$H_{m}=\Omega X_{qc},\quad I_{m}=JZ_{qc}\ .$ (181)

The qubit couplers are extended objects that have small mutual inductances
where they overlap:

$V_{c}=\sum_{i>j}f_{ij}Z_{qc,i}Z_{qc,j}\ .$ (182)

Finally, each qubit of the simulator has a local field:

$V_{q}=h_{i}^{c}Z_{q,i}+t_{i}^{c}X_{q,i}\ .$ (183)

It will turn out that $t,f,h$ are all $\sim 1/n$, while $J,\Omega$ is $O(1)$.
The smallness of $f$ ensures that if the extended coupler qubit is a flux
qubit, its inductance is more than enough to account for all those $f$
mutuals: imagine a long coil split into $n$ segments, and each brought in
proximity with the corresponding coil of another coupler qubit. Two coils can
wind together on a chip. The bare Hamiltonian is just the qubit couplers and
their couplings to the qubit:

$H_{0}=\sum_{i}JZ_{q,i}Z_{qc,i}+\Omega X_{qc,i}\ .$ (184)

The perturbation includes the noise in the bare Hamiltonian, as well as the
remaining terms. As discussed in the general proof, they can also have noise,
but we’ll just use upper bounds on them except for the first-order error. The
full perturbation is:

$V=V_{q}+V_{c}+\sum_{i}\delta J_{i}Z_{q,i}Z_{qc,i}+\delta\Omega_{i}X_{qc,i}\
.$ (185)

To apply our result, we define the logical subspace $P$ in the bare
Hamiltonian. Out of the 4 states of each qubit-coupler pair, we choose 2 as
follows. Note that $Z_{q}$ is the integral of motion. For each of the two
eigenvalues $\pm 1$ of $Z_{q}$, choose the corresponding ground state. We will
not use the state itself, only the projector. Let $P_{0},P_{1}$ be the
projectors onto the qubit states in the computational basis (with $\pm 1$
eigenvalues of $Z_{q}$). The projector onto (in other words, pure state
density matrix of) the corresponding coupler state is given by:

$\rho_{0,1}=\frac{1}{2}(1-\frac{\Omega X_{qc}\pm
JZ_{qc}}{\sqrt{\Omega^{2}+J^{2}}})\ .$ (186)

The total subspace of interest is

$P_{0}\rho_{0}+P_{1}\rho_{1}\ ,$ (187)

for this pair, and defining

$\Pi_{i}=|\tilde{0}_{i}\rangle\langle\tilde{0}_{i}|+|\tilde{1}_{i}\rangle\langle\tilde{1}_{i}|=P_{0,i}\rho_{0,i}+P_{1,i}\rho_{1,i}\
,$ (188)

the subspace of interest of the whole system becomes

$P=\prod_{i}\Pi_{i}=\prod_{i}\sum_{m_{i}}|\tilde{m}_{i}\rangle\langle\tilde{m}_{i}|\
.$ (189)

Here we defined the basis within the subspace by
$\prod_{i}|\tilde{m}_{i}\rangle$. If we treat it as the computational basis
corresponding to bitstrings $m_{1}\dots m_{n}$, we can denote the
corresponding Pauli matrices acting within the subspace as
$\tilde{X},~{}\tilde{Z}$. For simplicity of notation, augment them by zeros
outside the subspace so that all the matrices act in the original space. The
qubit $Z$ matrices coincide with $\tilde{Z}$ within the subspace
$PZ=P\tilde{Z}$, so we’ll use them interchangeably. The qubit $X$ is not as
trivial, it contains out-of-subspace transitions, and there is a nontrivial
overlap coefficient that appears in front of $\tilde{X}$ when $X$ is expressed
via it, as we will see below. Note that the gap between this subspace and the
rest is $\Delta=\sqrt{\Omega^{2}+J^{2}}$. The quantities required for our
lemma can be computed using Eq. (186):

$\displaystyle\chi_{i,j}=\text{tr}\rho_{1}Z_{qc}=\frac{J}{\sqrt{J^{2}+\Omega^{2}}}\
,\quad
F=\sqrt{\text{tr}\rho_{0}\rho_{1}}=\frac{\Omega}{\sqrt{J^{2}+\Omega^{2}}}\
,\quad i_{i,j}=\|PZ_{qc}\|=1\ .$ (190)

Setting the gap $\Delta=\sqrt{\Omega^{2}+J^{2}}=1$, the Hamiltonian defined
above takes the form

$H_{m}=\sqrt{1-J^{2}}X,\quad I_{m}=JZ\ .$ (191)

The coupling operator $I_{i,j}=Z_{qc,i}$ independent of $j$. The list of the
values required by our theorem is as follows:

$\chi_{i,j}=J,~{}F=\sqrt{1-J^{2}},~{}i_{i,j}=1\ .$ (192)

The errors $\delta_{J}\geq\|P\delta J_{i}Z_{q,i}Z_{qc,i}\|=|\delta J_{i}|$ and
$\delta_{H}\geq\|P\delta\Omega_{i}X_{q,i}Z_{qc,i}\|=|\delta\Omega_{i}|$ are
all just control errors of the 1 and 2 qubit terms, that we consider being
$\delta$ just as the errors in $f,h,t$. Plugging it all in the theorem, we
get:

$\delta\leq\text{max}_{J}\frac{(s\epsilon)^{2}\Delta_{V}(3+\sqrt{1-J^{2}}+sJ^{2})^{-1}}{12\cdot
7n(1+\sqrt{1-J^{2}}^{-1}+sJ^{-2})^{2}}\ .$ (193)

From this, we obtain the bound on $\delta$ for all values of $s\geq
1.5,~{}\Delta_{V}\leq\Delta=1,~{}J\in[0,1],~{}\epsilon\leq 7/16$:

$\delta\leq\frac{\epsilon^{2}}{12\cdot
7n(1+\sqrt{1-J^{2}}^{-1}+1.5J^{-2})}\leq\frac{1}{3\cdot 2^{9}n}\ .$ (194)

The reduction factor we use:

$\alpha_{o}=\frac{s\epsilon\Delta_{V}}{3\cdot
7n(1+\sqrt{1-J^{2}}^{-1}+sJ^{-2})^{2}}\ ,$ (195)

is bounded as:

$\alpha_{o}=\frac{J^{2}}{3\cdot 2^{4}n(1+\sqrt{1-J^{2}}^{-1}+sJ^{-2})}\ .$
(196)

For $v$, we can just use $\|V\|$:

$\displaystyle\|V\|\leq
n(\delta_{J}+\delta_{H})+\alpha_{o}(1+F^{-1}+s\frac{1}{|\chi_{i,j},\chi_{j,i}|}))\leq\frac{1}{3\cdot
2^{8}}+\frac{1}{3\cdot 2^{4}}\leq 0.023\ .$ (197)

Thus $\Delta_{V}=0.95\Delta=0.95$ can be used.

Investigating Eq. (193) numerically, we get $10^{-7}$ required control
precision from this simplified formula for $n=40,~{}s=3,~{}\epsilon=0.1$.
Using more careful embedding and bounds discussed in the Appendix H improves
this result by 2 orders of magnitude.

Several checks need to be made before one can use the solution above. First,
note that $f_{ij}\sim\alpha\ll 1/n$, therefore each individual interaction
between mediators can be realized just as a mutual inductance between the
corresponding portions of their inductors (for flux qubit mediators). If we
needed $f_{ij}\sim 1$, it would be unrealistic to fit so many interactions
onto a single qubit. Finally, $\alpha F^{-1}\leq 1$ which is simple here since
$F$ is $n$-independent.

###  E.2 results for LC circuit

We now move on to the Harmonic oscillator (LC-circuit) mediator, which would
be a warmup for the transmission line. Let’s first define the Hamiltonian:

$H_{m}=a^{\dagger}a,\quad I_{m}=J(a+a^{\dagger})\ .$ (198)

This $I_{m}$ results in a shift $\tilde{a}=a+ZJ$, where $\tilde{a}$ defines
the new ground state. The coupling operator $I_{i,j}=a_{i}+a_{i}^{\dagger}$
independent of $j$. The quantities needed for the lemma can be computed to be:

$\Delta=1,\quad\chi=2J,\quad i_{m}=1\ .$ (199)

The computation of $F$ is nontrivial: in the coordinate basis corresponding to
$x=(a+a^{\dagger})/\sqrt{2}$, the Hamiltonian becomes:

$H_{m}+ZI_{m}=\frac{x^{2}+p^{2}-1}{2}+\sqrt{2}JZx=\frac{\tilde{x}^{2}+p^{2}-1}{2}-J^{2}\
,$ (200)

where $\tilde{x}=x+\sqrt{2}JZ$. The wavefunctions of the groundstates for the
two values of $Z$ are:

$\psi_{0}(x,Z)=\frac{1}{\pi^{1/4}}e^{-\frac{(x+\sqrt{2}JZ)^{2}}{2}}\ .$ (201)

Their overlap is:

$F=\int_{-\infty}^{\infty}\psi_{0}(x,1)\psi_{0}(x,-1)dx=e^{-2J^{2}}\ .$ (202)

The errors in $J$ are $\delta_{I}$ directly, while the meaning of $\delta_{H}$
is more complex. We have various errors in the form
$x,x^{2},p,p^{2},\\{x,p\\}$. For a single mediator, the error operator is:

$\displaystyle\delta H_{m}=$
$\displaystyle\delta_{x}x+\delta_{x2}x^{2}+\delta_{p}p+\delta_{p2}p^{2}+\delta_{px}\\{p,x\\}=$
$\displaystyle=$
$\displaystyle\delta_{x}(\tilde{x}-\sqrt{2}JZ)+\delta_{x2}(\tilde{x}^{2}-2\sqrt{2}\tilde{x}JZ+2J^{2})+\delta_{p}p+\delta_{p2}p^{2}+\delta_{px}(\\{p,\tilde{x}\\}-2\sqrt{2}JZp)\
,$ $\displaystyle\|P\delta H_{m}\|\leq$
$\displaystyle\frac{(|\delta_{x}|+|\delta_{x2}|2\sqrt{2}J)+(|\delta_{p}|+|\delta_{px}|2\sqrt{2}J)}{\sqrt{2}}+\sqrt{2}|\delta_{px}|+\frac{\sqrt{3}(|\delta_{x2}|+|\delta_{p2}|)}{2}+|\delta_{x2}|2J^{2}+|\delta_{x}|\sqrt{2}J\leq$
$\displaystyle\leq$
$\displaystyle\frac{3|\delta_{x}|+|\delta_{p}|}{\sqrt{2}}+(2+\sqrt{2})|\delta_{px}|+\frac{(8+\sqrt{3})|\delta_{x2}|+\sqrt{3}|\delta_{p2}|}{2}\leq\delta_{H}\
,$

since $J\leq 1$. The theorem will work for any bound $\delta_{H}\geq\|P\delta
H_{m}\|$, but for the purposes of constraining $\Delta_{V}$ we define one such
bound $\delta_{H}$ as:

$\delta_{H}=\text{max}\left(2.2|\delta_{x}|+0.8|\delta_{p}|+3.5|\delta_{px}|+4.9|\delta_{x2}|+0.9|\delta_{p2}|\right)\
.$ (203)

We now use the above definitions to compute the contribution to the value
$v=\|V\|_{c}$ used in $\Delta_{V}$, where $\|\cdot\|_{c}$ is defined in Eq.
25. Specifically we will compute a bound $C$ on $\|\sum_{i}\delta
H_{m,i}\|_{c}$, which will require us to prove:

$-C(I+H_{0})\leq\sum_{i}\delta H_{m,i}\leq\|\sum_{i}C(I+H_{0})\ .$ (204)

If each $\delta H_{m,i}$ (we omit the index $i$ when describing individual
mediator with the Hamiltonian $H_{m}+ZI_{m}+E_{0}$ where we choose the energy
shift $E_{0}=J^{2}$ s.t. the ground state energy is exactly zero) obeys

$-aI-b(H_{m}+ZI_{m}+E_{0})\leq\delta H_{m}\leq aI+b(H_{m}+ZI_{m}+E_{0})\ ,$
(205)

then the norm of the sum obeys:

$-naI-bH_{0}\leq\sum_{i}\delta H_{m,i}\leq naI+bH_{0}\ ,$ (206)

since $H_{0}=\sum_{i}H_{m,i}+Z_{i}I_{m,i}+E_{0}$. The identity term will
almost always dominate in the bound $C=$max$(na,b)$. Still, we compute both.
First note that:

$\displaystyle|x|\leq\frac{1+x^{2}}{2}\leq 1+H_{m},\quad|p|\leq 1+H_{m}\
,\quad\pm\\{x,p\\}\leq x^{2}+p^{2}\leq 1+2H_{m}\ ,\quad x^{2}\leq
1+2H_{m},\quad p^{2}\leq 1+2H_{m}\ .$

The absolute value on operators possesses the properties $A\leq|A|$ and
$-A\leq|A|$. By transitivity if $|A|\leq B$ then $A\leq B$ and $-A\leq B$. The
full Hamiltonian of one mediator is not just $H_{m}=(x^{2}+p^{2}-1)/2$, but
$H_{m,i}+ZI_{m,i}+E_{0}=(p^{2}+(x+\sqrt{2}JZ)^{2}/2=(p^{2}+\tilde{x}^{2}-1)/2$.
A more careful calculation follows:

$\displaystyle\pm
x=\pm\tilde{x}\mp\sqrt{2}JZ\leq\sqrt{2}J+\frac{1+\tilde{x}^{2}}{2}\leq
1+\sqrt{2}J+H_{m}+ZI_{m}+E_{0}\ ,\quad\pm
Z\tilde{x}\leq\frac{1+\tilde{x}^{2}}{2}\leq 1+H_{m}+ZI_{m}+E_{0}\ .$

The other two terms containing $x$ are bounded as follows:

$\displaystyle\pm\\{x,p\\}=\mp
2\sqrt{2}JZp\pm\\{\tilde{x},p\\}\leq\sqrt{2}J+(\sqrt{2}J+1)(\tilde{x}^{2}+p^{2})=2\sqrt{2}J+1+2(\sqrt{2}J+1)(H_{m}+ZI_{m}+E_{0})\
,$ (207) $\displaystyle
x^{2}=(\tilde{x}-\sqrt{2}JZ)^{2}=\tilde{x}^{2}-2\sqrt{2}JZ\tilde{x}+2J^{2}\leq
2J^{2}+2\sqrt{2}J+1+(2+2\sqrt{2}J)(H_{m}+ZI_{m}+E_{0})\ .$ (208)

This allows us to determine $a$ and $b$:

$\displaystyle a\leq$
$\displaystyle\left(1+\sqrt{2}J\right)|\delta_{x}|+|\delta_{p}|+|\delta_{p}|^{2}+\left(2J^{2}+2\sqrt{2}J+1\right)|\delta_{x2}|+(2\sqrt{2}J+1)|\delta_{px}|\leq$
$\displaystyle\leq$
$\displaystyle~{}2.5|\delta_{x}|+|\delta_{p}|+|\delta_{p2}|+5.9|\delta_{x2}|+3.9|\delta_{px}|\leq
1.25\delta_{H}\ ,$ $\displaystyle b\leq$
$\displaystyle|\delta_{x}|+|\delta_{p}|+(\sqrt{2}J+2)|\delta_{xp}|+(2+2\sqrt{2}J)|\delta_{x2}|+2|\delta_{p2}|\leq|\delta_{x}|+|\delta_{p}|+3.5|\delta_{xp}|+4.9|\delta_{x2}|+2|\delta_{p2}|\leq
2.3\delta_{H}\ .$

Using this we find:

$\displaystyle\|\sum_{i}\delta
H_{m,i}\|_{c}\leq\text{max}(na,b)\leq\text{max}(1.25\delta_{H}n,2.3\delta_{H})\leq
1.25\delta_{H}n\ .$ (209)

If instead of $x\leq\frac{1+x^{2}}{2}$ we had used
$x\leq\frac{\lambda+\lambda^{-1}x^{2}}{2}$, then optimizing $\lambda$ would
have balanced the powers of $n$ between two arguments of the maximum at least
for some of the terms, but we will find that even this weak bound will be
enough for our purposes. We see that even though we did not specify $\|\delta
H_{m}\|_{c}$ in the definition of $\delta_{H}$, a multiple of it bounds it.
The same occurs for $\|\delta
I_{m}\|_{c}\leq\sqrt{2}(1+\sqrt{2}J)\delta_{I}n\leq 3.5\delta_{I}n$. The
quantities used in the theorem are:

$i_{i,j}=\chi_{i,j}+i_{m}=1+2J,\quad\frac{i_{i,j}}{\chi_{i,j}}=\frac{1}{2J}+1\
.$ (210)

Plugging it in we get the following:

$\displaystyle\delta_{H}+\delta_{I}+\delta(1+e^{-2J^{2}}+s(2J)^{2})\leq\frac{(s\epsilon)^{2}\Delta_{V}}{12\cdot
7n(1+e^{2J^{2}}+s(\frac{1}{2J}+1)^{2})^{2}}\ .$ (211)

Since $\Delta_{V}\leq\Delta$, and the full range of parameters is $s\geq
1.5,~{}\epsilon\leq 7/16,~{}J\leq 1$, we establish:

$\delta_{H}+\delta_{I}\leq\frac{c}{2^{6}\cdot 3^{5}n}\ .$ (212)

The theorem uses

$\alpha_{o}=\frac{s\epsilon\Delta_{V}}{3\cdot
7n(1+e^{2J^{2}}+s(\frac{1}{2J}+1)^{2})^{2}}\ .$ (213)

Which can be bounded as

$\alpha_{o}\leq\frac{1}{3n2^{4}(\frac{1}{2J}+1)^{2}(1+e^{2J^{2}}+s(\frac{1}{2J}+1)^{2})}\
.$ (214)

Now we can compute $v$:

$\displaystyle v\leq
n\left(1.25\delta_{H}+3.5\delta_{I}+2\alpha_{o}+\frac{2s\alpha_{o}}{2\text{min}{\chi_{i,j}\chi_{j,i}}}\right)\
.$ (215)

The inequality on $v$ is obtained using the bound on $\alpha_{o}$ and
$\delta_{H}+\delta_{I}$:

$\displaystyle v\leq\frac{3.5\cdot 7}{2^{6}\cdot
3^{5}}+\frac{\left(1+\frac{s}{4J^{2}}\right)}{3\cdot
2^{4}(\frac{1}{2J}+1)^{2}(1+e^{2J^{2}}+s(\frac{1}{2J}+1)^{2})}\ .$

The last term is bounded as:

$\frac{4\left(\frac{J^{2}}{s}+\frac{1}{4}\right)}{3\cdot
2^{4}(\frac{1}{\sqrt{2J}}+\sqrt{2J})^{4}}\leq\frac{\frac{2}{3}+\frac{1}{4}}{3\cdot
2^{6}}\leq 0.0048\ .$ (216)

The upper bound on $v$ is then

$v\leq 0.0016+0.0048\leq 0.0064\ .$ (217)

In other words, taking $\Delta_{V}=0.9936\Delta$ is justified in the entire
region of solutions.

We see that even though qubit is a nonlinear circuit element, and the harmonic
oscillator is a linear circuit element, there’s only a minute difference
between them in this gadget.

###  E.3 The main result for the transmission line

Now let’s consider the transmission line. Specifically, we use the open
boundary conditions transmission line of $n$ sites with the qubit attached at
the start. This circuit is analyzed in App. F, resulting in the following
Hamiltonian:

$\displaystyle
H_{m,i}=\sum_{l}^{n}p_{ci,l}^{2}+x_{ci,1}^{2}+x_{ci,n}^{2}+\sum_{l=2}^{n}(x_{ci,l}-x_{ci,l-1})^{2}\
,\quad I_{m,i}=Jx_{ci,1}\ .$ (218)

The coefficient $J\leq 1$. The other $n-1$ chains have a mutual inductance at
inductors $2\dots n$ in order. Specifically, for the chain $i$, the coupling
to chain $j$ is at a location $r_{i}(j)=j+1$ for $j<i$ and $r_{i}(j)=j$ for
$j>i$. The coupling operator is:

$I_{i,j}=x_{ci,r_{i}(j)}-x_{ci,r_{i}(j)-1}\ .$ (219)

Unfortunately, there is no way to couple locally to $x_{ci,r_{i}(j)}$ only.
For the computation of the parameters needed for the theorem, see Appendix G.
The results are:

$\displaystyle\Delta=2~{}\text{sin}\frac{\pi}{2(n+1)}~{},\quad\chi=\frac{J}{n+1}\
,\quad i_{i,j}\leq\sqrt{2}\ ,\quad
F^{-1}=\text{exp}\frac{J^{2}}{4(n+1)}\sum_{k=1}^{n}\frac{\cos^{2}\frac{k\pi}{2(n+1)}}{\sin\frac{k\pi}{2(n+1)}}\
.$ (220)

The expression for $F$ is bounded by:

$F^{-1}\leq e^{J^{2}(0.5+\text{ln}n)/4}\ .$ (221)

Since $J\leq 1$, this is upper bounded by $e^{1/8}n^{1/4}$.

The meaning of errors $\delta_{H},\delta_{I}$ is as follows. If $\delta_{zx}$
is the error in the specification of $J$, then $\|\delta
I_{m}\|=|\delta_{zx}|\|Px_{ci,1}\|\leq|\delta_{zx}|(J+\sqrt{(\text{ln}n+1)/2})$
from Eq. (271,277) and we define
$\delta_{I}=\text{max}|\delta_{zx}|(J+\sqrt{(\text{ln}n+1)/2})$ (In the main
text, we use $J=1$ and a simpler expression $1+\sqrt{\text{ln}n}\geq
1+\sqrt{(\text{ln}n+1)/2}$ for our range $n\geq 4$). The corresponding
contribution to $v$ from a single mediator is:

$\pm x_{ci,1}\leq\frac{1+J}{2}-E_{0}+\frac{1}{2}(H_{m,i}+Z_{i}I_{m,i}+E_{0})\
,$ (222)

where $E_{0}$ is given in App. G.1. The leading contribution from the sum is
then:

$\displaystyle\|\sum_{i}x_{ci,1}\|_{c}\leq(\frac{1+J}{2}-E_{0})n\leq(1.5+n)n\
,\quad\|\sum_{i}\delta
I_{m,i}\|_{c}\leq(1.5+n)n\text{max}|\delta_{j}|=\frac{(1.5+n)n\delta_{I}}{J+\sqrt{(\text{ln}n+1)/2}}\
.$ (223)

As for the errors $\delta H_{m,i}$, while it is straightforward to include
terms like $\sum_{l}\delta_{p,il}p_{ci,l}$ and the quadratic terms as we did
for the LC circuit mediator, we will consider only the terms $\delta
H_{m,l}=\sum_{l}^{n-1}\delta_{x,il}(x_{ci,l}-x_{ci,l+1})$ for simplicity:

$\displaystyle\|P\delta
H_{m,i}\|\leq\sum_{l}^{n-1}|\delta_{x,il}|\|P((x_{ci,l}-x_{ci,l+1})\|\leq\sum_{l}^{n-1}|\delta_{x,il}|i_{i,l}\leq\sqrt{2}\sum_{l}^{n-1}|\delta_{x,il}|\
.$ (224)

We define: $\delta_{H}=\sqrt{2}(n-1)\text{max}|\delta_{x,il}|$ Note that any
definition of control errors the bound $\delta_{H}$ will contain a factor of
$n$: $\delta_{H}=n\text{max}\|PO_{ci,l}\|$ where $O_{ci,l}$ are some local
error operators (we have used $O_{ci,l}=I_{i,l}$). This means that
$\delta_{H}=\sqrt{2}(n-1)\delta_{H,\text{loc}}$ where $\delta_{H,\text{loc}}$
is the constant local error, the maximum of the terms $|\delta_{x,il}|$. That
would be the most optimistic estimate for our purposes. Now we use:

$\pm I_{i,j}\leq\frac{1}{2}-E_{0}+\frac{1}{2}(H_{m,i}+Z_{i}I_{m,i}+E_{0})\ ,$
(225)

to bound the contribution to $v$ as:

$\|\sum_{i}\delta
H_{m,i}\|_{c}\leq\left(\frac{1}{2}-E_{0}\right)\frac{n\delta_{H}}{\sqrt{2}}\leq\frac{n(1+n)\delta_{H}}{\sqrt{2}}\
.$ (226)

We now use the theorem:

$\displaystyle\delta_{H}+\delta_{I}+\delta(1+F+J^{2}s(n+1)^{-2})\leq\frac{\Delta_{V}(s\epsilon)^{2}}{12\cdot
7n(1+F^{-1}+2sJ^{-2}(n+1)^{2})^{2}}\ .$ (227)

Before computing $\Delta_{V}$, we note that the $J^{-2}(n+1)^{2}$ term is the
leading term in the $n\to\infty$ limit, while $F^{-1}\leq n^{1/4}$ is
subleading. Due to this, we take possibly suboptimal $J=1$ for our final
result that was reported in Sec. III. We will now bound
$\delta_{H}+\delta_{I}$ in the whole range:

$\displaystyle\delta_{H}+\delta_{I}\leq\frac{7\Delta s^{2}}{2^{10}\cdot
3n(1+F^{-1}(J)+2sJ^{-2}(n+1)^{2})^{2}}\leq\frac{7\pi}{2^{12}\cdot
3n(n+1)^{5}J^{-4}}\leq\frac{7\pi}{2^{12}\cdot 3n(n+1)^{5}}\ .$ (228)

The value of $\alpha_{o}$ is:

$\alpha_{o}=\frac{\Delta_{V}s\epsilon}{3\cdot
7n(1+F^{-1}(J)+2sJ^{-2}(n+1)^{2})^{2}}\ .$ (229)

It is bounded as:

$\displaystyle\alpha_{o}\leq\frac{\Delta}{3\cdot
2^{5}nJ^{-2}(n+1)^{2}(1+F^{-1}(J)+3J^{-2}(n+1)^{2})}\leq\frac{\pi
J^{2}}{9\cdot 2^{5}n(n+1)^{5}}\ .$ (230)

We note that

$\displaystyle\|\sum_{ij}\frac{J_{ij}^{*}}{\chi_{i,j}\chi_{j,i}}I_{i,j}I_{j,i}\|_{c}=2^{-1}(n+1)^{2}J^{-2}\|\sum_{ij}J_{ij}^{*}(I_{i,j}^{2}+I_{j,i}^{2})\|_{c}\leq
2^{-1}(n+1)^{2}J^{-2}(-E_{0}n)\leq 2^{-1}(n+1)^{2}J^{-2}(0.5+n)n\ .$

The expression for $v$ is:

$\displaystyle v\leq
n\left(\frac{(1.5+n)\delta_{I}}{J+\sqrt{(\text{ln}n+1)/2}}+\frac{(n+1)\delta_{H}}{\sqrt{2}}+\alpha_{o}\left(2+\frac{(n+1)^{2}J^{-2}(0.5+n)n}{2n}\right)\right)\leq$
$\displaystyle\leq(n+1)n(\delta_{I}+\delta_{H})+\frac{\pi(2n+(n+1)^{2}(0.5+n)n/2)}{9\cdot
2^{5}n(n+1)^{5}}\leq\frac{7\pi}{2^{12}\cdot
3(n+1)^{4}}+\frac{\pi(2+0.5(n+1)^{2}(0.5+n))}{9\cdot 2^{5}(n+1)^{5}}\leq$
$\displaystyle\leq\left(\frac{7\pi}{2^{12}\cdot
3(n+1)^{3}}+\frac{\pi(2+0.5(n+1)^{2}(0.5+n))}{9\cdot
2^{5}(n+1)^{4}}\right)\sin\frac{\pi}{2(n+1)}\leq 0.00052\Delta\ ,$

for $n\geq 4$. Since $\Delta\leq 2\sin(\pi/10)$, we obtain:

$\Delta-\Delta_{V}=v(1+\Delta)\leq 0.0009\Delta\ .$ (231)

In other words, it is sufficient to take $\Delta_{V}=0.9991\Delta$. We also
note that the correction $\Delta-\Delta_{V}$ has $n^{-1}\Delta$ scaling. The
resulting scaling of errors is $n^{-6}$ for most errors and $n^{-7}$ for the
errors in the implementation of the transmission line $\delta_{H,\text{loc}}$.
The full form of the constraint on $\delta$’s is:

$\displaystyle\sqrt{2}(n-1)\delta_{H,\text{loc}}+\delta_{I}+\delta(1+F+\frac{s}{n^{2}})\leq
2~{}\text{sin}\frac{\pi}{2(n+1)}\frac{0.9991(s\epsilon)^{2}}{12\cdot
7n(1+F^{-1}+2s(n+1)^{2})^{2}}\ .$ (232)

In Sec. III we present a stronger inequality from which the above follows:

$\displaystyle\sqrt{2}n\delta_{H,\text{loc}}+\delta_{1}(1+\sqrt{\text{ln}n})+3\delta\leq\frac{0.011\epsilon^{2}}{n(n+1)^{5}}\
,$ (233)

where $\delta_{1}$ is the maximum of the error terms $|\delta_{zx,i}|$. We can
also use the freedom in $\Delta_{V}$ as one of the ways to simplify $\alpha$.
Indeed, let $\Delta_{V}^{*}\leq\Delta_{V}$ be the new value defined as
follows:

$\Delta_{V}^{*}=\frac{\pi\Delta_{V}}{(n+1)\Delta(\frac{J^{2}(1+F^{-1}(J))}{2s(n+1)^{2}}+1)^{2}},\quad\alpha_{o}=\frac{\Delta_{V}^{*}\epsilon
J^{4}}{12\cdot 7ns(n+1)^{4}}\ .$ (234)

We can then use $\Delta_{V}^{*}$ in the expression on $\delta^{\prime}$s and
strengthen the factor to a constant (for $n\geq 4,s\geq 1.5,J\in[0,1]$, we use
$(\frac{J^{2}(1+F^{-1}(J))}{2s(n+1)^{2}}+1)^{-2}\geq 0.934$):

$\displaystyle\sqrt{2}n\delta_{H,\text{loc}}+\delta_{1}(1+\sqrt{\text{ln}n})+3\delta\leq\frac{0.01\epsilon^{2}}{n(n+1)^{5}}\
.$ (235)

Which is the main reported result together with the simplified expression for
$\alpha_{o}$ for $J=1$ and
$\Delta_{V}^{*}=0.934*0.9991\pi/(n+1)=0.933\pi/(n+1)$:

$\alpha_{o}=\frac{0.035\epsilon}{ns(n+1)^{5}}\ .$ (236)

##  Appendix F Transmission line circuit description

Figure 4: A transmission line where every local inductance $L$ has a small
mutual inductance $M$ with another circuit

Here we explain how a Hamiltonian of $x_{i}$ and $p_{i}$ is derived from a
circuit description of the collection of transmission lines, and why coupling
is to the difference $x_{i}-x_{i-1}$

The simplest model of a transmission line we use is illustrated in Fig. 4. It
is made of two parallel wires, and for each portion of the length of those
wires with an inductance $L$ there’s a capacitance $C$ that connects the
wires. The boundary conditions are that the ends of the transmission line are
connected and there’s one extra inductance so the circuit is reflection-
symmetric. We first investigate the mutual inductance $M$ between two
inductors with inductance $L$. For currents $\mathcal{I}_{u},\mathcal{I}_{d}$
going through them, the flux energy is given by:

$E_{\Phi}=\frac{L}{2}(\mathcal{I}_{d}^{2}+\mathcal{I}_{u}^{2})+M\mathcal{I}_{d}\mathcal{I}_{u}\
.$ (237)

The flux through each of the elements is:

$\Phi_{d}=L\mathcal{I}_{d}+M\mathcal{I}_{u},\quad\Phi_{u}=L\mathcal{I}_{u}+M\mathcal{I}_{d}\
.$ (238)

Expressing the energy in terms of flux:

$E_{\Phi}=\frac{L(\Phi_{u}^{2}+\Phi_{d}^{2})-2M\Phi_{u}\Phi_{d}}{2(L^{2}-M^{2})}\
.$ (239)

Now let’s consider the full gadget, with $j$’th transmission line having
inductance $L_{j,i}$ at position $i=1..n$, and the coupling position
$r_{j}(i)$ defined in Eq. (3). The flux energy is:

$E_{\Phi,\text{tot}}=\sum_{i=1}^{n}\sum_{j=1}^{n+1}\frac{L_{i,j}}{2}\mathcal{I}_{i,j}^{2}+\sum_{i>j>1}^{n}M_{ij}\mathcal{I}_{i,r_{i}(j)}\mathcal{I}_{j,r_{j}(i)}+M_{q}\sum_{i=1}^{n}\mathcal{I}_{i,1}\mathcal{I}_{qi}\
.$ (240)

All the inductors except for the one at $j=n+1$ are coupled to something in
our construction. The one at $j=0$ is coupled to a qubit and all the middle
ones are coupled to different transmission lines by tunable couplers $M_{ij}$.
In hardware, a tunable coupler is a complicated circuit itself, but we can
consider the mutual inductance coupling to its inductance constant, and the
tunability coming from the properties of its design and applied controls. We
do not need to go into details of its implementation for our construction, we
only observe that as long as the required $M$ is not bigger than $L$, it is
feasible to realize. At the very least, one can always directly fabricate the
required values without tunability allowed. Switching from currents to fluxes
yields:

$E_{\Phi,\text{tot}}=\sum_{i=1}^{n}\left(\frac{1}{2L_{i,1}}\Phi_{i,1}^{2}+\frac{1}{2L_{i,n+1}}\Phi_{i,n+1}^{2}+\sum_{j=2}^{n}\frac{L_{i,j}}{2(L_{i,j}^{2}-M_{i,j}^{2})}\Phi_{i,j}^{2}\right)+\sum_{i>j>1}^{n}\frac{-M_{i,j}}{(L_{i,j}^{2}-M_{i,j}^{2})}\Phi_{i,r_{i}(j)}\Phi_{j,r_{j}(i)}-\frac{M_{q}}{L_{i,1}}\sum_{i=1}^{n}\Phi_{i,1}\mathcal{I}_{qi}\
.$ (241)

To obtain the starting point of our construction in the main text, we need to
enforce (introducing dimensionless units):

$E_{\Phi,\text{TL
}j}=\sum_{i=1}^{n+1}\Phi_{j,i}^{2}+\sum_{i>j>1}f_{ij}\Phi_{j,r_{j}(i)}\Phi_{i,r_{i}(j)}+\sum_{i=1}^{n}j\Phi_{i,1}Z_{i}\
.$ (242)

We choose each $L_{i,j},M_{i,j}$ as solutions of the equations:

$L_{i,1}=L_{i,n+1}=1,\quad
M_{q}=J|\mathcal{I}_{qi}|,\quad\frac{L_{i,j}}{2(L_{i,j}^{2}-M_{i,j}^{2})}=1|_{j=2\dots
n}\quad\frac{-M_{i,j}}{L_{i,j}^{2}-M_{i,j}^{2}}=f_{ij}|_{j=2\dots n}\ .$ (243)

As there is a closed flux loop, we enforce $\sum_{k=1}^{n+1}\Phi_{i,k}=0$,
which allows to express $\Phi_{i,n+1}=-\sum_{k=1}^{n}\Phi_{i,k}$. We next note
that the voltages across the capacitors are:

$Cq_{j,i}=\sum_{k=1}^{i}\dot{\Phi}_{j,k}\ .$ (244)

It is these sums that make it impossible to couple to the variable conjugate
to $q$ locally. The best we can do is to couple to a difference as follows.
Introducing new variables $x_{cj,l}$ such that
$\Phi_{j,l}=x_{cj,l}-x_{cj,l-1}$ for $i>1$, and $\Phi_{j,1}=x_{cj,1}$, we
express the sums as:

$\sum_{k=1}^{l}\Phi_{j,k}=x_{cj,l},\quad\Phi_{j,n+1}=-x_{cj,n}\ .$ (245)

We define the interaction operators
$I_{i,j}=\Phi_{i,r_{i}(j)}=x_{ci,r_{i}(j)}-x_{ci,r_{i}(j)-1}$. The total
energy is given by:

$E_{\text{tot}}=\sum_{j=1}^{n}\left(\sum_{l=1}^{n}\frac{\dot{x}_{j,l}^{2}}{2C}+x_{cj,1}^{2}+x_{cj,n}^{2}+\sum_{l=2}^{n}(x_{cj,l}-x_{cj,l-1})^{2}\right)+\sum_{i>j>1}f_{ij}I_{i,j}I_{j,i}+\sum_{i=1}^{n}Jx_{ci,1}Z_{i}\
.$ (246)

Finally, we find the conjugate variables $p_{i,l}$ for $x_{ci,l}$ to be
$p_{i,l}=\dot{x}_{i,l}/C$. Setting $C=2$ results in the form of the
Hamiltonian that was the starting point of our construction. The steps we take
here appear slightly different from the common approach to circuit
quantization [31], but are equivalent to it.

##  Appendix G Transmission line calculation

The Hamiltonian derived in the previous section can be split into mediators
(individual transmission lines) coupled to their respective qubits, and
perturbation. The terms in the unperturbed Hamiltonian are:

$\displaystyle
H_{m,i}=\sum_{l}^{n}p_{ci,l}^{2}+2x_{ci,l}^{2}-2\sum_{l}^{n-1}x_{ci,l}x_{ci,l+1}$
(247) $\displaystyle I_{m}=Jx_{ci,1}\ .$ (248)

The perturbation includes the other $n-1$ couplings are at sites $2\dots n$ in
order. Specifically, for the chain $i$, the coupling to chain $j$ is at a
location $r_{i}(j)=j+1$ for $j<i$ and $r_{i}(j)=j$ for $j>i$. The coupling
operator is:

$I_{i,j}=x_{ci,r_{i}(j)}-x_{ci,r_{i}(j)-1}\ .$ (249)

To find quantities required the theorem in Sec. VI, we will start with finding
the spectrum of $H_{m}$. We can use the eigenvectors $v_{k}$ of the $n\times
n$ tridiagonal matrix $T_{ij}=\delta_{i,j+1}+\delta_{i,j-1}$ (open boundary
conditions) and their corresponding eigenvalues $e_{k}$, indexed by $k=1\dots
n$:

$v_{k,m}=\sqrt{\frac{2}{n+1}}\sin\frac{km\pi}{n+1},\quad m=1\dots n,\quad
e_{k}=2\cos\frac{k\pi}{n+1}\ .$ (250)

It is normalized:

$\sum_{k}v_{k,j}^{2}=1\ .$ (251)

With this we can introduce the change of variables (omitting the subscript
$ci$):

$p_{ci,l}=\sum_{k}v_{k,l}p_{k},\quad x_{ci,l}=\sum_{k}v_{k,l}x_{k}\ .$ (252)

It preserves the commutation relation
$[p_{ci,l},x_{ci,l^{\prime}}]=-i\delta_{ll^{\prime}}$. The resulting
Hamiltonian is:

$H_{m,i}=\sum_{k}^{n}p_{k}^{2}+(2-2\cos\frac{k\pi}{n+1})x_{k}^{2}+JZ_{i}x_{k}\sqrt{\frac{2}{n+1}}\sin\frac{k\pi}{n+1}\
.$ (253)

Or in terms of creation-annihilation operators:

$\displaystyle p_{k}=i\sqrt{\omega_{k}/2}(a_{k}-a_{k}^{\dagger}),\quad
x_{k}=\frac{1}{\sqrt{2\omega_{k}}}(a_{k}+a_{k}^{\dagger})$ (254)
$\displaystyle\omega_{k}=\sqrt{2(1-\cos\frac{k\pi}{n+1})}\ .$ (255)

Define the combination of the two transformations:

$x_{n}=\sum_{k}g_{k}^{n}(a_{k}+a_{k}^{\dagger}),\quad
g^{n}_{k}=\frac{1}{\sqrt{\omega_{k}}}\sqrt{\frac{1}{n+1}}\sin\frac{kn\pi}{n+1}\
.$ (256)

The linear term becomes

$\displaystyle
H_{m,i}=\sum_{k}\omega_{k}(a_{k}^{\dagger}a_{k}+0.5)+JZ(g_{k}^{1}a_{k}+g_{k}^{1*}a_{k}^{\dagger})\
.$ (257)

The shift in creation-annihilation operators diagonalizes the Hamiltonian

$H_{m,i}=\sum_{k}\omega_{k}(a_{k}+\frac{Jg_{k}^{1*}}{\omega_{k}})^{\dagger}(a_{k}+\frac{Jg_{k}^{1*}}{\omega_{k}})+\frac{\omega_{k}}{2}-\frac{J^{2}|g_{k}^{1}|^{2}}{\omega_{k}}\
,$ (258)

which suggests the magnitude of the shift in $a$:

$a_{k}=\tilde{a}_{k}-\frac{Jg_{k}^{1*}}{\omega_{k}}\ .$ (259)

The groundstate is annihilated by the new operators:

$\tilde{a}_{k}|g_{b_{i}}\rangle|_{b_{i}=1}=0\ .$ (260)

In terms of $\tilde{a}$, the relevant operators are:

$\displaystyle
H_{m,i}+E_{0}=\sum_{k}\omega_{k}\tilde{a}_{k}^{\dagger}\tilde{a}_{k},\quad\omega_{k}=2\sin\frac{k\pi}{2(n+1)},\quad
E_{0}=\sum_{k}-\frac{\omega_{k}}{2}+\frac{J^{2}|g_{k}^{1}|^{2}}{\omega_{k}}$
(261) $\displaystyle I_{i,j}=x_{ci,r_{i}(j)}-x_{ci,r_{i}(j)-1},\quad
x_{ci,r}=\sum_{k=1}^{n}g_{k}^{r}a_{k}+g_{k}^{r*}a_{k}^{\dagger}-\frac{J}{\omega_{k}}(g_{k}^{r}g_{k}^{1*}+g_{k}^{r*}g_{k}^{1})\
.$ (262)

In the following sections, we will compute the energy shift $E_{0}$ as well as
the following:

$\displaystyle\chi_{i,j}=\langle
g_{b_{i}}|I_{i,j}|g_{b_{i}}\rangle|_{b_{i}=1}\ ,\quad\|PI_{i,j}\|\leq i_{i,j}\
,\quad F=\langle g_{b_{i}=1}|g_{b_{i}=-1}\rangle\ .$ (263)

###  G.1 Energy shift

We need to know the exact energy shift such that the ground state is at zero
energy. The sums involved in $E_{0}$ are:

$\displaystyle-
E_{0}=\sum_{k}\frac{\omega_{k}}{2}-J^{2}\frac{|g_{k}^{1}|^{2}}{\omega_{k}}=\sum_{k}\sin\frac{k\pi}{2(n+1)}-J^{2}\frac{\sin^{2}\frac{k\pi}{n+1}}{4(n+1)\sin^{2}\frac{k\pi}{2(n+1)}}=$
(264)
$\displaystyle=\frac{1}{2}\left(\frac{\cos\frac{\pi}{4(n+1)}}{\sin\frac{\pi}{4(n+1)}}-1\right)-\frac{J^{2}}{n+1}\sum_{k}\cos^{2}\frac{k\pi}{2(n+1)}=\frac{1}{2}\left(\frac{\cos\frac{\pi}{4(n+1)}}{\sin\frac{\pi}{4(n+1)}}-1\right)-\frac{J^{2}n}{2(n+1)}\
.$ (265)

This expression is $\sim n$ due to the first term. It is bounded as:

$-E_{0}\leq n+\frac{1-J^{2}(1+n^{-1})^{-1}}{2}\ .$ (266)

###  G.2 Susceptibility

For the calculation of $\chi_{ij}$ we’ll need to find expectation values of
$x_{ci,r}$ at various positions $r$:

$\displaystyle\langle
g_{b_{i}}|x_{ci,r}|g_{b_{i}}\rangle|_{b_{i}=1}=-\sum_{k}\langle
g_{b_{i}}|\frac{J}{\omega_{k}}(g_{k}^{r}g_{k}^{1*}+g_{k}^{r*}g_{k}^{1})|g_{b_{i}}\rangle|_{b_{i}=1}=-\sum_{k}\frac{J(g_{k}^{r}g_{k}^{1*}+g_{k}^{r*}g_{k}^{1})}{\omega_{k}}=J\zeta_{r}\
.$

In the limit of $J\to 0$, the expression $\zeta_{r}$ is the linear response
susceptibility. It can be shown that the Kubo formula gives exactly this
expression, in other words, the linear response here coincides with the exact
response. In our notation, $\chi_{i,j}$ includes $J$ as a factor, and is the
difference:

$\chi_{i,j}=J(\zeta_{r_{i}(j)}-\zeta_{r_{i}(j)-1})\ .$ (267)

More generally, define

$\zeta_{r,s}=-\sum_{k}\frac{(g_{k}^{r}g_{k}^{s*}+g_{k}^{r*}g_{k}^{s})}{\omega_{k}}\
,$ (268)

such that $\zeta_{r}=\zeta_{r,1}$. Plugging in the expressions for $g$ and
$\omega$:

$\displaystyle\zeta_{r,s}=-\frac{2}{n+1}\sum_{k=1}^{n}\frac{\sin\frac{kr\pi}{n+1}\sin\frac{ks\pi}{n+1}}{2(1-\cos\frac{\pi
k}{n+1})}\ .$

The susceptibility can be computed using the change of variables
$z=$exp$(k\pi/2(n+1))$ and the formula for geometric series:
$1-z^{a+1}=(1-z)\sum_{i=0}^{a}z^{i}$. The result is:

$\zeta_{r,s}=-\frac{1}{n+1}((n+1)\text{min}(r,s)-rs)\ .$ (269)

For $\chi_{i,j}$ we get, independent of $j$:

$\chi_{i,j}=J(\zeta_{1,r_{i}(j)}-\zeta_{1,r_{i}(j)-1})=J/(n+1)\ .$ (270)

This concludes the calculation of $\chi$. Note that the response of the
variable $x_{ci,1}$ corresponding to the flux in the inductor connected to the
qubit is

$\langle
g_{b_{i}}|x_{ci,1}|g_{b_{i}}\rangle|_{b_{i}=1}=J\zeta_{1}=-\frac{Jn}{n+1}\ ,$
(271)

and the response in the last inductor is

$-\langle
g_{b_{i}}|x_{ci,n}|g_{b_{i}}\rangle|_{b_{i}=1}=-J\zeta_{n}=\frac{J}{n+1}\ .$
(272)

Flux cancellation is satisfied as expected of the definition of $x$: $\langle
g_{b_{i}}|x_{ci,1}|g_{b_{i}}\rangle|_{b_{i}=1}-\langle
g_{b_{i}}|x_{ci,n}|g_{b_{i}}\rangle|_{b_{i}=1}+\sum_{j=2}^{n}\chi_{i,j}=0$ The
fluxes in the unperturbed inductors spread evenly to cancel the perturbation-
induced flux in the first inductor.

###  G.3 Projected coupling

We will choose $i_{i,j}=|\chi_{i,j}|+i_{m}$ that upper bounds
$|\chi_{i,j}|+\|PI_{i,j}Q\|$. Below we will show that the bound on any
individual $\|Px_{ci,r}Q\|$ is:

$\|Px_{ci,r}Q\|\leq\sqrt{\sum_{k}{g_{k}^{r2}}}\leq\sqrt{\frac{1}{\pi}(1+\text{ln}n)+\left(\frac{\pi}{2}-1\right)\frac{n}{2\pi^{2}(n+1)}}\
.$ (273)

Recall that:

$\displaystyle\omega_{k}=2\sin\frac{k\pi}{2(n+1)}\ ,\quad
g^{n}_{k}=\frac{1}{\sqrt{\omega_{k}}}\sqrt{\frac{1}{n+1}}\sin\frac{kn\pi}{n+1}\
.$ (274)

Plugging that in:

$\displaystyle\sqrt{\sum_{k}|g_{k}^{r}|^{2}}=\sqrt{\frac{1}{n+1}\sum_{k}\frac{\sin^{2}kr\pi/(n+1)}{\omega_{k}}}\
,$ (275)

and:

$\sum_{k}g_{k}^{r2}=\frac{1}{2(n+1)}\sum_{k}\frac{\sin^{2}kr\pi/(n+1)}{\sin
k\pi/2(n+1)}\ .$ (276)

A simple way to upper bound this is to use $\sin^{2}x\leq 1$ in the numerator
and $\sin x\leq 2x/\pi$ in the denominator. This costs a total factor of $\pi$
in the leading term compared to the true scaling.

$\sum_{k}g_{k}^{r2}\leq\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k}\leq\frac{1}{2}(1+\text{ln}n)\
.$ (277)

A tighter bound with the correct leading term can be obtained by using $1/\sin
x\leq(1+(\frac{\pi}{2}-1)(\frac{2x}{\pi})^{2})/x$ for the denominator:

$\displaystyle\sum_{k}g_{k}^{r2}\leq\sum_{k=1}^{n}\frac{1}{\pi
k}(1+(\frac{\pi}{2}-1)(\frac{2}{\pi})^{2})(\frac{\pi
k}{2(n+1)})^{2})\leq\frac{1}{\pi}(1+\text{ln}n)+(\frac{\pi}{2}-1)(\frac{2}{\pi})^{2}\frac{n}{8(n+1)}\
.$ (278)

We will not use the above bound directly, as the difference can be bounded
better:

$\displaystyle\|P(x_{ci,r}-x_{ci,r-1})Q\|\leq\sqrt{\sum_{k}(g_{k}^{r}-g_{k}^{r-1})^{2}}\leq\sqrt{\frac{1}{n+1}\sum_{k}\frac{(\sin\frac{kr\pi}{n+1}-\sin\frac{k(r-1)\pi}{n+1})^{2}}{2\sin\frac{k\pi}{2(n+1)}}}\
.$ (279)

In the numerator, we get
$4\cos^{2}\frac{k(r-1/2)\pi}{n+1}\sin^{2}\frac{k\pi}{2(n+1)}$, so the $\sin$
cancels with the denominator. The range of $r,n$ is $n\geq 2,~{}2\leq r\leq
n$. We compute $n=2$ separately and use $n\geq 3$ to derive the bound for all
the other values. We will use the following:

$\sum_{k=1}^{n}\sin\frac{k\pi}{2(n+1)}=\frac{1}{2}\left(\frac{\cos\frac{\pi}{4(n+1)}}{\sin\frac{\pi}{4(n+1)}}-1\right)\
.$ (280)

Plugging that in:

$\displaystyle\|P(x_{ci,r}-x_{ci,r-1})Q\|\leq\sqrt{\frac{2}{n+1}\sum_{k}\cos^{2}\frac{k(r-1/2)\pi}{n+1}\sin\frac{k\pi}{2(n+1)}}\leq\sqrt{\frac{2}{n+1}\sum_{k}\sin\frac{k\pi}{2(n+1)}}$
(281)
$\displaystyle=\sqrt{\frac{1}{n+1}\left(\frac{\cos\frac{\pi}{4(n+1)}}{\sin\frac{\pi}{4(n+1)}}-1\right)}\leq\sqrt{\frac{1}{n+1}\frac{1}{\eta\frac{\pi}{4(n+1)}}}\leq\sqrt{\frac{4}{\eta\pi}}\leq
1.14\ ,$ (282)

where and $\eta=\frac{\sin\pi/16}{\pi/16}\approx 0.9936$ is a factor we can
use for $n\geq 3$. The bound $i_{i,j}$ is:

$\frac{J}{n+1}+\|P(x_{ci,r}-x_{ci,r-1})Q\|\leq\sqrt{2}=i_{i,j}\ ,$ (283)

which can be shown for all $n,r$ in the allowed range: for $n\geq 3$, $J<1$ we
can use $1/4+1.14\leq\sqrt{2}$, while for $n=2$ we just compute
$\sqrt{\frac{2}{n+1}\sum_{k}\sin\frac{k\pi}{2(n+1)}}=\sqrt{\frac{1}{3}(\frac{1}{2}+\frac{\sqrt{3}}{2})}<0.7$
and $0.7+\frac{1}{3}\leq\sqrt{2}$. which concludes the calculation of
$i_{i,j}$.

###  G.4 Overlap

Using the Hamiltonian (Eq. (253)) of individual transmission line written in
terms of $p_{k},x_{k}$, as well as $\omega_{k},g_{k}^{1}$:

$H_{m,i}=\sum_{k}^{n}p_{k}^{2}+\omega_{k}^{2}x_{k}^{2}+JZ_{i}x_{k}\sqrt{2\omega_{k}}g_{k}^{1}\
,$ (284)

the overlap is defined as:

$F=\langle g_{b_{i}=1}|g_{b_{i}=-1}\rangle=\prod F_{k}\ ,$ (285)

where $|g_{b_{i}}\rangle$ is the ground state of transmission line $H_{m,i}$
with $Z_{i}=b_{i}$ treated as a number. $F_{k}$ is defined as the overlap of
$Z_{i}=\pm 1$ ground states of each individual term in the sum over $k$:

$\displaystyle H_{k\pm}=p_{k}^{2}+\omega_{k}^{2}x_{k}^{2}\pm
J\sqrt{2\omega_{k}}g_{k}^{1}x_{k}=p_{k}^{2}+\omega_{k}^{2}\left(x_{k}\pm\frac{Jg_{k}^{1}}{\omega_{k}^{3/2}\sqrt{2}}\right)^{2}+\text{const}\
.$ (286)

The ground state wavefunctions are:

$\psi_{k\pm}(x)=\left(\frac{\omega_{k}}{\pi}\right)^{\frac{1}{4}}\text{exp}(-\omega_{k}(x\pm
x_{0k})^{2}/2)=\frac{\text{exp}(-\omega_{k}(x\pm
x_{0k})^{2}/2)}{\left(\int_{-\infty}^{\infty}dx\text{exp}(-\omega_{k}x^{2})\right)^{\frac{1}{2}}}\
,$ (287)

where $x_{0k}=\frac{Jg_{k}^{1}}{\sqrt{2}\omega_{k}^{3/2}}$. The overlap is
then:

$F_{k}=\frac{\int e^{-\omega_{k}((x-x_{0k})^{2}+(x+x_{0k})^{2})/2}dx}{\int
e^{-\omega_{k}x^{2}}dx}=e^{-\omega_{k}x_{0k}^{2}}=\text{exp}-\frac{J^{2}|g_{k}^{1}|^{2}}{2\omega_{k}^{2}}\
,$ (288)

and the total overlap:

$F=\text{exp}-\frac{J^{2}}{2}\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}\
.$ (289)

The sum that appears in the overlap:

$\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}=\frac{1}{8(n+1)}\sum_{k=1}^{n}\frac{\sin^{2}\frac{k\pi}{n+1}}{\sin^{3}\frac{k\pi}{2(n+1)}}=\frac{1}{2(n+1)}\sum_{k=1}^{n}\frac{\cos^{2}\frac{k\pi}{2(n+1)}}{\sin\frac{k\pi}{2(n+1)}}=\frac{1}{2(n+1)}\sum_{k=1}^{n}\frac{1}{\sin\frac{k\pi}{2(n+1)}}-\sin\frac{k\pi}{2(n+1)}\
.$ (290)

We have already used the second sum in the previous section:

$\frac{2(n+1)\cos\frac{\pi}{16}}{\pi}-1/2\leq\sum_{k=1}^{n}\sin\frac{k\pi}{2(n+1)}=\frac{1}{2}\left(\frac{\cos\frac{\pi}{4(n+1)}}{\sin\frac{\pi}{4(n+1)}}-1\right)\leq\frac{2(n+1)}{\eta\pi}\
,$ (291)

again for $n\geq 3$ and $\eta=\frac{\sin\pi/16}{\pi/16}\approx 0.9936$. $n=2$
case can be checked independently. The remaining sum can be bounded as
follows:

$(n+1)\frac{2}{\pi}\text{ln}n\leq\sum_{k=1}^{n}\frac{1}{\frac{k\pi}{2(n+1)}}\leq\sum_{k=1}^{n}\frac{1}{\sin\frac{k\pi}{2(n+1)}}\leq\sum_{k=1}^{n}\frac{1}{\frac{2}{\pi}\frac{k\pi}{2(n+1)}}\leq(n+1)(1+\text{ln}n)\
.$ (292)

The bound on $\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}$ is:

$\frac{1}{\pi}\text{ln}n-\frac{1}{\eta\pi}\leq\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}\leq\frac{1}{2}(1+\text{ln}n)-\frac{\cos\frac{\pi}{16}}{\pi}+\frac{1}{4(n+1)}\
.$ (293)

For $n=2$ we find $\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}\approx
0.298$ and the above is satisfied as $-0.1\leq 0.298\leq 0.62$. We can
slightly weaken the bounds to simplify the expressions:

$\frac{1}{\pi}\text{ln}n-\frac{1}{3}\leq\sum_{k}\frac{|g_{k}^{1}|^{2}}{\omega_{k}^{2}}\leq\frac{1}{2}\text{ln}n+\frac{1}{4}\
,$ (294)

where we confirmed the inequality directly for $n=2,3$, and it is weaker than
the previous version for $n\geq 4$.

For the theorem, we will need the inverse overlap $F^{-1}$. Its exact value
is:

$F^{-1}=\text{exp}\frac{J^{2}}{4(n+1)}\sum_{k=1}^{n}\frac{1}{\sin\frac{k\pi}{2(n+1)}}-\sin\frac{k\pi}{2(n+1)}\
,$ (295)

and the bounds are:

$n^{\frac{J^{2}}{2\pi}}e^{-\frac{J^{2}}{6}}\leq F^{-1}\leq
n^{\frac{J^{2}}{4}}e^{\frac{J^{2}}{8}}\ .$ (296)

Since $J\leq 1$, the power of $n$ is at most $\frac{1}{4}$. This concludes the
calculation of the overlap.

##  Appendix H 40 qubit example, full calculation

Figure 5: Left: the specific instance of $n=40,~{}s=4$ random graph that we
study. The colors represent a partition into 20 qubit subsystems that has the
least interactions between subsystems. Right: An embedding of this problem
into D-Wave Pegasus architecture. White links correspond to the interactions
between qubits and their mediators. Blue links encode the problem
interactions.

We would like to repeat the calculation one more time for a specific instance
of $n=40$, degree $2s=4$ that requires a minimum of $16$ interactions to be
cut to split the graph into two $20$ qubit partitions. The problem graph and
the corresponding gadget with single qubit couplers as mediators are
illustrated in Fig. 5. Since the allowed Hamiltonian error is $\epsilon ns$,
neglecting the $16$ interactions of strength $\leq 1$ between subsystems would
correspond to $\epsilon=0.1$ of error. We take this value as the target
precision for our gadget and check what quantum hardware capabilities are
needed for the operation at that target precision. We assume that a Pegasus
graph is used [20]. An embedding (see Fig. 5, right) exists where $18$ pairs
of hardware qubits are used to represent $18$ logical qubits, and the
remaining $22$ are represented by a single hardware qubit. One of each of
those pairs will be the coupler in our construction. We would also need to
count the required qubit-coupler and coupler-coupler interactions (the qubit
coupler is different in that it has $f_{ij}\sim 1/\chi$ instead of
$1/\chi^{2}$). We choose the coupler out of the pair to minimize the number of
those interactions and maximize the direct ones. In fact, it is possible to
choose the couplers so that no coupler-coupler interactions are needed. The
minimal number of qubit-coupler interactions is $31$. We also note that the
degree of the Pegasus graph is $15$, and each interaction has some control
error, even if we are not using them. We will assume that the effect of the
unused interactions is incorporated into the local field error.

We note that classically one can neglect the 16 interactions and simulate each
20 qubit subsystem via an exact diagonalization. There is, however, no out-of-
the-box algorithm for $\epsilon<0.1$. Of course, the classical state of the
art is much better than that, and it is not our goal to review it here. We
still consider beating the out-of-the-box classical algorithm an important
milestone towards being competitive with the state of the art in the future.

The Hamiltonian of the $n=40,~{}2s=4$ gadget for this graph is as follows.
Some qubits of the gadget are coupled to a qubit coupler. The bare Hamiltonian
is just the qubit couplers and their couplings to the qubit:

$H_{0}=\sum_{i\in vq}JZ_{q,i}Z_{qc,i}+\sqrt{1-J^{2}}X_{qc,i},\quad\Delta=1\ .$
(297)

Here $vq$ is the set of indices of qubits with a coupler. The qubit couplers
are extended objects that are coupled to one or a few other qubits directly.
Let’s denote the set of these couplings $v$:

$V_{c}=\sum_{i,j\in v}f_{ij}Z_{qc,i}Z_{q,j}\ .$ (298)

Each qubit of the simulator has a local field, and some of the qubits are
coupled directly. Let’s denote the set of direct couplings $vd$:

$V_{q}=\sum_{i}h_{i}^{c}Z_{q,i}+t_{i}^{c}X_{q,i}+\sum_{i,j\in
vd}J_{ij}^{c}Z_{q,i}Z_{q,j}\ .$ (299)

Note that $vq$ is the set of the unique first elements of $v$. The
perturbation also includes the noise of strength $\delta$ in the bare
Hamiltonian. Note that the terms that are part of the perturbation are also
considered noisy with the same strength $\delta$, but we assume that their
noisy values don’t exceed the range of their ideal values. The total
perturbation is:

$V=V_{q}+V_{c}+\sum_{i\in
vq}\delta_{Zi}Z_{q,i}Z_{qc,i}+\delta_{Xi}X_{qc,i},\quad|\delta_{Zi}|,~{}|\delta_{Xi}|\leq\delta\
.$ (300)

Compare this with the general definition of precision used previously:

$\displaystyle\|P\delta H_{m}\|\leq\delta_{H},~{}\|P\delta
I_{m}\|\leq\delta_{I},\quad\delta H_{m}=\delta_{X}\cdot
X_{qc},~{}\delta_{I}=\delta_{Z}\cdot Z_{qc}\ .$ (301)

Note that for any unitary $U$ and any projector $P$ the norm $\|PU\|=1$. In
particular, $\|PX_{q}\|=\|PZ_{q}\|=\|PZ_{qc}Z_{q}\|=\|PX_{qc}\|=1.$ Thus
$\delta_{H},\delta_{J}=\delta$ for this system. The properties of the qubit
coupler have been found before in Eq. (192):

$\chi_{i,j}=J,~{}F=\sqrt{1-J^{2}},~{}i_{i,j}=1\ .$ (302)

The required $f_{ij}=\alpha J_{i,j}/\chi_{i,j},~{}J_{ij}^{c}=\alpha
J_{ij},~{}h_{i}^{c}=\alpha h_{i},~{}t_{i}^{c}=\alpha F^{-1}t_{i}$ for qubits
with a coupler or $\alpha t_{i}$ for qubits without a coupler. The proof of
the general theorem allows for a minor modification: instead of every qubit
coupling to $s$ others, we now have $|vq|$ qubits out of the total $n$
coupling to $s=1$ others through the mediator, and $|vd|$ direct couplings are
included in the perturbation.

The bound on the error $r=H_{\text{targ}}-H_{\text{eff}}$ is given by:

$\displaystyle\|r\|\leq$
$\displaystyle\delta\left(\sum_{i}\|PZ_{i}P\|+\|PX_{i}P\|+\sum_{i\in
vq}\|PZ_{q,i}Z_{qc,i}P\|+\|PX_{qc,i}P\|+\right.$ (303)
$\displaystyle\left.+\sum_{i,j\in vd}\|PZ_{q,i}Z_{q,j}P\|+\sum_{ij\in
vq}\|PZ_{qc,i}Z_{q,j}P\|\right)+2c\frac{\|PV\|^{2}}{\Delta-(1+\Delta)\|V\|}\
.$ (304)

In this section, we use the value $c=3.5$ unless otherwise specified. It would
be convenient to renormalize $\frac{c}{\Delta-(1+\Delta)\|V\|}\leq c^{*}$. We
will try $c^{*}=1.01c$, and then check if the resulting
$\|V\|=2\delta|vq|+(n+|vq|F^{-1}+|vd|+|v|J^{-1}+sn-|v|)\alpha$ satisfies the
inequality.

Using

$\|PX_{q\in
vq}P\|=\sqrt{1-J^{2}}=F,\quad\|PZ_{qc}P\|=J,\quad\|PX_{qc}P\|=\sqrt{1-J^{2}},\quad\|PZ_{q}P\|=\|PX_{q\in
vd}P\|=1,\quad\|PZ_{qc}Z_{qc}P\|=J\ .$

The first term is just:

$(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta\ .$ (305)

The $\|PV\|$ contains:

$\displaystyle\|PV\|\leq\delta\sum_{i\in
vq}(\|PZ_{q,i}Z_{qc,i}\|+\|PX_{qc,i}\|)+\sum_{i}|h_{i}|+|t_{i}|+\sum_{i,j\in
vd}\|PJ_{i,j}Z_{q,i}Z_{q,j}\|+\sum_{ij\in vq}\|Pf_{ij}Z_{qc,i}Z_{q,j}\|$ (306)
$\displaystyle\leq
2\delta|vq|+\alpha(n+|vq|F^{-1}+|vd|+\frac{|v|}{J}+(sn-|v|))\ .$ (307)

Plugging the expressions above into Eq. (304), we get a quadratic inequality
on $\alpha$:

$\displaystyle(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta+2c^{*}(2\delta|vq|+\alpha(n+|vq|F^{-1}+|vd|+\frac{|v|}{J}+(sn-|v|)))^{2}\leq\alpha\epsilon
ns\ .$ (308)

We arrive at the same form as in the proof:

$nG_{1}(\alpha+G_{2})^{2}+G_{4}-\alpha s\epsilon\leq 0\ ,$ (309)

where now:

$\displaystyle
G_{1}=2c^{*}(n+|vq|F^{-1}+|vd|+\frac{|v|}{J}+(sn-|v|))^{2}/n^{2}$ (310)
$\displaystyle
G_{2}=\frac{2\delta|vq|}{(n+|vq|F^{-1}+|vd|+\frac{|v|}{J}+(sn-|v|))}$ (311)
$\displaystyle
G_{4}=(J+F)\delta\frac{|vq|}{n}+(1+\frac{|vq|F+|vd|+|v|J+(ns-|v|)}{n})\delta\
.$ (312)

For the $n=40,~{}2s=4$ problem we study here the values
$|vq|=18,~{}|vd|=n-|vq|=22,~{}|v|=31$.

We use the full expression from the proof of the theorem:

$s\epsilon=\text{min}2(nG_{1}G_{2}+\sqrt{nG_{1}G_{4}+n^{2}G_{1}^{2}G_{2}^{2}})\
.$ (313)

For $\epsilon=0.1$, this is satisfied for $J=0.76,~{}\delta=4.1\cdot 10^{-7}$.
We also check that for this value of $\delta$ the chosen $c^{*}=1.01c$
satisfies $\frac{c}{\Delta-(1+\Delta)\|V\|}=1.006c\leq c^{*}$, and that
$x=2\|PV\|/(\Delta-(1+\Delta)\|V\|)=0.0064<1/16$.

Now we will show that a bigger precision is also allowed, using the version of
the theorem that contains $\|PVQ\|(\|V-QVQ\|)$ instead of $2\|PV\|$. For that,
we need to calculate $\|V-QVQ\|$ and $\|PVQ\|$. For each system of a qubit and
its qubit coupler, $P$ and $Q$ are given by

$P=P_{0}\rho_{0}+P_{1}\rho_{1},\quad
Q=P_{0}q_{0}+P_{1}q_{1},\quad\rho_{0,1}=\frac{1}{2}(1-\sqrt{1-J^{2}}X_{qc}\pm
JZ_{qc}),\quad q_{0,1}=1-\rho_{0,1}=\frac{1}{2}(1+\sqrt{1-J^{2}}X_{qc}\mp
JZ_{qc})\ .$

The terms in $V$ are $X_{q},Z_{q},X_{qc},Z_{qc}Z_{q},Z_{q}Z_{q}$, and the
norms of the required projections (obtained in Mathematica [19]) are:

$\displaystyle\|PX_{q\in
vq}Q\|=\|\rho_{0}q_{1}\|=J,\quad\|X_{q}-QX_{q}Q\|=a_{J}=\sqrt{\frac{1}{2}(1+J^{2}+\sqrt{1+2J^{2}-3J^{4}})}$
(314) $\displaystyle\|PZ_{q}Q\|=0,\quad\|Z_{q}-QZ_{q}Q\|=1$ (315)
$\displaystyle\|PX_{qc}Q\|=J,\quad\|X_{qc}-QX_{qc}Q\|=b_{J}=\frac{1}{2}(\sqrt{1-J^{2}}+\sqrt{1+3J^{2}})$
(316) $\displaystyle q,qc\in v:\quad$
$\displaystyle\|PZ_{qc}Z_{q}Q\|=(1-J^{2})^{\frac{1}{4}},\quad\|Z_{qc}Z_{q}-QZ_{qc}Z_{q}Q\|=c_{J}=\frac{1}{2}(J+\sqrt{4-3J^{2}})\
.$ (317)

With these, the norms $\|PVQ\|$ and $\|V-QVQ\|$ take the form:

$\displaystyle\|PVQ\|\leq\delta|vq|((1-J^{2})^{\frac{1}{4}}+J)+\alpha|vq|\frac{J}{\sqrt{1-J^{2}}}$
(318)
$\displaystyle\|V-QVQ\|\leq\delta|vq|(c_{J}+b_{J})+\alpha(n+|vq|\frac{a_{J}}{\sqrt{1-J^{2}}}+|vd|+\frac{|v|}{J}+(sn-|v|))\
.$ (319)

The inequality on $\alpha$ is:

$\displaystyle(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta+$
$\displaystyle
c^{*}(\delta|vq|((1-J^{2})^{\frac{1}{4}}+J)+\alpha|vq|\frac{J}{\sqrt{1-J^{2}}})(\delta|vq|(c_{J}+b_{J})+\alpha(n+|vq|\frac{a_{J}}{\sqrt{1-J^{2}}}+|vd|+\frac{|v|}{J}+(sn-|v|)))\leq\alpha\epsilon
ns\ ,$

where we now used the exact adjustment of the gap $c^{*}=c/(1-2\|V\|)$. We
will collect it into the form:

$G_{1}(\alpha+G_{2})(\alpha+G_{3})+G_{4}-\alpha sn\epsilon\leq 0\ ,$ (320)

where

$\displaystyle
G_{1}=c^{*}|vq|\frac{J}{\sqrt{1-J^{2}}}(n+|vq|\frac{a_{J}}{\sqrt{1-J^{2}}}+|vd|+\frac{|v|}{J}+(sn-|v|))$
(321) $\displaystyle
G_{2}=\delta|vq|((1-J^{2})^{\frac{1}{4}}+J)\frac{\sqrt{1-J^{2}}}{|vq|J}$ (322)
$\displaystyle
G_{3}=\frac{\delta|vq|(c_{J}+b_{J})}{n+|vq|\frac{a_{J}}{\sqrt{1-J^{2}}}+|vd|+\frac{|v|}{J}+(sn-|v|)}$
(323) $\displaystyle G_{4}=(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta\
.$ (324)

The quadratic polynomial in $\alpha$ has coefficients:

$G_{1}\alpha^{2}+(G_{1}(G_{2}+G_{3})-sn\epsilon)\alpha+G_{4}+G_{1}G_{2}G_{3}\leq
0\ .$ (325)

Searching for the zero of:

$(G_{1}(G_{2}+G_{3})-sn\epsilon)^{2}-4G_{1}G_{4}-4G_{1}^{2}G_{2}G_{3}=0\ ,$
(326)

unfortunately, leads to $x=\|V-QvQ\|/(\Delta-(1+\Delta)\|V\|)>1/16$, so
instead we set $x=1/16$ and seek the numbers $J,\delta,\alpha$ that satisfy
both $x=1/16$ and
$G_{1}\alpha^{2}+(G_{1}(G_{2}+G_{3})-sn\epsilon)\alpha+G_{4}+G_{1}G_{2}G_{3}=0$.
The largest $\delta$ we were able to find numerically is $0.9\cdot 10^{-5}$,
with $\alpha=0.000256,~{}J=0.37$.

Let’s try to repeat the calculation using the finite-dimensional Lemma.

The inequality on $\alpha$ now contains
$\|V\|=2\delta|vq|+\alpha(n+|vq|F^{-1}+|vd|+|v|J^{-1}+sn-|v|)$ as follows:

$\displaystyle(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta+$
$\displaystyle
c(\delta|vq|((1-J^{2})^{\frac{1}{4}}+J)+\alpha|vq|\frac{J}{\sqrt{1-J^{2}}})(2\delta|vq|+\alpha(n+|vq|F^{-1}+|vd|+|v|J^{-1}+sn-|v|))\leq\alpha\epsilon
ns\ .$

We will collect it into the form:

$G_{1}(\alpha+G_{2})(\alpha+G_{3})+G_{4}-\alpha sn\epsilon\leq 0\ ,$ (327)

where

$\displaystyle
G_{1}=c|vq|\frac{J}{\sqrt{1-J^{2}}}(n+|vq|F^{-1}+|vd|+|v|J^{-1}+sn-|v|)$ (328)
$\displaystyle
G_{2}=\delta|vq|((1-J^{2})^{\frac{1}{4}}+J)\frac{\sqrt{1-J^{2}}}{|vq|J}$ (329)
$\displaystyle G_{3}=\frac{2\delta|vq|}{n+|vq|F^{-1}+|vd|+|v|J^{-1}+sn-|v|}$
(330) $\displaystyle G_{4}=(J+F)\delta|vq|+(n+|vq|F+|vd|+|v|J+(ns-|v|))\delta\
.$ (331)

The solution is again on the boundary given by

$\|V\|=\frac{1}{16},\quad
G_{1}\alpha^{2}+(G_{1}(G_{2}+G_{3})-sn\epsilon)\alpha+G_{4}+G_{1}G_{2}G_{3}=0\
.$ (332)

We numerically find that the control precision $\delta=1.1\cdot 10^{-5}$ is
allowed at $J=0.42$. The corresponding $\alpha=0.000302$.

This noise level is not unrealistic when compared to the reported control
noise [18] that is between $10^{-2}$ and $10^{-3}$ depending on the specific
point in the parameter space. Recall that our rigorous bounds are not tight
and a more noisy simulator can still output accurate answers, just without
theoretical guarantees.