# The quantum switch is uniquely defined by its action on unitary operations

Qingxiuxiong Dong<EMAIL_ADDRESS>Department of Physics, Graduate
School of Science, The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan
Marco Túlio Quintino<EMAIL_ADDRESS>Institute for Quantum Optics and
Quantum Information (IQOQI), Austrian Academy of Sciences, Boltzmanngasse 3,
A-1090 Vienna, Austria  Vienna Center for Quantum Science and Technology
(VCQ), Faculty of Physics, University of Vienna, Boltzmanngasse 5, A-1090
Vienna, Austria  Department of Physics, Graduate School of Science, The
University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan  Akihito Soeda
<EMAIL_ADDRESS>Department of Physics, Graduate School of Science,
The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan  Mio Murao
<EMAIL_ADDRESS>Department of Physics, Graduate School of Science,
The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan  Trans-scale Quantum
Science Institute, The University of Tokyo, Bunkyo-ku, Tokyo 113-0033, Japan

###### Abstract

The quantum switch is a physical process that creates a coherent control
between different unitary operations which is often described as a process
which transforms a pair of unitary operations $(U_{1},U_{2})$ into a
controlled unitary operation that coherently applies them in different orders
as $|{0}\rangle\\!\langle{0}|\otimes
U_{1}U_{2}+|{1}\rangle\\!\langle{1}|\otimes U_{2}U_{1}$. This description,
however, does not directly define its action on non-unitary operations. The
action of quantum switch on non-unitary operations is then chosen to be a
“natural” extension of its action on unitary operation. Since, in general, the
action of a process on non-unitary operations is not uniquely determined by
its action on only unitary operations, in principle, there could be a set of
inequivalent extensions of quantum switch for non-unitary operations. In this
paper, we prove that there is a unique way to extend the actions of quantum
switch to non-unitary operations. In other words, contrary to the general
case, the action of quantum switch on non-unitary operations is completely
determined by its action on unitary operations. We also discuss the general
problem of when the complete description of a quantum process is uniquely
determined by its action on unitary operations and identify a set of single-
slot processes which are completely defined by their action on unitary
operations.

##  1 Introduction

The quantum switch [1] is a higher-order quantum operation [2, 3, 4, 5] that
transforms a pair of input unitary operations into a “coherent superposition
of different orders” of the input operations. It is an example of a higher-
order quantum operation that cannot be implemented within the standard quantum
circuit model, but there is no fundamental principle that forbids its
implementation in quantum mechanics. For that, quantum switch is a well-
studied higher-order quantum operation in the context of indefinite causal
order [1, 6, 7], for revealing how the causal structure may affect the quantum
information processing in quantum mechanics. It is also proposed as a resource
for discriminating non-signaling channels [8], quantum computation [9], and
quantum communication [10, 11, 12, 13, 14, 15]. Also, although we only
consider quantum switch that acts on a pair of input operations, the concept
of quantum switch can be generalized to multiple input operations as studied
in Ref. [9].

The action of quantum switch is to produce a coherent control of causal
orders. In Ref. [1], the action of quantum switch is originally defined on
unitary operations, and then generalized to quantum operations beyond unitary
operations. That is, given two unitary operations
$\widetilde{\mbox{$\mathcal{U}$}}_{1}$ and
$\widetilde{\mbox{$\mathcal{U}$}}_{2}$ characterized by the unitary operators
$U_{1}$ and $U_{2}$ as input operations111 In this work, we denote quantum
operations and maps with a tilde, and supermaps with a double tilde. In this
way a unitary operation $\widetilde{\mbox{$\mathcal{U}$}}$ described by the
operator $U$ is defined via $\widetilde{\mbox{$\mathcal{U}$}}(\rho)=U\rho
U^{\dagger}$., quantum switch transforms them into the coherent control of
different orders, namely, the unitary operation given by

$\displaystyle(U_{1},U_{2})\mapsto|{0}\rangle\\!\langle{0}|_{C}\otimes
U_{2}U_{1}+|{1}\rangle\\!\langle{1}|_{C}\otimes U_{1}U_{2},$ (1)

by using each of $U_{1}$ and $U_{2}$ only once. If the control qubit is in
$|{0}\rangle_{C}$, $U_{1}$ is applied on the target system first, and followed
by $U_{2}$, and if the control qubit is in $|{1}\rangle_{C}$, $U_{2}$ is
applied first followed by $U_{1}$. If the control qubit is in a superposition
of $|{0}\rangle$ and $|{1}\rangle$, then the two different orders should be
“superposed” in a coherent way.

The action of quantum switch for unitary operations is defined in a natural
way, but its action for general quantum operations is not trivial from the
definition for unitary operations. In the definition of quantum switch [1],
the action on general quantum operations is chosen to be the most “natural”
one based on its Kraus representation [16] as follows: given two quantum
operations represented by the Kraus operators $\\{K_{i}\\}_{i}$ and
$\\{L_{j}\\}_{j}$, respectively, the output operation of quantum switch is
given by the Kraus operators $\\{W_{ij}\\}_{ij}$ defined by

$\displaystyle W_{ij}:=|{0}\rangle\\!\langle{0}|_{C}\otimes
L_{j}K_{i}+|{1}\rangle\\!\langle{1}|_{C}\otimes K_{i}L_{j}.$ (2)

Following this definition, different results have been established. One
interesting result is that quantum switch can help in enhancing certain
communication tasks [11, 12, 13]. In particular, if two input operations are
the depolarizing channels, which do not transfer any information, the output
operation is not depolarizing channel anymore, and can be used for
transferring information. The action of quantum switch on the depolarizing
channels looks different from that on unitary operations, that is, coherently
controlled two differently causally ordered operations, since the composition
of two depolarizing channels is a depolarizing channel. The definition of the
quantum switch for general operations presented in Eq.(2) is widely used in
various studies, but before this paper, it was not clear if the definition in
Eq.(2) follows uniquely by the behavior of the quantum switch on unitary
operators described in Eq.(1).

Before proceeding we present a simple example to show that, in general, the
behavior of a quantum process for unitary operations does not uniquely define
its action on general operations. Our example is illustrated in terms of
quantum circuits shown in Fig. 1 and consists in presenting two processes
which despite being different, have identical action on unitary operations.
This example follows from the fact that since unitary operations are unital,
that is, $\widetilde{\mbox{$\mathcal{U}$}}(I)=I$, if
$\widetilde{\mbox{$\mathcal{D}$}}(\rho):=I/d\,{\mathrm{Tr}}\,\rho$ is the
completely depolarizing channel and $\widetilde{\mbox{$id$}}(\rho)=\rho$ is
the identity channel, we have that

$\displaystyle\widetilde{\mbox{$\mathcal{D}$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{D}$}}$
$\displaystyle=\widetilde{\mbox{$id$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{D}$}}=\widetilde{\mbox{$\mathcal{D}$}}.$
(3)

But, when considering non-unital channels $\widetilde{\mbox{$\Lambda$}}$, that
is $\widetilde{\mbox{$\Lambda$}}(I)\neq I$, the identity in Eq. (3) does not
hold. In particular, if we take the quantum operation
$\widetilde{\mbox{$\Lambda$}}(\rho)=|{0}\rangle\\!\langle{0}|\,{\mathrm{Tr}}\,\rho$
which replaces the input state $\rho$ by the quantum state
$|{0}\rangle\\!\langle{0}|$, we obtain

$\displaystyle\widetilde{\mbox{$\mathcal{D}$}}\circ\widetilde{\mbox{$\Lambda$}}\circ\widetilde{\mbox{$\mathcal{D}$}}=\widetilde{\mbox{$\mathcal{D}$}}$
(4)
$\displaystyle\widetilde{\mbox{$id$}}\circ\widetilde{\mbox{$\Lambda$}}\circ\widetilde{\mbox{$\mathcal{D}$}}=\widetilde{\mbox{$\Lambda$}}.$
(5)

Although simple, this example can be used to show that almost all higher-order
quantum operations share this property, namely, the action on unitary
operations does not uniquely define the action on non-unitary operations. In
particular, for any fixed circuit operation, we can construct two new circuits
by considering probabilistic mixtures of the fixed operations with the
identity channel or the depolarizing channel. For any non-trivial probability
mixture, the same argument used in our previous example shows that these two
new circuits may act identically on unitary operations, but they act
differently for non-unital operations. Moreover, this argument does not rely
on quantum circuits and can be used for any general process (potentially with
indefinite causal order [1]) which transforms quantum operations. We just need
to recognize that the two different circuits presented in Fig. 1 are valid
processes, and by performing any non-trivial convex combinations, one can
always construct examples of two higher-order transformations which the action
on unitary operations does not uniquely define the action on general.

$\forall$ all unitary operations $\widetilde{\mbox{$\mathcal{U}$}}$: but
$\exists$ non-unitary operations such as $\widetilde{\mbox{$\Lambda$}}$:

Figure 1: Two quantum circuits implementing higher-order quantum operations
that transform an arbitrary unitary operation into the depolarizing channel.
Here $\widetilde{\mbox{$\mathcal{U}$}}$ denotes the input unitary operation,
$\widetilde{\mbox{$\mathcal{D}$}}$ denotes the depolarizing channel, and
$\widetilde{\mbox{$id$}}$ denotes the identity channel. While the actions of
both of them are equivalent on unitary operations, their actions on some
quantum operations, e.g., the quantum operation
$\widetilde{\mbox{$\Lambda$}}(\rho)=|{0}\rangle\\!\langle{0}|\,{\mathrm{Tr}}\,\rho$
which replaces the input state $\rho$ by the quantum state
$|{0}\rangle\\!\langle{0}|$, are different.

Our main question then is, does there exist a higher-order quantum operation
which respects the constraints of the quantum switch on unitary operations
presented in Eq. (1), but does not respect the constraints of the quantum
switch for general operations presented in Eq. (2)?

It is worth mentioning that the difficulty of the problem follows the fact
that general quantum operations cannot be written as a linear combination of
unitary operations. For instance, a non-unital channel can never be written as
a linear combination of unitary operations. However, linearity is not the only
condition in which higher-order operations should respect. It is crucial to
respect a notion of complete positivity. That is, since quantum operations
should be completely positive (CP), it is inevitable to impose that processes
transforming quantum operations should be completely CP preserving [2, 3, 4,
11, 6, 7].

When dealing with the “lower-order case” where one considers quantum
operations which transform quantum states into quantum states, linearity
itself is enough to ensure that the action of quantum operations on pure
quantum states uniquely defines the action of this operation on any quantum
state. This holds because the linear space spanned by pure quantum states
forms the whole set of self-adjoint operators. An analogous problem then would
be to define the action of a quantum operation for a restricted set of pure
states and then analyze how the CP condition restricts the extension of this
operation to the set of all quantum states in a non-trivial way, problem which
is discussed in Refs. [17, 18, 19, 20]. We also remark Ref. [21] which
investigates when the marginals of a linear map uniquely determine its whole
action and Ref. [22] which analyses when compatible tuples of measurements
admit a unique joint measurement.

In this paper, we analyze how the action of higher-order processes on unitary
quantum operations restricts its action on general quantum operations. In
particular, we show that the constraints of linearity and completely CP
preserving, required by such quantum processes, ensure that the action of
quantum switch on general quantum operations uniquely defines their action on
unitary operations. We then consider the general problem of understanding the
restrictions imposed by the completely CP preserving property and apply our
proof techniques to problems which are not directly related to the quantum
switch. In particular, we identify a class of single slot processes which,
similarly to the switch, their action on unitary operations uniquely defines
their action on general operations. This class of single input-operation
processes includes identity supermaps, unitary transposition, and qubit
unitary complex conjugation. This result strengthens the theoretical
background for the studies on quantum switch, and provides a physical meaning
to the experimental realization of quantum switch because the assumptions we
pose are necessary to promise the action of quantum switch on general quantum
operations.

This paper is organized as follows. In Sec. 2, we provide the statement of our
main theorem. We also introduce the Choi representation, and state the main
theorem in terms of the Choi operators. In Sec. 3, we show the consequences
that can be obtained only from the linearity, which are also used in later
proofs. In Sec. 4, we provide a similar result on the supermaps with a single
input operation instead of quantum switch. The conclusion is given in Sec. 5.
In Appendix A, we provide the proofs for the results from the linearity stated
in Sec. 3. In Appendix B, we provide the proof of the result on the supermaps
with a single input operation stated in Sec. 4, and in Appendix C, we provide
the proof of the result on quantum switch stated in Sec. 2. We provide the
proof for quantum switch in the latter Appendix because while the technique
for both proofs is similar, the former one is much simpler. In Appendix D and
E, we provide the detailed calculations that are used in Appendix C.

##  2 The Uniqueness of Quantum Switch

In this paper, we derive the suitable action of quantum switch on arbitrary
quantum operations from its action on unitary operations. In particular, we
derive the action of quantum switch from the following three conditions:

  1. 1.

The action of quantum switch on unitary operations is given by Eq. (1).

  2. 2.

Quantum switch is a linear supermap.

  3. 3.

Quantum switch is a completely CP preserving supermap.

The precise conditions will be presented in the following of this section. The
second condition is equivalent to that quantum switch uses each input
operation once, and the third condition is a necessary condition for a
supermap to be (probabilistically) implementable [3, 4]. Note that since we do
not assume any other linear constraint known as the causal condition for
quantum switch, these conditions do not exclude the possibility of a
probabilistic action of quantum switch on certain input operations.

######  Theorem 1.

If the action of a 2-slot supermap on unitary operations is given by Eq. (1)
for arbitrary input unitary operation $U_{1}$ and $U_{2}$, and is
implementable by using each input operation once, then its action on arbitrary
quantum operations is uniquely determined by the Kraus operators Eq. (2).

Figure 2: Conceptual figure of a two-slot supermap which may not respect a
definite causal order such as the quantum switch. Two-slot supermaps transform
a pair of quantum operations as input and output a quantum operation from
${\mathcal{P}}$ to ${\mathcal{F}}$. In general, the action of a supermap on
unitary operations does not uniquely define its action on non-unitary
operations.

In order to prove Theorem 1, we use the Choi representation of supermaps [2,
3, 23, 24]. That is, since the Choi-Jamiołkowski isomorphism is a bijection,
we prove the uniqueness of the extension of quantum switch by proving that the
corresponding Choi operator $W$ is uniquely determined.

Here we first introduce the Choi representation of linear maps [23, 24], which
will be useful to represent quantum operations. Then we discuss how to
represent supermaps that describe higher-order quantum operations as linear
operators via the Choi representation [2, 4]. For a linear map
$\widetilde{\mbox{$\Lambda$}}:{\mathcal{L}}({\mathcal{I}})\to{\mathcal{L}}({\mathcal{O}})$,
the corresponding Choi operator is given by

$\displaystyle
J_{\mathcal{A}}:=\sum_{ij}|{i}\rangle\\!\langle{j}|\otimes\widetilde{\mbox{$\Lambda$}}(|{i}\rangle\\!\langle{j}|)\in{\mathcal{L}}({\mathcal{I}}\otimes{\mathcal{O}}),$
(6)

where $\\{|{i}\rangle\\}$ is an orthonormal basis for ${\mathcal{I}}$. The
condition that a map $\widetilde{\mbox{$\Lambda$}}$ is completely positive
(CP) corresponds to the positivity of $J_{\mathcal{A}}$ as
$J_{\mathcal{A}}\geq 0$. In this paper, since unitary operations play an
important role, we also denote a unitary operation
$\widetilde{\mbox{$\mathcal{U}$}}$ with the corresponding unitary operator
$U$, and its Choi operator as
$J_{U}:=\sum_{ij}|{i}\rangle\\!\langle{j}|\otimes\widetilde{\mbox{$\mathcal{U}$}}(|{i}\rangle\\!\langle{j}|)$.

A linear supermap, also simply referred as a supermap, is linear
transformations on maps and are useful to represent higher-order
transformations. The quantum switch has a particular structure of a two-slot
process hence its supermap may has the form of

$\small{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}:[{\mathcal{L}}({\mathcal{I}}_{1})\to{\mathcal{L}}({\mathcal{O}}_{1})\otimes{\mathcal{L}}({\mathcal{I}}_{2})\to{\mathcal{L}}({\mathcal{O}}_{2})]\to{\mathcal{L}}({\mathcal{P}})\to{\mathcal{L}}({\mathcal{F}}),$
(7)

where

$\displaystyle{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}(\alpha\widetilde{\mbox{$\mathcal{A}$}}_{1}\otimes\widetilde{\mbox{$\mathcal{A}$}}_{2}+\beta\widetilde{\mbox{$\mathcal{B}$}}_{1}\otimes\widetilde{\mbox{$\mathcal{B}$}}_{2})$
$\displaystyle\quad=\alpha{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}(\widetilde{\mbox{$\mathcal{A}$}}_{1}\otimes\widetilde{\mbox{$\mathcal{A}$}}_{2})+\beta{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}(\widetilde{\mbox{$\mathcal{B}$}}_{1}\otimes\widetilde{\mbox{$\mathcal{B}$}}_{2})$
(8)

for arbitrary input maps $\widetilde{\mbox{$\mathcal{A}$}}_{1}$ ,
$\widetilde{\mbox{$\mathcal{A}$}}_{2}$,
$\widetilde{\mbox{$\mathcal{B}$}}_{1}$, $\widetilde{\mbox{$\mathcal{B}$}}_{2}$
and arbitrary $\alpha,\beta\in\mathbb{C}$. In Eq. (7),
${\mathcal{I}}_{k},{\mathcal{O}}_{k}\cong\mathbb{C}^{d}$ for $k=1,2$ are the
Hilbert spaces of the two input operations,
${\mathcal{P}},{\mathcal{F}}\cong\mathbb{C}^{2d}$ denote the Hilbert spaces of
the output operation as Fig. 2 shows. Since linear maps can be represented by
operators with the Choi isomorphism, supermaps may be viewed as maps action on
Choi operators. Hence, by taking the Choi representation of this induced map,
we can also represent a supermap as in Eq. (7) by an operator
${W\in{\mathcal{L}}({\mathcal{P}}\otimes{\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{F}})}$,
also if $A\in{\mathcal{L}}({\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1})$ and
$B\in{\mathcal{L}}({\mathcal{I}}_{2}\otimes{\mathcal{O}}_{2})$ are the Choi
operators of quantum operations, the Choi operator of the output operation is
given by [2, 4, 1]

${\mathrm{Tr}}_{{\mathcal{I}}_{1}{\mathcal{O}}_{1}{\mathcal{I}}_{2}{\mathcal{O}}_{2}}\Big{[}W(I_{\mathcal{P}}\otimes
A_{{\mathcal{I}}_{1}{\mathcal{O}}_{1}}\otimes
B_{{\mathcal{I}}_{2}{\mathcal{O}}_{2}}\otimes I_{\mathcal{F}})^{t}\Big{]},$
(9)

where $t$ stands for transposition in the computational basis.

A supermap
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}:[{\mathcal{L}}({\mathcal{I}})\to{\mathcal{L}}({\mathcal{O}})]\to{\mathcal{L}}({\mathcal{P}})\to{\mathcal{L}}({\mathcal{F}}),$
is CP preserving if it transforms CP maps into CP maps. Also, a supermap is
completely CP preserving if all of its trivial extensions transforms CP maps
into CP maps, that is, for every CP linear map which may make use of an
auxiliary space
$\widetilde{\mbox{$\mathcal{C}$}}:{\mathcal{L}}({\mathcal{I}}\otimes\mathcal{H}_{\text{aux}})\to{\mathcal{L}}({\mathcal{O}}\otimes\mathcal{H}_{\text{aux}})$,
the output-map
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}\otimes{\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$id$}}$}}}}(\widetilde{\mbox{$\mathcal{C}$}})$
is also CP, where
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$id$}}$}}}}:{\mathcal{L}}(\mathcal{H}_{\text{aux}})\to{\mathcal{L}}(\mathcal{H}_{\text{aux}})$
is the identity supermap on the auxiliary space defined by
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$id$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$id$}}$}}}}(\widetilde{\mbox{$\mathcal{A}$}})=\widetilde{\mbox{$\mathcal{A}$}}$
for every map $\widetilde{\mbox{$\mathcal{A}$}}$. In Choi operator
representation completely CP preserving condition of
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{W}$}}$}}}}$
is simply given by the positivity constraint $W\geq 0$ [2, 4], similar to the
CP condition for quantum operation case in Choi representation. If, in
addition we impose that a supermap corresponds to a superchannel, that is, a
deterministic quantum supermap, extra linear constraints should be imposed.
For the case of general two-slot superchannels which may not have a definite
causal order these constraints constitute in imposing that the supermap
transforms pair of independent quantum channels into a quantum channel [1, 7,
6]. We note however that such linear constraints are not used in the proof of
the uniqueness of the extension of quantum switch. For an introduction on
deterministic quantum transformations we suggest the introductory sections of
Ref. [25] and [26] and references therein.

We denote the Choi operator of quantum switch defined by the Kraus operators
given in Ref. [1] as
$W_{0}\in{\mathcal{L}}({\mathcal{P}}\otimes{\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{F}})$.
In particular, this operator can be calculated as
$W_{0}=|{W_{0}}\rangle\\!\langle{W_{0}}|$ with

$\displaystyle|{W_{0}}\rangle$
$\displaystyle=|{00}\rangle_{{\mathcal{P}}_{C}{\mathcal{F}}_{C}}|{I}\rangle\\!\rangle_{{\mathcal{P}}_{T}{\mathcal{I}}_{1}}|{I}\rangle\\!\rangle_{{\mathcal{O}}_{1}{\mathcal{I}}_{2}}|{I}\rangle\\!\rangle_{{\mathcal{O}}_{2}{\mathcal{F}}_{T}}$
$\displaystyle\quad+|{11}\rangle_{{\mathcal{P}}_{C}{\mathcal{F}}_{C}}|{I}\rangle\\!\rangle_{{\mathcal{P}}_{T}{\mathcal{I}}_{2}}|{I}\rangle\\!\rangle_{{\mathcal{O}}_{2}{\mathcal{I}}_{1}}|{I}\rangle\\!\rangle_{{\mathcal{O}}_{1}{\mathcal{F}}_{T}}$
$\displaystyle=\sum_{i,j,k=0}^{d-1}(|{ijjkik00}\rangle+|{jkijik11}\rangle),$
(10)

where $|{I}\rangle\\!\rangle:=\sum_{i}|{ii}\rangle$ and the Hilbert spaces for
the vector in the last line is in the order of
${\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{O}}_{2}\otimes{\mathcal{P}}_{T}\otimes{\mathcal{F}}_{T}\otimes{\mathcal{P}}_{C}\otimes{\mathcal{F}}_{C}$.
Note that the normalization of the Choi operators is given by
${\mathrm{Tr}}W={\mathrm{Tr}}W_{0}=2d^{3}$, and each elements of the matrix
$W_{0}$ is 0 or 1.

In this paper, we show that the only possible $W$ satisfies $W=W_{0}$ under
the following assumptions: when input operations are unitary operations
denoted by $J=J_{U_{1}}\otimes J_{U_{2}}$, the action of quantum switch is
given by $W_{0}$ as222When we write ${\mathrm{Tr}}_{in}W_{0}J^{t}$, we are
suppressing the identity operators, that is, in the full notation we have
${{\mathrm{Tr}}_{in}\big{[}W_{0}(I_{\mathcal{P}}\otimes J^{t}\otimes
I_{\mathcal{F}})\big{]}}$. To avoid a heavy notation, whenever clear from
context, we will suppress the identity operator in these trace relations.
${\mathrm{Tr}}_{in}W_{0}J^{t}$, where
$\mathcal{H}_{in}:={\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{I}}_{2}$
stands for the total input-channel space of the switch; the linearity of the
supermap, which is automatically satisfied as the linearity is required when
the Choi operator is defined; and the positivity $W\geq 0$. In terms of Choi
representation, Theorem 1 is equivalent to the following Lemma.

######  Lemma 1.

Let $W_{0}=|{W_{0}}\rangle\\!\langle{W_{0}}|$ where $|{W_{0}}\rangle$ is
defined by Eq. (10). If there exists $W\geq 0$ such that
${\mathrm{Tr}}_{in}[W(J_{U_{1}}\otimes
J_{U_{2}})^{t}]={\mathrm{Tr}}_{in}[W_{0}(J_{U_{1}}\otimes J_{U_{2}})^{t}]$
holds for all ${U_{1}}$ and ${U_{2}}$, then $W=W_{0}$ holds.

The main proof of Lemma 1 is given in Appendix C. Here we provide a sketch of
the proof. We first consider the linear span of the Choi operator of unitary
operations, as the action of a supermap on this linear span is uniquely
determined due to the linearity. In Sec. 3 and Appendix A, we provide a set of
operators that are in the linear span. In Appendix C, we consider the
consequences from the positivity of the Choi operator $W\geq 0$, which also
relies on the results on the linear span. Remark that the Choi operator of
quantum switch given by Eq. (10) has a simple form, namely, it is a matrix
with a submatrix filled with only 1 and the remaining being 0. The proof can
be separated into two parts. In the first part, we determine all of the
diagonal elements which are 1, where the main calculations are separated into
Appendix D. After determining all of the diagonal elements, the remaining is
to show that the submatrix with the diagonal elements being 1 is filled with
1. In the second part of the proof, we evaluate the element-wise 1-norm, i.e.,
the sum of absolute values of all elements in the computational basis, and
show that it is equal to the number of elements of the submatrix, where the
main calculations are separated into Appendix E.

##  3 The Linear Span of the Choi Operators of Unitary Operations

While the positivity (of the Choi operator) is a key feature for seeking the
unique extension of quantum switch, we first focus on the linearity in this
section. From the linearity, the action on any input operations that is a
linear combination of unitary operations is given in the same way, that is,
for any $J\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$ where

$\displaystyle\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$
$\displaystyle\quad:=\\{O\mid
O=\sum_{i_{1},i_{2}}c_{i_{1}i_{2}}(J_{U_{i_{1}}}\otimes
J_{U_{i_{2}}}),c_{i_{1}i_{2}}\in\mathbb{C}\\},$ (11)

the action of quantum switch on $J$ is given by
${\mathrm{Tr}}_{in}W_{0}J^{t}$. For that, we consider the operators in
$\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$. Since
$\mathrm{span}\\{J_{U_{1}}\otimes
J_{U_{2}}\\}=\mathrm{span}\\{J_{U}\\}\otimes\mathrm{span}\\{J_{U}\\}$ holds,
it is enough to consider the operators in $\mathrm{span}\\{J_{U}\\}$.

We first summarize the results in this section: the following operators are in
$\mathrm{span}\\{J_{U}\\}$.

  1. 1.

$|{ii}\rangle\\!\langle{jj}|,|{ij}\rangle\\!\langle{ji}|\in\mathrm{span}\\{J_{U}\\}$
for $i\neq j$.

  2. 2.

$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$
when $i,j,i^{\prime},j^{\prime}$ are pairwise different.

  3. 3.

$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$ if
only two of $i,j,i^{\prime},j^{\prime}$ are the same and others are different.

  4. 4.

$\sum_{j=i+k}|{ij}\rangle\\!\langle{ij}|\in\mathrm{span}\\{J_{U}\\}$ for
$k=0,\ldots,d-1$.

  5. 5.

$(|{ij}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ji}|),(|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|)\in\mathrm{span}\\{J_{U}\\}$
for $i\neq j$.

The proof is given in Lemmas 3-7.

We remark that the set of the operators listed here does not span the whole
space of $\mathrm{span}\\{J_{U}\\}$ unless $d=2$. There are only
$2d(d-1)+d(d-1)(d-2)(d-3)+6d(d-1)(d-2)+d+2d(d-1)=d(d^{3}-3d+3)$ elements here,
but the dimension of $\mathrm{span}\\{J_{U}\\}$ is $(d^{2}-1)^{2}+1$ (Lemma 3
of Ref. [27]), which is strictly greater than $d(d^{3}-3d+3)$ for $d>2$. As we
show in Appendix C that Lemma 1 can be proved by only considering the action
of quantum switch on these operators, requiring the action of quantum switch
on all unitary operations is not necessary in general to uniquely determine
the action of quantum switch on general quantum operations. Instead, requiring
the action of quantum switch on a restricted set of unitary operations is
enough for uniquely determining the action on the rest of unitary operations
and general quantum operations.

Another important fact is that the action of quantum switch on the
depolarizing channels is determined only by the linearity, because

$\displaystyle\frac{I}{d}\otimes\frac{I}{d}\in\mathrm{span}\\{J_{U_{1}}\otimes
J_{U_{2}}\\}$ (12)

holds, where $\frac{I}{d}$ is the Choi operator of the depolarizing channel.
This is also obvious because the depolarizing channel is a probabilistic
mixture of uniformly random unitary operations. Thus, the results of Refs.
[11, 12, 13] can be obtained with only the assumption of the linearity. In
contrast, if we apply quantum switch to non-unital quantum operations, the
action may not be determined from only the linearity.

##  4 Uniqueness of single input supermaps

In this section, we provide a similar theorem on the supermaps with a single
input operation instead of quantum switch.

Consider the operations
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}$
which acts on as the identity supermap on unitary operations, that is, for
every $\widetilde{\mbox{$\mathcal{U}$}}$ we have
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}(\widetilde{\mbox{$\mathcal{U}$}})=\widetilde{\mbox{$\mathcal{U}$}}$.
We show that by imposing that
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}$
is completely CP preserving,
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}$
also behaves as the identity supermap for any arbitrary input operation
$\widetilde{\mbox{$\Lambda$}}$, that is,
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}(\widetilde{\mbox{$\Lambda$}})=\widetilde{\mbox{$\Lambda$}}$.
In other words, the Choi operator of
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}$
is uniquely determined by

$\displaystyle
C_{0}=\sum_{i,j,k,l=1}^{d}|{ij}\rangle\\!\langle{kl}|\otimes|{ij}\rangle\\!\langle{kl}|\in{\mathcal{L}}({\mathcal{I}}\otimes{\mathcal{O}}\otimes{\mathcal{P}}\otimes{\mathcal{F}}).$
(13)

######  Theorem 2.

If the action of a 1-slot supermap on unitary operations is given by
$\widetilde{\mbox{$\mathcal{U}$}}\mapsto\widetilde{\mbox{$\mathcal{U}$}}$ for
an arbitrary input unitary operation $\widetilde{\mbox{$\mathcal{U}$}}$, and
is implementable by using the input operation once, then its action on
arbitrary quantum operations is uniquely given by
$\widetilde{\mbox{$\Lambda$}}\mapsto\widetilde{\mbox{$\Lambda$}}$ for an
arbitrary input quantum operation $\widetilde{\mbox{$\Lambda$}}$.

Similarly, Theorem 2 is equivalent to the following Lemma.

######  Lemma 2.

Let $C_{0}$ be the Choi operator of the identity supermap defined by Eq. (13).
If there exists $C\geq 0$ such that
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C_{0}J_{U}^{t})$
holds for all ${U}$, then $C=C_{0}$ holds.

The proof of Lemma 2 is given in Appendix B.

While Theorem 2 only holds supermaps which act as the identity on unitary
operations, this result can be easily extended to cases to supermaps which act
on unitary operations as
$\widetilde{\mbox{$\mathcal{U}$}}\mapsto\widetilde{\mbox{$\mathcal{B}$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{A}$}}$,
where $\widetilde{\mbox{$\mathcal{A}$}}$ and
$\widetilde{\mbox{$\mathcal{B}$}}$ are fixed unitary operations. This class of
supermaps is precisely the set of reversibility preserving one-slot
superchannels (also known as one-partite pure processes [7]) which do not
modify the dimension of the input-operation [25].

######  Corollary 1.

If the action of a 1-slot supermap on unitary operations is given by
$\widetilde{\mbox{$\mathcal{U}$}}\mapsto\widetilde{\mbox{$\mathcal{B}$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{A}$}}$
for an arbitrary input unitary operation $\widetilde{\mbox{$\mathcal{U}$}}$
and two fixed unitary operations $\widetilde{\mbox{$\mathcal{A}$}}$ and
$\widetilde{\mbox{$\mathcal{B}$}}$, and is implementable by using the input
operation once, then its action on arbitrary quantum operations is uniquely
given by
$\widetilde{\mbox{$\Lambda$}}\mapsto\widetilde{\mbox{$\mathcal{B}$}}\circ\widetilde{\mbox{$\Lambda$}}\circ\widetilde{\mbox{$\mathcal{A}$}}$
for an arbitrary input quantum operation $\widetilde{\mbox{$\Lambda$}}$.

###### Proof.

Corollary 1 can be reduced to Theorem 2 as follows. Consider the condition for
unitary operations

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}J_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}_{0}J_{U}^{t}).$
(14)

By introducing $C=(A_{{\mathcal{I}}}\otimes
B_{{\mathcal{F}}})^{\dagger}C^{\prime}(A_{{\mathcal{I}}}\otimes
B_{{\mathcal{F}}})$, the l.h.s. can be written as

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}J_{U}^{t})$
$\displaystyle={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[(A_{\mathcal{I}}\otimes
B_{\mathcal{F}})C(A_{\mathcal{I}}\otimes B_{\mathcal{F}})^{\dagger}J_{U}^{t}]$
$\displaystyle=B_{\mathcal{F}}\\{{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(A_{\mathcal{I}}^{\dagger}J_{U}^{t}A_{\mathcal{I}})]\\}B_{\mathcal{F}}^{\dagger}$
$\displaystyle=B_{\mathcal{F}}\\{{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(J_{UA^{*}})^{t}]\\}B_{\mathcal{F}}^{\dagger},$
(15)

whereas the r.h.s. can be written as

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}_{0}J_{U}^{t})=B_{\mathcal{F}}\\{{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C_{0}(J_{UA^{*}})^{t}]\\}B_{\mathcal{F}}^{\dagger}.$
(16)

Since $UA^{*}$ is also a unitary operator and spans the whole linear span of
$\mathrm{span}\\{J_{U}\\}$, the uniqueness of $C^{\prime}$ is equivalent to
the uniqueness of the identity supermap $C$. ∎

Corollary 1 can be used to show that the process leading to qubit unitary
complex conjugation analyzed in [28] is uniquely defined by its action on
unitary operations. This holds true because
$\widetilde{\mbox{$\mathcal{U}$}}^{*}=\widetilde{\mbox{$\mathcal{Y}$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{Y}$}}$,
where $\widetilde{\mbox{$\mathcal{Y}$}}$ denotes the Pauli Y unitary
operation.

Another important supermap that can be proven to be unique is the unitary
transposition [29, 26]. Here it is interesting to remark that unitary
transposition cannot be implemented deterministic and its probability of
success is limited by $1/d^{2}$. Note however that our proofs only rely on
linearity and the fact that quantum supermaps should be completely CP
preserving and do not make use of normalization or causal constraints. Due to
that, we can also tackle probabilistic quantum transformations such as unitary
transposition.

######  Corollary 2 (Unitary transposition).

Consider the 1-slot supermap on unitary operations that transforms a unitary
operator $U$ into its transpose operator $U^{T}$ with with some fixed success
probability $p$. More formally, let $C\geq 0$ be the Choi operator of a
supermap which respects
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ_{U}^{t})=pJ_{U^{T}}$ for all
unitary operators $U$. The action of $C$ on a general operation with Choi
operator $J_{\Lambda}\in\mathcal{L}(\mathcal{I}\otimes\mathcal{O})$ is
necessarily given by
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ_{\Lambda}^{t})=p\,FJ_{\Lambda}F$
where $F$ denotes the flip (swap) operator defined by
$F|{\psi\phi}\rangle=|{\phi\psi}\rangle$ for all $|{\phi}\rangle$ and
$|{\psi}\rangle$.

###### Proof.

A possible Choi operator for unitary transposition is given by
$C^{\prime}_{0}=F_{{\mathcal{I}}{\mathcal{O}}}C_{0}F_{{\mathcal{I}}{\mathcal{O}}}^{\dagger}$.
Now, consider the condition for unitary operations

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}J_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}_{0}J_{U}^{t}).$
(17)

In this case, by introducing
$C=(F_{{\mathcal{I}}{\mathcal{O}}})^{\dagger}C^{\prime}(F_{{\mathcal{I}}{\mathcal{O}}})$,
the l.h.s. can be written as

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}J_{U}^{t})$
$\displaystyle={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[(F_{{\mathcal{I}}{\mathcal{O}}})C(F_{{\mathcal{I}}{\mathcal{O}}})^{\dagger}J_{U}^{t}]$
$\displaystyle={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(F_{{\mathcal{I}}{\mathcal{O}}})^{\dagger}J_{U}^{t}(F_{{\mathcal{I}}{\mathcal{O}}})]$
$\displaystyle={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(J_{U^{t}})^{t}]$
(18)

whereas the r.h.s. can be written as

$\displaystyle{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C^{\prime}_{0}J_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(J_{U^{t}})^{t}].$
(19)

Since $U^{t}$ is also a unitary operator and spans the whole linear span of
$\mathrm{span}\\{J_{U}\\}$, the uniqueness of $C^{\prime}$ is equivalent to
the uniqueness of the identity supermap $C$. ∎

######  Remark 1.

In general, for a given action of a supermap on unitary operations, an
extension of the domain of the supermap is not necessarily unique under the
three assumptions: the action on unitary operation; the linearity; and the
positivity. Thus, it is natural to consider what extra condition may lead to
the uniqueness. The results of Corollary 1 shows the Choi operator of the
supermap
$\widetilde{\mbox{$\mathcal{U}$}}\mapsto\widetilde{\mbox{$\mathcal{B}$}}\circ\widetilde{\mbox{$\mathcal{U}$}}\circ\widetilde{\mbox{$\mathcal{A}$}}$
is a rank-1 operator. One could wonder that, if there exists a rank-1 positive
semidefinite operator extension, then there is a unique extension.
Unfortunately, this hypothesis is not enough for the uniqueness of the
extension as we will now show.

Consider a supermap transforming all two-dimensional unitary operations to the
two-dimensional unitary operation $I$, the identity channel, with some
probability. That is, there exists a Choi operator $C_{0}$ such that
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C_{0}J_{U}^{t})\propto J_{I}$. For
$0\leq p\leq 1$, we define the Choi operator
$C_{p}\in\mathcal{L}\big{(}({\mathcal{I}}\otimes{\mathcal{O}})\otimes({\mathcal{P}}\otimes{\mathcal{F}})\big{)}$

$\displaystyle C_{p}:=\begin{pmatrix}1&p&p&1\\\ p&1&1&p\\\ p&1&1&p\\\
1&p&p&1\end{pmatrix}\otimes\phi^{+},$ (20)

where $\phi^{+}$ denotes the maximally entangled state which is proportional
to the Choi operator of the identity channel. We can check that for any $0\leq
p\leq 1$, the positivity constraint $C_{p}\geq 0$ is respected and
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C_{p}J_{U}^{t})\propto J_{I}$. But
note however that when $p=1$, $C_{p}$ is a rank-1 operator. Hence, the
existence of a rank-1 is not sufficient for proving the uniqueness of the
corresponding supermap on unitary operations.

##  5 Conclusion and Future Scope

In this paper we have addressed the question of whether the description of a
higher-order process is uniquely determined by its action on unitary
operations. We proved that, differently than general processes, the action of
the quantum switch on unitary operations uniquely defines its action on
general quantum operations. Our proof follows from linearity and the fact that
every higher-order quantum process transforming quantum operations must
respect the completely CP preserving property. We have also shown this
uniqueness property hold for different classes of higher-order quantum
processes which include unitary identity supermap, unitary transposition [26],
and qubit unitary complex conjugate [28].

In order to tackle more general processes, it would be useful to find a simple
criterium to identify which processes are uniquely defined by its action on
unitary operations. Here particular cases of interest would be to analyze the
$n$-slot quantum switch [9], and processes which may be viewed as
generalizations of the switch, such as the ones which can be described by
coherent control of causal order [25, 30, 31].

We remark that when multiple uses of an input operation are allowed, the
action on unitary operation cannot be uniquely determined for general quantum
operations in general. In particular, if a higher-order quantum operation can
be implemented with $k$ uses of an input operation, then if more than $k$ uses
is allowed, the uniqueness does not hold because it is always possible to
discard a certain number of input operations. Even though, the completely CP
preserving condition may be used to restrict the form of the supermap.

Another interesting future direction would be to understand how the
constraints of normalization would limit the possible extensions of quantum
processes. In this work we have obtained our results simply by arguments based
on positivity, but the normalization constraints for deterministic
transformations or linear constraints arising from causal relations [4] may
considerably restrict the possibilities of extending the action of processes
to non-unitary operations. We note that, when extending lower-order operations
from a restricted set of quantum states, example 5 of Ref. [19] shows that
requiring the transformation to be deterministic may be a severe constraint.
We then expect that stronger claims may be proven by imposing that a higher-
order process should be deterministic.

As we show here, it is possible to strongly restrict the form of a supermap
given only its action on unitary operations, and such a technique may reduce
the difficulty on the analysis of the performance of higher-order quantum
operation, since it is enough to perform an optimization on a restricted set
of supermaps. We hope that questions, answers, and the methods covered in this
paper contribute to our understanding on how positivity conditions restrict
higher-order quantum operations and that this work paves the way for
understanding the extension of general quantum processes.

## Acknowledgments

We acknowledge Simon Milz for comments on the first version of this
manuscript. This work was supported by MEXT Quantum Leap Flagship Program
(MEXT QLEAP) Grant No. JPMXS0118069605, JPMXS0120351339, Japan Society for the
Promotion of Science (JSPS) KAKENHI Grant No. 17H01694, 18H04286, 18K13467,
21H03394 and Advanced Leading Graduate Course for Photon Science (ALPS). M. T.
Quintino acknowledges the Austrian Science Fund (FWF) through the SFB project
“BeyondC” (sub-project F7103), a grant from the Foundational Questions
Institute (FQXi) Fund and a grant from the John Templeton Foundation (sub-
project F7103) as part of the The Quantum Information Structure of Spacetime
(QISS) Project (qiss.fr). The opinions expressed in this publication are those
of the authors and do not necessarily reflect the views of the John Templeton
Foundation. This project has received funding from the European Union’s
Horizon 2020 research and innovation programme under the Marie Skłodowska-
Curie grant agreement No 801110 and the Austrian Federal Ministry of
Education, Science and Research (BMBWF). It reflects only the authors’ view,
the EU Agency is not responsible for any use that may be made of the
information it contains.

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##  Appendix A The Elements in the Span of the Choi Operator of Unitary
Operations

######  Lemma 3.

$|{ii}\rangle\\!\langle{jj}|,|{ij}\rangle\\!\langle{ji}|\in\mathrm{span}\\{J_{U}\\}$
holds for $i\neq j$.

###### Proof.

Since only the linear span is considered, we use unnormalized quantum states
for convenience. Consider the following maximally entangled state

$\displaystyle|{\psi_{\theta,\phi}}\rangle=|{ii}\rangle+e^{i\theta}|{jj}\rangle+e^{i\phi}|{KK}\rangle,$
(21)

where $|{KK}\rangle:=\sum_{k\neq i,j}|{kk}\rangle$, then
$|{\psi_{\theta,\phi}}\rangle\\!\langle{\psi_{\theta,\phi}}|=J_{U}$ for some
$U$ and
$|{\psi_{\theta,\phi}}\rangle\\!\langle{\psi_{\theta,\phi}}|\in\mathrm{span}\\{J_{U}\\}$.
Since

$\displaystyle|{\psi_{\theta,\phi}}\rangle\\!\langle{\psi_{\theta,\phi}}|$
$\displaystyle=|{ii}\rangle\\!\langle{ii}|+|{jj}\rangle\\!\langle{jj}|+|{KK}\rangle\\!\langle{KK}|$
(22)
$\displaystyle+e^{-i\theta}|{ii}\rangle\\!\langle{jj}|+e^{-i\phi}|{ii}\rangle\\!\langle{KK}|+e^{i(\theta-\phi)}|{jj}\rangle\\!\langle{KK}|$
(23)
$\displaystyle+e^{i\theta}|{jj}\rangle\\!\langle{ii}|+e^{i\phi}|{KK}\rangle\\!\langle{ii}|+e^{-i(\theta-\phi)}|{KK}\rangle\\!\langle{jj}|,$
(24)

we obtain

$\displaystyle\int_{0}^{2\pi}d\theta\int_{0}^{2\pi}d\phi\
e^{i\theta}|{\psi_{\theta,\phi}}\rangle\\!\langle{\psi_{\theta,\phi}}|=|{ii}\rangle\\!\langle{jj}|\in\mathrm{span}\\{J_{U}\\}$
(25)

for all $i\neq j$. The same calculation with the maximally entangled state

$\displaystyle|{\psi^{\prime}_{\theta,\phi}}\rangle=|{ij}\rangle+e^{i\theta}|{ji}\rangle+e^{i\phi}|{KK}\rangle,$
(26)

leads to $|{ij}\rangle\\!\langle{ji}|\in\mathrm{span}\\{J_{U}\\}\mbox{ for all
}i\neq j$. ∎

######  Lemma 4.

$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$
holds for $i\neq j\neq i^{\prime}\neq j^{\prime}$.

###### Proof.

Consider the maximally entangled state

$\displaystyle|{\psi_{\theta,\phi}}\rangle=|{ij}\rangle+e^{i\theta}|{i^{\prime}j^{\prime}}\rangle+e^{i\phi}|{KK}\rangle$
(27)

where $|{KK}\rangle:=|{ji}\rangle+|{j^{\prime}i^{\prime}}\rangle+\sum_{k\neq
i,j,i^{\prime}j^{\prime}}|{kk}\rangle$. Then the same calculation of Lemma 3
proves
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$. ∎

######  Lemma 5.

$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$
holds if only two of $i,j,i^{\prime},j^{\prime}$ are the same and others are
different.

###### Proof.

There are 6 cases that only two of $i,j,i^{\prime},j^{\prime}$ are the same
and others are different. If
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$,
then the conjugate transpose also satisfies
$|{i^{\prime}j^{\prime}}\rangle\\!\langle{ij}|\in\mathrm{span}\\{J_{U}\\}$.
Thus, it is enough to consider the following four cases with $i\neq j\neq k$.

$\displaystyle|{ii}\rangle\\!\langle{jk}|$
$\displaystyle\in\mathrm{span}\\{J_{U}\\}$ (28)
$\displaystyle|{ij}\rangle\\!\langle{jk}|$
$\displaystyle\in\mathrm{span}\\{J_{U}\\}$ (29)
$\displaystyle|{ij}\rangle\\!\langle{ik}|$
$\displaystyle\in\mathrm{span}\\{J_{U}\\}$ (30)
$\displaystyle|{ij}\rangle\\!\langle{kj}|$
$\displaystyle\in\mathrm{span}\\{J_{U}\\}$ (31)

For each of them, consider the following maximally entangled states, and the
same calculation of Lemma 3 proves that they are in
$\mathrm{span}\\{J_{U}\\}$, respectively.

  1. 1.

$|{\psi_{\theta,\phi}}\rangle=|{ii}\rangle+e^{i\theta}|{jk}\rangle+e^{i\phi}|{LL}\rangle$
with $|{LL}\rangle:=|{kj}\rangle+\sum_{l\neq i,j,k}|{ll}\rangle$.

  2. 2.

$|{\psi_{\theta,\phi}}\rangle=|{ij}\rangle+e^{i\theta}|{jk}\rangle+e^{i\phi}|{LL}\rangle$
with $|{LL}\rangle:=|{ki}\rangle+\sum_{l\neq i,j,k}|{ll}\rangle$.

  3. 3.

$|{\psi_{\theta,\phi}}\rangle=|{i}\rangle(|{j}\rangle+e^{i\theta}|{k}\rangle)+e^{i\phi}|{LL}\rangle$
with
$|{LL}\rangle:=|{j}\rangle(|{j}\rangle-e^{i\theta}|{k}\rangle)+|{ki}\rangle+\sum_{l\neq
i,j,k}|{ll}\rangle$.

  4. 4.

$|{\psi_{\theta,\phi}}\rangle=(|{i}\rangle+e^{i\theta}|{k}\rangle)|{j}\rangle+e^{i\phi}|{LL}\rangle$
with
$|{LL}\rangle:=(|{i}\rangle-e^{i\theta}|{k}\rangle)|{i}\rangle+|{jk}\rangle+\sum_{l\neq
i,j,k}|{ll}\rangle$.

∎

######  Lemma 6.

$\sum_{j=i+k}|{ij}\rangle\\!\langle{ij}|\in\mathrm{span}\\{J_{U}\\}$ holds for
$k=0,\ldots,d-1$.

###### Proof.

Consider the maximally entangled state

$\displaystyle|{\psi_{\theta_{i}}}\rangle=\sum_{i}e^{\theta_{i}}|{i,i+k}\rangle.$
(32)

Note that we consider the $d$-dimensional system and $|{i+k}\rangle$ indicates
$|{i+k\ (\mathrm{mod}\ d)}\rangle$. Then we obtain

$\displaystyle\int_{0}^{2\pi}\prod_{i}d\theta_{i}\
|{\psi_{\theta_{i}}}\rangle\\!\langle{\psi_{\theta_{i}}}|=\sum_{i}|{i,i+k}\rangle\\!\langle{i,i+k}|=\sum_{j=i+k}|{ij}\rangle\\!\langle{ij}|\in\mathrm{span}\\{J_{U}\\}.$
(33)

∎

######  Lemma 7.

$(|{ij}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ji}|),(|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|)\in\mathrm{span}\\{J_{U}\\}$
holds for $i\neq j$.

###### Proof.

Consider the maximally entangled state

$\displaystyle|{\psi^{1}_{\theta,\phi}}\rangle$
$\displaystyle=(|{i}\rangle+e^{i\theta_{2}}|{j}\rangle)(|{i}\rangle+e^{i\theta_{2}}|{j}\rangle)+e^{i\theta_{1}}(|{i}\rangle-e^{i\theta_{2}}|{j}\rangle)(|{i}\rangle-e^{i\theta_{2}}|{j}\rangle)+e^{i\phi}|{KK}\rangle$
(34) $\displaystyle|{\psi^{2}_{\theta,\phi}}\rangle$
$\displaystyle=(|{i}\rangle+e^{i\theta_{2}}|{j}\rangle)(|{i}\rangle-e^{i\theta_{2}}|{j}\rangle)+e^{i\theta_{1}}(|{i}\rangle-e^{i\theta_{2}}|{j}\rangle)(|{i}\rangle+e^{i\theta_{2}}|{j}\rangle)+e^{i\phi}|{KK}\rangle$
(35)

where $|{KK}\rangle:=\sum_{k\neq i,j}|{kk}\rangle$. Here

$\displaystyle\int_{0}^{2\pi}d\theta_{1}d\theta_{2}d\phi
e^{i\theta_{1}}e^{-i\theta_{2}}|{\psi^{1}_{\theta,\phi}}\rangle\\!\langle{\psi^{1}_{\theta,\phi}}|$
$\displaystyle=|{ij}\rangle\\!\langle{ii}|+|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|-|{jj}\rangle\\!\langle{ji}|$
(36) $\displaystyle\int_{0}^{2\pi}d\theta_{1}d\theta_{2}d\phi
e^{i\theta_{1}}e^{-i\theta_{2}}|{\psi^{2}_{\theta,\phi}}\rangle\\!\langle{\psi^{2}_{\theta,\phi}}|$
$\displaystyle=-|{ij}\rangle\\!\langle{ii}|+|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|+|{jj}\rangle\\!\langle{ji}|,$
(37)

holds, and by considering the sum and difference, we obtain

$\displaystyle(|{ij}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ji}|),(|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|)\in\mathrm{span}\\{J_{U}\\}.$
(38)

∎

##  Appendix B The proof of Lemma 2

Consider the identity supermap
${\mathchoice{\widetilde{\hbox{\set@color$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\textstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}{\widetilde{\hbox{\set@color$\scriptscriptstyle\widetilde{\mbox{$\mathcal{C}$}}$}}}}$
that transforms an input unitary operation $\widetilde{\mbox{$\mathcal{U}$}}$
on ${\mathcal{I}}\otimes{\mathcal{O}}\cong\mathbb{C}^{d}\otimes\mathbb{C}^{d}$
to $\widetilde{\mbox{$\mathcal{U}$}}$ on
${\mathcal{P}}\otimes{\mathcal{F}}\cong\mathbb{C}^{d}\otimes\mathbb{C}^{d}$.
We show that this supermap transforms an arbitrary input operation
$\widetilde{\mbox{$\Lambda$}}$ on ${\mathcal{I}}\otimes{\mathcal{O}}$ to
itself on ${\mathcal{P}}\otimes{\mathcal{F}}$ by assuming the completely CP
preserving property. That is, the Choi operator of the identity supermap $C$
satisfies

$\displaystyle
C=\sum_{i,j,k,l=1}^{d}|{ij}\rangle\\!\langle{kl}|\otimes\widetilde{\mbox{$\mathcal{C}$}}(|{ij}\rangle\\!\langle{kl}|)\in{\mathcal{L}}({\mathcal{I}}\otimes{\mathcal{O}}\otimes{\mathcal{P}}\otimes{\mathcal{F}}),$
(39)

where $\widetilde{\mbox{$\mathcal{C}$}}$ denotes the corresponding map of $C$,
i.e.,
$\widetilde{\mbox{$\mathcal{C}$}}(J)={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ^{t})$
for $J\in{\mathcal{L}}({\mathcal{I}}\otimes{\mathcal{O}})$, and we show that
this Choi operator is uniquely determined as

$\displaystyle
C_{0}=\sum_{i,j,k,l=1}^{d}|{ij}\rangle\\!\langle{kl}|\otimes|{ij}\rangle\\!\langle{kl}|\in{\mathcal{L}}({\mathcal{I}}\otimes{\mathcal{O}}\otimes{\mathcal{P}}\otimes{\mathcal{F}})$
(40)

under the following assumptions: its action on unitary operations is
determined as
${\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C_{0}J_{U}^{t})$
for an arbitrary unitary operation $U$; and it is a positive operator $C\geq
0$. Note that the action of $\widetilde{\mbox{$\mathcal{C}$}}$ is given by
$\widetilde{\mbox{$\mathcal{C}$}}(J_{U})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(CJ_{U}^{t})={\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}(C_{0}J_{U}^{t})=J_{U}$
by assumption. We first consider the diagonal elements of $C$, then consider
the off-diagonal elements of $C$.

(First part: diagonal elements of $C$) Since
$\sum_{i,j=1}^{d}|{ij}\rangle\\!\langle{ij}|=I\in\mathrm{span}\\{J_{U}\\}$,
the action of $\widetilde{\mbox{$\mathcal{C}$}}$ is determined as
$\widetilde{\mbox{$\mathcal{C}$}}(I)=I$, and we obtain

$\displaystyle\sum_{i,j=1}^{d}|{ij}\rangle\\!\langle{ij}|=\widetilde{\mbox{$\mathcal{C}$}}(\sum_{i,j=1}^{d}|{ij}\rangle\\!\langle{ij}|)$
$\displaystyle=\sum_{i,j=1}^{d}\widetilde{\mbox{$\mathcal{C}$}}(|{ij}\rangle\\!\langle{ij}|).$
(41)

Thus, for
$|{ijkl}\rangle\in{\mathcal{I}}\otimes{\mathcal{O}}\otimes{\mathcal{P}}\otimes{\mathcal{F}}$

$\displaystyle\sum_{i,j=1}^{d}\langle{ijkl}|C|{ijkl}\rangle$
$\displaystyle=\sum_{i,j=1}^{d}\langle{kl}|{\mathrm{Tr}}\,(C|{ij}\rangle\\!\langle{ij}|^{t})|{kl}\rangle=\sum_{i,j=1}^{d}\langle{kl}|\widetilde{\mbox{$\mathcal{C}$}}(|{ij}\rangle\\!\langle{ij}|)|{kl}\rangle$
$\displaystyle=\langle{kl}|\sum_{i,j=1}^{d}\widetilde{\mbox{$\mathcal{C}$}}(|{ij}\rangle\\!\langle{ij}|)|{kl}\rangle=\langle{kl}|\big{(}\sum_{i,j=1}^{d}|{ij}\rangle\\!\langle{ij}|\big{)}|{kl}\rangle=1,$
(42)

holds for all $k,l$, and we obtain

$\displaystyle{\mathrm{Tr}}\,C=\sum_{i,j,k,l=1}^{d}\langle{ijkl}|C|{ijkl}\rangle$
$\displaystyle=d^{2}.$ (43)

Since every diagonal element of a positive semidefinite operator is non-
negative, Eq. (42) shows that every diagonal element of $C$, say $a$,
satisfies $0\leq a\leq 1$, and Eq. (43) shows that the sum of all diagonal
elements of $C$ is equal to $d^{2}$.

Since
$|{ij}\rangle\\!\langle{ji}|,|{ii}\rangle\\!\langle{jj}|\in\mathrm{span}\\{J_{U}\\}$
holds for all $i\neq j$ from Lemma 3, the action of
$\widetilde{\mbox{$\mathcal{C}$}}$ is determined as

$\displaystyle\widetilde{\mbox{$\mathcal{C}$}}(|{ij}\rangle\\!\langle{ji}|)=|{ij}\rangle\\!\langle{ji}|,\quad\widetilde{\mbox{$\mathcal{C}$}}(|{ii}\rangle\\!\langle{jj}|)=|{ii}\rangle\\!\langle{jj}|,$
(44)

and the Choi operator $C$ satisfies

$\displaystyle\langle{ijij}|C|{jiji}\rangle=1,\quad\langle{iiii}|C|{jjjj}\rangle=1.$
(45)

Since $C\geq 0$ all principal minors of $C$ are non-negative. Consider the
following two-dimensional submatrices of $C$

$\displaystyle\begin{pmatrix}\langle{ijij}|C|{ijij}\rangle&\langle{ijij}|C|{jiji}\rangle\\\
\langle{jiji}|C|{ijij}\rangle&\langle{jiji}|C|{jiji}\rangle\end{pmatrix},\quad\begin{pmatrix}\langle{iiii}|C|{iiii}\rangle&\langle{iiii}|C|{jjjj}\rangle\\\
\langle{jjjj}|C|{iiii}\rangle&\langle{jjjj}|C|{jjjj}\rangle\end{pmatrix},$
(46)

with $i\neq j$. Eq. (45) indicates that all the off-diagonal elements of these
matrices are 1, and the diagonal elements are between 0 and 1 due to Eq. (42).
Then these matrices are positive semi-definite only if

$\displaystyle\langle{ijij}|C|{ijij}\rangle=1,\quad\langle{iiii}|C|{iiii}\rangle=1$
(47)

for all $i\neq j$.

Since the sum of all diagonal elements is equal to $d^{2}$ due to Eq. (43),
and Eq. (47) provides the $d(d-1)+d=d^{2}$ elements that are equal to 1, the
other diagonal elements of $C$ are equal to 0. For positive semi-definite
matrices, if a diagonal element is 0, all elements in that row or column are
also 0. Thus, we have

$\displaystyle\langle{i_{1}j_{1}k_{1}l_{1}}|C|{i_{2}j_{2}k_{2}l_{2}}\rangle\neq
0$ (48)

only if $(i_{1},j_{1},k_{1},l_{1})=(i,j,i,j)$ and
$(i_{2},j_{2},k_{2},l_{2})=(k,l,k,l)$ for some $i,j,k,l$.

(Second part: off-diagonal elements of $W$) The second part of the proof is to
show that all the elements which are non-zero are equal to 1. Consider a Choi
operator $J\in\mathrm{span}\\{J_{U}\\}$ with $\langle{ij}|J|{kl}\rangle\neq 0$
for all $i\neq j,k\neq l$, e.g., the Choi operator of the Fourier transform.
Since $J\in\mathrm{span}\\{J_{U}\\}$, the equality

$\displaystyle\langle{ij}|\widetilde{\mbox{$\mathcal{C}$}}(J)|{kl}\rangle$
$\displaystyle=\langle{ij}|J|{kl}\rangle$ (49)

holds for all $i\neq j,k\neq l$. On the other hand, the action determined by
the Choi operator $C$ is given by

$\displaystyle\langle{ij}|\widetilde{\mbox{$\mathcal{C}$}}(J)|{kl}\rangle$
$\displaystyle=\langle{ij}|{\mathrm{Tr}}_{{\mathcal{I}}{\mathcal{O}}}[C(J^{t}_{{\mathcal{I}}{\mathcal{O}}}\otimes
I_{{\mathcal{P}}{\mathcal{F}}})]|{kl}\rangle$ (50)
$\displaystyle={\mathrm{Tr}}(\langle{ij}|C|{kl}\rangle J^{t})$ (51)
$\displaystyle=\sum_{i^{\prime},j^{\prime},k^{\prime},l^{\prime}}\langle{i^{\prime}j^{\prime}ij}|C|{k^{\prime}l^{\prime}kl}\rangle\langle{k^{\prime}l^{\prime}}|J^{t}|{i^{\prime}j^{\prime}}\rangle$
(52)
$\displaystyle=\langle{ijij}|C|{klkl}\rangle\langle{kl}|J^{t}|{ij}\rangle$
(53) $\displaystyle=\langle{ijij}|C|{klkl}\rangle\langle{ij}|J|{kl}\rangle,$
(54)

where the fourth equality holds due to Eq. (47). Therefore, by considering the
condition that both Eq. (49) and Eq. (54) hold, we obtain
$\langle{ijij}|C|{klkl}\rangle=1$ for all $i\neq j,k\neq l$. Similarly,
consider a Choi operator $J\in\mathrm{span}\\{J_{U}\\}$ with
$\langle{ii}|J|{kl}\rangle\neq 0$ for all $i$ and $k\neq l$, e.g., the Choi
operator of the Fourier transform, we obtain $\langle{iiii}|C|{klkl}\rangle=1$
for all $i$ and $k\neq l$. Similar calculations show that
$\langle{ijji}|C|{kkkk}\rangle=1$ for $i\neq j$ and all $k$, and
$\langle{iiii}|C|{kkkk}\rangle=1$ for all $i$ and $k$. Thus, we uniquely
determine all the elements of $C$ and obtain $C=C_{0}$.

##  Appendix C The Proof of Lemma 1

The Choi operator $W_{0}$ of quantum switch that we prove is given by Eq.
(10). For
$J=|{ijkl}\rangle\\!\langle{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|\in{\mathcal{L}}({\mathcal{I}}_{1}\otimes{\mathcal{O}}_{1}\otimes{\mathcal{I}}_{2}\otimes{\mathcal{O}}_{2})$,
the action of $W_{0}$ can be evaluated as

$\displaystyle{\mathrm{Tr}}_{in}(W_{0}J^{t})=\langle{ijkl}||{W_{0}}\rangle\\!\langle{W_{0}}||{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\rangle$
$\displaystyle=\delta_{jk}\delta_{j^{\prime}k^{\prime}}|{il00}\rangle\\!\langle{i^{\prime}l^{\prime}00}|+\delta_{il}\delta_{i^{\prime}l^{\prime}}|{kj11}\rangle\\!\langle{k^{\prime}j^{\prime}11}|$
$\displaystyle\quad+\delta_{jk}\delta_{i^{\prime}l^{\prime}}|{il00}\rangle\\!\langle{k^{\prime}j^{\prime}11}|+\delta_{il}\delta_{j^{\prime}k^{\prime}}|{kj11}\rangle\\!\langle{i^{\prime}l^{\prime}00}|.$
(55)

Note that for $J\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$,
${\mathrm{Tr}}_{in}(WJ^{t})={\mathrm{Tr}}_{in}(W_{0}J^{t})$ holds by
assumption.

Here each element of $W_{0}$ is 0 or 1. In order to prove Theorem 1, it is
enough to prove that each element of also $W$ satisfies $|(W)_{ij}|=0\mbox{ or
}1$. This is because if an arbitrary $W$ satisfies $|(W)_{ij}|=0\mbox{ or }1$
but $W\neq W_{0}$, then some elements of $\epsilon W+(1-\epsilon)W_{0}$, which
is also a candidate for $W$, must have a value satisfying $0<|(W)_{ij}|<1$,
which contradicts with the assumption that $W$ satisfies $|(W)_{ij}|=0\mbox{
or }1$.

The rest of the proof is divided into two parts: the first part considers the
diagonal elements of $W$, and the second part considers the off-diagonal
elements of $W$.

(First part: diagonal elements of $W$) Consider
$J=|{ijkl}\rangle\\!\langle{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|$. For
all $i,j,k,l$, there exist $i^{\prime},j^{\prime},k^{\prime},l^{\prime}$ such
that $J\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$. That is, if $i=j$
then let $i^{\prime}=j^{\prime}(\neq i)$ so that
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$,
and in this case, we relabel the variables $i^{\prime},j^{\prime}$ as $j$.
Similarly, if $i\neq j$ then let $i^{\prime}=j,j^{\prime}=i$ so that
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U}\\}$.
(See Sec. 3 or Lemma 3) The same arguments hold for $k,l$. We then obtain the
followings for $i\neq j$ and $k\neq l$:

$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}(|{ijkl}\rangle\\!\langle{jilk}|)=(\delta_{jk}|{il00}\rangle+\delta_{il}|{kj11}\rangle)(\delta_{il}\langle{jk00}|+\delta_{jk}\langle{li11}|)$
(56)
$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}(|{ijkk}\rangle\\!\langle{jill}|)=(\delta_{jk}|{ik00}\rangle+\delta_{ik}|{kj11}\rangle)(\delta_{il}\langle{jl00}|+\delta_{jl}\langle{li11}|)$
(57)
$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}(|{iikl}\rangle\\!\langle{jjlk}|)=(\delta_{ik}|{il00}\rangle+\delta_{il}|{ki11}\rangle)(\delta_{jl}\langle{jk00}|+\delta_{jk}\langle{lj11}|)$
(58)
$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}(|{iikk}\rangle\\!\langle{jjll}|)=(\delta_{ik}|{ik00}\rangle+\delta_{ik}|{ki11}\rangle)(\delta_{jl}\langle{jl00}|+\delta_{jl}\langle{lj11}|),$
(59)

where $\widetilde{\mbox{$\mathcal{W}$}}$ denotes the corresponding map of $W$,
i.e., $\widetilde{\mbox{$\mathcal{W}$}}(J)={\mathrm{Tr}}_{in}WJ^{t}$. Also,
consider that $I\otimes I\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$, we
obtain

$\displaystyle{\mathrm{Tr}}\,\widetilde{\mbox{$\mathcal{W}$}}(I\otimes
I)={\mathrm{Tr}}({\mathrm{Tr}}_{in}W)={\mathrm{Tr}}W=2d^{3}.$ (60)

By considering Eqs. (56) – (60) and the positivity $W\geq 0$, we obtain the
following elements of $W$ with $i\neq j,k\neq l$

$\displaystyle\langle{ijklil00}|W|{ijklil00}\rangle=1$ $\displaystyle\text{
for }j=k$ (61) $\displaystyle\langle{ijklkj11}|W|{ijklkj11}\rangle=1$
$\displaystyle\text{ for }i=l$ (62)
$\displaystyle\langle{ijkkik00}|W|{ijkkik00}\rangle=1$ $\displaystyle\text{
for }j=k$ (63) $\displaystyle\langle{ijkkkj11}|W|{ijkkkj11}\rangle=1$
$\displaystyle\text{ for }i=k$ (64)
$\displaystyle\langle{iiklil00}|W|{iiklil00}\rangle=1$ $\displaystyle\text{
for }i=k$ (65) $\displaystyle\langle{iiklki11}|W|{iiklki11}\rangle=1$
$\displaystyle\text{ for }i=l$ (66)
$\displaystyle\langle{iikkik00}|W|{iikkik00}\rangle=1$ $\displaystyle\text{
for }i=k$ (67) $\displaystyle\langle{iikkki11}|W|{iikkki11}\rangle=1$
$\displaystyle\text{ for }i=k,$ (68)

or equivalently,

$\displaystyle\langle{ijjlil00}|W|{ijjlil00}\rangle$ $\displaystyle=1\text{
for }i\neq j,j\neq l$ (69)
$\displaystyle\langle{ijkikj11}|W|{ijkikj11}\rangle$ $\displaystyle=1\text{
for }i\neq j,k\neq i$ (70)
$\displaystyle\langle{ijjjij00}|W|{ijjjij00}\rangle$ $\displaystyle=1\text{
for }i\neq j$ (71) $\displaystyle\langle{ijiiij11}|W|{ijiiij11}\rangle$
$\displaystyle=1\text{ for }i\neq j$ (72)
$\displaystyle\langle{iiilil00}|W|{iiilil00}\rangle$ $\displaystyle=1\text{
for }i\neq l$ (73) $\displaystyle\langle{iikiki11}|W|{iikiki11}\rangle$
$\displaystyle=1\text{ for }i\neq k$ (74)
$\displaystyle\langle{iiiiii00}|W|{iiiiii00}\rangle$ $\displaystyle=1$ (75)
$\displaystyle\langle{iiiiii11}|W|{iiiiii11}\rangle$ $\displaystyle=1.$ (76)

The proof is given in Appendix D. The number of terms in 1st, 3rd, 5th, and
7th lines (or 2nd, 4th, 6th, 8th lines) is given by
$d(d-1)^{2},d(d-1),d(d-1),d$ respectively, and the total number of terms in
these 8 lines is $2d^{3}$. Since ${\mathrm{Tr}}\,W=2d^{3}$, all the other
diagonal terms of $W$ are 0, that is,
$\langle{\psi}|W|{\psi^{\prime}}\rangle=0$ for all
$|{\psi}\rangle,|{\psi^{\prime}}\rangle$ not of these forms.

######  Remark 2.

Here we illustrate the key idea of the proof in Appendix D with a simpler
example. Consider a linear map
$\widetilde{\mbox{$\mathcal{W}$}}:{\mathcal{L}}(\mathbb{C}^{2})\to{\mathcal{L}}(\mathbb{C}^{2})$
and the corresponding Choi operator $W\in{\mathcal{L}}(\mathbb{C}^{4})$
satisfying

$\displaystyle\widetilde{\mbox{$\mathcal{W}$}}(|{i}\rangle\\!\langle{j}|)$
$\displaystyle=|{i}\rangle\\!\langle{j}|\quad\mbox{for}\quad(i,j)=(0,1)\text{
and }\;(i,j)=(1,0)$ (77) $\displaystyle{\mathrm{Tr}}\,W$ $\displaystyle=2$
(78) $\displaystyle W$ $\displaystyle\geq 0,$ (79)

then the matrix $W$ has the form of

$\displaystyle W=\begin{pmatrix}a&b&0&1\\\ c&d&0&0\\\ 0&0&e&f\\\ 1&0&g&h\\\
\end{pmatrix}$ (80)

with $a+d+e+h=2$. From the positivity $W\geq 0$, the determinant of all its
minors has to be non-negative, hence we obtain $ah-1\geq 0$ and $a,d,e,h\geq
0$. By combining these constraints we can verify that

$\displaystyle 2$ $\displaystyle=a+d+e+h$ (81) $\displaystyle\geq
2\sqrt{ah}+e+d,\quad$ $\displaystyle\left[\text{since
}\frac{ah}{2}\geq\sqrt{ah}\right]$ (82) $\displaystyle\geq 2+e+d,\quad$
$\displaystyle\left[\text{since }ah\geq 1\right]$ (83) $\displaystyle\geq
2.\quad$ $\displaystyle\left[\text{since }e\geq 0,\;d\geq 0\right]$ (84)

Hence, we obtain $a+h=2\sqrt{ah}$ as a necessary condition. Thus, we obtain
$a=d=1$ and the diagonal terms are uniquely determined as
$(a,d,e,h)=(1,0,0,1)$.

(Second part: off-diagonal elements of $W$) In this part, we show that every
element of $W$ satisfies $|(W)_{ij}|=0\mbox{ or }1$. From the positivity, if a
diagonal element is 0, then all the elements of that row or column is 0.
Therefore, we can assume that $W$ is given by

$\displaystyle{\mathrm{Tr}}_{in}WJ^{t}=\langle{ijkl}||{W}\rangle\\!\langle{W}||{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\rangle$
$\displaystyle=a_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\delta_{jk}\delta_{j^{\prime}k^{\prime}}|{il00}\rangle\\!\langle{i^{\prime}l^{\prime}00}|$
$\displaystyle\quad+b_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\delta_{il}\delta_{i^{\prime}l^{\prime}}|{kj11}\rangle\\!\langle{k^{\prime}j^{\prime}11}|$
$\displaystyle\quad+c_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\delta_{jk}\delta_{i^{\prime}l^{\prime}}|{il00}\rangle\\!\langle{k^{\prime}j^{\prime}11}|$
$\displaystyle\quad+d_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\delta_{il}\delta_{j^{\prime}k^{\prime}}|{kj11}\rangle\\!\langle{i^{\prime}l^{\prime}00}|$
(85)

with
$|a_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}|,|b_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}|,|c_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}|,|d_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}|\leq
1$. (cf. Eq. (55)) As we mentioned, it is enough to show that all of the
absolute values of them (with the non-vanishing Kronecker deltas) are 1.
Considering that there are at most $(2d^{3})^{2}$ non-zero elements, this is
also equivalent to show that $\|W\|_{1}=(2d^{3})^{2}$, where $\|W\|_{1}$
denotes the element-wise 1-norm (i.e., the sum of absolute values of all
elements in the computational basis). Since $\|W\|_{1}\leq(2d^{3})^{2}$ is
trivial, we prove $\|W\|_{1}\geq(2d^{3})^{2}$ in the following.

######  Remark 3.

Before proceeding to the proof, we briefly explain the key idea of the proof
of this part by adding an extra assumption which makes the proof much simpler.
Let
$J=\sum_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}|{ijkl}\rangle\\!\langle{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|$,
we have

$\displaystyle\|W\|_{1}=\sum_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\|\langle{ijkl}|W|{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\rangle\|_{1}\geq\|\sum_{ijkli^{\prime}j^{\prime}k^{\prime}l^{\prime}}\langle{ijkl}|W|{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\rangle\|_{1}=\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}.$
(86)

Assume that $J\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$ (which is the
extra assumption and is not true for the case of the proof), then the r.h.s.
can be evaluated using ${\mathrm{Tr}}_{in}WJ^{t}={\mathrm{Tr}}_{in}W_{0}J^{t}$
as

$\displaystyle\|W\|_{1}\geq\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=\|{\mathrm{Tr}}_{in}W_{0}J^{t}\|_{1}=(2d^{3})^{2},$
(87)

which completes the proof.

In order to evaluate $\|W\|_{1}$, we first provide a grouping of
$|{ijkl}\rangle\\!\langle{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|=|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\otimes|{kl}\rangle\\!\langle{k^{\prime}l^{\prime}}|$,
which covers all possible elements and also satisfies that each group is in
$\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$. See Sec. 3 for the proof that
they are in $\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$. We first classify
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|$ into the following groups.
($i,j,i^{\prime},j^{\prime}$ are assumed to be different by default)

  1. 1.

If all $i,j,i^{\prime},j^{\prime}$ are different, or if only two of them are
the same, or $i=j,i^{\prime}=j^{\prime}$ or $i=j^{\prime},j=i^{\prime}$, then
we use
$|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\in\mathrm{span}\\{J_{U_{1}}\otimes
J_{U_{2}}\\}$.

  2. 2.

If $i=i^{\prime},j=j^{\prime}$ or $i=i^{\prime}=j=j^{\prime}$, then we use
$\sum_{j=i+k}|{ij}\rangle\\!\langle{ij}|\in\mathrm{span}\\{J_{U_{1}}\otimes
J_{U_{2}}\\}$.

  3. 3.

If $i=j=j^{\prime}$ or $i=j=j^{\prime}$ or $i=i^{\prime}=j^{\prime}$ or
$j=i^{\prime}=j$, then we use
$(|{ij}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ji}|),(|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|)\in\mathrm{span}\\{J_{U_{1}}\otimes
J_{U_{2}}\\}$.

We denote the three groups as $G_{1}$, $G_{2}$ and $G_{3}$ respectively. To
make the statement clear, we provide the definitions explicitly as follows:

$\displaystyle G_{1}$
$\displaystyle=\\{|{ij}\rangle\\!\langle{i^{\prime}j^{\prime}}|\mid i\neq
j\neq i^{\prime}\neq j^{\prime},$ $\displaystyle\qquad\mbox{ or }i=j,i\neq
i^{\prime}\neq j^{\prime}\quad\mbox{ (and the remaining 5 patterns)},$
$\displaystyle\qquad\mbox{ or }i=j,i^{\prime}=j^{\prime},i\neq i^{\prime},$
$\displaystyle\qquad\mbox{ or }i=j^{\prime},j=i^{\prime},i\neq j\\},$ (88)
$\displaystyle G_{2}$
$\displaystyle=\\{\sum_{j=i+k}|{ij}\rangle\\!\langle{ij}|\mid
k=0,\ldots,d-1\\},$ (89) $\displaystyle G_{3}$
$\displaystyle=\\{|{ij}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ji}|,|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|\mid
i\neq j\\}.$ (90)

The number of elements of each group is given by

$\displaystyle|G_{1}|$ $\displaystyle=d(d-1)(d-2)(d-3)+6d(d-1)(d-2)+2d(d-1)$
$\displaystyle=d(d-1)(d^{2}+d-4),$ (91) $\displaystyle|G_{2}|$
$\displaystyle=d,$ (92) $\displaystyle|G_{3}|$ $\displaystyle=2d(d-1).$ (93)

Note that $|G_{1}|+d|G_{2}|+2|G_{3}|=d^{4}$ holds. The same grouping is chosen
for $|{kl}\rangle\\!\langle{k^{\prime}l^{\prime}}|$. Since the three groups
have no overlap between each other, we obtain

$\displaystyle\|W\|_{1}$ $\displaystyle\geq\sum_{a,b=1}^{3}\sum_{J\in
G_{a}\otimes G_{b}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}.$ (94)

Moreover, since all $J$ considered here satisfies
$J\in\mathrm{span}\\{J_{U_{1}}\otimes J_{U_{2}}\\}$, we can evaluate the
r.h.s. with ${\mathrm{Tr}}_{in}WJ^{t}={\mathrm{Tr}}_{in}W_{0}J^{t}$.

The summation $\sum_{J\in G_{a}\otimes G_{b}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
for each $a,b$ is evaluated as follows. See Appendix E for calculations. Note
that it is enough to consider only 6 cases that satisfy $a\leq b$ because of
the symmetry between the two systems.

$\displaystyle\sum_{J\in G_{1}\otimes G_{1}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=2d(d-1)(2d^{4}+2d^{3}-18d^{2}+11d+8)$ (95)
$\displaystyle\sum_{J\in G_{1}\otimes G_{2}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=2d(d-1)(2d^{2}-d-4)$ (96) $\displaystyle\sum_{J\in G_{2}\otimes
G_{2}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=2d^{2}(d+1)$ (97)
$\displaystyle\sum_{J\in G_{1}\otimes G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=4d(d-1)(4d^{2}-5d-2)$ (98) $\displaystyle\sum_{J\in
G_{2}\otimes G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=4d(d-1)(d+2)$ (99) $\displaystyle\sum_{J\in G_{3}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=16d^{2}(d-1).$ (100)

By using these equations, $\|W\|_{1}$ is evaluated as

$\displaystyle\|W\|_{1}\geq(2d^{3})^{2},$ (101)

which completes the proof of Lemma 1.

##  Appendix D Evaluation of the Diagonal Terms in the Proof of Lemma 1

Here we provide the proof of Eqs. (61) – (68) from Eqs. (56) – (60) and the
positivity $W\geq 0$. Let

$\displaystyle S_{1}=\\{|{ijklil00}\rangle\mid i\neq j,k\neq
l,j=k\\}\cup\\{|{ijklkj11}\rangle\mid i\neq j,k\neq l,i=l\\}$ (102)
$\displaystyle S_{2}=\\{|{ijkkik00}\rangle\mid i\neq j,j=k\\}$ (103)
$\displaystyle S_{3}=\\{|{ijkkkj11}\rangle\mid i\neq j,i=k\\}$ (104)
$\displaystyle S_{4}=\\{|{iiklil00}\rangle\mid k\neq l,i=k\\}$ (105)
$\displaystyle S_{5}=\\{|{iiklki11}\rangle\mid k\neq l,i=l\\}$ (106)
$\displaystyle S_{6}=\\{|{iikkik00}\rangle\mid i=k\\}$ (107) $\displaystyle
S_{7}=\\{|{iikkki11}\rangle\mid i=k\\},$ (108)

and it is enough to prove that

$\displaystyle\prod_{|{\psi}\rangle\in S_{a}}\langle{\psi}|W|{\psi}\rangle\geq
1$ (109)

holds for $a=1,\ldots,7$. This is because if Eq. (109) holds, the inequality

$\displaystyle 2d^{3}\geq\sum_{a=1}^{7}\sum_{|{\psi}\rangle\in
S_{a}}\langle{\psi}|W|{\psi}\rangle\geq\sum_{a=1}^{7}|S_{a}|\left(\prod_{|{\psi}\rangle\in
S_{a}}\langle{\psi}|W|{\psi}\rangle\right)^{\frac{1}{|S_{a}|}}\geq\sum_{a=1}^{7}|S_{a}|=2d^{3}$
(110)

also holds, and by considering the condition for equality, we obtain

$\displaystyle\langle{\psi}|W|{\psi}\rangle=1$ (111)

for all $|{\psi}\rangle\in S_{a}$ for all $a$. Here we only prove Eq. (109)
for $a=1,2,6$. The case of $a=4$ can be proved in the same way as the case of
$a=2$ by considering the symmetry between the two input operations. The rest
cases $a=3,5,7$ can be proved in the same way as $a=2,4,6$, respectively, by
considering the symmetry between the action of the control qubit being 0 and
1.

(Case 1: $S_{1}$) In this case, we show the following inequality holds

$\displaystyle\prod_{|{\psi}\rangle\in S_{1}}\langle{\psi}|W|{\psi}\rangle$
$\displaystyle=\left(\prod_{i\neq j,k\neq
l,j=k}\langle{ijklil00}|W|{ijklil00}\rangle\right)\left(\prod_{i\neq j,k\neq
l,i=l}\langle{ijklkj11}|W|{ijklkj11}\rangle\right)$
$\displaystyle=\left(\prod_{i\neq k,k\neq
l}\langle{ikklil00}|W|{ikklil00}\rangle\right)\left(\prod_{i\neq k,k\neq
l}\langle{kilkli11}|W|{kilkli11}\rangle\right)$ $\displaystyle=\prod_{i\neq
k,k\neq
l}\langle{ikklil00}|W|{ikklil00}\rangle\langle{kilkli11}|W|{kilkli11}\rangle\geq
1.$ (112)

Let $|{\psi_{ikl}^{0}}\rangle:=|{ikklil00}\rangle$ and
$|{\psi_{ikl}^{1}}\rangle:=|{kilkli11}\rangle$, then
$\langle{\psi_{ikl}^{0}}|W|{\psi_{ikl}^{1}}\rangle=1$ holds for arbitrary
$i,k,l$ satisfying $i\neq k,k\neq l$ because of Eq. (56). By considering the
positivity of the submatrix

$\displaystyle\begin{pmatrix}\langle{\psi_{ikl}^{0}}|W|{\psi_{ikl}^{0}}\rangle&\langle{\psi_{ikl}^{0}}|W|{\psi_{ikl}^{1}}\rangle\\\
\langle{\psi_{ikl}^{1}}|W|{\psi_{ikl}^{0}}\rangle&\langle{\psi_{ikl}^{1}}|W|{\psi_{ikl}^{1}}\rangle\end{pmatrix}=\begin{pmatrix}\langle{\psi_{ikl}^{0}}|W|{\psi_{ikl}^{0}}\rangle&1\\\
1&\langle{\psi_{ikl}^{1}}|W|{\psi_{ikl}^{1}}\rangle\end{pmatrix}\geq 0,$ (113)

we obtain the inequality

$\displaystyle\langle{\psi_{ikl}^{0}}|W|{\psi_{ikl}^{0}}\rangle\langle{\psi_{ikl}^{1}}|W|{\psi_{ikl}^{1}}\rangle\geq
1,$ (114)

which completes the proof of Eq. (112).

(Case 2: $S_{2}$) In this case, we show the following inequality holds

$\displaystyle\prod_{|{\psi}\rangle\in S_{3}}\langle{\psi}|W|{\psi}\rangle$
$\displaystyle=\prod_{i\neq j,j=k}\langle{ijkkik00}|W|{ijkkik00}\rangle$
$\displaystyle=\prod_{i\neq
k}\langle{\psi_{ik}}|W|{\psi_{ik}}\rangle=\prod_{i<k}\langle{\psi_{ik}}|W|{\psi_{ik}}\rangle\langle{\psi_{ki}}|W|{\psi_{ki}}\rangle\geq
1,$ (115)

where we define $|{\psi_{ik}}\rangle:=|{ikkkik00}\rangle$ for
$i,k=0,\ldots,d-1$. Since $\langle{\psi_{ik}}|W|{\psi_{ki}}\rangle=1$ holds
arbitrary $i,k$ satisfying $i\neq k$ because of Eq. (57), we consider the
positivity of the submatrix

$\displaystyle\begin{pmatrix}\langle{\psi_{ik}}|W|{\psi_{ik}}\rangle&\langle{\psi_{ik}}|W|{\psi_{ki}}\rangle\\\
\langle{\psi_{ki}}|W|{\psi_{ik}}\rangle&\langle{\psi_{ki}}|W|{\psi_{ki}}\rangle\end{pmatrix}=\begin{pmatrix}\langle{\psi_{ik}}|W|{\psi_{ik}}\rangle&1\\\
1&\langle{\psi_{ki}}|W|{\psi_{ki}}\rangle\end{pmatrix}\geq 0,$ (116)

and obtain

$\displaystyle\langle{\psi_{ik}}|W|{\psi_{ik}}\rangle\langle{\psi_{ki}}|W|{\psi_{ki}}\rangle\geq
1$ (117)

for any $i\neq k$, which completes the proof of Eq. (115).

(Case 3: $S_{6}$) In this case, we show the following inequality holds

$\displaystyle\prod_{|{\psi}\rangle\in
S_{7}}\langle{\psi}|W|{\psi}\rangle=\prod_{i=k}\langle{iikkik00}|W|{iikkik00}\rangle=\prod_{k}\langle{kkkkkk00}|W|{kkkkkk00}\rangle\geq
1.$ (118)

Let $|{\psi_{k}}\rangle:=|{kkkkkk00}\rangle$ for $k=0,\ldots,d-1$, then
$\langle{\psi_{k}}|W|{\psi_{k^{\prime}}}\rangle=1$ holds for $k\neq
k^{\prime}$ because of Eq. (59). By considering the positivity of the
submatrix

$\displaystyle\begin{pmatrix}\langle{\psi_{0}}|W|{\psi_{0}}\rangle&\langle{\psi_{0}}|W|{\psi_{1}}\rangle&\cdots&\langle{\psi_{0}}|W|{\psi_{d-1}}\rangle\\\
\langle{\psi_{1}}|W|{\psi_{0}}\rangle&\langle{\psi_{1}}|W|{\psi_{1}}\rangle&\ddots&\langle{\psi_{1}}|W|{\psi_{d-1}}\rangle\\\
\vdots&\ddots&\ddots&\vdots\\\
\langle{\psi_{d-1}}|W|{\psi_{0}}\rangle&\langle{\psi_{d-1}}|W|{\psi_{1}}\rangle&\cdots&\langle{\psi_{d-1}}|W|{\psi_{d-1}}\rangle\end{pmatrix}\geq
0,$ (119)

where all of the off-diagonal elements are 1, we obtain Eq. (118).

In total, we uniquely determined the diagonal elements of $W$.

##  Appendix E Evaluation of the Summation in the Proof of Lemma 1

We evaluate the summation $\sum_{J\in G_{a}\otimes
G_{b}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ for each $a,b$ in the following. Note
that it is enough to consider only 6 cases that satisfy $a\leq b$ because of
the symmetry between the two systems. In the following, when we take summation
on $J$, it also indicates the summation on the corresponding variables in the
form $|{ijkl}\rangle\\!\langle{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}|$.

(Case 1: $G_{1}\otimes G_{1}$) In this case, the equality

$\displaystyle\sum_{J\in G_{1}\otimes G_{1}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=\sum_{J\in G_{1}\otimes
G_{1}}\|\langle{ijkl}||{W}\rangle\\!\langle{W}||{i^{\prime}j^{\prime}k^{\prime}l^{\prime}}\rangle\|_{1}$
(120) $\displaystyle=\sum_{J\in G_{1}\otimes
G_{1}}(\|\delta_{jk}\delta_{j^{\prime}k^{\prime}}|{il00}\rangle\\!\langle{i^{\prime}l^{\prime}00}|\|_{1}+\|\delta_{il}\delta_{i^{\prime}l^{\prime}}|{kj11}\rangle\\!\langle{k^{\prime}j^{\prime}11}|\|_{1}$
$\displaystyle\qquad\qquad+\|\delta_{jk}\delta_{i^{\prime}l^{\prime}}|{il00}\rangle\\!\langle{k^{\prime}j^{\prime}11}|\|_{1}+\|\delta_{il}\delta_{j^{\prime}k^{\prime}}|{kj11}\rangle\\!\langle{i^{\prime}l^{\prime}00}|\|_{1}),$
(121)

holds. Moreover, the summation over the 1st and 2nd terms or the 3rd and 4th
terms are equal because of the symmetry, and thus we only evaluate the 1st and
3rd terms here.

For the 1st term of Eq. (121), the equality

$\displaystyle\sum_{J\in G_{1}\otimes
G_{1}}\|\delta_{jk}\delta_{j^{\prime}k^{\prime}}|{il00}\rangle\\!\langle{i^{\prime}l^{\prime}00}|\|_{1}=\sum_{J\in
G_{1}\otimes G_{1}}\delta_{jk}\delta_{j^{\prime}k^{\prime}}$ (122)

holds. In order to evaluate the summation, we consider the following two
cases. If $j\neq j^{\prime}$, there are
$d(d-1)(d-2)(d-3)+5d(d-1)(d-2)+2d(d-1)$ possible choices on
$(i,j,i^{\prime},j^{\prime})$ and for each choice of
$(i,j,i^{\prime},j^{\prime})$, there are only $(d-2)(d-3)+5(d-2)+2$ possible
choices on $(k,l,k^{\prime},l^{\prime})$ because $k=j,k^{\prime}=j^{\prime}$
is necessary for the non-vanishing Kronecker delta. Similarly, if
$j=j^{\prime}$, there are $d(d-1)(d-2)$ possible choices on
$(i,j,i^{\prime},j^{\prime})$ and for each choice of
$(i,j,i^{\prime},j^{\prime})$, there are $(d-1)(d-2)$ possible choices on
$(k,l,k^{\prime},l^{\prime})$. For the 3rd term of Eq. (121), the equality

$\displaystyle\sum_{J\in G_{1}\otimes
G_{1}}\|\delta_{jk}\delta_{i^{\prime}l^{\prime}}|{il00}\rangle\\!\langle{k^{\prime}j^{\prime}11}|\|_{1}=\sum_{J\in
G_{1}\otimes G_{1}}\delta_{jk}\delta_{i^{\prime}l^{\prime}}$ (123)

holds. Similarly, if $j\neq i^{\prime}$, there are
$d(d-1)(d-2)(d-3)+5d(d-1)(d-2)+d(d-1)$ possible choices on
$(i,j,i^{\prime},j^{\prime})$ and for each choice of
$(i,j,i^{\prime},j^{\prime})$, there are $(d-2)(d-3)+5(d-2)+1$ possible
choices on $(k,l,k^{\prime},l^{\prime})$. If $j=i^{\prime}$, there are
$d(d-1)(d-2)+d(d-1)$ possible choices on $(i,j,i^{\prime},j^{\prime})$ and for
each choice of $(i,j,i^{\prime},j^{\prime})$, there are $(d-1)(d-2)+(d-1)$
possible choices on $(k,l,k^{\prime},l^{\prime})$.

In total, we obtain

$\displaystyle\sum_{J\in G_{1}\otimes G_{1}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=2\sum_{J\in G_{1}\otimes
G_{1}}\delta_{jk}\delta_{j^{\prime}k^{\prime}}+2\sum_{J\in G_{1}\otimes
G_{1}}\delta_{jk}\delta_{i^{\prime}l^{\prime}}$ (124)
$\displaystyle=2d(d-1)(2d^{4}+2d^{3}-18d^{2}+11d+8).$ (125)

(Case 2: $G_{1}\otimes G_{2}$) In this case, the equality

$\displaystyle\sum_{J\in G_{1}\otimes G_{2}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=\sum_{J\in
G_{1}}\sum_{m=0}^{d-1}\|\sum_{k=0}^{d-1}\langle{ijk,k+m}||{W}\rangle\\!\langle{W}||{i^{\prime}j^{\prime}k,k+m}\rangle\|_{1}$
(126) $\displaystyle=\sum_{J\in
G_{1}}\sum_{m=0}^{d-1}\|\sum_{k=0}^{d-1}(\delta_{jk}\delta_{j^{\prime}k}|{i,k+m,00}\rangle\\!\langle{i^{\prime},k+m,00}|+\delta_{i,k+m}\delta_{i^{\prime},k+m}|{kj11}\rangle\\!\langle{kj^{\prime}11}|$
$\displaystyle\qquad+\delta_{jk}\delta_{i^{\prime},k+m}|{i,k+m,00}\rangle\\!\langle{kj^{\prime}11}|+\delta_{i,k+m}\delta_{j^{\prime}k}|{kj11}\rangle\\!\langle{i^{\prime},k+m,00}|)\|_{1}$
(127) $\displaystyle=\sum_{J\in
G_{1}}\sum_{m,k}(\delta_{jk}\delta_{j^{\prime}k}+\delta_{i,k+m}\delta_{i^{\prime},k+m}+\delta_{jk}\delta_{i^{\prime},k+m}+\delta_{i,k+m}\delta_{j^{\prime}k})$
(128)

holds. The summation over the 1st and 2nd terms or the 3rd and 4th terms are
equal, and thus we only evaluate the 1st and 3rd terms here.

The summation over the 1st term of Eq. (128) can be evaluated as
$d^{2}(d-1)(d-2)$, because for each $k,m=0,\ldots,d-1$, there are $(d-1)(d-2)$
possible choices on $(i,j,i^{\prime},j^{\prime})$ due to $j=j^{\prime}=k$. The
summation over the 3rd term of Eq. (128) can be evaluated as follows: for each
$k=0,\ldots,d-1$, if $m=0$, there are $(d-1)(d-2)+(d-1)$ possible choices on
$(i,j,i^{\prime},j^{\prime})$, and if $m\neq 0$, there are
$(d-2)(d-3)+5(d-2)+1$ possible choices on $(i,j,i^{\prime},j^{\prime})$.

In total, we obtain

$\displaystyle\sum_{J\in G_{1}\otimes G_{2}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=2\\{d^{2}(d-1)(d-2)+d[(d-1)(d-2)+(d-1)]$
$\displaystyle\qquad+d(d-1)[(d-2)(d-3)+5(d-2)+1]\\}$ (129)
$\displaystyle=2d(d-1)(2d^{2}-d-4).$ (130)

(Case 3: $G_{2}\otimes G_{2}$) In this case, the summation is evaluated as

$\displaystyle\sum_{J\in G_{2}\otimes G_{2}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=\sum_{n,m=0}^{d-1}\sum_{i,k=0}^{d-1}(\delta_{i+n,k}\delta_{i+n,k}+\delta_{i,k+m}\delta_{i,k+m}+\delta_{i+n,k}\delta_{i,k+m}+\delta_{i,k+m}\delta_{i+n,k})$
(131)
$\displaystyle=2\sum_{n,m=0}^{d-1}\sum_{i,k=0}^{d-1}(\delta_{i+n,k}\delta_{i+n,k}+\delta_{i+n,k}\delta_{i,k+m})$
(132) $\displaystyle=2(d^{3}+d^{2})=2d^{2}(d+1).$ (133)

(Case 4: $G_{1}\otimes G_{3}$) Here we consider $G_{3}$ with only the elements
of the form $|{kl}\rangle\\!\langle{kk}|-|{ll}\rangle\\!\langle{lk}|$. The
elements of the other form can be evaluated in the same way since the two
input systems are symmetric. We denote this group as $G_{3}^{\prime}$, that
is,

$\displaystyle G_{3}^{\prime}$
$\displaystyle=\\{|{kl}\rangle\\!\langle{kk}|-|{ll}\rangle\\!\langle{lk}|\mid
k\neq l\\}$ (134)

and the following equality holds

$\displaystyle\sum_{J\in G_{1}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=2\sum_{J\in G_{1}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}.$ (135)

In this case, the summation can be evaluated as

$\displaystyle\sum_{J\in G_{1}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=\sum_{J\in
G_{1}}\sum_{k\neq
l}\|\langle{ijkl}||{W}\rangle\\!\langle{W}||{i^{\prime}j^{\prime}kk}\rangle-\langle{ijll}||{W}\rangle\\!\langle{W}||{i^{\prime}j^{\prime}lk}\rangle\|_{1}$
(136) $\displaystyle=\sum_{J\in G_{1}}\sum_{k\neq
l}\|(\delta_{jk}\delta_{j^{\prime}k}|{il00}\rangle\\!\langle{i^{\prime}k00}|-\delta_{jl}\delta_{j^{\prime}l}|{il00}\rangle\\!\langle{i^{\prime}k00}|)$
$\displaystyle\qquad\qquad+(\delta_{il}\delta_{i^{\prime}k}|{kj11}\rangle\\!\langle{kj^{\prime}11}|-\delta_{il}\delta_{i^{\prime}k}|{lj11}\rangle\\!\langle{lj^{\prime}11}|)$
$\displaystyle\qquad\qquad+(\delta_{jk}\delta_{i^{\prime}k}|{il00}\rangle\\!\langle{kj^{\prime}11}|-\delta_{jl}\delta_{i^{\prime}k}|{il00}\rangle\\!\langle{lj^{\prime}11}|)$
$\displaystyle\qquad\qquad+(\delta_{il}\delta_{j^{\prime}k}|{kj11}\rangle\\!\langle{i^{\prime}k00}|-\delta_{il}\delta_{j^{\prime}l}|{lj11}\rangle\\!\langle{i^{\prime}k00}|)\|_{1}$
(137) $\displaystyle=\sum_{J\in G_{1}}\sum_{k\neq
l}(\|\delta_{jk}\delta_{j^{\prime}k}|{il00}\rangle\\!\langle{i^{\prime}k00}|-\delta_{jl}\delta_{j^{\prime}l}|{il00}\rangle\\!\langle{i^{\prime}k00}|\|_{1}$
$\displaystyle\qquad\qquad+\|\delta_{il}\delta_{i^{\prime}k}|{kj11}\rangle\\!\langle{kj^{\prime}11}|-\delta_{il}\delta_{i^{\prime}k}|{lj11}\rangle\\!\langle{lj^{\prime}11}|\|_{1}$
$\displaystyle\qquad\qquad+\|\delta_{jk}\delta_{i^{\prime}k}|{il00}\rangle\\!\langle{kj^{\prime}11}|-\delta_{jl}\delta_{i^{\prime}k}|{il00}\rangle\\!\langle{lj^{\prime}11}|\|_{1}$
$\displaystyle\qquad\qquad+\|\delta_{il}\delta_{j^{\prime}k}|{kj11}\rangle\\!\langle{i^{\prime}k00}|-\delta_{il}\delta_{j^{\prime}l}|{lj11}\rangle\\!\langle{i^{\prime}k00}|\|_{1})$
(138) $\displaystyle=\sum_{J\in G_{1}}\sum_{k\neq
l}(\delta_{jk}\delta_{j^{\prime}k}+\delta_{jl}\delta_{j^{\prime}l}+2\delta_{il}\delta_{i^{\prime}k}+\delta_{jk}\delta_{i^{\prime}k}+\delta_{jl}\delta_{i^{\prime}k}+\delta_{il}\delta_{j^{\prime}k}+\delta_{il}\delta_{j^{\prime}l})$
(139) $\displaystyle=2\sum_{J\in G_{1}}\sum_{k\neq
l}(\delta_{jk}\delta_{j^{\prime}k}+\delta_{il}\delta_{i^{\prime}k}+\delta_{jk}\delta_{i^{\prime}k}+\delta_{jl}\delta_{i^{\prime}k}),$
(140)

where the third equality holds because the states of control qubits are
different, the forth equality holds because $k\neq l$, and the last equality
holds because the summation over the 1st and 2nd, 4th and 7th, 5th and 6th
terms are the same, respectively. For each $(k,l)$, by counting the number of
possible choices of $(i,j,i^{\prime},j^{\prime})$, we obtain

$\displaystyle\sum_{J\in G_{1}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$
$\displaystyle=2d(d-1)\sum_{J\in
G_{1}}(\delta_{jk}\delta_{j^{\prime}k}+\delta_{il}\delta_{i^{\prime}k}+\delta_{jk}\delta_{i^{\prime}k}+\delta_{jl}\delta_{i^{\prime}k}),$
(141) $\displaystyle=2d(d-1)\times\\{[(d-1)(d-2)]+[(d-2)(d-3)+5(d-2)+2]$
$\displaystyle\qquad+[(d-1)(d-2)+(d-1)]+[(d-2)(d-3)+5(d-2)+1]\\}$ (142)
$\displaystyle=2d(d-1)(4d^{2}-5d-2),$ (143)

and thus, we obtain

$\displaystyle\sum_{J\in G_{1}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=4d(d-1)(4d^{2}-5d-2).$ (144)

(Case 5: $G_{2}\otimes G_{3}$) As in the case 4, we consider $G_{3}^{\prime}$
instead, and using the following equality:

$\displaystyle\sum_{J\in G_{2}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=2\sum_{J\in G_{2}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}.$ (145)

In this case, the summation can be evaluated as

$\displaystyle\sum_{J\in G_{2}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=\sum_{k\neq
l}\sum_{m=0}^{d-1}\|\sum_{i=0}^{d-1}(\langle{i,i+m,kl}||{W}\rangle\\!\langle{W}||{i,i+m,kk}\rangle-\langle{i,i+m,ll}||{W}\rangle\\!\langle{W}||{i,i+m,lk}\rangle)\|_{1}$
(146) $\displaystyle=\sum_{k\neq
l}\sum_{m=0}^{d-1}\|\sum_{i=0}^{d-1}\delta_{i+m,k}\delta_{i+m,k}|{il00}\rangle\\!\langle{ik00}|-\delta_{i+m,l}\delta_{i+m,l}|{il00}\rangle\\!\langle{ik00}|\|_{1}$
$\displaystyle\quad+\|\sum_{i}\delta_{il}\delta_{ik}|{k,i+m,11}\rangle\\!\langle{k,i+m,11}|-\delta_{il}\delta_{ik}|{l,i+m,11}\rangle\\!\langle{l,i+m,11}|\|_{1}$
$\displaystyle\quad+\|\sum_{i}\delta_{i+m,k}\delta_{ik}|{il00}\rangle\\!\langle{k,i+m,11}|-\delta_{i+m,l}\delta_{ik}|{il00}\rangle\\!\langle{l,i+m,11}|\|_{1}$
$\displaystyle\quad+\|\sum_{i}\delta_{il}\delta_{i+m,k}|{k,i+m,11}\rangle\\!\langle{ik00}|-\delta_{il}\delta_{i+m,l}|{l,i+m,11}\rangle\\!\langle{ik00}|\|_{1}$
(147) $\displaystyle=\sum_{k\neq
l}\sum_{m,i=0}^{d-1}(\delta_{i+m,k}\delta_{i+m,k}+\delta_{i+m,l}\delta_{i+m,l}+\delta_{il}\delta_{ik}+\delta_{il}\delta_{ik}$
$\displaystyle\qquad\qquad\qquad+\delta_{i+m,k}\delta_{ik}+\delta_{i+m,l}\delta_{ik}+\delta_{il}\delta_{i+m,k}+\delta_{il}\delta_{i+m,l})$
(148) $\displaystyle=d(d-1)\times(d+d+0+0+1+1+1+1)$ (149)
$\displaystyle=2d(d-1)(d+2).$ (150)

Here, the second equality holds because the states of control qubits are
different, the third equality holds because $k\neq l$, and the fourth equality
holds by counting the number of possible choices of
$(i,j,i^{\prime},j^{\prime})$ for each $(k,l)$. In total, we obtain

$\displaystyle\sum_{J\in G_{2}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=4d(d-1)(d+2).$ (151)

(Case 6: $G_{3}\otimes G_{3}$) In this case, we divide the first $G_{3}$ into
two groups, i.e., $G_{3}^{\prime}$ and $G_{3}^{\prime\prime}$, where
$G_{3}^{\prime}$ defined by Eq. (134) is the one used in cases 4 and 5, and
$G_{3}^{\prime\prime}$ is defined as the remaining of $G_{3}$, that is,

$\displaystyle G_{3}^{\prime\prime}$
$\displaystyle=\\{|{ji}\rangle\\!\langle{ii}|-|{jj}\rangle\\!\langle{ij}|\mid
i\neq j\\}.$ (152)

Here we evaluate both $G_{3}^{\prime}\otimes G_{3}^{\prime}$ and
$G_{3}^{\prime\prime}\otimes G_{3}^{\prime}$, and the result can be obtained
by

$\displaystyle\sum_{J\in G_{3}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=2(\sum_{J\in G_{3}^{\prime}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}+\sum_{J\in
G_{3}^{\prime\prime}\otimes G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}).$
(153)

We first consider the summation over $G_{3}^{\prime}\otimes G_{3}^{\prime}$,
which is evaluated as

$\displaystyle\sum_{J\in G_{3}^{\prime}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=\sum_{i\neq
j}\sum_{k\neq
l}\|\langle{ijkl}||{W}\rangle\\!\langle{W}||{iikk}\rangle-\langle{ijll}||{W}\rangle\\!\langle{W}||{iilk}\rangle$
$\displaystyle\qquad\qquad-\langle{jjkl}||{W}\rangle\\!\langle{W}||{jikk}\rangle+\langle{jjll}||{W}\rangle\\!\langle{W}||{jilk}\rangle\|_{1}$
(154) $\displaystyle=\sum_{i\neq j,k\neq
l}\|(\delta_{jk}\delta_{ik}-\delta_{jl}\delta_{il})|{il00}\rangle\\!\langle{ik00}|-(\delta_{jk}\delta_{ik}-\delta_{jl}\delta_{il})|{jl00}\rangle\\!\langle{jk00}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{il}\delta_{ik}-\delta_{jl}\delta_{jk})|{kj11}\rangle\\!\langle{ki11}|-(\delta_{il}\delta_{ik}-\delta_{jl}\delta_{jk})|{lj11}\rangle\\!\langle{li11}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{jk}\delta_{ik}|{il00}\rangle\\!\langle{ki11}|-\delta_{jl}\delta_{ik}|{il00}\rangle\\!\langle{li11}|-\delta_{jk}\delta_{jk}|{jl00}\rangle\\!\langle{ki11}|+\delta_{jl}\delta_{jk}|{jl00}\rangle\\!\langle{li11}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{il}\delta_{ik}|{kj11}\rangle\\!\langle{ik00}|-\delta_{il}\delta_{il}|{lj11}\rangle\\!\langle{ik00}|-\delta_{jl}\delta_{ik}|{kj11}\rangle\\!\langle{jk00}|+\delta_{jl}\delta_{il}|{lj11}\rangle\\!\langle{jk00}|\|_{1}$
(155) $\displaystyle=\sum_{i\neq j,k\neq
l}3\delta_{jk}\delta_{ik}+3\delta_{jl}\delta_{il}+3\delta_{il}\delta_{ik}+3\delta_{jl}\delta_{jk}+2\delta_{jl}\delta_{ik}+\delta_{jk}\delta_{jk}+\delta_{il}\delta_{il}$
(156) $\displaystyle=\sum_{i\neq j,k\neq
l}2\delta_{jl}\delta_{ik}+\delta_{jk}\delta_{jk}+\delta_{il}\delta_{il}$ (157)
$\displaystyle=2d(d-1)+2d(d-1)^{2}=2d^{2}(d-1),$ (158)

where the second equality holds because the states of control qubits are
different, and the third equality holds because $i\neq j$ and $k\neq l$.
Similarly, the summation over $G_{3}^{\prime\prime}\otimes G_{3}^{\prime}$ can
be evaluated as

$\displaystyle\sum_{J\in G_{3}^{\prime\prime}\otimes
G_{3}^{\prime}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}$ $\displaystyle=\sum_{i\neq
j,k\neq
l}\|\langle{jikl}||{W}\rangle\\!\langle{W}||{iikk}\rangle-\langle{jill}||{W}\rangle\\!\langle{W}||{iilk}\rangle$
$\displaystyle\qquad\qquad-\langle{jjkl}||{W}\rangle\\!\langle{W}||{ijkk}\rangle+\langle{jjll}||{W}\rangle\\!\langle{W}||{ijlk}\rangle\|_{1}$
(159) $\displaystyle=\sum_{i\neq j,k\neq
l}\|(\delta_{ik}\delta_{ik}-\delta_{il}\delta_{il}-\delta_{jk}\delta_{jk}+\delta_{jl}\delta_{jl})|{jl00}\rangle\\!\langle{ik00}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{jl}\delta_{ik}|{ki11}\rangle\\!\langle{ki11}|-\delta_{jl}\delta_{ik}|{li11}\rangle\\!\langle{li11}|-\delta_{jl}\delta_{ik}|{kj11}\rangle\\!\langle{kj11}|+\delta_{jl}\delta_{ik}|{lj11}\rangle\\!\langle{lj11}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{ik}\delta_{ik}|{jl00}\rangle\\!\langle{ki11}|-\delta_{il}\delta_{ik}|{jl00}\rangle\\!\langle{li11}|-\delta_{jk}\delta_{ik}|{jl00}\rangle\\!\langle{kj11}|+\delta_{jl}\delta_{ik}|{jl00}\rangle\\!\langle{lj11}|)\|_{1}$
$\displaystyle\qquad+\|(\delta_{jl}\delta_{ik}|{ki11}\rangle\\!\langle{ik00}|-\delta_{jl}\delta_{il}|{li11}\rangle\\!\langle{ik00}|-\delta_{jl}\delta_{jk}|{kj11}\rangle\\!\langle{ik00}|+\delta_{jl}\delta_{jl}|{lj11}\rangle\\!\langle{ik00}|\|_{1}$
(160) $\displaystyle=\sum_{i\neq j,k\neq
l}2\delta_{ik}\delta_{ik}+\delta_{il}\delta_{il}+\delta_{jk}\delta_{jk}+2\delta_{jl}\delta_{jl}+4\delta_{jl}\delta_{ik}+\delta_{il}\delta_{ik}+\delta_{jk}\delta_{ik}+2\delta_{jl}\delta_{ik}+\delta_{jl}\delta_{il}+\delta_{jl}\delta_{jk}$
(161) $\displaystyle=\sum_{i\neq j,k\neq
l}4\delta_{jl}\delta_{ik}+2\delta_{ik}\delta_{ik}+2\delta_{jl}\delta_{jl}+2\delta_{jl}\delta_{ik}+\delta_{il}\delta_{il}+\delta_{jk}\delta_{jk}$
(162) $\displaystyle=d(d-1)[4+2(d-1)+2(d-1)+2+(d-1)+(d-1)]$ (163)
$\displaystyle=6d^{2}(d-1)$ (164)

where the second equality holds because the states of control qubits are
different. For the third equality, the expansion can be done for all but the
first term using $i\neq j$ and $k\neq l$. For the first term, there exist some
patterns where more than one terms of the Kronecker deltas are non-zero, for
example, $\delta_{ik}\delta_{ik}$ and $\delta_{jl}\delta_{jl}$ can be non-zero
simultaneously. However, since they have same sign, the expansion is possible
and the third equality holds.

In total, we obtain

$\displaystyle\sum_{J\in G_{3}\otimes
G_{3}}\|{\mathrm{Tr}}_{in}WJ^{t}\|_{1}=16d^{2}(d-1).$ (165)

Finally, we have evaluated all terms in Eq. (94), and we obtain

$\displaystyle\|W\|_{1}\geq(2d^{3})^{2},$ (166)

which completes the proof of Lemma 1.