# Quantum Polarization of Qudit Channels

Ashutosh Goswami1 Mehdi Mhalla2 Valentin Savin3
1 Univ. Grenoble Alpes, Grenoble INP, LIG, F-38000 Grenoble, France
2 Univ. Grenoble Alpes, CNRS, Grenoble INP, LIG, F-38000 Grenoble, France
3 Univ. Grenoble Alpes, CEA, LETI, F-38054 Grenoble, France

###### Abstract

We provide a generalization of quantum polar codes to quantum channels with
qudit-input, achieving the symmetric coherent information of the channel. Our
scheme relies on a channel combining and splitting construction, where a two-
qudit unitary randomly chosen from a unitary 2-design is used to combine two
instances of a qudit-input channel. The inputs to the synthesized bad channels
are frozen by sharing EPR pairs between the sender and the receiver, so our
scheme is entanglement assisted. Using the fact that the generalized two-qudit
Clifford group forms a unitary 2-design, we conclude that the channel
combining operation can be chosen from this set. Moreover, we show that
polarization also happens for a much smaller subset of two-qudit Cliffords,
which is not a unitary 2-design. Finally, we show how to decode the proposed
quantum polar codes on Pauli qudit channels.

##  1 Introduction

In classical information theory, polar codes are the first explicit
construction provably achieving the symmetric capacity of any discrete
memoryless channel [1]. The construction is based on the recursive application
of a channel combining and splitting procedure. It first combines two
instances of the transmission channel, using a controlled-NOT gate as channel
combiner, and then splits the combined channel into two virtual channels,
referred to as good and bad channels. Applied recursively $n$ times, the above
procedure yields $N=2^{n}$ virtual channels. These virtual channels exhibit a
polarization property, in the sense that they tend to become either completely
noisy or noiseless, as $N$ goes to infinity. Polar coding consists of
efficient encoding and decoding algorithms that take effective advantage of
the channel polarization property.

Polar codes have been generalized to classical-quantum channels with binary
and non-binary classical input in [2, 3]. For the transmission of quantum
information over quantum channels with qubit-input, two approaches have been
considered in the literature. The first approach is based on CSS-like
constructions, which essentially exploit polarization in either amplitude or
phase basis [4, 5, 6]. The second approach relies on a purely quantum
polarization construction [7, 8], where the synthesized virtual channels tend
to become either completely noisy or noiseless as quantum channels, not merely
in one basis. This approach uses a randomized channel combining, employing a
random two-qubit Clifford unitary as channel combiner.

In this work, we extend the work in [7] to the case of quantum channels with
qudit-input. To the best of our knowledge, this is the first generalization of
polar codes to qudit-input channels. First, we show that purely quantum
polarization (in the sense of [7]) happens for any qudit-input quantum
channel, using as channel combiner a random two-qudit unitary, chosen from a
unitary 2-design. Further, we provide a simple proof of the fact that the
generalized two-qudit Clifford group forms a unitary 2-design, therefore the
channel combining operation can be randomly chosen from this set. Moreover,
when the qudit dimension $d$ is a prime, we show that polarization happens for
a subset of two-qudit Clifford unitaries containing only $d^{4}+d^{2}-2$
elements, which is not a unitary 2-design. Hence, unitary 2-designs are not
necessary for the quantum polarization of qudit-input channels.

To exploit the above polarization property, the inputs to the synthesized
noisy channels are frozen by presharing EPR pairs between the sender and the
receiver. Hence, our polar coding scheme is entanglement assisted. Finally, we
consider the case of Pauli qudit channels. Similarly to [7], we associate a
classical counterpart channel to a Pauli qudit channel. Then, we show that a
quantum polar code on a Pauli qudit channel yields a classical polar code on
the classical counterpart channel. Hence, we show that Pauli errors can be
identified by decoding the polar code on the classical counterpart channel,
using classical polar decoding.

The paper is organized as follows. Section 2 provides the basic definitions
needed for quantum polarization. Section 3 contains our main polarization
results for qudit-input quantum channels. Section 4 discusses the decoding of
our quantum polar codes on Pauli qudit channels.

##  2 Preliminaries

We consider $d$-dimensional quantum systems, referred to as qudits, where
$d\geq 2$ is fixed throughout the paper. We denote by $\rho_{A}$ a quantum
state (i.e., density matrix) of a quantum system $A$. When no confusion is
possible, we shall discard the quantum system from the notation. For a
bipartite quantum state $\rho_{AB}$, we shall denote by
$\rho_{B}:=\operatorname{Tr}_{A}(\rho_{AB})$ the quantum state of the system
$B$, obtained by tracing out the system $A$. The identity matrix is denoted by
either $\mathbbm{1}$ or $I$, with the former notation used for quantum states,
and the latter for quantum operators. Throughout the paper, logarithm is taken
in base $d$.

######  Definition 1 (von Neumann entropy).

(a) The von Neumann entropy of a quantum state $\rho$ is defined as

$H(\rho):=-\operatorname{Tr}\left(\rho\log\rho\right).$

(b) The conditional von Neumann entropy of a bipartite quantum state
$\rho_{AB}$ is defined as

$H(A|B)_{\rho_{AB}}=H(\rho_{AB})-H(\rho_{B}).$

######  Definition 2 (Conditional sandwiched Rényi entropy of order 2).

Let $\rho_{AB}$ be a quantum state. Then,

$\tilde{H}^{\downarrow}_{2}(A|B)_{\rho}:=-\log\operatorname{Tr}\left[\rho_{B}^{-\frac{1}{2}}\rho_{AB}\rho_{B}^{-\frac{1}{2}}\rho_{AB}\right].$

######  Definition 3 (Petz-Rényi entropy of order $\frac{1}{2}$).

Let $\rho_{AB}$ be a quantum state. Then,

$H^{\uparrow}_{\frac{1}{2}}(A|B)_{\rho}:=2\log\sup_{\sigma_{B}}\operatorname{Tr}\left[\rho_{AB}^{\frac{1}{2}}\sigma^{\frac{1}{2}}_{B}\right],$

where the supremum is taken over all quantum states $\sigma_{B}$.

We consider quantum channels $\mathcal{W}_{A^{\prime}\rightarrow B}$, with
qudit input system $A^{\prime}$, and output system $B$ of arbitrary dimension.
When no confusion is possible, we shall discard the channel input and output
systems from the notation. An EPR pair on two-qudit systems $A$ and
$A^{\prime}$ is the quantum state
$\Phi_{AA^{\prime}}:=\mathchoice{{\left\lvert\Phi_{AA^{\prime}}\right\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}\mathchoice{{\left\langle\Phi_{AA^{\prime}}\right\rvert}}{{\langle\Phi_{AA^{\prime}}\rvert}}{{\langle\Phi_{AA^{\prime}}\rvert}}{{\langle\Phi_{AA^{\prime}}\rvert}}$,
with
$\mathchoice{{\left\lvert\Phi_{AA^{\prime}}\right\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}:=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}\mathchoice{{\left\lvert
i\right\rangle}}{{\lvert i\rangle}}{{\lvert i\rangle}}{{\lvert
i\rangle}}_{A}\mathchoice{{\left\lvert i\right\rangle}}{{\lvert
i\rangle}}{{\lvert i\rangle}}{{\lvert i\rangle}}_{A^{\prime}}$. Given a
quantum channel $\mathcal{W}_{A^{\prime}\rightarrow B}$, we denote by
$\mathcal{W}(\Phi_{AA^{\prime}}):=(I_{A}\varotimes\mathcal{W})(\Phi_{AA^{\prime}})$
the quantum state on the $AB$ system obtained by applying $\mathcal{W}$ on the
$A^{\prime}$-half of the EPR pair $\Phi_{AA^{\prime}}$.

######  Definition 4 (Symmetric coherent information).

Let $\mathcal{W}_{A^{\prime}\rightarrow B}$ be a channel with qudit input
$A^{\prime}$ and output system $B$ of arbitrary dimension. The symmetric
coherent information of $\mathcal{W}$ is defined as the coherent information
of the channel for a uniformly distributed input, that is

$I(\mathcal{W}):=-H(A|B)_{\mathcal{W}(\Phi_{AA^{\prime}})}\in[-1,1].$

We further introduce the following parameter of a quantum channel, which can
be seen as the quantum counterpart of the classical Bhattacharyya parameter
[7], and which we refer to as the “Rényi-Bhattacharyya” parameter.

######  Definition 5 (Rényi-Bhattacharyya parameter).

Let $\mathcal{W}_{A^{\prime}\rightarrow B}$ be a channel with qudit input
$A^{\prime}$ and output system $B$ of arbitrary dimension. Then,

$R(\mathcal{W}):=d^{H^{\uparrow}_{\frac{1}{2}}(A|B)_{\mathcal{W}(\Phi_{AA^{\prime}})}}=d^{-\tilde{H}^{\downarrow}_{2}(A|E)_{\mathcal{W}^{c}(\Phi_{AA^{\prime}})}}\in\left[\tfrac{1}{d},d\right],$

where $\mathcal{W}^{c}$ denotes the complementary channel associated with
$\mathcal{W}$ [9], and the equality
$H^{\uparrow}_{\frac{1}{2}}(A|B)_{\mathcal{W}(\Phi_{AA^{\prime}})}=-\tilde{H}^{\downarrow}_{2}(A|E)_{\mathcal{W}^{c}(\Phi_{AA^{\prime}})}$
follows from [10, Theorem 2].

We will also need the definitions of the generalized (qudit) Pauli and
Clifford groups [11, 12], and unitary $2$-designs [13].

######  Definition 6 (Generalized Pauli Group).

(a) The Pauli operators $X$ and $Z$ for a qudit quantum system are defined as
$X=\sum_{j=0}^{d-1}\mathchoice{{\left\lvert j\right\rangle}}{{\lvert
j\rangle}}{{\lvert j\rangle}}{{\lvert j\rangle}}\mathchoice{{\left\langle
j\oplus 1\right\rvert}}{{\langle j\oplus 1\rvert}}{{\langle j\oplus
1\rvert}}{{\langle j\oplus 1\rvert}}$, and
$Z=\sum_{j=0}^{d-1}\omega^{j}\mathchoice{{\left\lvert j\right\rangle}}{{\lvert
j\rangle}}{{\lvert j\rangle}}{{\lvert j\rangle}}\mathchoice{{\left\langle
j\right\rvert}}{{\langle j\rvert}}{{\langle j\rvert}}{{\langle j\rvert}}$,
where $\oplus$ denotes the sum modulo $d$, and
$\omega=e^{\frac{2\pi\imath}{d}}$.

(b) The generalized Pauli group on one qudit is defined as
$\mathcal{P}_{d}^{1}:=\\{\omega^{\lambda}P_{r,s}\mid\lambda,r,s=0,\dots,d-1\\}$,
where $P_{r,s}:=X^{r}Z^{s}$.

(c) The generalized Pauli group on $n$ qudits is defined as
$\mathcal{P}_{d}^{n}:=\mathcal{P}_{d}^{1}\varotimes\mathcal{P}_{d}^{1}\varotimes\cdots\varotimes\mathcal{P}_{d}^{1}$.

It is easily seen that $X^{d}=Z^{d}=I$ and $XZ=\omega ZX$, hence
$\mathcal{P}_{d}^{1}$ is indeed a group. Applying the commutation relation
$XZ=\omega ZX$ appropriately many times, we have that

$P_{r,s}P_{t,u}=\omega^{ru-st}P_{t,u}P_{r,s}.$ (1)

######  Definition 7 (Generalized Clifford Group).

The Clifford group $\mathcal{C}_{d}^{n}$ is the unitary group on $n$ qudits
that takes $\mathcal{P}_{d}^{n}$ to $\mathcal{P}_{d}^{n}$ by conjugation.

Let $\mathcal{U}(d^{n})$ be the set of unitary operators on $n$ qudits, and
$\mathcal{W}_{n}$ be a quantum channel with $n$-qudit input. The twirling of
$\mathcal{W}_{n}$ with respect to $\mathcal{U}(d^{n})$ is defined as the
quantum channel that maps a $n$-qudit quantum state $\rho$ to $\int
U^{\dagger}\mathcal{W}_{n}(U\rho U^{\dagger})Ud\eta$, where
$U\in\mathcal{U}(d^{n})$ is randomly chosen according to the Haar measure
$\eta$. The twirling of $\mathcal{W}_{n}$ with respect to a finite subset
$\mathcal{U}\subset\mathcal{U}(d^{n})$ is defined as the quantum channel
acting as
$\rho\mapsto\frac{1}{|\mathcal{U}|}\sum_{U\in\mathcal{U}}U^{\dagger}\mathcal{W}_{n}(U\rho
U^{\dagger})U$.

######  Definition 8 (Unitary 2-Design).

A finite subset $\mathcal{U}\subset\mathcal{U}(d^{n})$ is said to form a
unitary 2-design if it satisfies the following, for all $n$-qudit input
quantum channels $\mathcal{W}_{n}$, and all $n$-qudit quantum states $\rho$:

$\frac{1}{|\mathcal{U}|}\sum_{U\in\mathcal{U}}U^{\dagger}\mathcal{W}_{n}(U\rho
U^{\dagger})U=\int U^{\dagger}\mathcal{W}_{n}(U\rho U^{\dagger})Ud\eta.$ (2)

##  3 Quantum Polarization of Qudit Channels

###  3.1 Main polarization results

Throughout this section ${\cal W}_{A^{\prime}\rightarrow B}$ denotes a quantum
channel with qudit input, and arbitrary dimension output. Our quantum
polarization scheme is based on the channel combining and splitting operations
depicted in the following figure.

$\mathcal{W}$$\mathcal{W}$$C$$A^{\prime}_{1}$$A^{\prime}_{2}$$B_{1}$$B_{2}$
(a) Combined channel

$\mathcal{W}$$\mathcal{W}$$C$$A^{\prime}_{1}$$\frac{\mathbbm{1}_{A^{\prime}_{2}}}{d}$$B_{1}$$B_{2}$
(b) Bad channel $\mathcal{W}_{C}^{(0)}$

$\mathcal{W}$$\mathcal{W}$$C$$A^{\prime}_{2}$$B_{1}$$B_{2}$$\Phi_{A_{1}A^{\prime}_{1}}$$A_{1}$
(c) Good channel $\mathcal{W}_{C}^{(1)}$

Figure 1: Channel combining and splitting. (a) combined channel: a two-qudit
unitary $C$ is applied on the two inputs. (b) bad channel: we input a totally
mixed state into the second input. (c) good channel: we input half of an EPR
pair into the first input, and the other half becomes the output $A_{1}$.

First, two instances of ${\cal W}$ are combined, by entangling their inputs
through a two-qudit unitary $C$. The combined channel is then split into one
bad and one good channel. The bad channel $\mathcal{W}_{C}^{(0)}$ is a channel
from $A^{\prime}_{1}$ to $B_{1}B_{2}$ that acts as
$\mathcal{W}_{C}^{(0)}(\rho)\break=\mathcal{W}^{\varotimes
2}\left(C(\rho\varotimes\frac{\mathbbm{1}_{A^{\prime}_{2}}}{d})C^{\dagger}\right)$,
where $\frac{\mathbbm{1}_{A^{\prime}_{2}}}{d}$ is the completely mixed state.
The good channel $\mathcal{W}_{C}^{(1)}$ is a channel from $A^{\prime}_{2}$ to
$A_{1}B_{1}B_{2}$ that acts as
$\mathcal{W}_{C}^{(1)}(\rho)=\mathcal{W}^{\varotimes
2}\left(C(\Phi_{A_{1}A^{\prime}_{1}}\varotimes\rho)C^{\dagger}\right)$, where
$\Phi_{A_{1}A^{\prime}_{1}}$ is an EPR pair.

The polarization construction is obtained by recursively applying the above
channel combining and splitting operations, while choosing $C$ randomly from
some finite set of unitaries, denoted by ${\cal U}\subset{\cal U}(d^{2})$. To
accommodate the random choice of $C\in{\cal U}$, a classical description of
$C$ is included as part of the output of the bad and good channels. Hence, for
$i=0,1$, we define:

${\cal W}^{(i)}(\rho)=\frac{1}{|{\cal U}|}\sum_{C\in{\cal
U}}\mathchoice{{\left\lvert C\right\rangle}}{{\lvert C\rangle}}{{\lvert
C\rangle}}{{\lvert C\rangle}}\mathchoice{{\left\langle
C\right\rvert}}{{\langle C\rvert}}{{\langle C\rvert}}{{\langle
C\rvert}}\varotimes{\cal W}_{C}^{(i)}(\rho),$ (3)

where $\\{\mathchoice{{\left\lvert C\right\rangle}}{{\lvert C\rangle}}{{\lvert
C\rangle}}{{\lvert C\rangle}}\\}_{C\in{\cal U}}$ is an orthogonal basis of
some auxiliary system. Applying twice the transformation ${\cal
W}\mapsto\left({\cal W}^{(0)},{\cal W}^{(1)}\right)$, we get channels ${\cal
W}^{(i_{1}i_{2})}:=\left({\cal W}^{(i_{1})}\right)\,\\!^{(i_{2})}$, where
$(i_{1}i_{2})\in\\{00,01,10,11\\}$. In general, after $n$ levels or recursion,
we obtain $2^{n}$ channels:

${\cal W}^{(i_{1}\dots i_{n})}:=\left({\cal W}^{(i_{1}\dots
i_{n-1})}\right)\,\\!^{(i_{n})},\ \forall(i_{1}\dots i_{n})\in\\{0,1\\}^{n}.$
(4)

The quantum polarization theorem below states that the symmetric coherent
information of the synthesized channels ${\cal W}^{(i_{1}\dots i_{n})}$
polarizes, meaning that it goes to either $-1$ or $+1$ as $n$ goes to infinity
(except possibly for a vanishing fraction of channels), provided that ${\cal
U}$ is a unitary $2$-design. The second theorem states that polarization also
happens when ${\cal U}$ is taken to be the generalized Clifford group on two
qudits, ${\cal C}_{d}^{2}$, or some specific subset of it.

######  Theorem 9.

Let $\mathcal{U}$ be a unitary 2-design. For any qudit-input quantum channel
${\cal W}$, let $\break\left\\{{\cal W}^{(i_{1}\dots i_{n})}:(i_{1}\dots
i_{n})\in\\{0,1\\}^{n}\right\\}$ be the set of channels defined in (4), with
channel combining unitary $C$ randomly chosen from ${\cal U}$. Then, for any
$\delta>0$,

$\displaystyle\lim_{n\rightarrow\infty}\frac{\\#\\{(i_{1}\dots
i_{n})\in\\{0,1\\}^{n}:I\left({\cal W}^{(i_{1}\dots
i_{n})}\right)\in(-1+\delta,1-\delta)\\}}{2^{n}}=0$

and furthermore,

$\displaystyle\lim_{n\rightarrow\infty}\frac{\\#\left\\{(i_{1},\dots,i_{n})\in\\{0,1\\}^{n}:I(\mathcal{W}^{(i_{1},\dots,i_{n})})\geqslant
1-\delta\right\\}}{2^{n}}=\frac{I(\mathcal{W})+1}{2}$

######  Theorem 10.

(a) The generalized Clifford group on two qudits, ${\cal C}_{d}^{2}$, is a
unitary $2$-design. Thus, polarization happens when the channel combining
unitary $C$ is randomly chosen from ${\cal C}_{d}^{2}$.

(b) If $d$ is prime, there exists a subset ${\cal U}\subset{\cal C}_{d}^{2}$,
of size $|{\cal U}|=d^{4}+d^{2}-2$, which is not a unitary $2$-design, and
such that polarization happens when the channel combining unitary $C$ is
randomly chosen from ${\cal U}$.

We note that part (a) of Theorem 10 may be inferred from Lemmas 1, 2 and 3 in
[14]. We will give an alternative and more elementary proof in Section 3.3, by
generalizing the proof from [13] to the qudit case.

###  3.2 Proof of Theorem 9 (quantum polarization)

To prove the polarization theorem, we essentially need three ingredients, as
follows.

  1. 1.

For any two-qudit unitary $C$, the total symmetric coherent information is
preserved under channel combining and splitting, that is, $I({\cal
W}_{C}^{(0)})+I({\cal W}_{C}^{(1)})=2I({\cal W})$. We omit the proof of this,
as the proof given in [8, Lemma 10] for qubit-input channels remains valid in
the qudit case, with minor adjustments.

  2. 2.

The symmetric coherent information $I({\cal W})$ approaches $\\{-1,+1\\}$
values if and only if the Rényi-Bhattacharyyia parameter $R({\cal W})$
approaches $\\{d,1/d\\}$ values. This follows from Lemma 11, below.

  3. 3.

Taking the good channel yields a guaranteed improvement of the average Rényi-
Bhattacharyya parameter, in the sense of Lemma 12, below.

The proof of Theorem 9 then follows by using [8, Lemma 7], similar to the
proof of quantum polarization for qubit-input channels in [8].

######  Lemma 11.

Let $\mathcal{W}_{A^{\prime}\rightarrow B}$ be a channel with qudit input.
Then,

  1. 1.

$R(\mathcal{W})\leqslant\frac{1}{d}+\delta\Rightarrow I(\mathcal{W})\geqslant
1-\log(1+d\delta)$.

  2. 2.

$R(\mathcal{W})\geqslant d-\delta\Rightarrow
I(\mathcal{W})\leqslant-1+2\sqrt{\frac{\delta}{d}}+\frac{\sqrt{d}+\sqrt{\delta}}{\sqrt{d}}h\left(\frac{\sqrt{\delta}}{\sqrt{d}+\sqrt{\delta}}\right)$,
where $h(\cdot)$ denotes the binary entropy function.

###### Proof.

We prove first 1). For $\rho_{AB}=\mathcal{W}(\Phi_{AA^{\prime}})$, we have
that

$\frac{1}{d}+\delta\geqslant
R(\mathcal{W})=d^{H^{\uparrow}_{\frac{1}{2}}(A|B)_{\rho}}\geqslant
d^{H(A|B)_{\rho}}=d^{-I(\mathcal{W})},$

where we have used $H^{\uparrow}_{\frac{1}{2}}(A|B)_{\rho}\geqslant
H(A|B)_{\rho}$ for the second inequality, which follows from the monotonically
decreasing property of the conditional Petz-Rényi entropy with respect to its
order [15, Theorem 7]. Hence, $I(\mathcal{W})\geqslant 1-\log(1+d\delta)$.

We now turn to point 2). We have that

$\displaystyle d-\delta$ $\displaystyle\leqslant R(\mathcal{W})\leqslant
R(\mathcal{W})$
$\displaystyle=\max_{\sigma_{B}}\operatorname{Tr}\left[\rho^{\frac{1}{2}}_{AB}\sigma^{\frac{1}{2}}_{B}\right]^{2}$
$\displaystyle=d\max_{\sigma_{B}}\operatorname{Tr}\left[\sqrt{\rho_{AB}}\sqrt{\frac{\mathbbm{1}_{A}}{d}\varotimes\sigma_{B}}\right]^{2}$
$\displaystyle\leqslant
d\max_{\sigma_{B}}\left\|\sqrt{\rho_{AB}}\sqrt{\frac{\mathbbm{1}_{A}}{d}\varotimes\sigma_{B}}\right\|_{1}^{2}$
(5)
$\displaystyle=d\max_{\sigma_{B}}F\left(\rho_{AB},\frac{\mathbbm{1}_{A}}{d}\varotimes\sigma_{B}\right)^{2}$
(6)

Using the Fuchs-van de Graaf inequalities [16], we get that there exists a
$\sigma_{B}$ such that
$\break\frac{1}{2}\left\|\rho_{AB}-\frac{\mathbbm{1}_{A}}{d}\varotimes\sigma_{B}\right\|_{1}\leqslant\sqrt{\frac{\delta}{d}}$.
We are now in a position to use the Alicki-Fannes-Winter [17, Lemma 2]
inequality, which states that

$\displaystyle\left|H(A|B)_{\rho}-1\right|\leqslant
2\sqrt{\frac{\delta}{d}}+\frac{\sqrt{d}+\sqrt{\delta}}{\sqrt{d}}h\left(\frac{\sqrt{\delta}}{{\sqrt{d}+\sqrt{\delta}}}\right).$

This concludes the proof of the lemma. ∎

######  Lemma 12.

Let $\mathcal{W}_{A^{\prime}\rightarrow B}$ be a channel with qudit input.
Then,

$\mathbb{E}_{C}R\left(\mathcal{W}_{C}^{(1)}\right)=\frac{d}{d^{2}+1}\left(1+R(\mathcal{W})^{2}\right)\leq
R(\mathcal{W}),$

where $\mathbb{E}_{C}$ denotes the expectation operator, $C$ is the channel
combining unitary, chosen uniformly at random from a unitary $2$-design ${\cal
U}$. Moreover, equality happens if and only if $R(\mathcal{W})\in\\{1/d,d\\}$.

###### Proof.

Let $\mathcal{W}^{c}_{A^{\prime}\rightarrow E}$ and
$(\mathcal{W}_{C}^{(1)})_{{{A^{\prime}_{2}\to E_{1}E_{2}}}}^{c}$ be the
complementary channel associated with $\mathcal{W}_{A^{\prime}\rightarrow B}$
and the good channel $\mathcal{W}^{(1)}_{C_{A^{\prime}_{2}\to
A_{1}B_{1}B_{2}}}$, respectively. The complementary of the good channel acts
as
$(\mathcal{W}_{C}^{(1)})^{c}(\rho)=(\mathcal{W}^{c}\varotimes\mathcal{W}^{c})\left(C\left(\frac{\mathbbm{1}_{A^{\prime}_{1}}}{d}\varotimes\rho\right)C^{\dagger}\right)$
(see [8, Appendix A] for a proof). Therefore,
$R(\mathcal{W}_{C}^{(1)})=d^{-\tilde{H}^{\downarrow}_{2}(A_{2}|E_{1}E_{2})_{\rho}}$,
where
$\rho_{A_{2}E_{1}E_{2}}=(\mathcal{W}_{C}^{(1)})^{c}(\Phi_{A_{2}A^{\prime}_{2}})$.
Note that
$\rho_{E_{1}E_{2}}=\mathcal{W}^{c}\left(\frac{\mathbbm{1}}{d}\right)\varotimes\mathcal{W}^{c}\left(\frac{\mathbbm{1}}{d}\right)$,
which is independent of $C$. To compute the expected value of
$R(\mathcal{W}_{C}^{(1)})$ with respect to $C$, we proceed as follows.

$\displaystyle\mathbb{E}_{C}d^{-\tilde{H}^{\downarrow}_{2}(A_{2}|E_{1}E_{2})_{\rho}}$
$\displaystyle=\mathbb{E}_{C}\operatorname{Tr}\left[\left(\rho_{E_{1}E_{2}}^{-\frac{1}{4}}\rho_{A_{2}E_{1}E_{2}}\rho_{E_{1}E_{2}}^{-\frac{1}{4}}\right)^{2}\right]$
$\displaystyle=\mathbb{E}_{C}\operatorname{Tr}\left[\left(\rho_{E_{1}E_{2}}^{-\frac{1}{4}}(\mathcal{W}^{c}\varotimes\mathcal{W}^{c})\left(C\left(\frac{\mathbbm{1}_{A^{\prime}_{1}}}{d}\varotimes\Phi_{A_{2}A^{\prime}_{2}}\right)C^{\dagger}\right)\rho_{E_{1}E_{2}}^{-\frac{1}{4}}\right)^{2}\right].$

Note that this is basically the same calculation as in [18, Equation (3.32)]
(there, $U$ is chosen according to the Haar measure over the full unitary
group, but all that is required is a unitary 2-design). However, we will not
make the simplifications after (3.44) and (3.45) in [18], but will instead
keep all the terms. We therefore get
$\mathbb{E}_{C}d^{-\tilde{H}^{\downarrow}_{2}(A_{2}|E_{1}E_{2})_{\rho}}=\alpha\operatorname{Tr}\left[(\frac{\mathbbm{1}_{A_{2}}}{d})^{2}\right]+\beta\operatorname{Tr}\left[(\frac{\mathbbm{1}_{A_{1}^{\prime}}}{d}\varotimes\Phi_{A_{2}A^{\prime}_{2}})^{2}\right]=\frac{1}{d}\alpha+\frac{1}{d}\beta$,
where
$\alpha=\frac{d^{4}}{d^{4}-1}-\frac{d^{2}}{d^{4}-1}d^{-\tilde{H}^{\downarrow}_{2}(A_{1}A_{2}|E_{1}E_{2})_{\omega}}$,
$\beta=\frac{d^{4}}{d^{4}-1}d^{-\tilde{H}^{\downarrow}_{2}(A_{1}A_{2}|E_{1}E_{2})_{\omega}}-\frac{d^{2}}{d^{4}-1}$,
and
$\omega_{A_{1}A_{2}E_{1}E_{2}}:=(\mathcal{W}^{c}\varotimes\mathcal{W}^{c})(\Phi_{A_{1}A^{\prime}_{1}}\varotimes\Phi_{A_{2}A^{\prime}_{2}})$.
Hence,

$\displaystyle\mathbb{E}_{C}d^{-\tilde{H}^{\downarrow}_{2}(A_{2}|E_{1}E_{2})_{\rho}}$
$\displaystyle=\frac{d}{d^{2}+1}+\frac{d}{d^{2}+1}d^{-\tilde{H}^{\downarrow}_{2}(A_{1}A_{2}|E_{1}E_{2})_{\omega}}$
$\displaystyle=\frac{d}{d^{2}+1}(1+R(\mathcal{W})^{2}),$

where the second equality follows from
$d^{-\tilde{H}^{\downarrow}_{2}(A_{1}A_{2}|E_{1}E_{2})_{\omega}}=R(\mathcal{W})^{2}$
using the fact that conditional sandwiched Rényi entropy of order 2 is
additive with respect to tensor-product states. It is easily seen that the
function $f(R)=\frac{d}{d^{2}+1}(1+R^{2})$ is a convex function satisfying
$f(R)=R$ for $R\in\\{\frac{1}{d},d\\}$ and $f(R)<R$ for $R\in(\frac{1}{d},d)$.
∎

###  3.3 Proof of Theorem 10

Proof of part (a). It is shown in [13, Theorem 1] (see also [19]) that the
Clifford group on $n$-qubits forms a unitary $2$-design for any $n\geq 1$.
Here, we generalize the proof from [13] to the qudit case, and for $n=2$. We
need to prove that the Clifford group $\mathcal{C}_{d}^{2}$ satisfies the
Definition 8. For this, it is sufficient to prove (2), with
$\mathcal{U}=\mathcal{C}_{d}^{2}$, for two-qudit input quantum channels of the
form $\mathcal{W}_{2}(\rho):=A\rho B$ (since any quantum channel is a convex
combination of quantum channels of this form).

We first consider the twirling of $\mathcal{W}_{2}$ with respect to the
Clifford group $\mathcal{C}_{d}^{2}$. Since the Pauli group
$\mathcal{P}_{d}^{2}$ is a normal subgroup of $\mathcal{C}_{d}^{2}$, we may
chose a subset $\bar{\mathcal{C}}_{d}^{2}\subset\mathcal{C}_{d}^{2}$
containing one representative for each equivalence class in the quotient group
$\mathcal{C}_{d}^{2}/\mathcal{P}_{d}^{2}$. Thus, any element of
$\mathcal{C}_{d}^{2}$ can be uniquely written as a product $CP$, where
$C\in\bar{\mathcal{C}}_{d}^{2}$, and $P\in\mathcal{P}_{d}^{2}$. Therfore, in
order to twirl $\mathcal{W}_{2}$ with respect to $\mathcal{C}_{d}^{2}$, we may
first twirl it with respect to $\mathcal{P}_{d}^{2}$, then twirl again the
obtained channel with respect to $\bar{\mathcal{C}}_{d}^{2}$.

The elements of $\mathcal{P}_{d}^{2}$ have the form
$\omega^{\lambda}P_{r,s}\varotimes P_{r^{\prime},s^{\prime}}$, with
$\lambda,r,s,r^{\prime},s^{\prime}=0,\dots,d-1$. Hence, twirling
$\mathcal{W}_{2}$ with respect to $\mathcal{P}_{d}^{2}$ gives a quantum
channel, denoted $\mathcal{W}_{2}^{\prime}$, defined below

$\displaystyle\mathcal{W}_{2}^{\prime}(\rho)$
$\displaystyle:=\frac{1}{d^{5}}\sum_{\lambda,r,s,r^{\prime},s^{\prime}}\left(\omega^{\lambda}P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)^{\dagger}A\left(\omega^{\lambda}P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)\rho\left(\omega^{\lambda}P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)^{\dagger}B\left(\omega^{\lambda}P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right),$
$\displaystyle=\frac{1}{d^{4}}\sum_{r,s,r^{\prime},s^{\prime}}(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})A\left(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})B\left(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right).$ (7)

The last equality from the above shows that it is actually enough to twirl
$\mathcal{W}_{2}$ with respect to the subset
$\bar{\mathcal{P}}_{d}^{2}:=\left\\{P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\mid r,s,r^{\prime},s^{\prime}=0,\dots,d-1\right\\}$,
obtained by omitting phase factors. Since $\bar{\mathcal{P}}_{d}^{2}$ forms an
operator basis (for two-qudit operators), we may write
$A=\sum_{r,s,r^{\prime},s^{\prime}}\alpha(r,s,r^{\prime},s^{\prime})P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}$, and
$B=\sum_{r,s,r^{\prime},s^{\prime}}\beta(r,s,r^{\prime},s^{\prime})P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}$. The following two lemmas are proven in Appendix A
and Appendix B, respectively.

######  Lemma 13.

The quantum channel $\mathcal{W}_{2}^{\prime}$, obtained by twirling
$\mathcal{W}_{2}$ with respect to $\bar{\mathcal{P}}_{d}^{2}$, is a Pauli
channel satisfying the following

$\displaystyle\mathcal{W}_{2}^{\prime}(\rho)$
$\displaystyle=\sum_{r,s,r^{\prime},s^{\prime}}\gamma_{r,s,r^{\prime},s^{\prime}}\left(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger}),$ (8)

where
$\gamma_{r,s,r^{\prime},s^{\prime}}:=\omega^{rs+r^{\prime}s^{\prime}}\alpha(r,s,r^{\prime},s^{\prime})\beta(-r,-s,-r^{\prime},-s^{\prime})$
and $-x$ denotes the additive inverse of $x$ modulo $d$.

######  Lemma 14.

The quantum channel obtained by twirling $\mathcal{W}_{2}^{\prime}$ with
respect to $\bar{\mathcal{C}}_{d}^{2}$, is the quantum channel
$\mathcal{W}_{2}^{\prime\prime}$ acting as

$\displaystyle\mathcal{W}_{2}^{\prime\prime}(\rho)=\frac{\operatorname{Tr}(AB)}{d^{4}}\mathbbm{1}\varotimes\mathbbm{1}+\frac{d^{2}\operatorname{Tr}(A)\operatorname{Tr}(B)-\operatorname{Tr}(AB)}{d^{2}(d^{4}-1)}\left(\rho-\frac{1}{d^{2}}\mathbbm{1}\varotimes\mathbbm{1}\right).$
(9)

Now, the quantum channel $\mathcal{W}_{2}^{\prime\prime}$ from (9) is the
twirling of $\mathcal{W}_{2}$ with respect to $\mathcal{C}_{d}^{2}$. To
conclude that $\mathcal{C}_{d}^{2}$ is a unitary 2-design, we need to show
that twirling $\mathcal{W}_{2}$ with respect to $\mathcal{U}(d^{2})$ yields
the same channel, which follows from [20].

Proof of part (b). We will need the following two lemmas. The first is
basically the same as [8, Lemma 14] and the proof can be easily generalized.
The second is proven in Appendix C.

######  Lemma 15.

Consider $C,C^{\prime}\in\mathcal{C}_{d}^{2}$, such that
$C^{\prime}=C(C_{1}\varotimes C_{2})$, for some
$C_{1},C_{2}\in\mathcal{C}_{d}^{1}$. Then, $C^{\prime}$ and $C^{\prime\prime}$
yield the same Rényi-Bhattacharya parameter for both good and bad channels,
i.e., following equalities hold,

  * 1)

$R(\mathcal{W}_{C}^{(0)})=R(\mathcal{W}_{C^{\prime}}^{(0)}).$

  * 2)

$R(\mathcal{W}_{C}^{(1)})=R(\mathcal{W}_{C^{\prime}}^{(1)}).$

######  Lemma 16.

If $d$ is a prime number, $|\mathcal{C}_{d}^{1}|=d^{3}(d^{2}-1)$ and
$|\mathcal{C}_{d}^{2}|=d^{8}(d^{4}-1)(d^{2}-1)$.

We are now in a position to prove the part (b) of the theorem. The group
$\mathcal{C}_{d}^{2}$ can be decomposed into left cosets with respect to the
subgroup
$\mathcal{C}_{d}^{1}\varotimes\mathcal{C}_{d}^{1}\subset\mathcal{C}_{d}^{2}$.
From Lemma 15, it follows that any two elements in the same left coset, when
used as channel combiners, yield the same Rényi-Bhattacharyya parameter for
both good and bad channels. Therefore, polarization also happens for any
subset ${\cal L}\subset\mathcal{C}_{d}^{2}$, containing one representative of
each left coset (since $\mathbb{E}_{C\in{\cal
L}}R(\mathcal{W}_{C}^{(1)})=\mathbb{E}_{C\in\mathcal{C}_{d}^{2}}R(\mathcal{W}_{C}^{(1)})$,
thus the guaranteed improvement of the average Rényi-Bhattacharyya parameter,
in the sense of Lemma 12, still holds when $C$ is randomly chosen from ${\cal
L}$). Using Lemma 16, the number of cosets of
$\mathcal{C}_{d}^{1}\varotimes\mathcal{C}_{d}^{1}$ in $\mathcal{C}_{d}^{2}$ is
equal to
$\frac{|\mathcal{C}_{d}^{2}|}{|\mathcal{C}_{d}^{1}\varotimes\mathcal{C}_{d}^{1}|}=d^{4}+d^{2}$,
therefore ${\cal L}$ contains $d^{4}+d^{2}$ representatives, two of which may
be chosen to be the identity ($I$) and the swap ($S$) operators. Since
$R(\mathcal{W}_{I}^{(1)})=R(\mathcal{W}_{S}^{(1)})=R(\mathcal{W})\geq\mathbb{E}_{C\in{\cal
L}}R(\mathcal{W}_{C}^{(1)})$, we may further remove $I$ and $S$ from ${\cal
L}$, thus getting a subset
$\mathcal{L}^{\prime}:=\mathcal{L}\setminus\\{I,S\\}$ containing
$d^{4}+d^{2}-2$ elements, which still ensures polarization of qudit-input
quantum channels. From [21, 22], we know that a set of unitaries in dimension
$\delta$ can only form a unitary 2-design if it has at least
$\delta^{4}-2\delta^{2}+2$ elements. As we consider a two-qudit system
(dimension $\delta=d^{2}$), a unitary 2-design would have at least
$d^{8}-2d^{4}+2$ two-qudit unitaries, which is clearly bigger than
$d^{4}+d^{2}-2$. Hence, the set $\mathcal{L}^{\prime}$ is not a unitary
$2$-design. This completes the poof of the part (b). ∎

One may try to further reduce the size of ${\cal L}^{\prime}$, by considering
the action of the swap gate $S$. Indeed, it can be seen that the two
equalities from Lemma 15 also hold for two
$C,C^{\prime}\in\mathcal{C}_{d}^{2}$, such that $C^{\prime}=SC$ (see also [8,
Lemma 15]). Hence, if both $C$ and $C^{\prime}$ belong to ${\cal L}^{\prime}$,
one of them can be removed, while still ensuring polarization. Now,
multiplying by $S$ on the left induces a permutation on the left cosets of
$\mathcal{C}_{d}^{1}\varotimes\mathcal{C}_{d}^{1}$ in $\mathcal{C}_{d}^{2}$,
which in turn induces a permutation ${\cal
L}^{\prime}\stackrel{{\scriptstyle\sim}}{{\rightarrow}}{\cal L}^{\prime}$. In
the qubit case ($d=2$), this permutation has no fixed points, thus the size of
${\cal L}^{\prime}$ can be reduced by half. However, in general the above
permutation may have fixed points. We provide such an example in Appendix D,
where we show that for $d=5$, there exist $C\in\mathcal{C}_{d}^{2}$ and
$C_{1},C_{2}\in\mathcal{C}_{d}^{1}$, such that $SC=C(C_{1}\varotimes C_{2})$.

##  4 Quantum Polar codes on Pauli Qudit channels

In this section, we discuss the decoding of quantum polar codes on a Pauli
qudit channel. We shall assume that all channel combining unitaries are
Clifford unitaries.

A Pauli qudit channel $\mathcal{W}$ is defined as the quantum channel that
maps a qudit quantum state $\rho$ to $\sum_{r,s}a_{r,s}P_{r,s}\rho
P_{r,s}^{\dagger}$, where $a_{r,s}\geq 0$ with $\sum_{r,s}a_{r,s}=1$. Similar
to [8, Definition 17], we associate a classical channel with $\mathcal{W}$,
which is referred to as the classical counterpart of $\mathcal{W}$, and
denoted by $\mathcal{W}^{\\#}$. The classical counterpart $\mathcal{W}^{\\#}$
is a classical channel with input and output alphabet
$\bar{\mathcal{P}}_{d}^{1}:=\\{P_{r,s}\mid r,s=0,\dots,d-1\\}$, and transition
probabilities $\mathcal{W}^{\\#}(P_{r,s}\mid P_{t,u})=a_{v,w}$, where
$v=r+t\>(\text{mod }d)$ and $w=s+u\>(\text{mod }d)$. Consider now the channel
combining and splitting procedure on $\mathcal{W}$, where
$C\in\mathcal{C}_{d}^{2}$ is used to combine the two copies of $\mathcal{W}$.
Let
$\Gamma_{C}:\bar{\mathcal{P}}_{d}^{1}\varotimes\bar{\mathcal{P}}_{d}^{1}\mapsto\bar{\mathcal{P}}_{d}^{1}\varotimes\bar{\mathcal{P}}_{d}^{1}$
be the permutation induced by the conjugate action of $C$. We may define a
channel combining and splitting procedure on the classical
$\mathcal{W}^{\\#}$, using $\Gamma_{C}$ to combine the two copies of
$\mathcal{W}^{\\#}$. Similarly to [8], we may prove (but the proof is omitted
here) that the Pauli qudit channel $\mathcal{W}$ and its classical counterpart
$\mathcal{W}^{\\#}$ polarize simultaneously, in the sense of [8, Proposition
$20$ and Corollary $21$], under their respective channel combining and
splitting procedure. As a consequence, to a quantum polar code on the Pauli
qudit channel $\mathcal{W}$, we may associate a classical polar code on
$\mathcal{W}^{\\#}$, then exploit classical polar decoding in order to decode
Pauli errors, as explained below (see also [8, Section 6]). Let $\mathbf{P}$
denote the unitary corresponding to a quantum polar code of length $N$ qudits
(see also [8, Section 5]), and $\mathbf{P}^{\\#}$ the linear map corresponding
to the classical polar code. To perform decoding, we first apply
$\mathbf{P}^{\dagger}$ on the $N$-qudit channel output, that is, the encoded
quantum state corrupted by some Pauli error, say
$E\in(\bar{\mathcal{P}}_{d}^{1})^{\varotimes N}$ (we may omit phase factors).
Hence, applying $\mathbf{P}^{\dagger}$ brings it back to the original (un-
encoded) state, which is however corrupted by a Pauli error
$E^{\prime}\in(\bar{\mathcal{P}}_{d}^{1})^{\varotimes N}$, such that
$\mathbf{P}^{\\#}(E^{\prime})=E$. We are now in position to decode
$E^{\prime}$, provided that we have been given the errors corresponding to the
noisy virtual channels. We know that the inputs to the noisy channels are
halves of preshared EPR pairs. Hence, we may perform projective measurements
on the preshared EPR pairs, with respect to the generalized Bell basis
$\\{I\varotimes
P_{r,s}\mathchoice{{\left\lvert\Phi_{AA^{\prime}}\right\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}{{\lvert\Phi_{AA^{\prime}}\rangle}}|P_{r,s}\in\bar{\mathcal{P}}_{d}^{1}\\}$,
which give us the errors, i.e., the $E^{\prime}$ components, on the noisy
virtual channels, as desired. Finally, we may decode the classical polar code
to determine $E^{\prime}$, and subsequently apply $E^{\prime\dagger}$ to
return the system to the original quantum state.

##  5 Conclusion and perspectives

The goal of this work has been to generalize the purely quantum polarization
construction to higher dimensional quantum systems. We have introduced the
necessary definitions and worked out the proof of quantum polarization,
assuming the channel combining unitary is randomized over (1) an unitary
2-design, (2) the two-qudit Clifford group, or (3) a smaller subset of two-
qudit Cliffords. Using Clifford channel combining unitaries is important, as
we showed it allows reducing the decoding problem to a classical polar code
decoding, for qudit Pauli channels. However, we note that the reliability of
the classical polar code decoding also depends on the speed of polarization
[1]. We believe that fast polarization properties can also be generalized to
the qudit case, although we leave this here as an open question.

## Acknowledgements

This research was supported in part by the “Investissements d’avenir”
(ANR-15-IDEX-02) program of the French National Research Agency. Ashutosh
Goswami acknowledges the European Union’s Horizon 2020 research and innovation
programme, under the Marie Skłodowska Curie grant agreement No 754303.

##  Appendix A Proof of Lemma 13

Recall that $\bar{\mathcal{P}}_{d}^{2}=\left\\{P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\mid r,s,r^{\prime},s^{\prime}=0,\dots,d-1\right\\}$
is the subset of two-qudit Pauli, without phase factors. Hence, twirling of
$\mathcal{W}_{2}$ with respect to $\bar{\mathcal{P}}_{d}^{2}$ gives

$\mathcal{W}_{2}^{\prime}(\rho)=\frac{1}{d^{4}}\sum_{r,s,r^{\prime},s^{\prime}}(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})A\left(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})B\left(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}\right)$ (10)

Since $\bar{\mathcal{P}}_{d}^{2}$ forms an operator basis, we may write

$\displaystyle A$
$\displaystyle=\sum_{r,s,r^{\prime},s^{\prime}}\alpha(r,s,r^{\prime},s^{\prime})P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}},$ (11) $\displaystyle B$
$\displaystyle=\sum_{r,s,r^{\prime},s^{\prime}}\beta(r,s,r^{\prime},s^{\prime})P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}$ (12)

Substituting $A$ and $B$ in the above equation, we get

$\displaystyle\mathcal{W}_{2}^{\prime}(\rho)$
$\displaystyle=\frac{1}{d^{4}}\sum_{t,u,t^{\prime},u^{\prime}}\>\sum_{v,w,v^{\prime},w^{\prime}}\alpha(t,u,t^{\prime},u^{\prime})\beta(v,w,v^{\prime},w^{\prime})\kappa,$
(13) $\displaystyle\text{where }\kappa$
$\displaystyle:=\sum_{r,r^{\prime},s,s^{\prime}}(P_{r,s}^{\dagger}P_{t,u}P_{r,s})\varotimes(P_{r^{\prime},s^{\prime}}^{\dagger}P_{t^{\prime},u^{\prime}}P_{r^{\prime},s^{\prime}})\rho\break(P_{r,s}^{\dagger}P_{v,w}P_{r,s})\varotimes(P_{r^{\prime},s^{\prime}}^{\dagger}P_{v^{\prime},w^{\prime}}P_{r^{\prime},s^{\prime}}).$
(14)

From (1), we have that $P_{t,u}P_{r,s}=\omega^{-ru+st}P_{r,s}P_{t,u}$. Then,
we may write

$\displaystyle\kappa$ $\displaystyle=k(P_{t,u}\varotimes
P_{t^{\prime},u^{\prime}})\rho(P_{v,w}\varotimes P_{v^{\prime},w^{\prime}})$
(15) $\displaystyle\text{with }k$
$\displaystyle:=\sum_{r,s}\omega^{-r(u+w)+s(v+t)}\sum_{r^{\prime},s^{\prime}}\omega^{-r^{\prime}(u^{\prime}+w^{\prime})+s^{\prime}(v^{\prime}+t^{\prime})}.$
(16)

When $u+w=v+t=0\ (\text{mod }d)$, we have
$\sum_{r,s}\omega^{-r(u+w)+s(v+t)}=d^{2}$. When either $u+v\neq 0\ (\text{mod
}d)$ or $t+w\neq 0\ (\text{mod }d)$, we have
$\sum_{r,s}\omega^{-r(u+w)+s(v+t)}=\frac{(\omega^{-d}-1)(\omega^{d}-1)}{(\omega^{-1}-1)(\omega-1)}=0$.
Therefore,

$k=\begin{cases}d^{4},&\text{when
}u+w=v+t=u^{\prime}+w^{\prime}=v^{\prime}+t^{\prime}=0\ (\text{mod }d)\\\
0,&\text{otherwise }\end{cases}$ (17)

The condition $u+w=v+t=0\ (\text{mod }d)$ implies that
$P_{t,u}P_{v,w}=X^{t}Z^{u}X^{v}Z^{w}=\omega^{-uv}I$. Using $t=-v\ (\text{mod
}d)$, we have that $P_{v,w}=\omega^{tu}P_{t,u}^{\dagger}$. Plugging $\kappa$
into (13), we get

$\displaystyle\mathcal{W}_{2}^{\prime}(\rho)$
$\displaystyle=\sum_{t,u,t^{\prime},u^{\prime}}\gamma_{t,u,t^{\prime},u^{\prime}}(P_{t,u}\varotimes
P_{t^{\prime},u^{\prime}})\rho(P_{t,u}^{\dagger}\varotimes
P_{t^{\prime},u^{\prime}}^{\dagger}),$ (18) $\displaystyle\text{where
}\gamma_{t,u,t^{\prime},u^{\prime}}$
$\displaystyle:=\omega^{tu+t^{\prime}u^{\prime}}\alpha(t,u,t^{\prime},u^{\prime})\beta(-t,-u,-t^{\prime},-u^{\prime}).$
(19)

Hence, $\mathcal{W}_{2}^{\prime}$ is a qudit Pauli channel, as desired. ∎

##  Appendix B Proof of Lemma 14

Recall that $\bar{\mathcal{C}}_{d}^{2}\subset\mathcal{C}_{d}^{2}$ is a subset
containing one representative for each equivalence class in the quotient group
$\mathcal{C}_{d}^{2}/\mathcal{P}_{d}^{2}$. Twirling of
$\mathcal{W}_{2}^{\prime}$ with respect to $\bar{\mathcal{C}}_{d}^{2}$ gives

$\mathcal{W}_{2}^{\prime\prime}(\rho)=\sum_{t,u,t^{\prime},u^{\prime}}\gamma_{t,u,t^{\prime}u^{\prime}}\frac{1}{|\bar{\mathcal{C}}_{d}^{2}|}\sum_{C\in\bar{\mathcal{C}}_{d}^{2}}C^{\dagger}(P_{t,u}\varotimes
P_{t^{\prime},u^{\prime}})C\rho C^{\dagger}(P_{t,u}^{\dagger}\varotimes
P_{t^{\prime},u^{\prime}}^{\dagger})C.$ (20)

We know that the conjugate action of the entire set
$\bar{\mathcal{C}}_{d}^{2}$ maps any $P_{t,u}\varotimes
P_{t^{\prime},u^{\prime}}\neq I\varotimes I$ to all $d^{4}-1$ two-qudit Paulis
excluding $I\varotimes I$, an equal number of times. In other words,
$P_{t,u}\varotimes P_{t^{\prime},u^{\prime}}\neq I\varotimes I$ gets mapped to
a Pauli $P_{r,s}\varotimes P_{r^{\prime},s^{\prime}}\neq I\varotimes I$,
$\frac{|\bar{\mathcal{C}}_{d}^{2}|}{d^{4}-1}$ times. Further, $I\varotimes I$
is always mapped to $I\varotimes I$. Therefore, we have that

$\displaystyle\mathcal{W}_{2}^{\prime\prime}(\rho)$
$\displaystyle=\gamma_{0,0,0,0}\rho+\frac{1}{d^{4}-1}\gamma^{\prime}\sum_{(r,s,r^{\prime},s^{\prime})\neq(0,0,0,0)}(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}})\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger}),$ (21) $\displaystyle\text{where
}\gamma^{\prime}$
$\displaystyle:=\sum_{(t,u,t^{\prime},u^{\prime})\neq(0,0,0,0)}\gamma_{t,u,t^{\prime},u^{\prime}}.$
(22)

Using the following three identities, we can easily transform (21) into the
form of (9).

  1. 1.

$\displaystyle\gamma_{0,0,0,0}=\frac{\text{Tr}(A)\text{Tr}(B)}{d^{4}}$.

  2. 2.

$\displaystyle\sum_{t,u,t^{\prime},u^{\prime}}\gamma_{t,u,t^{\prime},u^{\prime}}=\frac{\text{Tr}(AB)}{d^{2}}$.

  3. 3.

$\displaystyle\sum_{r,s,r^{\prime},s^{\prime}}(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}})\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})=d^{2}I\varotimes I$.

Proof of identity 1) We have that
$\gamma_{0,0,0,0}=\alpha(0,0,0,0)\beta(0,0,0,0)$. Also,

$\text{Tr}(P_{r,s})=\begin{cases}d,&\text{when }P_{r,s}=I\\\
0,&\text{otherwise }\end{cases}$

Using (11) and (12), we get $\text{Tr}(A)=\alpha(0,0,0,0)d^{2}$ and
$\text{Tr}(B)=\beta(0,0,0,0)d^{2}$. Hence,
$\gamma_{0,0,0,0}=\frac{\text{Tr}(A)\text{Tr}(B)}{d^{4}}$.

Proof of identity 2) We have,

$\displaystyle\text{Tr}(AB)$
$\displaystyle=\sum_{t,u,t^{\prime},u^{\prime}}\>\sum_{v,w,v^{\prime},w^{\prime}}\alpha(t,u,t^{\prime},u^{\prime})\beta(v,w,v^{\prime},w^{\prime})\text{Tr}(P_{t,u}P_{v,w})\text{Tr}(P_{t^{\prime},u^{\prime}}P_{v^{\prime},w^{\prime}})$
$\displaystyle=\sum_{t,u,t^{\prime},u^{\prime}}d^{2}\omega^{tu+t^{\prime}u^{\prime}}\alpha(t,u,t^{\prime},u^{\prime})\beta(-t,-u,-t^{\prime},-u^{\prime})$
$\displaystyle=d^{2}\sum_{t,u,t^{\prime},u^{\prime}}\gamma_{t,u,t^{\prime},u^{\prime}}.$

Proof of identity 3) Let
$\rho=\sum_{r,s,r^{\prime},s^{\prime}}\rho_{r,s,r^{\prime},s^{\prime}}P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}}$. Since $\rho$ is a density matrix, we have
$\rho_{0,0,0,0}=\frac{\operatorname{Tr}(\rho)}{d^{2}}=\frac{1}{d^{2}}$. Hence,

$\displaystyle\sum_{r,s,r^{\prime},s^{\prime}}(P_{r,s}\varotimes
P_{r^{\prime},s^{\prime}})\rho(P_{r,s}^{\dagger}\varotimes
P_{r^{\prime},s^{\prime}}^{\dagger})$
$\displaystyle=\sum_{r,s,r^{\prime},s^{\prime}}\>\sum_{t,u,t^{\prime},u^{\prime}}\rho_{t,u,t^{\prime},u^{\prime}}(P_{r,s}P_{t,u}P_{r,s}^{\dagger})\varotimes(P_{r^{\prime},s^{\prime}}P_{t^{\prime},u^{\prime}}P_{r^{\prime},s^{\prime}}^{\dagger})$
$\displaystyle=\sum_{t,u,t^{\prime},u^{\prime}}\rho_{t,u,t^{\prime},u^{\prime}}\left(\sum_{r,s,r^{\prime},s^{\prime}}\omega^{-st+ru}\omega^{-s^{\prime}t^{\prime}+r^{\prime}u^{\prime}}\right)P_{t,u}\varotimes
P_{t^{\prime},u^{\prime}}$ $\displaystyle=d^{4}\rho_{0,0,0,0}I\varotimes I$
$\displaystyle=d^{2}I\varotimes I.$

We get (9) from (21) by using the above identities, while also substituting
the notation $\mathbbm{1}$ for the identity matrix $I$, as it denotes a
quantum state here. ∎

##  Appendix C Proof of Lemma 16

Consider the one-qudit Clifford group $\mathcal{C}_{d}^{1}$. We count first
the permutations generated by $\mathcal{C}_{d}^{1}$ on
$\bar{\mathcal{P}}_{d}^{1}:=\\{P_{r,s}|r,s=0,\dots,d-1\\}$, and later we will
accommodate the phase factors. Any Clifford $C\in\mathcal{C}_{d}^{1}$ is
uniquely determined by its conjugate action on the generators of the Pauli
group, $X$ and $Z$. Suppose that $C$ maps $X\mapsto P_{r,s}$ and $Z\mapsto
P_{t,u}$ via its conjugate action, where $P_{r,s},P_{t,u}\neq I$. On the one
hand, since commutation relations are preserved under unitary conjugation,
$P_{r,s}$ and $P_{t,u}$ must satisfy $P_{r,s}P_{t,u}=\omega P_{t,u}P_{r,s}$.
On the other hand, from (1), we have that $P_{r,s}P_{t,u}=\omega^{ru-
st}P_{t,u}P_{r,s}$. Therefore, $r,u,s,t$ must be such that $ru-st=1\
(\text{mod d})$. We fix $r,s$ and solve for $t,u$. Since $P_{r,s}\neq I$, it
follows that either $r$ or $s$ is non-zero. Without loss of generality, we may
assume that $r\neq 0$. Since $d$ is a prime number, $r$ is invertible under
multiplication modulo $d$. Therefore, for any $t\in\\{0,\dots,d-1\\}$, there
exists a unique $u:=r^{-1}(1+st)\ (\text{mod }d)$, satisfying $ru-st=1$.
Hence, there are exactly $d$ choices for the $t,u$ pair. Since we have
$d^{2}-1$ choices for the $r,s$ pair, it follows that there are $d(d^{2}-1)$
pairs of Paulis, $P_{r,s}$ and $P_{t,u}$, such that $P_{r,s}P_{t,u}=\omega
P_{t,u}P_{r,s}$. Taking into account the phase factors,
$\omega^{\lambda},\lambda\in\\{0,\dots,d-1\\}$, it follows that
$\mathcal{C}_{d}^{1}$ has $d^{3}(d^{2}-1)$ elements.

We now count the number of elements in $\mathcal{C}_{d}^{2}$. The two-qudit
Pauli group $\mathcal{P}_{d}^{2}$ is generated by a set of four Paulis
$I\varotimes X,I\varotimes Z,X\varotimes I$ and $Z\varotimes I$, and any
Clifford $C\in\mathcal{C}_{d}^{2}$ is uniquely determined by its conjugate
action on these four generators. The commutation relations between the four
generators are illustrated in Fig. 2.

$I\varotimes Z$$I\varotimes X$$Z\varotimes I$$X\varotimes I$ Figure 2:
Connected Paulis satisfy $AB=\omega BA$, with $A$ is the Pauli on the top row,
and $B$ the Pauli on the bottom row. Paulis that are not connected commute.

Consider a mapping $I\varotimes X\mapsto A$, $I\varotimes Z\mapsto B$,
$X\varotimes I\mapsto A^{\prime}$, $Z\varotimes I\mapsto B^{\prime}$, where
$A,B,A^{\prime},B^{\prime}\in\bar{\mathcal{P}}_{d}^{2}$, that preserves all
the commutation relations between generators. Pauli $I\varotimes X$ can be
mapped to any two-qudit Pauli $A\neq I\varotimes I$, so there are $d^{4}-1$
choices for $A$. It is not very difficult to see that for any $A\neq
I\varotimes I$ there are $d^{3}$ choices for $B$ such that $AB=\omega BA$.
Further, there are $d(d^{2}-1)$ pairs of two-qudit Paulis $A^{\prime}$ and
$B^{\prime}$, which commute with both $A$ and $B$, and satisfy
$A^{\prime}B^{\prime}=\omega B^{\prime}A^{\prime}$. Therefore, we have
$d^{4}(d^{4}-1)(d^{2}-1)$ possible permutations on
$\bar{\mathcal{P}}_{d}^{2}$, which satisfy all the commutation relations.
Taking into account the phase factors, it follows that $\mathcal{C}_{d}^{2}$
has $d^{8}(d^{4}-1)(d^{2}-1)$ elements. ∎

##  Appendix D Example of left coset fixed by the swap gate

We consider $d=5$. Let $C_{1}=I$ be the identity, and
$C_{2}^{\prime}\in\mathcal{C}_{d}^{1}$ be such that it maps $X\mapsto X^{4}$
and $Z\mapsto Z^{4}$, via conjugation. Since $X^{4}Z^{4}=\omega Z^{4}X^{4}$,
$C_{2}^{\prime}$ is indeed a one-qudit Clifford. We define
$C_{2}=C_{2}^{\prime}X^{2}Z^{2}$. Further, let $C\in\mathcal{C}_{d}^{2}$, such
that its conjugate action generates the following permutation on the
generators of $\mathcal{P}_{d}^{2}$,

$\displaystyle I\varotimes X$ $\displaystyle\mapsto X^{4}Z\varotimes XZ^{4},$
$\displaystyle I\varotimes Z$ $\displaystyle\mapsto XZ\varotimes X^{4}Z^{4},$
$\displaystyle X\varotimes I$ $\displaystyle\mapsto X^{4}Z\varotimes X^{4}Z,$
$\displaystyle Z\varotimes I$ $\displaystyle\mapsto XZ\varotimes XZ.$

Using (1), it is easily seen that the above permutation preserves all the
commutation relations between the generators. Now, the conjugate actions of
$SC$ and $C(C_{1}\varotimes C_{2})$ generate the same permutation on
$\mathcal{P}_{d}^{2}$. Therefore, $SC=C(C_{1}\varotimes C_{2})$.

## References

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[1em]1em1em

# Taylor Expansion of
homogeneous functions

Joachim Paulusch and Sebastian Schlütter

(July 7, 2021, this version: August 27, 2024)

###### Abstract

We derive the Taylor polynomial of a function which is $m$-times continuously
differentiable and positive homogeneous of order $m$. The Taylor polynomial in
$a$ for $f(b)$ of order $m$ in general is a polynomial of order $m$ in $b-a$.
If the given function is positive homogeneous of order $m$, the Taylor
polynomial is a polynomial in $b$ rather than $b-a$, and the order of all
terms is $m$.

The result can be applied to powers of homogeneous functions of order $1$ as
well.

We consider Taylor polynomials of homogeneous functions. Derivatives of higher
order are denoted by

$d^{m}f(x;t)=\sum_{k\in\\{1,\ldots,n\\}^{m}}D_{k}f(x)t_{k_{1}}\cdots
t_{k_{m}}.$ (1)

Let $U\subseteq\mathbb{R}^{n}$ be open and $f:U\to\mathbb{R}$ have continuous
partial derivatives of order $m$ in $U$. Let $a,b\in U$ and the line segment
connecting $a$ and $b$ in $U$. The Taylor polynomial for $f(b)$ of order $m$
at $a$ is given by

$T^{(m)}f(a;b-a)=f(a)+\sum_{k=1}^{m}\frac{1}{k!}d^{k}f(a;b-a).$ (2)

######  Theorem 1.

Let $U\subseteq\mathbb{R}^{n}$ be open and $f:U\to\mathbb{R}$ be positive
homogeneous111I.e. $f(\lambda x)=\lambda^{m}f(x)$ for all $x\in U$,
$\lambda>0$ with $\lambda x\in U$. This gives rise to a unique extension of
$f$ to the set $\\{\lambda x\,|\,x\in U,\lambda>0\\}$. In case $0\in U$, this
extends $f$ to $\mathbb{R}^{n}$. of degree $m$ and have continuous partial
derivatives of order $m$ in $U$. Let $a,b\in U$ and the line segment
connecting $a$ and $b$ in $U$. The Taylor polynomial for $f(b)$ of order $m$
at $a$ is given by

$T^{(m)}f(a;b-a)=\frac{1}{m!}\,d^{m}f(a;b).$ (3)

This means that the Taylor polynomial is a polynomial in $b$ rather than $b-a$
and the order of all terms is $m$, i.e. a homogeneous polynomial of order $m$.
Theorem 1 immediately implies

######  Corollary 2.

Let $U\subseteq\mathbb{R}^{n}$ be open and $f:U\to\mathbb{R}$ be positive
homogeneous of degree $1$ and have continuous partial derivatives of order
$\ell$ in $U$. Let $a,b\in U$ and the line segment connecting $a$ and $b$ in
$U$. Then, for all $1\leq m\leq\ell$, we have

$T^{(m)}f^{m}(a;b-a)=\frac{1}{m!}\,d^{m}f^{m}(a;b).$ (4)

######  Remark 3.

Note that in case $0\in U$ it can be shown that a homogeneous function $f$ of
order $m$ actually is a homogeneous polynomial of order $m$, cf. [1, p. 175].
A typical example not covered by this but Theorem 1, or Corollary 2
respectively, is

$f(x)=\sqrt{x^{\text{T}}R\,x},$ (5)

wherein $R$ is a positive semidefinite matrix. This can be applied to models
for the aggregation of risk, c.f. [2], where the assertion is shown for $m=2$.

###### Proof of Theorem 1.

To facilitate the handling of many dimensions we will use Einstein’s summation
convention, i.e. we write

$\sum_{k\in\\{1,\ldots,n\\}^{m}}D_{k}f(x)t_{k_{1}}\dots
t_{k_{m}}=D_{\mu_{1}\dots\mu_{m}}f(x)t^{\mu_{1}}\dots t^{\mu_{m}}.$ (6)

Note that Euler’s Theorem on homogeneous functions [3, p. 6] implies

$D_{\mu_{1}\dots\mu_{k}}f(a)a^{\mu_{k}}=\left(m-(k-1)\right)D_{\mu_{1}\dots\mu_{k-1}}f(a),$
(7)

as $D_{\mu_{1}\dots\mu_{k-1}}f$ is homogeneous of order
$\left(m-(k-1)\right)$. Moreover, by continuously differentiability, we have
the symmetry

$D_{\mu_{1}\dots\mu_{k}}f(a)=D_{\mu_{\sigma(1)}\dots\mu_{\sigma(k)}}f(a),$ (8)

for all $1\leq k\leq m$ and permutation $\sigma$. Taylor’s formula reads in
Einstein notation

$T^{(m)}f(a;b-a)=f(a)+\sum_{k=1}^{m}\frac{1}{k!}D_{\mu_{1}\dots\mu_{k}}f(a)(b-a)^{\mu_{1}}\cdots(b-a)^{\mu_{k}}.$
(9)

We use (7) to write

$D_{\mu_{1}\dots\mu_{k}}f(a)=\frac{1}{(m-k)!}D_{\mu_{1}\dots\mu_{m}}f(a)a^{\mu_{k+1}}\cdots
a^{\mu_{m}},$ (10)

that is (using the symmetry of the derivatives)

$\begin{split}&T^{(m)}f(a;b-a)=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)\sum_{k=0}^{m}{m\choose
k}(b-a)^{\mu_{1}}\cdots(b-a)^{\mu_{k}}a^{\mu_{k+1}}\cdots a^{\mu_{m}}\\\
&\quad=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)\sum_{k=0}^{m}{m\choose
k}a^{\mu_{k+1}}\cdots a^{\mu_{m}}\sum_{p=0}^{k}(-1)^{p}{k\choose
p}b^{\mu_{1}}\cdots b^{\mu_{k-p}}a^{\mu_{k-p+1}}\cdots a^{\mu_{k}}\\\
&\quad=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)\sum_{k=0}^{m}\sum_{p=0}^{k}(-1)^{p}{m\choose
k}{k\choose p}b^{\mu_{1}}\cdots b^{\mu_{k-p}}a^{\mu_{k-p+1}}\cdots
a^{\mu_{m}}.\end{split}$ (11)

To collect powers of $b$, we write $q=k-p$ and obtain

$\begin{split}&T^{(m)}f(a;b-a)=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)\sum_{q=0}^{m}\sum_{k=q}^{m}(-1)^{k-q}{m\choose
k}{k\choose k-q}b^{\mu_{1}}\cdots b^{\mu_{q}}a^{\mu_{q+1}}\cdots
a^{\mu_{m}}\\\
&\quad=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)\sum_{q=0}^{m}b^{\mu_{1}}\cdots
b^{\mu_{q}}a^{\mu_{q+1}}\cdots a^{\mu_{m}}\sum_{k=q}^{m}(-1)^{k-q}{m\choose
k}{k\choose k-q}.\end{split}$ (12)

For $q=m$, the inner sum over $k$ equals 1. So let $0\leq q\leq m-1$. We write
$p=k-q$, $r=m-q$ and derive for $1\leq r\leq m$

$\begin{split}&\sum_{k=q}^{m}(-1)^{k-q}{m\choose k}{k\choose
k-q}=\sum_{p=0}^{m-q}(-1)^{p}{m\choose p+q}{p+q\choose p}\\\
&\quad=\sum_{p=0}^{r}(-1)^{p}{m\choose p+m-r}{p+m-r\choose
p}=\sum_{p=0}^{r}(-1)^{p}\frac{m!}{p!\,(m-r)!\,(r-p)!}\\\ &\quad={m\choose
r}\sum_{p=0}^{r}(-1)^{p}{r\choose p}={m\choose r}(1-1)^{r}=0.\end{split}$ (13)

Going back to equation (12), this implies

$T^{(m)}f(a;b-a)=\frac{1}{m!}D_{\mu_{1}\dots\mu_{m}}f(a)b^{\mu_{1}}\cdots
b^{\mu_{m}},$ (14)

as was to be shown. ∎

## References

  * [1] Serge Lang: Calculus of Several Variables. Third Edition, Springer (1987)
  * [2] Joachim Paulusch and Sebastian Schlütter: Sensitivity-implied tail-correlation matrices. Journal of Banking & Finance 134 (2022): 106333.
  * [3] Dirk Tasche: Risk contributions and performance measurement. Preprint (1999)