The asymmetric quantum cloning region

Abstract

Quantum cloning is a fundamental protocol of quantum information theory. Perfect universal quantum cloning is prohibited by the laws of quantum mechanics, only imperfect copies being reachable. Symmetric quantum cloning is concerned with case when the quality of the clones is identical. In this work, we study the general case of $1 \rightarrow N$ asymmetric cloning, where one asks for arbitrary qualities of the clones. We characterize, for all Hilbert space dimensions and number of clones, the set of all possible clone qualities. This set is realized as the nonnegative part of the unit ball of a newly introduced norm, which we call the $\mathcal Q$-norm. We also provide a closed-form expression for the quantum cloner achieving a given clone quality vector. Our analysis relies on the Schur–Weyl duality and on the study of the spectral properties of partially transposed permutation operators.

Appendices

Appendix A: Structure of permutation operators

This appendix is devoted to the presentation of technical results on permutations operator. Numerous results or remarks are illustrated by graphical calculus.

1.1 graphical calculus

To illustrate the technical results of this section, we employ a graphical method for representing tensors, which is based on Penrose’s graphical notation [37]. Comparable approaches have been established in the context of tensor network states and categorical quantum information theory, as detailed in [2, 13, 44].

In this diagrammatic notation, tensors are depicted as boxes and wires.

The wires are labeled with indices, and a box represents the tensor’s value corresponding to those specific indices.

Specifically, a vector is represented by a box with a single wire extending to the left, labeled by the coordinate index. Conversely, a dual vector has its wire directed to the right.

Note that the left and right directions follow the convention that corresponds to the standard right-to-left composition in linear algebra.

Tensor diagrams can be combined using two methods: tensor product and tensor contraction. The tensor product combines two diagrams either vertically or horizontally.

Tensor contraction combines two diagrams by taking the sum across all common indices.

Specifically, we recover the scalar product.

The matrix product.

And the matrix trace.

Scalars multiply the diagrams and are displayed adjacent to the diagram.

In conclusion, the transposition can be visually illustrated by exchanging the “input” and “output” wires of a box representing a matrix. The partial transposition involves only the permutation of wires corresponding to specific spaces.

1.2 Contributions of permutations

In Sect. 3, we only focus on permutations where 0 is not a fixed point. This is motivated by the following remarks

Let $T: \mathcal {M}_d \rightarrow {(\mathcal {M}_d)}^{\otimes N}$ be a $\mathcal {U}(d)$-covariant quantum channel, and $C_T$ its Choi matrix, then from Proposition 3.2 we have for all $\rho \in \mathcal {D}^1_d$

$$\begin{aligned} T(\rho )&= \hbox {Tr}_0 \Big [ C_T \big ( \rho ^\textsf{T}\otimes I^{\otimes N} \big ) \Big ] \\&= \hbox {Tr}_0 \bigg [ \sum _{\sigma \in \mathfrak {S}_{N + 1}} \beta _\sigma \cdot \Pi _\sigma ^{\mathsf {\Gamma }} \big ( \rho ^\textsf{T}\otimes I^{\otimes N} \big ) \bigg ], \end{aligned}$$

for some $\beta _\sigma \in \mathbb {C}$ (see Example A.3 for the case $N=2$).

Let us justify that $\sigma \in \mathfrak {S}_{N + 1}$ such that $\sigma (0) = 0$ do not contribute to the performance of the cloning map, whereas the $(N + 1)$ cycles contribute the most to the copies. Indeed, for a permutation $\sigma \in \mathfrak {S}_{N + 1}$ such that $\sigma (0) = 0$ we have $\Pi _\sigma ^{\mathsf {\Gamma }} = \Pi _\sigma$ and for all $\rho \in \mathcal {D}^1_d$

$$\begin{aligned} \hbox {Tr}_1 \Big [ \Pi _\sigma ^{\mathsf {\Gamma }} \big ( \rho ^\textsf{T}\otimes I^{\otimes N} \big ) \Big ]&= \hbox {Tr}_0 \Big [ \Pi _\sigma \big ( \rho ^\textsf{T}\otimes I^{\otimes N} \big ) \Big ] \\&= \frac{\hbox {Tr}\rho }{d} \hbox {Tr}_0 \Big [ \Pi _\sigma \; I^{\otimes (N + 1)} \Big ] \\&= \frac{1}{d} \cdot \hbox {Tr}_0 \Pi _\sigma \end{aligned}$$

whose contributions consist in deteriorating the quality of the cloning, since the quantity $\hbox {Tr}_1 \sigma$ does not depend on $\rho$. Now, if $\sigma \in \mathfrak {S}_{N + 1}$ is a $(N + 1)$ cycle, the marginals of

$$\begin{aligned} T_{\scriptscriptstyle \text {perfect}} : \mathcal {M}_d&\longrightarrow {(\mathcal {M}_d)}^{\otimes N} \\ \rho&\longmapsto \hbox {Tr}_0 \Big [ \Pi _\sigma ^{\mathsf {\Gamma }} \big ( \rho ^\textsf{T}\otimes I^{\otimes N} \big ) \Big ] \end{aligned}$$

are all equals to identity, i.e., ${(T_{\scriptscriptstyle \text {perfect}})}_i(\rho ) = \rho$ for all $\rho \in \mathcal {M}_d$. This would be the case of a perfect quantum cloning map, but the operator $T_{\scriptscriptstyle \text {perfect}}$ are of course not completely positive operator due to the no-cloning theorem (see Example A.4 for computations with graphical calculus).

More generally, let $T: \mathcal {M}_d \rightarrow {(\mathcal {M}_d)}^{\otimes N}$ be a $\mathcal {U}(d)$-covariant quantum channel. From Proposition 3.2, the marginals $T_i$ are for all $\rho \in \mathcal {D}^1_d$

$$\begin{aligned} T_i(\rho ) = p_i \! \cdot \! \rho + (1 - p_i)\frac{I}{d} \end{aligned}$$

where the $p_i$ are the sums of the $\frac{N!}{2}$ coefficients $\beta _\sigma$ of the permutations contributing for this copy, that is, the $\sigma$ such that i is in the orbit of 0 under the action of $\sigma$. The following examples illustrate some results in terms of graphical calculus.

Example A.1

(Partition of $\Sigma _{a,b}, N = 3$) The permutation set $\big \{ \Pi ^{\mathsf {\Gamma }}_\sigma \; \big | \;\sigma \in \mathfrak {S}_4, \quad \sigma (0) \ne 0 \big \}$ is partitioned in

Example A.2

(Partially transposed permutations, $N = 3$) On $\mathfrak {S}_{4}$, the $(0 \, i)$-partially transposed permutations are

Example A.3

(Choi Matrix $N = 2$) Let $T: \mathcal {M}_d \rightarrow \mathcal {M}_d \otimes \mathcal {M}_d$ be a $\mathcal {U}(d)$-covariant quantum channel, then its Choi matrix $C_T$ is:

Example A.4

(Contributions of quantum cloning) Let $T: \mathcal {M}_d \rightarrow \mathcal {M}_d \otimes \mathcal {M}_d$ be a $\mathcal {U}(d)$-covariant quantum channel, such that its Choi matrix $C_T$ is the partially transposed cycle $(1 \, 2 \, 3)$:

Then, the two marginals are

and

1.3 Proof of Lemma lem:struct

In this subsection, we prove the key Lemma 3.4, which gives the structure of partially permutations operators and action on vectors. We illustrate as well with graphical calculus.

Lemma A.5

Let $1 \le a,b \le N$, for any $\sigma \in \Sigma _{a,b}$, there exists a unique $\hat{\sigma } \in \mathfrak {S}_{N - 1}$ such that the partially transposed permutation operator $\sigma ^{\mathsf {\Gamma }}$ is

$$\begin{aligned} \Pi _\sigma ^{\mathsf {\Gamma }} = \Pi _{(1 \, a)} \, (\omega _{(0,1)} \otimes \Pi _{\hat{\sigma }}) \; \Pi _{(1 \, b)}. \end{aligned}$$

Proof

Let $1 \le a,b,c,d \le N$, and $\sigma \in \Sigma _{a,b}$, then $\Pi _{(c \, a)} \; \Pi _\sigma ^{\mathsf {\Gamma }} \; \Pi _{(d \, b)}$ is a partially transposed permutation operator from $\Sigma _{c,d}$. Let $\sigma ' \in \Sigma _{1,1}$ be the permutation such that

$$\begin{aligned} \Pi _{(1 \, a)} \; \Pi _\sigma ^{\mathsf {\Gamma }} \; \Pi _{(1 \, b)} = \Pi _{\sigma '}^{\mathsf {\Gamma }}. \end{aligned}$$

There is a unique $\hat{\sigma } \in \mathfrak {S}_{N - 1}$, such that

$$\begin{aligned} \Pi _{\sigma '}. = \Pi _{(0,1)} \otimes \Pi _{\hat{\sigma }}, \end{aligned}$$

and finally

$$\begin{aligned} \Pi _{\sigma '}^{\mathsf {\Gamma }}. = \omega _{(0,1)} \otimes \Pi _{\hat{\sigma }}, \end{aligned}$$

where the permutation operator $\Pi _{\hat{\sigma }}$ is

$$\begin{aligned} \Pi _{\hat{\sigma }} = \frac{1}{d} \cdot \hbox {Tr}_{0,1} \big [ \Pi _{(1 \, a)} \; \Pi _\sigma ^{\mathsf {\Gamma }} \; \Pi _{(1 \, b)} \big ]. \end{aligned}$$

$\square$

Example A.6

Let $\sigma = (0 \, 3 \, 2) \in \mathfrak {S}_4$. Its partial transposition operator

can be decomposed into

Recall that for two integers i, j, we denote by $(i \,: \, j)$ the permutation $\mathfrak {S}_n$ defined by

$$\begin{aligned} (i \, : \, j) = {\left\{ \begin{array}{ll} \big ( (i - 1) \, (i - 2) \, \cdots \, j \big ) &{}\text {if } 0 \le j< i \le n \\ \big ( i \, (i + 1) \, \cdots \, (j - 1) \big ) &{}\text {if } 0 \le i < j \le n \\ (0)(1) \cdots (n - 1) &{}\text {otherwise} \end{array}\right. }. \end{aligned}$$

That is, the $|i - j|$ cycle of $\mathfrak {S}_n$ between i and j.

Example A.7

The non-trivial $(i \,: \, j)$-cycles of $\mathfrak {S}_3$ are:

Lemma A.8

Let $1 \le a,b,c \le N$, and let $\sigma \in \Sigma _{a,b}$. Let $\phi \in \mathcal {H}^{\otimes (N - 1)}$, then

$$\begin{aligned} \Pi _\sigma ^{\mathsf {\Gamma }}&(\left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle ) \\&= {\left\{ \begin{array}{ll} d \cdot \left| \Omega \right\rangle _{(0,a)} \otimes \big ( \Pi _{(0 \, : \, (a - 1))} \circ \Pi _{\hat{\sigma }}\circ \Pi _{((b - 1) \, : \, 0)} \big ) \left| \phi \right\rangle &{}\text {if } b = c \\[1em] \left| \Omega \right\rangle _{(0,a)} \otimes \big ( \Pi _{(0 \, : \, (a - 1))} \circ \Pi _{\hat{\sigma }} \circ \Pi _{((b - 1) \, : \, 0)} \circ \Pi _{((c - 1) \, : \, (b - 1))} \big ) \left| \phi \right\rangle &{}\text {if } b \ne c. \end{array}\right. } \end{aligned}$$

Proof

From Lemma A.5, we can write $\Pi _\sigma ^{\mathsf {\Gamma }} = \Pi _{(1 \, a)} \, (\omega _{(0,1)} \otimes \Pi _{\hat{\sigma }}) \; \Pi _{(1 \, b)}$, or equivalently

$$\begin{aligned} \Pi _\sigma ^{\mathsf {\Gamma }} = \Pi _{(1 \, a)} \, (I^{\otimes 2} \otimes \Pi _{\hat{\sigma }}) \, \big ( \omega _{(0,1)} \otimes I^{\otimes (N - 1)} \big ) \; \Pi _{(1 \, b)}. \end{aligned}$$

Together with $\omega \left| \Omega \right\rangle = d \cdot \left| \Omega \right\rangle$, we have the following relations:

$$\begin{aligned} \big ( \omega _{(0,1)} \otimes I^{\otimes (N - 1)} \big ) \; \Pi _{(1 \, b)} = \Pi _{(1 \, b)} \; \big ( \omega _{(0,b)} \otimes I^{\otimes (N - 1)} \big ), \end{aligned}$$

and

$$\begin{aligned} \Pi _{(1 \, b)} \big ( \left| \Omega \right\rangle _{(0,b)} \otimes \left| \phi \right\rangle \big ) = \left| \Omega \right\rangle \otimes \Pi _{((b - 1) \, : \, 0)} \left| \phi \right\rangle . \end{aligned}$$

If $b = c$, then

$$\begin{aligned}&\big ( \omega _{(0,1)} \otimes I^{\otimes (N - 1)} \big ) \; \Pi _{(1 \, b)} \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big )\\&= \Pi _{(1 \, b)} \; \big ( \omega _{(0,b)} \otimes I^{\otimes (N - 1)} \big ) \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big ) \\&= \Pi _{(1 \, b)} \big ( \omega _{(0,b)} \left| \Omega \right\rangle _{(0,b)} \otimes \left| \phi \right\rangle \big ) \\&= d \cdot \Pi _{(1 \, b)} \big ( \left| \Omega \right\rangle _{(0,b)} \otimes \left| \phi \right\rangle \big ) \\&= d \cdot \big ( \left| \Omega \right\rangle \otimes \Pi _{((b - 1) \, : \, 0)} \left| \phi \right\rangle \big ), \end{aligned}$$

such that finally $\Pi _\sigma ^{\mathsf {\Gamma }} \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big ) = d \cdot \Pi _{(1 \, a)} \Big [ \left| \Omega \right\rangle _{(0,1)} \otimes \big ( \Pi _{\hat{\sigma }} \circ \Pi _{((b - 1) \,: \, 0)} \circ \big ) \left| \phi \right\rangle \Big ]$. If $b \ne c$, we have the following relation:

$$\begin{aligned} \big ( \omega _{(0,b)} \otimes I^{\otimes (N - 1)} \big ) \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big ) = \left| \Omega \right\rangle _{(0,b)} \otimes \Pi _\sigma \left| \phi \right\rangle , \end{aligned}$$

for some $\sigma \in \mathfrak {S}_{N-1}$. Note that $\left| \phi \right\rangle$ is permuted according to $\sigma$ which is a cycle whose order depends on whether $c<b$ or $b<c$. That is,

$$\begin{aligned} \big ( \omega _{(0,b)} \otimes I^{\otimes (N - 1)} \big ) \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big ) = \left| \Omega \right\rangle _{(0,b)} \otimes \Pi _{((c - 1) \, : \, (b - 1))} \left| \phi \right\rangle . \end{aligned}$$

Then,

$$\begin{aligned} \big ( \omega _{(0,1)} \otimes I^{\otimes (N - 1)} \big ) \; \Pi _{(1 \, b)} \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big )&= \Pi _{(1 \, b)} \; \big ( \omega _{(0,b)} \otimes I^{\otimes (N - 1)} \big ) \big ( \left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle \big ) \\&= \Pi _{(1 \, b)} \big ( \left| \Omega \right\rangle _{(0,b)} \otimes \Pi _{((c - 1) \, : \, (b - 1))} \left| \phi \right\rangle \big ) \\&= \left| \Omega \right\rangle _{(0,1)} \otimes \big ( \Pi _{((b - 1) \, : \, 0)} \circ \Pi _{((c - 1) \, : \, (b - 1))} \big ) \left| \phi \right\rangle , \end{aligned}$$

such that finally 

$$\Pi _\sigma ^{\mathsf {\Gamma }}(\left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle ) = \Pi _{(1 \, a)} \big [ \left| \Omega \right\rangle _{(0,1)} \otimes \big ( \Pi _{\hat{\sigma }} \circ \Pi _{((b - 1) \,: \, 0)} \circ \Pi _{((c - 1) \,: \, (b - 1))} \big ) \left| \phi \right\rangle \Big ].$$

To conclude, from the relation

$$\begin{aligned} \Pi _{(1 \, a)} \big ( \left| \Omega \right\rangle _{(0,1)} \otimes \left| \phi \right\rangle \big ) = \left| \Omega \right\rangle _{(0,a)} \otimes \Pi _{(0 \, : \, (a - 1))} \left| \phi \right\rangle , \end{aligned}$$

we obtain

$$\begin{aligned}&\Pi _\sigma ^{\mathsf {\Gamma }}(\left| \Omega \right\rangle _{(0,c)} \otimes \left| \phi \right\rangle ) \\&= {\left\{ \begin{array}{ll} d \cdot \left| \Omega \right\rangle _{(0,a)} \otimes \big ( \Pi _{(0 \, : \, (a - 1))} \circ \Pi _{\hat{\sigma }}\circ \Pi _{((b - 1) \, : \, 0)} \big ) \left| \phi \right\rangle &{}\text {if } b = c \\[1em] \left| \Omega \right\rangle _{(0,a)} \otimes \big ( \Pi _{(0 \, : \, (a - 1))} \circ \Pi _{\hat{\sigma }} \circ \Pi _{((b - 1) \, : \, 0)} \circ \Pi _{((c - 1) \, : \, (b - 1))} \big ) \left| \phi \right\rangle &{}\text {if } b \ne c. \end{array}\right. } \end{aligned}$$

$\square$

Example A.9

Let $\sigma = (0 \, 3 \, 1 \, 2) \in \mathfrak {S}_4$, and let $\phi \in \mathcal {H}^{\otimes 2}$, then

In order to simplify the next equations, we use the notations $\Pi _{\hat{\sigma }(a,b,c)}$ and $\Pi _{b,c}$, defined by:

$$\begin{aligned} \Pi _{\hat{\sigma }(a,b,c)} = \Pi _{(0 \, : \, (a - 1))} \circ \Pi _{\hat{\sigma }} \circ \Pi _{b,c}, \end{aligned}$$

and

$$\begin{aligned} \Pi _{b,c} = {\left\{ \begin{array}{ll} \Pi _{((b - 1) \, : \, 0)} &{}\text {if } b = c \\ \Pi _{((b - 1) \, : \, 0)} \circ \Pi _{((c - 1) \, : \, (b - 1))} &{}\text {if } b \ne c \end{array}\right. }. \end{aligned}$$

The following lemma concerns the image and the kernel of operators $\Pi _\sigma ^{\mathsf {\Gamma }}$.

Lemma A.10

For any $1 \le a,b \le N$ and any $\sigma \in \Sigma _{a,b}$,

$$\begin{aligned} {{\,\textrm{Im}\,}}\Pi _\sigma ^{\mathsf {\Gamma }}&= {{\,\textrm{Span}\,}}\Big \{ \left| \Omega \right\rangle _{(0,a)} \otimes \left| \phi \right\rangle \; \Big | \; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \} \\ {{\,\textrm{Ker}\,}}\Pi _\sigma ^{\mathsf {\Gamma }}&\supseteq {{\,\textrm{Span}\,}}{\Big \{ \left| \Omega \right\rangle _{(0,k)} \otimes \left| \phi \right\rangle \; \Big | \; 1 \le k \le N, \; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \}}^\perp . \end{aligned}$$

Proof

Let $1 \le a,b \le N$, let $\sigma \in \Sigma _{a,b}$, and let $v \in {{\,\textrm{Im}\,}}\Pi _\sigma ^{\mathsf {\Gamma }}$, then $v = \left| \Omega \right\rangle _{(0,a)} \otimes \left| \phi \right\rangle$ with $\phi \in \mathcal {H}^{\otimes (N - 1)}$, from Lemma A.5, and

$${{\,\textrm{Im}\,}}\Pi _\sigma ^{\mathsf {\Gamma }} \subseteq {{\,\textrm{Span}\,}}\Big \{ \left| \Omega \right\rangle _{(0,a)} \otimes \left| \phi \right\rangle \; \Big | \; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \}.$$

Let $\phi \in \mathcal {H}^{\otimes (N - 1)}_d$, and define . Then from Lemma A.8, $\Pi _\sigma ^{\mathsf {\Gamma }}(u) = \left| \Omega \right\rangle _{(0,a)} \otimes \left| \phi \right\rangle$, and

$$\begin{aligned} {{\,\textrm{Im}\,}}\Pi _\sigma ^{\mathsf {\Gamma }} \supseteq {{\,\textrm{Span}\,}}\Big \{ \left| \Omega \right\rangle _{(0,a)} \otimes \left| \phi \right\rangle \; \Big | \; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \}. \end{aligned}$$

Since on finite-dimensional spaces, the kernel of a linear map M is equal to the orthogonal complement of the image of adjoint $M^*$, i.e., ${{\,\textrm{Ker}\,}}M = {\big ( {{\,\textrm{Im}\,}}M^* \big )}^\perp$, and because the adjoint of $\Pi _\sigma ^{\mathsf {\Gamma }}$ is a partially transposed permutation operator $\Pi _{\sigma ^*}^{\mathsf {\Gamma }}$ with $\sigma ^* \in \Sigma _{b,a}$, we have

$$\begin{aligned} {{\,\textrm{Ker}\,}}\Pi _\sigma ^{\mathsf {\Gamma }}&= {{\,\textrm{Span}\,}}{\Big \{ \left| \Omega \right\rangle _{(1,b)} \otimes \left| \phi \right\rangle \; \Big | \; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \}}^\perp \\&\supseteq {{\,\textrm{Span}\,}}{\Big \{ \left| \Omega \right\rangle _{(1,k)} \otimes \left| \phi \right\rangle \; \Big | \; 1 \le k \le N,\; \phi \in \mathcal {H}^{\otimes (N - 1)} \Big \}}^\perp \end{aligned}$$

$\square$

1.4 Partial trace of partially transposed permutation operators

Lemma A.11

Let $1 \le a,b \le N$, then

$$\begin{aligned} \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,a}} \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ]&= (N - 1)! \; \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] \cdot I \\ \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ]&= \frac{(N - 1)!}{d} \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] \cdot I, \end{aligned}$$

where the orthogonal projector onto the trivial representation subspace of $\mathfrak {S}_{N - 1}$ is defined by $P^+_{\mathfrak {S}_{N - 1}} = \frac{1}{(N - 1)!} \sum _{\sigma \in \mathfrak {S}_{N - 1}} \Pi _\sigma$.

Proof

For the first equation, we have from Lemma A.5

where $\Pi _{(0 \, (a - 1))}$ is a permutation operator of $\mathfrak {S}_N$. For the second equation, since the partial transposition is a linear operator, we have the equality

$$\begin{aligned} \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ] = {\bigg ( \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma \bigg ] \bigg )}^{\mathsf {\Gamma }}. \end{aligned}$$

For any $\sigma \in \Sigma _{a,b}$, the relation

$$\begin{aligned} (\bar{U} \otimes U^{\otimes N}) \, \Pi _\sigma ^{\mathsf {\Gamma }} \, (\bar{U} \otimes U^{\otimes N}) = \Pi _\sigma ^{\mathsf {\Gamma }} \end{aligned}$$

holds for all $U \in \mathcal {U}(d)$, such that $\hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \big [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma ^{\mathsf {\Gamma }} \Big ]$ is a multiple of the identity, and

$$\begin{aligned} \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ] = \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma \bigg ] = c \cdot I, \end{aligned}$$

with $c = \frac{1}{d} \hbox {Tr}\big [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma \Big ]$. For a permutation $\sigma \in \mathfrak {S}_n$, we write $\# \sigma$ the number of disjoint cycles of $\sigma$. Then,

$$\begin{aligned} \hbox {Tr}\big [ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma \Big ]&= \sum _{\sigma \in \Sigma _{a,b}} d^{\# \sigma } \\&= \sum _{\sigma \in \Sigma _{a,a}} d^{\# [\sigma \circ (a \, b)]}. \end{aligned}$$

Let $\sigma \in \Sigma _{a,a}$ and write $\sigma = c_1 \circ \cdots \circ c_k$ the decomposition of $\sigma$ in disjoint cycles $c_i = (s^i_1 \, \cdots \, s^i_{l_i})$. If it exists $i \in [k]$ such that $a,b \in c_i$, then $c_i \circ (a \, b)$ can be decomposed in two disjoint cycles. Otherwise, if there exist $i,j \in [k]$ such that $a \in c_i$ and $b \in c_j$, then $c_i \circ c_j \circ (a \, b)$ can be decomposed in one disjoint cycle. Finally,

$$\begin{aligned} \# \big [ \sigma \circ (a \, b) \Big ] = {\left\{ \begin{array}{ll} \# \sigma + 1 &{}\text {if } \exists i\in [k] \text { s.t. } a,b \in c_i \\ \# \sigma - 1 &{}\text {otherwise} \end{array}\right. }. \end{aligned}$$

But since $\sigma \in \Sigma _{a,a}$ and $b \ne a$, in the decomposition of $\sigma$ in disjoint cycles, a is in the cycle $(0 \, a)$, and $\# \big [ \sigma \circ (a \, b) \Big ] = \# \sigma - 1$. That is

$$\begin{aligned} \hbox {Tr}\left[ \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma \right] = \frac{1}{d} \hbox {Tr}\left[ \sum _{\sigma \in \Sigma _{a,a}} \Pi _\sigma \right] . \end{aligned}$$

$\square$

1.5 Scalar product of vectors in $\mathcal {V}^\lambda$

Lemma A.12

Let $\lambda _1, \lambda _2$ be two irreducible representations of $\mathfrak {S}_{N - 1}$, let $\textbf{v}_1, \textbf{v}_2$ be two normalized vectors in the irreducible subspace $\lambda _1$ and $\lambda _2$ of $\mathfrak {S}_{N - 1}$, and let $1 \le k,l \le N$. Then,

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,l)} \otimes \textbf{v}_2 \rangle = {\left\{ \begin{array}{ll} d \cdot \langle \textbf{v}_1|\textbf{v}_2\rangle &{}\textrm{if } k = l \\ \left\langle \textbf{v}_1\right| \Pi _{((l - 1) \, : \, (k - 1))} \left| \textbf{v}_2\right\rangle &{}\textrm{if } k \ne l. \end{array}\right. } \end{aligned}$$

As a consequence,

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,k)} \otimes \textbf{v}_1\rangle = d \end{aligned}$$

and if $\lambda _1$ and $\lambda _2$ are distinct,

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,l)} \otimes \textbf{v}_2\rangle = 0. \end{aligned}$$

In particular, the vector spaces ${{\,\textrm{Span}\,}}\mathcal {V}^\lambda$, defined in Eq. (3.3), are in orthogonal direct sum.

Proof

Since both $\textbf{v}_1$ and $\textbf{v}_2$ are normalized, we have $\langle \textbf{v}_i|\textbf{v}_i\rangle = 1$ for all $i \in \{1, 2\}$. Then, from $\langle \Omega |\Omega \rangle = d$ we have

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,k)} \otimes \textbf{v}_1\rangle&= \langle \Omega |\Omega \rangle \otimes \langle \textbf{v}_1|\textbf{v}_1\rangle \\&= d. \end{aligned}$$

When $k \ne l$, from the equation $\left\langle \Omega _{(0, k)} \otimes \phi \right| \left| \Omega _{(0, l)} \otimes \psi \right\rangle = \left\langle \phi \right| \Pi _{((l - 1) \,: \, (k - 1))} \left| \psi \right\rangle$ for any $\phi , \psi \in \mathcal {H}^{\otimes (N - 1)}$, we have

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,l)} \otimes \textbf{v}_1\rangle = \left\langle \textbf{v}_1\right| \Pi _{((l - 1) \, : \, (k - 1))} \left| \textbf{v}_1\right\rangle . \end{aligned}$$

When we look at $\lambda _1, \lambda _2$, two distinct irreducible representations of $\mathfrak {S}_{N - 1}$, the scalar product becomes

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}_1|\Omega _{(0,l)} \otimes \textbf{v}_2 \rangle = {\left\{ \begin{array}{ll} \langle \Omega |\Omega \rangle \otimes \langle \textbf{v}_1|\textbf{v}_2\rangle &{}\text {if } k = l \\ \left\langle \textbf{v}_1\right| \Pi _{((l - 1) \, : \, (k - 1))} \left| \textbf{v}_2\right\rangle &{}\text {if } k \ne l. \end{array}\right. } \end{aligned}$$

Note that $\Pi _{((l - 1) \,: \, (k - 1))} \left| \textbf{v}_2\right\rangle$ lives in the irreducible subspace of $\mathfrak {S}_{N - 1}$ associated with $\lambda _2$, since the irreducible representations are stable by any representation of permutations. Then, both scalar products are null because the irreducible spaces are orthogonal each other.$\square$

Remark A.13

In the particular case of the trivial representation, we have for all $1 \le k,l \le N$ and all normalized $\textbf{v}$ in the irreducible subspace $\vee ^{(N - 1)}(\mathcal {H})$,

$$\begin{aligned} \langle \Omega _{(0,k)} \otimes \textbf{v}|\Omega _{(0,k)} \otimes \textbf{v}\rangle&= d \\ \langle \Omega _{(0,k)} \otimes \textbf{v}|\Omega _{(0,l)} \otimes \textbf{v}\rangle&= 1. \end{aligned}$$

From Lemma A.8, the action of the permutation operators $\Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}}$ on the vectors $\left| \Omega \right\rangle _{(0,k)} \otimes \left| \textbf{v}\right\rangle$, for all $1 \le k \le N$, is

$$\begin{aligned} \Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}} \big ( \left| \Omega \right\rangle _{(0,k)} \otimes \left| \textbf{v}\right\rangle \big ) = {\left\{ \begin{array}{ll} d \cdot \left| \Omega \right\rangle _{(1,a)} \otimes \Pi ^\lambda _{\hat{\sigma },a,b,b} \left| \textbf{v}\right\rangle &{}\text { if } b = k\\ \left| \Omega \right\rangle _{(1,a)} \otimes \Pi ^\lambda _{\hat{\sigma },a,b,k} \left| \textbf{v}\right\rangle &{}\text { if } b \ne k, \end{array}\right. } \end{aligned}$$

(31)

with $\Pi ^\lambda _\sigma = P^\lambda \, \Pi _\sigma \, P^\lambda$, and $P^\lambda$ the projector onto the irreducible subspace $\lambda$.

For any permutation operator $\Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}}$, we know from Lemma A.10 that the $V^{\lambda }$’s form a generators of the orthogonal complement of the kernel of $\Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}}$. However, in general they do not define a basis. We can extract a subset of linearly independent vectors to form a basis. In this basis, $\Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}}$ can be block diagonalized due to Eq. (31), since for all irreducible representation $\lambda$, and any vector v in $\mathcal {V}^{\lambda }$, we have

$$\begin{aligned} \Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}} (v) \in {{\,\textrm{Span}\,}}\mathcal {V}^{\lambda }. \end{aligned}$$

Example A.14

($N = 3$) Let $x_1, x_2, x_3 \in \mathbb {R}_+$, and define

$$\begin{aligned} S = x_1 \cdot \Pi ^{\mathsf {\Gamma }}_{(0 \, 1)} + x_2 \cdot \Pi ^{\mathsf {\Gamma }}_{(0 \, 2)} + x_3 \cdot \Pi ^{\mathsf {\Gamma }}_{(0 \, 3)}, \end{aligned}$$

that is

Then, an eigenvector for the largest eigenvalue of R is

$$\begin{aligned} \chi = \beta _1 \cdot \left| \Omega \right\rangle _{(0,1)} \otimes \left| \textbf{v}\right\rangle + \beta _2 \cdot \left| \Omega \right\rangle _{(0,2)} \otimes \left| \textbf{v}\right\rangle + \beta _3 \cdot \left| \Omega \right\rangle _{(0,3)} \otimes \left| \textbf{v}\right\rangle , \end{aligned}$$

that is,

for some coefficients $\beta _i$, and $\textbf{v} = \displaystyle \sum _{1 \le i \le j < d} \left| ij\right\rangle$.

Appendix B: Some results in linear algebra

In the proofs, we use simple linear algebra results that are not mainstream. We gather them in this appendix for the convenience of the reader.

The first one is a consequence of [9], for which we provide a simple proof.

Lemma B.1

For all $1 \le i, j \le n$, let $M_{ij}$ be some matrices in $\mathcal {M}_m$, and define the block matrix

and the matrix . We then have

$$\begin{aligned} \Vert M\Vert \le \Vert M'\Vert . \end{aligned}$$

Proof

Consider v and w two unit vectors such that $\Vert M\Vert = \left\langle v\right| M \left| w\right\rangle$, we can decompose $v=\displaystyle \sum \nolimits _{i=1}^n \vert v_i\rangle \otimes \vert i\rangle$ with $v_i\in \mathbb C^m$ and we define $v'=\displaystyle \sum \nolimits ^n_{i=1} \Vert v_i\Vert \cdot \vert i\rangle$. Since we consider the Euclidian norm, we have obviously $\Vert v'\Vert =\Vert v\Vert =1$. We define $w'$ similarly and then have

$$\begin{aligned} \Big | \! \left\langle v\right| M \left| w\right\rangle \! \Big |&= \bigg | \sum _{ij} \left\langle v_i\right| M_{ij} \left| w_j\right\rangle \bigg | \\&\le \sum _{ij} \Vert v_i\Vert \, \Vert M_{ij}\Vert \, \Vert w_j\Vert \quad (\text {Cauchy-Schwarz}) \\&= \sum _{ij} \Vert v_i\Vert \, \Vert w_j\Vert \, \big \langle i \Big | M^{'} \Big | j \big \rangle \\&=\big \langle v' \Big | M^{'} \Big | w' \big \rangle \\&\le \Vert v'\Vert \, \Vert w'\Vert \, || M^{'} ||\\&\le || M^{'} ||. \end{aligned}$$

which ends the proof.$\square$

The following lemma is useful in the paper when we relate the spectrum of $S_x$ with the one of $\tilde{S}$

Lemma B.2

Let $A\in \mathcal M$ be an operator, and let $n\ge m$ let $\{ f_1, \ldots , f_n \}$ be a set of vectors such that $\mathbb C^m = {{\,\textrm{Span}\,}}\{ f_1, \ldots , f_n \}$. We suppose that for all $i=1,\ldots ,n$ there exists $a_{ij}$, $j=1,\ldots ,n$ such that

$$\begin{aligned} Af_i=\sum _{i,j=1}^n a_{ij}f_j \end{aligned}$$

Denote $\tilde{A}$ the $n\times n$ matrix with coefficients $(a_{ij})$. Then, 

$$\begin{aligned} {{\,\textrm{Spec}\,}}{A} \subseteq {{\,\textrm{Spec}\,}}{\tilde{A}}. \end{aligned}$$

Proof

Let F be the matrix composed of the vectors $f_i$, $i=1,\ldots ,n$ written in column. Then, F is a $m\times n$ matrix of rank m since $\mathbb C^m=Vect\{ f_1, \ldots , f_n \}$. By definition of $\tilde{A}$, we have

$$AF=F\tilde{A}$$

Furthermore, for all scalar $\alpha$

$$\begin{aligned} (A - \alpha \, I_m) F&= A F - \alpha \, I_m \, F \\&= F \tilde{A} - \alpha \, F\\&=F(\tilde{A}-\alpha I_n) \end{aligned}$$

Now using the fact that $rank(AB)\le min(rank(A),rank(B))$ for all matrices A and B, if $\alpha \in Spec A$ since $\hbox {rank}(F)=m$

$$\begin{aligned} \hbox {rank}(F(\tilde{A}-\alpha I_n))=\hbox {rank}((A - \alpha \, I_m) F)<m. \end{aligned}$$

Now, since $n\ge m$, this necessarily implies that $\hbox {rank}(\tilde{A}-\alpha I_n)<n$ which says that $\alpha \in Spec \tilde{A}$ $\square$

Remark B.3

Note that the matrix $\tilde{A}$ is not uniquely defined and depends on a choice of the way of writing $Af_i=\sum _{i,j=1}^n a_{ij}f_j$, but the results concerning the inclusion are true for any form of the matrix $\tilde{A}$. Of course the other inclusion is not true in general unless $n=m$ and $\{ f_1, \ldots , f_n \}$ is linearly independent (but this is trivial in this case).

Remark B.4

In the lemma, the condition $Af_i=\sum _{i,j=1}^n a_{ij}f_j$ means that A leaves $V= {{\,\textrm{Span}\,}}\{ f_1, \ldots , f_n \}$ invariant and we can rephrase this lemma by considering the restriction to A on V and then saying that ${{\,\textrm{Spec}\,}}{A_{\vert V}} \subseteq {{\,\textrm{Spec}\,}}{\tilde{A}}.$ which is what we use in the body of the paper.

Appendix C: The $\mathcal Q$-norm

Lemma C.1

For all $x\in \mathbb R^N$, we have

$$\begin{aligned} \frac{1}{d}\Vert x\Vert _1\le \lambda _{\max }(S_x)\le d\Vert x\Vert _1. \end{aligned}$$

In particular, the $\Vert \cdot \Vert _{\mathcal Q}$ quantity is nonnegative. The first inequality is saturated if and only if the matrix $S_x$ is a (nonnegative) multiple of the identity. The second inequality is saturated if and only if x has at most one nonzero entry.

Proof

For the first inequality, note that

$$\begin{aligned} \lambda _{\max }(S_x) \ge \frac{\hbox {Tr}S_x}{d^{N+1}} = \frac{ \sum ^N_{i=1} d |x_i| \cdot d^{N-1}}{d^{N+1}} = \frac{\Vert x\Vert _1}{d}. \end{aligned}$$

The inequality above is saturated if and only if the eigenvalues of $S_x$ are identical.

For the second inequality, we use the subadditivity of the $\lambda _{\max }$ functional:

$$\begin{aligned} \lambda _{\max }(S_x) \le \sum ^N_{i=1} \lambda _{\max }\left( |x_i| \cdot \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \right) = \sum ^N_{i=1} d |x_i| = d\Vert x\Vert _1. \end{aligned}$$

The inequality above is saturated if and only if the matrices $|x_i| \cdot \omega _{(0,i)} \otimes I$ have a common largest eigenvector, which can happen only if the support of x has size 0 or 1.$\square$

Lemma C.2

For all $x, y \in \mathbb {R}^N_+$, $\Vert x + y\Vert _{\mathcal {Q}} \le \Vert x\Vert _{\mathcal {Q}} + \Vert y\Vert _{\mathcal {Q}}$.

Proof

This follows from the subadditivity of the $\lambda _{\max }$ functional: For $x, y \in \mathbb {R}^N_+$, we have

$$\begin{aligned} \lambda _{\max } \bigg [ \sum ^{N}_{i = 1}&(x_i + y_i) \cdot \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \bigg ] \\&\le \lambda _{\max } \bigg [ \sum ^{N}_{i = 1} x_i \cdot \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \bigg ] + \lambda _{\max } \bigg [ \sum ^{N}_{i = 1} y_i \cdot \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \bigg ]. \end{aligned}$$

$\square$

Lemma C.3

For all $t \in [0,1]^N$ and $x \in \mathbb {R}^N_+$, $\Vert t \cdot x\Vert _{\mathcal {Q}} \le \Vert x\Vert _{\mathcal {Q}}$.

Proof

Let us show that for $0 \le x \le y$ (meaning that $x_i\le y_i$, $i=1,\ldots ,N$), we have $\Vert x\Vert _{\mathcal {Q}} \le \Vert y\Vert _{\mathcal {Q}}$. Let $\chi = \sum _k \beta _k \left| \Omega _{(0,k)}\right\rangle \otimes \left| \textbf{v}\right\rangle$ be a normalized eigenvector corresponding to the largest eigenvalue $\lambda _{\max }(S_x)$, from Theorem 3.13, that is

$$\begin{aligned} \lambda _{\max }(S_x) = \left\langle \chi \Big | \sum ^{N}_{i = 1} x_i \cdot \big ( \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \big ) \Big | \chi \right\rangle . \end{aligned}$$

Note that due to the Perron–Frobenius theorem, $\beta _i > 0$ for all $i=1,\ldots ,N$. Let us point out two facts. First using Lemma 3.6

$$\begin{aligned} \left\langle \chi \right| \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \left| \chi \right\rangle&= \sum ^N_{k = 1} \sum ^N_{l = 1} \beta _k \beta _l \cdot \left\langle \Omega _{(0,k)} \otimes \textbf{v}\right| \big ( \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \big ) \left| \Omega _{(0,l)} \otimes \textbf{v}\right\rangle \\&= \sum ^N_{k = 1} \beta _k \cdot \big \langle \Omega _{(0,k)} \otimes \textbf{v} \Big | d \beta _i \cdot \Omega _{(0,i)} \otimes \textbf{v} + \sum ^N_{\begin{array}{c} l = 1 \\ l \ne i \end{array}} \beta _l \cdot \Omega _{(0,i)} \otimes \textbf{v} \big \rangle \\&= \sum ^N_{k = 1} \beta _k \cdot \big \langle \Omega _{(0,k)} \otimes \textbf{v} \Big | (d - 1) \beta _i \cdot \Omega _{(0,i)} \otimes \textbf{v} + \sum ^N_{l = 1} \beta _l \cdot \Omega _{(0,i)} \otimes \textbf{v} \big \rangle \\&= {\bigg ( (d - 1) \beta _i + \sum _k \beta _k \bigg )}^2 \end{aligned}$$

Second

$$\begin{aligned} 1&=\langle \chi |)\rangle \\&= \sum _{i\ne j}\beta _i\beta _j+d\sum _i\beta _i^2\\&= \left( \sum _{i=1}^N\beta _i\right) ^2+(d-1)\sum _{i=1}^N\beta _i^2 \end{aligned}$$

Now using that $\left( \sum _{i=1}^N\beta _i\right) ^2\ge \sum _{i=1}^N\beta _i^2$ we get that $1\le d\left( \sum _{i=1}^N\beta _i\right) ^2$ then since $\beta _i\ge 0$

$$\begin{aligned} \left\langle \chi \right| \omega _{(0,i)} \otimes I^{\otimes (N - 1)} \left| \chi \right\rangle \ge \left( \sum _{i=1}^N\beta _i\right) ^2\ge \frac{1}{d} \end{aligned}$$

This way since $y\ge x$, we have that $\lambda _{max}(S_y)\ge \langle \chi ,S_y\chi \rangle$ then

$$\begin{aligned} \lambda _{max}(S_y)-\lambda _{max}(S_x)&\ge \langle \chi \vert S_y-S_x\vert \chi \rangle \\&=\sum _{i=1}^N(y_i-x_i)\langle \chi \vert \omega _{(0,i)} \otimes I^{\otimes (N - 1)}\vert \chi \rangle \\&=\sum _{i=1}^N(y_i-x_i){\bigg ( (d - 1) \beta _i + \sum _k \beta _k \bigg )}^2\\&\ge \sum _{i=1}^N(y_i-x_i)\frac{1}{d} \end{aligned}$$

Then,

$$\begin{aligned} \Vert y\Vert _{\mathcal {Q}}=\lambda _{max}(S_y)-\frac{1}{d} \Vert y\Vert \ge \lambda _{max}(S_x)-\frac{1}{d} \Vert x\Vert =\Vert x\Vert _{\mathcal {Q}} \end{aligned}$$

which was the desired result.$\square$

Appendix D: Optimal cloning

Lemma D.1

Let some positive real numbers ${(\beta _i)}_{1 \le i \le N}$. The operator

$$\begin{aligned} \widetilde{C}_{T_\beta }=\sum _{\begin{array}{c} 1 \le a,b \le N \\ \sigma \in \Sigma _{a,b} \end{array}} \frac{\beta _a \beta _b}{(N - 1)!} \Pi _\sigma ^{\mathsf {\Gamma }} \end{aligned}$$

is an orthogonal projection if and only if the condition

$$\begin{aligned} (d - 1) \sum ^N_{i = 1} \beta ^2_i + {\bigg ( \sum ^N_{i = 1} \beta _i \bigg )}^2 = 1 \end{aligned}$$

(32)

is satisfied. In this case, the operator

$$\begin{aligned} C_{T_\beta } = \frac{d}{\hbox {Tr}P^+_{\mathfrak {S}_N}} \frac{N + d - 1}{N} \cdot \widetilde{C}_{T_\beta }, \end{aligned}$$

is a positive operator, such that $\hbox {Tr}_{\scriptscriptstyle [\![ 1,N]\!]} C_{T_\beta } = I$

Proof

Let some positive real numbers ${(\beta _i)}_{1 \le i \le N}$, then

$$\begin{aligned} {\big ( \widetilde{C}_{T_\beta } \big )}^2&= \sum _{\begin{array}{c} 1 \le a,b \le N \\ 1 \le c,d \le N \end{array}} \, \sum _{\begin{array}{c} \sigma \in \Sigma _{a,b} \\ \tau \in \Sigma _{c,d} \end{array}} \frac{\beta _a \beta _b \beta _c \beta _d}{{(N-1)!}^2} \, \Pi _\sigma ^{\mathsf {\Gamma }} \Pi _\tau ^{\mathsf {\Gamma }} \\&= \sum _{1 \le a,d \le N} \left[ d \sum _{\begin{array}{c} 1 \le b,c \le N \\ b = c \end{array}} \sum _{\sigma \in \Sigma _{a,d}} \frac{\beta _a \beta _b \beta _c \beta _d}{{(N-1)!}^2} \, \Pi _\sigma ^{\mathsf {\Gamma }} + \sum _{\begin{array}{c} 1 \le b,c \le N \\ b \ne c \end{array}} \sum _{\sigma \in \Sigma _{a,d}} \frac{\beta _a \beta _b \beta _c \beta _d}{{(N-1)!}^2} \, \Pi _\sigma ^{\mathsf {\Gamma }} \right] . \end{aligned}$$

The projection condition ${\big ( \widetilde{C}_{T_\beta } \big )}^2 = \widetilde{C}_{T_\beta }$ is equivalent to

$$\begin{aligned} \beta _a \beta _b&= d \sum ^N_{i = 1} \beta _a \beta _i \beta _i \beta _b + \sum _{\begin{array}{c} 1 \le i,j \le N \\ i \ne j \end{array}} \beta _a \beta _i \beta _j \beta _b \\&= (d - 1) \sum ^N_{i = 1} \beta _a \beta _i \beta _i \beta _b + \sum _{1 \le i,j \le N} \beta _a \beta _i \beta _j \beta _b \\&= \beta _a \bigg ( (d - 1) \sum ^N_{i = 1} \beta _i \beta _i + \sum _{1 \le i,j \le N} \beta _i \beta _j \bigg ) \beta _b \\&= \beta _a \bigg ( (d - 1) \sum _{i=1}^N\ \beta ^2_i + {\bigg ( \sum _i \beta _i \bigg )}^2 \bigg ) \beta _b. \end{aligned}$$

Since (17) is assumed to be satisfied, then $\widetilde{C}_{T_\beta }$ is an orthogonal projection and hence a positive operator. This way $C_{T_\beta }$ is positive. We have now from Lemma A.11

$$\begin{aligned} \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \big ( \widetilde{C}_{T_\beta } \big )&= \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \bigg [ \sum _{\begin{array}{c} 1 \le a,b \le N \\ \sigma \in \Sigma _{a,b} \end{array}} \frac{\beta _a \beta _b}{(N - 1)!} \sigma ^{\mathsf {\Gamma }} \bigg ] \\&= \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] \bigg ( \sum _i \beta _i \beta _i + \frac{1}{d} \sum _{i \ne j} \beta _i \beta _j \bigg ) \cdot I \\&= \frac{\hbox {Tr}P^+_{\mathfrak {S}_{N - 1}}}{d} \bigg [ (d - 1) \sum _i \beta ^2_i + {\bigg ( \sum _i \beta _i \bigg )}^2 \bigg ] \cdot I \\&= \frac{\hbox {Tr}P^+_{\mathfrak {S}_{N - 1}}}{d} \cdot I \end{aligned}$$

Then,

$$\begin{aligned} \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} (C_{T_\beta })&= \hbox {Tr}_{\scriptscriptstyle [\![ 1,N ]\!]} \big [ \frac{d}{\hbox {Tr}P^+_{\mathfrak {S}_N}} \frac{N + d - 1}{N} \cdot \widetilde{C}_{T_\beta } \Big ] \\&= \frac{\hbox {Tr}P^+_{\mathfrak {S}_{N - 1}}}{\hbox {Tr}P^+_{\mathfrak {S}_N}} \frac{N + d - 1}{N} \cdot I \\&= I, \end{aligned}$$

where the last equation comes from $\hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] = \frac{N}{N + d - 1} \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N}} \Big ]$.$\square$

Lemma D.2

For any $1 \le a,b \le N$, and any $\sigma \in \Sigma _{a,b}$, the permutation operator $\sigma ^{\mathsf {\Gamma }}$ is the Choi matrix a linear map $T_{\mu ,\nu }: \mathcal {M}_d \rightarrow {(\mathcal {M}_d)}^{\otimes N}$ defined by

$$\begin{aligned} T_{\mu ,\nu }(X) = \Pi _\mu \big ( X \otimes I^{\otimes (N - 1)} \big ) \Pi _\nu \end{aligned}$$

for some permutations $\mu$ and $\nu$ in $\mathfrak {S}_N$ such that $\mu (0) = a - 1$ and $\nu (b - 1) = 0$.

As a consequence for some fixed $1 \le a,b \le N$,

$$\begin{aligned} \sum _{\sigma \in \Sigma _{a,b}} \Pi _\sigma ^{\mathsf {\Gamma }} = \frac{1}{(N - 1)!} C_{T_{a,b}} \end{aligned}$$

where

$$\begin{aligned} T_{a,b}(X) = \sum _{\begin{array}{c} \scriptstyle \mu ,\nu \scriptstyle \in \mathfrak {S}_N \\ \scriptstyle \mu (0) \scriptstyle = a - 1 \\ \scriptstyle \nu (b - 1) \scriptstyle = 0 \end{array}} \Pi _\mu \big ( X \otimes I^{\otimes (N - 1)} \big ) \Pi _\nu . \end{aligned}$$

Proof

Let $1 \le a,b \le N$, and $\sigma \in \Sigma _{a,b}$. Then from Lemma A.5, we have

$$\begin{aligned} \Pi _\sigma ^{\mathsf {\Gamma }} = \Pi _{(1 \, a)} \, (\omega \otimes \Pi _{\hat{\sigma }}) \; \Pi _{(1 \, b)}, \end{aligned}$$

such that on $X \in \mathcal {M}_d$, the partial trace yields

And the result holds for $\Pi _\mu = \Pi _{(0 \, (a - 1))}$ and $\Pi _\nu = (I \otimes \Pi _{\hat{\sigma }}) \circ \Pi _{(0 \, (b - 1))}$. $\square$

Lemma D.3

Let an orthogonal projector $\widetilde{C}_{T_\beta }$ for some positive real number ${(\beta _i)}_{1 \le i \le N}$ satisfying Eq. (17). Then for any $1 \le i \le N$, we have

$$\begin{aligned} \widetilde{C}_{T_\beta } \, \omega _{(0,i)} = \frac{1}{(N - 1)!} \sum _{\begin{array}{c} 1 \le a \le N \\ \sigma \in \Sigma _{a,i} \end{array}} \beta _a \bigg ( (d - 1) \beta _i + \sum _{1 \le b \le N} \beta _b \bigg ) \Pi _\sigma ^{\mathsf {\Gamma }}. \end{aligned}$$

Let Choi matrix $C_{T_\beta }$ for some positive real numbers ${(\beta _i)}_{1 \le i \le N}$ satisfying Eq. (17). Then for any $1 \le i \le N$, we have

$$\begin{aligned} \hbox {Tr}\big [ C_{T_\beta } \, \omega _{(0,i)} \Big ] = d {\bigg ( (d - 1) \beta _i + \sum ^N_{j = 1} \beta _j \bigg )}^2. \end{aligned}$$

Proof

Since we are summing on all permutations $\Pi ^{\mathsf {\Gamma }}_{\sigma _{a,b}}$, a direct calculation gives us

$$\begin{aligned} \widetilde{C}_{T_\beta } \, \omega _{(0,i)}&= \frac{1}{(N - 1)!} \sum _{\begin{array}{c} 1 \le a,b \le N \\ \sigma \in \Sigma _{a,b} \end{array}} \beta _a \beta _b \cdot \Pi _\sigma ^{\mathsf {\Gamma }} \, \omega _{(0,i)} \\&= \frac{1}{(N - 1)!} \sum _{1 \le a \le N} \beta _a \bigg ( (d - 1) \sum _{\sigma \in \Sigma _{a,i}} \beta _i \cdot \sigma ^{\mathsf {\Gamma }} + \sum _{\begin{array}{c} 1 \le b \le N \\ \sigma \in \Sigma _{a,i} \end{array}} \beta _b \cdot \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ) \\&= \frac{1}{(N - 1)!} \sum _{\begin{array}{c} 1 \le a \le N \\ \sigma \in \Sigma _{a,i} \end{array}} \beta _a \bigg ( (d - 1) \beta _i + \sum _{1 \le b \le N} \beta _b \bigg ) \Pi _\sigma ^{\mathsf {\Gamma }}. \end{aligned}$$

And then we look at the action of the Choi matrix $C_{T_\beta }$ onto some $\omega _{(0,i)}$ and take the trace.

Lemma A.11 yields

$$\begin{aligned} \hbox {Tr}\big [ C_{T_\beta } \, \omega _{(0,i)} \Big ]&= \frac{d (N + d - 1)}{N \, \hbox {Tr}\! \big [ P^+_{\mathfrak {S}_N} \Big ]} \sum ^N_{a = 1} \frac{\beta _a}{(N - 1!)} \bigg ( (d - 1) \beta _i + \sum ^N_{b = 1} \beta _b \bigg ) \hbox {Tr}\bigg [ \sum _{\sigma \in \Sigma _{a,i}} \Pi _\sigma ^{\mathsf {\Gamma }} \bigg ] \\&= \frac{d (N + d - 1)}{N \, \hbox {Tr}\! \big [ P^+_{\mathfrak {S}_N} \Big ]} \bigg ( (d - 1) \beta _i + \sum ^N_{b = 1} \beta _b \bigg ) \bigg ( (d - 1) \beta _i + \sum ^N_{a = 1} \beta _a \bigg ) \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] \\&= \frac{d (N + d - 1)}{N \, \hbox {Tr}\! \big [ P^+_{\mathfrak {S}_N} \Big ]} {\bigg ( (d - 1) \beta _i + \sum ^N_{j = 1} \beta _j \bigg )}^2 \hbox {Tr}\big [ P^+_{\mathfrak {S}_{N - 1}} \Big ] \\&= d {\bigg ( (d - 1) \beta _i + \sum ^N_{j = 1} \beta _j \bigg )}^2. \end{aligned}$$

$\square$