# Tensorization of the strong data processing inequality for quantum chi-
square divergences

Yu Cao<EMAIL_ADDRESS>Department of Mathematics, Duke University, Box
90320, Durham NC 27708, USA  Jianfeng Lu<EMAIL_ADDRESS>Department of
Mathematics, Duke University, Box 90320, Durham NC 27708, USA  Department of
Physics and Department of Chemistry, Duke University, Box 90320, Durham NC
27708, USA

###### Abstract

It is well-known that any quantum channel $\mathcal{E}$ satisfies the data
processing inequality (DPI), with respect to various divergences, e.g.,
quantum $\chi^{2}_{\kappa}$ divergences and quantum relative entropy. More
specifically, the data processing inequality states that the divergence
between two arbitrary quantum states $\rho$ and $\sigma$ does not increase
under the action of any quantum channel $\mathcal{E}$. For a fixed channel
$\mathcal{E}$ and a state $\sigma$, the divergence between output states
$\mathcal{E}(\rho)$ and $\mathcal{E}(\sigma)$ might be strictly smaller than
the divergence between input states $\rho$ and $\sigma$, which is
characterized by the strong data processing inequality (SDPI). Among various
input states $\rho$, the largest value of the rate of contraction is known as
the SDPI constant. An important and widely studied property for classical
channels is that SDPI constants tensorize. In this paper, we extend the
tensorization property to the quantum regime: we establish the tensorization
of SDPIs for the quantum $\chi^{2}_{\kappa_{1/2}}$ divergence for arbitrary
quantum channels and also for a family of $\chi^{2}_{\kappa}$ divergences
(with $\kappa\geq\kappa_{1/2}$) for arbitrary quantum-classical channels.

Keywords:  strong data processing inequality, tensorization, quantum chi-
square divergence

##  1 Introduction

In information theory, the data processing inequality (DPI) has been an
important property for divergence measures to possess operational meaning. For
instance, DPI has been proved for quantum $\chi^{2}_{\kappa}$ divergences (see
e.g., [12, Thm. II.14] or [21, Thm. 4]), among other divergences. More
explicitly, for any quantum channel $\mathcal{E}$ and for all quantum states
$\rho,\sigma\in\mathfrak{D}_{n}$, we have

$\chi^{2}_{\kappa}\left(\mathcal{E}(\rho)\mid\mid\mathcal{E}(\sigma)\right)\leq\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right).$
(1)

In the above, $\kappa$ is a real-valued positive function (see (7) below); the
definition of $\chi^{2}_{\kappa}$ divergences will be postponed to § 2.1, as
it involves some technicalities.

Compared with the DPI, the strong data processing inequality (SDPI)
quantitatively and more precisely characterizes the extent that quantum states
contract under the channel $\mathcal{E}$ [1, 18, 16, 17]. Given any
$(\mathcal{E},\sigma)$-pair where $\mathcal{E}$ is any quantum channel and
$\sigma\in\mathfrak{D}^{+}_{n}$ is any full-rank quantum state
($\mathfrak{D}^{+}_{n}$ is the space of strictly positive density matrices on
a $n$-dimensional Hilbert space), if there is a constant
$\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E},\sigma\right)\in[0,1)$ such that

$\chi^{2}_{\kappa}\left(\mathcal{E}(\rho)\mid\mid\mathcal{E}(\sigma)\right)\leq\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E},\sigma\right)\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right),\
\forall\rho\in\mathfrak{D}_{n},$ (2)

then the quantum channel $\mathcal{E}$ is said to satisfy the _strong data
processing inequality_ (SDPI) for the quantum $\chi^{2}_{\kappa}$ divergence
and the smallest constant
$\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E},\sigma\right)$ such that (2) holds
is called the _SDPI constant_. Evidently,

$\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E},\sigma\right)=\sup_{\rho\in\mathfrak{D}_{n}:\
\rho\neq\sigma}\frac{\chi^{2}_{\kappa}\left(\mathcal{E}(\rho)\mid\mid\mathcal{E}(\sigma)\right)}{\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right)}.$
(3)

Many applications of SDPIs can be found in e.g., [17, Sec. 2.3] and [18, Sec.
v@].

It is common in quantum information theory to consider high-dimensional
quantum channels, formed by the tensor product of low-dimensional quantum
channels. Except for very special cases, in general, obtaining SDPI constants
for high-dimensional quantum channels can be rather challenging, even
numerically. It is desirable if one could reduce the problem of calculating
the SDPI constant for a (global) high-dimensional quantum channel, to
calculating the SDPI constants of low-dimensional quantum channels. For a
specific divergence (e.g., quantum $\chi^{2}_{\kappa}$ divergence in this
work), if the SDPI constant for the high-dimensional channel is the maximum
value of SDPI constants for these low-dimensional channels, we say that the
SDPI constant for this divergence satisfies the _tensorization property_.

Our main result in this work is that the SDPI constant for $\chi^{2}_{\kappa}$
tensorizes, summarized in the following theorem.

######  Theorem 1.

Consider $N$ finite-dimensional quantum systems whose Hilbert spaces are
$\mathcal{H}_{j}$ with dimension $n_{j}$ ($1\leq j\leq N$) and consider any
density matrix $\sigma_{j}\in\mathfrak{D}^{+}_{n_{j}}$ and any quantum channel
$\mathcal{E}_{j}$ acting on $\mathcal{H}_{j}$, such that for all $1\leq j\leq
N$, $\mathcal{E}_{j}(\sigma_{j})\in\mathfrak{D}^{+}_{n_{j}}$. If either of the
followings holds

  1. (i)

$\kappa=\kappa_{1/2}$;

  2. (ii)

$\kappa\geq\kappa_{1/2}$ and $\mathcal{E}_{j}$ are quantum-classical (QC)
channels;

then we have the tensorization of the SDPI constant for the quantum
$\chi^{2}_{\kappa}$ divergence, i.e.,

$\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{1}\otimes\mathcal{E}_{2}\otimes\cdots\otimes\mathcal{E}_{N},\sigma_{1}\otimes\sigma_{2}\otimes\cdots\otimes\sigma_{N}\right)=\max_{1\leq
j\leq N}\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{j},\sigma_{j}\right).$ (4)

######  Remark.

  1. (i)

The function $\kappa_{1/2}(x):=x^{-1/2}$ is a special example of weight
functions. There are some properties that only the quantum
$\chi_{\kappa_{1/2}}^{2}$ divergence possesses (see e.g., Lemma 7 (ii)); in
addition, $\chi_{\kappa_{1/2}}^{2}$ is tightly connected to the sandwiched
Rényi divergence of order $2$ [13].

There is a whole family of $\kappa_{\alpha}$ parameterized by
$\alpha\in[0,1]$, satisfying the condition $\kappa_{\alpha}\geq\kappa_{1/2}$;
see Example 3 for details; in § 2.2, we also present other examples of
$\kappa(x)$ such that $\kappa\geq\kappa_{1/2}$. The notion of QC channel will
be recalled in § 2.6.

  2. (ii)

These assumptions only provide sufficient conditions for the tensorization of
SDPIs to hold, and it is an interesting open question to further investigate
weaker conditions. In addition, it is also an interesting open question
whether the tensorization of SDPIs holds for (quasi) relative entropies and
the geodesic distances [12, 8]. We shall leave these questions to future
research.

The tensorization property in the classical regime has been well studied and
widely used; see e.g., [1, 22, 18]. For SDPI constants, the tensorization
property was proved in [18, Thm<EMAIL_ADDRESS>for any $\Phi$-divergence, denoted by
$D_{\Phi}\left(\nu\mid\mid\mu\right):=\mathbb{E}_{\mu}\Big{[}\Phi(\frac{\,\mathrm{d}\nu}{\,\mathrm{d}\mu})\Big{]}-\Phi(1),$
provided that the associated $\Phi$-entropy is sub-additive and homogeneous.
As a remark, the $\Phi$-divergence includes the relative entropy (with
$\Phi(x)=x\log(x)$) and the classical $\chi^{2}$ divergence (with
$\Phi(x)=(x-1)^{2}$) as special instances. The tensorization of SDPI constants
associated with the classical relative entropy has been applied to study the
lower bounds of Bayes risk [24].

Establishing tensorization in the quantum regime seems to be more challenging
and our understanding is much limited. Recently, the tensorization technique
has been developed for the quantum hypercontractivity of qubit system [11],
reversed hypercontractivity [7, 3], $2$-log-Sobolev constant [10, 3], as well
as the quantum maximal correlation [2]. For the tensorization of the quantum
(reversed) hypercontractivity and log-Sobolev constants, all existing works,
as far as we know, focus exclusively on reversible (or even more special)
quantum Markov semigroups (i.e., Lindblad equations).

We would like to briefly mention and highlight the proof techniques used for
Theorem 1. The first main ingredient is to formulate the SDPI constant as the
second largest eigenvalue of a certain operator (see Lemma 10); similar
results have been obtained in e.g., [5, 18, 20, 12, 8]. This result
immediately leads into the proof of the case (i). The second main ingredient
is to bound $\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)$ above by
$\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}$ (see Lemma 12), whose
proof uses Petz recovery map [14] as the bridge. This relation together with
special properties of $\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)$
leads into the proof of case (ii).

### Related techniques to quantify the loss of information

Apart from the DPI and the SDPI, there are other concepts used to characterize
the contraction of quantum states under the action of noisy channels. For
instance, one widely studied quantity is the _contraction coefficient_

$\eta_{\chi^{2}_{\kappa}}(\mathcal{E}):=\sup_{\sigma\in\mathfrak{D}^{+}_{n}}\
\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma).$ (5)

The contraction coefficient is very similar to the SDPI constant. However,
compared with the SDPI constant, the contraction coefficient for various
divergence measures has been much more extensively studied in the literature,
see e.g., [20, 15, 12, 21, 8] for the quantum case, and see e.g., [6] for the
classical case. The bijection maps that preserve the quantum
$\chi^{2}_{\kappa_{\alpha}}$ divergence (see Example 3 about the family
$\kappa_{\alpha}$) have been characterized in [4], which complements the study
of the contraction of quantum states. There are other tools based on the
functional perspective, including quantum (reverse) hypercontractivity and
related quantum functional inequalities [10, 11, 7, 19, 3].

### Contribution

We summarize new results obtained in this work, as follows:

  1. (i)

Our main result is Theorem 1, which establishes the tensorization of SDPI
constants, under certain assumptions: for the quantum
$\chi^{2}_{\kappa_{1/2}}$ divergence, the tensorization of SDPI constants
holds for general quantum channels; for the quantum $\chi_{\kappa}^{2}$
divergence with $\kappa\geq\kappa_{1/2}$, the tensorization holds for any
quantum-classical channel.

  2. (ii)

Along the analysis of the SDPI, we also establish a connection between the
SDPI constant associated with $\kappa_{1/2}$ and a variant of quantum maximal
correlations; see Theorem 18 for details.

  3. (iii)

To use the tensorization property, we need to understand the SDPI constants
for local channels, i.e., we need to compute
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{j},\sigma_{j})$ for $1\leq j\leq N$.
Motivated by this, we study the SDPI constants for special qubit channels in §
5. We notice that there is a particular QC channel $\mathcal{E}$ associated
with a fixed $\sigma\in\mathfrak{D}^{+}_{2}$ such that the largest value of
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\approx 1$ for
$\kappa=\kappa_{\min}$, while
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\approx 0$ for
$\kappa=\kappa_{\max}$ (however, $\sigma$ is close to a singular matrix); see
§ 5.1 for details. This extreme example shows the high dependence of SDPI
constants on the choice of $\kappa$, which magnifies the difference between
the quantum SDPI constant and its classical analog, because there is only one
SDPI constant for the classical $\chi^{2}$ divergence.

This paper is organized as follows. In § 2, we provide some preliminary
results, in particular, we recall the eigenvalue formalism of the SDPI
constant. In § 3, we prove Theorem 1 and in § 4, we study the connection
between the SDPI constant and the quantum maximal correlation. In § 5, we
consider SDPI constants for qubit channels and study the dependence of
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ on $\sigma$ and $\kappa$. § 6
concludes the paper with some additional remarks.

##  2 Preliminaries

This section contains preliminary results that we will use to prove the
tensorization of the strong data processing inequality, Theorem 1. In
particular, we will present two variational formulations of SDPI constants,
and discuss the relation between various SDPI constants.

Notations. We shall consider finite dimensional systems only, i.e., the
Hilbert space $\mathcal{H}\cong\mathbb{C}^{n}$. Let $\mathbb{M}_{n}$,
$\mathfrak{D}_{n}$, $\mathfrak{D}^{+}_{n}$, $\mathbb{H}_{n}$ be the space of
linear operators, density matrices, strictly positive density matrices and
Hermitian matrices on $\mathcal{H}$, respectively. Let $\mathbb{M}_{n}^{0}$
and $\mathbb{H}_{n}^{0}$ be the space of traceless elements of
$\mathbb{M}_{n}$ and $\mathbb{H}_{n}$, respectively. Denote the $n$-by-$n$
identity matrix by $\mathbb{I}_{n}$ (acting on $\mathcal{H}$); let
$\mathcal{I}_{n}$ be the identity operator acting on $\mathbb{M}_{n}$. If the
Hilbert space
$\mathcal{H}=\mathcal{H}_{1}\otimes\mathcal{H}_{2}\otimes\cdots\otimes\mathcal{H}_{N}$,
and $\mathcal{H}_{j}$ has the dimension $n_{j}$ (for $1\leq j\leq N$), then
the space of linear operators on $\mathcal{H}$ is denoted by
$\mathbb{M}_{n_{1}\times n_{2}\times\cdots\times n_{N}}$; the same convention
applies similarly to other spaces, e.g., $\mathbb{H}_{n_{1}\times
n_{2}\times\cdots\times n_{N}}$. As a reminder, following the above notation
convention, $\mathbb{H}^{0}_{n_{1}\times
n_{2}}\neq\mathbb{H}^{0}_{n_{1}}\otimes\mathbb{H}^{0}_{n_{2}}$.

Let $\left\langle\cdot,\cdot\right\rangle$ denote a generic inner product on
$\mathbb{M}_{n}$; the Hilbert-Schmidt inner product is defined as
$\left\langle
A,B\right\rangle_{HS}:=\operatorname{Tr}\left(A^{\dagger}B\right)$. For any
positive semidefinite operator $\mathscr{T}$ on $\mathbb{M}_{n}$, define the
sesquilinear form $\left\langle A,B\right\rangle_{\mathscr{T}}:=\left\langle
A,\mathscr{T}(B)\right\rangle_{HS}$ and the semi-norm $\left\lVert
A\right\rVert_{\mathscr{T}}:=\sqrt{\left\langle
A,A\right\rangle_{\mathscr{T}}}$ for all $A,\ B\in\mathbb{M}_{n}$; when
$\mathscr{T}$ is strictly positive, the sesquilinear form becomes an inner
product and the semi-norm becomes a norm.

For convenience, for any $A,B\in\mathbb{M}_{n}$, we denote

$\displaystyle A\\#B:=L_{A}R_{B},$ (6)

where $L_{A}$ and $R_{B}$ are left and right multiplication of $A$ and $B$,
respectively; in other words, $(A\\#B)(X)=AXB$.

###  2.1 Quantum $\chi^{2}_{\kappa}$ divergences

Throughout this work, we consider the quantum $\chi_{\kappa}^{2}$ divergence,
introduced in [21, Def. 1]. Let us introduce a set $\mathcal{K}$,

$\displaystyle\mathcal{K}:=\left\\{\kappa:(0,\infty)\rightarrow(0,\infty)|-\kappa\text{
is operator monotone},\kappa(1)=1,x\kappa(x)=\kappa(x^{-1})\right\\}.$ (7)

As a remark, it is easy to check that $\kappa_{1/2}(x)\equiv x^{-1/2}$ is in
the family $\mathcal{K}$.

######  Definition 2 (Quantum $\chi^{2}_{\kappa}$ divergence).

For any $\kappa\in\mathcal{K}$, define the quantum $\chi^{2}_{\kappa}$
divergence between quantum states $\rho,\sigma\in\mathfrak{D}_{n}$ by

$\displaystyle\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right)$
$\displaystyle:=\left\langle\rho-\sigma,\Omega_{\sigma}^{\kappa}(\rho-\sigma)\right\rangle_{HS}$
(8)

when $\text{supp}(\rho)\subset\text{supp}(\sigma)$; otherwise, set
$\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right)=\infty$. The operator
$\Omega_{\sigma}^{\kappa}$ above is given by

$\displaystyle\Omega_{\sigma}^{\kappa}:=R_{\sigma}^{-1}\kappa(L_{\sigma}R_{\sigma}^{-1})\equiv
L_{\sigma}^{-1}\kappa(R_{\sigma}L_{\sigma}^{-1}).$ (9)

The second equality comes from the assumption that
$x\kappa(x)=\kappa(x^{-1})$. As a remark, when $\sigma$ is not a full-rank
density matrix, $\Omega_{\sigma}^{\kappa}$ can still be well-defined on the
support of $\sigma$.

Essentially, the operator $\Omega_{\sigma}^{\kappa}$ is a non-commutative way
to multiply $\sigma^{-1}$. Properties of the operator
$\Omega_{\sigma}^{\kappa}$ will be further discussed in § 2.3.

Next, let us introduce the non-commutative way to multiply $\sigma$. Define
the weight operator $\Gamma_{\sigma}$

$\displaystyle\Gamma_{\sigma}:=\sigma^{1/2}\\#\sigma^{1/2}.$ (10)

Note that the operator $\Gamma_{\sigma}$ is completely positive, with the
Kraus operator $\sigma^{1/2}$ and
$\Omega_{\sigma}^{\kappa_{1/2}}=(\Gamma_{\sigma})^{-1}$. For any
$\kappa\in\mathcal{K}$, let us define a generalization of the operator
$\Gamma_{\sigma}$

$\displaystyle\mho_{\sigma}^{\kappa}:=$ $\displaystyle
L_{\sigma}\kappa(L_{\sigma}R_{\sigma}^{-1})\equiv
L_{\sigma}\kappa(\sigma\\#\sigma^{-1})$ (11) $\displaystyle=$
$\displaystyle\Gamma_{\sigma}\circ\Omega_{\sigma}^{\kappa}\circ\Gamma_{\sigma}.$
(12)

Notice that $\mho_{\sigma}^{\kappa_{1/2}}=\Gamma_{\sigma}$.

###  2.2 Examples of $\kappa(x)$

In this subsection, we provide three examples of $\kappa$ such that
$\kappa\geq\kappa_{1/2}$ (satisfying one of the conditions in Theorem 1). More
examples can be found in [9, Sec. 4.2] and [8, Sec. (III)].

######  Example 3 (Quantum $\chi^{2}_{\kappa_{\alpha}}$ divergence).

An important family of the quantum $\chi^{2}_{\kappa}$ divergence is the
quantum $\chi^{2}_{\kappa_{\alpha}}$ divergence, with the parameter
$\alpha\in[0,1]$ and

$\displaystyle\kappa_{\alpha}(x)=\frac{1}{2}\left(x^{-\alpha}+x^{\alpha-1}\right).$
(13)

  1. (i)

The case $\alpha=1/2$ is very special: $\kappa_{1/2}(x)=x^{-1/2}$ and
$\Omega_{\sigma}^{\kappa_{1/2}}={\sigma^{-1/2}}\\#{\sigma^{-1/2}}$ is
completely positive with the Kraus operator $\sigma^{-1/2}$. In fact,
$\kappa_{1/2}$ is the only one in $\mathcal{K}$ such that for any $\sigma$,
both $\Omega_{\sigma}^{\kappa}$ and
$\left(\Omega_{\sigma}^{\kappa}\right)^{-1}$ are completely positive [9,
Theorem 3.5].

  2. (ii)

We can immediately verify that $\kappa_{\alpha}=\kappa_{1-\alpha}$ and for any
fixed $x\in(0,\infty)$, $\kappa_{\alpha}(x)$ is monotonically decreasing with
respect to $\alpha\in[0,1/2]$; thus $\kappa_{\alpha}(x)\geq\kappa_{1/2}(x)$.

More results about this family of the quantum $\chi^{2}_{\kappa_{\alpha}}$
divergence (also called _mean $\alpha$-divergence_) could be found in [21].

######  Example 4 (Wigner-Yanase-Dyson).

Another family of $\kappa^{WYD}_{\beta}$ (see e.g., [8, 9]) corresponds to the
Wigner-Yanase-Dyson metric, and it is parameterized by $\beta\in[-1,2]$,

$\kappa^{WYD}_{\beta}(x):=\frac{1}{\beta(1-\beta)}\frac{(1-x^{\beta})(1-x^{1-\beta})}{(1-x)^{2}},\qquad
x\in(0,1)\cup(1,\infty).$ (14)

When $x=1$, ${\kappa}^{WYD}_{\beta}(x)$ is simply set as $1$ or is defined by
taking the limit $x\rightarrow 1$ in the above equation. In general, finding
all possible $\beta\in[-1,2]$ such that $\kappa^{WYD}_{\beta}\geq\kappa_{1/2}$
seems to be slightly technical; however, at least, for a few special choices
of $\beta$, e.g., when $\beta=1.5$
($\kappa^{WYD}_{1.5}(x)\equiv\kappa_{1/2}(x)+\frac{x^{-1/2}(\sqrt{x}-1)^{4})}{3(1-x)^{2}}$)
and $\beta=2$ ($\kappa^{WYD}_{2}(x)\equiv\frac{1+x}{2x}$) we can easily check
that ${\kappa}^{WYD}_{\beta}\geq\kappa_{1/2}$ for these two cases.

######  Example 5 (The largest possible $\kappa$).

The largest $\kappa\in\mathcal{K}$ is $\kappa_{\max}:=\frac{1+x}{2x}$ (see
e.g., [8, Eq. (11)]). It is obvious that $\kappa_{\max}\geq\kappa_{1/2}$.

As a remark, ${\kappa}^{WYD}_{2}$ in the family of Wigner-Yanase-Dyson metric
is exactly the maximum one.

###  2.3 Basic properties of operators $\Omega_{\sigma}^{\kappa}$ and
$\mho_{\sigma}^{\kappa}$

We list without proof some elementary while useful properties of the operator
$\Omega_{\sigma}^{\kappa}$. Recall the assumption that
$x\kappa(x)=\kappa(x^{-1})$, which is used below in the proof of
$\Omega_{\sigma}^{\kappa}$ being Hermitian-preserving.

######  Lemma 6.

Suppose $\sigma\in\mathfrak{D}^{+}_{n}$ and its eigenvalue decomposition
$\sigma=\sum_{j=1}^{n}s_{j}\left\lvert s_{j}\right\rangle\left\langle
s_{j}\right\rvert$. Then

  1. (i)

The operator $\Omega_{\sigma}^{\kappa}$ can be decomposed as

$\displaystyle\Omega_{\sigma}^{\kappa}=\sum_{j,m=1}^{n}\kappa\left(\frac{s_{j}}{s_{m}}\right)\frac{1}{s_{m}}\
{\left\lvert s_{j}\right\rangle\left\langle s_{j}\right\rvert}\\#{\left\lvert
s_{m}\right\rangle\left\langle s_{m}\right\rvert}.$ (15)

For any Hermitian matrix $A\in\mathbb{M}_{n}$,

$\displaystyle\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}\equiv\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}=\sum_{j,m=1}^{n}\kappa\left(\frac{s_{j}}{s_{m}}\right)\frac{1}{s_{m}}\left\lvert\left\langle
s_{j}\right\rvert A\left\lvert s_{m}\right\rangle\right\rvert^{2}\geq 0.$ (16)

Thus, $\Omega_{\sigma}^{\kappa}$ is a strictly positive operator with respect
to the Hilbert-Schmidt inner product, and the inner product
$\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma}^{\kappa}}$ is well-
defined.

  2. (ii)

$\Omega_{\sigma}^{\kappa}$ is Hermitian-preserving.

  3. (iii)

We have $\Omega_{\sigma}^{\kappa}(\sigma)=\mathbb{I}_{n}$. Thus for any
$A\in\mathbb{M}_{n}$,

$\displaystyle\left\langle
A,\sigma\right\rangle_{\Omega_{\sigma}^{\kappa}}=\left\langle
A,\mathbb{I}_{n}\right\rangle_{HS},\qquad\left\langle\sigma,A\right\rangle_{\Omega_{\sigma}^{\kappa}}=\left\langle\mathbb{I}_{n},A\right\rangle_{HS}=\operatorname{Tr}(A).$
(17)

In particular, for any density matrix $\rho\in\mathfrak{D}_{n}$,
$\left\langle\rho,\sigma\right\rangle_{\Omega_{\sigma}^{\kappa}}=\left\langle\sigma,\rho\right\rangle_{\Omega_{\sigma}^{\kappa}}=1$.

Then let us consider the properties of $\Omega_{\sigma}^{\kappa}$ for a
composite system.

######  Lemma 7.

  1. (i)

Consider $\sigma_{1}\in\mathfrak{D}^{+}_{n_{1}}$ and
$\sigma_{2}\in\mathfrak{D}^{+}_{n_{2}}$. Then for any $A\in\mathbb{M}_{n_{1}}$
and $B\in\mathbb{M}_{n_{2}}$, we have

$\displaystyle\Omega_{\sigma_{1}\otimes\sigma_{2}}^{\kappa}(A\otimes\sigma_{2})=\Omega_{\sigma_{1}}^{\kappa}(A)\otimes\mathbb{I}_{n_{2}}\qquad\Omega_{\sigma_{1}\otimes\sigma_{2}}^{\kappa}(\sigma_{1}\otimes
B)=\mathbb{I}_{n_{1}}\otimes\Omega_{\sigma_{2}}^{\kappa}(B).$ (18)

  2. (ii)

$\kappa_{1/2}$ is the only one in $\mathcal{K}$ such that for all
$\sigma_{1}\in\mathfrak{D}^{+}_{n_{1}}$ and
$\sigma_{2}\in\mathfrak{D}^{+}_{n_{2}}$, we have

$\displaystyle\Omega_{\sigma_{1}\otimes\sigma_{2}}^{\kappa}=\Omega_{\sigma_{1}}^{\kappa}\otimes\Omega_{\sigma_{2}}^{\kappa}.$
(19)

###### Proof.

Let us decompose
$\sigma_{1}=\sum_{j=1}^{n_{1}}\lambda_{j}\left\lvert\psi_{j}\right\rangle\left\langle\psi_{j}\right\rvert$
and
$\sigma_{2}=\sum_{m=1}^{n_{2}}\mu_{m}\left\lvert\phi_{m}\right\rangle\left\langle\phi_{m}\right\rvert$,
then $\sigma_{1}\otimes\sigma_{2}$ has an eigenvalue decomposition
$\sigma_{1}\otimes\sigma_{2}=\sum_{j,m}\lambda_{j}\mu_{m}\left\lvert\psi_{j}\right\rangle\left\langle\psi_{j}\right\rvert\otimes\left\lvert\phi_{m}\right\rangle\left\langle\phi_{m}\right\rvert$.

  1. (i)

By the decomposition of the operator $\Omega^{k}_{(\cdot)}$ in (15),

$\displaystyle\Omega_{\sigma_{1}\otimes\sigma_{2}}^{\kappa}=\sum_{j_{1},j_{2},m_{1},m_{2}}\kappa\left(\frac{\lambda_{j_{1}}\mu_{m_{1}}}{\lambda_{j_{2}}\mu_{m_{2}}}\right)\frac{1}{\lambda_{j_{2}}\mu_{m_{2}}}\left(\left\lvert\psi_{j_{1}}\right\rangle\left\langle\psi_{j_{1}}\right\rvert\otimes\left\lvert\phi_{m_{1}}\right\rangle\left\langle\phi_{m_{1}}\right\rvert\right)\\#\left(\left\lvert\psi_{j_{2}}\right\rangle\left\langle\psi_{j_{2}}\right\rvert\otimes\left\lvert\phi_{m_{2}}\right\rangle\left\langle\phi_{m_{2}}\right\rvert\right).$

Then by direct calculation,

$\displaystyle\Omega_{\sigma_{1}\otimes\sigma_{2}}^{\kappa}(A\otimes\sigma_{2})$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},m_{1},m_{2}}\kappa\left(\frac{\lambda_{j_{1}}\mu_{m_{1}}}{\lambda_{j_{2}}\mu_{m_{2}}}\right)\frac{1}{\lambda_{j_{2}}\mu_{m_{2}}}\mu_{m_{1}}\delta_{m_{1},m_{2}}\left(\left\lvert\psi_{j_{1}}\right\rangle\left\langle\psi_{j_{1}}\right\rvert
A\left\lvert\psi_{j_{2}}\right\rangle\left\langle\psi_{j_{2}}\right\rvert\right)\otimes\left\lvert\phi_{m_{1}}\right\rangle\left\langle\phi_{m_{1}}\right\rvert$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},m_{1}}\kappa\left(\frac{\lambda_{j_{1}}}{\lambda_{j_{2}}}\right)\frac{1}{\lambda_{j_{2}}}\left(\left\lvert\psi_{j_{1}}\right\rangle\left\langle\psi_{j_{1}}\right\rvert
A\left\lvert\psi_{j_{2}}\right\rangle\left\langle\psi_{j_{2}}\right\rvert\right)\otimes\left\lvert\phi_{m_{1}}\right\rangle\left\langle\phi_{m_{1}}\right\rvert$
$\displaystyle=$ $\displaystyle\
\Omega_{\sigma_{1}}^{\kappa}(A)\otimes\mathbb{I}_{n_{2}}.$

The other case can be similarly proved.

  2. (ii)

When $\kappa=\kappa_{1/2}$, by the fact that
$\Omega_{\sigma}^{\kappa_{1/2}}=(\Gamma_{\sigma})^{-1}$, we can immediately
see the tensorization (19). As for the other direction, from the assumption
that (19) holds and after some straightforward simplification, one could
obtain that
$\kappa\left(\frac{\lambda_{j_{1}}\mu_{m_{1}}}{\lambda_{j_{2}}\mu_{m_{2}}}\right)=\kappa\left(\frac{\lambda_{j_{1}}}{\lambda_{j_{2}}}\right)\kappa\left(\frac{\mu_{m_{1}}}{\mu_{m_{2}}}\right)$,
for all indices $j_{1},j_{2},m_{1},m_{2}$. Since $\sigma_{1}$ and $\sigma_{2}$
are arbitrary density matrices, we have $\kappa(xy)=\kappa(x)\kappa(y)$ for
all $x,y>0$; in particular, $1=\kappa(1)=\kappa(x)\kappa(x^{-1})$. Since
$\kappa\in\mathcal{K}$, we also have $x\kappa(x)=\kappa(x^{-1})$, which leads
into $\kappa(x)=x^{-1/2}=\kappa_{1/2}$.

∎

Similarly, we list without proof the following properties of
$\mho_{\sigma}^{\kappa}$; all properties can be easily verified by the
definition of $\mho_{\sigma}^{\kappa}$ in (11).

######  Lemma 8 (Operator $\mho_{\sigma}^{\kappa}$).

Suppose $\sigma\in\mathfrak{D}^{+}_{n}$ and its eigenvalue decomposition
$\sigma=\sum_{j=1}^{n}s_{j}\left\lvert s_{j}\right\rangle\left\langle
s_{j}\right\rvert$. Then

  1. (i)

the operator $\mho_{\sigma}^{\kappa}$ for any $\kappa\in\mathcal{K}$ has a
decomposition

$\displaystyle\mho_{\sigma}^{\kappa}=\sum_{j,m}s_{j}\kappa\left(\frac{s_{j}}{s_{m}}\right)\
{\left\lvert s_{j}\right\rangle\left\langle s_{j}\right\rvert}\\#{\left\lvert
s_{m}\right\rangle\left\langle s_{m}\right\rvert},$ (20)

thus $\mho_{\sigma}^{\kappa}$ is strictly positive with respect to the
Hilbert-Schmidt inner product;

  2. (ii)

the operator $\mho_{\sigma}^{\kappa}$ is Hermitian-preserving;

  3. (iii)

$\mho_{\sigma}^{\kappa}(\mathbb{I}_{n})=\sigma$.

###  2.4 Eigenvalue formalism of SDPI constants

The eigenvalue formalism of the quantum contraction coefficient can be found
in e.g. [20, 12, 8]; the classical analogous result can be found in e.g., [5,
18]. In this subsection, we concisely present this formalism, for the sake of
completeness.

Let us consider the ratio in the SDPI constant.

$\displaystyle\frac{\chi^{2}_{\kappa}\left(\mathcal{E}(\rho)\mid\mid\mathcal{E}(\sigma)\right)}{\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right)}$
$\displaystyle=\frac{\left\langle\rho-\sigma,\mathcal{E}^{\dagger}\circ\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}(\rho-\sigma)\right\rangle_{HS}}{\left\langle\rho-\sigma,\Omega_{\sigma}^{\kappa}(\rho-\sigma)\right\rangle_{HS}}=\frac{\left\langle\rho-\sigma,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(\rho-\sigma)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle\rho-\sigma,\rho-\sigma\right\rangle_{\Omega_{\sigma}^{\kappa}}},$
(21)

where we introduce

$\begin{split}\Upsilon_{\mathcal{E},\sigma}^{\kappa}&:=\left(\Omega_{\sigma}^{\kappa}\right)^{-1}\circ\mathcal{E}^{\dagger}\circ\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}.\end{split}$
(22)

Here are some properties of the operator
$\Upsilon_{\mathcal{E},\sigma}^{\kappa}$.

######  Lemma 9.

Assume that $\sigma,\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$.

  1. (i)

The operator $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is positive semidefinite
with respect to the inner product
$\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma}^{\kappa}}$.

  2. (ii)

$\Upsilon_{\mathcal{E},\sigma}^{\kappa}(\sigma)=\sigma$.

  3. (iii)

$\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is Hermitian perserving.

  4. (iv)

For any $A\in\mathbb{M}_{n}$, we have

$\displaystyle\left\langle
A,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(A)\right\rangle_{\Omega_{\sigma}^{\kappa}}\leq\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}.$

Therefore, the eigenvalue of $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is
bounded above by $1$.

###### Proof.

Part (i) is obvious from (22) and Lemma 6 (i). Part (ii) can be verified
directly by Lemma 6 (iii) and the fact that $\mathcal{E}$ is trace-preserving
(or equivalently $\mathcal{E}^{\dagger}$ is unital). As for part (iii), since
the quantum channel $\mathcal{E}$ is completely positive, it is thus also
Hermitian-preserving; so is $\mathcal{E}^{\dagger}$. By Lemma 6 (ii),
$\Omega_{\sigma}^{\kappa}$ is Hermitian-preserving, thus so is
$\left(\Omega_{\sigma}^{\kappa}\right)^{-1}$. Finally, since the composition
of two Hermitian-preserving operators is also Hermitian-preserving, we
conclude that $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is Hermitian-
preserving. Part (iv) is essentially the data processing inequality; see e.g.
[12, Thm. II.14] and [21, Thm. 4] for the proof. ∎

Then

$\begin{split}\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::sdpi_const}}}}}{{=}}}&\sup_{\rho\in\mathfrak{D}_{n}:\
\rho\neq\sigma}\frac{\chi^{2}_{\kappa}\left(\mathcal{E}(\rho)\mid\mid\mathcal{E}(\sigma)\right)}{\chi^{2}_{\kappa}\left(\rho\mid\mid\sigma\right)}\\\
\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::sdpi_ratio}}}}}{{=}}}&\sup_{\rho\geq
0:\
\rho\neq\sigma,\operatorname{Tr}(\rho)=1}\frac{\left\langle\rho-\sigma,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(\rho-\sigma)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle\rho-\sigma,\rho-\sigma\right\rangle_{\Omega_{\sigma}^{\kappa}}}\\\
=&\sup_{A\in\mathbb{H}_{n}^{0}:\ A\neq 0}\frac{\left\langle
A,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(A)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}}.\end{split}$ (23)

As one might observe, the last equation is closely connected to the eigenvalue
formalism of the operator $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$, which is
stated in the following lemma.

######  Lemma 10.

For $\sigma\in\mathfrak{D}^{+}_{n}$ and $\kappa\in\mathcal{K}$ and for any
quantum channel $\mathcal{E}$ such that
$\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$, let
$\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa})$ be the second largest
eigenvalue of $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ (defined in (22)). Then

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa}).$
(24)

###### Proof.

Since $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is positive semidefinite with
respect to the inner product
$\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma}^{\kappa}}$ from Lemma 9
(i), it admits a spectral decomposition with
$\Upsilon_{\mathcal{E},\sigma}^{\kappa}(V_{j})=\theta_{j}V_{j},\
\theta_{j}\geq 0,$ where $j=1,2,\cdots,n^{2}$ and $\\{V_{j}\\}_{j=1}^{n^{2}}$
is an orthonormal basis in the Hilbert space
$\left(\mathbb{M}_{n},\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma}^{\kappa}}\right)$.
Note that $\sigma$ is always an eigenvector of
$\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ from Lemma 9 (ii); without loss of
generality, let $V_{1}=\sigma$ and $\theta_{1}=1$. By the orthogonality of
$\\{V_{j}\\}_{j}$, we know
$0=\left\langle\sigma,V_{j}\right\rangle_{\Omega_{\sigma}^{\kappa}}=\operatorname{Tr}(V_{j})$
for $j\geq 2$. By Lemma 9 (iv), $\theta_{j}\leq 1$ for all $1\leq j\leq
n^{2}$; thus without loss of generality, assume $\theta_{j}$ are listed in
descending order and hence
$\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa})=\theta_{2}$. By rewriting
$A=\sum_{j=2}^{n^{2}}c_{j}V_{j}$ in (23) where $c_{j}\in\mathbb{C}$, we
immediately know that
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\leq\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa})$.

By the fact that $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ is Hermitian-
preserving (see Lemma 9 (iii)), $V_{2}^{\dagger}$ is also an eigenvector
associated with the eigenvalue
$\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa})$. Then we choose
$A\in\mathbb{H}_{n}^{0}$ in (23) by $\frac{V_{2}+V_{2}^{\dagger}}{2}$ or
$\frac{V_{2}-V_{2}^{\dagger}}{2i}$. Note that such an $A$ is also an
eigenvector of $\Upsilon_{\mathcal{E},\sigma}^{\kappa}$ with the eigenvalue
$\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa})$. Then
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\geq\left\langle
A,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(A)\right\rangle_{\Omega_{\sigma}^{\kappa}}/{\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}}=\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa}).$
∎

###  2.5 Another variational formalism of SDPI constants

Recall the definition of the operator $\mho_{\sigma}^{\kappa}$ from (12). In
Lemma 11 below, we provide another variational characterization of the SDPI
constant; essentially, it follows from the connection between the eigenvalue
formalism (as discussed in the last subsection) and the corresponding singular
value formalism. Its classical version is well-known and can be found in e.g.
the proof of [18, Thm<EMAIL_ADDRESS>This idea for quantum $\chi^{2}_{\kappa}$
divergences has appeared implicitly in [21, Thm. 9]; however, we don’t assume
$\sigma$ to be the stationary state of the quantum channel herein, compared
with [21].

######  Lemma 11.

Assume that quantum states
$\sigma,\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$. For any
$\kappa\in\mathcal{K}$,

$\displaystyle\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}=\max_{F,G}\left\lvert\left\langle\mathscr{K}(F),G\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}\right\rvert,$
(25)

where the operator $\mathscr{K}$ is defined by

$\displaystyle\mathscr{K}:=\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mathcal{E}\circ\Gamma_{\sigma},$
(26)

and the maximum is taken over all $F,G\in\mathbb{M}_{n}$ such that

$\displaystyle\left\langle\mathbb{I}_{n},F\right\rangle_{\mho_{\sigma}^{\kappa}}=\left\langle\mathbb{I}_{n},G\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=0,\qquad\left\lVert
F\right\rVert_{\mho_{\sigma}^{\kappa}}=\left\lVert
G\right\rVert_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=1.$ (27)

###### Proof of Lemma 11.

First, we rewrite Lemma 10 in the language of the relative density (whose
classical analog is the Radon–Nikodym derivative); specifically, to get the
third equality below, $A$ is replaced by $\Gamma_{\sigma}(A)$. By Lemma 10,

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\sup_{A\in\mathbb{M}_{n}^{0},\
A\neq 0}\frac{\left\langle
A,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(A)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}}=\sup_{A\in\mathbb{M}_{n}^{0},\
A\neq
0}\frac{\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}(A)\right\rangle_{HS}}{\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}}$ $\displaystyle=$
$\displaystyle\sup_{A\neq 0,\
\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\sigma}^{\kappa}}=0}\frac{\left\langle\mathcal{E}\circ\Gamma_{\sigma}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}\circ\Gamma_{\sigma}(A)\right\rangle_{HS}}{\left\langle\Gamma_{\sigma}(A),\Omega_{\sigma}^{\kappa}\circ\Gamma_{\sigma}(A)\right\rangle_{HS}}\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::mho_op_kappa}}}}}{{=}}}\sup_{A\neq
0,\
\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\sigma}^{\kappa}}=0}\frac{\left\langle\mathscr{K}(A),\mho_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathscr{K}(A)\right\rangle_{HS}}{\left\langle
A,A\right\rangle_{\mho_{\sigma}^{\kappa}}}$ $\displaystyle=$
$\displaystyle\sup_{A\neq 0,\
\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\sigma}^{\kappa}}=0}\frac{\left\langle
A,\left(\mho_{\sigma}^{\kappa}\right)^{-1}\circ\mathscr{K}^{\dagger}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathscr{K}(A)\right\rangle_{\mho_{\sigma}^{\kappa}}}{\left\langle
A,A\right\rangle_{\mho_{\sigma}^{\kappa}}}.$

As for the operator $\mathscr{K}$, it can be straightforwardly checked that

  * •

$\mathscr{K}$ is completely positive and unital
($\mathscr{K}(\mathbb{I}_{n})=\mathbb{I}_{n}$).

  * •

$\mathscr{K}^{\dagger}=\Gamma_{\sigma}\circ\mathcal{E}^{\dagger}\circ\Gamma_{\mathcal{E}(\sigma)}^{-1}$
is completely positive, trace-preserving, and
$\mathscr{K}^{\dagger}(\mathcal{E}(\sigma))=\sigma$.

  * •

Consider the following two Hilbert spaces $\mathscr{H}_{1}$ and
$\mathscr{H}_{2}$,

$\displaystyle\mathscr{H}_{1}$
$\displaystyle:=\left\\{A\in\mathbb{M}_{n}:\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\sigma}^{\kappa}}=0\right\\},\text{
equipped with the inner product
}\left\langle\cdot,\cdot\right\rangle_{\mho_{\sigma}^{\kappa}};$
$\displaystyle\mathscr{H}_{2}$
$\displaystyle:=\left\\{A\in\mathbb{M}_{n}:\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=0\right\\},\text{
equipped with the inner product
}\left\langle\cdot,\cdot\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}.$

Then we can readily verify that $\mathscr{K}$ is an operator from
$\mathscr{H}_{1}$ to $\mathscr{H}_{2}$, i.e., if
$\left\langle\mathbb{I}_{n},A\right\rangle_{\mho_{\sigma}^{\kappa}}=0$, then
$\left\langle\mathbb{I}_{n},\mathscr{K}(A)\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=0$.
The dual operator of $\mathscr{K}$, denoted by $\widetilde{\mathscr{K}}$, maps
from $\mathscr{H}_{2}$ to $\mathscr{H}_{1}$ and it is explicitly given by
$\widetilde{\mathscr{K}}=\left(\mho_{\sigma}^{\kappa}\right)^{-1}\circ\mathscr{K}^{\dagger}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}$.

Then, we have

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\sup_{A\neq 0,\
A\in\mathscr{H}_{1}}\frac{\left\langle
A,\widetilde{\mathscr{K}}\circ\mathscr{K}(A)\right\rangle_{\mho_{\sigma}^{\kappa}}}{\left\langle
A,A\right\rangle_{\mho_{\sigma}^{\kappa}}}.$

Let us denote the SVD decomposition of $\widetilde{\mathscr{K}}$ by
$\widetilde{\mathscr{K}}(\cdot)=\sum_{j}a_{j}\phi_{j}\left\langle\varphi_{j},\cdot\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}$
where $a_{j}\geq 0$, $\\{\phi_{j}\\}_{j}$ and $\\{\varphi_{j}\\}_{j}$ are
orthonormal basis of $\mathscr{H}_{1}$ and $\mathscr{H}_{2}$ respectively.
Then, easily we know
$\mathscr{K}(\cdot)=\sum_{j}a_{j}\varphi_{j}\left\langle\phi_{j},\cdot\right\rangle_{\mho_{\sigma}^{\kappa}}$
and that
$\widetilde{\mathscr{K}}\circ\mathscr{K}(\cdot)=\sum_{j}a_{j}^{2}\phi_{j}\left\langle\phi_{j},\cdot\right\rangle_{\mho_{\sigma}^{\kappa}}$.
Then $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ is simply the largest
value of $a_{j}^{2}$; namely,
$\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}$ is the largest singular
value of $\mathscr{K}$, and the result in Lemma 11 follows immediately. ∎

###  2.6 Comparison of SDPI constants

First, we provide a uniform lower bound of
$\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}$ for any
$\kappa\in\mathcal{K}$ in terms of
$\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)$ in Lemma 12, which is a
new result to the best of our knowledge. One of our corollaries in (30) can
also be derived by [8, Thm. 4.4] and [8, Thm. 5.3]. However, our approach to
show (30) is different from [8]: their result comes from comparing the
contraction coefficient $\eta_{\chi^{2}_{\kappa}}(\mathcal{E})$ with
$\eta_{\operatorname{Tr}}(\mathcal{E})$ (the contraction coefficient for trace
norm); we use the SDPI constant of the Petz recovery map as the bridge.
Second, we consider quantum-classical (QC) channels and provide the ordering
of SDPI constants for different $\kappa$ in Lemma 14; similar results have
appeared in [8, Prop. 5.5] for contraction coefficients.

######  Lemma 12.

For any quantum channel $\mathcal{E}$ and quantum state
$\sigma\in\mathfrak{D}^{+}_{n}$ such that
$\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$, we have

$\displaystyle\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)\leq\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\eta_{\chi^{2}_{\kappa}}(\mathcal{R}_{\mathcal{E},\sigma},\mathcal{E}(\sigma))}\leq\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)},$
(28)

where $\mathcal{R}_{\mathcal{E},\sigma}$ is the Petz recovery map, defined by

$\displaystyle\mathcal{R}_{\mathcal{E},\sigma}(A):=\sigma^{1/2}\mathcal{E}^{\dagger}(\mathcal{E}(\sigma)^{-1/2}A\mathcal{E}(\sigma)^{-1/2})\sigma^{1/2}\equiv\Gamma_{\sigma}\circ\mathcal{E}^{\dagger}\circ\Gamma^{-1}_{\mathcal{E}(\sigma)}(A),\qquad\forall
A\in\mathbb{M}_{n}$ (29)

mapping $\mathcal{E}(\sigma)$ to $\sigma$.

The followings are immediate consequences of the lemma above.

######  Corollary 13.

Under the same assumption as in Lemma 12,

  1. (i)

The SDPI constant associated with $\kappa_{1/2}$ for the pair
$(\mathcal{E},\sigma)$ equals the SDPI constant for the recovery map pair
$(\mathcal{R}_{\mathcal{E},\sigma},\mathcal{E}(\sigma))$, that is to say,

$\displaystyle\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)=\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{R}_{\mathcal{E},\sigma},\mathcal{E}(\sigma)).$

  2. (ii)

Further assume that for any $\sigma\in\mathfrak{D}^{+}_{n}$, we have
$\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$. Then, for the contraction
coefficient of the quantum channel $\mathcal{E}$, we have

$\displaystyle\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E})\leq\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E})}.$
(30)

###### Proof.

The first part comes from letting $\kappa=\kappa_{1/2}$ in (28) and the fact
that the Petz recovery map of $\mathcal{R}_{\mathcal{E},\sigma}$ is exactly
the channel $\mathcal{E}$; the second part comes from taking the supremum over
all $\sigma\in\mathfrak{D}^{+}_{n}$. ∎

###### Proof of Lemma 12.

It is straightforward to verify that $\mathcal{R}_{\mathcal{E},\sigma}$,
defined in (29), is a bona-fide quantum channel, mapping the quantum state
$\mathcal{E}(\sigma)$ back to $\sigma$. We can easily verify by definition
(22) and (29) that

$\displaystyle\Upsilon_{\mathcal{E},\sigma}^{\kappa_{1/2}}=\mathcal{R}_{\mathcal{E},\sigma}\circ\mathcal{E}.$
(31)

Recall from Lemma 10 that there exists a
$\lambda_{2}\equiv\lambda_{2}(\Upsilon_{\mathcal{E},\sigma}^{\kappa_{1/2}})=\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)$
and a traceless Hermitian matrix $V\in\mathbb{H}^{0}_{n}$ such that
$\Upsilon_{\mathcal{E},\sigma}^{\kappa_{1/2}}(V)=\lambda_{2}V$. Let
$\widetilde{V}:=\mathcal{E}(V)\in\mathbb{H}_{n}^{0}$. Then

$\displaystyle\left(\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma)\right)^{2}$
$\displaystyle=\lambda_{2}^{2}=\frac{\left\langle\Upsilon_{\mathcal{E},\sigma}^{\kappa_{1/2}}(V),\Upsilon_{\mathcal{E},\sigma}^{\kappa_{1/2}}(V)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle
V,V\right\rangle_{\Omega_{\sigma}^{\kappa}}}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Upsilon_op_recovery_map}}}}}{{=}}}\frac{\left\langle\mathcal{R}_{\mathcal{E},\sigma}(\widetilde{V}),\Omega_{\sigma}^{\kappa}\circ\mathcal{R}_{\mathcal{E},\sigma}(\widetilde{V})\right\rangle_{HS}}{\left\langle
V,V\right\rangle_{\Omega_{\sigma}^{\kappa}}}$
$\displaystyle=\frac{\left\langle\mathcal{R}_{\mathcal{E},\sigma}(\widetilde{V}),\Omega_{\mathcal{R}_{\mathcal{E},\sigma}(\mathcal{E}(\sigma))}^{\kappa}\circ\mathcal{R}_{\mathcal{E},\sigma}(\widetilde{V})\right\rangle_{HS}}{\left\langle\widetilde{V},\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\widetilde{V})\right\rangle_{HS}}\frac{\left\langle\widetilde{V},\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\widetilde{V})\right\rangle_{HS}}{\left\langle
V,V\right\rangle_{\Omega_{\sigma}^{\kappa}}}$
$\displaystyle\leq\eta_{\chi^{2}_{\kappa}}\left(\mathcal{R}_{\mathcal{E},\sigma},\mathcal{E}(\sigma)\right)\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma).$

The inequality in the last step follows from Lemma 10. Hence, we have proved
the first inequality in (28); the second inequality follows immediately from
the data processing inequality of the quantum $\chi^{2}_{\kappa}$ divergence.
∎

Next, we consider any quantum-classical (QC) channel $\mathcal{E}$, which
refers to a physical process in which one first performs a measurement
according to a POVM $\\{F_{j}\\}_{j=1}^{n}$ ($F_{j}\in\mathbb{M}_{n}$ are
positive semidefinite and $\sum_{j=1}^{n}F_{j}=\mathbb{I}_{n}$); then based on
the measurement outcome, one prepares a pure state, selected from a set
$\\{\psi_{j}\\}_{j=1}^{n}$ which also forms an orthonormal basis of
$\mathcal{H}$. More specifically,

$\displaystyle\mathcal{E}(A)=\sum_{j=1}^{n}\operatorname{Tr}\left(F_{j}A\right)\left\lvert\psi_{j}\right\rangle\left\langle\psi_{j}\right\rvert,\qquad\forall
A\in\mathbb{M}_{n}.$ (32)

Define a ratio $\mathsf{R}_{\mathcal{E},\sigma}^{\kappa}$ on
$\mathbb{H}^{0}_{n}$ by

$\displaystyle\mathsf{R}_{\mathcal{E},\sigma}^{\kappa}(A):=\frac{\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}(A)\right\rangle_{HS}}{\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}}\equiv\frac{\left\langle
A,\Upsilon_{\mathcal{E},\sigma}^{\kappa}(A)\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle
A,A\right\rangle_{\Omega_{\sigma}^{\kappa}}},\qquad\text{for
}A\in\mathbb{H}^{0}_{n},\ A\neq 0.$ (33)

######  Lemma 14.

Suppose $\kappa\geq\kappa_{1/2}$, $\mathcal{E}$ is a QC channel with
$F_{j}\neq 0$ for all $1\leq j\leq n$ and $\sigma\in\mathfrak{D}^{+}_{n}$.
Then

$\displaystyle\mathsf{R}_{\mathcal{E},\sigma}^{\kappa}(A)\leq\mathsf{R}_{\mathcal{E},\sigma}^{\kappa_{1/2}}(A),\qquad\forall
A\in\mathbb{H}_{n}^{0},\ A\neq 0.$ (34)

Consequently, we have

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\leq\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E},\sigma).$
(35)

###### Proof.

By (32) and (16), we can readily calculate that

$\displaystyle\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\mathcal{E}(A))\right\rangle_{HS}=\sum_{j=1}^{n}\frac{\left\lvert\operatorname{Tr}(F_{j}A)\right\rvert^{2}}{\operatorname{Tr}(F_{j}\sigma)},$
(36)

which is independent of $\kappa$. By (16), it is straightforward to observe
that when $\kappa\geq\kappa_{1/2}$, one has $\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}\geq\left\langle
A,\Omega_{\sigma}^{\kappa_{1/2}}(A)\right\rangle_{HS}$. Thus (34) follows
immediately; (35) follows from (34) by taking the supremum over all non-zero
$A\in\mathbb{H}^{0}_{n}$ (see (23)). ∎

##  3 Proof of Theorem 1

Setting up: First notice that it is sufficient to prove Theorem 1 for $N=2$.
The general case can be straightforwardly proved by mathematical induction on
$N$. Next, for the case $N=2$, one direction is trivial: suppose $\rho_{1}$
achieves the maximum in
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1},\sigma_{1})$; let
$\rho_{1,2}=\rho_{1}\otimes\sigma_{2}$ and by direct calculation,

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1}\otimes\mathcal{E}_{2},\sigma_{1}\otimes\sigma_{2})$
$\displaystyle\geq\frac{\chi^{2}_{\kappa}\left(\mathcal{E}_{1}\otimes\mathcal{E}_{2}(\rho_{1,2})\mid\mid\mathcal{E}_{1}\otimes\mathcal{E}_{2}(\sigma_{1}\otimes\sigma_{2})\right)}{\chi^{2}_{\kappa}\left(\rho_{1,2}\mid\mid\sigma_{1}\otimes\sigma_{2}\right)}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Omega_op_composite}}}}}{{=}}}\frac{\chi^{2}_{\kappa}\left(\mathcal{E}_{1}(\rho_{1})\mid\mid\mathcal{E}_{1}(\sigma_{1})\right)}{\chi^{2}_{\kappa}\left(\rho_{1}\mid\mid\sigma_{1}\right)}=\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{1},\sigma_{1}\right).$

Similarly, by choosing $\rho_{1,2}=\sigma_{1}\otimes\rho_{2}$ where $\rho_{2}$
achieves the maximum in
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{2},\sigma_{2})$, we have
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1}\otimes\mathcal{E}_{2},\sigma_{1}\otimes\sigma_{2})\geq\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{2},\sigma_{2}\right)$.
Therefore,

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1}\otimes\mathcal{E}_{2},\sigma_{1}\otimes\sigma_{2})\geq\max\left(\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{1},\sigma_{1}\right),\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{2},\sigma_{2}\right)\right).$

In the below, we shall prove the other direction, i.e.,

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1}\otimes\mathcal{E}_{2},\sigma_{1}\otimes\sigma_{2})\leq\max\left(\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{1},\sigma_{1}\right),\eta_{\chi^{2}_{\kappa}}\left(\mathcal{E}_{2},\sigma_{2}\right)\right)=:\eta_{\max}^{\kappa}.$
(37)

Notations:

Since we fix states $\sigma_{m}$ and channels $\mathcal{E}_{m}$ for $m=1,2$
throughout this section, let us denote
$\Upsilon^{\kappa}_{m}\equiv\Upsilon_{\mathcal{E}_{m},\sigma_{m}}^{\kappa}$
for simplicity of notation. By Lemma 10, $\Upsilon_{m}^{\kappa}$ has an eigen-
basis $\\{V^{\kappa,m}_{j}\\}_{j=1}^{n_{m}^{2}}$ associated with eigenvalue
$\\{\theta^{\kappa,m}_{j}\\}_{j=1}^{n_{m}^{2}}$ with respect to the inner
product $\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma_{m}}^{\kappa}}$
such that

$\displaystyle\Upsilon^{\kappa}_{m}(V_{j}^{\kappa,m})=\theta_{j}^{\kappa,m}V_{j}^{\kappa,m},\qquad
1\leq j\leq n_{m}^{2},$

where $V^{\kappa,m}_{1}=\sigma_{m}$, $\theta_{1}^{\kappa,m}=1$ and
$V_{j}^{\kappa,m}$ are Hermitian for all $1\leq j\leq n_{m}^{2}$, since from
Lemma 9 $\Upsilon^{\kappa}_{m}$ are both Hermitian-preserving positive
semidefinite operators. In addition, we know from Lemma 10 (or say Lemma 9
(iv)) that for both $m=1,2$,

$\displaystyle\theta_{j}^{\kappa,m}\leq\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{m},\sigma_{m})\leq\eta_{\max}^{\kappa},\qquad\text{for
any }2\leq j\leq n_{m}^{2}.$

For convenience, let $\sigma=\sigma_{1}\otimes\sigma_{2}$ and
$\mathcal{E}=\mathcal{E}_{1}\otimes\mathcal{E}_{2}$; let
$\Upsilon^{\kappa}\equiv\Upsilon_{\mathcal{E},\sigma}^{\kappa}$. For any index
pair $\textbf{J}=(j_{1},j_{2})$, define

$\displaystyle V_{\textbf{J}}^{\kappa}:=V_{j_{1}}^{\kappa,1}\otimes
V_{j_{2}}^{\kappa,2},\qquad\theta_{\textbf{J}}^{\kappa}:=\theta_{j_{1}}^{\kappa,1}\theta_{j_{2}}^{\kappa,2}.$

Case (i@): For $\kappa=\kappa_{1/2}$ and any quantum channel. From Lemma 7
part (ii), $\Omega_{\sigma}^{\kappa}$ tensorizes, thus
$\Upsilon^{\kappa}=\Upsilon^{\kappa}_{1}\otimes\Upsilon_{2}^{\kappa}$. Next,
we can straightforwardly verify that
$\left\\{V_{\textbf{J}}^{\kappa}\right\\}_{\textbf{J}}$ (for
$\textbf{J}=(j_{1},j_{2})$) is an orthonormal eigenbasis of
$\Upsilon^{\kappa}$ with respect to the inner product
$\left\langle\cdot,\cdot\right\rangle_{\Omega_{\sigma}^{\kappa}}$, and the
associated eigenvalues are
$\left\\{\theta_{\textbf{J}}^{\kappa}\right\\}_{\textbf{J}}$. The largest
eigenvalue of $\Upsilon^{\kappa}$ on the domain
$\text{span}(\sigma)^{\perp}\equiv\mathbb{M}_{n_{1}\times n_{2}}^{0}$ becomes
$\max_{\textbf{J}\neq(1,1)}\\{\theta_{\textbf{J}}^{\kappa}\\}=\eta_{\max}^{\kappa}$.
Therefore, by Lemma 10, we have
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\max_{\textbf{J}\neq(1,1)}\\{\theta_{\textbf{J}}^{\kappa}\\}=\max\left(\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1},\sigma_{1}),\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{2},\sigma_{2})\right).$
Thus we complete the proof of (37) for the case $\kappa_{1/2}$.

Case (ii@): For $\kappa\geq\kappa_{1/2}$ and QC channels. Let us decompose
$\textbf{A}\in\mathbb{H}^{0}_{n_{1}\times n_{2}}$ by
$\textbf{A}=\sum_{\textbf{J}}c_{\textbf{J}}V_{\textbf{J}}^{\kappa}$ where
$c_{\textbf{J}}\in\mathbb{R}$. From the constraint that
$\operatorname{Tr}(\textbf{A})=0$, we know $c_{(1,1)}=0$. Thus, we can rewrite
A by

$\displaystyle\textbf{A}=\sigma_{1}\otimes
A_{2}+A_{1}\otimes\sigma_{2}+\widetilde{A},$ (38)

where

$\left\\{\begin{split}A_{2}&=\sum_{\textbf{J}:\ j_{1}=1,\ 2\leq j_{2}\leq
n_{2}^{2}}c_{\textbf{J}}V^{\kappa,2}_{j_{2}};\\\ A_{1}&=\sum_{\textbf{J}:\
2\leq j_{1}\leq n_{1}^{2},\ j_{2}=1}c_{\textbf{J}}V^{\kappa,1}_{j_{1}};\\\
\widetilde{A}&=\sum_{\textbf{J}:\ j_{1}\neq 1,\ j_{2}\neq
1}c_{\textbf{J}}V_{\textbf{J}}.\end{split}\right.$ (39)

To prove (37), by (23), it is equivalent to prove that for all
$\textbf{A}\in\mathbb{H}_{n_{1}\times n_{2}}^{0}$ and $\textbf{A}\neq 0$, we
have

$\displaystyle\frac{\left\langle\textbf{A},\Upsilon^{\kappa}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle\textbf{A},\textbf{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}}\leq\eta_{\max}^{\kappa}.$
(40)

The next lemma shows that it is sufficient to consider A as $\widetilde{A}$.

######  Lemma 15.

If (40) holds for any
$\textbf{A}\in\mathbb{H}_{n_{1}}^{0}\otimes\mathbb{H}_{n_{2}}^{0}$, then (40)
holds for any $\textbf{A}\in\mathbb{H}_{n_{1}\times n_{2}}^{0}$.

Notice that
$\mathbb{H}_{n_{1}}^{0}\otimes\mathbb{H}_{n_{2}}^{0}\subset\mathbb{H}_{n_{1}\times
n_{2}}^{0}$. The proof of this lemma is postponed to the end of this section
and let us continue to complete the proof of Theorem 1. It is straightforward
to verify that when $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ are QC channels,
$\mathcal{E}=\mathcal{E}_{1}\otimes\mathcal{E}_{2}$ is also a QC channel for
the composite system. By Lemma 14, for any
$\textbf{A}\in\mathbb{H}^{0}_{n_{1}}\otimes\mathbb{H}^{0}_{n_{2}}$, we have

$\displaystyle\frac{\left\langle\textbf{A},\Upsilon^{\kappa}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}}{\left\langle\textbf{A},\textbf{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::monotonicity_sdpiratio}}}}}{{\leq}}}\frac{\left\langle\textbf{A},\Upsilon^{\kappa_{1/2}}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{{\kappa_{1/2}}}}}{\left\langle\textbf{A},\textbf{A}\right\rangle_{\Omega_{\sigma}^{{\kappa_{1/2}}}}}$
$\displaystyle\leq\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E}_{1},\sigma_{1})\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E}_{2},\sigma_{2})$
$\displaystyle\leq\max\left(\mathopen{\big{missing}}(\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E}_{1},\sigma_{1})\mathclose{\big{missing}})^{2},\mathopen{\big{missing}}(\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E}_{2},\sigma_{2})\mathclose{\big{missing}})^{2}\right)$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::chialphahalf_less_chihalfsquare}}}}}{{\leq}}}\max\left(\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{1},\sigma_{1}),\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{2},\sigma_{2})\right)=\eta_{\max}^{\kappa}.$

The second inequality comes from the observation that
$\Upsilon^{\kappa_{1/2}}$ is a positive semidefinite operator on the space
$\mathbb{H}_{n_{1}}^{0}\otimes\mathbb{H}_{n_{2}}^{0}$ with eigenvalues
$\theta^{\kappa_{1/2}}_{\textbf{J}}$; for $j_{1}\neq 1$ and $j_{2}\neq 1$,
recall from previous results that
$\theta^{{\kappa_{1/2}}}_{\textbf{J}}=\theta^{{\kappa_{1/2}},1}_{j_{1}}\theta^{{\kappa_{1/2}},2}_{j_{2}}\leq\eta_{\chi^{2}_{\kappa_{1/2}}}(\mathcal{E}_{1},\sigma_{1})\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E}_{2},\sigma_{2})$.
The last equation means (40) holds for all Hermitian
$\textbf{A}\in\mathbb{H}_{n_{1}}^{0}\otimes\mathbb{H}_{n_{2}}^{0}$ and by
Lemma 15, (40) holds for all Hermitian $\textbf{A}\in\mathbb{H}_{n_{1}\times
n_{2}}^{0}$. This completes the proof of Theorem 1.

###### Proof of Lemma 15..

For any Hermitian A in (38), we claim that

$\displaystyle\begin{split}\left\langle\textbf{A},\Upsilon^{\kappa}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}=\left\langle\sigma_{1}\otimes
A_{2},\Upsilon^{\kappa}(\sigma_{1}\otimes
A_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}&+\left\langle
A_{1}\otimes\sigma_{2},\Upsilon^{\kappa}(A_{1}\otimes\sigma_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}\\\
&+\left\langle\widetilde{A},\Upsilon^{\kappa}(\widetilde{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}.\end{split}$
(41)

To prove this, we need to show that all cross product terms in the expansion
of
$\left\langle\textbf{A},\Upsilon^{\kappa}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}$
vanish. For instance, consider any
$\textbf{B}\in\mathbb{M}_{n_{1}}\otimes\mathbb{M}_{n_{2}}$,

$\displaystyle\left\langle\textbf{B},\Upsilon^{\kappa}(\sigma_{1}\otimes
A_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::op_Upsilon}}}}}{{=}}}$
$\displaystyle\left\langle\textbf{B},\mathcal{E}^{\dagger}\circ\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}(\sigma_{1}\otimes
A_{2})\right\rangle_{HS}$ $\displaystyle=$
$\displaystyle\left\langle\textbf{B},\mathcal{E}^{\dagger}\circ\Omega_{\mathcal{E}(\sigma)}^{\kappa}\left(\mathcal{E}_{1}(\sigma_{1})\otimes\mathcal{E}_{2}(A_{2})\right)\right\rangle_{HS}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Omega_op_composite}}}}}{{=}}}$
$\displaystyle\left\langle\textbf{B},\mathcal{E}^{\dagger}\left(\mathbb{I}_{n_{1}}\otimes\Omega_{\mathcal{E}_{2}(\sigma_{2})}^{\kappa}\circ\mathcal{E}_{2}(A_{2})\right)\right\rangle_{HS}$
$\displaystyle=$
$\displaystyle\left\langle\textbf{B},\mathbb{I}_{n_{1}}\otimes\left(\mathcal{E}_{2}^{\dagger}\circ\Omega_{\mathcal{E}_{2}(\sigma_{2})}^{\kappa}\circ\mathcal{E}_{2}(A_{2})\right)\right\rangle_{HS}.$

If $\textbf{B}=A_{1}\otimes\sigma_{2}$ or $\textbf{B}=\widetilde{A}$, by
plugging the expression of $A_{1}$ or $\widetilde{A}$ into the last equation
and after expanding all terms, it is straightforward to verify that
$\left\langle\textbf{B},\Upsilon^{\kappa}(\sigma_{1}\otimes
A_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}=0$ for both choices of B. We
can apply similar arguments to
$\left\langle\textbf{B},\Upsilon^{\kappa}(A_{1}\otimes\sigma_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}$
for $\textbf{B}=\sigma_{1}\otimes A_{2}$ or $\textbf{B}=\widetilde{A}$.
Similarly, we have (or let
$\mathcal{E}=\mathcal{I}_{n_{1}}\otimes\mathcal{I}_{n_{2}}$ in (41))

$\begin{split}\left\langle\textbf{A},\textbf{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}&=\left\langle\sigma_{1}\otimes
A_{2},\sigma_{1}\otimes
A_{2}\right\rangle_{\Omega_{\sigma}^{\kappa}}+\left\langle
A_{1}\otimes\sigma_{2},A_{1}\otimes\sigma_{2}\right\rangle_{\Omega_{\sigma}^{\kappa}}+\left\langle\widetilde{A},\widetilde{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}\\\
&\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Omega_op_composite}}}}}{{=}}}\left\langle
A_{2},A_{2}\right\rangle_{\Omega_{\sigma_{2}}^{\kappa}}+\left\langle
A_{1},A_{1}\right\rangle_{\Omega_{\sigma_{1}}^{\kappa}}+\left\langle\widetilde{A},\widetilde{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}.\end{split}$
(42)

Let us simplify the term on the right hand side of (41). For instance,

$\displaystyle\left\langle\sigma_{1}\otimes
A_{2},\Upsilon^{\kappa}(\sigma_{1}\otimes
A_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}$ $\displaystyle=$
$\displaystyle\left\langle\mathcal{E}(\sigma_{1}\otimes
A_{2}),\Omega_{\mathcal{E}(\sigma)}^{\kappa}\circ\mathcal{E}(\sigma_{1}\otimes
A_{2})\right\rangle_{HS}$
$\displaystyle\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Omega_op_composite}}}}}{{=}}}$
$\displaystyle\left\langle\mathcal{E}_{1}(\sigma_{1})\otimes\mathcal{E}_{2}(A_{2}),\mathbb{I}_{n_{1}}\otimes\Omega_{\mathcal{E}_{2}(\sigma_{2})}^{\kappa}\circ\mathcal{E}_{2}(A_{2})\right\rangle_{HS}$
$\displaystyle=$
$\displaystyle\left\langle\mathcal{E}_{2}(A_{2}),\Omega_{\mathcal{E}_{2}(\sigma_{2})}^{\kappa}\circ\mathcal{E}_{2}(A_{2})\right\rangle_{HS}$
$\displaystyle=$ $\displaystyle\left\langle
A_{2},\Upsilon^{\kappa}_{2}(A_{2})\right\rangle_{\Omega_{\sigma_{2}}^{\kappa}}$
$\displaystyle\leq$ $\displaystyle\
\eta_{\chi^{2}_{\kappa}}(\mathcal{E}_{2},\sigma_{2})\left\langle
A_{2},A_{2}\right\rangle_{\Omega_{\sigma_{2}}^{\kappa}}$ $\displaystyle\leq$
$\displaystyle\ \eta_{\max}^{\kappa}\left\langle
A_{2},A_{2}\right\rangle_{\Omega_{\sigma_{2}}^{\kappa}}.$

Similarly,

$\displaystyle\left\langle
A_{1}\otimes\sigma_{2},\Upsilon^{\kappa}(A_{1}\otimes\sigma_{2})\right\rangle_{\Omega_{\sigma}^{\kappa}}\leq\eta_{\max}^{\kappa}\left\langle
A_{1},A_{1}\right\rangle_{\Omega_{\sigma_{1}}^{\kappa}}.$

Therefore, we have

$\displaystyle\begin{aligned}
\left\langle\textbf{A},\Upsilon^{\kappa}(\textbf{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}\leq&\eta_{\max}^{\kappa}\left(\left\langle
A_{2},A_{2}\right\rangle_{\Omega_{\sigma_{2}}^{\kappa}}+\left\langle
A_{1},A_{1}\right\rangle_{\Omega_{\sigma_{1}}^{\kappa}}\right)+\left\langle\widetilde{A},\Upsilon^{\kappa}(\widetilde{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}.\end{aligned}$
(43)

By comparing (42) and (43), to prove (40), it is sufficient to show

$\displaystyle\left\langle\widetilde{A},\Upsilon^{\kappa}(\widetilde{A})\right\rangle_{\Omega_{\sigma}^{\kappa}}\leq\eta_{\max}^{\kappa}\left\langle\widetilde{A},\widetilde{A}\right\rangle_{\Omega_{\sigma}^{\kappa}}.$
(44)

Thus we complete the proof of Lemma 15.

∎

##  4 Connection to the quantum maximal correlation

The SDPI constant for the classical $\chi^{2}$ divergence is closely connected
to the classical maximal correlation (see e.g., [18, Theorem iii@.2]). In the
proposition below, we provide a quantum analog of this relation when
$\kappa=\kappa_{1/2}$.

To begin with, we need to define the quantum maximal correlation. This concept
was previously proposed and studied in [2]. Since there is a whole family of
quantum $\chi^{2}_{\kappa}$ divergences, it is natural to imagine that there
could also exist a whole family of quantum maximal correlations, as a
straightforward generalization of [2].

######  Definition 16 ($\kappa$-quantum maximal correlation).

Consider any fixed $\kappa\in\mathcal{K}$ and Hilbert spaces $\mathcal{H}_{1}$
and $\mathcal{H}_{2}$ with dimensions $n_{1}$ and $n_{2}$ respectively. For
any bipartite quantum state $\rho_{1,2}$ on the composite system
$\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, denote the reduced density matrices
by $\rho_{1}$ and $\rho_{2}$ respectively (i.e.,
$\operatorname{Tr}_{2}(\rho_{1,2})=\rho_{1}$,
$\operatorname{Tr}_{1}(\rho_{1,2})=\rho_{2}$). Define the $\kappa$-quantum
maximal correlation $\mu_{\kappa}(\rho_{1,2})$ by

$\displaystyle\mu_{\kappa}(\rho_{1,2}):=\max\left\lvert\operatorname{Tr}\left(\rho_{1,2}F\otimes
G^{\dagger}\right)\right\rvert,$ (45)

where the maximum is taken over all $F\in\mathbb{M}_{n_{1}}$,
$G\in\mathbb{M}_{n_{2}}$ such that

$\displaystyle\left\langle\mathbb{I}_{n_{1}},F\right\rangle_{\mho_{\rho_{1}}^{\kappa}}=\left\langle\mathbb{I}_{n_{2}},G\right\rangle_{\mho_{\rho_{2}}^{\kappa}}=0,\qquad\left\lVert
F\right\rVert_{\mho_{\rho_{1}}^{\kappa}}=\left\lVert
G\right\rVert_{\mho_{\rho_{2}}^{\kappa}}=1.$ (46)

Technically, when $\rho_{1}$ is not a full-rank density matrix, the notation
$\left\langle\cdot,\cdot\right\rangle_{\mho_{\rho_{1}}^{\kappa}}$ should be
understood as a sesquilinear form, as we explained at the beginning of § 2 and
the operator $\mho_{\rho_{1}}^{\kappa}$ is still well-defined on the support
of $\rho_{1}$ via (20). By Lemma 8, we easily verify that
$\left\langle\mathbb{I}_{n_{1}},F\right\rangle_{\mho_{\rho_{1}}^{\kappa}}\equiv\operatorname{Tr}(\rho_{1}F)$
and
$\left\langle\mathbb{I}_{n_{2}},G\right\rangle_{\mho_{\rho_{2}}^{\kappa}}\equiv\operatorname{Tr}(\rho_{2}G)$.
When $\kappa(x)=1$ is a constant function, we recover the quantum maximal
correlation defined in [2]; in this case,
$\mho_{\sigma}^{\kappa(x)=1}=L_{\sigma}$; however, notice that this choice of
$\kappa$ is not included in the set $\mathcal{K}$ and the corresponding
operator $\mho_{\sigma}^{\kappa(x)=1}$ is not Hermitian-preserving.

######  Lemma 17 (Invariance of the $\kappa$-quantum maximal correlation under
local isometries).

Suppose $U:\mathcal{H}_{1}\rightarrow\widetilde{\mathcal{H}}_{1}$ and
$V:\mathcal{H}_{2}\rightarrow\widetilde{\mathcal{H}}_{2}$ are two isometries
(i.e., $U^{\dagger}U=\mathbb{I}_{{\dim(\mathcal{H}_{1})}}$ and
$V^{\dagger}V=\mathbb{I}_{{\dim(\mathcal{H}_{2})}}$), where
$\dim(\mathcal{H}_{1})\leq\dim(\widetilde{\mathcal{H}}_{1})$ and
$\dim(\mathcal{H}_{2})\leq\dim(\widetilde{\mathcal{H}}_{2})$. For any
bipartite quantum state $\rho_{1,2}$ on
$\mathcal{H}_{1}\otimes\mathcal{H}_{2}$, define
$\widetilde{\rho}_{1,2}:=(U\otimes V)\rho(U\otimes V)^{\dagger}$. We have

$\displaystyle\mu_{\kappa}(\rho_{1,2})=\mu_{\kappa}(\widetilde{\rho}_{1,2}).$
(47)

###### Proof.

By definition,

$\displaystyle\mu_{\kappa}(\widetilde{\rho}_{1,2})$
$\displaystyle=\max_{\widetilde{F},\widetilde{G}}\left\lvert\operatorname{Tr}\left(\widetilde{\rho}_{1,2}\
\widetilde{F}\otimes\widetilde{G}^{\dagger}\right)\right\rvert$
$\displaystyle=\max_{\widetilde{F},\widetilde{G}}\left\lvert\operatorname{Tr}\left(\rho_{1,2}\
(U^{\dagger}\widetilde{F}U)\otimes(V^{\dagger}\widetilde{G}^{\dagger}V)\right)\right\rvert=\max_{\widetilde{F},\widetilde{G}}\left\lvert\operatorname{Tr}\left(\rho_{1,2}F\otimes
G^{\dagger}\right)\right\rvert,$

where we define $F:=U^{\dagger}\widetilde{F}U$ and
$G:=V^{\dagger}\widetilde{G}V$. Denote the reduced density matrices of
$\rho_{1,2}$ as $\rho_{1}$ and $\rho_{2}$ respectively. Then the reduced
density matrices of $\widetilde{\rho}_{1,2}$ are given by
$\widetilde{\rho}_{1}:=U\rho_{1}U^{\dagger}$ and
$\widetilde{\rho}_{2}:=V\rho_{2}V^{\dagger}$ respectively. From (46), the
condition in the maximization is given by

$\displaystyle\operatorname{Tr}\left(\widetilde{\rho}_{1}\widetilde{F}\right)=\operatorname{Tr}\left(\widetilde{\rho}_{2}\widetilde{G}\right)=0,\qquad\operatorname{Tr}\left(\widetilde{F}^{\dagger}\mho_{\widetilde{\rho}_{1}}^{\kappa}(\widetilde{F})\right)=\operatorname{Tr}\left(\widetilde{G}^{\dagger}\mho_{\widetilde{\rho}_{2}}^{\kappa}(\widetilde{G})\right)=1.$

By (20), it could be readily shown that
$\mho_{\widetilde{\rho}_{1}}^{\kappa}(\cdot)=(U\\#U^{\dagger})\circ\mho_{\rho_{1}}^{\kappa}\circ\left(U^{\dagger}\\#U\right)$
and similarly for $\mho_{\widetilde{\rho}_{2}}^{\kappa}(\cdot)$. As a remark,
in this case, $\widetilde{\rho}_{1}$ and $\widetilde{\rho}_{2}$ might not be
strictly positive, then the decomposition in (20) only considers eigenstates
with respect to non-zero eigenvalues (i.e.,
$\mho_{\widetilde{\rho}_{1}}^{\kappa}$ is only defined on the support of
$\widetilde{\rho}_{1}$). Then, with direct calculation, one could verify that
the above four conditions are equivalent to

$\displaystyle\operatorname{Tr}\left(\rho_{1}F\right)=\operatorname{Tr}\left(\rho_{2}G\right)=0,\qquad\operatorname{Tr}\left(F^{\dagger}\mho_{\rho_{1}}^{\kappa}(F)\right)=\operatorname{Tr}\left(G^{\dagger}\mho_{\rho_{2}}^{\kappa}(G)\right)=1.$

Therefore, we know
$\mu_{\kappa}(\widetilde{\rho}_{1,2})\leq\mu_{\kappa}(\rho_{1,2})$. Since
$\widetilde{F}$ is a linear operator on a higher-dimensional Hilbert space
$\widetilde{\mathcal{H}}_{1}$ than $F$ on $\mathcal{H}_{1}$, for any such $F$,
there exists $\widetilde{F}$ such that $U^{\dagger}\widetilde{F}U=F$
(similarly for $G$); therefore the equality can be achieved and
$\mu_{\kappa}(\widetilde{\rho}_{1,2})=\mu_{\kappa}(\rho_{1,2})$. ∎

######  Theorem 18.

For a Hilbert space $\mathcal{H}$ with dimension $n$, suppose
$\sigma\in\mathfrak{D}^{+}_{n}$ and $\mathcal{E}$ is any quantum channel on
$\mathcal{H}$ such that the quantum state
$\mathcal{E}(\sigma)\in\mathfrak{D}^{+}_{n}$. Thus, $\sigma$ has an eigenvalue
decomposition $\sigma=\sum_{j=1}^{n}s_{j}\left\lvert
s_{j}\right\rangle\left\langle s_{j}\right\rvert.$ For the choice
$\kappa=\kappa_{1/2}$,

$\displaystyle\sqrt{\eta_{\chi^{2}_{{\kappa_{1/2}}}}(\mathcal{E},\sigma)}=\mu_{\kappa_{1/2}}(\rho_{1,2}),$
(48)

where the bipartite quantum state
$\rho_{1,2}:=(\mathbb{I}_{n}\otimes\mathcal{E})\left(\left\lvert\psi\right\rangle\left\langle\psi\right\rvert\right)$
and the wave function $\left\lvert\psi\right\rangle$ is any purification of
$\sigma$ on the system $\mathcal{H}\otimes\mathcal{H}$.

Recall that a pure state $\left\lvert\psi\right\rangle$ on
$\mathcal{H}\otimes\mathcal{H}$ is a purification of $\sigma$ if
$\operatorname{Tr}_{1}\left(\left\lvert\psi\right\rangle\left\langle\psi\right\rvert\right)=\sigma$
(see [23, Chap. 5]). The canonical choice of the purification
$\left\lvert\psi\right\rangle$ of $\sigma$ is

$\displaystyle\left\lvert\psi_{\text{c}}\right\rangle:=\sum_{j=1}^{n}\sqrt{s_{j}}\left\lvert
s_{j},s_{j}\right\rangle.$ (49)

###### Proof.

In the first step, we prove it for the choice
$\left\lvert\psi\right\rangle=\left\lvert\psi_{c}\right\rangle$; in the second
step, we extend the result to the general purification.

_Step (i@)_. By Lemma 11, we have

$\displaystyle\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}=\max_{F,G}\left\lvert\left\langle\mathcal{E}\circ\Gamma_{\sigma}(F),\widetilde{G}\right\rangle_{HS}\right\rvert=\max_{F,G}\left\lvert\left\langle\widetilde{G},\mathcal{E}\circ\Gamma_{\sigma}(F)\right\rangle_{HS}\right\rvert,$

where
$\widetilde{G}:=\left(\Gamma_{\mathcal{E}(\sigma)}\right)^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}(G)$.
Let us decompose $\mathcal{E}\circ\Gamma_{\sigma}(F)$ based on the eigenstates
of $\sigma$,

$\displaystyle\mathcal{E}\circ\Gamma_{\sigma}(F)=\sum_{j,m=1}^{n}\sqrt{s_{m}s_{j}}\left\langle
s_{m}\right\rvert F\left\lvert s_{j}\right\rangle\mathcal{E}(\left\lvert
s_{m}\right\rangle\left\langle s_{j}\right\rvert).$

Hence,

$\displaystyle\sqrt{\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)}$
$\displaystyle=\max_{F,G}\left\lvert\sum_{j,m=1}^{n}\sqrt{s_{m}s_{j}}\left\langle
s_{m}\right\rvert F\left\lvert
s_{j}\right\rangle\left\langle\widetilde{G},\mathcal{E}(\left\lvert
s_{m}\right\rangle\left\langle
s_{j}\right\rvert)\right\rangle_{HS}\right\rvert$
$\displaystyle=\max_{F,G}\left\lvert\sum_{j,m=1}^{n}\sqrt{s_{m}s_{j}}\left\langle
s_{j}\right\rvert\widetilde{F}\left\lvert
s_{m}\right\rangle\left\langle\widetilde{G},\mathcal{E}(\left\lvert
s_{m}\right\rangle\left\langle
s_{j}\right\rvert)\right\rangle_{HS}\right\rvert$
$\displaystyle=\max_{F,G}\left\lvert\operatorname{Tr}\left(\rho_{1,2}\widetilde{F}\otimes\widetilde{G}^{\dagger}\right)\right\rvert,$

where $\widetilde{F}=F^{T}$ and the superscript $T$ means transpose with
respect to the eigenstates of $\sigma$, i.e., $\left\langle
s_{j}\right\rvert\widetilde{F}\left\lvert s_{m}\right\rangle:=\left\langle
s_{m}\right\rvert F\left\lvert s_{j}\right\rangle$ for all $1\leq j,m\leq n$.
The last equality above can be verified directly by
$\rho_{1,2}=(\mathcal{I}_{n}\otimes\mathcal{E})(\left\lvert\psi_{c}\right\rangle\left\langle\psi_{c}\right\rvert)$.

Notice that from Lemma 11, the maximum is taken over all $F,G$ given in (27).
Hence, to prove Theorem 18, it remains to verify that conditions (27) for $F$
and $G$ are equivalent to conditions (46) for $\widetilde{F}$ and
$\widetilde{G}$. More specifically, we need to verify the following four
relations.

  1. (i)

$\left\langle\mathbb{I}_{n},F\right\rangle_{\mho_{\sigma}^{\kappa}}=\left\langle\mathbb{I}_{n},\widetilde{F}\right\rangle_{\mho_{\sigma}^{\kappa}}$.
Note that

$\displaystyle\left\langle\mathbb{I}_{n},F\right\rangle_{\mho_{\sigma}^{\kappa}}=\operatorname{Tr}(\sigma
F)=\sum_{j}s_{j}\left\langle s_{j}\right\rvert F\left\lvert
s_{j}\right\rangle=\sum_{j}s_{j}\left\langle
s_{j}\right\rvert\widetilde{F}\left\lvert
s_{j}\right\rangle=\operatorname{Tr}(\sigma\widetilde{F})=\left\langle\mathbb{I}_{n},\widetilde{F}\right\rangle_{\mho_{\sigma}^{\kappa}}.$

  2. (ii)

$\left\lVert
F\right\rVert_{\mho_{\sigma}^{\kappa}}=\left\lVert\widetilde{F}\right\rVert_{\mho_{\sigma}^{\kappa}}$.
Note that

$\displaystyle\left\lVert F\right\rVert_{\mho_{\sigma}^{\kappa}}^{2}$
$\displaystyle=\operatorname{Tr}\left(F^{\dagger}\mho_{\sigma}^{\kappa}(F)\right)\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::mho_decomp}}}}}{{=}}}\
\sum_{j,m}s_{j}\kappa\left(\frac{s_{j}}{s_{m}}\right)\left\lvert\left\langle
s_{j}\right\rvert F\left\lvert s_{m}\right\rangle\right\rvert^{2}$
$\displaystyle=\sum_{j,m}s_{j}\kappa\left(\frac{s_{j}}{s_{m}}\right)\left\lvert\left\langle
s_{m}\right\rvert\widetilde{F}\left\lvert
s_{j}\right\rangle\right\rvert^{2}\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::set_K}}}}}{{=}}}\sum_{j,m}s_{m}\kappa\left(\frac{s_{m}}{s_{j}}\right)\left\lvert\left\langle
s_{m}\right\rvert\widetilde{F}\left\lvert s_{j}\right\rangle\right\rvert^{2}$
$\displaystyle=\operatorname{Tr}\left(\mathopen{\big{missing}}(\widetilde{F}\mathclose{\big{missing}})^{\dagger}\mho_{\sigma}^{\kappa}(\widetilde{F})\right)=\left\lVert\widetilde{F}\right\rVert^{2}_{\mho_{\sigma}^{\kappa}}.$

  3. (iii)

$\left\langle\mathbb{I}_{n},G\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=\left\langle\mathbb{I}_{n},\widetilde{G}\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}$.
Note that

$\displaystyle\left\langle\mathbb{I}_{n},\widetilde{G}\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}$
$\displaystyle=\left\langle\mathcal{E}(\sigma),\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}(G)\right\rangle_{HS}=\left\langle\mathcal{E}(\sigma),G\right\rangle_{HS}=\left\langle\mathbb{I}_{n},G\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}.$

  4. (iv)

$\left\lVert
G\right\rVert_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}=\left\lVert\widetilde{G}\right\rVert_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}$.
Note that

$\displaystyle\left\lVert\widetilde{G}\right\rVert_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}^{2}$
$\displaystyle=\left\langle\widetilde{G},\mho_{\mathcal{E}(\sigma)}^{\kappa}\left(\widetilde{G}\right)\right\rangle_{HS}=\left\langle\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}(G),\mho_{\mathcal{E}(\sigma)}^{\kappa}\circ\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}(G)\right\rangle_{HS}$
$\displaystyle=\left\langle
G,\left(\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}\right)^{2}(G)\right\rangle_{\mho_{\mathcal{E}(\sigma)}^{\kappa}}.$

When $\kappa=\kappa_{1/2}$,
$\Gamma_{\mathcal{E}(\sigma)}^{-1}\circ\mho_{\mathcal{E}(\sigma)}^{\kappa}=\mathcal{I}_{n}$.
Thus the relation holds for this special choice of $\kappa$ and this is the
only place we employ this assumption.

_Step (ii@):_ We then extend the result from the canonical purification
$\left\lvert\psi_{c}\right\rangle$ to any purification
$\left\lvert\psi\right\rangle$ on the bipartite quantum system
$\mathcal{H}\otimes\mathcal{H}$. By [23, Theorem 5.1.1], there exists a
unitary (thus also isometry) $U:\mathcal{H}\rightarrow\mathcal{H}$ such that
$\left\lvert\psi\right\rangle=U\otimes\mathbb{I}_{n}\left\lvert\psi_{c}\right\rangle$.
Hence,
$(\mathcal{I}_{n}\otimes\mathcal{E})(\left\lvert\psi\right\rangle\left\langle\psi\right\rvert)=(U\otimes\mathbb{I}_{n})\left((\mathcal{I}_{n}\otimes\mathcal{E})(\left\lvert\psi_{c}\right\rangle\left\langle\psi_{c}\right\rvert)\right)(U\otimes\mathbb{I}_{n})^{\dagger}$.
By Lemma 17, the conclusion follows immediately. ∎

##  5 SDPI constants for special qubit channels

In this section, we will illustrate the dependence of SDPI constants on the
reference state $\sigma$ and the weight function $\kappa$, for several special
qubit channels. The dependence on $\sigma$ is one major difference between the
quantum SDPI framework and the quantum contraction coefficient approach. The
dependence on $\kappa$ is one major difference between the quantum SDPI
framework and its classical version: all quantum $\chi^{2}_{\kappa}$
divergences coincide for classical states $\rho$ and $\sigma$ (i.e., $\rho$
and $\sigma$ commute) and simply reduce to the classical $\chi^{2}$
divergence; in particular, classical $\chi^{2}$ divergence, as well as the
associated classical SDPI constant, does not depend on $\kappa$; however, the
SDPI constant for quantum $\chi^{2}_{\kappa}$ divergences might fluctuate
significantly between approximately $0$ and $1$ for various $\kappa$, in a
special example that we provide below.

Three Pauli matrices are denoted by $\sigma_{X},\sigma_{Y},\sigma_{Z}$.
Without loss of generality, assume
$\sigma=\frac{1}{2}\left(\mathbb{I}_{2}+s\sigma_{Z}\right)=\mathopen{\big{missing}}[\begin{smallmatrix}(1+s)/2&0\\\
0&(1-s)/2\end{smallmatrix}\mathclose{\big{missing}}]$ with $s\in[0,1)$,
because one can always choose the eigenbasis of $\sigma$ as the computational
basis; of course, the matrix representation of the quantum channel is changed,
by choosing such a specific computational basis.

###  5.1 QC channel

By the expression of QC channel (32) and by (36), we have for any
$A\in\mathbb{H}_{2}^{0}$ that

$\displaystyle\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\mathcal{E}(A))\right\rangle_{HS}=\sum_{j=1}^{2}\frac{\left\lvert\operatorname{Tr}(F_{j}A)\right\rvert^{2}}{\operatorname{Tr}(F_{j}\sigma)}=\left\lvert\operatorname{Tr}(F_{1}A)\right\rvert^{2}\left(\frac{1}{\operatorname{Tr}(F_{1}\sigma)}+\frac{1}{1-\operatorname{Tr}(F_{1}\sigma)}\right).$

The second equality comes from the fact that $F_{2}=\mathbb{I}_{2}-F_{1}$ and
$\operatorname{Tr}(A)=0$. Let us decompose
$A=a_{x}\sigma_{X}+a_{y}\sigma_{Y}+a_{z}\sigma_{Z}$ and
$F_{1}=f_{0}\mathbb{I}_{2}+f_{x}\sigma_{X}+f_{y}\sigma_{Y}+f_{z}\sigma_{Z}$;
notice that all coefficients for $A$ and $F_{1}$ are real numbers. Next,
rewrite the above equation by

$\displaystyle\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\mathcal{E}(A))\right\rangle_{HS}=\frac{4}{\operatorname{Tr}(F_{1}\sigma)\mathopen{\big{missing}}(1-\operatorname{Tr}(F_{1}\sigma)\mathclose{\big{missing}})}\left(f_{x}a_{x}+f_{y}a_{y}+f_{z}a_{z}\right)^{2}.$

From (16), we also have

$\displaystyle\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}=c_{s}(a_{x}^{2}+a_{y}^{2})+\frac{4}{1-s^{2}}a_{z}^{2},$

where

$\displaystyle
c_{s}:=\kappa\left(\frac{1+s}{1-s}\right)\frac{2}{1-s}+\kappa\left(\frac{1-s}{1+s}\right)\frac{2}{1+s}\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::set_K}}}}}{{=}}}\frac{4}{1-s}\kappa\left(\frac{1+s}{1-s}\right)\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::set_K}}}}}{{=}}}\frac{4}{1+s}\kappa\left(\frac{1-s}{1+s}\right).$
(50)

By the Cauchy–Schwarz inequality and the fact that $1-s^{2}>0$ and $c_{s}>0$,
we have

$\displaystyle\left(f_{x}a_{x}+f_{y}a_{y}+f_{z}a_{z}\right)^{2}\leq\left(\frac{f_{x}^{2}}{c_{s}}+\frac{f_{y}^{2}}{c_{s}}+\frac{f_{z}^{2}}{4/(1-s^{2})}\right)\left(c_{s}(a_{x}^{2}+a_{y}^{2})+\frac{4}{1-s^{2}}a_{z}^{2}\right).$

Hence, we know that

$\displaystyle\begin{split}\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)&=\frac{4}{\operatorname{Tr}(F_{1}\sigma)\mathopen{\big{missing}}(1-\operatorname{Tr}(F_{1}\sigma)\mathclose{\big{missing}})}\left(\frac{f_{x}^{2}}{c_{s}}+\frac{f_{y}^{2}}{c_{s}}+\frac{f_{z}^{2}}{4/(1-s^{2})}\right)\\\
&=\frac{4}{(f_{0}+sf_{z})(1-f_{0}-sf_{z})}\left(\frac{f_{x}^{2}}{c_{s}}+\frac{f_{y}^{2}}{c_{s}}+\frac{f_{z}^{2}}{4/(1-s^{2})}\right).\end{split}$
(51)

As we can observe, the SDPI constant
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ depends on $\kappa$ and the
parameter $s$ in a complicated way; however, it does not depend on the choice
of pure states in the post-measurement preparation in (32). In the following,
let us consider a few special choices of the POVM
$\\{F_{1},\mathbb{I}_{2}-F_{1}\\}$.

#### (High dependence on $\sigma$, for the quantum implementation of BSC)

If $F_{1}=\begin{bmatrix}1-\epsilon&0\\\ 0&\epsilon\end{bmatrix}$ (thus
$F_{2}=\begin{bmatrix}\epsilon&0\\\ 0&1-\epsilon\end{bmatrix}$) with
$\epsilon\in[0,1]$, then the channel $\mathcal{E}$ is exactly a quantum
implementation of the binary symmetric channel with crossover probability
$\epsilon$ (or BSC($\epsilon$) in short). Easily, we know $f_{0}=\frac{1}{2}$,
$f_{z}=\frac{1-2\epsilon}{2}$, $f_{x}=f_{y}=0$ and thus the SDPI constant can
be simplified as

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=(1-2\epsilon)^{2}\frac{1-s^{2}}{1-(1-2\epsilon)^{2}s^{2}}\leq(1-2\epsilon)^{2}.$
(52)

Notice that the SDPI constant in this case is independent of the choice of
$\kappa$; the upper bound comes from the fact that $\epsilon\in[0,1]$. When we
further let $s=0$, i.e., the reference state $\sigma$ has the distribution
Bern($\frac{1}{2}$), the SDPI constant achieves the upper bound
$(1-2\epsilon)^{2}$, which recovers [18, Example<EMAIL_ADDRESS>In Figure 1, we show
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ with respect to the parameter
$s$ in $\sigma$, for fixed $\epsilon=0.05$; the high dependence of
$\eta_{\chi^{2}_{\kappa}}$ on $s$ (i.e., on $\sigma$) can be clearly seen, for
this particular case.

#### (High dependence on $\kappa$).

If $F_{1}=\frac{1}{2}\left(\mathbb{I}_{2}+\xi\sigma_{X}\right)$ with
$\xi\in[-1,1]$, then $f_{0}=\frac{1}{2}$, $f_{x}=\frac{\xi}{2}$ and
$f_{y}=f_{z}=0$. Hence,

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\frac{4\xi^{2}}{c_{s}}\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::qc_channel_c_s}}}}}{{=}}}\frac{(1+s)\xi^{2}}{\kappa(\frac{1-s}{1+s})}\leq\xi^{2}.$
(53)

The inequality comes from the fact for any $\kappa\in\mathcal{K}$, we have
$\kappa(x)\geq\kappa_{\min}(x)\equiv\frac{2}{1+x}$ (see [8, Eq. (11)]). As one
could observe, even for this simple example, the dependence of
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ on $s$ and $\kappa$ is
nonlinear and slightly complicated. Similarly, by the fact that for any
$\kappa\in\mathcal{K}$, we have
$\kappa(x)\leq\kappa_{\max}(x)\equiv\frac{1+x}{2x}$ (see [8, Eq. (11)]), one
could immediately show that

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)\geq(1-s^{2})\xi^{2}.$
(54)

Notice that both upper and lower bounds in the above can be achieved for some
$\kappa\in\mathcal{K}$. When $s\approx 1$ and $\xi\approx 1$, the largest
value of $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ is approximately $1$,
while the smallest value is approximately $0$, which illustrates the high
dependence of $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ on the choice of
$\kappa$, for this extreme case. In Figure 2, we visualize the SDPI constant
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ with respect to various choices
of $\kappa$, for $\xi=s=0.95$; the high dependence of
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ on $\kappa$ can be clearly
observed.

Figure 1: The SDPI constant $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$
with respect to $s$, for BSC(0.05).

(a) $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ for the family of
$\kappa_{\alpha}$ in (13).

(b) $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ for the family of
$\kappa_{\beta}^{WYD}$ in (14).

Figure 2: The SDPI constant $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$
with respect to various choices of $\kappa$ for the QC channel $\mathcal{E}$
with $F_{1}=\frac{1}{2}\left(\mathbb{I}_{2}+\xi\sigma_{X}\right)$, $\xi=0.95$,
and $s=0.95$.

###  5.2 Depolarizing channel

The depolarizing channel on a qubit has the following form

$\displaystyle\mathcal{E}(\rho)=\epsilon\rho+(1-\epsilon)\operatorname{Tr}(\rho)\frac{\mathbb{I}_{2}}{2},$

for $\epsilon\in[0,1]$. It refers to a physical process in which for a given
input state $\rho$, one prepares $\rho$ with probability $\epsilon$ and
prepares the maximal mixed state $\frac{\mathbb{I}_{2}}{2}$ with probability
$1-\epsilon$. Easily, we know that
$\mathcal{E}(\sigma)=\frac{\mathbb{I}_{2}}{2}+\frac{s\epsilon}{2}\sigma_{Z}$,
and $\mathcal{E}(A)=\epsilon A$ for any $A\in\mathbb{H}_{2}^{0}$. Hence,

$\displaystyle\frac{\left\langle\mathcal{E}(A),\Omega_{\mathcal{E}(\sigma)}^{\kappa}(\mathcal{E}(A))\right\rangle_{HS}}{\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}}$
$\displaystyle=\epsilon^{2}\frac{\left\langle
A,\Omega_{\mathcal{E}(\sigma)}^{\kappa}(A)\right\rangle_{HS}}{\left\langle
A,\Omega_{\sigma}^{\kappa}(A)\right\rangle_{HS}}\mathrel{\stackrel{{\scriptstyle\makebox[0.0pt]{\mbox{\tiny\eqref{eqn::Omega_decomp}}}}}{{=}}}\epsilon^{2}\frac{\frac{4}{1-s^{2}\epsilon^{2}}a_{z}^{2}+c_{s\epsilon}(a_{x}^{2}+a_{y}^{2})}{\frac{4}{1-s^{2}}a_{z}^{2}+c_{s}(a_{x}^{2}+a_{y}^{2})}$
$\displaystyle=\epsilon^{2}\left(\frac{1-s^{2}}{1-s^{2}\epsilon^{2}}+\frac{c_{s\epsilon}-\frac{1-s^{2}}{1-s^{2}\epsilon^{2}}c_{s}}{\frac{4}{1-s^{2}}\frac{a_{z}^{2}}{a_{x}^{2}+a_{y}^{2}}+c_{s}}\right).$

If $c_{s\epsilon}-\frac{1-s^{2}}{1-s^{2}\epsilon^{2}}c_{s}\geq 0$, then

$\displaystyle\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)=\epsilon^{2}\frac{c_{s\epsilon}}{c_{s}}=\epsilon^{2}\frac{\kappa\left(\frac{1+s\epsilon}{1-s\epsilon}\right)/(1-s\epsilon)}{\kappa\left(\frac{1+s}{1-s}\right)/(1-s)}.$

For fixed $s$ and $\epsilon$, $\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$
might be largely affected by $\kappa$ as well.

##  6 Conclusion and outlook

In this paper, we provide a partial solution to the problem of the
tensorization of SDPIs for quantum channels in Theorem 1. In addition, we
extend the connection between the SDPI constant for classical $\chi^{2}$
divergence and the maximal correlation to the quantum region in Theorem 18.
For a particular QC channel $\mathcal{E}$ and a special quantum state
$\sigma$, we observe an extreme scenario, in which the SDPI constant
$\eta_{\chi^{2}_{\kappa}}(\mathcal{E},\sigma)$ ranges approximately from $0$
to $1$ for different $\kappa\in\mathcal{K}$. This implies that choosing
different $\kappa$ might largely affect the rate of contraction of quantum
channels. Our numerical experiments (not presented in the paper) conducted for
both qubit (i.e., $n=2$) and qudit (with $n=3$) systems show that the
tensorization property (4) seems to hold for any quantum channel
$\mathcal{E}$, any reference state $\sigma\in\mathfrak{D}^{+}_{n}$ and at
least a few weight functions $\kappa$ being tested (e.g.,
$\kappa_{\min}\equiv\frac{2}{1+x}$, $\kappa_{\max}\equiv\frac{1+x}{2x}$ and
the family $\kappa_{\alpha}$ with $\alpha=\frac{1}{4}$ and
$\alpha=\frac{3}{4}$). Proving such tensorization properties is an interesting
future work.

Finally, let us comment on the potential generalization of our approach, as
well as the limitation. As one might observe, provided that one could show
(34), the tensorization of SDPIs is an immediate consequence. However, it
seems to be challenging to characterize the class of quantum channels that
satisfy (34) in general and this is the reason why we restrict to QC channels
and the case $\kappa\geq\kappa_{1/2}$ in Theorem 1. In terms of the validity
of (34), we notice that when $\kappa\leq\kappa_{1/2}$ (e.g., $\kappa_{\min}$),
(34) does not hold even for QC channels. As mentioned above, numerical
experiments seem to suggest that the tensorization also holds for
$\kappa_{\min}$. Therefore, further understanding of the properties of quantum
$\chi^{2}_{\kappa}$ divergences is needed to extend our results.

## Acknowledgment

This work is supported in part by the US National Science Foundation via
grants DMS-1454939 and CCF-1910571 and by the US Department of Energy via
grant DE-SC0019449. We thank Iman Marvian and Henry Pfister for helpful
discussions. Iman Marvian pointed out the possible generalization of Theorem
18 from the canonical purification to any general purification. Henry Pfister
introduced us to the topic of the strong data processing inequality for
classical noisy channels. We also thank anonymous referees for helpful
suggestions.

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