needed to ensure its convergence. Here, we combine Theorem 4.1.1 with the
stabilizer de Finetti theorems 2.2.2 and 2.2.3, leading to an approximation of
a state where there is separability in the cut between $A$ and $B$, and $B$ is
approximately given by a convex combination of tensor powers of stabilizer
states (instead of arbitrary states).

######  Theorem 4.2.1 (Stabilizer de Finetti Theorem with Linear Constraints).

Let $\rho_{AB_{1}^{n}}$ be a quantum state on the Hilbert space
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}^{\otimes n}$ that commutes with the
action of $\mathcal{O}_{n}$ acting on the systems $B_{1}^{n}\equiv
B_{1}B_{2}\cdots B_{n}$, each of dimension $d_{B}^{r}$. Let $d_{B}$ be an odd
prime. Let $\Lambda_{A\rightarrow C_{A}}$ and $\Gamma_{B\rightarrow C_{B}}$ be
linear maps, and $X_{C_{A}}$ and $Y_{C_{B}}$ be operators such that the
following two linear constraints hold:

$\Lambda_{A\rightarrow
C_{A}}(\rho_{AB_{1}^{n}})=X_{C_{A}}\otimes\rho_{B_{1}^{n}},$
$\Gamma_{B_{n}\rightarrow C_{B}}(\rho_{B_{1}^{n}})=\rho_{B_{1}^{n-1}}\otimes
Y_{C_{B}}.$

Then, there exists a probability distribution $\\{p_{Z}(z)\\}_{z\in Z}$ and a
probability distribution $\\{p_{S}(\sigma_{B})\\}$ over the set of mixed
stabilizer states on $r$ qudits such that

$\Big{\|}\rho_{AB}-\sum_{z,\sigma_{B}}p_{Z}(z)p_{S}(\sigma_{B})\rho_{A|z}\otimes\sigma_{B}\Big{\|}_{\operatorname{tr}}\leq\epsilon(d_{B}^{r},d_{A},n)+\bar{\epsilon}(d_{B},r,n)$

where $\sigma_{B}$ are the mixed stabilizer states of $r$ qudits on
$\mathcal{H}_{B}={(\mathbbm{C}^{d_{B}})}^{\otimes r}$,
$\epsilon(d_{B}^{r},d_{A},n)$ defined in (4.1),

$\bar{\epsilon}(d_{B},r,n):=2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-1)}$
(4.2)

and

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z})=X_{C_{A}},\ \Gamma_{B_{n}\rightarrow
C_{B}}(\rho_{B|x})=Y_{C_{B}},$

$\Big{\|}\sum_{z}p_{Z}(z)\big{(}\rho_{B|z}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}\leq\bar{\epsilon}(d_{B},r,n).$

This theorem shows that a state that is partially invariant under the
stochastic orthogonal group introduced in (4) is approximately separable and
close to a convex combination of stabilizer states on one side. It is a direct
combination of two de Finetti like statements: On the one hand, it makes use
of the de Finetti theorem with additional linear constraints, Theorem 4.1.1;
on the other hand, it relies on the stabilizer de Finetti theorem, Theorem
2.2.2. The theorem with linear constraints ensures separability in the cut
between Alice’s and Bob’s subsystem, while the stabilizer de Finetti theorem
ensures closeness to stabilizer states on Bob’s side.

In Theorem 4.1.1, the marginal of a permutation invariant state is
approximated by a separable state where each part satisfies the linear
constraints directly. It is important to note that this is not entirely
analogous for stochastic orthogonal invariance: while Bob’s side is now
approximated by stabilizer states, these stabilizer states do not precisely
satisfy the linear constraints. Instead, they satisfy them approximately, with
an error corresponding to the bound of stabilizer de Finetti theorem. Notably,
this error is much smaller than the overall error for approximating separable
and partly stabilizer states.

To relate the two theorems to each other, consider the following definition:

######  Definition 11 (Post-Measurement Quantum State).

For positive operator valued measures (POVMs) $\\{\Pi_{z}\\}$ mapping from $C$
to classical system $Z$ and a state $\omega_{AC}$ with classical system $Z$,
the state after measurement of $z$ is given by:

$\omega_{A|z}=\dfrac{\operatorname{tr}_{Z}\Big{(}(\mathbbm{1}_{A}\otimes\Pi_{z})\omega_{AC}\Big{)}}{\operatorname{tr}_{AZ}\Big{(}(\mathbbm{1}_{A}\otimes\Pi_{z})\omega_{AC}\Big{)}}.$

With this definition, the two concepts - Theorem 4.1.1 and Theorem 2.2.2 \-
can be combined via the following observations, which will be integral to the
proof of Theorem 4.2.1:

######  Observation 1.

For all POVMs $\\{\Pi_{z}\\}$ mapping from $C$ to classical system $Z$ and
quantum states $\omega_{A|z}$, there exists a CPTP map $\mathcal{M}$ that acts
as follows on any quantum state $\rho_{BC}$:

$\mathcal{M}:\rho_{BC}\mapsto\sum_{z}\omega_{A|z}\otimes\operatorname{tr}_{C}(\Pi_{z}\rho_{BC})=\sum_{z}p_{Z}(z)\omega_{A|z}\otimes\rho_{B|z}$

with a probability distribution $p_{Z}(z)$.

###### Proof of Observation 1.

The map $\mathcal{M}$ can be interpreted as a composition of three maps:

$\mathcal{M}=\mathcal{M}_{3}\circ\mathcal{M}_{2}\circ\mathcal{M}_{1}$

The first map $\mathcal{M}_{1}$ acts as a measurement of the POVMs
$\\{\Pi_{z}\\}$. Applying it to a quantum state $\rho_{BC}$ yields

$\mathcal{M}_{1}:\rho_{BC}\mapsto\sum_{z}\operatorname{tr}_{C}(\Pi_{z}\rho_{BC})\otimes\ket{z}\bra{z}_{Z}.$

Clearly, this map is trace preserving and completely positive. The second map
recovers a specific purification of the state, introducing system $A$:

$\mathcal{M}_{2}\circ\mathcal{M}_{1}:\rho_{BC}\mapsto\sum_{z}\omega_{A|z}\otimes\operatorname{tr}_{C}(\mathbbm{1}_{B}\otimes\Pi_{z}\rho_{BC})\otimes\ket{z}\bra{z}_{Z}$

This map is completely positive because $\omega_{A|z}$, as defined in
Definition 11, is positive semidefinite - and trace preserving because
$\omega_{A|z}$ has unit trace. Lastly, the final map $\mathcal{M}_{3}$
corresponds to taking the trace over the classical system $Z$:

$\mathcal{M}_{3}\circ\mathcal{M}_{2}\circ\mathcal{M}_{1}:\rho_{BC}\mapsto\sum_{z}\omega_{A}\otimes\operatorname{tr}_{C}(\Pi_{z}\rho_{BC})$

Clearly, taking the trace over $Z$ is a completely positive and trace
preserving operation, making the composed map
$\mathcal{M}_{3}\circ\mathcal{M}_{2}\circ\mathcal{M}_{1}=\mathcal{M}$ a CPTP
map. ∎

######  Observation 2.

For all POVMs $\\{\Pi_{z}\\}$ mapping from $C$ to classical system $Z$, there
exists a CPTP map ${\mathcal{M}^{\prime}}$ that acts as follows on any quantum
state $\rho_{BC}$:

$\mathcal{M}^{\prime}:\rho_{BC}\mapsto\sum_{z}\operatorname{tr}_{C}(\Pi_{z}\rho_{BC})=\sum_{z}p_{Z}(z)\rho_{B|z}$

with a probability distribution $p_{Z}(z)$.

###### Proof of Observation 2.

The map $\mathcal{M}^{\prime}$ can be regarded as a composition of two of the
maps appearing in the proof of 1. Using the notation from there, it can be
rewritten as

$\mathcal{M}^{\prime}=\mathcal{M}_{3}\circ\mathcal{M}_{1}.$

As shown in the proof of 1, the maps $\mathcal{M}_{3}$ and $\mathcal{M}_{1}$
are CPTP, and thus their composition is also CPTP. ∎

Using Observations 1 and 2, Theorem 4.2.1 can be proven. This theorem admits
two interpretations: it can be regarded as a stabilizer de Finetti theorem
with additional linear constraints - or as a de Finetti theorem with linear
constraints with the additional constraint that some parts should be a
stabilizer state.

###### Proof of Theorem 4.2.1.

The proof follows the following general outline: Starting from the stabilizer
de Finetti theorem in 2.2.2 on Bob’s side, we find a measurement and a
purification that transform the theorem into a statement about the closeness
of separable states and states with stabilizer tensor powers on Bob’s side.
Then, the de Finetti theorem with linear constraints in 4.1.1 can be used to
connect separable states with the full state $\rho_{AB_{1}^{n}}$ via triangle
inequality, which leads to a theorem bounding the closeness of the full state
and states with stabilizer tensor powers on Bob’s side.

Since $\rho_{AB_{1}^{n}}$ is invariant under the stochastic orthogonal group
with representation $O_{n}$ acting on $B_{1}^{n}$, clearly
$\operatorname{tr}_{A}(\rho_{AB_{1}^{n}})=\rho_{B_{1}^{n}}$ is invariant under
$O_{n}$ acting on $B_{1}^{n}$. Because of this, the stabilizer de Finetti
theorem (Theorem 2.2.2) can be applied. Therefore, for any $k\leq n-1$, we
find:

$\Big{\|}\rho_{B_{1}^{k+1}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B}^{\otimes{k+1}}\Big{\|}_{\operatorname{tr}}\leq
2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-(k+1))}$ (4.3)

where $\sigma_{B}$ is a mixed stabilizer state on $\mathcal{H}_{B}$. Using
Observation 1, this can be connected to the de Finetti theorem for permutation
invariant states with linear constraints, Theorem 4.1.1. Choosing the systems
$A\rightarrow A$, $B\rightarrow B_{1}^{k}$ and $C\rightarrow B_{k+1}$, a
bitstring $z\rightarrow z_{1}^{k}$ and

$\rho_{A|z_{1}^{k}}=\frac{\operatorname{tr}_{B_{1}^{n}}(\mathbbm{1}_{A}\otimes\\{\Pi_{z_{1}^{k}}\otimes\mathbbm{1}_{B_{k+1}^{n}}\rho_{AB_{1}^{n}})}{\operatorname{tr}_{AB_{1}^{n}}(\mathbbm{1}_{A}\otimes\\{\Pi_{z_{1}^{k}}\otimes\mathbbm{1}_{B_{k+1}^{n}}\rho_{AB_{1}^{n}})}$

according to Definition 11 with POVMs $\\{\Pi_{z_{1}^{k}}\\}$, we find a CPTP
map $\mathcal{M}$ that acts in the following way:

$\mathcal{M}:X_{B_{1}^{k+1}}\rightarrow\sum_{z_{1}^{k}}\rho_{A|z_{1}^{k}}\otimes\operatorname{tr}_{B_{1}^{k}}(\Pi_{z_{1}^{k}}\otimes\mathbbm{1}_{B_{k+1}}X_{B_{1}^{k+1}})$

Applying this to the two states of interest in the distance relation above
yields

$\begin{split}\mathcal{M}(\rho_{B_{1}^{k+1}})&=\sum_{z_{1}^{k}}\rho_{A|z_{1}^{k}}\otimes\operatorname{tr}_{B_{1}^{k}}(\Pi_{z_{1}^{k}}\otimes\mathbbm{1}_{B_{k+1}}\rho_{B_{1}^{k+1}})\\\
&=\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\rho_{A|z_{1}^{k}}\otimes\rho_{B_{k+1}|z_{1}^{k}}\end{split}$

and

$\begin{split}\mathcal{M}(\sigma^{\otimes(k+1)})&=\sum_{z_{1}^{k}}\rho_{A|z_{1}^{k}}\otimes\operatorname{tr}_{B_{1}^{k}}(\Pi_{z_{1}^{k}}\otimes\mathbbm{1}_{B_{k+1}}\sigma^{\otimes(k+1)})\\\
&=\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\rho_{A|z_{1}^{k}}\otimes\sigma_{B_{k+1}}.\end{split}$

Inserting this into the trace distance relation (4.3), and using the fact that
$\mathcal{M}$ is trace-nonincreasing, yields

$\begin{split}2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-(k+1))}&\geq\Big{\|}\rho_{B_{1}^{k+1}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\Big{\|}_{\operatorname{tr}}\\\
&\geq\Big{\|}\mathcal{M}\big{(}\rho_{B_{1}^{k+1}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}\\\
&=\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\rho_{A|z_{1}^{k}}\otimes\big{(}\rho_{B_{k+1}|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}.\end{split}$
(4.4)

In total, we find a statement about separable states which are obtained from a
state with invariance under $\mathcal{O}_{n}$ being close to a mixture of
stabilizer states:

$\Big{\|}\sum_{{z}_{1}^{k}}p_{Z}(z_{1}^{k})\rho_{A|{z}_{1}^{k}}\otimes\big{(}\rho_{B_{k+1}|z_{1}^{k}}-\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}\leq
2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-(k+1))}$ (4.5)

The de Finetti theorem with linear constraints (Theorem 4.1.1) relates a
state’s closeness to such a separable state. It has to be noted that
permutations are a subgroup of the stochastic orthogonal group, which means
that $\rho_{B_{1}^{n}}$ is also invariant with respect to permutations of the
$B_{1}^{n}$ subsystems (but with dimension $d_{B}^{r}$, since each subsystem
consists of $r$ qudits). Therefore, given the linear constraints listed in the
theorem, there exists an $m\in[0,n-1]$, such that

$\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}\leq\min\Big{\\{}d_{B}^{2r}(d_{B}^{r}+1),18\sqrt{d_{A}d_{B}^{r}}\Big{\\}}\sqrt{\frac{2\ln(2)\ln(d_{A})}{n}}=\epsilon(d_{B}^{r},d_{A},n)$

with

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z_{1}^{m}})=X_{C_{A}},\
\Gamma_{B_{m+1}\rightarrow C_{B}}(\rho_{B_{m+1}|z_{1}^{m}})=Y_{C_{B}}.$

Choosing $k=m$ for the stabilizer de Finetti theorem and combining the two
statements (4.4) and (4.5) yields the following relation, which can be bound
using the triangle inequality:

$\displaystyle\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{m+1}})\Big{\|}_{\operatorname{tr}}$
$\displaystyle=\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}+\sum_{{z}_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|{z}_{1}^{m}}\otimes\big{(}\rho_{B_{m+1}|z_{1}^{m}}-\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{m+1}}\big{)}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\leq\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}+\Big{\|}\sum_{{z}_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|{z}_{1}^{m}}\otimes\big{(}\rho_{B_{m+1}|z_{1}^{m}}-\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{m+1}}\big{)}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\leq\epsilon(d_{B}^{r},d_{A},n)+2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-{(m+1)})}$

In total, we find that there exists an $m\in[0,n-1]$ such that

$\begin{split}&\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{m+1}})\Big{\|}_{\operatorname{tr}}\\\
&\leq\epsilon(d_{B}^{r},d_{A},n)+2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-{(m+1)})}.\end{split}$

Note that permutations are a subgroup of the stochastic orthogonal group,
which means that $\rho_{AB_{1}^{n}}$ is also invariant with respect to
permutations of the $B_{1}^{n}$ subsystems. Because of this, all t subsystems
$B_{1}^{n}$ are separately equal to some subsystem $B$ \- meaning
$B_{1}^{n}=B^{\otimes n}$. Therefore, $m$ can be chosen freely out of
$[0,n-1]$, including the best possible case in terms of the bound, $m=0$,
which leads to the desired statement in Theorem 4.2.1.

The additional resulting statements follow directly from Theorem 4.1.1, and
from combining (4.3) and the fact that there exists a trace preserving map
$\mathcal{M}^{\prime}$ (see Observation 2) transforming a state according to

$\mathcal{M}^{\prime}:X_{B_{1}^{k+1}}\mapsto\sum_{z_{1}^{k}}\operatorname{tr}_{Z_{1}^{k}}(\Pi_{z_{1}^{k}}X_{B_{1}^{k+1}}).$

Applied to the two states in (4.3), we obtain

$\begin{split}\mathcal{M}^{\prime}(\rho_{B_{1}^{k+1}})&=\sum_{z_{1}^{k}}\operatorname{tr}_{Z_{1}^{k}}(\Pi_{z_{1}^{k}}\rho_{B_{1}^{k+1}})\\\
&=\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\rho_{B_{k+1}|z_{1}^{k}}\end{split}$

and

$\begin{split}\mathcal{M}^{\prime}(\sigma^{\otimes(k+1)})&=\sum_{z_{1}^{k}}\operatorname{tr}_{Z_{1}^{k}}(\Pi_{z_{1}^{k}}\sigma^{\otimes(k+1)})\\\
&=\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\sigma_{B_{k+1}}.\end{split}$

Inserting this into (4.3), and using the fact that $\mathcal{M}^{\prime}$ is
CPTP, yields

$\begin{split}2d_{B}^{2(r+1)^{2}}d_{B}^{-\frac{1}{2}(n-(k+1))}&\geq\Big{\|}\rho_{B_{1}^{k+1}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\Big{\|}_{\operatorname{tr}}\\\
&\geq\Big{\|}\mathcal{M}^{\prime}\big{(}\rho_{B_{1}^{k+1}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}\\\
&=\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{B_{k+1}|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}.\end{split}$
(4.6)

∎

In analogy to the stabilizer de Finetti theorem, this theorem is restricted to
the cases where $d_{B}$ is an odd prime. However, one can also consider states
that are not invariant under the whole stochastic orthogonal group, but just
permutations and anti-identity, which leads to the de Finetti theorem for
qubits (Theorem 2.2.3). This can be used to obtain the following alternative
stabilizer de Finetti theorem with additional linear constraints for qubits:

######  Theorem 4.2.2 (Stabilizer de Finetti Theorem for Qubits with Linear
Constraints).

Let $\rho_{AB_{1}^{n}}$ be a quantum state on the Hilbert space
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}^{\otimes n}$ that commutes with the
action of all permutations and the anti-identity on
$B_{1}^{n}=B_{1}B_{2}\cdots B_{n}$. Let $\Lambda_{A\rightarrow C_{A}}$ and
$\Gamma_{B\rightarrow C_{B}}$ be linear maps, and $X_{C_{A}}$ and $Y_{C_{B}}$
be operators such that the following two linear constraints hold:

$\Lambda_{A\rightarrow
C_{A}}(\rho_{AB_{1}^{n}})=X_{C_{A}}\otimes\rho_{B_{1}^{n}},$
$\Gamma_{B_{n}\rightarrow C_{B}}(\rho_{B_{1}^{n}})=\rho_{B_{1}^{n-1}}\otimes
Y_{C_{B}}.$

Then, there exists a probability distribution $\\{p_{Z}(z)\\}_{z\in Z}$ and a
probability distribution $\\{p_{S}(\sigma_{B})\\}$ over the set of mixed
stabilizer states on $r$ qubits such that

$\Big{\|}\rho_{AB}-\sum_{z,\sigma_{B}}p_{Z}(z)p_{S}(\sigma_{B})\rho_{A|z}\otimes\sigma_{B}\Big{\|}_{\operatorname{tr}}\leq\epsilon(d_{B}^{r},d_{A},n)+\tilde{\epsilon}(d_{B},r,n)$

where $\sigma_{B}$ are the mixed stabilizer states of $r$ qudits on
$\mathcal{H}_{B}={(\mathbbm{C}^{d_{B}})}^{\otimes r}$,
$\epsilon(d_{B}^{r},d_{A},n)$ defined in (4.1),

$\tilde{\epsilon}(d_{B},r,n):=6\sqrt{2}\ 2^{r}\sqrt{\frac{1}{n}}$ (4.7)

and

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z})=X_{C_{A}},\ \Gamma_{B_{n}\rightarrow
C_{B}}(\rho_{B|x})=Y_{C_{B}},$

$\Big{\|}\sum_{z}p_{Z}(z)\big{(}\rho_{B|z}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}\leq\tilde{\epsilon}(d_{B},r,n).$

###### Proof of Theorem 4.2.2.

The proof for Theorem 4.2.2 has one additional step, but is otherwise
completely analogous to the proof of Theorem 4.2.1 with changed error
probability in instances where the stabilizer de Finetti theorem appears. The
additional step is neccessary because the stabilizer de Finetti theorem for
qubits only holds for $k\leq n$ being a multiple of six (using the notation
conventions from the proof of 4.2.1). Therefore, to facilitate chosing $k$
equal to $m$, $m$ must be a multiple of six as well. However, since a state
that is invariant under stochastic orthogonal transformations is also
invariant under permutations, the de Finetti theorem with linear constraints
in 4.1.1 holds for any $m$, so a multiple of six can be chosen. In particular,
$k=m=0$ can be chosen. ∎

Both of the above stabilizer de Finetti theorems with linear constraints are
symmetric in the choice of Alice and Bob (see also Remark 5.5 in [13]), which
means that analogous theorems can be stated approximating Alice’s side instead
of Bob’s, which might lead to a better error bound depending on $d_{A}$ and
$d_{B}$. Subsequently, this will lead to a hierarchy approximating the encoder
of a QEC procedure by stabilizer state tensor powers instead of the decoder in
the next section.

Furthermore, one could also simultaneously approximate Alice’s and Bob’s side
by stabilizer state tensor powers by expanding the above theorems, leading to
a slight change in the error. For Theorem 4.2.1, proof of this can be found in
Appendix C.

###  4.3 An SDP Hierarchy for Maximum Channel Fidelity with Optimal Clifford
Decoder

As mentioned before, de Finetti theorems with permutation invariance can be
used to design an SDP hierarchy for approximating separable states, which is a
problem that appears in many quantum information theory settings, but is in
general rather hard to solve. When using a de Finetti approximation, instead
of enforcing separability as a numerical constraint, permutation invariance
can be imposed, which is in general more tractable. As the approximation
becomes better with increasing number of systems $n$, this immediately
translates to a hierarchy for optimizing polynomial functions. However,
although the de Finetti theorem in Theorem 4.1.1 from [13] has (at first
glance) worse convergence than best known standard de Finetti theorems, it
additionally and crucially ensures that the right linear constraints are
separately satisfied on Alice’s and Bob’s systems.

Instead of permutation-based de Finetti theorems, a similar hierarchy could be
constructed for states with invariance under the stochastic orthogonal group
based on the de Finetti theorem with stabilizer states, Theorem 2.2.2, found
in [10]. This in itself is interesting for some problems, e.g. optimizing over
the convex hull of stabilizers, which could be relaxed to a hierarchy of
stochastic orthogonal invariant states. In previous approaches to this
problem, this kind of optimization was found to be a linear problem [69] in
principle (though containing some tedious enumeration), and it is therefore
not clear at all that an SDP relaxation would provide an improvement. However,
particularly because of the exponential convergence of the Stabilizer de
Finetti Theorem, it might. However, using Theorem 2.2.2 alone would not allow
us to impose the important additional constraints.

In Section 4.2, we have found that Theorem 4.2.1 implies that a state with
stochastic orthogonal invariance on one side (here: Bob’s) is approximately
separable in the cut between $A$ and $B$ and approximated by a convex
combination of stabilizer states on Bob’s side. Similarly to [13], where the
de Finetti theorem with linear constraints implies that separable states can
be approximated by the hierarchy given by (4.1.3), this theorem implies that
the stabilizer de Finetti theorem with linear constraints can be used to
obtain a hierarchy approximating separable and partly stabilizer states.

For now, we are interested in finding the best arbitrary encoder and Clifford
decoder, which is given by the optimization problem 4.1.4. Using Choi-
Jamiołkowski isomorphism, this can be translated to the following optimization
problem:

######  Optimization problem 4.3.1.

$\begin{split}F_{C}(N,d_{M})=\text{maximize}\
&d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}E_{A\tilde{A}}\otimes
D_{B\tilde{B}}\big{)}\Big{)}\\\ \text{subject to }&E_{A\tilde{A}}\geq 0,\
D_{B\tilde{B}}\geq 0\\\
&\operatorname{tr}_{\tilde{A}}(E_{A\tilde{A}})=\frac{\mathbbm{1}_{A}}{d_{A}},\
\operatorname{tr}_{\tilde{B}}(D_{B\tilde{B}})=\frac{\mathbbm{1}_{B}}{d_{B}^{r}}\\\
&D_{B\tilde{B}}=\sum_{\sigma_{B\tilde{B}}}p_{S}(\sigma_{B\tilde{B}})\sigma_{B\tilde{B}}\text{
with stabilizer states $\sigma_{B\tilde{B}}$}\\\ \end{split}$

To connect this optimization problem with Theorem 2.2.2, the systems $A$ and
$B$ appearing in the stabilizer de Finetti theorem with linear constraints
must be renamed to $A\rightarrow A\tilde{A}$ and $B\rightarrow B\tilde{B}$.
Then, the linear constraints appearing in the theorem translate to

$\operatorname{tr}_{\tilde{A}}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=\frac{\mathbbm{1}_{A}}{d_{A}}\otimes\rho_{(B\tilde{B})_{1}^{n}}$

and

$\operatorname{tr}_{\tilde{B}_{n}}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=\rho_{(B\tilde{B})_{1}^{n-1}}\otimes\frac{\mathbbm{1}_{B_{n}}}{d_{B}^{r}}$

which corresponds to choosing $C_{A}={A}$,
$\Lambda_{A\tilde{A}\rightarrow{A}}=\operatorname{tr}_{\tilde{A}}$ and
$X_{A}=\frac{\mathbbm{1}_{A}}{d_{A}}$, as well as $C_{B}={B}$,
$\Gamma_{B\tilde{B}\rightarrow{B}}=\operatorname{tr}_{\tilde{B}}$ and
$Y_{B}=\frac{\mathbbm{1}_{B}}{d_{B}^{r}}$ and inserting these choices in the
constraints of Theorem 4.2.1.

Now, Theorem 4.2.1 implies that a state
$\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}}$ that is invariant under the action of
$O_{n}$ on Bob’s side can be used to approximate a separable state where Bob’s
side of the state (in the above case $D_{B\tilde{B}}$) is a mixture of
stabilizer states. Recast as an optimization problem, level $n$ of the
hierarchy is given by:

######  Optimization problem 4.3.2.

$\begin{split}F_{C}^{(n)}(N,d_{M})=\text{maximize}\
&d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}\big{)}\Big{)}\\\
\text{subject to }&\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}}\geq 0,\
\operatorname{tr}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=1\\\
&\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}}\text{ is invariant under the action of
}O_{n}\\\
&\operatorname{tr}_{\tilde{A}}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=\frac{\mathbbm{1}_{A}}{d_{A}}\otimes\rho_{(B\tilde{B})_{1}^{n}}\\\
&\operatorname{tr}_{\tilde{B}_{n}}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=\rho_{A\tilde{A}(B\tilde{B})_{1}^{n-1}}\otimes\frac{\mathbbm{1}_{B_{n}}}{d_{B}^{r}}\\\
\end{split}$

The optimal value $F_{C}(N,d_{M})$ of problem 4.3.1 (which is ultimately what
we want to know) and the optimal value $F_{C}^{(n)}(N,d_{M})$ of the SDP 4.3.2
(which is more easily computable) correspond directly to the states inserted
in the trace distance in the stabilizer de Finetti theorem with linear
constraints, and thereby, their difference corresponds to the error bound in
the theorem. With increasing level $n$, this error decreases, and the values
$F_{C}^{(n)}(N,d_{M})$ approach the value $F_{C}(N,d_{M})$. The convergence of
the SDP hierarchy, meaning
$\lim_{n\rightarrow\infty}F_{C}^{(n)}(N,d_{M})=F_{C}^{(n)}(N,d_{M})$, thus
follows from Theorem 4.2.1.

######  Theorem 4.3.1 (Convergence of the hierarchy).

Considering the SDPs in (4.3.2) with optimal value $F_{C}^{(n)}(N,d_{M})$ for
some $n\geq 0$, we find that

$F_{C}^{(n+1)}(N,d_{M})\leq F_{C}^{(n)}(N,d_{M})$
$\lim_{n\rightarrow\infty}F_{C}^{(n)}(N,d_{M})=F_{C}(N,d_{M})$

where $F_{C}(N,d_{M})$ is the optimal value to the optimization problem
(4.3.1).

###### Proof of Theorem 4.3.1.

Because of Theorem 4.1 in [13], we know that convergence holds for the
hierarchy with permutation invariance. Replacing permutation invariance by
stochastic orthogonal invariance does not change their proof. Because the
constraints become more powerful, $F_{C}^{(n+1)}(N,d_{M})\leq
F_{C}^{(n)}(N,d_{M})$. Given a feasible solution $E_{A\tilde{A}}\otimes
D_{B\tilde{B}}$ to $F_{C}(N,d_{M})$, a feasible solution for
$F_{C}^{(n)}(N,d_{M})$ can easily be constructed via $E_{A\tilde{A}}\otimes
D_{B\tilde{B}}^{\otimes n}$. Given a feasible solution to
$F_{C}^{(n)}(N,d_{M})$, this is only approximately equal to a convex
combination of feasible solutions to $F_{C}(N,d_{M})$, where the approximation
error is given in form of the bound of the de Finetti theorem, which improves
with increasing $n$, and tends to zero in the limit $n\rightarrow\infty$. This
can easily be seen in the following calculation:

$\begin{split}&F_{C}^{(n)}(N,d_{M})-F_{C}(N,d_{M})\\\
&=d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}\big{)}\Big{)}-d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}E_{A\tilde{A}}\otimes
D_{B\tilde{B}}\big{)}\Big{)}\\\
&=d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}-E_{A\tilde{A}}\otimes
D_{B\tilde{B}}\big{)}\Big{)}\\\
&=d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}-E_{A\tilde{A}}\otimes\sum_{\sigma_{B\tilde{B}}}p_{S}(\sigma_{B\tilde{B}})\sigma_{B\tilde{B}}\big{)}\Big{)}\\\
\end{split}$

$F_{C}(N,d_{M})$ can be rewritten in terms of a convex combination, which
corresponds to adding classical shared randomness assistance, which does not
affect the optimal value of the fidelity [13]. Then, we find

$\begin{split}F_{C}^{(n)}(N,d_{M})-F_{C}(N,d_{M})&=d_{\tilde{A}}d_{B}^{r}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}-\sum_{z,\sigma_{B\tilde{B}}}p_{Z}(z)E_{A\tilde{A}|z}\otimes
p_{S}(\sigma_{B\tilde{B}})\sigma_{B\tilde{B}}\big{)}\Big{)}\\\
&=d_{\tilde{A}}d_{B}^{r}\big{(}\epsilon(d_{B}^{r},d_{A},n)+\bar{\epsilon}(d_{B},r,n)\big{)}\operatorname{tr}\Big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\Big{)}\\\
\end{split}$

with $\epsilon(d_{B}^{r},d_{A},n)$ given in (4.1) and
$\bar{\epsilon}(d_{B},r,n)$ given in (4.2).

Then, because $\lim_{n\rightarrow\infty}\epsilon(d_{B}^{r},d_{A},n)=0$ and
$\lim_{n\rightarrow\infty}\tilde{\epsilon}(d_{B},r,n)=0$, the above tends to
zero, and consequently the objective value of the optimization problem 4.3.2
converges to the optimal value of the optimization problem 4.3.1.∎

######  Remark.

In contrast to the permutation-based version, the linear constraints are not
fulfilled exactly, but approximately. If separability is achieved after a low
number of steps, stabilizer-ness cannot be guaranteed. However, since the
convergence towards stabilizer states is exponential, it is considerably
faster than the convergence towards separability.

###  4.4 Remark on numerical tests

One central goal of establishing a convergent hierarchy like the one in
Section 4.3 is to implement and apply it to some numerical computations. Most
naturally, the simila-rities to the convergent hierarchy in [13] could be
exploited to obtain a direct comparison between the maximum channel fidelity
for arbitrary encoder and decoder, $F(N,d_{M})$ and the maximum channel
fidelity with Clifford operations $F_{C}(N,d_{M})$. However, due to the high
number of parameters and time constraints, the numerical implementation was
not completed within this thesis.

As a preliminary exercise, we were interested in finding the optimal Clifford
decoder for the 3-qubit bitflip code. This means that separability between
encoder and decoder is given, and we are not making use of the hierarchy
above, but a simplified version of it to reduce the number of parameters. Even
for two copies (i.e. no orthogonal symmetry, just swapping the subsystems), a
home computer did not have sufficient memory for this task. However, the
problem was not yet optimized by using symmetry to reduce the number of
variables, which may make the problem more tractable.

For studies making use of the hierarchy, it must be noted that only some
particular combinations of the number of subsystems $n$, the local dimension
$d$ and the number of qudits $r$ that the Clifford unitaries act on could be
of interest. In the case of qubits, i.e. $d=2$, the stabilizer de Finetti
theorem for qubits only holds for $n\geq 6$. For qudits and combinations $d=3$
and $n=2,3$, the stochastic orthogonal group is equal to the permutation
group. Consequently, the first potentially interesting case may appear for
$d=3$, $n=4$.

All of these combinations already imply a large number of parameters. However,
the number of parameters can be somewhat reduced because of the symmetry of
the problem. Permutation invariant states can be defined in terms of fewer
parameters than regular states (using Clebsch-Gordan coefficients). In
addition, depending on the problem, the symmetry of the channel can be used to
simplify the optimization as well (as noted in [13]).

As with many SDP problems, something may also be gained from looking at the
corresponding dual problem. It may reduce the number of variables (which is
unlikely in our case), or show some structure that allows for a good guess at
a feasible solution. More importantly, it could be argued that the dual is the
more natural optimization for the original problem because the dual directly
gives an upper bound on the optimal value for any feasible solution, thereby
giving an upper bound to the upper bound on the maximum channel fidelity
achieved at this level. The primal problem may only give an upper bound of the
maximum channel fidelity if it attains the optimal value, as it is a
maximization rather than a minimization. In other words, any feasible solution
of the dual provides an upper bound on $F_{C}^{(n)}(N,d_{M})$, while
$F_{C}^{(n)}(N,d_{M})$ itself is an upper bound on $F_{C}(N,d_{M})$.

To obtain the dual to optimization problem 4.3.2, it is helpful to rephrase
the problem in terms of maps:

######  Optimization problem 4.4.1.

$\begin{split}F_{C}^{(n)}(N,d_{M})=\text{maximize}\
&d_{\tilde{A}}d_{B}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}\big{)}\Big{)}\\\
\text{subject to }&\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}}\geq 0,\
\operatorname{tr}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=1\\\
&\mathcal{U}_{O}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}})=0\ \forall
O\in\mathcal{O}_{n}\\\ &\mathcal{C}_{A}(\rho_{A(B\tilde{B})_{1}^{n}})=0\\\
&\mathcal{C}_{B_{n}}(\rho_{A\tilde{A}(B\tilde{B})_{1}^{n-1}B_{n}})=0\\\
\end{split}$

with the following maps:

$\mathcal{U}_{O}(x)=OxO^{\dagger}-x$
$\mathcal{C}_{A}(x)=\big{(}\mathcal{T}_{HW}^{A}\otimes\mathbbm{1}_{(B\tilde{B})_{1}^{n}}\big{)}x\big{(}(\mathcal{T}_{HW}^{A})^{\dagger}\otimes\mathbbm{1}_{(B\tilde{B})_{1}^{n}}\big{)}-x$
$\mathcal{C}_{B_{n}}(x)=\big{(}\mathbbm{1}_{A\tilde{A}(B\tilde{B})_{1}^{n-1}}\otimes\mathcal{T}_{HW}^{B_{n}}\big{)}x\big{(}\mathbbm{1}_{A\tilde{A}(B\tilde{B})_{1}^{n-1}}\otimes(\mathcal{T}_{HW}^{B_{n}})^{\dagger}\big{)}-x$

with $\mathcal{T}_{HW}^{A}$ being the twirling map from (3.17) applied to
subsystem $A$, and $\mathcal{T}_{HW}^{B_{n}}$ the twirling of subsystem
$B_{n}$.

Then, the dual of optimization problem 4.4.1 (and thereby 4.3.2) is given by
the following optimization problem:

######  Optimization problem 4.4.2.

$\begin{split}F_{C}^{(n)}(N,d_{M})=\text{minimize}\ &\lambda\\\ \text{subject
to
}&\big{(}J_{\tilde{A}B_{1}}^{N}\otimes\Phi_{A\tilde{B}_{1}}\big{)}\otimes\mathbbm{1}_{(B\tilde{B})_{2}^{n}}+\sum_{O}\mathcal{U}_{O}(P_{A\tilde{A}(B\tilde{B})_{1}^{n}}^{O})\\\
&\ \ \ \ \
+\mathcal{C}_{A}(Y_{A(B\tilde{B})_{1}^{n}})\otimes\mathbbm{1}_{\tilde{A}}+\mathcal{C}_{B_{n}}(Z_{A\tilde{A}(B\tilde{B})_{1}^{n-1}B})\otimes\mathbbm{1}_{\tilde{B}_{n}}\leq\lambda\mathbbm{1}_{A\tilde{A}(B\tilde{B})_{1}^{n}}\\\
\end{split}$

As to be expected, while the primal problem contains one SDP variable
($\rho_{A\tilde{A}(B\tilde{B})_{1}^{n}}$) and many constraints, the dual
contains many SDP variables ($\lambda$,
$P_{A\tilde{A}(B\tilde{B})_{1}^{n}}^{O}$, $Y_{A(B\tilde{B})_{1}^{n}}$,
$Z_{A\tilde{A}(B\tilde{B})_{1}^{n-1}B}$) and only one constraint. If the
constraints in the primal are relaxed to inequality constraints instead of
equalities, the dual contains additional constraints on the positivity of
$P_{A\tilde{A}(B\tilde{B})_{1}^{n}}^{O}$, $Y_{A(B\tilde{B})_{1}^{n}}$ and
$Z_{A\tilde{A}(B\tilde{B})_{1}^{n-1}B}$.

While numerical tests are possible in principle, and something can be gained
from exploiting the symmetry of the problem to reduce the number of
parameters, any reasonable test would still require a computer with the
ability to keep track of large numbers of parameters. First attempts were
already beyond the capabilities of a home computer. None-theless, with
additional time and effort, numerical results which allow direct comparison to
[13] are likely attainable.

## 5 Summary and Outlook

In accordance with the multitudes of applications relying on the original de
Finetti theorem with permutation invariance, the stabilizer de Finetti theorem
in 2.2.2 also has various interesting applications. Although the results are
similar to some degree, they are nonetheless interesting, for two reasons: on
the one hand, the stabilizer de Finetti theorem has exponential convergence,
and on the other hand, it leads to an approximation by stabilizer states
instead of arbitrary quantum states.

The convergence of the theorem makes it particularly interesting for QKD,
where a large number of subsystems must usually be traced out to lead to any
reasonable error bound on the communication - a notorious problem for real-
world applications. Using the postselection technique, it can be shown that
the security of general attacks can be inferred from the security of
collective attacks, with a multiplicative factor on the error which depends on
the dimension of a permutation-invariant subspace. Here, we have shown that
the postselection technique can be generalized to arbitrary symmetry groups,
and a bound on the diamond norm of the difference of two CPTP maps can be
obtained from it, where this bound depends chiefly on the dimension of the
invariant subspace. Then, a computation of the dimension of an invariant
subspace associated with the symmetry appearing in the stabilizer de Finetti
theorem - stochastic orthogonal symmetry - was performed, finding that the
resulting multiplicative factor for QKD applications is significantly smaller
and independent of the number of subsystems $n$.

The natural next step would be using this postselection theorem to obtain key
rates and achievable key lengths for known protocols. The permutation-symmetry
based postselection theorem has already been applied to determine key rates
for a variety of protocols, including qubits [70] and qudits [71] in the BB84
protocol, 6-state protocol [58] and $N$-party 6-state protocol [45]. However,
finding a protocol with the desired symmetry is not trivial. For BB84, the
technique of using uncertainty relations for QKD [72] outperforms most de
Finetti type considerations, and it is unlikely that our result will
constitute an improvement. The 6-state protocol, on the other hand, is a more
promising candidate. If the 6-state protocol has stochastic orthogonal
invariance, the corresponding key rates can be determined by replacing one
summand in the key lengths found in [58] and [45] \- precisely the summand
corresponding to the difference between collective and general attack key
lengths (corresponding to the dimension of the permutation-invariant
subspace). Furthermore, as another potential protocol, one might use the
protocol appearing in the context of continuous variable de Finetti theorems
with orthogonal symmetry [59] and restrict it to finite dimensions, but the
practicality of such an endeavour is unclear.

Because it leads to an approximation by tensor powers of stabilizer states,
the stabilizer de Finetti theorem is also interesting for the study of
applications tied to stabilizer codes and Clifford operations. Here, we find
that it can be employed for benchmarking the success of a QEC procedure in
terms of its maximum channel fidelity, where either the encoder or decoder (or
both) are Clifford operations. Computing maximum channel fidelity contains a
bilinear optimization for finding the optimal encoder and decoder for a given
noisy channel, which is in general not directly possible. However, it can be
approximated by a converging hierarchy of SDP relaxations, which we also find
to be possible for maximum channel fidelity with restriction to Clifford
decoders (or encoders, or both).

This directly leads to the next, most logical step: performing numerical tests
for some examples. As a preliminary, one could look at a simplified problem of
finding the optimal Clifford decoder for 3-qubit repitition code with
bitflips, before moving on to more complicated examples which can be compared
to the numerical results in [13]. Although we did take first steps in this
direction, full fledged numerical computations are outside the scope of this
work.

While this thesis focuses on two aspects of applications emerging from the
stabilizer de Finetti theorem, it could also be interesting to investigate
many others. In particular, it could be beneficial to look at other results
relying on the permutation-based de Finetti theorem and postsselection
technique, including applications pertaining to tomography [7] and Shannon
reverse coding [31].

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##  Appendix A Proof of the Bound on the Diamond Norm

###  A.1 Resolution of Identity

The postselection technique can only be applied for a symmetry group
$\mathcal{S}$ if the following condition, termed resolution of identity,
holds:

$\operatorname{Cond0}=\Big{\\{}\exists d_{\mathcal{S}}(\cdot)\
\operatorname{s.t.}\ \int\rho^{\otimes
n}{d}_{\mathcal{S}}(\rho)=\frac{1}{g_{n,d}}\mathbbm{1}_{{N}}\Big{\\}}$ (A.1)

Here, we state two necessary, but not sufficient conditions for Cond0, and
thereby for resolution of identity. Thus, if these conditions are met, it is
sufficient to consider the particular state $\tau_{T}$ when computing the
diamond norm, instead of performing an optimization over a large number of
states:

$\displaystyle\operatorname{Cond1}=\Big{\\{}\exists\ket{\phi^{(k)}}\in\mathcal{H}\otimes\mathcal{H}\
\operatorname{s.t.}\ \mathcal{N}\ \operatorname{is\ spanned\ by}\
\ket{\phi^{(k)}}^{\otimes n}\Big{\\}}$
$\displaystyle{\Longleftrightarrow}\operatorname{Cond2}=\Big{\\{}\exists
X^{(k)}\in L(\mathcal{H},\mathcal{H})\ \operatorname{s.t.\ the\ commutant\
of}\ \mathcal{S}\ \operatorname{is\ spanned\ by}\ (X^{(k)})^{\otimes
n}\Big{\\}}$

where $L(\mathcal{H},\mathcal{H})$ denotes the space of linear maps
$X^{(k)}:\mathcal{H}\mapsto\mathcal{H}$, and
$\mathcal{N}=(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}$.

These alternative formulations provide an advantage over the original
condition Cond0 because they are conditions in linear algebra that are less
specific to our particular problem.

Firstly, we show the equivalence of the two conditions Cond1 and Cond2 by
finding a one-to-one map between the two sets they describe.

###### Proof: Cond1 $\Longleftrightarrow$ Cond2.

Firstly, note the definition of the space $\mathcal{N}$:

$\mathcal{N}=(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}=\\{\ket{\Phi}\in(\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes n})|\ket{\Phi}=s\otimes\overline{s}\ket{\Phi}\
\forall s\in{\mathcal{S}}\\}$

and the definition of the commutant:

$S^{\prime}=\\{X\in L(\mathcal{H}^{\otimes n},\mathcal{H}^{\otimes
n})|X=sXs^{\dagger}\ \forall s\in{\mathcal{S}}\\}$

There is a one-to-one map between these two spaces such that the $\ket{\Phi}$
and the $X$ can be transformed into one another.

Let $\\{\ket{i}\\}$ be a basis of $\mathcal{H}$. Then,
$\\{\ket{i}\otimes\ket{j}\\}$ is a basis of $\mathcal{H}\otimes\mathcal{H}$.
The map we propose transforms vectors on two copies of $\mathcal{H}$ to linear
maps between the two spaces: $\mathcal{H}\otimes\mathcal{H}\rightarrow
L(\mathcal{H},\mathcal{H})$, and acts in the following way:
$\ket{i}\otimes\ket{j}\mapsto\ket{i}\bra{j}$.

First, let us expand $\ket{\Phi}$ in this basis and perform the mapping.

$\ket{\Phi}=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\ket{i_{1},...,i_{n}}\otimes\ket{j_{1},...,j_{n}}\mapsto\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\ket{i_{1},...,i_{n}}\bra{j_{1},...,j_{n}}\equiv
X$

Now we show that the restrictions for $\ket{\Phi}\in\mathcal{N}$ and $X\in
S^{\prime}$ translate into one another.

$\displaystyle\ket{\Phi}$
$\displaystyle=(s\otimes\overline{s})\ket{\Phi}=(s\otimes\overline{s})\Big{(}\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\ket{i_{1},...,i_{n}}\otimes\ket{j_{1},...,j_{n}}\Big{)}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\otimes\Big{(}\overline{s}\ket{j_{1},...,j_{n}}\Big{)}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\otimes$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\bigg{(}\sum_{\begin{subarray}{c}m_{1},...,m_{n}\\\
n_{1},...,n_{n}\end{subarray}}\Big{(}\bra{m_{1},...,m_{n}}\overline{s}\ket{n_{1},...,n_{n}}\Big{)}\ket{m_{1},...,m_{n}}\bra{n_{1},...,n_{n}}\bigg{)}\ket{j_{1},...,j_{n}}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\ j_{1},...,j_{n}\\\
m_{1},...,m_{n}\\\
n_{1},...,n_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\otimes$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\big{(}\bra{m_{1},...,m_{n}}\overline{s}\ket{n_{1},...,n_{n}}\big{)}\ket{m_{1},...,m_{n}}\big{(}\bra{n_{1},...,n_{n}}\ket{j_{1},...,j_{n}}\big{)}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\ j_{1},...,j_{n}\\\
m_{1},...,m_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\otimes\big{(}\bra{m_{1},...,m_{n}}\overline{s}\ket{j_{1},...,j_{n}}\big{)}\ket{m_{1},...,m_{n}}$
$\displaystyle\mapsto\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\\\
m_{1},...,m_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\big{(}\bra{m_{1},...,m_{n}}\overline{s}\ket{j_{1},...,j_{n}}\big{)}\bra{m_{1},...,m_{n}}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\ j_{1},...,j_{n}\\\
m_{1},...,m_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\big{(}\bra{j_{1},...,j_{n}}(\overline{s})^{T}\ket{m_{1},...,m_{n}}\big{)}\bra{m_{1},...,m_{n}}$
$\displaystyle=\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\big{(}s\ket{i_{1},...,i_{n}}\big{)}\Big{(}\bra{j_{1},...,j_{n}}\overline{s}^{\dagger}\big{(}\sum_{m_{1},...,m_{n}}\ket{m_{1},...,m_{n}}\big{)}\bra{m_{1},...,m_{n}}\Big{)}$
$\displaystyle=s\Big{(}\sum_{\begin{subarray}{c}i_{1},...,i_{n}\\\
j_{1},...,j_{n}\end{subarray}}\Phi_{i_{1},...,i_{n},j_{1},...,j_{n}}\ket{i_{1},...,i_{n}}\bra{j_{1},...,j_{n}}\Big{)}s^{\dagger}=sXs^{\dagger}=X$

Each $\ket{\Phi}\in\mathcal{N}$ therefore defines an $X\in S^{\prime}$, and
since the proposed map is invertible and one-to-one, each $X\in S^{\prime}$
defines one $\ket{\Phi}\in\mathcal{N}$.

Now, Cond1 implies

$\mathcal{N}=\\{\ket{\Phi}\in(\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes
n})|\ket{\Phi}=\sum_{k}c_{k}\ket{\phi^{(k)}}^{\otimes n}\\}.$

Let the map be applied to each summand of $\ket{\Phi}$ (i.e. for each $k$):

$\ket{\phi^{(k)}}^{\otimes n}=...\mapsto....=(X^{(k)})^{\otimes n}$

Therefore, for each possible $\ket{\Phi}\in\mathcal{N}$, we obtain

$\ket{\Phi}=\sum_{k}c_{k}\ket{\phi^{(k)}}^{\otimes
n}\mapsto\sum_{k}c_{k}(X^{(k)})^{\otimes n}=X,$

defining all possible $X\in S^{\prime}$. In consequence, the commutant is
spanned by $(X^{(k)})^{\otimes n}$, proving Cond1 $\Rightarrow$ Cond2.

Since the map is one-to-one and invertible, the other direction (Cond2
$\Rightarrow$ Cond1) is analogous with the inverse map $X\mapsto\ket{\Phi}$.

∎

Having proven that the two proposed conditions are equivalent, we move on to
show that Cond1 (and thus the equivalent Cond2) is a necessary condition for
the original statement Cond0. By contraposition, this can be phrased as:
Whenever Cond0 is true, Cond1 is true, which is the statement that will be
proven in the following.

###### Proof: Cond0 $\Rightarrow$ Cond1.

We assume that a measure $d_{\mathcal{S}}(\cdot)$ exists, and
$\int\rho^{\otimes n}{d}_{\mathcal{S}}(\rho)=\frac{1}{g_{n,d}}\mathbbm{1}_{N}$
holds.

Since we are only concerned with finite spaces, the integral can be written as
a sum:

$\int\rho^{\otimes
n}{d}_{\mathcal{S}}(\rho)\rightarrow\sum_{k}({\rho^{(k)}})^{\otimes
n}{d}_{\mathcal{S}}({\rho^{(k)}})$

As this is equal to the identity on $\mathcal{N}$ by assumption, this implies
that the space $\mathcal{N}$ is spanned by the $({\rho^{(k)}})^{\otimes n}$.
Because each ${\rho^{(k)}}$ is pure, there exists a vector $\ket{\phi^{(k)}}$
such that ${\rho^{(k)}}=\ket{\phi^{(k)}}\bra{\phi^{(k)}}$, and therefore
$({\rho^{(k)}})^{\otimes n}=(\ket{\phi^{(k)}}\bra{\phi^{(k)}})^{\otimes n}$.
Thus, $\exists\ket{\phi^{(k)}}$ such that $\mathcal{N}$ is spanned by
$\ket{\phi^{(k)}}^{\otimes n}$ (Cond1).

∎

Thus Cond1 (and the equivalent Cond2) are necessary, but not sufficient
conditions of the original statement Cond0, which in turn is a necessary and
sufficient condition for the postselection technique to be applicable for a
symmetry $\mathcal{S}$.

###  A.2 Preliminary Lemmata

The postselection theorem 3.2.1 states that the diamond norm of a difference
of $\mathcal{S}$-invariant CPTP maps, which entails an optimization problem
over all states within a certain space, can be bound using one specific state,
which is a purification of the de Finetti state given in (3.4). To simplify
the proof of this bound on the diamond norm in Theorem 3.2.1, three lemmata
are useful. First, Lemma A.2.1 establishes a connection between states on
$\mathcal{H}^{\otimes n}$ that are invariant under the symmetry $\mathcal{S}$
and a state with support on the invariant subspace $(\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes n})^{(s\otimes\overline{s})}$. This lemma will
appear in the proof of Lemma 3.2.1, where it is shown to be sufficient to
consider states with support on the invariant subspace
$\mathcal{N}=(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}$ for computing the diamond norm of an invariant
map. Thirdly, in the context of the diamond norm, we state and prove Lemma
3.2.2 which connects states with support on the invariant subspace
$\mathcal{N}$ and the de Finetti state from (3.4).

######  Lemma A.2.1.

Any state $\rho_{T}$ on $\mathcal{H}_{T}=\mathcal{H}^{\otimes n}$ that is
invariant under a symmetry group $\mathcal{S}$ admits a purification
$\rho_{TE}$ with support on $\mathcal{N}=(\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}\subseteq\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes n}=\mathcal{H}_{T}\otimes\mathcal{H}_{E}$.

###### Proof of Lemma A.2.1.

The state $\rho_{T}$ is $\mathcal{S}$-invariant, which means
$\rho_{T}=s\rho_{T}s^{\dagger}$ $\forall s\in\mathcal{S}$. Let $\Lambda$ be
the set of eigenvalues of $\rho_{T}$, and let
$\\{\ket{x}\\}_{x\in\mathcal{X}}$ be its eigenbasis. For each eigenvalue
$\lambda\in\Lambda$, consider the associated eigenvectors
$\\{\ket{x_{\lambda}}\\}_{x_{\lambda}\in\mathcal{X}_{\lambda}}$ (with
$\mathcal{X}_{\lambda}\subseteq\mathcal{X}$). Then, for each
$\lambda\in\Lambda$, $\rho_{T}\ket{x_{\lambda}}=\lambda\ket{x_{\lambda}}$
$\forall x_{\lambda}\in\mathcal{X}_{\lambda}$. With these eigenvalues and the
eigenbasis, we can write the state as follows:

$\rho_{T}=\sum_{\lambda\in\Lambda}\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\lambda\ket{x_{\lambda}}\bra{x_{\lambda}}$

Construct the following state for each $\lambda\in\Lambda$:

$\ket{\Phi_{\lambda}}=\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes{\ket{x_{\lambda}}}.$

Then, we claim that
$\ket{\Phi}\bra{\Phi}\in\mathfrak{S}(\mathcal{H}_{T}\otimes\mathcal{H}_{E})$
with $\mathcal{H}_{E}=\mathcal{H}_{T}$ and with

$\ket{\Phi}=\sum_{\lambda\in\Lambda}\sqrt{\lambda}\ket{\Phi_{\lambda}}$

is a purification of $\rho_{T}\in\mathfrak{S}(\mathcal{H}_{T})$ with the
desired characteristics.

To test this claim, we firstly show that $\ket{\Phi}\bra{\Phi}$ is indeed a
purification of $\rho_{T}$, which can be shown by a quick calculation:

$\operatorname{tr}_{E}\ket{\Phi}\bra{\Phi}=\operatorname{tr}_{E}(\sum_{\lambda,\lambda^{\prime}\in\Lambda}\sqrt{\lambda}\sqrt{\lambda^{\prime}}\sum_{\begin{subarray}{c}x_{\lambda}\in\mathcal{X}_{\lambda}\\\
x_{\lambda^{\prime}}\in\mathcal{X}_{\lambda^{\prime}}\end{subarray}}(\ket{x_{\lambda}}\otimes{\ket{x_{\lambda}}})(\bra{x_{\lambda^{\prime}}}\otimes{\bra{x_{\lambda^{\prime}}}}))$
$=\sum_{\lambda,\lambda^{\prime}\in\Lambda}\sum_{\begin{subarray}{c}x_{\lambda}\in\mathcal{X}_{\lambda}\\\
x_{\lambda^{\prime}}\in\mathcal{X}_{\lambda^{\prime}}\end{subarray}}\sqrt{\lambda}\sqrt{\lambda^{\prime}}\operatorname{tr}_{E}(\ket{x_{\lambda}}\bra{x_{\lambda^{\prime}}}\otimes{\ket{x_{\lambda}}}{\bra{x_{\lambda^{\prime}}}})=\sum_{\lambda\in\Lambda}\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\lambda\ket{x_{\lambda}}\bra{x_{\lambda}}=\rho_{T}$

Secondly, we show that $\ket{\Phi}\bra{\Phi}$ has support on $\mathcal{N}$.
Since $\ket{\Phi_{\lambda}}$ has a structure similar to a kind of “maximally
entangled state”, the following property holds for any operator $A$:

$\begin{split}\mathbbm{1}_{T}\otimes
A\ket{\Phi_{\lambda}}&=\mathbbm{1}_{T}\otimes
A\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes{\ket{x_{\lambda}}}=\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes{A\ket{x_{\lambda}}}\\\
&=\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes{\Big{(}\sum_{y_{\lambda}\in\mathcal{X}_{\lambda}}\ket{y_{\lambda}}\bra{y_{\lambda}}\Big{)}A\ket{x_{\lambda}}}\\\
&=\sum_{x_{\lambda},y_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes\ket{y_{\lambda}}\bra{y_{\lambda}}A\ket{x_{\lambda}}=\sum_{x_{\lambda},y_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\otimes\ket{y_{\lambda}}\bra{x_{\lambda}}A^{T}\ket{y_{\lambda}}\\\
&=\sum_{y_{\lambda}\in\mathcal{X}_{\lambda}}(\sum_{x_{\lambda}\in\mathcal{X}_{\lambda}}\ket{x_{\lambda}}\bra{x_{\lambda}})A^{T}\ket{y_{\lambda}}\otimes\ket{y_{\lambda}}=A^{T}\otimes\mathbbm{1}_{E}\sum_{y_{\lambda}\in\mathcal{X}_{\lambda}}\ket{y_{\lambda}}\otimes\ket{y_{\lambda}}=A^{T}\otimes\mathbbm{1}_{E}\ket{\Phi_{\lambda}}\end{split}$

With this property, we find for symmetry transformations $s\in\mathcal{S}$:

$s\otimes\overline{s}\ket{\Phi_{\lambda}}=s(\overline{s})^{T}\otimes\mathbbm{1}_{E}\ket{\Phi_{\lambda}}=ss^{\dagger}\otimes\mathbbm{1}_{E}\ket{\Phi_{\lambda}}=\ket{\Phi_{\lambda}}$

since we can assume $s$ to be a unitary representation with
$ss^{\dagger}=\mathbbm{1}_{T}$. By linearity, the above statement is also true
for a linear combination of $\ket{\Phi_{\lambda}}$, and thus for $\ket{\Phi}$.
In conclusion, $\ket{\Phi}\bra{\Phi}$ is a purification of $\rho_{T}$, and it
has support on $\mathcal{N}$.

∎

For the permutation group and previous versions of a quantum de Finetti
theorem, this lemma and its proof appear in [3, 35].

With this established, we can investigate the diamond norm. To begin, we move
from arbitrary states to states with support on the $\mathcal{S}$-invariant
subspace $\mathcal{N}$. The following lemma implies that it is sufficient to
consider such states when computing the diamond norm of a map that is
invariant under $\mathcal{S}$.

######  Lemma 3.2.1.

Let $\Delta$ be a linear map from $\operatorname{End}(\mathcal{H}^{\otimes
n})$ to $\operatorname{End}(\mathcal{H}^{\prime})$ that is invariant under the
symmetry $\mathcal{S}$. For any finite-dimensional space $\mathcal{M}$ and any
(arbitrary) density operator $\sigma_{TM}$, the following holds:

$\big{\|}(\Delta\otimes\mathbbm{1})\sigma_{TM}\big{\|}_{\operatorname{tr}}\leq\big{\|}(\Delta\otimes\mathbbm{1})\rho_{TE}\big{\|}_{\operatorname{tr}}$

where $\rho_{TE}$ is a state with support on
$\mathcal{N}=(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}$.

###### Proof of Lemma 3.2.1.

We introduce an additional space $\mathcal{L}$ with dimension $|\mathcal{S}|$
equal to the number of elements of the symmetry group $\mathcal{S}$. Then, we
can write its orthonormal basis as $\\{\ket{s}\\}_{s\in\mathcal{S}}$.

Using this, the following transformation of the state is possible:

$\big{\|}(\Delta\otimes\mathbbm{1}_{{M}})\sigma_{TM}\big{\|}_{\operatorname{tr}}=\big{\|}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\mathbbm{1}_{{L}})\big{\|}_{\operatorname{tr}}$
$=\big{\|}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}=\big{\|}\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{T{M}}\otimes\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}$

Since every $s$ is trace non-increasing, it can be inserted in the following
way:

$\big{\|}\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}=\big{\|}\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(s\circ\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\ket{s}\bra{s}_{L})\big{\|}_{\operatorname{tr}}$

By assumption, the CPTP map $\Delta$ is invariant under the symmetry
$\mathcal{S}$, meaning $\Delta=s\circ\Delta\circ s^{\dagger}\ \forall
s\in\mathcal{S}$. Thus, equivalently, $s\circ\Delta=\Delta\circ s$. Therefore:

$\big{\|}\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(s\circ\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}=\big{\|}\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(\Delta\circ
s\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}$

$=\big{\|}(\Delta\otimes\mathbbm{1}_{ML})\frac{1}{|\mathcal{S}|}\sum_{s\in\mathcal{S}}(s\otimes\mathbbm{1}_{ML})(\sigma_{TM}\otimes\ket{s}\bra{s}_{{L}})\big{\|}_{\operatorname{tr}}=\big{\|}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TML})\big{\|}_{\operatorname{tr}}$

Thus we have constructed a state $\sigma_{TML}$. For this state, its marginal
$\rho_{T}=\operatorname{tr}_{{M}{L}}(\sigma_{TML})$ is invariant under all
$s$, thus invariant under the symmetry group $\mathcal{S}$ (by construction).

For any state $\rho_{T}$ that is invariant under the symmetry group
$\mathcal{S}$, Lemma A.2.1 states that there exists a purification
$\rho_{TE}\in\mathfrak{S}(\mathcal{H}_{T}\otimes\mathcal{H}_{E})$,
$\mathcal{H}_{E}=\mathcal{H}_{T}^{\prime}$ with support on $\mathcal{N}$.

All purifications are equal up to an isometry, and thus there exists a CPTP
map $\mathcal{Z}:\operatorname{End}(\mathcal{H}^{\otimes
n})\rightarrow\operatorname{End}(\mathcal{M}\mathcal{L})$ such that
$\sigma_{TML}=(\mathbbm{1}_{T}\otimes Z)\rho_{TE}$.

Using this and the fact that $\mathcal{Z}$ is trace-preserving (and thus
trace-nonincreasing), we find:

$\big{\|}(\Delta\otimes\mathbbm{1}_{ML})(\sigma_{TML})\big{\|}_{\operatorname{tr}}=\big{\|}(\Delta\otimes\mathbbm{1}_{ML})(\mathbbm{1}_{T}\otimes\mathcal{Z})\rho_{TE}\big{\|}_{\operatorname{tr}}\leq\big{\|}(\Delta\otimes\mathbbm{1}_{ML})\rho_{TE}\big{\|}_{\operatorname{tr}}$

In summary:

$\big{\|}(\Delta\otimes\mathbbm{1}_{{M}})\sigma_{TM}\big{\|}_{\operatorname{tr}}\leq\big{\|}\
(\Delta\otimes\mathbbm{1}_{E})\rho_{TE}\big{\|}_{\operatorname{tr}}$

∎

As a second step in bounding the diamond norm, we establish a connection
between states with support on the $\mathcal{S}$-invariant subspace
$(\mathcal{H}^{\otimes n})^{s}$, and a purification $\tau_{TEN}$ of our
specific de Finetti state $\tau_{T}$ (3.4).

Here, the specific structure of $\tau_{T}$ becomes important and the
assumption of resolution of identity for $\tau_{TE}$, as discussed in Section
A.1, is required.

######  Lemma 3.2.2.

Suppose we have a state $\rho_{TE}$ with support on the subspace
$\mathcal{N}=(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}\subseteq\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes n}$. For any such state, there exists a linear
completely positive trace-nonincreasing map
$\mathcal{C}:\operatorname{End}(\mathcal{N})\to\mathbbm{C}$ such that

$\rho_{TE}=g_{n,d}(\mathbbm{1}_{TE}\otimes\mathcal{C})(\tau_{TEN})$

with
$\operatorname{tr}_{\mathcal{N}}\tau_{TEN}=\tau_{TE}=\frac{1}{g_{n,d}}\mathbbm{1}_{N}$
and $g_{n,d}=\dim(\mathcal{N})$.

###### Proof of Lemma 3.2.2.

Let $\\{\ket{i}\\}_{i}$ be an eigenbasis of $\rho_{TE}$. Since $\tau_{TEN}$ is
a purification of $\tau_{TE}\propto\mathbbm{1}_{TEN}$, it can be written as
$\tau_{TEN}=\ket{\Psi}\bra{\Psi}$ with a pure state
$\ket{\Psi}=\frac{1}{\sqrt{g_{n,d}}}\sum_{i}\ket{i}\otimes\ket{i}$ with

$g_{n,d}=\dim(\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes
n})^{(s\otimes\overline{s})}=\dim(\mathcal{N}).$ (A.2)

Let
$\mathcal{C}:\sigma_{\mathcal{N}}\mapsto\operatorname{tr}(\sigma_{\mathcal{N}}\rho^{T}_{\mathcal{N}}$).
Then we find

$g_{n,d}(\mathbbm{1}_{TE}\otimes\mathcal{C})(\tau_{TEN})=(\mathbbm{1}_{TE}\otimes\mathcal{C})\sum_{i,j}(\ket{i}\otimes\ket{i})(\bra{j}\otimes\bra{j})=\sum_{i,j}(\ket{i}\bra{j}\otimes\mathcal{C}(\ket{i}\bra{j})$
$=\sum_{i,j}(\ket{i}\bra{j}\otimes\operatorname{tr}(\ket{i}\bra{j}\rho^{T}_{\mathcal{N}})=\sum_{i,j}\ket{i}\bra{j}(\bra{j}\rho^{T}_{\mathcal{N}}\ket{i})=\sum_{i,j}\ket{i}\bra{j}(\rho^{T}_{\mathcal{N}})_{j,i}=\sum_{i,j}\ket{i}\bra{j}\otimes(\rho_{TE})_{i,j}=\rho_{TE}.$

Thus it is demonstrated that for any $\rho_{TE}$ on $(\mathcal{H}^{\otimes
n}\otimes\mathcal{H}^{\otimes n})^{(s\otimes\overline{s})}$, a map
$\mathcal{C}$ exists such that the Lemma holds.

∎

######  Remark.

This proof can be interpreted as a teleportation. Since
$\tau_{TE}\propto\mathbbm{1}_{N}$, its purification will be a maximally
entangled state in the corresponding space. This maximally entangled state can
then be used as a resource to teleport the state $\rho$ from $\mathcal{N}$ to
$\mathcal{H}^{\otimes n}\otimes\mathcal{H}^{\otimes n}$.

##  Appendix B Proof of the de Finetti Theorem with Linear Constraints

In this part of the appendix, the de Finetti theorem with linear constraints
from Section 4.2 will be proven:

######  Theorem 4.1.1 (De Finetti Theorem with Linear Constraints).

Let $\rho_{AB_{1}^{n}}$ be a quantum state that is permutation invariant with
respect to permutations of the $n$ subsystems $B_{1}^{n}$, let
$\Lambda_{A\rightarrow C_{A}}$ and $\Gamma_{B\rightarrow C_{B}}$ be linear
maps, and $X_{C_{A}}$ and $Y_{C_{B}}$ be operators such that the following two
linear constraints hold:

$\Lambda_{A\rightarrow
C_{A}}(\rho_{AB_{1}^{n}})=X_{C_{A}}\otimes\rho_{B_{1}^{n}},$
$\Gamma_{B_{n}\rightarrow C_{B}}(\rho_{B_{1}^{n}})=\rho_{B_{1}^{n-1}}\otimes
Y_{C_{B}}.$

Then, there exists an $m\in[0,n-1]$ and a probability distribution
$\\{p_{Z}(z_{1}^{m})\\}_{z_{1}^{m}\in Z}$ such that

$\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}\leq\epsilon(d_{B},d_{A},n)$

with

$\epsilon(d_{B},d_{A},n):=\min\Big{\\{}d_{B}^{2}(d_{B}+1),18\sqrt{d_{A}d_{B}}\Big{\\}}\sqrt{\frac{2\ln(2)\ln(d_{A})}{n}}$

and

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z_{1}^{m}})=X_{C_{A}},\
\Gamma_{B_{m+1}\rightarrow C_{B}}(\rho_{B_{m+1}|z_{1}^{m}})=Y_{C_{B}}.$

To prove this theorem, there are three helpful preliminary lemmata. All of
these lemmata are given and proven in either [13] or [73]. However, the first
lemma, Lemma B.0.1, is stated to be requiring permutation invariance, which we
have found to be an unnecessary assumption that can be omitted. We give here a
more detailed proof than the original.

The first lemma connects the entire system’s quantum state $\rho_{AZ_{1}^{n}}$
with post-measurement states (conditioned on some measurement outcomes), and
provides a bound on the expectation value of their trace distance.

######  Lemma B.0.1 (See [13], Lemma 3.1).

Let $\rho_{AZ_{1}^{n}}$ be a quantum state with the $Z_{1}^{n}$ systems
classical. Then, there exists $0\leq m<n-1$, such that

$\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AZ_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}\leq\frac{2\ln(2)\ln(d_{A})}{n}.$

###### Proof of Lemma B.0.1.

Using the fact that the quantum relative entropy of two quantum states is
bounded by the dimension of a subsystem, we can find a bound on the
expectation value of quantum relative entropy of the quantum state
$\rho_{AZ_{1}^{n}}$ and a separable, conditional state
$\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}}$, before employing
Pinzker’s inequality to relate this to a bound on the distance of these two
quantum states.

Quantum relative entropy is defined for arbitrary quantum states $\rho$ and
$\sigma$ in the following way:

$D(\rho||\sigma):=\operatorname{tr}\big{(}\rho(\log(\rho)-\log(\sigma)\big{)}$

and is bound by the dimensions of the systems.

We compare the quantum state $\rho_{AZ_{1}^{n}}$ and a related separable state
$\operatorname{tr}_{Z_{1}^{n}}(\rho_{AZ_{1}^{n}})\otimes\operatorname{tr}_{A}(\rho_{AZ_{1}^{n}})=\rho_{A}\otimes\rho_{Z_{1}^{n}}$,
to find that the following always holds:

$D(\rho_{AZ_{1}^{n}}\|\rho_{A}\otimes\rho_{Z_{1}^{n}})\leq\log(d_{A}).$

The quantum relative entropy above can be recast as a sum over the quantum
relative entropy of different subsystems.

$D(\rho_{AZ_{1}^{n}}\|\rho_{A}\otimes\rho_{Z_{1}^{n}})=\sum_{m=0}^{n-1}\Big{(}D(\rho_{AZ_{1}^{m+1}}\|\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-D(\rho_{AZ_{1}^{m}}\|\rho_{A}\otimes\rho_{Z_{1}^{m}})\Big{)}\leq\log(d_{A})$

which can be related to a bound on one individual term of the sum: Because the
$Z$-systems are classical, each term is positive, and therefore there exists
an $m\in[0,n-1]$ such that

$D(\rho_{AZ_{1}^{m+1}}\|\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-D(\rho_{AZ_{1}^{m}}\|\rho_{A}\otimes\rho_{Z_{1}^{m}})\leq\frac{\log(d_{A})}{n}.$

Using the definition of the quantum relative entropy, we can write this out in
terms of traces:

$\displaystyle
D(\rho_{AZ_{1}^{m+1}}\|\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-D(\rho_{AZ_{1}^{m}}\|\rho_{A}\otimes\rho_{Z_{1}^{m}})$
$\displaystyle=\operatorname{tr}\Big{(}\rho_{AZ_{1}^{m+1}}(\log(\rho_{AZ_{1}^{m+1}})-\log(\rho_{A}\otimes\rho_{Z_{1}^{m+1}})\Big{)}-\operatorname{tr}\Big{(}\rho_{AZ_{1}^{m}}(\log(\rho_{AZ_{1}^{m}})-\log(\rho_{A}\otimes\rho_{Z_{1}^{m}})\Big{)}$
$\displaystyle=\operatorname{tr}\Big{(}\rho_{AZ_{1}^{m+1}}\big{(}\log(\rho_{AZ_{1}^{m+1}})-\log(\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-\log(\rho_{AZ_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}-\log(\rho_{A}\otimes\rho_{Z_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}\big{)}\Big{)}$

Now, two characteristics of the matrix logarithm are needed: On the one hand,
if two matrices $A$ and $B$ commute, their logarithm is additive:
$\log(AB)=\log(A)+\log(B)$. On the other hand,
$\log(A\otimes\mathbbm{1})=\log(A)\otimes\mathbbm{1}$. Using these two
identities, we obtain

$\displaystyle
D(\rho_{AZ_{1}^{m+1}}\|\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-D(\rho_{AZ_{1}^{m}}\|\rho_{A}\otimes\rho_{Z_{1}^{m}})=$
$\displaystyle\operatorname{tr}\Big{(}\rho_{AZ_{1}^{m+1}}\big{(}\log(\rho_{AZ_{1}^{m+1}})-\log(\rho_{A})\otimes\mathbbm{1}_{Z_{1}^{m+1}}-\mathbbm{1}_{A}\otimes\log(\rho_{Z_{1}^{m+1}})-\log(\rho_{AZ_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
-\log(\rho_{A})\otimes\mathbbm{1}_{Z_{1}^{m+1}}-\mathbbm{1}_{A}\otimes\log(\rho_{Z_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}\big{)}\Big{)}.$

Some of these terms cancel. In addition, we can make use of the fact that the
$Z$ systems are classical, which means the state can be written as

$\rho_{AZ_{1}^{m+1}}=\sum_{z_{1}^{m+1}}p_{Z}(z_{1}^{m+1})\rho_{A|z_{1}^{m+1}}\otimes\ket{z_{1}^{m}}\bra{z_{1}^{m}}=\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\rho_{AZ_{m+1}|z_{1}^{m+1}},$

and its logarithm is

$\log(\rho_{AZ_{1}^{m+1}})=\sum_{z_{1}^{m}}\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\big{(}\log(p_{Z}(z_{1}^{m}))\mathbbm{1}_{AZ_{m+1}}+p_{Z}(z_{1}^{m})\log(\rho_{AZ_{m+1}|z_{1}^{m}})\big{)}.$

Inserting this, we arrive at the following:

$\displaystyle
D(\rho_{AZ_{1}^{m+1}}\|\rho_{A}\otimes\rho_{Z_{1}^{m+1}})-D(\rho_{AZ_{1}^{m}}\|\rho_{A}\otimes\rho_{Z_{1}^{m}})$
$\displaystyle=\operatorname{tr}\bigg{(}\Big{(}\sum_{z_{1}^{m+1}}p_{Z}(z_{1}^{m+1})\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\rho_{AZ_{m+1}|z_{1}^{m+1}}\Big{)}\Big{(}\sum_{\tilde{z}_{1}^{m+1}}\ket{\tilde{z}_{1}^{m}}\bra{\tilde{z}_{1}^{m}}\otimes\big{(}\log(p_{Z}(\tilde{z}_{1}^{m})\otimes\mathbbm{1}_{AZ_{m+1}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
+p_{Z}(\tilde{z}_{1}^{m})\log(\rho_{AZ_{m+1}|\tilde{z}_{1}^{m}})-\log(p_{Z}(\tilde{z}_{1}^{m})\otimes\mathbbm{1}_{AZ_{m+1}}-p_{Z}(\tilde{z}_{1}^{m})\mathbbm{1}_{A}\otimes\log(\rho_{Z_{m+1}|\tilde{z}_{1}^{m}})$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
-\log(p_{Z}(\tilde{z}_{1}^{m})\otimes\mathbbm{1}_{AZ_{m+1}}-p_{Z}(\tilde{z}_{1}^{m})\log(\rho_{A|\tilde{z}_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}+\log(p_{Z}(\tilde{z}_{1}^{m})\otimes\mathbbm{1}_{AZ_{m+1}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
+p_{Z}(\tilde{z}_{1}^{m})\mathbbm{1}_{A}\otimes\log(\mathbbm{1}_{AZ_{m+1}})\big{)}\Big{)}\bigg{)}$
$\displaystyle=\operatorname{tr}\bigg{(}\sum_{z_{1}^{m+1}}p(z_{m+1}|z_{1}^{m})\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\Big{(}\rho_{AZ_{m+1}|z_{1}^{m+1}}\big{(}p_{Z}(z_{1}^{m})\log(\rho_{AZ_{m+1}|z_{1}^{m}})-p_{Z}(z_{1}^{m})\mathbbm{1}_{A}\otimes\log(\rho_{Z_{m+1}|z_{1}^{m}})$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \
-p_{Z}(z_{1}^{m})\log(\rho_{A|z_{1}^{m}})\otimes\mathbbm{1}_{Z_{m+1}}\big{)}\Big{)}\bigg{)}$
$\displaystyle=\operatorname{tr}\bigg{(}\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m+1})p_{Z}(z_{1}^{m})\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\Big{(}\rho_{AZ_{m+1}|z_{1}^{m+1}}\big{(}\log(\rho_{AZ_{m+1}|z_{1}^{m}})-\log(\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}})\big{)}\Big{)}\bigg{)}$
$\displaystyle=\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\operatorname{tr}\bigg{(}\sum_{z_{m+1}}p_{Z}(z_{1}^{m+1})p_{Z}(z_{1}^{m})\ket{z_{1}^{m}}\bra{z_{1}^{m}}\otimes\Big{(}\rho_{AZ_{m+1}|z_{1}^{m+1}}\big{(}\log(\rho_{AZ_{m+1}|z_{1}^{m}})-\log(\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}})\big{)}\Big{)}\bigg{)}$
$\displaystyle=\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\operatorname{tr}\Big{(}\rho_{AZ_{m+1}|z_{1}^{m+1}}\big{(}\log(\rho_{AZ_{m+1}|z_{1}^{m}})-\log(\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}})\big{)}\Big{)}$
$\displaystyle=\mathbb{E}_{z_{1}^{m}}\Big{\\{}D(\rho_{AZ_{m+1}|z_{1}^{m}}\|\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}})\Big{\\}}\leq\frac{\log(d_{A})}{n}$

Thereby, we obtain a bound on the quantum relative entropy of the states we
want to compare. Now, we can use Pinsker’s inequality [73] to relate the
quantum relative entropy to the trace distance of the states via

$D(\rho||\sigma)\geq\frac{1}{2\ln(2)}\big{(}\operatorname{tr}(|\sigma-\rho|)\big{)}^{2}=\frac{1}{2\ln(2)}\|\rho-\sigma\|_{\operatorname{tr}}^{2}.$

Then, we obtain the statement of the claim for a distance between density
operators. ∎

To facilitate the use of Lemma B.0.1, we need a means to connect the physical
systems $B_{1}^{n}$ to classical systems $Z_{1}^{n}$, which is done via
measurement of the physical systems. There are two competing strategies to
move between the systems, which relate to different changes in the trace
distance (called measurement distortion).

######  Lemma B.0.2 (See [13], Lemma 3.2, and [73], Lemma 16).

There exists a product mea-surement $M_{A}\otimes M_{B}$ with finitely many
outcomes such that for any Hermitian and traceless operator $\xi_{AB}$, we
have

$\Big{\|}(M_{A}\otimes
M_{B})\xi_{AB}\Big{\|}_{\operatorname{tr}}\geq\frac{1}{18\sqrt{d_{A}d_{B}}}\Big{\|}\xi_{AB}\Big{\|}_{\operatorname{tr}}.$

This lemma has been proven in [73], Section 3.1.

######  Lemma B.0.3 (See [13], Lemma 3.3).

Consider a state two-design on $B$, i.e. a set of rank-one projectors
$\\{P_{z}\\}$ such that $\frac{1}{n}\sum_{z=1}^{Z}P_{z}\otimes
P_{z}=\frac{2P_{Symm}}{d_{B}(d_{B}+1)}$, where $P_{Symm}$ is the projector on
the symmetric subspace of $B\otimes B$. Let $M_{B}$ be the measurement defined
by

$M_{B}(x)=\sum_{z}\frac{d_{B}}{Z}\operatorname{tr}(P_{z}x)\ket{z}\bra{z}.$

Then, for any Hermitian opertor $\xi_{AB}$,

$\Big{\|}(\mathbbm{1}_{A}\otimes
M_{B})\xi_{AB}\Big{\|}_{\operatorname{tr}}\geq\frac{1}{d_{B}^{2}(d_{B}+1)}\Big{\|}\xi_{AB}\Big{\|}_{\operatorname{tr}}.$

This lemma has been proven in [13], Section 3.

It depends on the subsystem’s underlying dimensions which of these distortions
has a greater impact; we are interested in the minimum of the two.

Having established these three lemmata, we can state the proof for the
original de Finetti statement with linear constraints, Theorem 4.1.1.

###### Proof of Theorem 4.1.1.

This proof has three key steps: First, we will show that bounding the left-
hand side is related to bounding an expectation value of the squared trace
distance of two states. Then, Lemmas B.0.1, B.0.2 and B.0.3 are employed to
bound this expectation value. Lastly, it remains to be shown that the linear
constraints are obeyed.

For some $m\in[0,n-1]$, note that

$\begin{split}\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}&=\Big{\|}\rho_{AB_{m+1}}-\mathbb{E}_{z_{1}^{m}}\Big{\\{}\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\\}}\Big{\|}_{\operatorname{tr}}\\\
&\leq\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}\Big{\\}}\\\
&\leq\sqrt{\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}}.\end{split}$

This follows from the fact that
$\mathbb{E}_{z_{1}^{m}}\rho_{AB_{m+1}|z_{1}^{m}}=\rho_{AB_{m+1}}$, the
convexity of the norm, and the convexity of the square function. Note that a
bound on this expectation value of squared trace distance will result in a
bound on the original statement, so we will now be bounding this expression
instead.

To make use of Lemma B.0.1, which provides a bound on an expectation value of
squared trace distance, the system on Bob’s side must be classical. To this
end, a measurement needs to be performed to map the $(m+1)$-th system from
$B_{m+1}$ to $Z_{m+1}$ at the cost of some distortion. Here, either Lemma
B.0.2 or Lemma B.0.3 could be applied, at different costs.

In the first case, using Lemma B.0.2, we obtain

$\displaystyle\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle\leq(18\sqrt{d_{A}d_{B}})^{2}\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}(M_{A}\otimes
M_{B_{m+1}})\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle=(18\sqrt{d_{A}d_{B}})^{2}\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AZ_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle\leq(18\sqrt{d_{A}d_{B}})^{2}{\frac{2\ln(2)\ln(d_{A})}{n}}.$

In the second case, using Lemma B.0.3, we obtain:

$\displaystyle\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle\leq(d_{B}^{2}(d_{B}+1))^{2}\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}(\mathbbm{1}_{A}\otimes
M_{B_{m+1}})\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle=(d_{B}^{2}(d_{B}+1))^{2}\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AZ_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{Z_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle\leq(d_{B}^{2}(d_{B}+1))^{2}{\frac{2\ln(2)\ln(d_{A})}{n}}.$

Since we are interested in the best possible upper bound, it should be as
small as possible; which upper bound is smaller depends on the underlying
dimensions $d_{A}$ and $d_{B}$, which means that the best possible bound can
be chosen depending on the setting of interest. In general, we take the
minimum of the two possibilities:

$\displaystyle\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}$
$\displaystyle\leq\min\Big{\\{}(d_{B}^{2}(d_{B}+1))^{2},(18\sqrt{d_{A}d_{B}})^{2}\Big{\\}}{\frac{2\ln(2)\ln(d_{A})}{n}}$

Lastly, the two previous steps are combined in taking the square root of the
above expectation value to obtain

$\begin{split}&\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}\\\
&\leq\sqrt{\mathbb{E}_{z_{1}^{m}}\Big{\\{}\Big{\|}\rho_{AB_{m+1}|z_{1}^{m}}-\rho_{A|z_{1}^{m}}\otimes\rho_{B_{m+1}|z_{1}^{m}}\Big{\|}_{\operatorname{tr}}^{2}\Big{\\}}}\\\
&\leq\min\Big{\\{}d_{B}^{2}(d_{B}+1),18\sqrt{d_{A}d_{B}}\Big{\\}}\sqrt{\frac{2\ln(2)\ln(d_{A})}{n}}\\\
&=\epsilon(d_{B},d_{A},n).\end{split}$

Finally, it must be checked that the additional linear constraints are upheld,
by showing that the initial condition $\Lambda_{A\rightarrow
C_{A}}(\rho_{AB_{1}^{n}})=X_{C_{A}}\otimes\rho_{B_{1}^{n}}$ implies
$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z_{1}^{m}})=X_{C_{A}}$. We find

$\begin{split}\Lambda_{A\rightarrow
C_{A}}(\rho_{A|z_{1}^{m}})&=\Lambda_{A\rightarrow
C_{A}}\bigg{(}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\bigg{)}\\\
&=\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}\Lambda_{A\rightarrow
C_{A}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}})\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\\\
&=\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}X_{C_{A}}\otimes\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\\\
&=X_{C_{A}}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}})\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\\\
&=X_{C_{A}}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})(\operatorname{tr}_{A}(\rho_{AB_{1}^{n}}))\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\\\
&=X_{C_{A}}\dfrac{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}})\big{)}}{\operatorname{tr}_{AZ_{1}^{m}B_{m+1}^{n}}\big{(}(\mathbbm{1}_{A}\otimes\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{AB_{1}^{n}}\big{)}}\\\
&=X_{C_{A}}\end{split}$

and analogous for $\Gamma_{B_{m+1}\rightarrow C_{B}}$. Note that this is the
first and only time that the assumption of permutation invariance is needed:
to specify that $\Gamma_{B_{m+1}\rightarrow C_{B}}$ acts in the same way as
$\Gamma_{B_{n}\rightarrow C_{B}}$ for any $m$. Then,

$\begin{split}\Gamma_{B_{m+1}\rightarrow
C_{B}}(\rho_{B_{m+1}|z_{1}^{m}})&=\Gamma_{B\rightarrow
C_{B}}\bigg{(}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+2}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{B_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}\bigg{)}\\\
&=\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+2}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\Gamma_{B\rightarrow
C_{B}}(\rho_{B_{1}^{n}})\big{)}}{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}\\\
&=\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+2}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{m}}\otimes
Y_{C_{B}}\otimes\rho_{B_{m+2}^{n}})\big{)}}{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}\\\
&=Y_{C_{B}}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+2}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})(\operatorname{tr}_{B_{m+1}}(\rho_{B_{1}^{n}}))\big{)}}{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}\\\
&=Y_{C_{B}}\dfrac{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}})\big{)}}{\operatorname{tr}_{Z_{1}^{m}B_{m+1}^{n}}\big{(}(\Pi_{z_{1}^{m}}\otimes\mathbbm{1}_{B_{m+1}^{n}})\rho_{B_{1}^{n}}\big{)}}\\\
&=Y_{C_{B}}.\end{split}$

∎

##  Appendix C Stabilizer de Finetti Theorem with Linear Constraints with
Stochastic Orthogonal Invariance on Both Sides

It may happen that one is interested in a problem where Alice and Bob both
have access to a system with stochastic orthogonal invariance, or where two
systems should each be approximated by Clifford operations. Therefore, one
could also be interested in studying maximum channel fidelity for such cases
via an SDP hierarchy:

######  Optimization problem C.0.1.

$\begin{split}F_{CC}(N,d_{M})=\text{maximize}\
&F_{s}\Big{(}\Phi_{\tilde{B}R},\big{(}(\mathcal{D}_{B\rightarrow\tilde{B}}\circ\mathcal{N}_{\tilde{A}\rightarrow{B}}\circ\mathcal{E}_{A\rightarrow\tilde{A}})\otimes\mathbbm{1}_{R}\big{)}(\Phi_{AR})\Big{)}\\\
\text{subject to
}&\mathcal{E}_{A\rightarrow\tilde{A}},\mathcal{D}_{B\rightarrow\tilde{B}}\text{
are Clifford channels}\end{split}$

Therefore, and for completeness’ sake, it should also be mentioned that the
states in previous stabilizer de Finetti theorems with linear constraints can
easily be extended to be approximated by stabilizer states on both sides,
using the triangle inequality, which still preserves separability in the cut
between Alice and Bob.

######  Theorem C.0.1 (Stabilizer de Finetti Theorem with Linear Constraints,
with stochastic orthogonal invariance on both sides).

Let $\rho_{A_{1}^{n}B_{1}^{n}}$ be a quantum state on the Hilbert space
$\mathcal{H}_{A}^{\otimes n}\otimes\mathcal{H}_{B}^{\otimes n}$ that commutes
with the action of $O_{n}$ acting on the systems $A_{1}^{n}=A_{1}A_{2}\cdots
A_{n}$, and $B_{1}^{n}=B_{1}B_{2}\cdots B_{n}$, respectively.
$\mathcal{H}_{A}$ is a Hilbert space containing $r_{A}$ qudits with local
dimension $d_{A}$, and $\mathcal{H}_{B}$ contains $r_{B}$ qudits with local
dimension $d_{B}$. Let $\Lambda_{A\rightarrow C_{A}}$ and
$\Gamma_{B\rightarrow C_{B}}$ be linear maps, and $X_{C_{A}}$ and $Y_{C_{B}}$
be operators such that the following two linear constraints hold:

$\Lambda_{A\rightarrow
C_{A}}(\rho_{AB_{1}^{n}})=X_{C_{A}}\otimes\rho_{B_{1}^{n}},$
$\Gamma_{B_{n}\rightarrow C_{B}}(\rho_{B_{1}^{n}})=\rho_{B_{1}^{n-1}}\otimes
Y_{C_{B}}.$

Then, there exists a probability distribution $\\{p_{Z}(z)\\}_{z\in Z}$ and
probability distributions $\\{p_{S_{A}}(\sigma_{A})\\}$ and
$\\{p_{S_{B}}(\sigma_{B})\\}$ over the set of mixed stabilizer states on $r$
qudits on the respective spaces such that

$\displaystyle\Big{\|}\rho_{AB}-\sum_{z,\sigma_{A},\sigma_{B}}p_{Z}(z)p_{S_{A}}(\sigma_{A})p_{S_{B}}(\sigma_{B})\sigma_{A}\otimes\sigma_{B}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \
\leq\min\Big{\\{}\epsilon(d_{B}^{r},d_{A}^{r},n),\epsilon(d_{A}^{r},d_{B}^{r},n)\Big{\\}}+\bar{\epsilon}(d_{A},r_{A},n)+\bar{\epsilon}(d_{B},r_{B},n)$

where $\sigma_{A}$ are the mixed stabilizer states of $r_{A}$
$d_{A}$-dimensional qudits on
$\mathcal{H}_{A}={(\mathbbm{C}^{d_{A}})}^{\otimes r}$ and $\sigma_{B}$ are the
mixed stabilizer states of $r_{B}$ $d_{B}$-dimensional qudits on
$\mathcal{H}_{B}={(\mathbbm{C}^{d_{B}})}^{\otimes r}$.
$\epsilon(d_{B}^{r},d_{A}^{r},n)$ and $\epsilon(d_{A}^{r},d_{B}^{r},n)$ are
defined in (4.1), and $\bar{\epsilon}(d_{A},r_{A},n)$ and
$\bar{\epsilon}(d_{B},r_{B},n)$ are defined in (4.2). In addition,

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z})=X_{C_{A}},\ \Gamma_{B_{n}\rightarrow
C_{B}}(\rho_{B|z})=Y_{C_{B}},$

$\Big{\|}\sum_{z}p_{Z}(z)\big{(}\rho_{A|z}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A}\big{)}\Big{\|}_{\operatorname{tr}}\leq\bar{\epsilon}(d_{A},r_{A},n),$
$\Big{\|}\sum_{z}p_{Z}(z)\big{(}\rho_{B|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}\leq\bar{\epsilon}(d_{B},r_{B},n).$

###### Proof of Theorem C.0.1.

With all the constraints in place, it is possible to apply Theorem 4.2.1. Note
that $\dim(\mathcal{H}_{A})=d_{A}^{r_{A}}$ and
$\dim(\mathcal{H}_{B})=d_{B}^{r_{B}}$. Therefore, there exists an
$m\in[0,n-1]$ such that

$\begin{split}&\Big{\|}\rho_{AB_{m+1}}-\sum_{z_{1}^{m}}p_{Z}(z_{1}^{m})\rho_{A|z_{1}^{m}}\otimes\sum_{\sigma}p_{S}(\sigma)\sigma_{B_{m+1}})\Big{\|}_{\operatorname{tr}}\\\
&\leq\epsilon(d_{B}^{r_{B}},d_{A}^{r_{A}},n)+2d_{B}^{2(r_{B}+1)^{2}}d_{B}^{-\frac{1}{2}(n-{(m+1)})}\end{split}$
(C.1)

with $\sigma_{B}$ being the mixed stabilizer states of $r$ qudits on
$\mathcal{H}_{B}={(\mathbbm{C}^{d_{B}})}^{\otimes r_{B}}$, and

$\Lambda_{A\rightarrow C_{A}}(\rho_{A|z})=X_{C_{A}},\ \Gamma_{B_{n}\rightarrow
C_{B}}(\rho_{B|x})=Y_{C_{B}},$

$\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{B|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}\leq
2d_{B}^{2(r_{B}+1)^{2}}d_{B}^{-\frac{1}{2}(n-(m+1)}.$

In comparison with the one-sided version of this theorem, there is now an
additional constraint on Alice’s side: $\rho_{A_{1}^{n}B_{1}^{n}}$ commutes
with the action of $O_{n}$ acting on the systems $A=A_{1}^{n}=A_{1}A_{2}\cdots
A_{n}$, which directly implies that
$\rho_{A_{1}^{n}}=\operatorname{tr}_{B_{1}^{n}}(\rho_{A_{1}^{n}B_{1}^{n}})$ is
invariant under the stochastic orthogonal group. Therefore, the stabilizer de
Finetti theorem in Theorem 2.2.2, holds for the subsystems $A_{1}^{n}$:

$\Big{\|}\rho_{A_{1}^{k+1}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A}^{\otimes(k+1)}\Big{\|}_{\operatorname{tr}}\leq
2d_{A}^{2(r_{A}+1)^{2}}d_{A}^{-\frac{1}{2}(n-(k+1))}$ (C.2)

Using Observation 2, and the same argumentation as in the final steps of the
proof of Theorem 4.2.1 (see (4.6)), there exists a CPTP map
$\mathcal{M}^{\prime}$, which transforms (C.2) into

$\begin{split}2d_{A}^{2(r_{A}+1)^{2}}d_{A}^{-\frac{1}{2}(n-(k+1))}&\geq\Big{\|}\rho_{A_{1}^{k+1}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\Big{\|}_{\operatorname{tr}}\\\
&\geq\Big{\|}\mathcal{M}^{\prime}\big{(}\rho_{A_{1}^{k+1}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}\\\
&=\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{A_{k+1}|z_{1}^{k}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}.\end{split}$
(C.3)

It is integral that the measurement contained in the maps used in (C.2) and
here should be the same, to ensure that the resulting probability distribution
$p_{Z}$ is the same.

Then, (LABEL:both-proof-step1) and (C.3) can be combined via the triangle
inequality:

$\displaystyle\Big{\|}\rho_{AB}-\sum_{z,\sigma_{A},\sigma_{B}}p_{Z}(z)p_{S_{A}}(\sigma_{A})p_{S_{B}}(\sigma_{B})\sigma_{A}\otimes\sigma_{B}\Big{\|}_{\operatorname{tr}}$
$\displaystyle=\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{B|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\big{)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\Big{(}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{A_{k+1}|z_{1}^{k}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\big{)}\Big{)}\otimes\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\leq\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{B|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\Big{\|}\Big{(}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{A_{k+1}|z_{1}^{k}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\big{)}\Big{)}\otimes\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\leq\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{B|z_{1}^{k}}-\sum_{\sigma_{B}}p_{S_{B}}(\sigma_{B})\sigma_{B}\big{)}\Big{\|}_{\operatorname{tr}}+\Big{\|}\sum_{z_{1}^{k}}p_{Z}(z_{1}^{k})\big{(}\rho_{A_{k+1}|z_{1}^{k}}-\sum_{\sigma_{A}}p_{S_{A}}(\sigma_{A})\sigma_{A_{k+1}}\big{)}\Big{\|}_{\operatorname{tr}}$
$\displaystyle\leq\epsilon(d_{B}^{r_{B}},d_{A}^{r_{A}},n)+2d_{B}^{2(r_{B}+1)^{2}}d_{B}^{-\frac{1}{2}(n-{(m+1)})}+2d_{A}^{2(r_{A}+1)^{2}}d_{A}^{-\frac{1}{2}(n-(k+1))}$

Because permutations are a subgroup of stochastic orthogonal group, we can
choose $m=k=0$.

Here, it has to be noted that the proof could also be done the other way
around, switching out $A$ and $B$, and first using the orthogonal invariance
of $A$ to get a statement about separability and closeness to convex
combinations of stabilizer states, and then using the invariance on the $B$
side. If this was switched, the dimensions $d_{A}$ and $d_{B}$ would also be
switched in the bound. In total, we are interested in the minimum bound, given
by taking the minimum over all the available options. Therefore, there are
four terms in the final statement of which one must choose the minimum.

Finally, the linear constraint on Alice’s side can be obtained using
Observation 1, in analogy to the final step in the proof of 4.2.1. ∎

Using Theorem C.0.1, the hierarchy described in (4.3.2) can be extended to a
hierarchy where the stochastic orthogonal invariance of both Alice’s and Bob’s
side implies the closeness to a separable state with stabilizer states on both
sides.

######  Optimization problem C.0.2.

$\begin{split}\text{maximize}\
&d_{\tilde{A}}d_{B}^{r_{B}}\operatorname{tr}\Big{(}\big{(}J_{\tilde{A}B}^{N}\otimes\Phi_{A\tilde{B}}\big{)}\big{(}\rho_{A\tilde{A}B\tilde{B}}\big{)}\Big{)}\\\
\text{subject to }&\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n}}\geq 0,\
\operatorname{tr}(\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n}})=1\\\
&\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n}}\text{ is invariant under the
action of }O_{n}\text{ with respect to }A_{1}^{n}\text{ and }B_{1}^{n}\\\
&\operatorname{tr}_{\tilde{A_{1}}}(\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n}})=\frac{\mathbbm{1}_{A_{1}}}{d_{A}^{r_{A}}}\otimes\rho_{(A\tilde{A})_{2}^{n}(B\tilde{B})_{1}^{n}}\\\
&\operatorname{tr}_{\tilde{B}_{n}}(\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n}})=\rho_{(A\tilde{A})_{1}^{n}(B\tilde{B})_{1}^{n-1}}\otimes\frac{\mathbbm{1}_{B_{n}}}{d_{B}^{r_{B}}}\\\
\end{split}$

In the one-sided hierarchy, only the decoder (or, symmetrically, only the
encoder) is approximated by Choi matrices of stabilizer states, which are
Clifford operations. With stochastic orthogonal invariance on both sides, both
the encoder and the decoder are approximated by Clifford operations.

It is also possible to extend stabilizer de Finetti theorem for qubits
(Theorem 4.2.2) such that both Alice’s and Bob’s part of the state are
invariant under all permutations and anti-identity, leading to an
approximation for qubits by stabilizer state tensor powers with separability
between Alice and Bob.

### Acknowledgement

For this thesis to come into existence, many parties had to communicate over
large distances via a complicated protocol. The goal of the protocol was to
hand in this thesis. The state of my brain was prepared via local interaction
in Cologne, transmitted to Zurich, transformed via local operations, and then
(surprisingly) sent to Constance to work on the desired output. During this
process, I have interacted with many people without whom the output would have
been considerably more noisy.

I owe particular thanks to Professor David Gross for letting me join his
group, getting me started on this project and offering me valuable support and
life advice. In addition, I want to extend my thanks to the whole Gross group,
especially Felipe Montealegre-Mora for helping me understand some
representation theory stuff and Mariami Gachechiladze for trying to understand
QKD with me.

I also owe many thanks to Professor Renato Renner for accepting me into his
group, NCCR QSIT for giving me the opportunity to come to Zurich for one
semester, and the whole Renner group, in particular Joe Renes for working with
me on the SDP hierarchy. I am especially grateful to have had the opportunity
to participate in the NCCR QSIT Winter School and General Meeting, where I
learned a lot.

Furthermore, I want to thank my family and friends who supported me locally
and remotely (especially in the final month) throughout my whole studies. The
output of this protocol also benefitted greatly from error correction
performed by Joe Renes, Felipe Monetalegre-Mora, Wolfgang Belzig and Carsten
Speckmann.

I also want to thank the Bonn-Cologne Graduate School for support of my master
studies, Petra Neubauer-Günther for support and advice pertaining to the
feasibility of this joint project, and everyone who facilitated this exchange.

### Eidesstattliche Erklärung

Hiermit versichere ich an Eides statt, dass ich die vorliegende Arbeit
selbstständig und ohne die Benutzung anderer als der angegebenen Hilfsmittel
angefertigt habe. Alle Stellen, die wörtlich oder sinngemäß aus
veröffentlichten oder nicht veröffentlichten Schriften entnommen wurden, sind
als solche kenntlich gemacht. Die Arbeit ist in gleicher oder ähnlicher Form
oder auszugsweise im Rahmen einer anderen Prüfung noch nicht vorgelegt worden.
Ich versichere, dass die eingereichte elektronische Fassung der eingereichten
Druckfassung vollständig entspricht.

Ort, Datum |  | Unterschrift
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