$\Phi^{\ell}_{\operatorname{Tr}}$ or $\Phi^{\ell}_{\mathrm{UNOT}}$. Lemma E.6
in Section E.5 shows that both channels can be implemented using $O(n\log n)$
elementary gates.

Hence, the overall gate complexity of $\mathcal{A}_{\mathbf{t}}$ is
$O(n^{4}\log n)$ irrespective of the interpolation parameters $\mathbf{t}$.
While computing these parameters takes additional time, the linear program
(LP) is of size $O(n)$ and solving such program takes significantly less than
$O(n^{4})$ time, hence this does not affect the asymptotic scaling. ∎

##  5 Applications of the general framework

This section contains several applications of the general framework for
computing symmetric equivariant Boolean functions derived in Section 4.4.
First, in Section 5.1, we present exact numerical values of optimal fidelities
for all functions up to $n=7$ arguments, and some empirical observations
regarding these values. We also compute exact optimal fidelities for two
infinite families of functions: majority $\mathrm{MAJ}_{n}$ and parity
$\mathrm{PAR}_{n}$. In Section 5.2, we analyze the asymptotic behavior of the
optimal fidelity for the majority function. Finally, in Section 5.3 we relax
complete positivity and introduce the abstract notion of unitary-equivariant
Boolean functions.

###  5.1 Exact numerical results

For any symmetric and equivariant Boolean function
$f:\\{0,1\\}^{n}\to\\{0,1\\}$, Theorem 4.7 shows that the optimal fidelity for
computing $f$ in a basis-independent way is given by the solution to a linear
program of size $O(n)$, which can easily be solved using a computer. This
section lists several exact numerical results obtained in this way.

####  5.1.1 Detailed solutions for $n=1$ and $n=3$

For $n=1$, the linear program (LP) reads

$\max_{c,t_{0}}c\quad\big{\lparen}1-f(0)\big{\rparen}t_{0}+\frac{1+f(0)}{3}(1-t_{0})\geq
c,\quad 0\leq t_{0}\leq 1,$ (5.1)

while for $n=3$ it reads

$\displaystyle\max_{c,t_{0},t_{1}}c\quad$
$\displaystyle(1-f(0))t_{0}+\tfrac{1+3f(0)}{5}(1-t_{0})\geq c,$
$\displaystyle\frac{1}{3}\left\lparen\tfrac{2-f(1)}{3}t_{0}+\tfrac{2+f(1)}{5}(1-t_{0})\right\rparen+\frac{2}{3}\left\lparen(1-f(1))t_{1}+\tfrac{1+f(1)}{3}(1-t_{1})\right\rparen\geq
c,$ (5.2) $\displaystyle 0\leq t_{0}\leq 1,\quad 0\leq t_{1}\leq 1.$

Solutions of these linear programs for all symmetric equivariant Boolean
functions $f$ with $n=1$ and $n=3$ arguments are summarized in Table 2, and
the Choi matrices of the corresponding quantum channels are listed in Section
G.1.

Function | $\bar{f}$ | $F_{f}$ | $t_{0}$ | $t_{1}$ | $c_{0}$ | $c_{1}$
---|---|---|---|---|---|---
$\mathrm{ID}$ | $0$ | $1$ | $1$ | – | $1$ | –
$\mathrm{NOT}$ | $1$ | $2/3$ | $0$ | – | $2/3$ | –
$\mathrm{MAJ}_{3}$ | $00$ | $8/9$ | $1$ | $1$ | $1$ | $8/9$
$\mathrm{PAR}_{3}$ | $01$ | $3/5$ | $1/2$ | $0$ | $3/5$ | $3/5$
$\mathrm{NPAR}_{3}$ | $10$ | $4/5$ | $0$ | $1$ | $4/5$ | $4/5$
$\mathrm{NMAJ}_{3}$ | $11$ | $29/45$ | $0$ | $0$ | $4/5$ | $29/45$
Table 2: Optimal fidelities $F_{f}$ and the corresponding interpolation
parameters $t_{0},t_{1}$ for all symmetric equivariant Boolean functions with
$n=1$ and $n=3$ arguments from Table 1. Here $\bar{f}$ stands for the truth
table of $\bar{f}(\lvert x\rvert):=f(x)$ and $c_{h}$ denotes the fidelity
achieved on inputs of Hamming weight $h\in\\{0,\dotsc,\lfloor n/2\rfloor\\}$.
The Choi matrices corresponding to these channels are listed in Section G.1.

####  5.1.2 Optimal fidelities for all functions up to $n=7$

The optimal fidelities for all symmetric equivariant Boolean functions on up
to $7$ bits are as follows (for $n=1,3,5,7$):

$\left\\{\begin{array}[]{cc}0&1\\\
1&\frac{2}{3}\end{array}\right.\qquad\left\\{\begin{array}[]{cc}00&\frac{8}{9}\\\
01&\frac{3}{5}\\\ 10&\frac{4}{5}\\\
11&\frac{29}{45}\end{array}\right.\qquad\left\\{\begin{array}[]{cc}000&\frac{62}{75}\\\
001&\frac{4}{7}\\\ 010&\frac{5}{7}\\\ 011&\frac{95}{153}\\\
100&\frac{124}{153}\\\ 101&\frac{4}{7}\\\ 110&\frac{5}{7}\\\
111&\frac{331}{525}\end{array}\right.\qquad\left\\{\begin{array}[]{cc}0000&\frac{2888}{3675}\\\
0001&\frac{5}{9}\\\ 0010&\frac{2}{3}\\\ 0011&\frac{47}{78}\\\
0100&\frac{59}{78}\\\ 0101&\frac{5}{9}\\\ 0110&\frac{2}{3}\\\
0111&\frac{1141}{1845}\\\ 1000&\frac{1444}{1845}\\\ 1001&\frac{5}{9}\\\
1010&\frac{2}{3}\\\ 1011&\frac{47}{78}\\\ 1100&\frac{59}{78}\\\
1101&\frac{5}{9}\\\ 1110&\frac{2}{3}\\\
1111&\frac{6841}{11025}\end{array}\right.$

Each block corresponds to a choice of $n$, while each row within a block
corresponds to a different function $f$. The function is represented by a bit
string that indicates its values on inputs of Hamming weight $0,\dotsc,\lfloor
n/2\rfloor$.101010This determines the remaining values of $f$, including the
values on inputs of Hamming weight larger than $\lfloor n/2\rfloor$, thanks to
symmetry and equivariance of $f$. Equivalently, the bit string representing
$f$ is the _truth table_ $\bar{f}(0)\bar{f}(1)\dots\bar{f}(\lfloor
n/2\rfloor)$ of the function $\bar{f}(\lvert x\rvert):=f(x)$. For example,
$\mathrm{MAJ}_{7}$ corresponds to the sequence $0000$ while $\mathrm{PAR}_{7}$
corresponds to $0101$.

Careful inspection of these values reveals a certain pattern. For example, for
$n=7$ all functions whose truth table ends with $10$ have fidelity $2/3$,
while those that end with $01$ have fidelity $5/9$. Similarly, suffixes $100$
and $011$ lead to fidelities $59/78$ and $47/78$, respectively. More
generally, the optimal fidelity $F_{f}$ of $f$ depends only on the final digit
$\bar{f}(\lfloor n/2\rfloor)$ and the size of the gap around $n/2$ in the
truth table of $\bar{f}$. The truth table of $\bar{f}$ ends either with a
string of $k$ zeroes or $k$ ones:

$\displaystyle*\dots*1\underbrace{0\dots 0}_{k}$
$\displaystyle*\dots*0\underbrace{1\dots 1}_{k}.$ (5.3)

This number $k$, together with the last digit $\bar{f}(\lfloor n/2\rfloor)$,
seems to completely determine the optimal fidelity $F_{f}$.111111This is an
empirical observation. We leave it for future work to establish this formally
and to determine the exact dependence on $k$. This observation is useful for
displaying the optimal fidelities in a more compact way (see Fig. 5).
Moreover, it reminds of the behavior of the regular quantum query complexity
$\mathsf{Q}(f)$ of symmetric Boolean functions established in [BBC+01, Theorem
3.3]. It is an interesting open problem to determine whether there is a closer
relationship between the optimal fidelity $F_{f}$ and the quantum query
complexity $\mathsf{Q}(f)$ of all equivariant Boolean functions $f$.

$0$$1$$0$$1$$\displaystyle\frac{2}{3}$$0$$\displaystyle\frac{5}{9}$$1$$0$$1$$\displaystyle\frac{59}{78}$$0$$\displaystyle\frac{47}{78}$$1$$0$$\displaystyle\frac{2888}{3675}$$1$$\displaystyle\frac{1444}{1845}$$0$$\displaystyle\frac{1141}{1845}$$1$$\displaystyle\frac{6841}{11025}$
Figure 5: Optimal fidelities of all symmetric and equivariant $7$-bit Boolean
functions, arranged in a tree. To determine $F_{f}$ for a given function $f$,
walk down the tree while reading the truth table of $\bar{f}$ backwards. For
example, if the truth table ends with $\dots 10$ then the fidelity is $2/3$,
while the suffix $\dots 011$ leads to fidelity $47/78$.

####  5.1.3 Optimal fidelities for majority

We have also calculated fidelities for $\mathrm{MAJ}_{n}$ for larger values of
$n=1,3,5,\dotsc$:

$1,\frac{8}{9},\frac{62}{75},\frac{2888}{3675},\frac{15014}{19845},\frac{117548}{160083},\frac{13848922}{19324305},\frac{5816048}{8281845},\frac{183562382}{265939245},\frac{30465827276}{44801898141},\frac{6378478534}{9503432939},\dotsc$

which, in decimal, are

$1.,0.888889,0.826667,0.785850,0.756563,0.734294,0.716658,0.702265,0.690242,0.680012,\dotsc$

These and larger values can also be obtained by computing $g(\frac{n+1}{2})$
where $g$ is given by the following recursive formula (see proof in Appendix
F):

lemmarecursion Let $n\geq 1$ be odd. The optimal fidelity for the $n$-qubit
quantum majority vote is $F_{\mathrm{MAJ}}(n)=g(\frac{n+1}{2})$ where

$\displaystyle g(1)$ $\displaystyle:=1,\quad g(2):=8/9,$ $\displaystyle g(m)$
$\displaystyle:=\frac{2m}{(2m-1)^{2}(2m+1)}\Big{[}\big{\lparen}2m(4m-7)+5\big{\rparen}g(m-1)-4(m-1)(m-2)g(m-2)+1\Big{]}.$

####  5.1.4 Optimal fidelities for parity

Finally, we have also computed fidelities for the parity function
$\mathrm{PAR}_{n}$, $n=1,3,5,\dotsc$:

$1,\frac{3}{5},\frac{5}{7},\frac{5}{9},\frac{7}{11},\frac{7}{13},\frac{3}{5},\frac{9}{17},\frac{11}{19},\frac{11}{21},\frac{13}{23},\frac{13}{25},\frac{5}{9},\frac{15}{29},\frac{17}{31},\frac{17}{33},\frac{19}{35},\frac{19}{37},\frac{7}{13},\frac{21}{41},\dotsc$

Based on these numerical results, we conjecture that

$F_{\mathrm{PAR}_{n}}=\frac{2\lceil\frac{n+1}{4}\rceil+1}{n+2}=1/2+O(1/n).$
(5.4)

###  5.2 Asymptotic fidelity of quantum majority

Armed with our general result, Theorem 4.7, we are able to calculate the
optimal fidelity that can be achieved for the majority function, hence proving
Fig. 3. Recall that for odd $n$, the majority function evaluates to 0 on all
$x$ such that $|x|\leq n/2$, and to 1 on all $x$ such that $|x|>n/2$. By
Theorem 4.7, the maximal fidelity that can be achieved is

$\min_{0\leq h\leq\frac{n-1}{2}}\sum_{k=0}^{h}p_{k}(h)\left\lparen
t_{k}\frac{n-h-k}{n-2k}+(1-t_{k})\frac{h-k+1}{n-2k+2}\right\rparen.$ (5.5)

Observe that as $n-h-k\geq h-k+1$ for all $h\leq(n-1)/2$, the term weighted by
$t_{k}$ in eq. 5.5 is larger than the term weighted by $1-t_{k}$, for all $k$.
So we can assume that $t_{k}=1$ for all $k$. Then the expression for the
optimal worst-case fidelity is $F_{\mathrm{MAJ}}(n):=\min_{0\leq
h\leq\frac{n-1}{2}}F_{n}(h)$ where

$F_{n}(h):=\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}\cdot\frac{n-h-k}{n-2k}.$
(5.6)

While this will not be directly used in the proofs below, we show in Lemma F.2
of Appendix F that $F_{n}(h)$ is strictly decreasing in
$h\in\\{0,\dotsc,\frac{n-1}{2}\\}$. This implies that the minimum in the
definition of $F_{\mathrm{MAJ}}(n)$ is achieved at $h=\frac{n-1}{2}$ and
proves the intuitive claim that, as the number of minority states increases,
it becomes harder to output the correct majority state.

*

###### Proof.

For simplicity, let us write $F(h)$ instead of $F_{n}(h)$. We now rewrite eq.
5.6 as

$F(h)=\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}\left\lparen
1-\frac{h-k}{n-2k}\right\rparen=1-\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}\cdot\frac{h-k}{n-2k}.$
(5.7)

We will consider two regimes. In the first, $h$ is arbitrary and we seek a
lower bound on $F(h)$. Here it will be helpful to parametrize
$h=(n-1)/2-\delta$, giving

$F(h)=1-\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}\cdot\frac{1}{2}\left\lparen
1-\frac{2\delta+1}{n-2k}\right\rparen=\frac{1}{2}+\frac{1}{2}\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}\cdot\frac{2\delta+1}{n-2k}.$
(5.8)

Since
$\binom{n}{k}-\binom{n}{k-1}=\binom{n}{k}\big{\lparen}1-\frac{k}{n-k+1}\big{\rparen}$,

$F(h)=\frac{1}{2}+\frac{2\delta+1}{2}\sum_{k=0}^{h}\frac{\binom{n}{k}}{\binom{n}{h}}\frac{n-2k+1}{(n-k+1)(n-2k)}.$
(5.9)

Since the last fraction is between $1/n$ and $4/n$ for all $k$ between 0 and
$h\leq(n-1)/2$,

$\frac{1}{2}+\frac{2\delta+1}{2n\binom{n}{h}}\sum_{k=0}^{h}\binom{n}{k}\leq
F(h)\leq\frac{1}{2}+\frac{2(2\delta+1)}{n\binom{n}{h}}\sum_{k=0}^{h}\binom{n}{k}.$
(5.10)

Fidelity with no promise on $|x|$. To complete the bound in this case, we
consider $\delta\leq\sqrt{n}$ and $\delta>\sqrt{n}$ separately. In the former
case, as $\delta=o(n^{2/3})$ we have [SF14]

$\binom{n}{h}=\binom{n}{n/2}e^{-(n-2h)^{2}/n}(1+o(1))=\binom{n}{n/2}e^{-(2\delta+1)^{2}/n}(1+o(1))\leq
C\frac{2^{n}}{\sqrt{n}},$ (5.11)

while $\sum_{k=0}^{h}\binom{n}{k}\geq C^{\prime}2^{n}$ (for example, via the
normal approximation to the binomial distribution), for some universal
constants $C$, $C^{\prime}$. So

$F(h)\geq\frac{1}{2}+\frac{\sqrt{n}}{2Cn2^{n}}C^{\prime}2^{n}=\frac{1}{2}+\frac{C^{\prime\prime}}{\sqrt{n}}$
(5.12)

for some new universal constant $C^{\prime\prime}$. If $\delta>\sqrt{n}$, as
the sum in eq. 5.10 is clearly larger than $\binom{n}{h}$,
$F(h)>1/2+(2\sqrt{n}+1)/(2n)>1/2+1/\sqrt{n}$.

To see that this is close to tight, consider the case $h=(n-1)/2$, for which
an upper bound $F(h)\leq 1/2+O(1/\sqrt{n})$ holds by eq. 5.10.

Fidelity with a promise on $|x|$. If we assume that $h:=|x|\leq n/3$, from eq.
5.7 we have that in this regime

$\displaystyle F(h)$ $\displaystyle\geq
1-\frac{3}{2n}\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}(h-k)=1-\frac{3}{2n}\left(h-\frac{1}{\binom{n}{h}}\sum_{k=0}^{h}\left(\binom{n}{k}-\binom{n}{k-1}\right)k\right)$
(5.13)
$\displaystyle=1-\frac{3}{2n}\left(h-\frac{1}{\binom{n}{h}}\sum_{k=0}^{h}k\binom{n}{k}+\frac{1}{\binom{n}{h}}\sum_{k=0}^{h}k\binom{n}{k-1}\right)$
(5.14)
$\displaystyle=1-\frac{3}{2n}\left(h-h-\frac{1}{\binom{n}{h}}\sum_{k=0}^{h-1}k\binom{n}{k}+\frac{1}{\binom{n}{h}}\sum_{k=0}^{h-1}(k+1)\binom{n}{k}\right)$
(5.15) $\displaystyle=1-\frac{3}{2n\binom{n}{h}}\sum_{k=0}^{h-1}\binom{n}{k}.$
(5.16)

As $\binom{n}{k}\geq 2\binom{n}{k-1}$ for $k\leq n/3$ (via
$\binom{n}{k}/\binom{n}{k-1}=(n-k+1)/k$), we have
$\sum_{k=0}^{h-1}\binom{n}{k}\leq 2\binom{n}{h}$ and hence

$F(h)\geq 1-\frac{3}{2n}.$ (5.17)

Running the same argument through for an upper bound, and modifying
appropriately to start with

$F(h)\leq
1-\frac{1}{n}\sum_{k=0}^{h}\frac{\binom{n}{k}-\binom{n}{k-1}}{\binom{n}{h}}(h-k),$
(5.18)

we obtain

$F(h)\leq 1-\frac{1}{n\binom{n}{h}}\sum_{k=0}^{h-1}\binom{n}{k}\leq
1-\frac{\binom{n}{h-1}}{n\binom{n}{h}}=1-\frac{h}{n(n-h+1)}.$ (5.19)

Fixing $h=\lfloor n/3\rfloor$, we get

$F(h)\leq 1-\frac{\lfloor n/3\rfloor}{n(n-\lfloor n/3\rfloor+1)}\leq
1-\frac{n/3-1}{n(2n/3+2)}=1-\Theta(1/n),$ (5.20)

as desired. ∎

###  5.3 Unitary-equivariant Boolean functions

Can a Boolean function have a continuous symmetry? And not just any continuous
symmetry, but a unitary one? At first glance these questions may seem absurd
since Boolean functions are inherently discrete, however they are implicit in
the premise of our work. Indeed, the main question posed and answered in this
work can be rephrased as follows: to what extent can a Boolean function be
extended to a quantum channel with continuous symmetries? In this section we
briefly elaborate on this perspective, leaving a more in-depth exploration for
future work.

Theorem 4.6 shows that any symmetric and equivariant Boolean function can be
computed with optimal fidelity by the template algorithm
$\mathcal{A}_{\mathbf{t}}$, for some choice of the interpolation parameters
$\mathbf{t}$ that can be found using the linear program stated in Theorem 4.7.
While no function (except for the trivial $1$-argument identity function
$\mathrm{ID}$) can be computed with fidelity one, we can relax the requirement
that the linear map
$\Phi:\mathrm{L}(\mathbb{C}^{2^{n}})\to\mathrm{L}(\mathbb{C}^{2})$ for
computing it is a quantum channel. In particular, we can drop the requirement
that $\Phi$ is completely positive and ask whether now perfect fidelity can be
achieved. This is equivalent to relaxing the requirement that the Choi matrix
$J(\Phi)$ is positive semidefinite. In the context of Theorem 4.7, this
corresponds to dropping the constraints $0\leq t_{k}\leq 1$ in the linear
program (LP). To achieve perfect fidelity, we also replace the “$\geq c$”
constraint by “$=1$”, which reduces the problem to a linear system of
equations. Since this system involves a lower-triangular matrix, it has a
unique solution that can be found by eliminating one variable at a time. Based
on numerical investigations, we believe this solution to be

$t_{k}:=\frac{n-2k}{2(n-2k+1)}-\frac{k(n-2k)}{2(n-2k+1)^{2}}(-1)^{\bar{f}(k-1)}+\frac{(n-k+1)(n-2k+2)}{2(n-2k+1)^{2}}(-1)^{\bar{f}(k)},$
(5.21)

for all $k\in\\{0,\dotsc,\lfloor n/2\rfloor\\}$. We leave it for future work
to verify this.

This suggests that for any symmetric and equivariant Boolean function
$f:\\{0,1\\}^{n}\to\\{0,1\\}$ there is a unique linear map
$\Phi_{f}:\mathrm{L}(\mathbb{C}^{2^{n}})\to\mathrm{L}(\mathbb{C}^{2})$ that is
permutation-invariant and unitary-equivariant, and agrees with $f$ on all
inputs. This map $\Phi_{f}$ can be thought of as a _unitary-equivariant
extension_ of $f$ or, alternatively, as a “ _unitary-equivariant Boolean
function_ ” in its own right, i.e., a Boolean function with continuous unitary
symmetries.

######  Definition 5.1.

Let $f:\\{0,1\\}^{n}\to\\{0,1\\}$ be a symmetric and equivariant Boolean
function. Its _unitary-equivariant extension_ is the linear map
$\Phi_{f}:\mathrm{L}(\mathbb{C}^{2^{n}})\to\mathrm{L}(\mathbb{C}^{2})$ such
that

$\displaystyle\Phi_{f}(U^{\otimes n}|{x}\rangle\langle{x}|U^{\dagger\otimes
n})$ $\displaystyle=U|{f(x)}\rangle\langle{f(x)}|U^{\dagger},$
$\displaystyle\forall x$ $\displaystyle\in\\{0,1\\}^{n},\quad\forall
U\in\mathrm{U}(2),$ (5.22) $\displaystyle\Phi_{f}(P(\pi)\rho
P(\pi)^{\dagger})$ $\displaystyle=\Phi(\rho),$ $\displaystyle\forall\rho$
$\displaystyle\in\mathrm{L}(\mathbb{C}^{2^{n}}),\quad\forall\pi\in\mathrm{S}_{n}.$
(5.23)

It is not obvious at all that such map $\Phi_{f}$ exists and is unique. In
fact, it is not even obvious that a map satisfying eq. 5.22 is linear! To
illustrate that this definition is not vacuous, we provide a simple procedure
for computing the Choi matrix of the map $\Phi_{f}$.

First, let $\rho(\vec{r})$ denote the state in eq. 2.3. For any
$s\in\\{0,1\\}^{n}$, let

$\rho(s,\vec{r}):=\bigotimes_{i=1}^{n}\rho\big{\lparen}(-1)^{s_{i}}\vec{r}\big{\rparen}$
(5.24)

denote the $n$-qubit state corresponding to the input string $s$ in an
arbitrary basis. Then the unitary-equivariant extension of $f$ has Choi matrix
$J\in\mathrm{L}(\mathbb{C}^{2^{n+1}})$ such that

$\displaystyle\operatorname{Tr}_{2}\big{[}J\cdot(I_{2}\otimes\rho(s,\vec{r})^{\mathsf{T}})\big{]}$
$\displaystyle=\rho\big{\lparen}(-1)^{f(s)}\vec{r}\big{\rparen},$
$\displaystyle\forall s$
$\displaystyle\in\\{0,1\\}^{n},\quad\forall\vec{r}\in\mathbb{S}^{2},$ (5.25)
$\displaystyle\big{\lparen}I_{2}\otimes
P(\pi)\big{\rparen}J\big{\lparen}I_{2}\otimes P(\pi)\big{\rparen}^{\dagger}$
$\displaystyle=J,$ $\displaystyle\forall\pi$ $\displaystyle\in\mathrm{S}_{n},$
(5.26)

where $\mathbb{S}^{2}:=\\{(x,y,z)\in\mathbb{R}^{3}:x^{2}+y^{2}+z^{2}=1\\}$
denotes the Bloch sphere. Reducing eq. 5.25 modulo the polynomial
$x^{2}+y^{2}+z^{2}-1$ and then comparing the coefficients at all monomials
$x^{i}y^{j}z^{k}$ produces a system of linear equations in the matrix entries
of $J$. Combining this with the linear system (5.26) produces a system that
uniquely determines the Choi matrix $J$.121212This is an empirical
observation.

Note that for the sake of efficiency in eq. 5.25 it is enough to consider only
one $s\in\\{0,1\\}^{n}$ for each Hamming weight $|s|\in\\{0,\dotsc,\lfloor
n/2\rfloor\\}$. Similarly, it is enough to impose eq. 5.26 only for
transpositions $\pi=(i,i+1)$ with $i\in\\{1,\dotsc,n-1\\}$ since they generate
$\mathrm{S}_{n}$.

We provide explicit Choi matrices for all symmetric and unitary-equivariant
Boolean functions with $n=1$ and $n=3$ arguments in Section G.2.

## Acknowledgements

We acknowledge support from the QuantERA ERA-NET Cofund in Quantum
Technologies implemented within the European Union’s Horizon 2020 Programme
(QuantAlgo project), EPSRC grants EP/R043957/1 and EP/T001062/1, and EPSRC
Early Career Fellowship EP/L021005/1. This project has received funding from
the European Research Council (ERC) under the European Union’s Horizon 2020
research and innovation programme (grant agreement No. 817581). No new data
were created during this study. MO was supported by an NWO Vidi grant (Project
No. VI.Vidi.192.109).

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##  Appendix A Graphical notation

In this appendix, we introduce graphical notation and prove some basic
identities that will be useful in Appendix B. For a much more thorough
introduction to tensor networks and graphical notation, see [Bia19].

###  A.1 Vectorization

Throughout this section, let us fix some dimension $d\geq 1$.

######  Definition A.1 (Vectorization).

The _vectorization_ of a matrix

$A=\sum_{\begin{subarray}{c}i\in[{d_{\mathrm{out}}}]\\\
j\in[{d_{\mathrm{in}}}]\end{subarray}}A_{ij}|{i}\rangle\langle{j}|\in\mathrm{L}(\mathbb{C}^{d_{\mathrm{in}}},\mathbb{C}^{d_{\mathrm{out}}})$
(A.1)

is defined131313We define $\langle{A}|$ as $|{A}\rangle^{\mathsf{T}}$ instead
of $|{A}\rangle^{\dagger}$ since in graphical notation transposition is more
natural than conjugate transposition. We use this convention only in the
context of matrix vectorization, not for general “ket” vectors. as follows:

$\displaystyle|{A}\rangle$
$\displaystyle:=\sum_{\begin{subarray}{c}i\in[{d_{\mathrm{out}}}]\\\
j\in[{d_{\mathrm{in}}}]\end{subarray}}A_{ij}|{i}\rangle|{j}\rangle,$
$\displaystyle\langle{A}|:=|{A}\rangle^{\mathsf{T}}$
$\displaystyle=\sum_{\begin{subarray}{c}i\in[{d_{\mathrm{out}}}]\\\
j\in[{d_{\mathrm{in}}}]\end{subarray}}A_{ij}\langle{i}|\langle{j}|.$ (A.2)

######  Proposition A.2.

For any $A,B\in\mathrm{L}(\mathbb{C}^{d})$,

  1. 1.

$\langle{B}|{A}\rangle=\operatorname{Tr}\big{\lparen}AB^{\mathsf{T}}\big{\rparen}$,

  2. 2.

$\operatorname{SWAP}|{A}\rangle=|{A^{\mathsf{T}}}\rangle$,

  3. 3.

$(B\otimes I_{d})|{A}\rangle=|{BA}\rangle$,

  4. 4.

$\big{\lparen}I_{d}\otimes\langle{B}|\big{\rparen}\big{\lparen}|{A}\rangle\otimes
I_{d}\big{\rparen}=AB$,

where $\operatorname{SWAP}$ denotes the two-qudit swap operation:
$\operatorname{SWAP}|{i}\rangle|{j}\rangle=|{j}\rangle|{i}\rangle$, for all
$i,j\in[d]$.

###### Proof.

The first three identities are quite straightforward:

$\displaystyle\langle{B}|{A}\rangle$
$\displaystyle=\sum_{i,j\in[d]}B_{ij}A_{ij}=\sum_{i,j\in[d]}A_{ij}(B^{\mathsf{T}})_{ji}=\operatorname{Tr}\big{\lparen}AB^{\mathsf{T}}\big{\rparen},$
(A.3) $\displaystyle\operatorname{SWAP}|{A}\rangle$
$\displaystyle=\sum_{i,j\in[d]}A_{ij}\operatorname{SWAP}\big{\lparen}|{i}\rangle|{j}\rangle\big{\rparen}=\sum_{i,j\in[d]}A_{ij}|{j}\rangle|{i}\rangle=|{A^{\mathsf{T}}}\rangle,$
(A.4) $\displaystyle\big{\lparen}B\otimes I_{d}\big{\rparen}|{A}\rangle$
$\displaystyle=\sum_{i,j\in[d]}A_{ij}\left\lparen\sum_{k,l\in[d]}B_{kl}|{k}\rangle\langle{l}|{i}\rangle\right\rparen|{j}\rangle=\sum_{k,j\in[d]}\left\lparen\sum_{i\in[d]}B_{ki}A_{ij}\right\rparen|{k}\rangle|{j}\rangle=|{BA}\rangle.$
(A.5)

The third identity can be obtained as follows:

$\displaystyle\big{\lparen}I_{d}\otimes\langle{B}|\big{\rparen}\big{\lparen}|{A}\rangle\otimes
I_{d}\big{\rparen}$
$\displaystyle=\sum_{i,j\in[d]}B_{ij}\big{\lparen}I_{d}\otimes\langle{i}|\otimes\langle{j}|\big{\rparen}\sum_{k,l\in[d]}A_{kl}\big{\lparen}|{k}\rangle\otimes|{l}\rangle\otimes
I_{d}\big{\rparen}$ (A.6)
$\displaystyle=\sum_{i,j,k,l\in[d]}A_{kl}B_{ij}|{k}\rangle\langle{i}|{l}\rangle\langle{j}|$
(A.7)
$\displaystyle=\Bigg{\lparen}\sum_{k,l\in[d]}A_{kl}|{k}\rangle\langle{l}|\Bigg{\rparen}\Bigg{\lparen}\sum_{i,j\in[d]}B_{ij}|{i}\rangle\langle{j}|\Bigg{\rparen}$
(A.8) $\displaystyle=AB,$ (A.9)

which completes the proof. ∎

The intuition behind these identities is best captured by _graphical
notation_. Let us convert basic linear algebra operations involving matrices
$A,B\in\mathrm{L}(\mathbb{C}^{d})$ into the following diagrams:

$A\;=$$A$$A^{\mathsf{T}}\;=$$A$$|{A}\rangle\;=$$\langle{B}|\;=$$A$$B$$\operatorname{Tr}(A)\;=$$A$$A\otimes
B\;=$$A$$B$$A\cdot B\;=$$A$$B$$I\;=$$\operatorname{SWAP}\;=$

Then the identities in Proposition A.2 have the following straightforward
proofs:

1.2.3.4.$=$$A$$B$$A$$B^{\mathsf{T}}$$=$$A$$A^{\mathsf{T}}$$=$$B$$A$$BA$$=$$A$$B$$A$$B$

Item 4 of Proposition A.2 can be generalized by chaining together any number
of vectorized operators. The resulting chain has one of three possible types,
depending on where it starts and ends: it may start on one side but end on the
other (see the diagram for item 4), it may start and end on the same side (see
the diagram for $|{A}\rangle$), or it may be a closed loop (see the diagram
for $\operatorname{Tr}(A)$). These chains describe different types of objects:
a matrix, a vector, and a scalar, respectively. The following proposition
expresses this object as the matrix product of the operators involved.

######  Proposition A.3.

For any $k\geq 0$ and
$A_{1},\dotsc,A_{k+1},B_{1},\dotsc,B_{k+1}\in\mathrm{L}(\mathbb{C}^{d})$,

$\displaystyle\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k}}\rangle\otimes
I_{d}\big{\rparen}$ $\displaystyle=A_{1}B_{1}\dotsb A_{k}B_{k},$ (A.10)
$\displaystyle\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\otimes
I_{d}\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k+1}}\rangle\big{\rparen}$
$\displaystyle=|{A_{1}B_{1}\dotsb A_{k}B_{k}A_{k+1}}\rangle,$ (A.11)
$\displaystyle\langle{B_{k+1}}|\operatorname{SWAP}\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\otimes
I_{d}\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k+1}}\rangle\big{\rparen}$
$\displaystyle=\operatorname{Tr}\big{\lparen}A_{1}B_{1}\dotsb
A_{k+1}B_{k+1}\big{\rparen}.$ (A.12)

###### Proof.

The $k=0$ case of the first identity is trivial since $I_{d}I_{d}=I_{d}$.
Assuming the identity holds for $k-1$, we get

$\displaystyle\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k-1}}|\otimes\langle{B_{k}}|\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k-1}}\rangle\otimes|{A_{k}}\rangle\otimes
I_{d}\big{\rparen}$ (A.13)
$\displaystyle=\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k-1}}|\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k-1}}\rangle\otimes
I_{d}\big{\rparen}\big{\lparen}I_{d}\otimes\langle{B_{k}}|\big{\rparen}\big{\lparen}|{A_{k}}\rangle\otimes
I_{d}\big{\rparen}$ (A.14) $\displaystyle=A_{1}B_{1}\dotsb A_{k-1}B_{k-1}\cdot
A_{k}B_{k},$ (A.15)

where we used the inductive assumption and item 4 from Proposition A.2. To
prove the second identity, we combine the previous identity with item 3 from
Proposition A.2:

$\displaystyle\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\otimes
I_{d}\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k+1}}\rangle\big{\rparen}$
(A.16)
$\displaystyle=\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\otimes
I_{d}\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k}}\rangle\otimes
I_{d}\otimes I_{d}\big{\rparen}|{A_{k+1}}\rangle$ (A.17)
$\displaystyle=\big{\lparen}A_{1}B_{1}\dotsb A_{k}B_{k}\otimes
I_{d}\big{\rparen}|{A_{k+1}}\rangle$ (A.18) $\displaystyle=|{A_{1}B_{1}\dotsb
A_{k}B_{k}A_{k+1}}\rangle.$ (A.19)

For the third identity, note from eq. A.2 and item 2 of Proposition A.2 that

$\langle{B_{k+1}}|\operatorname{SWAP}=\big{\lparen}\operatorname{SWAP}^{\mathsf{T}}\langle{B_{k+1}}|^{\mathsf{T}}\big{\rparen}^{\mathsf{T}}=\big{\lparen}\operatorname{SWAP}|{B_{k+1}}\rangle\big{\rparen}^{\mathsf{T}}=\big{\lparen}|{B_{k+1}^{\mathsf{T}}}\rangle\big{\rparen}^{\mathsf{T}}=\langle{B_{k+1}^{\mathsf{T}}}|.$
(A.20)

Combining this with the previous identity,

$\displaystyle\langle{B_{k+1}}|\operatorname{SWAP}\big{\lparen}I_{d}\otimes\langle{B_{1}}|\otimes\dotsb\otimes\langle{B_{k}}|\otimes
I_{d}\big{\rparen}\big{\lparen}|{A_{1}}\rangle\otimes\dotsb\otimes|{A_{k+1}}\rangle\big{\rparen}$
(A.21) $\displaystyle=\langle{B_{k+1}^{\mathsf{T}}}|{A_{1}B_{1}\dotsb
A_{k}B_{k}A_{k+1}}\rangle$ (A.22)
$\displaystyle=\operatorname{Tr}\big{\lparen}A_{1}B_{1}\dotsb
A_{k+1}B_{k+1}\big{\rparen},$ (A.23)

where the last equality follows from item 1 of Proposition A.2. ∎

The identities in Proposition A.3 also have straightforward graphical proofs.
For example, the first identity amounts to

$A_{1}$$B_{1}$$\ldots$$\ldots$$\vdots$$A_{k}$$B_{k}$$=$$\ldots$$A_{1}$$B_{1}$$A_{k}$$B_{k}$

###  A.2 Singlet chains

In Appendix B, we will be particularly interested in the special case when all
operators involved describe singlets. For this it is useful to introduce the
matrix

$S:=|{0}\rangle\langle{1}|-|{1}\rangle\langle{0}|=\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}.$ (A.24)

The _singlet state_ $|{\Psi^{-}}\rangle$ is proportional to the vectorization
of $S$:

$|{\Psi^{-}}\rangle=\frac{1}{\sqrt{2}}\big{\lparen}|{0}\rangle|{1}\rangle-|{1}\rangle|{0}\rangle\big{\rparen}=\frac{1}{\sqrt{2}}|{S}\rangle=\frac{1}{\sqrt{2}}\;\leavevmode\hbox
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}\definecolor{pgffillcolor}{rgb}{0,0.8,0}{}\pgfsys@moveto{0.0pt}{48.3696pt}\pgfsys@moveto{-45.52458pt}{48.3696pt}\pgfsys@lineto{45.52458pt}{48.3696pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss} {}{{}}{}{{}{}}{{}}{}
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}\definecolor{pgffillcolor}{rgb}{0,0.8,0}{}\pgfsys@moveto{113.81104pt}{48.3696pt}\pgfsys@moveto{68.28645pt}{48.3696pt}\pgfsys@lineto{159.33562pt}{48.3696pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}
{{}}{}{{}}{}{{}{}}{{}}{} {{}{}}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{2.0pt}\pgfsys@invoke{
}\color[rgb]{0.8,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.8,0,0}\pgfsys@color@rgb@stroke{0.8}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0.8}{0}{0}\pgfsys@invoke{
}\definecolor{pgffillcolor}{rgb}{0.8,0,0}{}\pgfsys@moveto{113.81104pt}{48.3696pt}\pgfsys@moveto{83.46146pt}{48.3696pt}\pgfsys@lineto{144.16061pt}{48.3696pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.8pt}\pgfsys@invoke{ } \par {}{{}}{}{{}}{}{{}{}}{{}}{}
{{{}{}}{}}{} {{{}{}}{}}{}
{{{}{}}{}}{}{{}{}{{}}}{{}{}{{}}}{}{}{{}{}{{}}}{{}{}{{}}}{}{}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{-25.0pt}{-9.0pt}\pgfsys@lineto{16.0pt}{-9.0pt}\pgfsys@curveto{20.97063pt}{-9.0pt}{25.0pt}{-4.97063pt}{25.0pt}{0.0pt}\pgfsys@lineto{25.0pt}{0.0pt}\pgfsys@curveto{25.0pt}{4.97063pt}{20.97063pt}{9.0pt}{16.0pt}{9.0pt}\pgfsys@lineto{-25.0pt}{9.0pt}\pgfsys@stroke\pgfsys@invoke{
} {{}{}}{{}}{}{{}{}}{{}}{}
{{}{}}{{}}{}{}{}{}{{}}{}\pgfsys@beginscope\pgfsys@invoke{
}\definecolor[named]{pgffillcolor}{rgb}{1,1,1}\pgfsys@color@gray@fill{1}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{9.0pt}\pgfsys@moveto{-10.0pt}{-1.0pt}\pgfsys@moveto{-10.0pt}{-1.0pt}\pgfsys@lineto{-10.0pt}{19.0pt}\pgfsys@lineto{10.0pt}{19.0pt}\pgfsys@lineto{10.0pt}{-1.0pt}\pgfsys@closepath\pgfsys@moveto{10.0pt}{19.0pt}\pgfsys@fillstroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{{{}{}}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{
}{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} {
}{{{{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-3.35416pt}{5.77084pt}\pgfsys@invoke{
}\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{$S$}}
}}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \par
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ (A.25)

Note that $S$ and $|{\Psi^{-}}\rangle$ are anti-symmetric: $S^{\mathsf{T}}=-S$
and $\operatorname{SWAP}|{\Psi^{-}}\rangle=-|{\Psi^{-}}\rangle$ (the two
identities are related via item 2 of Proposition A.2). Since
$|{cA}\rangle=c|{A}\rangle$ for any $c\in\mathbb{C}$ and
$A\in\mathrm{L}(\mathbb{C}^{d})$,

$|{S^{\mathsf{T}}}\rangle=-|{S}\rangle.$ (A.26)

The matrix $S$ behaves like the complex number $i$ since

$\displaystyle S^{2}$ $\displaystyle=-I,$ $\displaystyle S^{3}$
$\displaystyle=-S,$ $\displaystyle S^{4}$ $\displaystyle=I,$ $\displaystyle
S^{\mathsf{T}}$ $\displaystyle=-S,$ (A.27)

where the transposition is analogous to complex conjugation. Consequently, for
any integer $k$,

$\displaystyle S^{2k}$ $\displaystyle=(-1)^{k}I,$ $\displaystyle S^{2k+1}$
$\displaystyle=(-1)^{k}S.$ (A.28)

The following corollary is a direct consequence of Propositions A.3 and A.28.

######  Corollary A.4.

For any $k\geq 0$,

$\displaystyle\big{\lparen}I\otimes\langle{S}|^{\otimes
k}\big{\rparen}\big{\lparen}|{S}\rangle^{\otimes k}\otimes I\big{\rparen}$
$\displaystyle=S^{2k}=(-1)^{k}I,$ (A.29)
$\displaystyle\big{\lparen}I\otimes\langle{S}|^{\otimes k}\otimes
I\big{\rparen}\big{\lparen}|{S}\rangle^{\otimes k+1}\big{\rparen}$
$\displaystyle=|{S^{2k+1}}\rangle=(-1)^{k}|{S}\rangle,$ (A.30)
$\displaystyle\langle{S}|\operatorname{SWAP}\big{\lparen}I\otimes\langle{S}|^{\otimes
k}\otimes I\big{\rparen}\big{\lparen}|{S}\rangle^{\otimes k+1}\big{\rparen}$
$\displaystyle=\operatorname{Tr}\big{\lparen}S^{2k+2}\big{\rparen}=2(-1)^{k+1}.$
(A.31)

Moreover, if $j$ instances of $|{S}\rangle$ are replaced by
$|{S^{\mathsf{T}}}\rangle=-|{S}\rangle$, the final value of each expression
picks up an extra factor of $(-1)^{j}$.

##  Appendix B Properties of the vectors $|{(\lambda,w,i)}\rangle$

In this appendix, we prove several results concerning the vectors
$|{(\lambda,w,i)}\rangle$ defined in Section 3.3. These vectors constitute a
particularly nice choice of _Schur basis_ for qubits, and define the Schur
transform $U_{\mathrm{Sch}}$ via eq. 3.19, which is used in the generic pre-
processing procedure described in Section 3.4. This appendix is concerned with
verifying that $|{(\lambda,w,i)}\rangle$ is indeed a Schur basis. The main
results proved here are as follows:

  * •

Lemma B.1: the permutation operators $P(\pi)$ are block-diagonal in the Schur
basis,

  * •

Lemma B.7: the vectors $|{(\lambda,w,i)}\rangle$ form an orthonormal basis,

  * •

Lemma B.8: the subspaces
$\mathcal{Q}_{\lambda,i}:=\operatorname{span}\big{\\{}|{(\lambda,w,i)}\rangle:w\in[m_{\lambda}]\big{\\}}$
are invariant under matrices of the form $M^{\otimes n}$, for any
$M\in\mathrm{L}(\mathbb{C}^{2})$,

  * •

Lemma B.9: the action of $M^{\otimes n}$ on $\mathcal{Q}_{\lambda,i}$ is given
by the corresponding $\lambda$-irrep $Q_{\lambda}(M)$ defined in eq. C.39;
this is the same as saying that we have the so-called Gelfand–Tsetlin basis on
the unitary register in eq. 3.6 (see Section C.1 for more details).

Along the way we also establish several technical results that we state as
propositions.

Recall the following notation from Sections 3.1 and 3.3. For any partition
$\lambda=(\lambda_{1},\lambda_{2})\mathbin{\vdash}n$ and Hamming weight
$w\in[m_{\lambda}]:=\\{0,\dotsc,\lambda_{1}-\lambda_{2}\\}$, we define an
$n$-qubit state

$|{(\lambda,w,0)}\rangle:=|{s_{\lambda_{1}-\lambda_{2}}(w)}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes\lambda_{2}},$
(B.1)

where $|{s_{\ell}(w)}\rangle$ is the symmetric state on $\ell$ qubits with
Hamming weight $w$,

$|{s_{\ell}(w)}\rangle:=\binom{\ell}{w}^{-1/2}\sum_{\begin{subarray}{c}x\in\\{0,1\\}^{\ell}\\\
\lvert x\rvert=w\end{subarray}}|{x}\rangle,$ (B.2)

and $|{\Psi^{-}}\rangle:=(|{01}\rangle-|{10}\rangle)/\sqrt{2}$ is the singlet
state. More generally, recall that the dimension $d_{\lambda}$ of the
permutation register is given in eq. 3.5. Then, for any $i\in[d_{\lambda}]$,
we define

$|{(\lambda,w,i)}\rangle:=\sum_{\pi\in\mathrm{S}_{n}}\alpha_{\pi}^{\lambda,i}P(\pi)|{(\lambda,w,0)}\rangle,$
(B.3)

for some coefficients $\alpha_{\pi}^{\lambda,i}\in\mathbb{R}$, where
$P(\pi)\in\mathrm{U}(2^{n})$ permutes $n$ qubits according to the permutation
$\pi\in\mathrm{S}_{n}$, see eq. 2.1.

###  B.1 Block-diagonalization of permutations

The goal of this section is to establish that the permutation operators
$P(\pi)$ are block-diagonal in the Schur basis $|{(\lambda,w,i)}\rangle$, with
blocks labeled by partitions $\lambda$ and Hamming weights $w$. More
precisely, we will establish eq. 3.6 which says that

$\hat{P}(\pi):=U_{\mathrm{Sch}}P(\pi)U_{\mathrm{Sch}}^{\dagger}=\bigoplus_{\lambda\mathbin{\vdash}n}I_{m_{\lambda}}\otimes
P_{\lambda}(\pi),$ (B.4)

where $U_{\mathrm{Sch}}$ denotes the Schur transform and
$P_{\lambda}(\pi)\in\mathrm{L}(\mathbb{C}^{d_{\lambda}})$ are some matrices.
Note that we will only determine the block structure of $\hat{P}(\pi)$ but not
the actual matrix entries of $P_{\lambda}(\pi)$.141414We cannot determine the
matrix entries unambiguously since eq. B.3 does not fully specify the states
$|{(\lambda,w,i)}\rangle$ with $i\neq 0$. Luckily, for the purpose of this
paper, the matrix entries of $P_{\lambda}(\pi)$ will not be necessary.

The following lemma establishes the block structure of $P(\pi)$ in the Schur
basis (it relies on Propositions B.4 and B.2 that are proved subsequently).

######  Lemma B.1.

Let $n\geq 1$. For any permutation $\pi\in\mathrm{S}_{n}$, partitions
$\lambda,\mu\mathbin{\vdash}n$, Hamming weights $w\in[m_{\lambda}]$ and
$v\in[m_{\mu}]$, and indices $i,j\in[d_{\lambda}]$,

$\langle{(\lambda,w,i)}|P(\pi)|{(\mu,v,j)}\rangle=\delta_{\lambda\mu}\delta_{wv}\langle{i}|P_{\lambda}(\pi)|{j}\rangle,$
(B.5)

for some operator $P_{\lambda}(\pi)\in\mathrm{L}(\mathbb{C}^{d_{\lambda}})$
that does not depend on the Hamming weights $w$ and $v$.

###### Proof.

Substituting $\langle{(\lambda,w,i)}|$ and $|{(\mu,v,j)}\rangle$ from eq. B.3
and expanding both sides,

$\displaystyle\langle{(\lambda,w,i)}|P(\pi)|{(\mu,v,j)}\rangle$
$\displaystyle=\langle{(\lambda,w,0)}|\left\lparen\sum_{\sigma\in\mathrm{S}_{n}}\alpha_{\sigma}^{\lambda,i}P(\sigma)^{\dagger}\right\rparen
P(\pi)\left\lparen\sum_{\tau\in\mathrm{S}_{n}}\alpha_{\tau}^{\mu,j}P(\tau)\right\rparen|{(\mu,v,0)}\rangle$
(B.6)
$\displaystyle=\sum_{\sigma,\tau\in\mathrm{S}_{n}}\alpha_{\sigma}^{\lambda,i}\alpha_{\tau}^{\mu,j}\langle{(\lambda,w,0)}|P(\sigma^{-1}\pi\tau)|{(\mu,v,0)}\rangle.$
(B.7)

  * •

If $\lambda\neq\mu$, all terms vanish thanks to Proposition B.4, giving us the
first delta function $\delta_{\lambda\mu}$.

  * •

If $\lambda=\mu$ but $w\neq v$, all terms vanish since
$\langle{(\lambda,w,0)}|$ and $P(\sigma^{-1}\pi\tau)|{(\lambda,v,0)}\rangle$
expanded in the standard basis consist of terms with different Hamming
weights, namely $w+\lambda_{2}$ and $v+\lambda_{2}$. This gives us the second
delta function $\delta_{wv}$.

  * •

Finally, if $\lambda=\mu$ and $w=v$, each term in eq. B.7 is of the form

$\alpha_{\sigma}^{\lambda,i}\alpha_{\tau}^{\lambda,j}\langle{(\lambda,w,0)}|P(\sigma^{-1}\pi\tau)|{(\lambda,w,0)}\rangle,$
(B.8)

which does not depend on $w$ according to Proposition B.2. Hence the sum in
eq. B.7 depends only on $\lambda,\pi,i,j$, so we can write it as
$\langle{i}|P_{\lambda}(\pi)|{j}\rangle$ for some
$P_{\lambda}(\pi)\in\mathrm{L}(\mathbb{C}^{d_{\lambda}})$.

Since this covers all cases, the proof is complete. ∎

The following proposition establishes that the value of
$\langle{(\lambda,w,0)}|P(\pi)|{(\lambda,w,0)}\rangle$ does not depend on $w$
and hence can be written as $\langle{0}|P_{\lambda}(\pi)|{0}\rangle$ for some
matrix $P_{\lambda}(\pi)$.

######  Proposition B.2.

Let $n\geq 1$ and fix any partition
$\lambda=(\lambda_{1},\lambda_{2})\mathbin{\vdash}n$ and a permutation
$\pi\in\mathrm{S}_{n}$. Then the value of

$\langle{(\lambda,w,0)}|P(\pi)|{(\lambda,w,0)}\rangle$ (B.9)

does not depend on $w\in[m_{\lambda}]$.

###### Proof.

Since $(U\otimes U)|{\Psi^{-}}\rangle=|{\Psi^{-}}\rangle$ for any
$U\in\mathrm{SU}(2)$, we see from eq. B.1 that

$U^{\otimes
n}|{(\lambda,w,0)}\rangle=U^{\otimes\ell}|{s_{\ell}(w)}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes
r}$ (B.10)

where $\ell:=\lambda_{1}-\lambda_{2}$ and $r:=\lambda_{2}$. By combining
Lemmas C.2 and D.2, we know that $U^{\otimes\ell}$ acts irreducibly in the
$(\ell+1)$-dimensional symmetric subspace
$\mathrm{Sym}(\ell):=\operatorname{span}\\{|{s_{\ell}(w)}\rangle:w\in\\{0,\dotsc,\ell\\}\\}$.
According to Burnside’s theorem (see [LR04] for a simple proof), the matrix
algebra corresponding to this action coincides with the set of all
$(\ell+1)\times(\ell+1)$ complex matrices. Hence any linear transformation on
$\mathrm{Sym}(\ell)$ can be expressed as a linear combination of
$U^{\otimes\ell}$, for some choice of $U\in\mathrm{SU}(2)$.

Consider a unitary $O$ that cyclically permutes the values
$w\in\\{0,\dotsc,\ell\\}$:

$O|{s_{\ell}(w)}\rangle:=O|{s_{\ell}(w\oplus 1)}\rangle,$ (B.11)

where addition is modulo $\ell+1$. Note that this specifies $O$ only on the
$(\ell+1)$-dimensional subspace $\mathrm{Sym}(\ell)$ of the ambient space
$(\mathbb{C}^{2})^{\otimes\ell}$. Thanks to Burnside’s theorem, we can write

$O=\sum_{i}\alpha_{i}U_{i}^{\otimes\ell},$ (B.12)

for some $\alpha_{i}\in\mathbb{C}$ and $U_{i}\in\mathrm{SU}(2)$, which extends
$O$ to the rest of $(\mathbb{C}^{2})^{\otimes\ell}$ while still agreeing with
eq. B.11. We can extend this further to the whole of
$(\mathbb{C}^{2})^{\otimes n}$ by letting

$\tilde{O}:=\sum_{i}\alpha_{i}U_{i}^{\otimes n}.$ (B.13)

Note that $\tilde{O}$ acts on the states $|{(\lambda,w,0)}\rangle$ as follows:

$\displaystyle\tilde{O}|{(\lambda,w,0)}\rangle$
$\displaystyle=\sum_{i}(\alpha_{i}U_{i}^{\otimes\ell}\otimes U_{i}^{\otimes
2r})\left\lparen|{s_{\ell}(w)}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes
r}\right\rparen$ (B.14)
$\displaystyle=\sum_{i}\alpha_{i}U_{i}^{\otimes\ell}|{s_{\ell}(w)}\rangle\otimes
U_{i}^{\otimes 2r}|{\Psi^{-}}\rangle^{\otimes r}$ (B.15)
$\displaystyle=O|{s_{\ell}(w)}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes r}$
(B.16) $\displaystyle=|{s_{\ell}(w\oplus
1)}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes r}$ (B.17)
$\displaystyle=|{(\lambda,w\oplus 1,0)}\rangle,$ (B.18)

for any $w\in\\{0,\dotsc,\ell\\}$. Furthermore, since
$O^{\dagger}|{s_{\ell}(w\oplus 1)}\rangle=|{s_{\ell}(w)}\rangle$, we can run
the same calculation backwards and conclude that
$\tilde{O}^{\dagger}|{(\lambda,w\oplus 1,0)}\rangle=|{(\lambda,w,0)}\rangle$
for all $w\in\\{0,\dotsc,\ell\\}$. Since $\tilde{O}^{\dagger}$ is a linear
combination of $(U_{i}^{\dagger})^{\otimes n}$, we see that
$\tilde{O}^{\dagger}P(\pi)=P(\pi)\tilde{O}^{\dagger}$ for all
$\pi\in\mathrm{S}_{n}$. Hence,

$\displaystyle\langle{(\lambda,w\oplus 1,0)}|P(\pi)|{(\lambda,w\oplus
1,0)}\rangle$
$\displaystyle=\langle{(\lambda,w,0)}|\tilde{O}^{\dagger}P(\pi)\tilde{O}|{(\lambda,w,0)}\rangle$
(B.19)
$\displaystyle=\langle{(\lambda,w,0)}|P(\pi)\tilde{O}^{\dagger}\tilde{O}|{(\lambda,w,0)}\rangle$
(B.20) $\displaystyle=\langle{(\lambda,w,0)}|P(\pi)|{(\lambda,w,0)}\rangle.$
(B.21)

Since the above holds for any $w$, the value of
$\langle{(\lambda,w,0)}|P(\pi)|{(\lambda,w,0)}\rangle$ is independent of $w$.
∎

The second ingredient that was used in Lemma B.1 to reveal the block structure
of $P(\pi)$ in the Schur basis was Proposition B.4, which shows that the
quantity $\langle{(\lambda,w,0)}|P(\pi)|{(\mu,v,0)}\rangle$ vanishes when
$\lambda\neq\mu$. To establish this, we first need to introduce some
additional terminology that will allow us to make use of the graphical
notation introduced in Appendix A.

Fix some $n\geq 1$ and consider integers $\ell_{1},\ell_{2},r_{1},r_{2}\geq 0$
such that $\ell_{1}+2r_{1}=\ell_{2}+2r_{2}=n$, and let
$x\in\\{0,1\\}^{\ell_{1}}$ and $y\in\\{0,1\\}^{\ell_{2}}$ be arbitrary bit
strings. Fix any permutation $\pi\in\mathrm{S}_{n}$ and consider the
expression

$\langle{x}|\otimes\langle{\Psi^{-}}|^{\otimes r_{1}}\cdot
P(\pi)\cdot|{y}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes r_{2}}.$ (B.22)

where $|{\Psi^{-}}\rangle$ denotes the singlet state. We associate to such
expression a _diagram_ (see examples in Fig. 6) that depicts the qubits of
both states by two columns of circles and the permutation $\pi$ by lines
connecting these circles. The bits of strings $x$ and $y$ are depicted by gray
circles labeled by $x_{i}$ and $y_{j}$, respectively, while the singlets are
depicted by arcs that connect the corresponding two qubits. Since the singlet
state $|{\Psi^{-}}\rangle=(|{01}\rangle-|{10}\rangle)/\sqrt{2}$ is not
symmetric under exchanging the two qubits, we include an arrow pointing from
its first qubit to its second qubit:

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{\pgfsys@beginscope\pgfsys@invoke{
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}\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{42.67912pt}{8.53586pt}\pgfsys@invoke{
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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} {{}{}}{{}}{}
{{}{}}{}{}{}{}{}{{}}\pgfsys@moveto{42.67912pt}{5.69041pt}\pgfsys@lineto{42.67912pt}{8.78136pt}\pgfsys@stroke\pgfsys@invoke{
}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@transformcm{0.0}{1.0}{-1.0}{0.0}{42.67912pt}{8.78136pt}\pgfsys@invoke{
}\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}}{{}}}} }\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope}{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}
{{{}}}
}{{}{}}{{}{}}{}{}{{}}{{}}{{}{}}{}{}{{}}{{}}{{}{}}{{}{}}{{}{}}{{}{}}{{{{}{}{{}}
}}{{}} {{}} } \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}
\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope \par
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}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\;.$ (B.23)

Flipping the arrow introduces an overall minus sign since $|{\Psi^{-}}\rangle$
is anti-symmetric.

Singlets in Fig. 6 are chained together by permutation $\pi$, forming paths
between the gray circles (ignore the arrow directions for now). These paths
effectively pair up the elements of the set
$\\{x_{1},\dotsc,x_{\ell_{1}},y_{1},\dotsc,y_{\ell_{2}}\\}$ representing the
bits of $x$ and $y$. Since each bit is paired up with exactly one other bit,
we call this a _matching_. For example, the four diagrams in Fig. 6 produce
the following matchings:

(a), (b) $\displaystyle\\{(x_{1},y_{3}),(x_{2},y_{1}),(x_{3},y_{2})\\},$ (c)
$\displaystyle\\{(x_{1},y_{2}),(x_{2},x_{3}),(y_{1},y_{3})\\},$ (d)
$\displaystyle\\{(x_{1},y_{3}),(x_{2},y_{1}),(x_{3},y_{2}),(x_{4},x_{5})\\}.$

(a)$x_{1}$$x_{2}$$x_{3}$$y_{1}$$y_{2}$$y_{3}$(b)$x_{1}$$x_{2}$$x_{3}$$y_{1}$$y_{2}$$y_{3}$(c)$x_{1}$$x_{2}$$x_{3}$$y_{1}$$y_{2}$$y_{3}$(d)$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}$$y_{1}$$y_{2}$$y_{3}$
Figure 6: Four different instances of eq. B.22 represented in graphical
notation. Different permutations $\pi$ result in different matchings among the
bits of $x$ and $y$ induced by singlet chains. Each singlet
$|{\Psi^{-}}\rangle=(|{01}\rangle-|{10}\rangle)/\sqrt{2}$ is denoted by an
arrow pointing from its first qubit to the second, see eq. B.23. The left two
matchings (a) and (b) are proper since the bits of $x$ are bijectively matched
with the bits of $y$, while the right two matchings (c) and (d) are improper,
since two bits of $x$ or $y$ are matched.

We call a matching _proper_ if each $x_{i}$ is matched to some $y_{j}$ (e.g.,
matchings (a) and (b)); otherwise we call it _improper_ (e.g., matchings (c)
and (d)). Note that an improper matching always pairs up two bits of $x$ or
two bits of $y$. In particular, if the strings $x$ and $y$ have different
lengths, such as in case (d), the matching will always be improper.

To evaluate eq. B.22, we will use the following restatement of Corollary A.4
in terms of the normalized singlet state
$|{\Psi^{-}}\rangle=|{S}\rangle/\sqrt{2}$.

######  Corollary B.3.

For any $k\geq 0$,

$\displaystyle\big{\lparen}I\otimes\langle{\Psi^{-}}|^{\otimes
k}\big{\rparen}\big{\lparen}|{\Psi^{-}}\rangle^{\otimes k}\otimes
I\big{\rparen}$ $\displaystyle=\frac{1}{2^{k}}(-1)^{k}I,$ (B.24)
$\displaystyle\big{\lparen}I\otimes\langle{\Psi^{-}}|^{\otimes k}\otimes
I\big{\rparen}\big{\lparen}|{\Psi^{-}}\rangle^{\otimes k+1}\big{\rparen}$
$\displaystyle=\frac{1}{2^{k}}(-1)^{k}|{\Psi^{-}}\rangle,$ (B.25)
$\displaystyle\langle{\Psi^{-}}|\operatorname{SWAP}\big{\lparen}I\otimes\langle{\Psi^{-}}|^{\otimes
k}\otimes I\big{\rparen}\big{\lparen}|{\Psi^{-}}\rangle^{\otimes
k+1}\big{\rparen}$ $\displaystyle=\frac{1}{2^{k}}(-1)^{k+1}.$ (B.26)

Moreover, if $j$ singlets are in the opposite direction, i.e.,
$|{\Psi^{-}}\rangle$ is replaced by
$\operatorname{SWAP}|{\Psi^{-}}\rangle=-|{\Psi^{-}}\rangle$, the final value
of each expression picks up an extra factor of $(-1)^{j}$.

The following result shows that certain matrix entries of $P(\pi)$ in the
Schur basis vanish whenever the matching resulting from the corresponding two
basis vectors is improper.

######  Proposition B.4.

Let $n\geq 1$ and fix any partitions
$\lambda=(\lambda_{1},\lambda_{2})\mathbin{\vdash}n$ and
$\mu=(\mu_{1},\mu_{2})\mathbin{\vdash}n$, and a permutation
$\pi\in\mathrm{S}_{n}$. If the diagram corresponding to

$\langle{(\lambda,w,0)}|P(\pi)|{(\mu,v,0)}\rangle$ (B.27)

yields an improper matching, the expression vanishes irrespective of the
Hamming weights $w\in[m_{\lambda}]$ and $v\in[m_{\mu}]$. In particular, it
vanishes whenever $\lambda\neq\mu$.

###### Proof.

If $\lambda\neq\mu$ then $\lambda_{1}-\lambda_{2}\neq\mu_{1}-\mu_{2}$, so
there will be more bits on one side of the diagram than the other (e.g., case
(d) in Fig. 6), so a proper matching between them is impossible.

To establish the main claim, we first write $\lambda=(r_{1}+\ell_{1},r_{1})$
and $\mu=(r_{2}+\ell_{2},r_{2})$ for some integers
$\ell_{1},\ell_{2},r_{1},r_{2}\geq 0$ such that
$\ell_{1}+2r_{1}=\ell_{2}+2r_{2}=n$ (see Fig. 4). Next, recall from eqs. B.1
and B.2 that

$\langle{(\lambda,w,0)}|=\binom{\ell_{1}}{w}^{-1/2}\sum_{\begin{subarray}{c}x\in\\{0,1\\}^{\ell_{1}}\\\
\lvert x\rvert=w\end{subarray}}\langle{x}|\otimes\langle{\Psi^{-}}|^{\otimes
r_{1}},$ (B.28)

and similarly for $|{(\mu,v,0)}\rangle$. Hence, we can expand (B.27) as
follows:

$\binom{\ell_{1}}{w}^{-1/2}\binom{\ell_{2}}{v}^{-1/2}\sum_{\begin{subarray}{c}x\in\\{0,1\\}^{\ell_{1}}\\\
\lvert
x\rvert=w\end{subarray}}\sum_{\begin{subarray}{c}y\in\\{0,1\\}^{\ell_{2}}\\\
\lvert y\rvert=v\end{subarray}}\langle{x}|\otimes\langle{\Psi^{-}}|^{\otimes
r_{1}}\cdot P(\pi)\cdot|{y}\rangle\otimes|{\Psi^{-}}\rangle^{\otimes r_{2}}.$
(B.29)

Each term in this expansion can be represented by one of the diagrams
discussed before. Since the wiring of the diagram depends only on $\pi$ and
the number of singlets $r_{1},r_{2}$, it is in fact the same diagram for all
terms, except with different values of the bit strings $x$ and $y$ substituted
in the gray circles (see examples in Fig. 6). Note that this expression can be
factorized over all paths and cycles in the diagram. Hence, to show that it
vanishes, it suffices to find a single path with a vanishing contribution.

By assumption, the diagram corresponding to eq. B.29 yields an improper
matching, meaning that two bits on the same side, say $x_{1}$ and $x_{2}$, are
matched with each other. Hence, the number of singlets (or arrows) on the path
between them is odd. According to eq. B.25 in Corollary B.3, such path is
equivalent to a single singlet. More specifically, eq. B.25 allows us to
replace a path of $2k+1$ singlets by a contracted path consisting only of
$(-1)^{k+j}|{\Psi^{-}}\rangle/2^{k}$, where $j$ is the number of singlets on
the original path that point in the direction opposite to the singlet on the
contracted path.151515This can also be seen by iteratively contracting two
singlets at a time using the identities displayed in Fig. 7. For example,

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