_0 \bigg{(} N (2\pi  N)^k \mathcal{F}_\textup{max}^{(0)}(V_B) + \sum_{p\in \zz \backslash \{0\}} 	\frac{(2\pi N)^{k}}{16\pi^4  p^2}\bigg( c_{03}+c_{04}k +c_{05}k^2  \bigg)\\
		&=  2  n^{k+1} 	A_0 (2\pi N)^{k} \bigg{(} N \mathcal{F}_\textup{max}^{(0)}(V_B) +	\frac{1}{16\pi^4}\frac{2 \pi^2}{6}\bigg( c_{03}+c_{04} \me^{\ln k} +c_{05}\me^{2\ln k}  \bigg)\bigg)\\
		&\leq  2  n^{k+1} 	A_0 (2\pi N)^{k} \bigg{(} N \mathcal{F}_\textup{max}^{(0)}(V_B) +	\frac{1}{48\pi^2}\bigg( c_{03}+c_{04} \me^{ k} +c_{05}\me^{2 k}  \bigg)\bigg),\label{eq:up bound mult deov intermed}
	\ea
	for $k\in\nno$.
	Thus recalling from Lemma \ref{lem:Normalization A0} that $A_0$ is upper bounded by $\underline{A}_0$ which is only a function of $N$, from Eq. \eqref{eq:up bound mult deov intermed}, it follows that there exists a coefficient $C_0$ which only depends on $N$ such that Eq. \eqref{eq:main eq lemma dev bound} holds.
\end{proof}
\subsubsection{Determining a parametrization of $n$ in terms of $d$ such that $\epsilon(T_0,d)\rightarrow0$ quicker than any polynomial in $d$.}

\iffalse

Recall Theorem E.1. in \cite{WSO16}:
where, in terms of order in $b$, $|t|$, $d$ and $\sigma$,
\be
\epsilon(t,d)=
	|t| \frac{d}{T_0}\!\left(\bo\left( \frac{\sigma^3}{\bar\upsilon \sigma^2/d+1}\right)^{1/2}\!\!+\bo\left(\frac{d^2}{\sigma^2}+b\right)\right) \exp\left(-\frac{\pi}{4}\frac{\alpha_0^2}{\left(\frac{d}{\sigma^2}+\bar\upsilon\right)^2} \left(\frac{d}{\sigma}\right)^2 \right)+\bo\left(|t|\frac{d^2}{\sigma^2}+1\right)\me^{-\frac{\pi}{4}\frac{d^2}{\sigma^2}}+\bo\left( \me^{-\frac{\pi}{2}\sigma^2} \right),
\ee
if $\mathcal{N}\geq 8$ and $\bar \upsilon\geq 0$, and where $\alpha_0\in(0,1]$, $b$, $\mathcal{N}$ is defined in Def. 8, and  $\bar{\upsilon}$ in Def. 9 of \cite{WSO16}. Specifically,
\be \label{eq: mathcal N def}
\mathcal N= \Bigg\lfloor \frac{\pi \alpha_0^2}{2\left(\bar\upsilon+\frac{d}{\sigma^2}\right)^2}\left(\frac{d}{\sigma}\right)^2\Bigg\rfloor.
\ee
\fi

 Let us start by introducing the constraint
\ba
\frac{d}{\bar{\upsilon} \sigma} &=d^{\epsilon_5},\quad \text{for some fixed constant } 0<{\epsilon_5}. \label{eq:epsilon 5 def}
\ea
It will also be useful to introduce the variable $\epsilon_6$ which is a function of $d$ and $\sigma$ via
\ba\sigma &= d^{\epsilon_6},
\ea
where $\epsilon_6$ uniformly bounded to the interval\footnote{By ``uniformly bounded'', it is meant that $\epsilon_5<\lim_{d\rightarrow \infty}\epsilon_6<1$} $\epsilon_5<\epsilon_6<1.$
We can re-write Eq. \eqref{eq:mathcal N def} as
\be
\mathcal N= \Bigg\lfloor \alpha_0^2\frac{\pi }{2}\,\frac{d^{2-2\epsilon_6}}{d^{2-2\epsilon_5-2\epsilon_6}+d^{2-4\epsilon_6}+d^{2-\epsilon_5-3\epsilon_6}}\Bigg\rfloor= \Bigg\lfloor \alpha_0^2\frac{\pi }{2}\,\frac{1}{d^{-2\epsilon_5}+d^{-2\epsilon_6}+d^{-\epsilon_5-\epsilon_6}}\Bigg\rfloor.
\ee
Thus recalling that $\alpha_0\in(0,1]$ is a fixed constant, it follows that $\mathcal{N}=\infty$ and $\bar\upsilon\geq 0$ in $d \rightarrow \infty$ limit. Hence Eqs.  \eqref{eq:mathcal N contraints} are satisfied for sufficiently large $d$  and Eq. \eqref{eq:epsilon t d} holds for the parameterisations considered in this proof in the limit $d\rightarrow\infty$.
We can now re-write eq. \eqref{eq:epsilon t d} as
\be\label{eq:epsilon t d re write}
\epsilon(t,d)=
|t| \frac{d}{T_0}\!\left(\bo\left( \frac{\sigma^3}{\sigma d^{-\epsilon_5}+1}\right)^{1/2}\!\!+\bo\left(\frac{d^2}{\sigma^2}+b\right)\right) \exp\left(-\frac{\pi}{4}\frac{\alpha_0^2}{\left(1+d^{\epsilon_5}/\sigma\right)^2} d^{2\epsilon_5} \right)+\bo\left(|t|\frac{d^2}{\sigma^2}+1\right)\me^{-\frac{\pi}{4}\frac{d^2}{\sigma^2}}+\bo\left( \me^{-\frac{\pi}{2}\sigma^2} \right).
\ee
thus since (as we will soon show), $b$ grows at most polynomially in $d$, $\epsilon(t,d)$ decays quicker than any polynomial in $d$ in the limit $d\rightarrow\infty$. We will now workout the implications of Eq. \eqref{eq:epsilon 5 def}. From Eq. \eqref{eq:b def eq 0} and the relation $-\mi \delta \,\bar{V}_0(x)=V_0(x)$ stated in Eq. \eqref{eq:pod eqs def}, we have that $b$ is any non-negative number satisfying
\be\label{eq:b def eq}
b\geq\; \sup_{k\in\nnp}\left(2\max_{x\in[0,2\pi]} \left|  \delta \bar{V}_0^{(k-1)}(x) \right|\,\right)^{1/k}.
\ee
Thus from Lemma \ref{lem: devs grow like powers}, it follows
\be
\sup_{k\in\nnp}\left(2\max_{x\in[0,2\pi]} \left| \delta \bar{V}_0^{(k-1)}(x) \right|\,\right)^{1/k}\leq \sup_{k\in\nnp}\left(2\delta n^k C_0^k \right)^{1/k}=2 \delta n C_0,
\ee
hence we will set $b= 2\delta n C_0$. Using definition the of $\bar{\upsilon}$ (Eq. \eqref{eq:upsilon def}), we have
\be
d^{\epsilon_5}= \frac{d}{\bar{\upsilon} \sigma}= \frac{d}{\sigma} \frac{\ln(\pi \alpha_0\sigma^2)}{\pi \alpha_0\kappa}\frac{1}{b}=\frac{d}{\sigma} \frac{\ln(\pi \alpha_0\sigma^2)}{2\pi C_0\alpha_0\kappa}\frac{1}{\delta n},
\ee
with $\kappa=0.792$ from which we find the constraint on $n$,
\be\label{eq:n in terms of d def}
n= \frac{\ln(\pi \alpha_0\sigma^2)}{2\pi C_0\alpha_0\kappa} \frac{d^{1-\epsilon_5}}{\delta \sigma}.
\ee

\subsubsection{Determining the constant $N$ and parametrization of $\delta$ in terms of $d$ such that conditions of Corollary \ref{corrolary w lims} are satisfied}
From Eq. \eqref{eq:tilde ep V def} and using elementary properties of the potential $\bar{V}_0$ in Eq. \eqref{eq:def of rapped pot} (namely, integrates to unity over one period and is symmetric w.r.t. $x_0$, i.e. that $\bar V_0(x+x_0)=\bar V_0(-x+x_0)$ we find
\ba\label{eq:tilde ep V v2}
\tilde\epsilon_V&=1-\int_{-x_{vr}}^{x_{vr}} dx\, \bar{V}_0(x+x_0)\\
&= 1- \bigg{(} \int_{-\pi}^{\pi} dx\, \bar{V}_0(x+x_0)-\int_{-\pi }^{-x_{vr}} dx\, \bar{V}_0(x+x_0)-\int_{x_{vr}}^{\pi} dx\, \bar{V}_0(x+x_0) \bigg{)}\\
&= 1- \bigg{(} 1-\int_{x_{vr}}^{\pi} dx\, \bar{V}_0(-x+x_0)-\int_{x_{vr}}^{\pi} dx\, \bar{V}_0(x+x_0) \bigg{)}\\
&= 2\int_{x_{vr}}^{\pi} dx\, \bar{V}_0(x+x_0)\\
&\leq 2(\pi-x_{vr}) \max_{x\in[x_{vr},\pi]}\big\{ \bar{V}_0(x+x_0) \big\}\\
&\leq 2 A_0 n (\pi-x_{vr}) \sum_{p=-\infty}^\infty \max_{x\in[x_{vr},\pi]}\big\{| V_B(n x-2\pi p n)| \big\}+2\frac{(\pi-x_{vr})}{\delta d^2}\\
&\leq  2 \underline{A}_0 n (\pi-x_{vr}) \sum_{p=-\infty}^\infty\max_{x\in[x_{vr},\pi]}\big\{|\pi (n x-2\pi p n)|^{-2N} \big\}+2\frac{(\pi-x_{vr})}{\delta d^2}\\
&\leq  2 \underline{A}_0 n (\pi-x_{vr})\bigg{(} (n\pi x_{vr})^{-2 N}+ (2\pi^2 n)^{-2N}\Big{(}\sum_{p\in\zz\backslash \{0\}} p^{-2N} \Big) \bigg{)}+2\frac{(\pi-x_{vr})}{\delta d^2}\\
&\leq  2 \underline{A}_0  \pi\bigg{(} \frac{n}{(n \pi x_{vr})^{2 N}}+ \frac{2n\zeta(2N)}{(2\pi^2 n)^{2N}} \bigg{)}+\frac{2\pi}{\delta d^2}\\
&\leq  2 \underline{A}_0  \pi(1+\zeta(2N))\frac{n}{(n \pi x_{vr})^{2 N}}+\frac{2\pi}{\delta d^2}.
\ea
Now recall Corollary \ref{corrolary w lims}. One of the conditions needed for this Corollary to be satisfied is $x_{vr}=o\left(\gamma\right)$. Recalling the bound
\be
d^{\eta/2} \frac{\sigma}{d}-\frac{2}{d} \leq \gamma \leq d^{\eta/2} \frac{\sigma}{d},
\ee
which was derived in Eqs \eqref{eq:up bound gamma even}, \eqref{eq:bound for gamma d over sigma odd}), we will find that the following parametrization of $x_{vr}$ and $\delta$ allows us to achieve this scaling and the other conditions introduced in Corollary \ref{corrolary w lims}.

Let
\ba \label{eq:x vr limit contraint}
&\pi x_{vr}=d^{\epsilon_7} \frac{\sigma}{d},\quad\quad \quad
 \delta=d^{\epsilon_8},&
\ea
for some fixed constants $0<\epsilon_7<\eta/2$, $\,0<\epsilon_8$.
We will now show that indeed Eq. \eqref{eq:tilde ep V v2} permits the existence of constants $N$, $\epsilon_7$, $\epsilon_8$ such that Eqs. \eqref{eq:x vr limit contraint} are satisfied.

Substituting expressions for $\pi x_{vr}$ and $n$ from Eq. \eqref{eq:x vr limit contraint}, and Eq. \eqref{eq:n in terms of d def} into \eqref{eq:tilde ep V v2}, we find,
\ba
\delta \tilde{\epsilon}_V &\leq  2 \underline{A}_0  \pi(1+\zeta(2N))\frac{\delta n}{(n \pi x_{vr})^{2 N}}+\frac{2\pi}{d^2}\\
&=  2 \underline{A}_0  \pi(1+\zeta(2N))\delta \frac{\ln(\pi \alpha_0 \sigma^2)}{2\pi C_0\alpha_0\kappa}\frac{d^{1-\epsilon_5}}{\delta \sigma} \left(\frac{2\pi C_0\alpha_0\kappa}{\ln(\pi \sigma^2)}\frac{\delta \sigma}{d^{1-\epsilon_5}}\frac{1}{d^{\epsilon_7}}\frac{d}{\sigma}  \right)^{2N}+\frac{2\pi}{d^2}\\
&=  2 \underline{A}_0  \pi(1+\zeta(2N))\left( \frac{2\pi C_0\alpha_0\kappa}{\ln(\pi\alpha_0\sigma^2)}\right)^{2N-1} \frac{d^{1-\epsilon_5+(\epsilon_8+\epsilon_5-\epsilon_7)2N}}{\sigma}+\frac{2\pi}{d^2}.
\ea
Now define a constant $\epsilon_9$ such that $0<\epsilon_9< \eta$, and impose the constraints $\epsilon_7>\epsilon_5+\epsilon_8$ and
\be\label{eq:inter proof bound ep 9}
\frac{d^{1-\epsilon_5+(\epsilon_8+\epsilon_5-\epsilon_7)2N}}{\sigma}\leq d^{\epsilon_9} \left(\frac{\sigma}{d}\right)^2.
\ee
Recalling $\sigma=d^{\epsilon_6}$, it follows from Eq. \eqref{eq:inter proof bound ep 9},
\ba
-3\epsilon_6-\epsilon_9+2&\leq -1+\epsilon_5+\underbrace{(\epsilon_7-\epsilon_8-\epsilon_5)}_{>0} 2N\\
\implies & N\geq \frac{-3\epsilon_6-\epsilon_9-\epsilon_5+3}{2(\epsilon_7-\epsilon_8-\epsilon_5)}.
\ea
Therefore, since $-\epsilon_5>-\epsilon_6$, we set the constant $N$ to be
\be
 N:=\left\lceil\frac{3-4\epsilon_5-\epsilon_9}{2(\epsilon_7-\epsilon_8-\epsilon_5)}\right\rceil> \frac{3-\epsilon_5-\epsilon_9-3\epsilon_6}{2(\epsilon_7-\epsilon_8-\epsilon_5)},
\ee
with $3-4\epsilon_5-\epsilon_9>0$ and $\lceil\cdot\rceil$ denoting the \textit{Ceiling function}. Thus from Eqs. \eqref{eq:tilde ep V v2}, \eqref{eq:inter proof bound ep 9}, it follows
\be
\delta\tilde{\epsilon}_V\leq 2 \underline{A}_0  \pi(1+\zeta(2N))\left( \frac{2\pi C_0\alpha_0\kappa}{\ln(\pi\alpha_0\sigma^2)}\right)^{2N-1} d^{\epsilon_9} \left(\frac{\sigma}{d}\right)^2+\frac{2\pi}{d^2}.
\ee
Thus all the conditions for Corollary \ref{corrolary w lims} are satisfied as long as all the constraints on the epsilons we have introduced can simultaneously be satisfied. We will show this in the proof of Theorem \ref{Thm:R lower quant bound}.

\subsection{Final Theorem}\label{Final Theorem}
\begin{theorem}\label{Thm:R lower quant bound}
	Consider the setup described in Section \ref{sec:setup} for \wso~Clocks and potential functions $\bar V_0$ introduced in Section \ref{Overview of the quasi-ideal clock} with constant $\alpha_0\in(0,1]$, and $k_0=0$. For all fixed constants $0<\eta\leq1$, and $\sigma$ satisfying
		\be\label{eq:baby constraint 2 thrm}
	d^{\eta/2}\leq \sigma<d,
	\ee
	 the clock accuracy ${R_1}$ is lower bounded by
	\be \label{eq:main thrm eq}
	{R_1}\geq \frac{d^{2-\eta}}{\sigma^2}+\lo\left(\frac{d^{2-\eta}}{\sigma^2}\right),
	\ee
	in the large $d$ limit.
\end{theorem}

\begin{proof}
	The bound
	\be \label{eq:main thrm eq almost}
	{R_1}\geq \frac{1}{12}\frac{d^{2-\eta}}{\sigma^2}+\lo\left(\frac{d^{2-\eta}}{\sigma^2}\right),
	\ee
	follows directly from Corollary \ref{corrolary w lims} and the results from Section \ref{Showing that the limit requirements}, so long as the constraints on the epsilon terms can all be simultaneously satisfied. We will check this here. The constraints introduced are:
	\ba \label{eq:epsilons contrains 1}
	&0<\epsilon_5<\epsilon_6<1,&\quad &0<\epsilon_7<\frac{\eta}{2},& \quad & 0<\epsilon_8,& \quad &\epsilon_5+\epsilon_8<\epsilon_7,& \quad & 0<\epsilon_9<\eta,& 0<3-4\epsilon_5-\epsilon_9,&
	\ea
	and
	\be  \label{eq:epsilons contrains 2}
	\frac{4}{\sigma}< d^{\eta/2}\leq \frac{d}{\sigma}.
	\ee
	Let $\epsilon_7=\eta/4,$ $\epsilon_5=\epsilon_8=\eta/16,$ and $\epsilon_9=\eta/2.$ As long as
	\be \label{eq:baby constraint}
	\frac{\eta}{16}< \epsilon_6<1-\frac{\eta}{6},\quad\quad 0<\eta<4
	\ee
	are satisfied, all constraints are met in Eq. \eqref{eq:epsilons contrains 1}. Similarly, Eq. \eqref{eq:epsilons contrains 2} is satisfied if
	\ba
	\frac{\ln 4}{\ln d}-\frac{\eta}{2}<\epsilon_6\leq 1-\frac{\eta}{2}.
	\ea
	Hence Eqs. \eqref{eq:epsilons contrains 1}, \eqref{eq:baby constraint} are satisfied for sufficiently large $d$ if
	\be \label{eq:baby constraint 2}
	\eta< \epsilon_6\leq 1-\frac{\eta}{2},
	\ee
	for $\eta>0$. Now make the substitution $\eta=\eta'/2$. Eq. \eqref{eq:main thrm eq almost} becomes
	\be\label{eq:R  wioth constant factor}
	{R_1}\geq \frac{d^{3\eta'/4}}{12}\left(\frac{d^{2-\eta'}}{\sigma^2}+\lo\left(\frac{d^{2-\eta'}}{\sigma^2}\right)\right)\geq \frac{d^{2-\eta'}}{\sigma^2}+\lo\left(\frac{d^{2-\eta'}}{\sigma^2}\right),
	\ee
	with constraint
	\be \label{eq:baby constraint 3}
	d^{\frac{\eta'}{2}}< \sigma\leq d^{1-\frac{\eta'}{4}},\quad \eta'>0,
	\ee
	where we have used Eq. \eqref{eq:baby constraint 2} and recalled $\sigma=d^{\epsilon_6}$. Finally, note that w.l.o.g., we can replace the upper bound on $\sigma$ in Eq. \eqref{eq:baby constraint 3} by $< d$. This is because of two observations: 1) by choosing $\eta'$ arbitrarily close to zero, $\sigma$ is upper bounded by a number arbitrarily close to $d$. 2) The lower bound on ${R_1}$ in Eq. \eqref{eq:R  wioth constant factor} is monotonically decreasing w.r.t. $\eta'$. To finalise the proof, we simply re-name $\eta'$ by $\eta$.
\end{proof}


TITLE: Notable examples of the square root idea in complexity analysis
QUESTION [15 upvotes]: There are a number of algorithms and data structures which exploit the idea that $\max \left\{k, n/k\right\}$ gets its minimum value at $k=\sqrt n$. Common examples include

baby-step giant-step algorithm for computing discrete logarithm in $O(\sqrt n)$,
static 2D orthogonal range counting in $O(\sqrt n)$ time and $O(n)$ memory,
priority queue with EXTRACT-MIN in $O(\sqrt[k] n)$ and DECREASE-KEY in $O(1)$,
colouring a 3-colourable graph with $O(\sqrt n)$ colours in polynomial time,

just to name a few.
While such algorithms often are suboptimal, they are easy to understand by students and good to quickly show that naive bounds aren't optimal. Also, square-root-idea data structures are sometimes more practical than their binary tree based counterparts because of cache friendliness (not considering cache-oblivious techniques). That's why I give a nice bit of attention to this topic while teaching.
I'm interested in more distinctive examples of this kind. So I'm looking for any (preferably elegant) algorithms, data structures, communication protocols etc which analysis relies on the square root idea. Their asymptotics do not need to be optimal.

REPLY [10 votes]: Chazelle, Liu, and Magen's paper Sublinear Geometric Algorithms (STOC 2003, SICOMP 2006) has several clever applications of the following random sampling trick.  Variations of the same trick were previously used by Gärtner and Welzl [DCG 2001], who cite the first edition of CLR (1990).
Suppose we are given a sorted circular linked list of numbers, stored in a contiguous block of memory.  That is, we have two arrays $Key[1..n]$ and $Next[1..n]$, where

$Key[1..n]$ stores a set of $n$ numbers in arbitrary order;
If $Key[i]$ is the largest number in the set, then $Key[Next[i]]$ is the smallest number in the set; otherwise, $Key[Next[i]]$ is the smallest number in the set that is larger than $Key[i]$.

Then we can find the successor of a given number $x$ in $O(\sqrt{n})$ expected time as follows:

Choose a random sample of $\sqrt{n}$ elements of the array $Key$.  Let $Key[j]$ be the largest sample that is smaller than $x$ (or the largest sample, if all samples are greater than $x$).
Follow $Next$ pointers from $Key[j]$ until we see a number greater than or equal to $x$ (after wrapping around if all samples were larger than $x$).

A relatively simple application of Yao's lemma implies that the $O(\sqrt{n})$ expected time bound is optimal.  Any deterministic algorithm for this problem requires $\Omega(n)$ time in the worst case.


TITLE: How do Peano's arithmetical axioms guarantee that we can construct the natural number set?
QUESTION [1 upvotes]: I'm probably not understanding but I don’t see how the axioms can guarantee total construction of the natural number set.
The successor function as well as the axiom of induction guarantee that the natural number set will not just infinitely loop over including the same element, and that it will grow to infinite cardinality.
But isn’t it possible that it can go
{0, 1, 5, 6, 7, 8, 9, ...} thereby leaving “holes” in the set?
Again, probably a misguided view of sets, but I feel like this can only be solved by doing a union between all the (infinitely many) generated sets from all the different trials/worlds, in order to completely populate the ultimate natural number set?
Or my other thinking is that the successor function must be defined for all the different algebraic presentations of all the natural numbers, and then the sets produced by it must undergo a union.
But then how does the successor function even equate "next" with +1. It's like using a word in its own definition, and then abstracting it to "next" in order to obfuscate the cyclical nature of the definition. It even feels like the axioms mix up Sets and Series.
Please help me smooth over this confusion.

REPLY [0 votes]: $0$, $S(0)$, $S(S(0))$ are terms that we can indeed interpret in any way we want ... and indeed the function symbols $s$, $+$, and $\times$ can likewise be interpreted in many different ways.
For example, I could say that $0$ denotes 'apples', $S(0)$ denotes 'bananas', $S(S(0))$ denotes 'Luxemburg', etc.
So you're right that the domain is not at all forced to be the natural numbers and their typical operators of successor, addition, and multiplication.
However, given that from the Peano Axioms we can prove things like:
$\forall x \ x \times 0 = 0$
$\forall x \ x \times S(0) = x$
$\forall x \ x \times S(S(0)) = x + x$
etc.
you will find that 'apples' behaves just like $0$, that 'bananas' behaves just like $1$, and that 'Luxemburg' behaves just like $2$.  That is, any model of the eano Axioms will have to behave just like the natural numbers: we say that the models are isomporph. So ... you might as well treat $0$ as $0$, $S(0)$ as $1$, etc.


TITLE: Evaluating an expression using snake oil and convolutions gives different answers
QUESTION [2 upvotes]: I have to evaluate this expression $\sum \limits_{k=0}^n(-1)^k\binom{n}{k}\binom{m+n-k}{n-k}$ using snake oil and convolutions. The problem is that I got two different results, could you help me to find the mistake?
(Notation: $[x^n]$ is the operator "take the coefficient of $x^n$.")
Convolution:
$$
\begin{eqnarray*}
\sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k}
&=& [x^n] \left( \sum_k(-1)^k \binom{n}{k}x^k \right) \left( \sum\binom{m+k}{k}x^k \right)
\\ &=& [x^n] (1-x)^n \frac{1}{(1-x)^{m+1}}
\\ &=& [x^n] (1-x)^{n-m-1}
\\ &=& (-1)^n\binom{n-m-1}{n}
\end{eqnarray*}
$$
Snake oil:
$$
\begin{eqnarray*}
\sum_m \sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m
&=&
\sum_{k=0}^n (-1)^k \binom{n}{k} \sum_m \binom{m+n-k}{n-k} x^m
\\ &=&
\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{(1-x)^{n-k}}
\\ &=&
\frac{1}{(1-x)^n} \sum_{k=0}^n (-1)^k \binom{n}{k}(1-x)^k
\\ &=&
\frac{x^n}{(1-x)^n}
\\ &=&
\sum_m \binom{m-1}{n-1} x^m ,
\end{eqnarray*}
$$
and so the value should be $\binom{m-1}{n-1}$.
Could you help me to find the mistake, please?

REPLY [4 votes]: I think there's an error in the "snake oil" part. You should have
$$\sum_{m = 0}^{\infty} \binom{m+n-k}{n-k} x^m = \frac{1}{(1-x)^{n+1-k}}$$
Which leads you to
$$\sum_{m = 0}^{\infty} \sum_{k = 0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m = \frac{x^n}{(1-x)^{n+1}} = \sum_{m = 0}^{\infty} \binom{m}{n} x^m$$
And finally, it's easy to check
$$\binom{m}{n} = (-1)^n \binom{n-m-1}{n}$$


TITLE: Positive Semi-Definite Matrix products
QUESTION [0 upvotes]: If A and C are positive definite and symmetric matrices and B is a positive semi-definite matrix and symmetric. Would the Product (assuming the dimensions match) i.e T=ABC be positive semi-definite? A sort of sub question in this same line of though that might be easier to justify is would T'=ABA be positive semi-definite. If this is true, could someone point me in the direction of some justification?
Thanks,

REPLY [1 votes]: For the latter case, consider
 $$x = A^{-1}y$$
Since $A$ is positive definite, it is also invertible which means that we can find $y$ in order to represent any vector $x$ this way. Now we have
$$x^T(ABA)x = y^T A^{-1} A B A A^{-1} y = y^T B y \ge 0 $$
since $B$ is positive semidefinite. This implies that $T'$ is positive semidefinite.
For the former case, I recommend reading through the answers to this question


TITLE: Simplified method for symmetric matrices
QUESTION [0 upvotes]: I am trying to find the determinant of the following  $4\times 4$ symmetric  matrix $A$ given as
$$A= \begin{pmatrix}
a & b & c & d \\
  b & c & d & a \\
  c & d & a & b \\
  d & a & b & c \end{pmatrix}$$
Any solution and help would be appreciated. Thanks.

REPLY [0 votes]: I'll explain the situation for $3 \times 3$ matrices, an you will have an idea of how it will work for larger matrices. It's a nice technique, though I don't know about the constant at the end.
The key:

Try to look for conditions on variables, that will make the determinant zero. That is, when we substitute these relations of variables into the determinant, maybe two rows become the same, or two columns become the same etc. Then, the remainder theorem will tell you that this determinant contains that relation, expressed as as a polynomial, as a factor.

I'll give you an example, by evaluating the $3 \times 3$ equivalent of this matrix:
$$
\begin{bmatrix}
a & b & c \\
b & c & a \\
c & a & b
\end{bmatrix}
$$
First of all, note that by doing $R_1 \to R_1 + R_2 + R_3$ :
$$
\begin{bmatrix}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{bmatrix}
$$
Now, if $a+b+c=0$, then the determinant will be zero. Hence, you see that $a+b+c$ is a factor of the given determinant.
Similarly, look at the original determinant. If $a=b=c$, then you can see that the determinant is zero. Hence, the determinant will be a multiple of $(a-b)^2 + (b-c)^2 + (c-a)^2$ also.
Now, since the degrees ($3$ and $3$) match,you can check that up to a constant, the determinant of the $3 \times 3$ case is $(a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2)$.
Similarly, in the four times four case:
1) Add all other rows to the first row, and note that if $a+b+c+d=0$ then the determinant is zero.
2) Similarly, if you let $c=a$ and $d=b$, then the first and third rows are the same, so you can check that $(c-a)^2 + (b-d)^2$ will divide the determinant.
3) Do $R_1 = R_1 -R_2 + R_3 -R_4$. This will give the first row as $[1,-1,1,-1] \times (a-b+c-d)$. Hence, if this quantity is zero, you get the determinant is zero, So this is another factor.
Finally, by comparing degrees, the product of these factors is up to a constant the determinant.
Last but not the least, since you want generalizations, I should compute this for a $5 \times 5$ determinant of that kind, and see where we can go.
So,  the $5 \times 5$ determinant will look like:
$$
\begin{bmatrix}
a & b & c & d & e \\
b & c & d & e & a \\
c & d & e & a & b \\
d & e & a & b & c \\
e & a &  b & c & d
\end{bmatrix}
$$
A look at the above matrix should tell you at least the following. Unfortunately Wolfram is hanging when I try to figure out the answer, but I want you to check at least the following:
1 ) $a+b+c+d+e$ is a factor of the determinant.
2) If $a=b=c=d=e$, then the determinant is zero, so $\frac 12 \sum_{x\neq y, (x,y) \in \{ a,b,c,d,e\}^2} (x-y)^2$ is a divisor of the given polynomial.
If not completely solve the question, at least this technique sheds light on the manner in which you can possibly approach larger matrices of this kind.
This technique is immensely useful in evaluating many symmetric, or patterned determinants up to a constant, especially the Cauchy formula for determinants of the form $\frac{1}{X_i + Y_j}$.


TITLE: Integration to find general solution of free particle
QUESTION [0 upvotes]: I was attempting problems in Griffiths Intro to QM when I came across the following:

A free particle has the initial wavefunction:
  $$\Psi(x, 0) = Ae^{-ax^2} \, .$$
  Find $\Psi(x, t)$.

I normalised the initial wavefunction and found $\phi(k)$, and I managed to get the following equation for $\Psi(x, t)$:
$$\Psi(x, t)
= \frac{1}{(2\pi a)^\frac{1}{4}} \int^{\infty}_{-\infty} \exp\left[ikx - k^2 \left( \frac{1}{2a} + \frac{\hbar t}{2m} \right) \right] \frac{dk}{2\pi} \, .$$
I got stuck after that so I looked at the answer booklet provided and my luck failed me. The answer went from the integral I provided immediately to the final result with no intermediate steps.
Any help would be much appreciated. It looks like it requires changing the integral to be with respect to x.

REPLY [1 votes]: The integral can be computed independent of $x$, in fact, it must, because $x$ must remain after the integration, otherwise, $\Psi(x,t)$ would no longer be a function of $x$.  Instead, you are integrating out the variable $k$.
Anyway, you can try writing down the argument for the exponential, and then completing the square.  Then you'll have something like $$e^{-C(k+A)^2 + B}$$in which case, $e^B$ can be factored out, and you can do a substitution $$u = k+A$$ $$du = dk$$ to make it a simple Gaussian integral.