$\displaystyle=\rho_{\lambda}+(1-\mathrm{Tr}(\rho_{\lambda}))|{\bf
0}_{\lambda}\rangle\langle{\bf 0}_{\lambda}|$ $\displaystyle=\rho_{\lambda}.$

Now

$S^{r}_{n}(\mathcal{N}_{r-1}(u,V_{\mu})\otimes\phi^{\mathfrak{z},r})=\bigoplus_{\lambda\in\Lambda_{1}}[\sigma^{r,n}\mathcal{N}_{r-1}(u,V_{\mu})](\lambda)\tilde{S}^{r}_{\lambda}(\phi^{\mathfrak{z},r})\otimes\frac{{\bf
1}_{\mathcal{K}_{\lambda}}}{\dim(\mathcal{K}_{\lambda})}.$

and with the notation
$\sigma^{r,n}\mathcal{N}^{u}_{\lambda}:=[\sigma^{r,n}\mathcal{N}_{r-1}(u,V_{\mu}))](\lambda)$
and
$q^{u,r,n}_{\lambda}:=\min(\sigma^{r,n}\mathcal{N}^{u}_{\lambda},p^{u,r,n}_{\lambda})$
we have

$\displaystyle
S_{n}(\phi^{\mathfrak{z},r}\otimes\mathcal{N}_{r-1}(u,V_{\mu}))-\rho^{\theta,r,n}$
$\displaystyle\quad=\bigoplus_{\lambda\in\Lambda_{1}}\left\\{q^{u,r,n}_{\lambda}(\tilde{S}^{r}_{\lambda}(\phi^{\mathfrak{z},r})-\rho^{\theta,n}_{\lambda})+(\sigma^{r,n}\mathcal{N}^{u}_{\lambda}-q^{u,r,n}_{\lambda})\tilde{S}^{r}_{\lambda}(\phi^{\mathfrak{z},r})-(p^{u,r,n}_{\lambda}-q^{u,r,n}_{\lambda})\rho^{\theta,r,n}_{\lambda}\right\\}\otimes\frac{{\bf
1}_{\mathcal{K}_{\lambda}}}{\dim(\mathcal{K}_{\lambda})}.$

Taking $L^{1}$ norms, and using that all $\phi$’s and $\rho$’s have trace $1$
and that channels (such as $\tilde{S}^{r}_{\lambda}$) are trace preserving, we
get the bound:

$\displaystyle\left\|S^{r}_{n}(\phi^{\mathfrak{z},r}\otimes\mathcal{N}_{r-1}(u,V_{\mu}))-\rho^{\theta,r,n}\right\|_{1}$
$\displaystyle\leq$
$\displaystyle\sum_{\lambda\in\Lambda_{1}}\left\|q^{u,r,n}_{\lambda}(\tilde{S}^{r}_{\lambda}(\phi^{\mathfrak{z},r})-\rho^{\theta,r,n}_{\lambda})\right\|_{1}+\sum_{\lambda\in\Lambda_{1}}\left|\sigma^{r,n}\mathcal{N}^{u}_{\lambda}-p^{u,r,n}_{\lambda}\right|$
$\displaystyle\leq$ $\displaystyle
2\sum_{\lambda\in\Lambda_{1}\cap\Lambda^{c}_{2}}q^{u,r,n}_{\lambda}+\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}\left\|\tilde{S}^{r}_{\lambda}(\phi^{\mathfrak{z},r})-\rho^{\theta,n}_{\lambda}\right\|_{1}+\left\|\sigma^{r,n}\mathcal{N}_{r-1}(u,V_{\mu})-p^{u,r,n}\right\|_{1}$
$\displaystyle\leq$ $\displaystyle
2\sum_{\lambda\in\Lambda_{1}\cap\Lambda^{c}_{2}}p^{u,r,n}_{\lambda}+\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}\left\|\phi^{\mathfrak{z},r}-T^{r}_{\lambda}(\rho^{\theta,n}_{\lambda})\right\|_{1}+\left\|\sigma^{n}\mathcal{N}_{r-1}(u,V_{\mu})-p^{u,r,n}\right\|_{1}.$

Now the first term is $O(n^{r^{2}}\exp(-n^{2\alpha-1}/2))$ by Lemma C.2 and
the third term is treated in the relation (82). For the second term one uses
Lemma C.3 through C.5 as in the proof of Lemma C.6. Thus we have the same
bound for the reverse channel i.e.

$\sup_{\theta\in\Theta_{n,r,\beta,\gamma}}||S^{r}_{n}(\Phi^{\theta,r})-\rho^{\theta,r,n}||_{1}=O(n^{-\kappa})$
(83)

∎

###### Proof of Theorem 5.1.

In the first step we use a $2$-design to construct a POVM which helps in
constructing the least squares estimator. A $2$-design is a finite set of rank
$1$ projectors $\\{|v_{i}\rangle\langle v_{i}|\\}_{i=1}^{m}$ which satisfies

$\frac{1}{m}\sum_{i=1}^{m}(|v_{i}\rangle\langle v_{i}|)^{\otimes
2}=\binom{d+1}{2}^{-1}P_{sym(2)}$

where $P_{sym(2)}$ is the projection operator onto the symmetric subspace of
$(\mathbb{C}^{d})^{\otimes 2}$. It can be easily shown by taking partial trace
that $\\{\frac{d}{m}|v_{i}\rangle\langle v_{i}|\\}_{i=1}^{m}$ forms a POVM.
When the state $\rho$ is measured using this POVM, we obtain outcomes from the
set $\\{1,\ldots,m\\}$ with the probability of observing $i$ given by
$\frac{d}{m}\langle v_{i}|\rho|v_{i}\rangle$. Now repeat this measurement on
$n$ states to get a set of outcomes $\\{i_{1},\ldots,i_{n}\\}$. Denote $n_{i}$
to be the number of times $i$ is observed and $f_{i}=\frac{n_{i}}{n}$ \- the
relative frequency of observing $i$. The least squares estimator is defined as
follows

$\hat{L}_{n}=\operatorname*{arg\,min}_{\rho}\sum_{i=1}^{m}\left(f_{i}-\frac{d}{m}\langle
v_{i}|\rho|v_{i}\rangle\right)^{2}.$

Let $\mathcal{M}$ be a linear map from $\mathcal{H}(\mathbb{C}^{d})$ (the
space of Hermitian operators from $\mathbb{C}^{d}$ to $\mathbb{C}^{d}$)
defined as follows

$\displaystyle\mathcal{M}:$
$\displaystyle\mathcal{H}(\mathbb{C}^{d})\rightarrow\mathbb{R}^{m}$
$\displaystyle\mathcal{M}(X)$ $\displaystyle=\frac{d}{m}\\{\langle
v_{i}|X|v_{i}\rangle\\}_{i=1}^{m},\quad X\in\mathcal{H}(\mathbb{C}^{d})$

The adjoint of $\mathcal{M}$ can be easily defined as follows

$\displaystyle\mathcal{M}^{*}:$
$\displaystyle\mathbb{R}^{m}\rightarrow\mathcal{H}(\mathbb{C}^{d})$
$\displaystyle\mathcal{M}^{*}(u)$
$\displaystyle=\frac{d}{m}\sum_{i=1}^{m}u_{i}|v_{i}\rangle\langle v_{i}|,\quad
u\in\mathbb{R}^{m}$

Then the least squares problem can restated as

$\hat{L}_{n}=\operatorname*{arg\,min}_{\rho}||f-\mathcal{M}(\rho)||^{2}$

where $f=\\{f_{i}\\}_{i=1}^{m}$. The closed form of the solution is well known
and is given by
$\hat{L}_{n}=(\mathcal{M}^{*}\mathcal{M})^{-1}\mathcal{M}^{*}(f)$. Explicit
computation yields (see [27] for details)

$\hat{L}_{n}=(d+1)\sum_{i=1}^{m}f_{i}|v_{i}\rangle\langle v_{i}|-I.$

One uses matrix concentration inequalities to show that $\hat{L}_{n}$
concentrates around $\rho$ with high probability. Equation (6) in [27] gives
the following rate of concentration

$P[||\hat{L}_{n}-\rho||\geq\epsilon]\leq de^{-\frac{3n\epsilon^{2}}{16d}}$

where $||.||$ denotes the operator norm. However, $\hat{L}_{n}$ may not be a
density matrix although it has trace 1. One projects $\hat{L}_{n}$ into the
space of density matrices to obtain what is called a projected least squares
estimator. However, as mentioned in the beginning we will take the route of
[12] to project into the space of spectrally thresholded density matrices
which would estimate the rank correctly. We define our estimator (called
physical estimator in [12]) as follows

$\tilde{\rho}_{n}=\operatorname*{arg\,min}_{\sigma\in
S(\epsilon)}||\sigma-\hat{L}_{n}||_{2}^{2}$

where $S(\epsilon)$ denotes the density matrices spectrally thresholded at
$2\epsilon$ i.e.

$S(\epsilon)=\\{\sigma:\sigma\text{ is a density matrix with eigenvalues
}\lambda_{j}\in\\{0\\}\cup(2\epsilon,1],j=1,2,\ldots,d\\}$

We show that $\tilde{\rho}_{n}$ concentrates around $\rho$ and also satisfies
the rank consistency condition.

The proof follows in the same way as [12] but we include it for the sake of
completion.

Let $\hat{\lambda}_{1}\geq\hat{\lambda_{2}}\geq\ldots\geq\hat{\lambda}_{d}$
denote the eigenvalues of $\hat{L}_{n}$. Let $k=0$ and define
$\tilde{\lambda}_{j}^{(0)}=\hat{\lambda}_{j}$ for $j=1,2\ldots,d$

for $k=1,2,\ldots,d$ do {

if $\tilde{\lambda}_{d-k+1}^{(k-1)}>2\epsilon$ then STOP

else

$\tilde{\lambda}_{d-k+1}^{(k)}=0$

$\tilde{\lambda}_{j}^{(k)}=\tilde{\lambda}_{j}^{(k-1)}+\frac{\tilde{\lambda}_{d-k+1}^{(k-1)}}{d-k}$
for $j=1,2,\ldots,d-k$

end if

end for

Note that if the algorithm stops at $k=\hat{k}$, then $\hat{r}=d-\hat{k}+1$ is
the rank of the estimator $\tilde{\rho}_{n}$ and the eigenvalues are given by
$\tilde{\lambda}_{j}=0$ for $j>\hat{r}$ and for $j\leq\hat{r}$

$\tilde{\lambda}_{j}=\hat{\lambda}_{j}+\frac{1}{\hat{r}}\sum_{j>\hat{r}}\hat{\lambda}_{j}.$

In particular if $\hat{L}_{n}=\hat{U}_{n}\hat{D}_{n}\hat{U}^{*}_{n}$, then
$\tilde{\rho}_{n}=\hat{U}_{n}\tilde{D}_{n}\hat{U}^{*}_{n}$ where

$\tilde{D}_{n}=diag(\tilde{\lambda}_{1},\ldots,\tilde{\lambda}_{\hat{r}},0,\ldots,0).$

Henceforth, we condition on the event $||\hat{L}_{n}-\rho||\leq\epsilon$ which
occurs with probability greater than $1-de^{-\frac{3n\epsilon^{2}}{16d}}$. By
Weyl perturbation inequality (see [9]),
$|\lambda_{j}-\hat{\lambda}_{j}|\leq\epsilon$ where
$\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{r}>6\epsilon$ (by
assumption) and $\lambda_{r+1}=\ldots\lambda_{d}=0$. Now if $\hat{r}>r$, then
for $j\leq\hat{r}$,

$\displaystyle|\tilde{\lambda}_{j}-\lambda_{j}|$
$\displaystyle=|\hat{\lambda}_{j}+\frac{1}{\hat{r}}\sum_{j>\hat{r}}\hat{\lambda}_{j}-\lambda_{j}|$
$\displaystyle\leq|\hat{\lambda}_{j}-\lambda_{j}|+\frac{1}{\hat{r}}|1-\sum_{j\leq\hat{r}}\hat{\lambda}_{j}|$
$\displaystyle\leq|\hat{\lambda}_{j}-\lambda_{j}|+\frac{1}{\hat{r}}\sum_{j\leq\hat{r}}|\lambda_{j}-\hat{\lambda}_{j}|$
$\displaystyle\leq 2\epsilon$

Since $\lambda_{\hat{r}}=0$ when $\hat{r}>r$, the above inequality implies
$\tilde{\lambda}_{\hat{r}}\leq\lambda_{\hat{r}}+2\epsilon=2\epsilon$, which is
a contradiction since $\hat{r}$ was chosen so that
$\tilde{\lambda}_{\hat{r}}>2\epsilon$. Now consider the case $\hat{r}<r$, so
that $\tilde{\lambda}_{r}=0$

$0=\tilde{\lambda}_{r}\geq\lambda_{r}-|\lambda_{r}-\tilde{\lambda}_{r}|\geq\lambda_{r}-|\lambda_{r}-\hat{\lambda}_{r}|-|\hat{\lambda}_{r}-\tilde{\lambda}_{r}|=\lambda_{r}-|\lambda_{r}-\hat{\lambda}_{r}|-|\hat{\lambda}_{r}|$
(84)

Now consider the step $k$ of the algorithm with $k=d-r+1$. Since
$\tilde{\lambda}_{r}=\tilde{\lambda}^{(d-r+1)}_{r}=0$,
$\tilde{\lambda}^{(d-r)}_{r}<2\epsilon$. A careful analysis of the algorithm
shows that at after step $d-r$, the sum of the last $d-r$ eigenvalues is
distributed equally among the first $r$ eigenvalues. Thus

$\tilde{\lambda}^{(d-r)}_{r}=\hat{\lambda}_{r}+\frac{1}{r}(1-\sum_{i=1}^{r}\hat{\lambda}_{r})=\hat{\lambda}_{r}+\frac{1}{r}\sum_{i=1}^{r}(\lambda_{r}-\hat{\lambda}_{r})<2\epsilon.$

This implies that
$\hat{\lambda}_{r}<2\epsilon+\frac{1}{r}\sum_{i=1}^{r}|\lambda_{r}-\hat{\lambda}_{r}|<3\epsilon$
.Since by the condition of the theorem $\lambda_{r}>6\epsilon$, (84) leads to

$0=\tilde{\lambda}_{r}>6\epsilon-\epsilon-3\epsilon=2\epsilon$

a contradiction. Thus with probability at least
$1-de^{-\frac{3n\epsilon^{2}}{16d}}$, $r=\hat{r}$.

To describe the concentration $\tilde{\rho}^{n}$, consider the following
diagonalizations of the states

$\rho=UDU^{*},\quad\hat{L}_{n}=\hat{U}_{n}\hat{D}_{n}\hat{U}_{n}^{*},\quad\tilde{\rho}^{n}=\hat{U}_{n}\tilde{D}_{n}\hat{U}_{n}^{*}$

and estimate the error of the projected estimator as follows:

$\displaystyle||\tilde{\rho}^{n}-\rho||_{2}$
$\displaystyle=||\hat{U}_{n}\tilde{D}_{n}\hat{U}_{n}^{*}-UDU^{*}||_{2}$
$\displaystyle\leq||\hat{U}_{n}\tilde{D}_{n}\hat{U}_{n}^{*}-\hat{U}_{n}D\hat{U}_{n}^{*}||_{2}+||\hat{U}_{n}D\hat{U}_{n}^{*}-UDU^{*}||_{2}$
$\displaystyle=||\tilde{D}_{n}-D||_{2}+||\hat{U}_{n}D\hat{U}_{n}^{*}-UDU^{*}||_{2}.$

For the first norm one has the following bound

$||\tilde{D}_{n}-D||_{2}=\sqrt{\sum_{k\leq\hat{r}}(\tilde{\lambda}_{j}-\lambda_{j})^{2}}\leq
2\epsilon\sqrt{r}.$

For the second norm first we give an upper bound in terms of the operator norm

$\displaystyle||\hat{U}_{n}D\hat{U}_{n}^{*}-UDU^{*}||$
$\displaystyle\leq||\hat{U}_{n}\hat{D}_{n}\hat{U}_{n}^{*}-\hat{U}_{n}D\hat{U}_{n}^{*}||+||\hat{U}_{n}\hat{D}_{n}\hat{U}_{n}^{*}-UDU||$
$\displaystyle=||\hat{D}_{n}-D||+||\hat{L}_{n}-\rho||\leq 2\epsilon.$

Both $\hat{U}_{n}D\hat{U}_{n}^{*}$ and $UDU^{*}$ are rank $r$ matrices, so
rank of $\hat{U}_{n}D\hat{U}_{n}^{*}-UDU^{*}$ is at most $2r$ and hence
$||\hat{U}_{n}D\hat{U}_{n}^{*}-UDU^{*}||_{2}\leq 2\epsilon\sqrt{2r}$. And thus
we obtain

$||\tilde{\rho}^{n}-\rho||_{2}^{2}\leq(2\sqrt{2}+2)^{2}r\epsilon^{2}\leq
25r\epsilon^{2}$

with probability at least $1-de^{-\frac{3n\epsilon^{2}}{16d}}$. ∎

###### Proof of Theorem 5.2.

Upper Bound

With $Z\sim\mathcal{N}_{r-1}(u,V_{\mu})$, as the estimator $\hat{u}$ the
classical part of the risk is

$R^{C}=E[\sum_{i=1}^{r-1}(u_{i}-Z_{i})^{2}+(\sum_{i=1}^{r-1}(u_{i}-Z_{i}))^{2}]=\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})$

where we have used the relation $\mu_{r}=1-\sum_{i=1}^{r-1}\mu_{i}$.

We start to compute the risk for the quantum part by first verifying that we
obtain normal random variables when we use the measurement $M$ on the shifted
thermal states. Denote $R_{t}=(-\infty,t_{1}]\times(-\infty,t_{2}]$. Let $X$
denote the 2 dimensional random variable generated when using measurement $M$
to estimate the state $\phi^{z_{0}}_{\beta}$. Then using the relation $P(X\in
R_{t})=\mathrm{Tr}(\phi^{z_{0}}_{\beta}M(R_{t}))$ we obtain,

$P(X\in
R_{t})=\mathrm{Tr}\left(\frac{e^{\beta}-1}{\pi}\int\exp{-(e^{\beta}-1)|z|^{2}})|z+z_{0}\rangle\langle
z+z_{0}|dz\frac{1}{2\pi}\int_{R_{t}}|\psi_{m}\rangle\langle\psi_{m}|dm\right).$

Changing from complex to two dimensional reals in the second integrals by
using $\mu=(\sqrt{2}Re(z),\\\ \sqrt{2}Im(z))$ and similarly obtaining
$\mu_{0}$ from $z_{0}$ we have

$\displaystyle P(X\in R_{t})$
$\displaystyle=\mathrm{Tr}\left(\frac{e^{\beta}-1}{2\pi}\int\exp(-(e^{\beta}-1)||\mu||^{2}/2)|\psi_{\mu+\mu_{0}}\rangle\langle\psi_{\mu+\mu_{0}}|d\mu\frac{1}{2\pi}\int_{R_{t}}|\psi_{m}\rangle\langle\psi_{m}|dm\right)$
$\displaystyle=\frac{e^{\beta}-1}{4\pi^{2}}\int_{R_{t}}\int\exp(-(e^{\beta}-1)||\mu||^{2}/2)|\langle\psi_{\mu+\mu_{0}}|\psi_{m}\rangle|^{2}d\mu
dm$
$\displaystyle=\frac{e^{\beta}-1}{4\pi^{2}}\int_{R_{t}}\int\exp(-(e^{\beta}-1)||\mu||^{2}/2)\exp(-\frac{1}{2}||\mu+\mu_{0}-m||^{2})d\mu
dm$
$\displaystyle=\frac{e^{\beta}-1}{4\pi^{2}}\int_{R_{t}}\int\exp(-\frac{1}{2}(e^{\beta}||\mu||^{2}-2\mu^{T}(m-\mu_{0})+||m-\mu_{0}||^{2}))d\mu
dm$
$\displaystyle=\frac{1-e^{-\beta}}{2\pi}\int_{R_{t}}\exp(-\frac{1}{2}(1-e^{-\beta})||m-\mu_{0}||^{2})dm.$

where the third equality is obtained by using the expansion (10). Thus the
generated random variable $X$ is a bivariate random variable with mean
$\mu_{0}$ and variance
$\frac{1}{1-e^{-\beta}}I_{2}=\frac{2\sigma^{2}_{\beta}+1}{2}I_{2}$ (since
$\sigma^{2}_{\beta}=\frac{\coth(\beta/2)}{2}$). A similar argument would yield
that $M_{c}$ (see Section 4) generates $cX$. As a special case when
$\beta=\infty$, $\phi_{\infty}=|0\rangle\langle 0|$ and thus the shifted
thermal state at infinite temperature is just a shifted pure state. In this
case the covariance matrix of $X$ is $I_{2}$.

Now consider the quantum part i.e.

$\phi^{\mathbf{\mathfrak{z}},r}=\bigotimes_{1\leq i<j\leq
r}\phi_{\beta_{ij}}^{\mathfrak{z}_{ij}}\otimes\bigotimes_{\begin{subarray}{c}1\leq
i\leq r\\\ r+1\leq j\leq d\end{subarray}}\phi_{\infty}^{\mathfrak{z}_{ij}}$

and estimation of the parameter $\\{\mathfrak{z}_{ij}\\}$ under the constraint
that $|\mathfrak{z}_{ij}|\leq n^{\beta}$. Changing from complex to two
dimensional reals (in this case write
$\xi_{ij}=(\sqrt{2}Re(\mathfrak{z}_{ij}),\sqrt{2}Im(\mathfrak{z}_{ij}))$) we
have the constraint set:

$\mathcal{Z}=\\{\\{\xi_{ij}\\}:|\xi_{ij}|\leq\sqrt{2}n^{\beta}\\}.$

Let $\kappa_{ij}=\mu_{i}-\mu_{j}$. Note that for the case $j\geq r+1$,
$\sigma^{2}_{\beta_{ij}}=1/2$ (since the state is pure or alternatively since
$\beta_{ij}=\ln(\mu_{i}/\mu_{j})$ and $\mu_{j}=0$ implies that
$\beta_{ij}=\infty$). Also reparametrize the the Gaussian state
$\phi^{\mathbf{\mathfrak{z}},r}$ as $\phi^{\mathbf{\xi},r}$. The risk for
estimation with the POVM $\bar{M}$ where
$\bar{M}=\bigotimes_{\begin{subarray}{c}{1\leq i\leq r}\\\ {i<j\leq
d}\end{subarray}}M$ is given by:

$\displaystyle R^{Q}_{\bar{M}}$
$\displaystyle=\int\sum_{\begin{subarray}{c}{1\leq i\leq r}\\\ {i<j\leq
d}\end{subarray}}\kappa_{ij}||\xi_{ij}-\hat{\xi}_{ij}||^{2}Tr(\phi^{\mathbf{\xi},r}\bar{M}(d\hat{\xi}))$
$\displaystyle=\sum_{\begin{subarray}{c}{1\leq i\leq r}\\\ {i<j\leq
d}\end{subarray}}\kappa_{ij}(2\sigma_{\beta_{ij}}^{2}+1).$

Thus when $\theta\in\Theta=\\{|u_{i}|\leq n^{\gamma}\text{ }\forall
i,|\mathfrak{z}_{kj}|\leq n^{\beta}\text{ }\forall 1\leq i\leq r,i<j\leq
d\\}$, the risk for the classical estimator $Z$ and the POVM $\bar{M}$ is
given by

$R_{Z,\bar{M}}(\theta)=\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq
d}\end{subarray}}\kappa_{ij}(2\sigma_{\beta_{ij}}^{2}+1).$ (85)

Note that
$2\sigma^{2}_{\beta_{ij}}+1=2(1-e^{-\beta_{ij}})^{-1}=2\mu_{i}/(\mu_{i}-\mu_{j})$
for $j\leq r$ and $2\sigma^{2}_{\beta_{ij}}+1=2$ for $j>r$. Plugging back the
expression for $\kappa_{ij}$ we obtain

$\limsup_{n\rightarrow\infty}\sup_{\theta\in\Theta}R_{Z,\bar{M}}(\theta)=\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}.$ (86)

since the risk in (85) is independent of $\theta$ and $n$.

Lower Bound

Consider the expression of the Bayes risk for estimating $\theta$ under the
following prior $\pi$

$\displaystyle u_{i}\sim$ $\displaystyle\mathcal{N}(0,n^{2\gamma_{0}})\text{
}\forall 1\leq i\leq r-1$ $\displaystyle\xi_{ij}\sim$
$\displaystyle\mathcal{N}_{2}(0,n^{2\beta_{0}}I_{2})\text{ }\forall 1\leq
i\leq r,i<j\leq d\ $

where the random variables are mutually independent and $\gamma_{0}<\gamma$
and $\beta_{0}<\beta$. Again we change the notation from complex to the real
i.e. we set
$\xi_{ij}=(\sqrt{2}Re(\mathfrak{z}_{ij}),\sqrt{2}Im(\mathfrak{z}_{ij}))$.
Since both the loss and the parameter space are separable into quantum and
classical parts we introduce the following notations

$\displaystyle\mathcal{L}(u,\hat{u})$
$\displaystyle=\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i})^{2}+(\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i}))^{2}$
$\displaystyle\mathcal{L}(\xi,\hat{\xi})$
$\displaystyle=\sum_{\begin{subarray}{c}{1\leq i\leq r}\\\ {i<j\leq
d}\end{subarray}}(\mu_{i}-\mu_{j})||\xi_{ij}-\hat{\xi}_{ij}||^{2}$
$\displaystyle\mathcal{U}$ $\displaystyle=\\{u:|u_{i}|\leq n^{\gamma}\text{
}\forall 1\leq i\leq r-1\\}$ $\displaystyle\mathcal{Z}$
$\displaystyle=\\{\xi:|\xi_{ij}|\leq n^{\beta}\text{ }\forall 1\leq i\leq
r,i<j\leq d\\}$

We denote $\pi_{1}$ as the prior on $u$ and $\pi_{2}$ the prior on $\xi$ and
by the structure of the prior we have
$\pi(d\theta)=\pi_{1}(du)\pi_{2}(d\xi)$.We give a lower bound to the maximum
risk by the average risk as follows:

$\displaystyle\sup_{\theta\in\Theta}R_{\hat{u},M}(\theta)$
$\displaystyle\geq\int_{\Theta}(R_{\hat{u}}^{C}(\theta)+R^{Q}_{M}(\theta))d\pi(\theta)$
$\displaystyle\geq\pi_{2}(\mathcal{Z})\inf_{\hat{u}\in\mathcal{U}}\int_{\mathcal{U}}E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)+\pi_{1}(\mathcal{U})\inf_{M:M(\mathcal{Z}^{c}=0)}\int_{\mathcal{Z}}E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)$
$\displaystyle=\pi_{2}(\mathcal{Z})\inf_{\hat{u}\in\mathcal{U}}(\int
E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)-\int_{\mathcal{U}^{c}}E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u))$
$\displaystyle+\pi_{1}(\mathcal{U})\inf_{M:M(\mathcal{Z}^{c}=0)}(\int
E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)-\int_{\mathcal{Z}^{c}}E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)).$
(87)

Since the parameters lie in $\mathcal{U}$ and $\mathcal{Z}$ respectively it is
enough to consider estimators supported on the parameter space themselves and
hence the second inequality follows.

Let $Y\sim\mathcal{N}_{d}(u,\Sigma_{1})$ and
$u\sim\mathcal{N}_{d}(0,\Sigma_{2})$, then $E[u|Y]$ is known to be the Bayes
estimator of $u$ under the loss $||u-\hat{u}||^{2}$. It can be easily shown
that $E[u|Y]=(I+\Sigma_{1}\Sigma^{-1}_{2})^{-1}Y$ and the Bayes risk is given
by

$E[||u-E[u|Y]||^{2}]=\mathrm{Tr}((\Sigma^{-1}_{1}+\Sigma^{-1}_{2})^{-1}).$

Using the above result we observe that

$\displaystyle\inf_{\hat{u}\in\mathcal{U}}\int
E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)\geq$
$\displaystyle\inf_{\hat{u}}\left(\int
E_{u}[\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i})^{2}]d\pi_{1}(u)+\int
E_{u}[(\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i}))^{2}]d\pi_{1}(u)\right)$
$\displaystyle\geq\inf_{\hat{u}}\int
E_{u}[\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i})^{2}]d\pi_{1}(u)+\inf_{w}\int
E_{u}[(\sum_{i=1}^{r-1}u_{i}-w)^{2}]d\pi_{1}(u)$
$\displaystyle=\mathrm{Tr}((\Sigma_{1}^{-1}+\Sigma_{2}^{-1})^{-1})+\frac{\sigma_{r}^{2}\tau_{r}^{2}}{\sigma_{r}^{2}+\tau_{r}^{2}}.$

Here $\Sigma_{1}=V(\mu)$, $\Sigma_{2}=n^{2\gamma_{0}}I_{r-1}$,
$\tau_{r}^{2}=(r-1)n^{2\gamma_{0}}$ and

$\sigma_{r}^{2}=\sum_{i=1}^{r-1}\mu_{i}(1-\mu_{i})-\sum_{i\neq
j}\mu_{i}\mu_{j}=\mu_{r}(1-\mu_{r}).$

Taking limit one obtains

$\liminf_{n\rightarrow\infty}\inf_{\hat{u}\in\mathcal{U}}\int
E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)\geq\mathrm{Tr}\;V(\mu)+\sigma_{r}^{2}=\sum_{i=1}^{r}\mu_{i}(1-\mu_{i}).$
(88)

For the quantum part the Bayes risk for a single component $\xi_{ij}$ with the
prior $\xi_{ij}\sim\sigma^{2}I_{2}$ is obtained by the covariant measurement
followed by an appropriate shrinkage (see section 4 for details) and the Bayes
risk is given by

$\inf_{M}\int\int||\xi_{ij}-\hat{\xi_{ij}}||^{2}\mathrm{Tr}\left(\phi^{\xi_{ij}}_{\beta_{ij}}M(d\hat{\xi_{ij}})\right)d\pi(\xi_{ij})=\frac{2\sigma^{2}(2\sigma^{2}_{\beta_{ij}}+1)}{2(\sigma^{2}+\sigma^{2}_{\beta_{ij}})+1}$

which is essentially equation (40). Thus we have

$\displaystyle\inf_{M:M(\mathcal{Z}^{c}=0)}\int
E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)\geq$
$\displaystyle\inf_{M}\int(\sum_{\begin{subarray}{c}{1\leq i\leq r}\\\
{i<j\leq
d}\end{subarray}}\kappa_{ij}||\xi_{ij}-\hat{\xi_{ij}}||^{2}\mathrm{Tr}[\phi^{\xi_{ij}}_{\beta_{ij}}M(d\hat{\xi_{ij}})])d\pi_{2}(\xi)$
$\displaystyle\geq$ $\displaystyle\sum_{\begin{subarray}{c}{1\leq i\leq r}\\\
{i<j\leq
d}\end{subarray}}\kappa_{ij}\frac{2n^{2\beta_{0}}(2\sigma^{2}_{\beta_{ij}}+1)}{2(n^{2\beta_{0}}+\sigma^{2}_{\beta_{ij}})+1}.$

Taking limit we obtain

$\liminf_{n\rightarrow\infty}\inf_{M:M(\mathcal{Z}^{c}=0)}\int
E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)\geq\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq
d}\end{subarray}}\kappa_{ij}(2\sigma^{2}_{\beta_{ij}}+1)=\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}$ (89)

the last equality following from the calculation preceding (86). We also have
the following lemma (the proof of which is deferred to Appendix D) which shows
that the second and fourth term of (87) goes to zero.

######  Lemma C.7.

Let $\hat{u}$ be an estimator of $u$ and suppose $\hat{u}\in\mathcal{U}$.
Similarly suppose we use the POVM $M$ to estimate $\xi$ where
$M(\mathcal{Z}^{c})=0$. Then we have

$\int_{\mathcal{U}^{c}}E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u),\int_{\mathcal{Z}^{c}}E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)\rightarrow
0.$

Finally we observe that $\pi_{1}(\mathcal{U}),\pi_{2}(\mathcal{Z})\rightarrow
1$, which together with (88),(89) and Lemma C.7 yields

$\liminf_{n\rightarrow\infty}\inf_{\hat{u},M}\sup_{\theta\in\Theta}R_{\hat{u},M}(\theta)\geq\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}$

when the terms are plugged in (87). ∎

###### Proof of Theorem 5.4.

Upper Bound

Split the sample into two parts $n_{1}=\lfloor n^{\delta_{0}}\rfloor+1$ for
some $0<\delta_{0}<1$ (we will describe the choice of $\delta_{0}$ in the next
step) and $n_{2}=n-n_{1}$. Let $\tilde{\rho}_{n}$ be the estimator obtained in
the first stage with the the sample $n_{1}$. Set
$25r\epsilon^{2}=n^{-1+2\epsilon_{0}}$ (where $\epsilon_{0}<\gamma$). By
Theorem 5.1

$P[||\rho-\tilde{\rho}^{n}||^{2}_{2}\geq n^{-1+2\epsilon_{0}}]\leq
de^{-3n^{-1+2\epsilon_{0}+\delta_{0}}/400rd}.$ (90)

since we have obtained $\tilde{\rho}^{n}$ using $n_{1}>n^{\delta_{0}}$. For
any $\epsilon_{0}$ we can choose $\delta_{0}$ such that
$2\epsilon_{0}+\delta_{0}>1$ and the concentration still holds. Further we
have

$P[r=\hat{r}]\geq 1-de^{-3n^{-1+2\epsilon_{0}+\delta_{0}}/400rd}.$

Now for the event
$A_{n}=\\{||\rho-\tilde{\rho}^{n}||^{2}_{2}<n^{-1+2\epsilon_{0}}\text{ and
}r=\hat{r}\\}$ we have,

$P[A_{n}]=1-2de^{-3n^{-1+2\epsilon_{0}+\delta_{0}}/400rd}.$ (91)

Recall the definition $\mathcal{L}(\theta,\hat{\theta})$ given in (43). The
risk can now be decomposed as follows

$\displaystyle E_{\rho}||\rho-\hat{\rho}^{n}||_{2}^{2}\leq$ $\displaystyle
4P[A^{c}_{n}]+E_{\rho}[||\rho-\hat{\rho}^{n}||^{2}\mathds{1}_{A_{n}}]$
$\displaystyle=$ $\displaystyle
4P[A^{c}_{n}]+E_{\rho_{\theta}}\left[\left(\frac{1}{n}\mathcal{L}(\theta,\hat{\theta})+O\left(\frac{||\theta||^{3},||\hat{\theta}_{n}||^{3}}{n^{3/2}}\right)\right)\mathds{1}_{A_{n}}\right].$

We have used the notation involving local parameters in the last line with
$\rho_{0}=\tilde{\rho}^{n}$ as the central state. Also the local parameter
estimate $\hat{\theta}_{n}$(and the corresponding state $\hat{\rho}_{n}$) is
formed using the second part of the sample $n_{2}$. Conditioned on $A_{n}$,
$||\theta||<n^{\epsilon_{0}}$ and we can restrict $\hat{\theta}_{n}$ to lie
within a radius of $2n^{\epsilon_{0}}$. This ensures that
$||\theta||<n^{\epsilon_{0}}<n^{\gamma}$ (choose $\beta<\gamma<1/9$) we can
use LAN. Secondly both $||\theta||^{3}$ and $||\hat{\theta}_{n}||^{3}$ are
less than $n^{3\gamma}<n^{1/2-\epsilon_{1}}$ for a sufficiently small
$\epsilon_{1}$ so that
$O(\frac{||\theta||^{3},||\hat{\theta}_{n}||^{3}}{n^{3/2}})=o(n^{-1})$. Thus
we obtain

$E_{\rho}n||\rho-\hat{\rho}^{n}||_{2}^{2}\leq
4nP[A^{c}_{n}]+E_{\rho_{\theta}}\left[\left(\mathcal{L}(\theta,\hat{\theta})\right)\mathds{1}_{A_{n}}\right]+o(1).$

Since $\mathcal{L}(\theta,\hat{\theta})=O(n^{2\epsilon_{0}})$ multiplying with
the asymptotic equivalence rate $n^{-\kappa}$ we have
$n^{2\epsilon_{0}-\kappa}=o(1)$ (we also choose $\epsilon_{0}<\kappa/2$).
Finally we use relations (28) and (30) to obtain

$E_{\rho_{\theta}}\left[E_{\rho_{\theta}}\left[\mathcal{L}(\theta,\hat{\theta}_{n})|\mathds{1}_{A_{n}}\right]\right]\leq
E_{\rho_{\theta}}\left[E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta}_{n})|\mathds{1}_{A_{n}}\right]\right]+o(1).$

Note that in this case $\Phi^{\theta,r}$ depends on the center
$\tilde{\rho}^{n}$ (obtained in the first step). Using relation (91) we
observe that $nP[A^{c}_{n}]=o(1)$ and hence

$E_{\rho}n||\rho-\hat{\rho}^{n}||_{2}^{2}\leq
E_{\rho_{\theta}}\left[E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta}_{n})|\mathds{1}_{A_{n}}\right]\right]+o(1).$
(92)

Since we have used $\tilde{\rho}^{n}$ as the centralizing state, if
$\tilde{\mu}_{1}>\ldots>\tilde{\mu}_{r}$ are the eigenvalues of
$\tilde{\rho}^{n}$, then with $\theta\in\Theta$ we have

$E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta}_{n})|\mathds{1}_{A_{n}}\right]=\sum_{i=1}^{r}\tilde{\mu}_{i}(1-\tilde{\mu}_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\tilde{\mu}_{i}$

by equation (85). Although we have used the second part of the sample $n_{2}$
to form the estimate $\hat{\theta}_{n}$, since $n_{2}=n(1+o(1))$ the same
asymptotics hold.

Finally let $\sigma$ be a rank-$r$ qudit with eigenvalues
$\mu_{1}>\ldots>\mu_{r}$, then if $||\rho-\sigma||_{2}<n^{-1/2+\varepsilon}$
for some $\varepsilon<\epsilon_{0}$ then $||\tilde{\rho}_{n}-\sigma||_{2}\leq
n^{-1/2+\epsilon_{0}}$ with high probability (see RHS of (90)). By the
equivalence of norms in the finite dimension the same claim can be made with
the operator norm i.e.

$P[||\sigma-\tilde{\rho}^{n}||^{2}\geq n^{-1/2+\varepsilon}]\leq
de^{-3n^{-1+2\epsilon_{0}+\delta_{0}}/400rd}.$

Using the above relation and Weyl perturbation inequality one can easily show
that $\tilde{\mu}_{i}=\mu_{i}+o_{p}(1)$ for all $i=1,\ldots,r-1$ and since
$\tilde{\mu}_{i}$ are bounded we have

$\limsup_{n}\sup_{||\rho-\sigma||_{2}<n^{-1/2+\varepsilon}}E_{\rho_{\theta}}\left[E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta}_{n})|\mathds{1}_{A_{n}}\right]\right]=\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}.$ (93)

Finally using relations (92) and (93) we obtain

$\limsup_{n}\sup_{||\rho-\sigma||_{2}<n^{-1/2+\varepsilon}}E_{\rho}n||\rho-\hat{\rho}^{n}||_{2}^{2}\leq\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}.$

Lower Bound

Again we observe that for $||\rho-\sigma||<n^{-1/2+\varepsilon}$ we can use
the local parametrization and write $\rho$ and $\hat{\rho}_{n}$ as
$\rho_{\theta}$ and $\rho_{\theta_{n}}$ respectively with the central state
now being $\sigma$. We have the following set of inequalities

$\displaystyle\inf_{\hat{\rho}^{n}}\sup_{||\rho-\sigma||<n^{-1/2+\varepsilon}}E_{\rho}n||\rho-\hat{\rho}^{n}||_{2}^{2}\geq$
$\displaystyle\inf_{\hat{\theta}^{n}}\sup_{||\theta||<n^{\varepsilon}}\mathcal{L}(\theta,\hat{\theta})+O\left(\frac{||\theta||^{3},||\hat{\theta}_{n}||^{3}}{n^{1/2}}\right)$
$\displaystyle\geq$
$\displaystyle\inf_{||\hat{\theta}||<n^{\varepsilon}}\sup_{||\theta||<n^{\varepsilon}}E_{\rho_{\theta}}\left[\mathcal{L}(\theta,\hat{\theta})+O\left(\frac{||\theta||^{3},||\hat{\theta}_{n}||^{3}}{n^{1/2}}\right)\right]$
$\displaystyle\geq$
$\displaystyle\inf_{||\hat{\theta}||<n^{\varepsilon}}\sup_{||\theta||<n^{\varepsilon}}E_{\rho_{\theta}}\left[\mathcal{L}(\theta,\hat{\theta})\right]-o(1)$
$\displaystyle\geq$
$\displaystyle\inf_{||\hat{\theta}||<n^{\varepsilon}}\sup_{||\theta||<n^{\varepsilon}}E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta})\right]-o(1)$
$\displaystyle\geq$
$\displaystyle\inf_{\hat{\theta}}\sup_{\theta\in\Theta_{n,\beta_{1},\gamma_{1}}}E_{\Phi^{\theta,r}}\left[\mathcal{L}(\theta,\hat{\theta})\right]-o(1)$
$\displaystyle\geq$
$\displaystyle\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}1\leq
i\leq r\\\ i<j\leq d\end{subarray}}2\mu_{i}-o(1).$

The first inequality follows since $||\theta||<n^{\varepsilon}$ implies
$||\rho-\sigma||_{2}<n^{-1/2+\varepsilon}$. Since $||\theta||<n^{\varepsilon}$
it is enough to consider $\hat{\theta}_{n}$ with
$||\hat{\theta}||_{n}<n^{\varepsilon}$ hence the second inequality follows.
For the third inequality we observe that one can choose $\varepsilon$ small
enough so that $3\varepsilon<1/2-\epsilon_{1}$ for some $\epsilon_{1}$ and
hence for the given restriction of $\theta$ and $\hat{\theta}_{n}$ the
remainder term is $o(1)$. Also for the same restriction of $\theta$ and
$\hat{\theta}_{n}$ within the ball of radius $n^{\varepsilon}$, the supremum
of the loss $\mathcal{L}(\theta,\hat{\theta}_{n})$ is $O(n^{2\varepsilon})$.
Let $\Theta_{n,\varepsilon}=\\{\theta:||\theta||<n^{\varepsilon}\\}$. Now fix
$0<\beta<\gamma<1/9$ which gives a rate of equivalence $n^{-\kappa}$ for some
$\kappa>0$ (see Theorem 3.3). Choose $\varepsilon$ small enough so that
$n^{2\varepsilon-\kappa}=o(1)$ and
$\Theta_{n,\varepsilon}\subset\Theta_{n,r,\beta,\gamma}$ so that we can use
LAN and using (28) and (31) we obtain the fourth inequality. Now choose
$\beta_{1}<\gamma_{1}<1/9$ so that
$\Theta_{n,r,\beta_{1},\gamma_{1}}\subset\Theta_{n,\varepsilon}$ and the fifth
inequality follows. We have also taken infimum over a larger class of
estimators without increasing the maximum risk. We use lower bound from
Theorem 5.2 to obtain the final inequality. Finally taking limits we obtain

$\liminf\inf_{\hat{\rho}^{n}}\sup_{||\rho-\sigma||_{2}<n^{-1/2+\varepsilon}}E_{\rho}n||\rho-\hat{\rho}^{n}||_{2}^{2}\geq\sum_{i=1}^{r}\mu_{i}(1-\mu_{i})+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}2\mu_{i}.$

∎

###### Proof of Theorem 6.1.

Upper Bound

Our aim is to show

$\sup_{\rho\in\Sigma_{n,\varepsilon}\left(\rho_{0}\right)}E_{\rho}|n^{1/2}\left(\bar{X}_{n}-\Psi\left(\rho\right)\right)|^{2}\rightarrow
V^{2}_{\rho_{0}}.$ (94)

Let $Y_{1,n},\ldots,Y_{n,n}$ be the array of independent r.v’s all having the
distribution of $n^{-1/2}\left(X_{A}-\Psi\left(\rho\right)\right)$ under
$\rho$; we have to check

$\displaystyle\sup_{\rho\in\Sigma_{n,\varepsilon}\left(\rho_{0}\right)}\sum_{i=1}^{n}EY_{n,i}^{2}$
$\displaystyle\rightarrow V_{\rho_{0}}^{2}.$

Since $Y_{1,n},\ldots,Y_{n,n}$ have identical distributions, this is
equivalent, for a r.v. $Y_{n,0}=X_{A}-\Psi\left(\rho\right)$ (distribution
under $\rho$)

$\displaystyle\sup_{\rho\in\Sigma_{n,\varepsilon}\left(\rho_{0}\right)}EY_{n,0}^{2}$
$\displaystyle\rightarrow V_{\rho_{0}}^{2}.$ (95)

Note that

$EY_{n,0}^{2}=Var_{\rho}X_{A}=\mathrm{Tr}\;A^{2}\rho-\left(\mathrm{Tr}\;A\rho\right)^{2}.$

Recall that $||A||$ is the operator norm of $A$. We have the following
inequalities

$\displaystyle\left|\mathrm{Tr}\;A^{2}\rho-\mathrm{Tr}\;A^{2}\rho_{0}\right|$
$\displaystyle=\left\|A^{2}\right\|\left\|\rho-\rho_{0}\right\|_{1}\leq\left\|A^{2}\right\|\gamma_{n}=o\left(1\right)$

where $\gamma_{n}=n^{-1/2+\varepsilon}$. Similarly we obtain

$\left|\mathrm{Tr}\;A\rho-\mathrm{Tr}\;A\rho_{0}\right|\leq\left\|A\right\|\gamma_{n}=o\left(1\right).$

The last two displays imply (95).

Lower Bound

We now have to establish the complement to (94), namely

$\liminf_{n}\inf_{\hat{\Psi}}\sup_{\rho\in\Sigma_{n,\varepsilon}\left(\rho_{0}\right)}E_{\rho}|n^{1/2}\left(\hat{\Psi}-\Psi\left(\rho\right)\right)|^{2}\geq
V^{2}_{\rho_{0}}.$ (96)

where $\inf_{\hat{\Psi}}$ denotes the infimum over all measurements and all
estimators $\hat{\Psi}$. We will construct, for a $d$-dimensional center state
$\rho_{0}$ (of rank $r$) and a parametrization of all $d$-dimensional nearby
states, a "least favorable parametric subfamily" so that the minimax risk of
estimating $\Psi\left(\rho\right)$ along this family will be lower bounded by
the minimax risk of estimating the parameter in the limiting model. Our
perturbation setup will be

$\rho_{t}=\rho_{0}+n^{-1/2}tH\text{, \ \
}t\in\left(-n^{\varepsilon},n^{\varepsilon}\right)$ (97)

where $H$ is self-adjoint with $\mathrm{Tr}\;H=0$, and satisfies

$\mathrm{Tr}\;AH=1.$ (98)

Then we have

$\displaystyle\Psi\left(\rho_{t}\right)$
$\displaystyle=\mathrm{Tr}\;A\rho_{t}=\mathrm{Tr}\;A\rho_{0}+n^{-1/2}t$
$\displaystyle=\Psi\left(\rho_{0}\right)+n^{-1/2}t$

so that $t$ is the appropriate parameter, estimation of which is equivalent to
estimating the functional $\Psi$. Recall that $\left\langle
A\right\rangle_{\rho_{0}}=\mathrm{Tr}(\rho_{0}A)$ for any operator $A$. For
the correct choice of $H$, set

$\displaystyle\tilde{A}$ $\displaystyle:=A-\left\langle
A\right\rangle_{\rho_{0}}\mathbf{1,}$ $\displaystyle\tilde{H}$
$\displaystyle:=\frac{1}{2}\left(\tilde{A}\rho_{0}+\rho_{0}\tilde{A}\right),$
$\displaystyle\hat{H}$
$\displaystyle:=\frac{\tilde{H}}{\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}}.$
(99)

With a choice $H=$ $\hat{H}$, the conditions on $H$ are fulfilled:

$\displaystyle\mathrm{Tr}\;\hat{H}$
$\displaystyle=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}^{-1}\mathrm{Tr}\;\tilde{H}=\left\langle
A^{2}\right\rangle_{\rho_{0}}^{-1}\mathrm{Tr}\;\tilde{A}\rho_{0}$
$\displaystyle=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}^{-1}\mathrm{Tr}\left(A-\left\langle
A\right\rangle_{\rho_{0}}\mathbf{1}\right)\rho_{0}$ $\displaystyle=0$

and for (98) we obtain

$\displaystyle\mathrm{Tr}\;A\hat{H}$
$\displaystyle=\mathrm{Tr}\;\left(A-\left\langle
A\right\rangle_{\rho_{0}}\mathbf{1}\right)\hat{H}=\mathrm{Tr}\;\tilde{A}\hat{H}$
$\displaystyle=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}^{-1}\mathrm{Tr}\;\tilde{A}\tilde{H}=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}^{-1}\mathrm{Tr}\;\tilde{A}^{2}\rho_{0}$
$\displaystyle=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}^{-1}\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}=1.$

We also note that

$\displaystyle V_{\rho_{0}}^{2}$
$\displaystyle=\mathrm{Var}_{\rho_{0}}\left(X_{A}\right)$
$\displaystyle=E_{\rho_{0}}\left(X_{A}-E_{\rho_{0}}X_{A}\right)^{2}$
$\displaystyle=\mathrm{Tr}\;\tilde{A}^{2}\rho_{0}=\left\langle\tilde{A}^{2}\right\rangle_{\rho_{0}}.$

We now proceed to find the limiting classical-quantum Gaussian model for this
choice of perturbation. It can be easily verified that up to first order terms
$\rho_{\vartheta,r}$ is equal to $\tilde{\rho}_{\vartheta,r}$ where
$\tilde{\rho}_{\vartheta,r}$ given as follows

$\tilde{\rho}_{\vartheta/\sqrt{n},r}:=\begin{bmatrix}\mu_{1}+\tilde{u}_{1}&\tilde{\zeta}_{1,2}^{*}&\dots&\tilde{\zeta}_{1,r}^{*}&\vline&\tilde{\zeta}^{*}_{1,r+1}&\tilde{\zeta}_{1,r+2}&\dots&\tilde{\zeta}^{*}_{1,d}\\\
\tilde{\zeta}_{1,2}&\mu_{2}+\tilde{u}_{2}&\ddots&\vdots&\vline&\tilde{\zeta}_{2,r+1}&\tilde{\zeta}^{*}_{2,r+2}&\ddots&\vdots\par\\\
\vdots&\ddots&\ddots&\tilde{\zeta}_{r-1,r}^{*}&\vline&\vdots&\ddots&\ddots&\tilde{\zeta}^{*}_{r-1,d}\\\
\tilde{\zeta}_{1,r}&\dots&\tilde{\zeta}_{r-1,r}&\mu_{r}-\sum_{i=1}^{r-1}\tilde{u}_{i}&\vline&\tilde{\zeta}^{*}_{r,r+1}&\dots&\tilde{\zeta}^{*}_{r,d-1}&\tilde{\zeta}^{*}_{r,d}\\\
\hline\cr\tilde{\zeta}_{1,r+1}&\tilde{\zeta}_{2,r+1}&\dots&\tilde{\zeta}_{r,r+1}&\vline&0&0&\dots&0\\\
\tilde{\zeta}_{1,r+2}&\tilde{\zeta}_{2,r+2}&\ddots&\vdots&\vline&0&0&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\tilde{\zeta}_{r,d-1}&\vline&\vdots&\ddots&\ddots&0\\\
\tilde{\zeta}_{1,d}&\dots&\tilde{\zeta}_{r-1,d}&\tilde{\zeta}_{r,d}&\vline&0&\dots&0&0\\\
\end{bmatrix}$ (100)

where $\tilde{u}=n^{-/2}u,\leavevmode\nobreak\ \tilde{\zeta}=n^{-1/2}\zeta$
and $u_{i}\in\mathbb{R},\leavevmode\nobreak\ \zeta_{j,k}\in\mathbb{C}$. From
(97) we obtain

$n^{-1/2}t\hat{H}=n^{-1/2}\begin{bmatrix}u_{1}&\zeta_{1,2}^{*}&\dots&\zeta_{1,r}^{*}&\vline&\zeta^{*}_{1,r+1}&\zeta_{1,r+2}&\dots&\zeta^{*}_{1,d}\\\
\zeta_{1,2}&u_{2}&\ddots&\vdots&\vline&\zeta^{*}_{2,r+1}&\zeta^{*}_{2,r+2}&\ddots&\vdots\par\\\
\vdots&\ddots&\ddots&\zeta_{r-1,r}^{*}&\vline&\vdots&\ddots&\ddots&\zeta^{*}_{r-1,d}\\\
\zeta_{1,r}&\dots&\zeta_{r-1,r}&-\sum_{i=1}^{r-1}u_{i}&\vline&\zeta^{*}_{r,r+1}&\dots&\zeta^{*}_{r,d-1}&\zeta^{*}_{r,d}\\\
\hline\cr\zeta_{1,r+1}&\zeta_{2,r+1}&\dots&\zeta_{r,r+1}&\vline&0&0&\dots&0\\\
\zeta_{1,r+2}&\zeta_{2,r+2}&\ddots&\vdots&\vline&0&0&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\zeta_{r,d-1}&\vline&\vdots&\ddots&\ddots&0\\\
\zeta_{1,d}&\dots&\zeta_{r-1,d}&\zeta_{r,d}&\vline&0&\dots&0&0\\\
\end{bmatrix}.$ (101)

Let $x=\langle A\rangle_{\rho_{0}}=\sum_{i=1}^{r}\mu_{i}A_{ii}$ and
$y=\langle\tilde{A}^{2}\rangle_{\rho_{0}}$. It can be easily verified that

$y=\mathrm{Tr}{\rho_{0}\tilde{A}^{2}}=\sum_{i=1}^{r}\mu_{i}(A_{ii}-x)^{2}+\sum_{\begin{subarray}{c}1\leq
i\leq r\\\ i<j\leq d\end{subarray}}|A_{ji}|^{2}(\mu_{i}+\mu_{j}).$ (102)

Also from (99) we obtain

$\hat{H}:=\frac{1}{y}\begin{bmatrix}B_{1,1}&B_{2,1}^{*}&\dots&B_{r,1}^{*}&\vline&C^{*}_{r+1,1}&C^{*}_{r+2,1}&\dots&C^{*}_{d,1}\\\
B_{2,1}&B_{2,2}&\ddots&\vdots&\vline&C^{*}_{r+1,2}&C^{*}_{r+2,2}&\ddots&\vdots\par\\\
\vdots&\ddots&\ddots&B_{r,r-1}^{*}&\vline&\vdots&\ddots&\ddots&C^{*}_{d,r-1}\\\
B_{r,1}&\dots&B_{r,r-1}&B_{r,r}&\vline&C^{*}_{r+1,r}&\dots&C^{*}_{d-1,r}&C^{*}_{d,r}\\\
\hline\cr C_{r+1,1}&C_{r+1,2}&\dots&C_{r+1,r}&\vline&0&0&\dots&0\\\
C_{r+2,1}&C_{r+2,2}&\ddots&\vdots&\vline&0&0&\ddots&\vdots\\\
\vdots&\ddots&\ddots&C_{d-1,r}&\vline&\vdots&\ddots&\ddots&0\\\
C_{d,1}&\dots&C_{d,r-1}&C_{d,r}&\vline&0&\dots&0&0\\\ \end{bmatrix}$ (103)

with the following relations

$\displaystyle B_{i,i}$ $\displaystyle=(A_{ii}-x)\mu_{i}\text{ for }1\leq
i\leq r$ $\displaystyle B_{j,i}$
$\displaystyle=\frac{A_{ji}}{2}(\mu_{i}+\mu_{j})\text{ for }i<j\leq r$
$\displaystyle C_{j,i}$ $\displaystyle=\frac{A_{ji}}{2}\mu_{i}\text{ for
}1\leq i\leq r,r+1\leq j\leq d.$

Comparing equations (101) and (103) we get

$u_{i}=t\frac{(A_{ii}-x)\mu_{i}}{y}\text{ for }1\leq i\leq
r,\quad\zeta_{i,j}=t\frac{A_{ji}}{2y}(\mu_{i}+\mu_{j})\text{ for }1\leq i\leq
r,i<j\leq r$ (104)

where $\mu_{j}=0$ for $j>r$. Also let
$\breve{u}_{i}=\frac{(A_{ii}-x)\mu_{i}}{y}$ and
$\breve{\zeta}_{i,j}=\frac{A_{ji}}{2y}(\mu_{i}+\mu_{j})$.

Recall that after defining
$\mathfrak{z}_{ij}=\zeta_{ij}/\sqrt{\mu_{i}-\mu_{j}}$ and replacing
$\vartheta$ by $\theta$, limiting model for $\rho^{\otimes
n}_{\theta/\sqrt{n}}$ is given by equation (21) and is given by

$\displaystyle\Phi^{\theta,r}=$
$\displaystyle\mathcal{N}_{r-1}(\tilde{u},V_{\mu})\otimes\bigotimes_{1\leq
i<j\leq
r}\phi_{\beta_{ij}}^{\mathfrak{z}_{ij}}\otimes\bigotimes_{\begin{subarray}{c}1\leq
j\leq r\\\ r+1\leq j\leq d\end{subarray}}\phi_{\infty}^{\mathfrak{z}_{ij}}$
$\displaystyle=$
$\displaystyle\mathcal{N}_{r-1}(t\breve{u},V_{\mu})\otimes\bigotimes_{1\leq
i<j\leq
r}\mathbb{N}_{2}\left(t\sqrt{\frac{2}{\mu_{i}-\mu_{j}}}(Re\breve{\zeta}_{ij},Im\breve{\zeta}_{ij}),\frac{\mu_{i}+\mu_{j}}{2(\mu_{i}-\mu_{j})}I_{2}\right)$
$\displaystyle\otimes\bigotimes_{\begin{subarray}{c}1\leq j\leq r\\\ r+1\leq
j\leq
d\end{subarray}}\mathbb{N}_{2}\left(t\sqrt{\frac{2}{\mu_{i}}}(Re\breve{\zeta}_{ij},Im\breve{\zeta}_{ij}),\frac{1}{2}I_{2}\right).$

Note that an alternate expression for the limiting state is given by
$\mathfrak{N}(t\tau,\mathscr{S})$ where $\Sigma$ is given as in (24) with
$\sigma^{2}_{\beta_{ij}}=\frac{\mu_{i}+\mu_{j}}{2(\mu_{i}-\mu_{j})}$ for
$1\leq i\leq r,i<j\leq d$ and $V=V_{\mu}$. Also the shift is given by

$\tau=\breve{u}\oplus{\bigoplus_{1\leq i\leq r,i<j\leq
d}}\sqrt{\frac{2}{\mu_{i}-\mu_{j}}}\left(Re\breve{\zeta}_{ij},Im\breve{\zeta}_{ij}\right).$

Denoting $\hat{\Psi}=\rho_{0}+n^{-1/2}\hat{t}H$, we have the following chain
of inequalities

$\displaystyle\inf_{\hat{\Psi}}\sup_{\rho\in\Sigma_{n,\varepsilon}(\rho_{0})}E_{\rho}|n^{1/2}(\hat{\Psi}-\Psi(\rho))|^{2}$
$\displaystyle\geq\inf_{\hat{t}}\sup_{|t|\leq
n^{\varepsilon}}E_{\rho_{t}}|\hat{t}-t|^{2}$
$\displaystyle\geq\inf_{|\hat{t}|\leq n^{\varepsilon}}\sup_{|t|\leq
n^{\varepsilon}}E_{\mathfrak{N}_{t}}|\hat{t}-t|^{2}+o(1)$

where $\mathfrak{N}_{t}:=\mathfrak{N}(t\tau,\mathscr{S})$. We observe that it
is the estimator $\hat{t}$ can also be restricted to a ball of radius
$n^{\varepsilon}$ without increasing the risk. Also for the same restriction
of $t$ and $\hat{t}_{n}$ the supremum of the the loss
$E_{\rho_{t}}|\hat{t}-t|^{2}$ is $O(n^{2\varepsilon})$ and multiplying by the
equivalence rate of the reverse channel we have $n^{2\varepsilon-\kappa}=o(1)$
(for sufficiently small $\varepsilon$) so that using (28) and (31) we obtain
the second inequality.

Now construct the following prior $\pi$ on $t$: $t\sim
N(0,n^{\varepsilon_{0}})$ with $\varepsilon_{0}<\varepsilon$. The minimax risk
can be lower bounded as follows:

$\inf_{|\hat{t}|\leq n^{\varepsilon}}\sup_{|t|\leq
n^{\varepsilon}}E_{\mathfrak{N}_{t}}|\hat{t}-t|^{2}\geq\inf_{|\hat{t}|\leq
n^{\varepsilon}}\int_{|t|\leq
n^{\varepsilon}}E_{\mathfrak{N}_{t}}|\hat{t}-t|^{2}d\pi(t).$

Using an argument similar to the proof of Lemma C.7 we can show that the RHS
is further bounded by $\inf_{|\hat{t}|\leq n^{\varepsilon}}\int
E_{\mathfrak{N}_{t}}|\hat{t}-t|^{2}d\pi(t)+o(1)$ and since taking a supremum
over a larger class of estimators (i.e. if $\hat{t}$ is allowed to be
unrestricted) decreases the risk, we obtain

$\inf_{\hat{\Psi}}\sup_{\rho\in\Sigma_{n,\varepsilon}(\rho_{0})}E_{\rho}|n^{1/2}(\hat{\Psi}-\Psi(\rho))|^{2}\geq\inf_{\hat{t}}\int
E_{\mathfrak{N}_{t}}|\hat{t}-t|^{2}d\pi(t)+o(1).$

By (74) the first term in RHS is given by
$\frac{\left(\tau^{T}\Sigma^{-1}\tau\right)^{-1}n^{\epsilon_{0}}}{\left(\tau^{T}\Sigma^{-1}\tau\right)^{-1}+n^{\epsilon_{0}}}$
and hence we obtain

$\liminf_{n}\inf_{\hat{\Psi}}\sup_{\rho\in\Sigma_{n,\varepsilon}(\rho_{0})}E_{\rho}|n^{1/2}(\hat{\Psi}-\Psi(\rho))|^{2}\geq(\tau^{T}\Sigma^{-1}\tau)^{-1}.$

Plugging in the value of $\Sigma$ and $\tau$ we obtain

$\displaystyle\tau^{T}\Sigma^{-1}\tau$
$\displaystyle=\breve{u}^{T}V^{-1}_{\mu}\breve{u}+\sum_{1\leq i<j\leq
r}\frac{4|\breve{\zeta}_{ij}|^{2}}{\mu_{i}+\mu_{j}}+\sum_{\begin{subarray}{c}1\leq
j\leq r\\\ r+1\leq j\leq
d\end{subarray}}\frac{4|\breve{\zeta}_{ij}|^{2}}{\mu_{i}}$
$\displaystyle=\breve{u}^{T}V^{-1}_{\mu}\breve{u}+\sum_{\begin{subarray}{c}1\leq
i\leq r\\\ i<j\leq
d\end{subarray}}\frac{|A_{ji}|^{2}}{y^{2}}(\mu_{i}+\mu_{j}).$

where the last line is obtained by plugging in the values of
$\breve{\zeta}_{ij}$ and letting $\mu_{j}=0$ for $j>r$.

By example 3.1 of [49]
$\left[V^{-1}_{\mu}\right]_{ij}=\mu^{-1}_{i}\delta_{ij}+(1-\sum_{i=1}^{r-1}\mu_{k})^{-1}$
and hence noting that $1-\sum_{i=1}^{r-1}\mu_{k}=\mu_{r}$ we have

$\displaystyle\breve{u}^{T}V^{-1}_{\mu}\breve{u}=$
$\displaystyle\frac{1}{y^{2}}\sum_{i,j=1}^{r-1}\mu_{i}\mu_{j}(A_{ii}-x)(A_{jj}-x)(\mu^{-1}_{i}\delta_{ij}+\mu_{r}^{-1})$
$\displaystyle=$
$\displaystyle\frac{1}{y^{2}}\left[\sum_{i=1}^{r-1}\mu_{i}(A_{ii}-x)^{2}+\mu_{r}^{-1}\sum_{i,j=1}^{r-1}A_{ii}A_{jj}\mu_{i}\mu_{j}-\mu_{r}^{-1}\sum_{i,j=1}^{r-1}xA_{jj}\mu_{i}\mu_{j}\right.$
$\displaystyle\left.-\mu_{r}^{-1}\sum_{i,j=1}^{r-1}A_{ii}x\mu_{i}\mu_{j}+\mu_{r}^{-1}\sum_{i,j=1}^{r-1}x^{2}\mu_{i}\mu_{j}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{y^{2}}\left[\sum_{i=1}^{r-1}\mu_{i}(A_{ii}-x)^{2}+\mu_{r}^{-1}(x-A_{rr}\mu_{r})^{2}-2x\mu_{r}^{-1}(1-\mu_{r})(x-A_{rr}\mu_{r})\right.$
$\displaystyle\left.+\mu_{r}^{-1}x^{2}(1-\mu_{r})^{2}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{y^{2}}\left[\sum_{i=1}^{r-1}\mu_{i}(A_{ii}-x)^{2}+\mu_{r}(A_{rr}-x)^{2}\right]$

Thus we obtain

$\tau^{T}\Sigma^{-1}\tau=\frac{1}{y^{2}}\sum_{i=1}^{r}\mu_{i}(A_{ii}-x)^{2}+\sum_{\begin{subarray}{c}1\leq
i\leq r\\\ i<j\leq
d\end{subarray}}\frac{|A_{ji}|^{2}}{y^{2}}(\mu_{i}+\mu_{j})$

Plugging back the expression of $y$ from equation (102) we observe that the
RHS of the above equation is $y^{-1}$ and hence we have

$\liminf_{n}\inf_{\hat{\Psi}}\sup_{\rho\in\Sigma_{n,\varepsilon}(\rho_{0})}E_{\rho}|n^{1/2}(\hat{\Psi}-\Psi(\rho))|^{2}\geq
y=V^{2}_{\rho_{0}}.$

∎

Remark: Note that if we had a multivariate Gaussian
$\mathcal{N}_{k}(t\tau,\Sigma)$ with $k=r-1+r(r-1)/2+r(d-r)$, then it can be
easily shown that the minimax risk for estimation of $t$ is
$(\tau^{T}\Sigma^{-1}\tau)^{-1}$. So the minimax risk for the classical model
and the classical-quantum model matches for this particular case where the
shift is indexed by $1$-dimensional parameter.

##  Appendix D Proof of technical lemmas

###  D.1 Auxiliary Lemma for channel construction

First we set some notations and prove a technical lemma (the equivalent of
Lemma 7.1 of [35]). We note that the some statements of the lemma differs
because of the low-rank structure. Although one can prove the statements by
suitably modifying Lemma 7.1 of [35] we include the proof of the entire lemma
mutatis mutandis for the convenience of the reader and some of the following
passages in this subsection will be quoted verbatim from the original Lemma
7.1 of [35]. However the explanations will be brief and the reader can refer
to the discussion preceding Lemma 7.1 for further clarification.

1) Let $l(c)$ be the length of the column $c$ in the Young diagram $\lambda$.
Note that there are $\lambda_{i}-\lambda_{i+1}$ columns such that $l(c)=i$.
Also, in our case $l(c)\leq r$ since the Young diagrams can only have $r$-rows
in our setup.

2) Recall that $f_{\mathbf{a}}=f_{\mathbf{a}(1)}\otimes\dots\otimes
f_{\mathbf{a}(n)}$ are the basis vectors and we associate a Young tableau
$t_{\bf a}$ with each $f_{\mathbf{a}}$ where the indices $\bf{a}(i)$ fill the
boxes of a diagram $\lambda$ row-wise. Let $t^{c}_{a}$ be the content of the
column $c$ of $t_{\bf a}$, i.e. the function

$t_{\bf a}^{c}:\\{1,\dots,l(c)\\}\rightarrow\\{1,\dots,d\\}$

which gives the column entries of the column $c$. For example, if $t_{\bf a}$
is given by the following semi-standard Young tableau

$1$&$2$$3$ $2$ $3$

we get the values:

$\displaystyle t_{\bf a}^{1}(1)$ $\displaystyle=1,$ $\displaystyle t_{\bf
a}^{1}(2)$ $\displaystyle=2,$ $\displaystyle t_{\bf a}^{2}(1)$
$\displaystyle=2,$ $\displaystyle t_{\bf a}^{2}(2)$ $\displaystyle=3,$
$\displaystyle t_{\bf a}^{3}(1)$ $\displaystyle=3.$

3) We also use the alternate notation $f_{\bf m}$ for $f_{\bf a}$ (with ${\bf
m}=\\{m_{i,j}:1\leq i\leq r,i<j\leq d\\}$, where $m_{i,j}$ is the number of
$j$’s in the row $i$ of $t_{\bf a}$) and that $y_{\lambda}f_{\bf
m}=q_{\lambda}p_{\lambda}f_{\bf m}$ spans $\mathcal{H}_{\lambda}$ where
$t_{\bf{a}}$ is a semi-standard tableau. Let $\mathcal{O}_{\lambda}({\bf m})$
be the orbit of $f_{\bf m}$ under the subgroup $\mathcal{R}_{\lambda}$ of row
permutations i.e. the set of vectors $f_{\mathbf{b}}$ which have exactly
$m_{i,j}$ boxes with $j$ in row $i$, and the rest are $i$. Note that a semi-
standard tableau ensures that if we permute among rows only, row $i$ has no
entries smaller than $i$. As a consequence of the orbit-stabilizer theorem we
have

$p_{\lambda}f_{\bf m}=\sum_{\sigma\in\mathcal{R}_{\lambda}}f_{{\bf
a}\circ\sigma}=\sum_{{f_{\bf b}}\in\mathcal{O}_{\lambda}({\bf
m})}\frac{\\#\mathcal{R}_{\lambda}}{\\#\mathcal{O}_{\lambda}({\bf m})}{f_{\bf
b}}.$ (105)

4) If a column contains two identical entries it does not contribute to
$y_{\lambda}f_{\bf a}$ since we are antisymmetrizing with $q_{\lambda}$. Thus
only tableaux $t_{\bf a}$ (not necessarily semistandard after the action of
$p_{\lambda}$) which do not have two equal entries in the same column are
important and we call them admissible ones and denote the collection of
associated vectors $f_{\bf a}$ by $\mathcal{V}$.

5) For any $f_{\bf a}\in\mathcal{O}_{\lambda}({\bf m})$ we define

$\Gamma(f_{\mathbf{a}}):=|{\bf m}|-\\#\\{1\leq c\leq\lambda_{1}:{t_{\bf
a}^{c}}\neq{\rm Id}^{c}\\},$

and denote by $\mathcal{V}^{\Gamma}({\bf m})$ the set of vectors $f_{\bf
a}\in\mathcal{O}_{\lambda}({\bf m})\bigcap\mathcal{V}$ with
$\Gamma(f_{a})=\Gamma$. Then we have

$\mathcal{O}_{\lambda}({\bf
m})\bigcap\mathcal{V}=\bigcup_{\Gamma\in\mathbb{N}}\mathcal{V}^{\Gamma}({\bf
m}).$

It can be easily seen that $\Gamma(f_{\bf a})\geq 0$ and is zero if and only
if each column ${t_{\bf a}^{c}}$ is either ${\rm Id}^{c}$ or of the form
${t_{\bf a}^{c}}(r)=j\delta_{r=i}+r\delta_{r\neq i}$ for some $i\leq l(c)<j$.
A tableau ${t_{\bf a}^{c}}$ of the latter form will be called an _$(i,j)$
-substitution_.

One then considers the following ‘algorithm’ to build all the possible
${f_{\bf a}}\in\mathcal{V}^{\Gamma}({\bf m})$.

_Algorithm_

For a given $({\bf m},\lambda)$ we generate a particular admissible $f_{\bf
a}\in\mathcal{O}_{\lambda}({\bf m})$ by selecting the $m_{i,j}$ boxes on row
$i$ which are filled with $j$, for all $1\leq i\leq r,i<j\leq d$. To ensure
admissibility we have to ensure that no column should have two boxes filled
with the same number. Start with $f_{\bf 0}:=f_{{\bf m}={\bf 0}}$ (row $i$ is
filled with $i$’s only), and with a set of $|{\bf m}|$ “bricks" containing
$m_{i,j}$ identical bricks labelled $(i,j)$, for $1\leq i\leq r,i<j\leq d$.
These bricks are placed sequentially i.e. for a particular $(i,j)$ brick we
replace the content of a box from row $i$ with $j$ (initially the content was
$i$). Once the bricks have been placed the content of each column c (i.e
$t_{\bf a}^{c}$) is uniquely defined by the set of bricks $\kappa$ which shall
be called a _column-modifier_.

For example if $\kappa=\\{(i,j),(f,l)\\}$ then the column has entries

$t_{\bf a}^{c}(k)=\left\\{\begin{array}[]{lll}j&\textrm{if $k=i$};\\\
l&\textrm{if $k=f$};\\\ k&\textrm{otherwise}.\end{array}\right.$

A column modifier of the form $\kappa(i,j)$ is called elementary and can only
be used in a column with $i\leq l(c)<j$ (otherwise the entry $j$ would appear
twice making the column inadmissible). Now, since the length of a column is at
most $r$ and all entries must be different, there are less than $d!$ (although
the upper bound is $d!/(d-r)!$ we use the $d!$ since a sharper bound does not
change the asymptotics) different types of column-modifiers. Note that a
column-modifier always increases the value of the modified cells, so that in
this case $t^{c}_{a}(\\{1,\dots,l(c)\\})\neq\\{1,\dots,l(c)\\}$.

To count the total number of ways the tableau can be formed we can count the
total number of ways the columns are modified. A given collection of column-
modifiers is uniquely determined by $\\{m_{\kappa},\kappa\\}$ where
$m_{\kappa}$ is the multiplicity of $\kappa$. We consider the following
procedure:

  1. I.

Choose $\Gamma$ bricks among our $|{\bf m}|$. As we have at most $rd$
different types of bricks there are at most $(rd)^{\Gamma}$ possibilities.

  2. II.

Recall that $|\textbf{m}|-\Gamma$ is the total number of modified columns. As
this is the exact number of bricks left after the first stage we can consider
them as a set of elementary column-modifiers. Indeed if all the column
modifiers were elementary, then $\Gamma=0$. If $\Gamma\neq 0$, we form non-
elementary column modifiers from elementary ones by sequentially adding each
of the $\Gamma$ bricks selected in the first stage, to some of the elementary
column-modifiers. At each step there are at most $d!$ different choices (of
column modifiers) to which we can attach the new brick. Hence, we have at most
$(d!)^{\Gamma}$ possibilities. At the end of stage II at least $\max\\{0,|{\bf
m}|-2\Gamma\\}$ of the column-modifiers are elementary, and we note that
$m_{\kappa(i,j)}\leq m_{i,j}$.

  3. III.

We apply the column-modifiers to the columns of $f_{\mathbf{0}}$, with the
constraints that no two modifiers are applied to the same column and the
resulting $f_{\bf{a}}$ is admissible.

While inserting the column modifiers we note that for a non-elementary column
modifier $\kappa$ there are less than n possibilities. In the case of an
elementary one of type $\kappa(i,j)$ we have the constraints that it can only
be inserted in a column with at least $i$ rows, and since the column cannot
contain another $j$ (to ensure admissibility) its length has to be smaller
than $j$. Noting that number of such columns is $\lambda_{i}-\lambda_{j}$, we
give an upper bound to the number of possibilities at stage three of the
algorithm as follows:

$\displaystyle\prod_{\kappa\neq\kappa({i,j})}\frac{n^{m_{\kappa}}}{m_{\kappa}!}\cdot\prod_{1\leq
i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{m_{\kappa(i,j)}}}{m_{\kappa(i,j)}!}$ (106)

the combinatorial factor $\prod_{\kappa}m_{\kappa}!$ ($m_{\kappa}$ being the
number of column modifiers of type $\kappa$) appearing to account for the
identical columns. If $\Gamma=0$, then all column modifiers are elementary and
for each $\kappa(i,j)$, the number of available columns is at least
$(\lambda_{i}-\lambda_{j}-\lvert{\bf
m}|)_{+}:=0\wedge\lambda_{i}-\lambda_{j}-\lvert{\bf m}|$. Thus we have the
following lower bound

$\displaystyle\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j}-|\mathbf{m}|)_{+}^{m_{i,j}}}{m_{i,j}!}.$
(107)

6) We collect all multiplicities in $E:=\\{m_{\kappa}^{\bf a}:\kappa\\},$
where $m_{\kappa}^{\bf a}$ is the number of columns with column-modifer
$\kappa$, noting that $\Gamma$ is a function of $E$ and the following relation
holds

$\Gamma(f_{\bf a})=|{\bf m}|-\sum_{\kappa}m_{\kappa}^{\bf a}.$

We note that the condition $\Gamma(f_{\bf a})=0$ identifies the multiplicity
set $E^{0}$. In particular, we have $m_{\kappa(i,j)}=m_{i,j}$ for all $1\leq
i\leq r,i<j\leq d$ and the other $m_{\kappa}=0$. In the general case, we
decompose the set $\mathcal{V}^{\Gamma}({\bf m})$ in terms of
$\mathcal{V}^{E}({\bf m})$ (the set of tableaux in $\mathcal{O}_{\lambda}({\bf
m})\bigcap\mathcal{V}$ with $E(f_{\bf a})=E$) as follows:

$\mathcal{V}^{\Gamma}({\bf m})=\bigcup_{E}\mathcal{V}^{E}({\bf m}).$ (108)

7) Finally to each column $c$ of $t_{\bf a}$ we associate the set of modified
entries: each element of this set is of the form $(i,+)$ or $(i,-)$ depending
on whether $i$ is added or deleted. It is easy to verify that if $t_{\bf a}$
is admissible, this modified set determines column-modifier $\kappa$
associated to $c$. and hence shall be denoted by $S(\kappa)$. For example
$S(\kappa(i,j))=\\{(i,-),(j,+)\\}$ and for $\kappa=\\{(i,j),(j,k)\\}$ we have
$S(\kappa)=\\{(i,-),(k,+)\\}$. We define the multiplicities $m_{S}^{\bf
a}=\sum_{\kappa:S(\kappa)=S}m_{\kappa}^{\bf a}$ and consider the set $F(f_{\bf
a}):=\\{m_{S}^{\bf a}:S\\}$ .

We are now ready to state the auxiliary lemma which will be crucial in proving
Lemma C.3 and Lemma C.4.

######  Lemma D.1.

  1. 1.

For any unitary operator $U\in M(\mathbb{C}^{d})$, for any basis vectors
${f_{\bf a}}$ and ${f_{\bf b}}$, we have

$\displaystyle\langle{f_{\bf a}}|q_{\lambda}U^{\otimes n}{f_{\bf b}}\rangle$
$\displaystyle=\prod_{1\leq c\leq\lambda_{1}}\det(U^{{t_{\bf a}^{c}},{t_{\bf
b}^{c}}}),$ (109)

where $U^{{t_{\bf a}^{c}},{t_{\bf b}^{c}}}$ is the ${l(c)}\times{l(c)}$ minor
of $U$ given by $[U^{{t_{\bf a}^{c}},{t_{\bf b}^{c}}}]_{i,j}=U_{{t_{\bf
a}^{c}}(i),{t_{\bf b}^{c}}(j)}$.

Under the assumptions

$\displaystyle|\mathbf{m}|$ $\displaystyle\leq n^{\eta},$ (110)
$\displaystyle\lambda$ $\displaystyle\in\Lambda_{2},$
$\displaystyle\inf_{i}|\mu_{i}-\mu_{i+1}|$ $\displaystyle\geq\delta,$
$\displaystyle\mu_{r}$ $\displaystyle\geq\delta,$
$\displaystyle\|\mathfrak{z}\|$ $\displaystyle\leq Cn^{\beta},\qquad\beta\leq
1/2.$

we have the following estimates with remainder terms uniform in the
eigenvalues $\\{\mu_{i}\\}$:

  1. 2.

The number of admissible ${f_{\bf a}}\in\mathcal{O}_{\lambda}({\bf m})$ with
$\Gamma({f_{\bf a}})=0$ is

$\\#\mathcal{V}^{0}({\bf m})=\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{m_{i,j}}}{m_{i,j}!}(1+O(n^{-1+2\eta}/\delta)).$
(111)

  2. 3.

Let $E:=\\{m_{\kappa}:\kappa\\}$ with $\Gamma(E)=\Gamma$. The number of
admissible ${f_{\bf a}}\in\mathcal{O}_{\lambda}({\bf m})$ with $E({f_{\bf
a}})=E$ is bounded by:

$\\#\mathcal{V}^{E}({\bf m})\leq n^{-\Gamma+\sum_{1\leq i\leq r,i<j\leq
d}(m_{i,j}-m_{\kappa({i,j})})}\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{m_{\kappa({i,j})}}}{m_{\kappa({i,j})}!}.$
(112)

  3. 4.

The number of admissible ${f_{\bf a}}\in\mathcal{O}_{\lambda}({\bf m})$ with
$\Gamma({f_{\bf a}})=\Gamma$ is bounded by:

$\\#\mathcal{V}^{\Gamma}({\bf m})\leq
C^{\Gamma}n^{-\Gamma}\delta^{-2\Gamma}|\mathbf{m}|^{2\Gamma}\prod_{1\leq i\leq
r,i<j\leq d}\frac{(\lambda_{i}-\lambda_{j})^{m_{i,j}}}{m_{i,j}!},$ (113)

for a constant $C=C(d)$.

  4. 5.

Let ${f_{\bf a}}\in\mathcal{V}^{\Gamma^{a}}(\mathbf{l})$, and consider
$\mathcal{V}^{\Gamma^{b}}({\bf m})\subset\mathcal{O}_{\lambda}(\mathbf{m})$
for some fixed $\Gamma^{b}$. Then:

$\displaystyle\left|\left\langle{f_{\bf a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
b}}\in\mathcal{V}^{\Gamma^{b}}({\bf m})}{f_{\bf
b}}\right\rangle\right|\leq\left\\{\begin{array}[]{cc}0&\mbox{if
}\Gamma^{b}\neq|\mathbf{m}|-|\mathbf{l}|+\Gamma^{a}\\\
(C|\mathbf{m}|)^{\Gamma^{b}}&\mbox{otherwise}\\\ \end{array}\right.,$ (116)

with $C=C(d)$.

  5. 6.

If ${f_{\bf a}}\in\mathcal{V}^{0}(\mathbf{m})$, then

$\displaystyle\left\langle{f_{\bf a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf b}}\right\rangle=1.$ (117)

  6. 7.

If $f_{\bf a}\in\mathcal{V}^{0}({\bf m})$ so that its set of elementary
column-modifiers is $E^{0}=\\{m_{\kappa(i,j)}=m_{i,j}\\}$, then

$\displaystyle\langle{f_{\bf
a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes n}{f_{\bf 0}}\rangle$
$\displaystyle=\exp\left(-\frac{\|\mathfrak{z}\|^{2}_{2}}{2}\right)\prod_{1\leq
i\leq r,i<j\leq
d}\left(\frac{\mathfrak{z}_{i,j}}{\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}}\right)^{m_{i,j}}r(n),$
(118)

where

$\|\mathfrak{z}\|^{2}_{2}=\sum_{1\leq i\leq r,i<j\leq
d}|\mathfrak{z}|^{2}_{ij}$

and the error factor is given by

$r(n)=1+O\left(n^{-1+2\beta+\eta}\delta^{-1},n^{-1/2+2\beta}\delta^{-1},n^{-1+2\beta+\alpha}\delta^{-1}\right).$

  7. 8.

If ${f_{\bf a}}\in\mathcal{V}^{E}({\bf m})$, so that its set of column-
modifiers is $E=\\{m_{\kappa}:\kappa\\}$ and $\Gamma(E)=\Gamma$, then

$\left|\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes
n}{f_{\bf
0}}\rangle\right|\leq\exp\left(-\frac{\|\mathfrak{z}\|^{2}_{2}}{2}\right)\left(\frac{C\|\mathfrak{z}\|}{\sqrt{n\delta}}\right)^{\Upsilon}\prod_{1\leq
i\leq r,i<j\leq
d}\left(\frac{\mathfrak{z}_{i,j}}{\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}}\right)^{m_{\kappa({i,j})}}r(n)$
(119)

where

$\Upsilon=-\Gamma+\sum_{1\leq i\leq r,i<j\leq d}(m_{i,j}-m_{\kappa({i,j})}),$

$C=C(d)$ a constant and $r(n)$ as in point 7 above.

  8. 9.

Under the further hypotheses that $\|\mathbf{z}\|\leq n^{\beta}$, $m_{i,j}\leq
2|\mathfrak{z}_{i,j}+z_{i,j}|n^{\beta+\epsilon}$ for some $\epsilon>0$, we
have:

$\displaystyle\left\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a}}\Bigg{|}q_{\lambda}U^{r}(\mathfrak{z}+\mathbf{z}/\sqrt{n}){f_{\bf
0}}\right\rangle$
$\displaystyle=\exp\left(-\frac{\|\mathfrak{z}+\mathbf{z}\|^{2}_{2}}{2}\right)\prod_{1\leq
i\leq r,i<j\leq
d}\frac{\big{(}(\mathfrak{z}_{i,j}+z_{i,j})(\sqrt{n}\sqrt{\mu_{i}-\mu_{j}})\big{)}^{m_{i,j}}}{m_{i,j}!}r(n),$
(120)

with

$r(n)=1+O\left(n^{-1+2\beta+\eta}\delta^{-1},n^{-1+2\beta+\alpha}\delta^{-1},n^{-1+2\eta}\delta^{-1},n^{-1+\alpha+\eta}\delta^{-1},\delta^{-3/2}n^{-1/2+3\beta+2\epsilon}\right).$

  9. 10.

Under the further hypotheses that $|\mathbf{l}|\leq|\mathbf{m}|$ and
$n^{1-3\eta}>2C/\delta^{2}$, where $C=C(d)$,

$\displaystyle\left|\left\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf
a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf b}}\right\rangle\right|$
$\displaystyle\leq(C|\mathbf{m}|)^{|\mathbf{m}|-|\mathbf{l}|}\prod_{1\leq
i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{l_{i,j}}}{l_{i,j}!}\left(\frac{C|\mathbf{l}|^{2}|\mathbf{m}|}{n\delta^{2}}\right)^{\Gamma^{a}_{\min}(\mathbf{l},\mathbf{m})}$
(121)

with

$\displaystyle\Gamma^{a}_{\min}(\mathbf{l},\mathbf{m})\geq\frac{\big{(}|\mathbf{l}-\mathbf{m}|+3|\mathbf{l}|-3|\mathbf{m}|\big{)}_{+}}{6}.$
(122)

  10. 11.

We have

$\displaystyle\left\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf b}}\right\rangle$
$\displaystyle=\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{m_{i,j}}}{m_{i,j}!}\big{(}1+O(n^{3\eta-1}/\delta)\big{)}.$
(123)

###### Proof.

_Proof of ( 109)._ Writing
$f_{\mathbf{a}}=f_{\mathbf{a}(1)}\otimes\ldots\otimes f_{\mathbf{a}(n)}$ (and
similarly for ${f_{\bf b}}$) we obtain:

$\displaystyle\langle{f_{\bf a}}|U^{\otimes n}{f_{\bf b}}\rangle$
$\displaystyle=\prod_{1\leq c\leq\lambda_{1}}\prod_{1\leq r\leq{l(c)}}\langle
f_{{t_{\bf a}^{c}}(r)}|Uf_{{t_{\bf b}^{c}}(r)}\rangle$
$\displaystyle=\prod_{1\leq c\leq\lambda_{1}}\prod_{1\leq
r\leq{l(c)}}U_{{t_{\bf a}^{c}}(r),{t_{\bf b}^{c}}(r)}.$

Since the subgroup of column permutations $\mathcal{C}_{\lambda}$ is the
product of the permutation groups of each column, each
$\sigma\in\mathcal{C}_{\lambda}$ can be decomposed as $\sigma=s_{1}\dots
s_{\lambda_{1}}$ with ${s_{c}}$ a permutation of column $c$ which transforms
${t_{\bf b}^{c}}(r)$ into ${t_{\bf b}^{c}}({s_{c}}(r))$. Then we have

$\displaystyle\langle{f_{\bf a}}|q_{\lambda}U^{\otimes n}{f_{\bf
b}}\rangle=\langle{f_{\bf a}}|U^{\otimes n}q_{\lambda}{f_{\bf b}}\rangle$
$\displaystyle=\sum_{\sigma\in\mathcal{C}_{\lambda}}\epsilon(\sigma)\prod_{1\leq
c\leq\lambda_{1}}\prod_{1\leq r\leq{l(c)}}U_{{t_{\bf a}^{c}}(r),{t_{\bf
b}^{c}}({s_{c}}(r))}$ $\displaystyle=\prod_{1\leq
c\leq\lambda_{1}}\sum_{{s_{c}}\in S_{c}}\epsilon({s_{c}})\prod_{1\leq
r\leq{l(c)}}U_{{t_{\bf a}^{c}}(r),{t_{\bf b}^{c}}({s_{c}}(r))}$
$\displaystyle=\prod_{1\leq c\leq\lambda_{1}}\det(U^{{t_{\bf a}^{c}},{t_{\bf
b}^{c}}}).$

_Proof of ( 111)._ To find the number of admissible ${f_{\bf a}}$ we multiply
the possibilities of the three stages of the algorithm. For $\Gamma({f_{\bf
a}})=0$ the first two stages yield $1$ as the combinatorial factor. Hence
$\\#\mathcal{V}^{0}$ is the number of possibilities at the third stage of the
algorithm. Thus from (106) we obtain the following upper bound for
$\\#\mathcal{V}^{0}({\bf m})$: $\prod_{1\leq i\leq r,i<j\leq
d}(\lambda_{i}-\lambda_{j})^{m_{i,j}}/m_{i,j}!$. Also, we note that from the
conditions (110) $\lambda_{i}-\lambda_{j}\geq\delta n/2$ and $|{\bf m}|\leq
n^{\eta}$, so that using (107) as a lower bound we obtain (111).

_Proof of ( 112)._ Since $E$ is fixed, the number of ${f_{\bf a}}$ in
$\mathcal{V}^{E}$ is given by the third stage of the algorithm. We then apply
(106) with $\sum_{\kappa}m_{\kappa}=|\mathbf{m}|-\Gamma$. Finally we neglect
the $m_{\kappa}!$ factors to obtain (112).

_Proof of ( 113)._ Recalling (108), we note that it is sufficient to give an
upper bound to $\\#\mathcal{V}^{E}$ since the first two stages of the
algorithm yield a combinatorial factor of at most $C^{\Gamma}$, with $C=C(d)$.
Since $\sum m_{\kappa({i,j})}\geq|\mathbf{m}|-2\Gamma$, we have

$\prod_{\kappa}m_{\kappa(i,j)}!\geq\prod_{1\leq i\leq r,i<j\leq
d}m_{i,j}!M^{-2\Gamma}$

where $M=\sup_{1\leq i\leq r,i<j\leq d}m_{i,j}\leq|\mathbf{m}|$. Note that
$\lambda_{i}-\lambda_{j}\geq\delta n/2$ and using equation ((112)) in
conjunction with the above result we obtain

$\\#\mathcal{V}^{E}\leq
n^{-\Gamma}\delta^{-2\Gamma}|\mathbf{m}|^{2\Gamma}\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{m_{i,j}}}{m_{i,j}!},\qquad\forall
E\leavevmode\nobreak\ {\rm with}\leavevmode\nobreak\ \Gamma(E)=\Gamma.$

Finally we multiply by the number of possible $E$ (i.e. $C^{\Gamma}$) to
obtain the desired result.

_Proof of ( 116)._ Denote by ${t_{\bf a}^{c}}([1,l(c)])$ the set of entries
$\\{{t_{\bf a}^{c}}(1),\dots,{t_{\bf a}^{c}}(l(c))\\}$ and consider equation
(109) with $U=\mathbf{1}$. Since both ${f_{\bf a}}$ and ${f_{\bf b}}$ are
tensor product of basis vectors, the scalar product $\langle{f_{\bf a}}\mid
q_{\lambda}{f_{\bf b}}\rangle$ is equal to $-1$ or $1$ if ${t_{\bf
a}^{c}}([1,l(c)])={t_{\bf b}^{c}}([1,l(c)])$ for all columns, and $0$
otherwise. Since a modified column cannot satisfy ${t_{\bf
a}^{c}}([1,l(c)])=[1,l(c)]$ (and the same for ${\bf b}$) the condition given
in the last line implies $\langle{f_{\bf a}}\mid q_{\lambda}{f_{\bf
b}}\rangle=0$ unless $t_{\bf{a}}$ and $t_{\bf{b}}$ have the same number of
modified columns. Since the total number of unmodified columns for ${f_{\bf
a}}$ and ${f_{\bf b}}$ are $|\mathbf{l}|-\Gamma^{a}$ and
$|\mathbf{m}|-\Gamma^{b}$ respectively, we obtain the first line of (116).

Now consider the case when $\Gamma^{b}=|\mathbf{m}|-|\mathbf{l}|+\Gamma^{a}$.
Since $|\langle{f_{\bf a}}\mid q_{\lambda}{f_{\bf b}}\rangle|\leq 1$, we can
bound the LHS of (116) by the number of non-zero inner products
$\langle{f_{\bf a}}\mid q_{\lambda}{f_{\bf b}}\rangle$. In other words we want
to count the number of $t_{\bf{b}}$ such that $t_{\bf a}^{c}([1,l(c)])=t_{\bf
b}^{c}([1,l(c)])$, or equivalently $S(\kappa_{\bf a}^{c})=S(\kappa_{\bf
b}^{c})$. For building the relevant ${f_{\bf b}}$, we can follow the algorithm
with the further condition that, at stage three the last condition (i.e.
$S(\kappa_{\bf a}^{c})=S(\kappa_{\bf b}^{c})$) is met.

The first two stages of the algorithm yield a $C^{\Gamma^{b}}$ factor. In the
third stage, for each $S$, we identify the column modifiers
$\kappa_{1},\dots,\kappa_{r(S)}$ such that $S(\kappa_{i})=S$ for all $1\leq
i\leq r(S)$. The total number of such objects is $m_{S}:=\sum_{i\leq
r(S)}m_{\kappa_{i}}$ and the total number of ways in which they can be
inserted to produce distinct tableaux is

$\left(\begin{array}[]{c}m_{S}\\\ m_{\kappa_{1}}\dots
m_{\kappa_{r(S)}}\end{array}\right).$

Recall that the total number of elementary column-modifiers $\sum_{1\leq i\leq
r,i<j\leq d}m_{\kappa(i,j)}$ can be lower bounded by $|{\bf m}|-2\Gamma^{b}$
and that each elementary column-modifier $\kappa(i,j)$ corresponds to a
different $S(\kappa(i,j))=\\{(i,-),(j,+)\\}$. Thus

$|{\bf m}|-2\Gamma^{b}\leq\sum_{1\leq i\leq r,i<j\leq
d}m_{\kappa(i,j)}\leq\sum_{S}\max_{\kappa:S(\kappa)=S}m_{\kappa}.$

Since

$\sum_{S}m_{S}=\sum_{\kappa}m_{\kappa}=|{\bf m}|-\Gamma^{b},$

we obtain

$\sum_{S}\left(m_{S}-\max_{\kappa:S(\kappa)=S}m_{\kappa}\right)\leq\Gamma^{b}.$

This implies

$\prod_{S}\left(\begin{array}[]{c}m_{S}\\\ m_{\kappa_{1}}\dots
m_{\kappa_{r(S)}}\end{array}\right)\leq|\mathbf{m}|^{\Gamma^{b}}.$

Again we multiply by the $C^{\Gamma^{b}}$ of the first two stages and get
(116).

_Proof of ( 117)._ From (116) the only non-zero contributions come from
$f_{\bf b}\in\mathcal{V}^{0}\subset\mathcal{O}_{\lambda}({\bf m})$. Since
$\Gamma^{b}=0$, the combinatorial factor from the two first stages of the
algorithm is $1$. Also since all the column modifiers are elementary we have
$m_{S}=m_{i,j}=m_{\kappa({i,j})}$ for all $S$ (i.e. $0$ for the non-elementary
ones) and the combinatorial factor is again one. In other words, the only
${f_{\bf b}}$ such that $\langle{f_{\bf a}}\mid q_{\lambda}{f_{\bf
b}}\rangle\neq 0$ is ${f_{\bf a}}$. Finally it is easy to check that
$\langle{f_{\bf a}}\mid q_{\lambda}{f_{\bf a}}\rangle=1$.

_Proof of ( 118)._ From (109) we have

$\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes n}{f_{\bf
0}}\rangle=\prod_{1\leq c\leq\lambda_{1}}{\rm det}(U^{t_{\bf a}^{c},{\rm
Id}^{c}}),\qquad U=U^{r}(\mathfrak{z}/\sqrt{n}).$

Now consider the following Taylor expansions. Entry-wise, for all $1\leq i\leq
r$ on the first line, and all $1\leq i\leq r,i<j\leq d$ on the second and
third lines:

$\displaystyle U^{r}_{i,i}(\mathfrak{z}/\sqrt{n})$ $\displaystyle=$
$\displaystyle 1-\frac{1}{2n}\sum_{i<j\leq
d}\frac{|\mathfrak{z}_{i,j}|^{2}}{\mu_{i}-\mu_{j}}-\frac{1}{2n}\sum_{1\leq
j<i}\frac{|\mathfrak{z}_{j,i}|^{2}}{\mu_{i}-\mu_{j}}+O(\|\mathfrak{z}\|^{3}n^{-3/2}\delta^{-3/2});$
$\displaystyle U^{r}_{i,j}(\mathfrak{z}/\sqrt{n})$ $\displaystyle=$
$\displaystyle-\frac{1}{\sqrt{n}}\frac{\mathfrak{z}_{i,j}^{*}}{\sqrt{\mu_{i}-\mu_{j}}}+O(\|\mathfrak{z}\|^{2}n^{-1}\delta^{-1});$
$\displaystyle U^{r}_{j,i}(\mathfrak{z}/\sqrt{n})$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{n}}\frac{\mathfrak{z}_{i,j}}{\sqrt{\mu_{i}-\mu_{j}}}+O(\|\mathfrak{z}\|^{2}n^{-1}\delta^{-1}).$

If $\mathfrak{z}=O(n^{\beta})$ and $\beta<1/2$, the remainder terms are
$O(n^{-3/2+3\beta}\delta^{-3/2})$ for the first line and
$O(n^{-1+2\beta}\delta^{-1})$ for the last two lines. We use the Taylor
expansions given above to find the dominating terms in the expansion $\det
A=\sum_{\sigma}\prod_{i}\epsilon(\sigma)A_{i,\sigma(i)}.$

Since ${f_{\bf a}}\in\mathcal{V}^{0}$, all ${t_{\bf a}^{c}}$ are either ${\rm
Id}^{c}$, or an _$(i,j)$ -substitution_. If ${t_{\bf a}^{c}}={\rm Id}^{c}$,
the summands with more than two non-diagonal terms are of the same order as
the remainder term, so we need to take into account only the identity and the
transpositions in the sum $\sum_{\sigma}\prod_{i}A_{i,\sigma(i)}$. Let
$l=l(c)$, then

$\upsilon(l):=\det(U^{{\rm Id}^{c},{\rm
Id}^{c}}(\mathfrak{z}/\sqrt{n}))=1-\frac{1}{2n}\sum_{\begin{subarray}{c}1\leq
i\leq l\\\ l+1\leq j\leq
d\end{subarray}}\frac{|\mathfrak{z}_{i,j}|^{2}}{\mu_{i}-\mu_{j}}+O(n^{-3/2+3\beta}\delta^{-3/2}).$

Consider now the case ${t_{\bf a}^{c}}\neq{\rm Id}^{c}$. Since ${t_{\bf
a}^{c}}(k)\geq k$ for all $k$, there exists a whole column of $U^{{t_{\bf
a}^{c}},{\rm Id}^{c}}$ whose entries are smaller in modulus than
$O(\|\mathfrak{z}\|/\sqrt{n\delta})=O(n^{-1/2+\beta}\delta^{-1})$. In
particular if ${t_{\bf a}^{c}}$ is an _$(i,j)$ -substitution_, then the only
summand that is of this order comes from the identity permutation. So that

$\upsilon(i,j):=\det(U^{{t_{\bf a}^{c}},{\rm
Id}^{c}}(\mathfrak{z}/\sqrt{n}))=\frac{\mathfrak{z}_{i,j}}{\sqrt{n(\mu_{i}-\mu_{j})}}+O(n^{-1+2\beta}\delta^{-1}).$
(124)

Note that each $1\leq i\leq r,i<j\leq d$ there are $m_{i,j}$ columns of the
type $(i,j)$-substitution. Out of the $\lambda_{l}-\lambda_{l+1}$ columns of
length $l=l(c)$, there are $\lambda_{l}-\lambda_{l+1}-R_{l}$ of the type ${\rm
Id}^{c}$, with $0\leq R_{l}\leq\lvert{\bf m}\rvert$. We are now ready to
expand (109) in terms of our estimates.

$\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes n}{f_{\bf
0}}\rangle=\prod_{l=1}^{r}\left(\upsilon(l))\right)^{\lambda_{l}-\lambda_{l+1}}\prod_{1\leq
i\leq r,i<j\leq
d}\left(\upsilon(i,j)\right)^{m_{i,j}}\prod_{l=1}^{r}(\upsilon(l))^{-R_{l}}.$
(125)

Now $\upsilon(l)=1+O(n^{-1+2\beta}\delta^{-1})$ and $R_{l}\leq|{\bf m}|\leq
n^{\eta}$, so the last product is $1+O(n^{-1+2\beta+\eta}\delta^{-1})$.
Similarly, since $\lambda\in\Lambda_{2}$ we have
$\lambda_{l}-\lambda_{l+1}=n(\mu_{l}-\mu_{l+1})+O(n^{\alpha})$. We need the
following lemma (see [35]) for an exponential approximation:

######  Lemma D.2.

If $x_{n}=O(n^{1/2-\epsilon})$, then

$\left(1+\frac{x_{n}}{n}\right)^{n}=\exp(x_{n})(1+O(n^{-\epsilon})).$

Using Lemma D.2 we estimate the first product in (125) as follows

$\displaystyle\prod_{l=1}^{r}\upsilon(l)^{\lambda_{l}-\lambda_{l+1}}$
$\displaystyle=\prod_{l=1}^{r}\exp\left(-\frac{1}{2}\sum_{\begin{subarray}{c}1\leq
i\leq l\\\ l+1\leq j\leq
d\end{subarray}}|\mathfrak{z}_{i,j}|^{2}\frac{\mu_{l}-\mu_{l+1}}{(\mu_{i}-\mu_{j})}\right)r(n)$
$\displaystyle=\exp\left(-\frac{1}{2}\sum_{\begin{subarray}{c}1\leq i\leq r\\\
i<j\leq
d\end{subarray}}\sum_{l=i}^{r\wedge(j-1)}|\mathfrak{z}_{i,j}|^{2}\frac{\mu_{l}-\mu_{l+1}}{(\mu_{i}-\mu_{j})}\right)r(n)$
$\displaystyle=\exp\left(-\frac{\|\mathfrak{z}\|^{2}_{2}}{2}\right)r(n),$
(126)

with

$\displaystyle r(n)$ $\displaystyle=$ $\displaystyle
1+O(n^{-1+\alpha+2\beta}\delta^{-1},n^{-1/2+2\beta}\delta^{-1}),$
$\displaystyle\|\mathfrak{z}\|^{2}_{2}$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}1\leq i\leq r\\\ i<j\leq
d\end{subarray}}|\mathfrak{z}_{i,j}|^{2}.$

Here we have used the relation that

$\sum_{l=i}^{r\wedge(j-1)}\frac{\mu_{l}-\mu_{l+1}}{(\mu_{i}-\mu_{j})}=1$

which is trivial for $j\leq r$ and for $j>r$ we have used the fact that both
$\mu_{j}$ and $\mu_{r+1}$ are $0$. We now estimate the middle product in the
RHS of (125) as

$\upsilon(i,j)^{m_{i,j}}=\left(\frac{\mathfrak{z}_{i,j}}{\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}}\right)^{m_{i,j}}\left(1+O\left(n^{-1+2\beta+\eta}\delta^{-1}\right)\right),$
(127)

where we have used that $|\mathbf{m}|\leq n^{\eta}$. From (125), (126) and
(127), we obtain (118).

_Proof of ( 119)._ We have the following analog of (125),

$\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes n}{f_{\bf
0}}\rangle=\prod_{l=1}^{r}\left(\upsilon(l))\right)^{\lambda_{l}-\lambda_{l+1}}\prod_{\kappa}\left(\upsilon(\kappa)\right)^{m_{\kappa}}\prod_{l=1}^{r}(\upsilon(l))^{-R_{l}}$

where $0\leq R_{l}\leq|{\bf m}|-\Gamma$ and $\upsilon(\kappa)$ is the
determinant of the minor of $U$ corresponding to the column where $\kappa$ is
applied. We further split the column-modifiers into elementary
($\kappa=\kappa(i,j)$) and non-elementary (say $\kappa=\kappa^{\prime})$ ones.
Then $\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes
n}{f_{\bf 0}}\rangle$ can be written as

$\prod_{l=1}^{r}\left(\upsilon(l))\right)^{\lambda_{l}-\lambda_{l+1}}\prod_{1\leq
i\leq r,i<j\leq
d}\left(\upsilon(i,j)\right)^{m_{\kappa(i,j)}}\prod_{l=1}^{r}(\upsilon(l))^{-R_{l}}\prod_{\kappa^{\prime}}\left(\upsilon(\kappa^{\prime})\right)^{m_{\kappa^{\prime}}}.$

The first three products on the RHS can be approximated like the terms in
(125). To give an upper bound to the fourth product, we note that if the
entries in the column have been modified in an admissible way, then ${t_{\bf
a}^{c}}(i)=j>l(c)$ for some $i$, so that $|\upsilon(\kappa)|\leq
C\|\mathfrak{z}\|/\sqrt{n\delta}$ for any $\kappa$, with some constant
$C=C(d)$.

Thus we have

$\left|\langle{f_{\bf a}}|q_{\lambda}U^{r}(\mathfrak{z}/\sqrt{n})^{\otimes
n}{f_{\bf
0}}\rangle\right|\leq\exp\left(-\frac{\|\mathfrak{z}\|_{2}^{2}}{2}\right)\left(\frac{C\|\mathfrak{z}\|}{\sqrt{n\delta}}\right)^{\sum_{\kappa^{\prime}}m_{\kappa^{\prime}}}\prod_{\begin{subarray}{c}1\leq
i\leq r\\\ i<j\leq
d\end{subarray}}\left(\frac{|\mathfrak{z}_{i,j}|}{\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}}\right)^{m_{\kappa(i,j)}}r(n).$
(128)

We obtain (119) by noting that the number of non-elementary modifiers is

$\Upsilon=\sum_{\kappa^{\prime}}m_{\kappa^{\prime}}=-\Gamma+\sum_{1\leq i\leq
r,i<j\leq d}(m_{i,j}-m_{\kappa(i,j)}).$

_Proof of ( 9)._

Recall that we identify $\mathbf{z}$ with the vector $(\mathbf{z},0)$. We
shall split the sum over $\mathcal{O}_{\lambda}({\bf m})$ into sub-sums using
the decomposition $\mathcal{O}_{\lambda}({\bf
m})\bigcap\mathcal{V}=\bigcup_{E}\mathcal{V}^{E}({\bf m})$, and compare each
sub-sum against $\mathcal{V}^{0}=\mathcal{V}^{E^{0}}$. Then we aggregate all
the $E$ corresponding to a particular $\Gamma$ (which results in a
multiplication by a combinatorial factor) and finally sum over different
$\Gamma$. From the given bounds on $\mathfrak{z}$ and $\mathbf{z}$ we note
that $\|\mathfrak{z}+\mathbf{z}\|=O(n^{\beta})$, so we can apply the previous
results with $\mathfrak{z}+\mathbf{z}$ instead of $\mathfrak{z}$.

Using (111) and (118) and recalling that $\lambda\in\Lambda_{2}$, we get:

$\displaystyle\left\langle\sum_{{f_{\bf a}}\in\mathcal{V}^{0}}{f_{\bf
a}}\Bigg{|}q_{\lambda}U^{r}(\mathfrak{z}+\mathbf{z}/\sqrt{n})^{\otimes
n}{f_{\bf 0}}\right\rangle$ $\displaystyle=$
$\displaystyle\exp\left(-\frac{\|\mathfrak{z}+\mathbf{z}\|^{2}_{2}}{2}\right)\prod_{1\leq
i\leq r,i<j\leq
d}\frac{\big{(}(\mathfrak{z}_{i,j}+z_{i,j})\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}\big{)}^{m_{i,j}}}{m_{i,j}!}r(n)$

with error factor

$r(n)=1+O\left(n^{-1+2\beta+\eta}\delta^{-1},n^{-1/2+2\beta}\delta^{-1},n^{-1+2\beta+\alpha}\delta^{-1},n^{-1+2\eta}\delta^{-1},n^{-1+\alpha+\eta}\delta^{-1}\right).$

For $E\neq E^{0}$ we combine (119) and (112) to obtain

$\displaystyle\left|\left\langle\sum_{{f_{\bf a}}\in\mathcal{V}^{E}}{f_{\bf
a}}\Bigg{|}q_{\lambda}U(\mathfrak{z}+\mathbf{z}/\sqrt{n}){f_{\bf
0}}\right\rangle\right|\cdot\left|\left\langle\sum_{{f_{\bf
a}}\in\mathcal{V}^{0}}{f_{\bf
a}}\Bigg{|}q_{\lambda}U(\mathfrak{z}+\mathbf{z}/\sqrt{n}){f_{\bf
0}}\right\rangle\right|^{-1}$ $\displaystyle\leq$ $\displaystyle
n^{-\Gamma}\prod_{1\leq i\leq r,i<j\leq
d}\left(\frac{\lambda_{i}-\lambda_{j}}{n}\right)^{m_{\kappa({i,j})}-m_{i,j}}\frac{m_{i,j}!}{m_{\kappa({i,j})}!}\left(\frac{\|\mathfrak{z}+\mathbf{z}\|}{\sqrt{\delta
n}}\right)^{-\Gamma}$ $\displaystyle\prod_{1\leq i\leq r,i<j\leq
d}\left(\frac{\sqrt{\delta
n}|\mathfrak{z}_{i,j}+z_{i,j}|}{\|\mathfrak{z}+\mathbf{z}\|\sqrt{n}\sqrt{\mu_{i}-\mu_{j}}}\right)^{m_{\kappa({i,j})}-m_{i,j}}r(n)$
$\displaystyle\leq$ $\displaystyle
O(n^{-\Gamma(1/2+\beta)})\delta^{-\Gamma/2}\prod_{1\leq i\leq r,i<j\leq
d:m_{i,j}\neq
0}\left(\frac{|\mathfrak{z}_{i,j}+z_{i,j}|\sqrt{\mu_{i}-\mu_{j}}}{m_{i,j}\|\mathfrak{z}+\mathbf{z}\|}\right)^{m_{\kappa(i,j)}-m_{i,j}}$
$\displaystyle\leq$ $\displaystyle
O\big{(}(2\delta^{-3/2}n^{-1/2+3\beta+2\epsilon})^{\Gamma}\big{)},$

with $O(\cdot)$ uniform in $\Gamma$. In the second inequality we have used

$m_{i,j}!/m_{\kappa(i,j)}!\leq
m_{i,j}^{m_{i,j}-m_{\kappa(i,j)}},\qquad\sum_{1\leq i\leq r,i<j\leq
d}(m_{\kappa({i,j})}-m_{i,j})\geq-2\Gamma,\qquad\lambda\in\Lambda_{2}$

and in the third inequality we have used

$m_{i,j}\leq
2|\mathfrak{z}_{i,j}+z_{i,j}|n^{\beta+\epsilon},\qquad\frac{|\mathfrak{z}_{i,j}+z_{i,j}|\sqrt{\mu_{i}-\mu_{j}}}{m_{i,j}\|\mathfrak{z}+\mathbf{z}\|}\leq
1.$

Again note that for a given $\Gamma$, there are at most $C^{\Gamma}$ different
$E$ such that $\Gamma(E)=\Gamma$ (where $C=C(d)$) corresponding to the
possible choices in the first two stages of the algorithm, . Hence, if $n$ is
large enough such that $2C\delta^{-3/2}n^{-1/2+3\beta+2\epsilon}<1$, we have:

$\displaystyle\left\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a}}\Bigg{|}q_{\lambda}U(\mathfrak{z}+\mathbf{z}/\sqrt{n}){f_{\bf
0}}\right\rangle=\sum_{\Gamma}\left\langle\sum_{{f_{\bf
a}}\in\mathcal{V}^{\Gamma}}{f_{\bf
a}}\Bigg{|}q_{\lambda}U(\mathfrak{z}+\mathbf{z}/\sqrt{n}){f_{\bf
0}}\right\rangle$
$\displaystyle=\Big{(}1+O(\delta^{-3/2}n^{-1/2+3\beta+2\epsilon})\Big{)}\exp\left(-\frac{\|\mathfrak{z}+\mathbf{z}\|^{2}_{2}}{2}\right)\prod_{1\leq
i\leq r,i<j\leq
d}\frac{\big{(}(\mathfrak{z}+z)_{i,j}(\sqrt{n}\sqrt{\mu_{i}-\mu_{j}})\big{)}^{m_{i,j}}}{m_{i,j}!}r(n)$
$\displaystyle=\exp\left(-\frac{\|\mathfrak{z}+\mathbf{z}\|^{2}_{2}}{2}\right)\prod_{1\leq
i\leq r,i<j\leq
d}\frac{\big{(}(\mathfrak{z}+z)_{i,j}(\sqrt{n}\sqrt{\mu_{i}-\mu_{j}})\big{)}^{m_{i,j}}}{m_{i,j}!}r_{2}(n)$

where the sum over $\Gamma$ was bounded using a geometric series and

$r_{2}(n)=1+O\left(n^{-1+2\beta+\eta}\delta^{-1},n^{-1+\alpha+\beta}\delta^{-1},n^{-1+2\eta}\delta^{-1},n^{-1+\alpha+\eta}\delta^{-1},\delta^{-3/2}n^{-1/2+3\beta+2\epsilon}\right).$

Thus we obtain (9).

_Proof of ( 121)._ We choose $\Gamma^{a}$ and $\Gamma^{b}$ satisfying the
condition $\Gamma^{b}-\Gamma^{a}=|{\bf m}|-|{\bf l}|$ to ensure that the inner
products in (116) are non-zero. By multiplying (113) and (116), we see that:

$\displaystyle\left|\left\langle\sum_{{f_{\bf
a}}\in\mathcal{V}^{\Gamma^{a}}(\mathbf{l})}{f_{\bf
a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf b}}\in\mathcal{V}^{\Gamma^{b}}({\bf
m})}{f_{\bf b}}\right\rangle\right|$
$\displaystyle\leq(C|\mathbf{m}|)^{\Gamma^{b}}\\!\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{l_{i,j}}}{l_{i,j}!}\left(\frac{C|\mathbf{l}|^{2}}{n\delta^{2}}\right)^{\Gamma^{a}}$
(129) $\displaystyle=(C|\mathbf{m}|)^{|\mathbf{m}|-|\mathbf{l}|}\prod_{1\leq
i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{l_{i,j}}}{l_{i,j}!}\left(\frac{C|\mathbf{l}|^{2}|\mathbf{m}|}{n\delta^{2}}\right)^{\Gamma^{a}}.$

Since $\Gamma^{b}$ is determined by $\Gamma^{a}$ it is sufficient to sum over
all possible $\Gamma^{a}$. If $n^{1-3\eta}>2C/\delta^{2}$, the dominating term
in the sum of bounds is that corresponding to the smallest possible
$\Gamma^{a}$. We aim to give a give a lower bound to $\Gamma^{a}$ under the
constraint that the inner product is non-zero.

As noted before the necessary condition for the inner product to be non-zero
is $m_{S}^{\bf a}=m_{S}^{\bf b}$. We consider a lower bound for
$\Gamma({f_{\bf a}})+\Gamma({f_{\bf b}})$ which will in turn give a lower
bound for $\Gamma({f_{\bf a}})$. We start with two tableaux $t_{\bf
a^{\prime}}\in\mathcal{O}_{\lambda}({\bf m})$ and $t_{\bf
b^{\prime}}\in\mathcal{O}_{\lambda}({\bf l})$ so that $\Gamma(f_{{\bf
a}^{\prime}})=\Gamma(f_{{\bf a}^{\prime}})=0$. Since $|{\bf m}|\leq n^{\eta}$
and $\lambda_{r}\geq\delta n+O(n^{\alpha})$, the tableau $t_{\bf a^{\prime}}$
can be chosen to have at most one modified box per column so the last
condition is met (and similarly for $t_{\bf b^{\prime}}$). We also ensure that
each of the modified columns of $t_{\bf a}$ (or $t_{\bf b}$) are also modified
in $t_{\bf a^{\prime}}$ (or $t_{\bf b^{\prime}}$). The idea is to count the
minimum number of ‘horizontal box shuffling’ operations necessary in order to
transform the Young tableau $t_{\bf a^{\prime}}$ into the tableau $t_{\bf a}$
and $t_{\bf b^{\prime}}$ into the given tableau $t_{\bf b}$ so that at the end
$m_{S}^{\bf a}=m_{S}^{\bf b}$.

The horizontal box shuffling is described as follows; at each step we
horizontally move one elementary column modifier $\kappa(i,j)$ of $t_{{\bf
a}^{\prime}}$ (or $t_{{\bf b}^{\prime}}$) into an already modified column,
with the aim of constructing $t_{\bf a}$ (or $t_{\bf b}$). Note that such
operation increases $\Gamma(f_{{\bf a}^{\prime}})+\Gamma(f_{{\bf
b}^{\prime}})$ by one and also changes $m_{S}^{{\bf a}^{\prime}}$ (or
$m_{S}^{{\bf b}^{\prime}}$) in the following way. The multiplicities
$m_{\\{(i,-),(j,+)\\}}$ and $m_{S_{0}}$ decrease by one, and
$m_{S_{0}+\\{(i,-),(j,+)\\}}$ increases by one. Here $S_{0}$ is the set of
modified entries for the column to which the box $(i,j)$ is moved. Hence the
distance $\sum_{S}|m_{S}^{{\bf a}^{\prime}}-m_{S}^{{\bf b}^{\prime}}|$
decreases by at most three. Since initially this quantity was equal to
$\sum_{1\leq i\leq r,i<j\leq d}|l_{i,j}-m_{i,j}|$ (because all the column
modifiers were elementary in $t_{{\bf a}^{\prime}}$ and $t_{{\bf
b}^{\prime}}$), we need at least $\sum_{1\leq i\leq r,i<j\leq
d}|{l}_{i,j}-{m}_{i,j}|/3$ such operations before reaching our goal
$m_{S}^{\bf a}=m_{S}^{\bf b}$. Hence we have $\Gamma({f_{\bf
a}})+\Gamma({f_{\bf b}})\geq|\mathbf{l}-\mathbf{m}|/3$.

Finally using $\Gamma^{b}-\Gamma^{a}=|{\bf m}|-|{\bf l}|$, we obtain
$\Gamma^{a}\geq(|\mathbf{l}-\mathbf{m}|+3|\mathbf{l}|-3|\mathbf{m}|)/6$ and
hence the desired result ((121)).

_Proof of ( 123)._ Since $\mathbf{l}=\mathbf{m}$, equations (111) and (117)
prove that the bound (129) is saturated when $\Gamma^{a}=0$, up to the error
factor $\left(1+O(n^{-1+2\eta}/\delta)\right)$. Hence the remainder term due
to the other $\Gamma$ consist in a geometric series with factor
$\left(\frac{C|\mathbf{m}|^{3}}{n\delta^{2}}\right)=O(n^{1-3\eta}/\delta^{2})$.

∎

###  D.2 Non-orthogonality issues and construction of $V^{r}_{\lambda}$

In this subsection we consider the basis vectors $|\bf{m}_{\lambda}\rangle$
and show that they are approximately orthogonal. We will use this result to
construct the isometry $V^{r}_{\lambda}$ and hence the channel
$T^{r}_{\lambda}$.

######  Lemma D.3.

Let $\mathbf{m}_{\lambda}$ and $\mathbf{l}_{\lambda}$ be two basis vectors
(corresponding to two semistandard Young tableaux with diagram $\lambda$) and
let $|{\bf m}|:=\sum_{1\leq i\leq r,i<j\leq d}m_{ij}$ and $|{\bf l}-{\bf
m}|:=\sum_{1\leq i\leq r,i<j\leq d}|l_{i,j}-m_{ij}|$.

If

$\sum_{j>i}m_{i,j}-\sum_{j<i}m_{j,i}\neq\sum_{j>i}l_{i,j}-\sum_{j<i}l_{j,i}$

for some $1\leq i\leq r$, then

$\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle=0.$

Otherwise, we derive an upper bound under the following conditions. We assume
that $\lambda_{i}-\lambda_{i+1}>\delta n$ for all $1\leq i\leq r-1$ and
$\lambda_{r}>\delta n$, for some $\delta>0$. Furthermore we assume
$|\mathbf{l}|\leq|\mathbf{m}|\leq n^{\eta}$ for some $\eta<1/3$ and that
$Cn^{3\eta-1}/\delta^{2}<1$ where $C=C(d)$ is a constant.

Then:

$\left|\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle\right|\leq(C^{\prime}n)^{-\eta(|{\bf
m}|-|{\bf l}|)/4}\,(C^{\prime}n)^{(9\eta-2)|{\bf m}-{\bf
l}|/12}\,\delta^{(|\mathbf{m}|-|\mathbf{l}|)/2-|\mathbf{m}-\mathbf{l}|/3}\,(1+O(n^{-1+3\eta}/\delta)).$
(130)

where $C^{\prime}=C^{\prime}(d,\eta)$ and the constant in the remainder term
depends only on $d$. The right side is of order less than $n^{(9\eta-2)|{\bf
m}-{\bf l}|/12}$ and converges to zero for $\eta<2/9$ when $n\to\infty$.

###### Proof.

We know that $\left|\mathbf{m}_{\lambda}\right\rangle$ is a linear combination
of $n$-tensor product vectors in which the basis vector $f_{i}$ appears
exactly $\lambda_{i}-\sum_{j>i}m_{i,j}+\sum_{j<i}m_{j,i}$ times for $1\leq
i\leq r$. As two tensor basis vectors are orthogonal if they do not have the
same number of $f_{i}$ in the decomposition, we get that
$\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle=0$ if
$\sum_{j>i}m_{i,j}-\sum_{j<i}m_{j,i}\neq\sum_{j>i}l_{i,j}-\sum_{j<i}l_{j,i}$
for any $1\leq i\leq r$.

Now consider the following expansion of the inner product,

$\displaystyle\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle$
$\displaystyle=\frac{\langle q_{\lambda}p_{\lambda}{f_{\bf
m}}|q_{\lambda}p_{\lambda}{f_{\bf l}}\rangle}{\sqrt{\langle
q_{\lambda}p_{\lambda}{f_{\bf m}}|q_{\lambda}p_{\lambda}{f_{\bf
m}}\rangle\langle q_{\lambda}p_{\lambda}{f_{\bf
l}}|q_{\lambda}p_{\lambda}{f_{\bf l}}\rangle}}$ (131)
$\displaystyle=\frac{\langle p_{\lambda}{f_{\bf
m}}|q_{\lambda}p_{\lambda}{f_{\bf l}}\rangle}{\sqrt{\langle p_{\lambda}{f_{\bf
m}}|q_{\lambda}p_{\lambda}{f_{\bf m}}\rangle\langle p_{\lambda}{f_{\bf
l}}|q_{\lambda}p_{\lambda}{f_{\bf l}}\rangle}}$ (132)

where we have used the fact that $q_{\lambda}$ is a projection (use (60) and
(61) except that we have only $r$ rows in our Young diagrams), up to a
constant factor. Now we decompose $p_{\lambda}f_{\bf m}$ and
$p_{\lambda}f_{\bf l}$ on orbits as in (105). Since the multiplicity of the
elements in the orbits are the same in numerator and denominator, we obtain:

$\displaystyle\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle=\frac{\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf a}}|q_{\lambda}\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf
b}}\rangle}{\sqrt{\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf a}}|q_{\lambda}\sum_{{f_{\bf
a^{\prime}}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a^{\prime}}}\rangle\langle\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf b}}|q_{\lambda}\sum_{{f_{\bf
b^{\prime}}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf
b^{\prime}}}\rangle}}.$ (133)

We use (123) for the denominator:

$\displaystyle\left\langle\sum_{{f_{\bf
a}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
a^{\prime}}}\in\mathcal{O}_{\lambda}(\mathbf{m})}{f_{\bf
a^{\prime}}}\right\rangle\Bigg{\langle}\sum_{{f_{\bf
b}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf
b}}\Bigg{|}q_{\lambda}\sum_{{f_{\bf
b^{\prime}}}\in\mathcal{O}_{\lambda}(\mathbf{l})}{f_{\bf
b^{\prime}}}\Bigg{\rangle}$ $\displaystyle=\prod_{1\leq i\leq r,i<j\leq
d}\frac{(\lambda_{i}-\lambda_{j})^{(m_{i,j}+l_{i,j})/2}}{\sqrt{m_{i,j}!\,l_{i,j}!}}(1+O(n^{3\eta-1}/\delta)),$

and the numerator is bounded as in (121). Then, under the assumption $|{\bf
m}|\geq|{\bf l}|$ we have

$\left|\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle\right|\leq(C|\mathbf{m}|)^{|\mathbf{m}|-|\mathbf{l}|}\left(\frac{C|\mathbf{m}|^{3}}{\delta^{2}n}\right)^{\Gamma_{min}}\cdot\prod_{1\leq
i\leq r,i<j\leq
d}(\lambda_{i}-\lambda_{j})^{(l_{i,j}-m_{i,j})/2}\sqrt{\frac{m_{i,j}!}{l_{i,j}!}}\cdot\left(1+O(n^{3\eta-1}/\delta)\right),$

where
$\Gamma_{min}=((|\mathbf{l}-\mathbf{m}|+3|\mathbf{l}|-3|\mathbf{m}|)/6)\wedge
0$.

The factorials can be bounded as

$\prod_{1\leq i\leq r,i<j\leq
d}\sqrt{\frac{m_{i,j}!}{l_{i,j}!}}\leq|\mathbf{m}|^{\sum(m_{i,j}-l_{i,j})_{+}/2}=|\mathbf{m}|^{(|\mathbf{m}-\mathbf{l}|+|\mathbf{m}|-|\mathbf{l}|)/4}.$

Since $|{\bf m}|\leq n^{\eta}$ and $Cn^{3\eta-1}/\delta^{2}<1$, we have

$\left(\frac{C|\mathbf{m}|^{3}}{\delta^{2}n}\right)^{\Gamma_{min}}\leq\left(\frac{Cn^{3\eta-1}}{\delta^{2}}\right)^{(|\mathbf{l}-\mathbf{m}|+3|\mathbf{l}|-3|\mathbf{m}|)/6}.$

Since $n\delta<\lambda_{i}-\lambda_{j}<n$ we have

$(\lambda_{i}-\lambda_{j})^{(l_{i,j}-m_{i,j})/2}\leq
n^{{(l_{i,j}-m_{i,j})/2}}\quad\text{ if }l_{i,j}\geq m_{i,j}$

and

$(\lambda_{i}-\lambda_{j})^{(l_{i,j}-m_{i,j})/2}\leq(n\delta)^{{(l_{i,j}-m_{i,j})/2}}\quad\text{
if }l_{i,j}<m_{i,j}$

so that we have

$\prod_{1\leq i\leq r,i<j\leq
d}(\lambda_{i}-\lambda_{j})^{(l_{i,j}-m_{i,j})/2}\leq n^{(|{\bf l}|-|{\bf
m}|)/2}\delta^{\sum(m_{i,j}-l_{i,j})_{+}/2}\leq(n\delta)^{(|{\bf l}|-|{\bf
m}|)/2}.$

The last inequality follows since $\delta<1$ and by assumption $|{\bf
l}|-|{\bf m}|\leq 0$.

The constant $C=C(d)$ can be replaced by another constant
$C^{\prime}=C^{\prime}(d,\eta)$ such that all powers of $n$ appear in the form
$(C^{\prime}n)^{\gamma}$. The above bound yields

$\displaystyle\left|\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle\right|\leq\delta^{(|\mathbf{m}|-|\mathbf{l}|)/2-|\mathbf{m}-\mathbf{l}|/3}(C^{\prime}n)^{-\eta(|{\bf
m}|-|{\bf l}|)/4}(C^{\prime}n)^{(9\eta-2)|{\bf m}-{\bf
l}|/12}(1+O(n^{-1+3\eta}/\delta))$

∎

We have the following lemma which shows that for a given
$\mathbf{m}_{\lambda}$, its overlap with $\mathbf{l}_{\lambda}$s is uniformly
small.

######  Lemma D.4.

Let $\eta<2/9$ and let $\mathbf{m}_{\lambda}$ be such that $|\mathbf{m}|\leq
n^{\eta}$. Assume as in Lemma D.3 that $\lambda_{i}-\lambda_{i+1}>\delta n$
for all $1\leq i\leq r-1$ and $\lambda_{r}>\delta n$, for some $\delta>0$, and
that $Cn^{3\eta-1}/\delta^{2}<1$ where $C=C(d)$ is a constant.

Then there exists a constant $C^{\prime\prime}=C^{\prime\prime}(d,\eta)$ such
that

$\sum_{\begin{subarray}{c}|\mathbf{l}|\leq n^{\eta}\\\
\mathbf{l}\neq\mathbf{m}\end{subarray}}\left|\langle\mathbf{m}_{\lambda}|\mathbf{l}_{\lambda}\rangle\right|\leq(C^{\prime\prime}n)^{(9\eta-2)/12}\delta^{-1/3}.$
(134)

###### Proof.

We break the sum into two parts ($|{\bf l}|\leq|{\bf m}|$ and $|{\bf l}|>|{\bf
m}|$), and we note it suffices to prove the statement under the additional
condition $|{\bf l}|\leq|{\bf m}|$ (otherwise we use (130) with the roles of
$\bf{l}$ and $\bf{m}$ exchanged).

We use (130) neglecting the terms containing $|{\bf m}|-|{\bf l}|$ in the
exponent which are less than $1$. Then the expression on the left side of
(134) is bounded from above by

$2\sum_{k\geq
1}N(k)\left[(C^{\prime}n)^{(9\eta-2)/12}\delta^{-1/3}\right]^{k}$

where $N(k)$ is the number of ${\bf l}$’s for which $|{\bf m}-{\bf l}|=k$.

Since there are at most $rd$ pairs $1\leq i\leq r,i<j\leq d$, there are at
most $(k+1)^{rd}$ different choices for the values $\\{|l_{i,j}-m_{i,j}|:1\leq
i\leq r,i<j\leq d\\}$ satisfying $\sum|l_{i,j}-m_{i,j}|=k$. Moreover, there
are $2^{rd}$ sign choices which fix ${\bf l}=\\{l_{i,j}\\}$ completely. Thus
$N(k)\leq(2(k+1))^{rd}\leq c^{k}$ for some constant $c$. Thus we have a
geometric series with the first term corresponding to the RHS of (134) and
hence the desired result is proved.

∎

We use this quasi-orthogonality to build an isometry
$V^{r}_{\lambda}:\mathcal{H}_{\lambda}\to\mathcal{F}^{r}$ which maps the
relevant finite-dimensional vectors $\left|{\bf m}_{\lambda}\right\rangle$
‘close’ to their Fock counterparts $\left|{\bf m}\right\rangle$, with
$|\mathbf{m}\rangle\in\mathcal{F}^{r}$. This is the aim of following two
lemmas.

######  Lemma D.5.

Let $A$ be a contraction (i.e. $A^{*}A\leq\bf{1}$) from a finite space
$\mathcal{H}$ to an infinite space $\mathcal{K}$. Then there is an
$R:\mathcal{H}\to\mathcal{K}$ such that $A+R$ is an isometry and ${\rm
Range}(A)\perp{\rm Range}(R)$.

As a consequence, for any unit vector $\phi$, we have
$\|R\phi\|^{2}=1-\|A\phi\|^{2}$.

###### Proof.

See [35]. ∎

######  Lemma D.6.

Let $\eta<2/9$. Suppose that $\lambda_{i}-\lambda_{i+1}\geq\delta n$ for all
$1\leq i\leq r$, with the convention $\lambda_{r+1}=0$. Then for
$n>n_{0}(\eta,\delta,d)$ there exists an isometry
$V^{r}_{\lambda}:\mathcal{H}_{\lambda}\to\mathcal{F}^{r}$ such that,
$V^{r}_{\lambda}|{\bf 0}_{\lambda}\rangle=|{\bf 0}\rangle$ and for
$|\mathbf{m}|\leq n^{\eta}$,

$\langle\mathbf{m}|V^{r}_{\lambda}=\frac{1}{\sqrt{1+(\tilde{C}n)^{(9\eta-2)/12}/\delta^{1/3}}}\left\langle\mathbf{m}_{\lambda}\right|$

where $\tilde{C}=\tilde{C}(\eta,d)$ is a particular constant. More precisely,
$n_{0}$ can be taken of the form $(C(d)/\delta^{2})^{1/(1-3\eta)}$.

###### Proof.

Let $A^{r}_{\lambda}:\mathcal{H}_{\lambda}\to\mathcal{F}^{r}$ be defined by

$A^{r}_{\lambda}:=\frac{1}{\sqrt{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}}\sum_{|\mathbf{l}|\leq
n^{\eta}}|\mathbf{l}\rangle\left\langle\mathbf{l}_{\lambda}\right|.$

where $|\mathbf{l}\rangle\in\mathcal{F}^{r}$. Then,

$\displaystyle A_{\lambda}^{r*}A^{r}_{\lambda}$
$\displaystyle=\frac{1}{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}\sum_{|\mathbf{l}|\leq
n^{\eta}}\left|\mathbf{l}_{\lambda}\right\rangle\left\langle\mathbf{l}_{\lambda}\right|$
$\displaystyle\leq\mathbf{1}_{\mathcal{H}_{\lambda}}.$

where the last inequality follows from the following argument. It is enough to
show that all eigenvalues of $A_{\lambda}^{r*}A^{r}_{\lambda}$ are smaller
than $1$. Let $\sum_{\bf m}c_{\bf m}\left|\mathbf{m}_{\lambda}\right\rangle$
be an eigenvector of $A_{\lambda}^{r*}A^{r}_{\lambda}$, and $a$ the
corresponding eigenvalue. Then by the linear independence of
$\left|\mathbf{m}_{\lambda}\right\rangle$ we get that for each ${\bf l}$

$\frac{1}{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}\sum_{|\mathbf{m}|\leq
n^{\eta}}\left\langle\mathbf{l}_{\lambda}|\mathbf{m}_{\lambda}\right\rangle
c_{\bf m}=ac_{\bf l}.$

If ${\bf l}_{0}$ is an index for which $|c_{\bf l}|$ is maximum, then by
taking absolute values on both sides and dividing by $|c_{\bf l_{0}}|$, we
obtain

$a\leq\frac{1}{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}\sum_{|\mathbf{m}|\leq
n^{\eta}}\left|\langle\mathbf{l}_{\lambda}|\mathbf{m}_{\lambda}\rangle\right|\leq
1$

where the last inequality follows from Lemma D.4. Now may apply Lemma D.5, and
find an $R^{r}_{\lambda}$ such that $A^{r}_{\lambda}+R^{r}_{\lambda}$ is an
isometry, and ${\rm Range}(R^{r}_{\lambda})\perp{\rm Range}(A^{r}_{\lambda})$,
so that $\langle\mathbf{m}|R_{\lambda}=0$. Defining
$V^{r}_{\lambda}:=A^{r}_{\lambda}+R^{r}_{\lambda}$, we have

$\displaystyle\langle\mathbf{m}|V^{r}_{\lambda}$ $\displaystyle=$
$\displaystyle\langle\mathbf{m}|(A^{r}_{\lambda}+R^{r}_{\lambda})$
$\displaystyle=$ $\displaystyle\langle\mathbf{m}|A^{r}_{\lambda}$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}}\langle\mathbf{m}|\sum_{|\mathbf{l}|\leq
n^{\eta}}|\mathbf{l}\rangle\left\langle\mathbf{l}_{\lambda}\right|$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+(Cn)^{(9\eta-2)/12}/\delta^{1/3}}}\left\langle\mathbf{m}_{\lambda}\right|.$

∎

###  D.3 Proof of the main lemmas

###### Proof of Lemma C.1.

To obtain upper bounds for the following expression

$\sup_{\theta\in\Theta}||\sum_{\lambda\in\Lambda_{1}}b_{\lambda}^{u,r,n}-\mathcal{N}_{r-1}(u,V_{\mu})||_{1}$

one first approximates $p_{\lambda}^{u,r,n}$ by a $r$ dimensional multinomial
distribution and then uses the classical channel $\tau_{\lambda}^{r,n}$ to map
the multinomial distribution to a distribution in $\mathbb{R}^{r-1}$ which
lies close to the multivariate normal. Thus we obtain the following variant of
equation (7.35) of [35]

$\left\|\mathcal{N}_{r-1}(u,V_{\mu})-\sum_{\lambda\in\Lambda_{1}}b_{\lambda}^{u,r,n}\right\|_{1}\leq\left\|p^{u,r,n}-M^{n}_{\mu^{u,n}_{1},\dots,\mu^{u,n}_{r}}\right\|_{1}+\left\|\mathcal{N}_{r-1}(u,V_{\mu})-\sum_{\lambda\in\Lambda_{1}}M^{n}_{\mu^{u,n}_{1},\dots,\mu^{u,n}_{r}}(\lambda)\tau_{\lambda}^{r,n}\right\|_{1},$
(135)

where $M^{n}_{\mu^{u,n}_{1},\dots,\mu^{u,n}_{r}}$ is the $r$-multinomial with
coefficients $\mu_{i}^{u,n}$. One then shows that both the terms in the RHS is
$O(n^{-1/2+\gamma}+n^{-1/4+\epsilon})$ for any $\epsilon>0$ and finally
obtain:

$\displaystyle\sup_{\theta\in\Theta}||\sum_{\lambda\in\Lambda_{1}}b_{\lambda}^{u,r,n}-\mathcal{N}_{r-1}(u,V_{\mu})||_{1}\leq
C_{r}(n^{-1/2+\gamma}+n^{-1/4+\epsilon})$

where $C_{r}$ is a constant which depends only on the rank $r$. Since the
proof is given in [35] we omit the details. ∎

###### Proof of Lemma C.2.

Note that

$p_{\lambda}^{u,r,n}=\prod_{1\leq i\leq
r}(\mu^{u,n}_{i})^{\lambda_{i}}\sum_{\mathbf{m}\in\mathcal{M}^{\lambda}_{r}}\prod_{1\leq
i<j\leq
r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\times\mathrm{dim}(\mathcal{K}_{\lambda}).$

where (see [19])

$\mathrm{dim}(\mathcal{K}_{\lambda})=\left(\begin{subarray}{c}n\\\ \\\
\lambda_{1},\lambda_{2},\ldots,\lambda_{r}\end{subarray}\right)\prod_{i=1}^{r}\frac{\lambda_{l}!\prod_{k=l+1}^{r}\lambda_{l}-\lambda_{k}+k-l}{(\lambda_{l}+r-l)!}.$

The rest of the proof proceeds as in [35]. One checks that

$\prod_{i=1}^{r}\frac{\lambda_{l}!\prod_{k=l+1}^{r}\lambda_{l}-\lambda_{k}+k-l}{(\lambda_{l}+r-l)!}\leq
1$

and

$\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\leq
1,\quad\sum_{m\in\mathcal{M}^{\lambda}_{r}}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\leq
n^{r^{2}},$

so that

$\sum_{\lambda\in\Lambda_{1}\cap\Lambda_{2}^{c}}p_{\lambda}^{u,r,n}\leq
n^{r^{2}}\sum_{\lambda\in\Lambda_{1}\cap\Lambda_{2}^{c}}\left(\begin{subarray}{c}n\\\
\\\
\lambda_{1},\lambda_{2},\ldots,\lambda_{r}\end{subarray}\right)\prod_{1\leq
i\leq r}(\mu^{u,n}_{i})^{\lambda_{i}}.$

Also note that

$\Lambda_{1}\cap\Lambda_{2}^{c}=\\{\lambda:\lambda_{k}=0\text{ for }k>r\text{
and }\exists i\text{ such that }|\lambda_{i}-n\mu_{i}|>n^{\alpha}\\}$

Thus we are summing a multinomial probability distribution over extreme
values. Since each coordinate of the multinomial is a binomial random variable
one uses Hoeffding inequality for Bernoulli random variables and union bound
to obtain

$\sum_{\lambda\in\Lambda_{1}\cap\Lambda_{2}^{c}}p_{\lambda}^{u,r,n}\leq
2rn^{r^{2}}\exp(-n^{2\alpha-1}/2)$

for large enough $n$. ∎

###### Proof of lemma C.3.

We start by considering the states

$\rho^{0,u,r,n}_{\lambda}=\frac{1}{N^{u,r,n}_{\lambda}}\sum_{\\{m_{i}\\}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\prod_{j=i+1}^{r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\sum_{m\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}}|\bar{\mathbf{m}}_{\lambda}\rangle\langle\bar{\mathbf{m}}_{\lambda}|$
$\phi^{0,r}=\sum_{\mathbf{m}_{1}\in\mathbb{N}^{r(r-1)/2}}\prod_{1\leq i<j\leq
r}\frac{\mu_{i}-\mu_{j}}{\mu_{i}}\left(\frac{\mu_{j}}{\mu_{i}}\right)^{m_{i,j}}|\mathbf{m}_{1}\rangle\langle\mathbf{m}_{1}|\otimes|\mathbf{0}_{2}\rangle\langle\mathbf{0}_{2}|$
(136)

Note that although $|\mathbf{m}_{\lambda}\rangle$ are eigenvectors of
$\rho^{0,u,r,n}_{\lambda}$, they are not orthogonal and hence the
decomposition is not straightforward. Let

$\mathcal{H}^{r}_{\lambda}=\text{ span
}\\{|\mathbf{m}_{\lambda}\rangle:\mathbf{m}\in\mathcal{M}_{r}^{\lambda}\\}$
(137)

We split $\mathcal{H}^{r}_{\lambda}$ into direct sum of two spaces i.e.

$\mathcal{H}^{r}_{\lambda}=\mathcal{H}^{r}_{\lambda,\eta}\bigoplus\bar{\mathcal{H}}^{r}_{\lambda,\eta}$

where

$\mathcal{H}^{r}_{\lambda,\eta}=\text{ span
}\\{|\mathbf{m}_{\lambda}\rangle\in\mathcal{H}^{r}_{\lambda}:|\mathbf{m}|\leq
n^{\eta}\\}.$

Also decompose
$\mathcal{F}^{r}=\mathcal{F}^{r}_{\eta}\bigoplus\mathcal{F}^{r\perp}_{\eta}$
with
$\\{|\mathbf{m}_{1}\rangle\otimes|\mathbf{0}_{2}\rangle:|\mathbf{m}_{1}|\leq
n^{\eta}\\}$ spanning the set $\mathcal{F}^{r}_{\eta}$. Define the projectors
into the subspaces
$\mathcal{H}^{r}_{\lambda,\eta},\bar{\mathcal{H}}^{r}_{\lambda,\eta}\mathcal{F}^{r}_{\eta},$
and $\mathcal{F}^{r\perp}_{\eta}$ One can show that the following
decomposition holds:

$\displaystyle||T^{r}_{\lambda}(\rho^{0,u,r,n}_{\lambda})-\phi^{0,r}||_{1}\leq
2||T^{r}_{\lambda}(P^{r}_{\lambda,\eta}\rho^{0,u,r,n}_{\lambda}P^{r}_{\lambda,\eta})-P^{r}_{\eta}\phi^{0,r}P^{r}_{\eta}||_{1}+2||P^{r\perp}_{\eta}\phi^{0,r}P^{r\perp}_{\eta}||_{1}.$
(138)

Being a geometric sum the second term of relation (138) can be shown to be
$\max_{1\leq i<j\leq r}(\mu_{i}/\mu_{j})^{n^{\eta}}=O(\exp(-c_{1}n^{\eta}))$
for some constant $c_{1}$. To deal with the first term note that
$|\mathbf{m}|\leq n^{\eta}$ and we have by Lemma D.6

$P(\\{m_{i}\\})=\sum_{\mathbf{m}\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}}|\bar{\mathbf{m}}_{\lambda}\rangle\langle\bar{\mathbf{m}}_{\lambda}|=\frac{1}{1+Cn^{(9\eta-2)/12}}\sum_{\mathbf{m}\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}}|\mathbf{m}_{\lambda}\rangle\langle\mathbf{m}_{\lambda}|+E(\\{m_{i}\\})$

and the positive remainder $E(\\{m_{i}\\})$ has trace norm

$\mathrm{Tr}(E(\\{m_{i}\\})=O(n^{(9\eta-2)/12}).\text{dim}(\mathcal{H}(\\{m_{i}\\}))$

where $\mathcal{H}(\\{m_{i}\\})$ is the space on which $P(\\{m_{i}\\})$
projects. Multiplying by the operator $P^{r}_{\lambda,\eta}$ on both sides we
obtain

$P^{r}_{\lambda,\eta}\rho^{0,u,r,n}_{\lambda}P^{r}_{\lambda,\eta}=\frac{1}{1+Cn^{(9\eta-2)/12}}\breve{\rho}^{0,u,r,n}_{\lambda}+\frac{1}{N^{u,r,n}_{\lambda}}\sum_{\\{m_{i}\\}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\prod_{j=i+1}^{r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}E(\\{m_{i}\\}).$
(139)

where the approximate state $\breve{\rho}^{0,u,r,n}_{\lambda}$ is given by

$\breve{\rho}^{0,u,r,n}_{\lambda}=\frac{1}{N^{u,r,n}_{\lambda}}\sum_{\\{m_{i}\\}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\prod_{j=i+1}^{r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}\\\
|\mathbf{m}|\leq
n^{\eta}\end{subarray}}|\mathbf{m}_{\lambda}\rangle\langle\mathbf{m}_{\lambda}|.$

Trace norm of the second term in (139) is given by

$\mathrm{Tr}\left(\frac{1}{N^{u,r,n}_{\lambda}}\sum_{\\{m_{i}\\}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\prod_{j=i+1}^{r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}E(\\{m_{i}\\})\right)=O(n^{9\eta-2/12})$

and since $T^{r}_{\lambda}$ is contractive we have

$||T^{r}_{\lambda}(P^{r}_{\lambda,\eta}\rho^{0,u,r,n}_{\lambda}P^{r}_{\lambda,\eta}-\tilde{\rho}^{0,u,r,n}_{\lambda})||_{1}\leq||P^{r}_{\lambda,\eta}\rho^{0,u,r,n}_{\lambda}P^{r}_{\lambda,\eta}-\tilde{\rho}^{0,u,r,n}_{\lambda}||_{1}=O(n^{9\eta-2/12}).$

Thus we need to give an upper bound to
$||T^{r}_{\lambda}(\breve{\rho}^{0,u,r,n}_{\lambda})-P^{r}_{\eta}\phi^{0,r}P^{r}_{\eta}||_{1}$
in order to give an upper bound to the RHS of (138). Since
$\mu_{i}^{u,n}=\mu_{i}+O(n^{-1/2+\gamma})$ for $|m|\leq n^{\eta}$ we have

$\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{ij}}=\left(\frac{\mu_{i}}{\mu_{j}}\right)^{m_{ij}}(1+O(n^{-1/2+\gamma+\eta}))$

and hence the from the definition of $N^{u,r,n}_{\lambda}$ we have

$\displaystyle\left[\frac{1}{N^{u,r,n}_{\lambda}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\right]^{-1}=$
$\displaystyle\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda},|\mathbf{m}|\leq
n^{\eta}}\prod_{1\leq i<j\leq
r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{ij}}+\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda},|\mathbf{m}|>n^{\eta}}\prod_{1\leq
i<j\leq r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{ij}}$
$\displaystyle=$
$\displaystyle\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda}}\prod_{1\leq
i<j\leq
r}\left(\frac{\mu_{j}}{\mu_{i}}\right)^{m_{ij}}(1+O(n^{-1/2+\gamma+\eta}))$
$\displaystyle-\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda},|\mathbf{m}|>n^{\eta}}\prod_{1\leq
i<j\leq
r}\left(\frac{\mu_{j}}{\mu_{i}}\right)^{m_{ij}}(1+O(n^{-1/2+\gamma+\eta}))$
$\displaystyle+\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda},|\mathbf{m}|>n^{\eta}}\prod_{1\leq
i<j\leq r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{ij}}.$

Again by a geometric sum argument the last two terms can easily be seen to be
of the order $O(\exp(-c_{2}n^{\eta}))$ for some constant $c_{2}$ and hence we
obtain

$\displaystyle\left[\frac{1}{N^{u,r,n}_{\lambda}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\right]^{-1}$
$\displaystyle=\sum_{\mathbf{m}\in\mathcal{M}^{r}_{\lambda}}\prod_{1\leq
i<j\leq
r}\left(\frac{\mu_{j}}{\mu_{i}}\right)^{m_{ij}}(1+O(n^{-1/2+\gamma+\eta}))$
$\displaystyle=\prod_{1\leq i<j\leq
r}\frac{\mu_{i}}{\mu_{i}-\mu_{j}}(1+O(n^{-1/2+\gamma+\eta}))$ (140)

By definition of $\mathcal{M}^{r}_{\lambda}$ $m_{i,j}=0$ unless $i,j\leq r$.
Using the fact that
$\||\psi\rangle\langle\psi|-|\psi^{\prime}\rangle\langle\psi^{\prime}|\|_{1}=2\sqrt{1-|\langle\psi|\psi^{\prime}\rangle|^{2}}$
(see (5)) holds for unit vectors ($\psi,\psi^{\prime}$) in conjunction with
Lemma D.6 we obtain

$\displaystyle||T^{r}_{\lambda}(|\mathbf{\mathbf{m}}_{\lambda}\rangle\langle\mathbf{\mathbf{m}}_{\lambda}|)-|\mathbf{m}_{1}\rangle\langle\mathbf{m}_{1}|\otimes|\mathbf{0}_{2}\rangle\langle\mathbf{0}_{2}|||_{1}$
$\displaystyle=||V^{r}_{\lambda}|\mathbf{m}_{\lambda}\rangle\langle\mathbf{m}_{\lambda}|V^{r*}_{\lambda}-|\mathbf{m}_{1}\rangle\langle\mathbf{m}_{1}|\otimes|\mathbf{0}_{2}\rangle\langle\mathbf{0}_{2}|)||_{1}$
$\displaystyle=O(n^{9\eta-2/24}).$

Note that we have ignored the $\delta$ since the minimum gap between the
eigenvalues is fixed. Consequently

$\displaystyle T^{r}_{\lambda}(\tilde{\rho}^{0,u,r,n}_{\lambda})$
$\displaystyle=\frac{1}{N^{u,r,n}_{\lambda}}\sum_{\\{m_{i}\\}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\prod_{j=i+1}^{r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}\\\
|\mathbf{m}|\leq
n^{\eta}\end{subarray}}T^{r}_{\lambda}(|\mathbf{m}_{\lambda}\rangle\langle\mathbf{m}_{\lambda}|)$
$\displaystyle=\frac{1}{N^{u,r,n}_{\lambda}}\prod_{i=1}^{r}(\mu^{u,n}_{i})^{\lambda_{i}}\sum_{\\{m_{i}\\}}\prod_{1\leq
i<j\leq
r}\left(\frac{\mu^{u,n}_{j}}{\mu^{u,n}_{i}}\right)^{m_{i,j}}\sum_{\begin{subarray}{c}m\in\mathcal{M}^{\lambda}_{r}\cap\\{m_{i}\\}\\\
|\mathbf{m}|\leq
n^{\eta}\end{subarray}}T^{r}_{\lambda}(|\mathbf{m}_{\lambda}\rangle\langle\mathbf{m}_{\lambda}|)$
$\displaystyle=\sum_{|\mathbf{m}_{1}|\leq n^{\eta}}\prod_{1\leq i<j\leq
r}\frac{\mu_{i}-\mu_{j}}{\mu_{i}}\left(\frac{\mu_{j}}{\mu_{i}}\right)^{m_{i,j}}|\mathbf{m}_{1}\rangle\langle\mathbf{m}_{1}|\otimes|\mathbf{0}_{2}\rangle\langle\mathbf{0}_{2}|+O(n^{-1/2+\gamma+\eta},n^{9\eta-2/24})$
$\displaystyle=P^{r}_{\eta}\phi^{0,r}P^{r}_{\eta}+O(n^{-1/2+\gamma+\eta},n^{9\eta-2/24})$

which was to be shown.

∎

###### Proof of Lemma C.4.

Recall that $|\mathbf{0}\rangle\langle\mathbf{0}|$ is the vacuum in the
multimode Fock space $\mathcal{F}^{r}$ where the number of modes is
$r(r-1)/2+r(d-r)$. Let us define

$D^{\mathfrak{z}_{1}+\mathbf{z},\mathfrak{z}_{2}}(|\mathbf{0}\rangle\langle\mathbf{0}|)=|\mathfrak{z}+\mathbf{z}\rangle\langle\mathfrak{z}+\mathbf{z}|.$

Although $\mathfrak{z}$ and $\mathbf{z}$ are of different dimension we can
identify $\mathbf{z}$ as $(\mathbf{z},0)$, so that
$|\mathfrak{z}+\mathbf{z}\rangle$ makes sense. Also note that
$|\mathfrak{z}+\mathbf{z}\rangle\langle\mathfrak{z}+\mathbf{z}|$ is a pure
state in the multimode Fock space $\mathcal{F}^{r}$.

Similarly define
$|\mathfrak{z}+\mathbf{z},\lambda\rangle=U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|\mathbf{0}_{\lambda}\rangle$
and hence

$T^{r}_{\lambda}\Delta_{\lambda}^{\mathfrak{z}+\mathbf{z},r,n}T^{r*}_{\lambda}(|\mathbf{0}\rangle\langle\mathbf{0}|)=V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle\langle\mathfrak{z}+\mathbf{z},\lambda|V^{r*}_{\lambda}.$

According to Lemma D.6, the coordinates of
$V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle$ in the Fock basis are
described by:

$\displaystyle\langle{\bf
m}|V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle=\left\\{\begin{array}[]{l}\langle{\bf
m}_{\lambda}|U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|{\bf
0}_{\lambda}\rangle(1+O(n^{(9\eta-2)/12}\delta^{-1/3}))\mbox{ if }|{\bf
m}|\leq n^{\eta};\vspace{2mm}\\\ \mbox{something not important if }|{\bf
m}|>n^{\eta}.\end{array}\right.$ (143)

To show that

$\sup_{\theta\in\Theta,||\mathbf{z}||\leq
n^{\beta}}\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}||(T^{r}_{\lambda}\Delta_{\lambda}^{\mathfrak{z}+\mathbf{z},r,n}T^{r*}_{\lambda}-D^{\mathfrak{z}+\mathbf{z}})(|\mathbf{0}\rangle\langle\mathbf{0}|)||_{1}\rightarrow
0$

and

$\sup_{\theta\in\Theta,||\mathbf{z}||\leq
n^{\beta}}\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}||(D^{\mathbf{z},0}-T^{r}_{\lambda}\Delta_{\lambda}^{\mathbf{z},r,n}T^{r*}_{\lambda})(|\mathbf{0}\rangle\langle\mathbf{0}|)||_{1}\rightarrow
0$

it is enough to show that the first limit goes to $0$ since the second one is
a special case of the first. Recalling (5) we observe that it is enough to
show

$\sup_{||\mathbf{z}||\leq
n^{\beta}}\sup_{\mathfrak{z}\in\Theta}\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}1-|\langle\mathfrak{z}+\mathbf{z}|V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle|=R(n)^{2}$

since
$V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle\langle\mathfrak{z}+\mathbf{z},\lambda|V^{*}_{\lambda}$
and $|\mathfrak{z}+\mathbf{z}\rangle\langle\mathfrak{z}+\mathbf{z}|$ are pure
states. Here $R(n)$ goes to $0$ (we will give an explicit expression of this
term later).

Recall that $\mathbf{m}=\\{m_{i,j}\\}_{1\leq i\leq r,i<j\leq d}$ and also note
that we need the full $\mathbf{m}$ here in sharp contrast with $\bf{m}_{1}$
(i.e. $\bf{m}\in\mathcal{M}^{r}_{\lambda}$) used in Lemma C.3. First we
decompose the inner product as follows

$\langle\mathfrak{z}+\mathbf{z}|V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle=\sum_{m}\langle\mathfrak{z}+\mathbf{z}|\bf{m}\rangle\langle\bf{m}|V^{r}_{\lambda}|\mathfrak{z}+\mathbf{z},\lambda\rangle$
(144)

Consider the following set

$\mathcal{M}:=\\{{\bf
m}:m_{i,j}\leq|(\mathfrak{z}+z)_{i,j}|^{2}n^{\epsilon},\quad 1\leq i\leq
r,i<j\leq d\\}.$ (145)

In particular, since $2\beta+\epsilon<\eta$ we have $\mathcal{M}\subset\\{{\bf
m}:|{\bf m}|\leq n^{\eta}\\}$.

We can upper bound the overlap of the coherent state with $|\bf{m}\rangle$s
where $m\notin\mathcal{M}$ as follows:

$\sum_{\mathbf{m}\not\in\mathcal{M}}|\langle\mathfrak{z}+\mathbf{z}|{\bf
m}\rangle|^{2}\leq\sum_{1\leq i\leq r,i<j\leq
d}\exp(-x_{i,j})\sum_{k>x_{i,j}n^{\epsilon}}\frac{x_{ij}^{k}}{k!}\leq
rdn^{-\beta},\qquad\text{ where }x_{i,j}=|\langle\mathfrak{z}+\mathbf{z}|{\bf
m}\rangle|^{2}_{i,j}.$ (146)

Each of the terms in the sum is a tail of Poisson distribution and is bounded
by $n^{-\epsilon n^{\beta}}$ if $x_{i,j}\geq 1$ and by $n^{-\beta}$ if
$x_{i,j}<1$.

From the first line of (143) we get

$\displaystyle\langle{\bf
m}|V^{r}_{\lambda}U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|{\bf
0}_{\lambda}\rangle$ $\displaystyle=\frac{\langle y_{\lambda}{f_{\bf
m}}|y_{\lambda}U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|f_{\bf
0}\rangle}{\sqrt{\langle y_{\lambda}{f_{\bf 0}}|y_{\lambda}{f_{\bf
0}}\rangle}\sqrt{\langle y_{\lambda}{f_{\bf m}}|y_{\lambda}{f_{\bf
m}}\rangle}}(1+O(n^{(9\eta-2)/12}\delta^{-1/3}))$ $\displaystyle=\frac{\langle
p_{\lambda}{f_{\bf
m}}|q_{\lambda}U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|f_{\bf
0}\rangle}{\sqrt{\langle p_{\lambda}{f_{\bf m}}|q_{\lambda}p_{\lambda}{f_{\bf
m}}\rangle}}(1+O(n^{(9\eta-2)/12}\delta^{-1/3})).$

The first line follows since $y_{\lambda}$ and
$U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})$ commutes and for the
second line we have used (61), (63) and (A).

We recall that $\mathcal{O}_{\lambda}({\bf m})$ is the orbit in
$(\mathbb{C}^{d})^{\otimes n}$ of ${f_{\bf m}}$ under $\mathcal{R}_{\lambda}$
and that we have the decomposition

$p_{\lambda}{f_{\bf m}}=\sum_{{f_{\bf a}}\in\mathcal{O}_{\lambda}({\bf
m})}\frac{\\#\mathcal{R}_{\lambda}}{\\#\mathcal{O}_{\lambda}({\bf m})}{f_{\bf
a}}.$

Using (9) and (123), we can write

$\displaystyle\langle{\bf
m}|V^{r}_{\lambda}U^{r}_{\lambda}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|{\bf
0}_{\lambda}\rangle$
$\displaystyle=\frac{\sum_{f_{\bf{a}}\in\mathcal{O}_{\lambda}({\bf m})}\langle
f_{\bf{a}}|q_{\lambda}U^{r}(\mathfrak{z}+\mathbf{z}/\sqrt{n})f_{\mathbf{0}}\rangle}{\sqrt{\sum_{f_{\bf{a}},f_{\bf{b}}\in\mathcal{O}_{\lambda}({\bf
m})}\langle
f_{\bf{a}}|q_{\lambda}f_{\bf{b}}\rangle}}(1+O(n^{(9\eta-2)/12}\delta^{-1/3})$
(147) $\displaystyle=e^{-\|\mathfrak{z}+\mathbf{z}\|^{2}_{2}/2}\prod_{1\leq
i\leq r,i<j\leq
d}\frac{(\mathfrak{z}+z)_{i,j}^{m_{i,j}}}{\sqrt{m_{i,j}!}}\left(\frac{n(\mu_{i}-\mu_{j})}{\lambda_{i}-\lambda_{j}}\right)^{m_{i,j}/2}r(n).$

The corresponding remainder term is

$r(n)=1+O\left(n^{(9\eta-2)/12}\delta^{-1/3},n^{-1+2\beta+\eta}\delta^{-1},n^{-1/2+3\beta+2\epsilon}\delta^{-3/2},n^{-1+\alpha+2\beta}\delta^{-1},n^{-1+\alpha+\eta}\delta^{-1},n^{-1+3\eta}\delta^{-1}\right).$

Since $\lambda\in\Lambda_{2}$ and the eigenvalues are separated by $\delta$ we
have
$\left(\frac{n(\mu_{i}-\mu_{j})}{\lambda_{i}-\lambda_{j}}\right)^{m_{i,j}/2}=1+O(n^{\alpha-1+\eta}/\delta)$
and the error can be absorbed in $r(n)$.

Thus, for $\mathbf{m}$ satisfying (145), we have:

$\langle{\bf m}|V^{r}_{\lambda}U^{r}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|{\bf
0}_{\lambda}\rangle=\langle{\bf m}|\mathfrak{z}+\mathbf{z}\rangle r(n).$

If $a_{\bf m}$ and $b_{\bf m}$ are the two sets of coefficients, such that
$\sum_{\bf m}|a_{\bf m}|^{2}=\sum_{\bf m}|b_{\bf m}|^{2}=1$, then

$1-\left|\sum_{\bf m}a_{\bf m}b_{\bf m}\right|\leq 1-\left|\sum_{{\bf
m}\in\mathcal{M}}a_{\bf m}b_{\bf m}\right|+\left|\sum_{{\bf
m}\notin\mathcal{M}}a_{\bf m}b_{\bf m}\right|\leq 2\left(1-\left|\sum_{{\bf
m}\in\mathcal{M}}a_{\bf m}b_{\bf m}\right|\right).$ (148)

We use (144) in conjunction with (146) and (148), we get

$\displaystyle
1-\left|\langle\mathfrak{z}+\mathbf{z}|V^{r}_{\lambda}U^{r}(\mathfrak{z}+\mathbf{z}/\sqrt{n})|{\bf
0}_{\lambda}\rangle\right|$ $\displaystyle=$ $\displaystyle
O\left(1-r(n),\sum_{\mathbf{m}\not\in\mathcal{M}}|\langle{\bf
m}|\mathfrak{z}+\mathbf{z}\rangle|^{2}\right)=R_{2}(n),$ (149)

with

$R_{2}(n)=O\left(n^{(9\eta-2)/12}\delta^{-1/3},n^{-1+2\beta+\eta}\delta^{-1},n^{-1/2+3\beta+2\epsilon}\delta^{-3/2},n^{-1+\alpha+2\beta}\delta^{-1},\right.\\\
\left.n^{-1+\alpha+\eta}\delta^{-1},n^{-1+3\eta}\delta^{-1},n^{-\beta}\right).$

Note that we are interested in the square root of the LHS of (149). Setting
$R(n)=\sqrt{R_{2}(n)}$ (144) and ignoring $\delta$ in the final statement
(since it is a fixed constant) we obtain the desired result. ∎

###### Proof of Lemma C.5.

Note that although $SU(d)$ forms a group the unitaries $U^{r}$ do not form a
subgroup. Instead we view the unitaries as elements of $SU(d)$ and the
corresponding $\Delta^{\mathbf{z},r,n}_{\lambda}$ as
$\Delta^{\mathbf{z},n}_{\lambda}$ and directly use Lemma 6.5 of [35] which
states that

$\sup_{||\mathbf{z}||\leq
n^{\beta}}\sup_{\theta\in\Theta_{n,\beta,\gamma}}\sup_{\lambda\in\Lambda_{n,\alpha}}||(\Delta_{\lambda}^{\mathfrak{z}+\mathbf{z},n}-\Delta_{\lambda}^{\mathfrak{z},n}\Delta_{\lambda}^{\mathbf{z},n})(|\mathbf{0}_{\lambda}\rangle\langle\mathbf{0}_{\lambda}|)||_{1}=R(n)$

Note that in our case also
$T^{r*}_{\lambda}(|\mathbf{0}\rangle\langle\mathbf{0}|)=|\mathbf{0}_{\lambda}\rangle\langle\mathbf{0}_{\lambda}|$.
Finally we identify $\Delta^{\mathbf{z},r,n}_{\lambda}$s as
$\Delta^{\mathbf{z},n}_{\lambda}$s as mentioned and since we are taking
supremum over restricted sets (for the coordinates of $\theta$ with $i,j\geq
r$ is $0$ and $\lambda$ is further restricted to $\Lambda_{1}$) we obtain

$\displaystyle\sup_{\theta\in\Theta,||\mathbf{z}||\leq
n^{\beta}}\sup_{\lambda\in\Lambda_{1}\cap\Lambda_{2}}||(\Delta_{\lambda}^{\mathfrak{z}+\mathbf{z},r,n}-\Delta_{\lambda}^{\mathfrak{z},r,n}\Delta_{\lambda}^{\mathbf{z},r,n})T^{r*}_{\lambda}(|\mathbf{0}\rangle\langle\mathbf{0}|)||_{1}$
$\displaystyle\leq\sup_{||\mathbf{z}||\leq
n^{\beta}}\sup_{\theta\in\Theta_{n,\beta,\gamma}}\sup_{\lambda\in\Lambda_{n,\alpha}}||(\Delta_{\lambda}^{\mathfrak{z}+\mathbf{z},n}-\Delta_{\lambda}^{\mathfrak{z},n}\Delta_{\lambda}^{\mathbf{z},n})(|\mathbf{0}_{\lambda}\rangle\langle\mathbf{0}_{\lambda}|)||_{1}=R(n).$

∎

###### Proof of Lemma C.7.

Define $u_{r}=\sum_{i=1}^{r-1}u_{i}$ and
$\hat{u}_{r}=\sum_{i=1}^{r-1}\hat{u}_{i}$ .

$\displaystyle\int_{\mathcal{U}^{c}}E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)$
$\displaystyle=\int_{\mathcal{U}^{c}}E_{u}\left[\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i})^{2}+(\sum_{i=1}^{r-1}(u_{i}-\hat{u}_{i}))^{2}\right]d\pi_{1}(u)$
$\displaystyle\leq
2\sum_{i=1}^{r}\int_{\mathcal{U}^{c}}E_{u}\left[(u^{2}_{i}+\hat{u}^{2}_{i})\right]d\pi_{1}(u)$
$\displaystyle\leq
2\sum_{i=1}^{r}\sqrt{E(u^{4}_{i})}\sqrt{\pi_{1}(\mathcal{U}^{c})}+rn^{2\gamma}\pi_{1}(\mathcal{U}^{c}).$

In the first part we have used Cauchy-Schwartz inequality and in the second
part we have used the fact that since $\hat{u}\in\mathcal{U}$ each component
$|\hat{u}_{i}|\leq n^{\gamma}$ for $i=1,\ldots,r-1$ and hence
$|\hat{u}_{r}|\leq rn^{\gamma}$.

Note that for $i=1,\ldots,r-1$,$u_{i}\sim N(0,n^{2\gamma_{0}})$ and hence
$Eu_{i}^{4}=3n^{4\gamma_{0}}$. It can be easily seen that $u_{r}\sim
N(0,(r-1)n^{2\gamma_{0}})$ and so $Eu_{r}^{4}=3(r-1)^{2}n^{2\gamma_{0}}$. Also
$P(|u_{i}|\geq n^{\gamma})\leq c\exp(-n^{2\gamma}/2n^{2\gamma_{0}})$ for
$i=1,\ldots,r-1$, so that by union bound we have $\pi(\mathcal{U}^{c})\leq
rc\exp(-n^{2\gamma}/2n^{2\gamma_{0}})$. Thus we have

$\int_{\mathcal{U}^{c}}E_{u}[\mathcal{L}(u,\hat{u})]d\pi_{1}(u)\leq
2\sum_{i=1}^{r}\sqrt{3r^{2}n^{4\gamma_{0}}}\sqrt{rc\exp(-n^{2\gamma}/2n^{2\gamma_{0}}}+rn^{2\gamma}rc\exp(-n^{2\gamma}/2n^{2\gamma_{0}})$

and the RHS of the above inequality tends to $0$ since $\gamma_{0}<\gamma$.

For the quantum part we have

$\int_{\mathcal{Z}^{c}}E_{\xi}[\mathcal{L}(\xi,\hat{\xi})]d\pi_{2}(\xi)=\int_{\mathcal{Z}^{c}}E_{\xi}\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq
d}\end{subarray}}(\mu_{i}-\mu_{j})|\xi_{ij}-\hat{\xi}_{ij}|^{2}d\pi_{2}(\xi).$

It is enough to show that
$\int_{\mathcal{Z}^{c}}E_{\xi}||\xi_{ij}-\hat{\xi}_{ij}||^{2}d\pi_{2}(\xi)\rightarrow
0$ for each pair of $i,j$. As in the classical part we have the following
inequalities:

$\displaystyle\int_{\mathcal{Z}^{c}}E_{\xi}|\xi_{ij}-\hat{\xi}_{ij}|^{2}d\pi_{2}(\xi)$
$\displaystyle\leq
2\int_{\mathcal{Z}^{c}}E_{\xi}\left[||\xi_{ij}||^{2}+||\hat{\xi}_{ij}||^{2}\right]d\pi_{2}(\xi)$
$\displaystyle\leq
2(\sqrt{E|(\xi_{ij})_{1}|^{4}}+\sqrt{E|(\xi_{ij})_{2}|^{4}})\sqrt{\pi_{2}(\mathcal{Z}^{c})}+4n^{2\beta}\pi_{2}(\mathcal{Z}^{c})$
$\displaystyle\leq
4\sqrt{3n^{4\beta_{0}}}\sqrt{rdc\exp(-n^{2\beta}/2n^{2\beta_{0}}}+4n^{2\beta}rdc\exp(-n^{2\beta}/2n^{2\beta_{0}}).$

Again the RHS goes to $0$ as $\beta_{0}<\beta$. ∎

###### Proof of Lemma 5.3.

Consider the first order approximation of the state $\rho_{\theta,r}$ i.e.
consider the expression in (100). Then we have,

$||\tilde{\rho}_{\theta^{(1)},r}-\tilde{\rho}_{\theta^{(2)},r}||^{2}_{2}=\sum_{i=1}^{r}(u^{(1)}_{i}-u^{(2)}_{i})^{2}+\sum_{\begin{subarray}{c}{1\leq
i\leq r}\\\ {i<j\leq d}\end{subarray}}|\zeta^{(1)}_{ij}-\zeta^{(2)}_{ij}|^{2}$

Replacing $u_{i}$ by $u_{i}/\sqrt{n}$ and $\zeta_{ij}$ by
$\mathfrak{z}_{ij}\sqrt{\frac{(\mu_{i}-\mu_{j})}{n}}$ we obtain the desired
result with the error term
$O\left(\frac{||\theta^{(1)}||^{3},||\theta^{(2)}||^{3}}{n^{3/2}}\right)$
corresponding to the error in first order approximation. ∎