Optimal simulation of three-qubit gates

0.17 MB·English

by  Nengkun Yu

Optimal simulation of three-qubit gates
3
Nengkun Yu and Mingsheng Ying
1
0 State Key Laboratory of Intelligent Technology and Systems,
2 Tsinghua National Laboratory for Information Science and Technology,

n Department of Computer Science and Technology,
a
Tsinghua University, Beijing 100084, China
J
6 Center for Quantum Computation and Intelligent Systems (QCIS),
1 Faculty of Engineering and Information Technology,

University of Technology, Sydney, NSW 2007, Australia
]
h
p January 17, 2013
-
t
n
a
Abstract
u
q
In this paper, we study the optimal simulation of three-qubit unitary
[
by using two-qubit gates. First, we give a lower bound on the two-qubit

1 gatescostofsimulatingamulti-qubitgate. Secondly,wecompletelychar-
v acterize the two-qubit gate cost of simulating a three-qubit controlled
7 controlled gate bygeneralizing our result on thecost of Toffoli gate. The
2 function of controlled controlled gate is simply a three-qubit controlled
7 unitary gate and can be intuitively explained as follows: the gate will
3
output the states of the two control qubit directly, and apply the given
1. one-qubit unitary u on the target qubit only if both the states of the
0 control are |1i. Previously, it is only known that five two-qubit gates
3 is sufficient for implementing such a gate [Sleator and Weinfurter, Phys.
1 Rev. Lett. 74, 4087 (1995)]. Our result shows that if the determinant
v: of u is 1, four two-qubit gates is achievable optimal. Otherwise, five is
i optimal. Thirdly, we show that five two-qubit gates are necessary and
X
sufficient for implementing the Fredkin gate(the controlled swap gate),
r which settles the open problem introduced in [Smolin and DiVincenzo,
a
Phys. Rev. A, 53, 2855 (1996)]. The Fredkin gate is one of the most
important quantum logic gates because it is universal alone for classical
reversible computation, and thus with little help, universal for quantum
computation. Before our work, a five two-qubit gates decomposition of
the Fredkin gate was already known, and numerical evidence of showing
fiveis optimal is found.
1 Introduction
A fundamental issue of several interacting systems is to quantify the strength
of this interaction. Particularly valuable are techniques that can compare in-
1
teractions of quite different types of system or particles and different physi-
cal manifestations of the interaction. In order to provide an intellectual back-
ground of comparing interaction strength of different physical systems, robust
characterizations is desirable. Quantum information theory has provided quite
new insights into this question. In particular, there has been considerable
progress in quantifying the strength of Hamiltonian and unitary interactions
[1, 2, 3, 4, 5, 6, 7, 8, 9] for bipartite system. The starting point was the the-
ory of entanglement of quantum states which quantifies how much nonclassical
correlation the state embodies. It would be quite interesting to extend these
resultstomultipartitesystem. Themostintriguingapproachmightbetostudy
how much bipartite correlation is needed to implement a multipartite correla-
tion. Inotherwords,howmanytwo-qubitunitaryisneededtosimulateagiven
multi-qubit gate? One obstacle standing in front of such desire is the fact that
multipartite entanglement is difficult to characterize. A possible direction is to
consider those symmetric gates.
This fundamental topic is clearly of interest to experimentalists who try to
createsystemsininteraction. Agreatchallengeinthecontemporaryscienceand
engineering is building a full-fledged quantum computer, which is essentially a
large quantum circuit consisting of basic quantum logical gates. In order to
accomplish a quantum algorithm, even in a small size, one has to implement
a relatively high level of control over the multi-qubit quantum system. It has
also been experimentally demonstrated that two-qubit gates can be realized
withhighfidelityusingthecurrenttechnology,forexample,two-qubitgatewith
superconductingquibtshavebeenpresentedwithfidelitieshigherthan90%[21].
Finding more efficient waysto implement quantum gates may allow small-scale
quantum computing tasks to be demonstrated on a shorter time scale.
Due to its significance in quantum computing, lots of efforts have been de-
votedtostudycorrelationofcontrolledunitary,see[10,11,12,13,14]asaquite
incomplete list. But no affirm result is known, even for some highly symmetric
three-qubit gate. Very recently, we have showed that five two-qubit gates are
optimalforimplementingaToffoligate[15]byemployingbasictechniquesfrom
quantum information.
Another gate that has received particular attention is the three qubit con-
ditional swap gate, or Fredkin gate. The Fredkin gate is of interest because it
is a universal gate for classical reversible computation[16], which means that
anylogicalorarithmeticoperationcanbeconstructedentirelyofFredkingates.
The quantum versionhas been used by Ekertand Macchiavelloto design a cir-
cuit for error correcting quantum computations with the symmetric subspace
methodof[17]. The experimentalandtheoreticalpursuitofefficientimplemen-
tation of the Fredkin gate using a sequence of single- and two-qubit gates has
quite a long history. A simple optical model to realize a reversible, potentially
error-freelogic gateaFredkingate is proposedin[18]. ChauandWilczek givea
specific six-gate construction of the Fredkin gate in [19]. An analytic five-gate
construction is presented and numerical tests suggest is minimal in [20].
In this paper, the technique introduced in [15] is used to deal with gener-
alized three-qubit controlled controlled gate and Fredkin gate. The two-qubit
2
gates cost on simulating three-qubit controlled controlled gate is completely
characterized. More precisely, it is showed that any controlled controlled gate
requires at least four two-qubit gates to simulate. If the determinant of the
controlled unitary is one, then four gates simulation is achievable. Otherwise,
five gates implementationis optimal. Later,we presenta theoreticalproofthat
a five two-bit gate is indeed the optimal implementation of the Fredkin gate.
2 Preliminaries and Notations
Note that any bipartite unitary U acting on a qubit system A and a general
AB
systemBissaidtobeacontrolled-gatewithcontrolonAifitcanbedecomposed
into the form of
U = 0 0 U + 1 1 U .
AB A A 0 A A 1
| ih |⊗ | ih |⊗
A controlled-controlled gate, acting on three quantum bits, namely A,B,
and C. Here A and B are control qubits, and C is the target qubit with
computational basis 0 , 1 for each qubit. Upon input abc , the gate will
{| i | i} | i
output the states of A and B directly, and apply U on the system C only if
both the states of A and B are 1 .
| i
In this letter, each three-qubit gate is regarded as a unitary transformation
performed on a tripartite system ABC, all the two-qubit gates employed to
implement three-qubit gate can be simply classified into three classes: class
-thegatesactingonthesubsystemAB,class -thegatesonBC,and
AB BC
K K
class - the gates on AC. Obviously, it is impossible that all the two-qubit
AC
K
gates used to simulate F belong to a single one of the three classes ,
ABC AB
K
, .
BC AC
K K
The validity of following propositions is showed in [15].
Proposition1. Anytwo-dimensional2 2statesubspacecontainssomeproduct
⊗
state.
Proposition 2. If U U is a three-qubit controlled unitary with control on
AB AC
A, where U and U , then there exist v ,v and w ,w
AB AB AC AC B1 B2 C1 C2
∈K ∈K
being one-qubit unitaries on and such that
B C
H H
U U = 0 0 v w + 1 1 v w .
AB AC B1 C1 B2 C2
| ih |⊗ ⊗ | ih |⊗ ⊗
3 Lower Bound for simulating general multi-qubit
gate
First, we give a general bound on the cost on the two-qubit gates for imple-
menting milti-qubit gates.
Theorem 3.1. Almost any n-qubit gate requires at least 4n−3n−1 two-qubit
⌈ 9 ⌉
unitaries to implement without ancilla.
3
Proof—: This theorem is proved by simply counting the degree of free-
dom(DOF).
Without loss of generality(wlog),we only consider quantum gates with unit
determinant, i.e, detU =1. It is well known that the DOF of n qubit unitary
−
with unit determinant is 4n 1. As a special case, the DOF of such two-qubit
−
unitary becomes 15. More precisely, it is proved in [8] that each two-qubit
unitary operation U can be expressed into form: U = (u u )U (v v ),
A B d A B
⊗ ⊗
where u , u , v , v are one-qubit unitary gates with unit determinant, and
A B A B
U =exp[i(α X X+α Y Y +α Z Z)], and X, Y, Z are Pauli matrices.
d x y z
⊗ ⊗ ⊗
Nowweassumethatarbitraryn-qubitgatesofsystem
HA1⊗HA2⊗···HAn
could be implemented by some circuit consisting of k two-qubit gates. Notice
that the structure of the circuit with five two-qubit gates is finite. For any
fixed structure, there are n gates on , respectively. It is easy to see that
i Ai
n n =2k. Therearen +1 localunHitaries on , the DOF forthis partis
i=1 i i HAi
3P(ni+1). Noticing the DOF of each two-qubit gate is 3, then the DOF of the
whole circuit is less or equal to
n
3 k+3 ( (n +1))=9k+3n.
i
× ×
Xi=1
It is easy to see that
4n 3n 1
9k+3n 4n 1 k − −
≥ − ⇒ ≥⌈ 9 ⌉
The proof of this theorem is complete.
Let n=3, we know that the simulation of a general three-qubit gate would
require at least 43−3×3−1 =6 two-qubitgates. The following nature question
⌈ 9 ⌉
arises,
Whether six two-qubit quantum gates are sufficient to generate any three-bit
quantum gate?
Unfortunately, we are not able to solve this problem. Instead, we can show
theoptimalimplementationofsomesymmetricthree-qubitgatesinthefollowing
sections.
4 Optimal simulation of controlled-controlled gate
Theparticular“controlled-controlled”gateisintroducedbyDeutschin[22]and
itisprovedthatsomeofsuchgateisuniversalforthefirsttime. Thatistheyare
adequate for constructing networks with any possible quantum computational
property.
It is proved in [13] that any controlled-controlled gate can be implemented
by using only five two-qubit gates circuit,
• • • •
=
• • •
U W W† W
4
The left circuit denotes “controlled-controlled-U” gate, W on the right hand
side is a unitary satisfying W2 =U.
To study the two-qubit gate cost for implementing the general “controlled-
controlled-U” gate, we only need to deal with diagonal U, i.e.,
1 0 0 0 0 0 0 0
 0 1 0 0 0 0 0 0 
0 0 1 0 0 0 0 0
 
 0 0 0 1 0 0 0 0 
V(θ ,θ )= .
1 2  0 0 0 0 1 0 0 0 
 
 0 0 0 0 0 1 0 0 
 
 0 0 0 0 0 0 eiθ1 0 
 
 0 0 0 0 0 0 0 eiθ2 
 
V(θ ,θ ) is regarded as a tripartite unitary with control on A and B.
1 2 ABC
It can be considered as a control unitary with control on A, B or C,
V(θ ,θ ) = 0 0 I + 1 1 R ,
1 2 ABC BC BC
| ih |⊗ | ih |⊗
where R= 00 00 + 01 01 +eiθ1 10 10 +eiθ2 11 11.
| ih | | ih | | ih | | ih |
Also, we can verify that V(θ ,θ ) is invariant under the permutation
1 2 ABC
of A and B, that is V(θ ,θ ) = S V(θ ,θ ) S with S denoting
1 2 ABC AB 1 2 ABC AB BC
the swap gate on A and B. These observations are used during the following
argument.
The following result from [15] is useful for our discussion.
Theorem 4.1. V(0,θ)=I (1 eiθ)111 111 requires five two-qubit gates to
− − | ih |
simulate, provide that eiθ =1.
6
One can easily verify the following equation
V(0,θ θ ) =V(θ ,θ ) W( θ ) =W( θ ) V(θ ,θ ) ,
2 1 ABC 1 2 ABC 1 AB 1 AB 1 2 ABC
− − −
where W(θ) = 00 00 + 01 01 + 10 10 +eiθ 11 11. Therefore, we can
| ih | | ih | | ih | | ih |
conclude that
Lemma 4.1. Any three qubit controlled controlled gate would require at least
four two-qubit gates to simulate.
Suppose eiθ1 = eiθ2. If such V(θ ,θ ) can be implemented by three or less
1 2
6
two-qubit gates, then four two-qubit gates or less can simulate some V(0,θ)=
I (1 eiθ)111 111, conflict from the above theorem.
− − | ih |
For V( θ,θ), we can find the following simulation circuit consisting of fout
−
two-qubit gates, therefore, it is optimal.
A
• •
B
U† U
C BC W BC W
5
where
e−iθ/2 0
W = .
(cid:18) 0 eiθ/2 (cid:19)
andU (W I )U† =diag eiθ/2,e−iθ/2,e−iθ/2,eiθ/2 . SuchU doesexist
BC C⊗ B BC { } BC
since the eigenvalues of W I are eiθ/2,e−iθ/2,e−iθ/2,eiθ/2 .
C B
⊗ { }
In order to study the rest case, we begin from the following special circuit.
Lemma 4.2. There is no U ,V and U ,V such that
AC AC AC BC BC BC
∈ K ∈ K
V(θ ,θ ) can be implemented in the following circuit with ei(θ1+θ2) = 1 and
1 2
6
eiθ1 =eiθ2,
6
B
V U
BC BC
C
V U
AC AC
A
Proof—: Invoking Lemma 4.1, we only need to study the case that all the
two-qubit gate are nonlocal. The circuit is
U U V V =V(θ ,θ ) ,
AC BC AC BC 1 2 ABC
then U U V is a control unitary with control on A. Moreover, for any
AC BC AC
input state 0 ψ , the A’s part state of the following state is is 0 .
| iA| iBC | iA
U U V i ψ .
AC BC AC | iA| iBC
After moving the local unitaries by invoking Proposition 1, we assume
V 0 0 = 0 0
AC| iA| iC | iA| iC
Thus, the A’s part’s state of the following state is 0 ,
| iA
U U V 0 y 0 =U U 0 y 0 .
AC BC AC| iA| iB| iC AC BC| iA| iB| iC
There are three cases about the states U y 0 :
BC| iB| iC
Case 1: There is some y such that U y 0 becomes entangled.
| 0iB BC| iB| iC
Assume there is 0<λ<1 such that
U y 0 =√λα 0 +√1 λ α⊥ 1 .
BC| 0iB| iC | iB| iC − B| iC
(cid:12) (cid:11)
Define χ =U U 0 0 z , we know tha(cid:12)t
| iABC AC BC| iA| iB| 0iC
χ =√λΦ α +√1 λΨ α⊥ ,
| iABC | iAC| iB − | iAC B
χ =λΦ +(1 λ)Ψ Φ =Ψ =(cid:12) 0(cid:11)0,
A A A A A (cid:12)
⇒ − ⇒ | ih |
where Φ = U 00 and Ψ = U 01 . Therefore, U is a control
| i AC| iAC | i AC| iAC AC
unitary with control on system A, so is V . We can assume that
AC
U = 0 0 I + 1 1 w ,
AC | ih |⊗ C | ih |⊗ C1
V = 0 0 I + 1 1 w .
AC | ih |⊗ C | ih |⊗ C2
6
We know that U V =I and
BC BC BC
w U w V =R U w U† =w† R =w† w† D,
C1 BC C2 BC BC ⇒ BC C2 BC C1 BC C1 ⊕ C1
where D =diag eiθ1,eiθ2 and
{ }
R= 00 00 + 01 01 +eiθ1 10 10 +eiθ2 11 11.
| ih | | ih | | ih | | ih |
eiϕ1,eiϕ2,eiϕ1,eiϕ2 are the eigenvalues of U w U† , where eiϕ1,eiϕ2
{ } BC C2 BC
aretheeigenvaluesofw . Itisdirecttoseethatw† cannothavetwoidentical
C2 C1
eigenvalues, then w† and w† D enjoys the same eigenvalues, that leas us to
C1 C1
that they have the same determinant.
det(w† )=det(w† D)=det(w† )det(D) det(D)=1 ei(θ1+θ2) =1
C1 C1 C1 ⇒ ⇒
Contradiction!
Otherwise, for any y , U y 0 is product.
| iB BC| iB| iC
Case 2: There is a β and a local unitary w on system C such that
| iB C
U y 0 = β w y , thus, U maps 0 to itself. Therefore,
BC| iB| iC | iB C| iC AC {| iA}⊗HC
U is a control unitary with control on A, so is V . The rest argument of
AC AC
this case is the same as case 1.
Case 3: There is a state on system C, wlog, says 0 , and a local unitary
| iC
v on system B such that U y 0 = v y 0 . Therefore, U is a
B BC| iB| iC B| iB| iC BC
control unitary with control on C. By moving this v to V , we make the
B BC
assumption that
U = 0 0 I + 1 1 u .
BC B B
| ih |⊗ | ih |⊗
Note that for any y , we have
| iB
V(θ ,θ ) 0 (V† y 0 ) = U U V 0 y 0 ,
1 2 ABC| iA BC| iB| iC AC BC AC| iA| iB| iC
= 0 (V† y 0 ) = U 0 y 0 .
⇒| iA BC| iB| iC AC| iA| iB| iC
Thus part B’s state of V† y 0 is y for all y , which means
BC| iB| iC | iB | iB ∈ HB
that there is γ suchthat V y γ = y 0 . Therefore,one canfind a
| iC BC| iB| iC | iB| iC
unitary w such that
C
V = 0 γ I + 1 γ⊥ v .
BC B B
| ih |⊗ | ih |⊗
In order to simplify the structure of the two-qubit gates, we observe that
VT VT UT UT =V(θ ,θ )T =V(θ ,θ ) ,
BC AC BC AC 1 2 ABC 1 2 ABC
hence also provides a simulation of V(θ ,θ ) . Now we consider the state
1 2 ABC
VT VT UT x 0 0 =VT VT x 0 0
BC AC BC| iA| iB| iC BC AC| iA| iB| iC
forany x . Theargumentofcases1and2excludesthefollowingpossibilities:
| iA
(i) there is some x suchthatVT x 0 is entangled,or (ii) there is a δ
| iA AC| iA| iC | iA
and a local unitary w on system C such that VT x 0 = δ w x .
C AC| iA| iC | iA C| iC
7
So the only possibility is that there is a state φ on system C, and a local
| iC
unitary w on system A such that VT x 0 =w x φ .
A AC| iA| iC A| iA| iC
According to V 0 0 = 0 0 , we can choose φ = 0 . Thus V
AC| iA| iC | iA| iC | i | i AC
is a controlled gate with control system C, i.e.,
V = 0 0 w + 1 1 v .
AC A A
| ih |⊗ | ih |⊗
By studying part B’s state of
U U V V 0 y 0 = 0 y 0 ,
AC BC AC BC| iA| iB| iC | iA| iB| iC
weseethat γ definedinV equalsto 0 or 1 ,uptosomeglobalphase.
| iC BC | iB | iB
Otherwise, assume that 0 = aγ +b γ⊥ for ab = 0. Then the state of
| iC | iC C 6
part B becomes a mixed state for genera(cid:12)(cid:12)l in(cid:11)put |0iA|yiB|0iC since uB is not
identityuptosomeglobalphaseandU isnonlocal. Forthecase γ = 0 ,
BC | iC | iC
weknowthatallthefourtwo-qubitgatesarecontrolledgatewithcontrolsystem
C, which implies that R defined in case 1 is a local unitary, a contradiction.
BC
For the case γ = 1 , let X be the NOT (flip) gate such that X 0 = 1
| iC | iC C | i | i
and X 1 = 0 , then one can verify that
| i | i
(U X )(X U X )(X V X )(X V )=V(θ ,θ ) .
AC C C BC C C AC C C BC 1 2 ABC
Then U X ,X U X ,X V X and X V are all controlledgate with
AC C C BC C C AC C C BC
control system C. This also leads us to the impossible conclusion that R is
BC
local. Impossible!
Now we are able to show that
Theorem 4.2. V(θ ,θ ) requires five two-qubit gates to simulate, provide that
1 2
ei(θ1+θ2) =1 and eiθ1 =eiθ2.
6 6
Proof—: If the four gates belong to two of the classes , , , the
AB AC BC
K K K
circuits that need to be considered are just U U V V = V(θ ,θ )
AC BC AC BC 1 2 ABC
and U U V V =V(θ ,θ ) . The previous one is studied in Lemma
AB BC AB BC 1 2 ABC
4.2. ThelatteroneisimpossiblebyusingV(0,θ θ ) =W( θ ) V(θ ,θ )
2 1 ABC 1 AB 1 2 ABC
− −
and Lemma 4.1 directly.
Otherwise, the four gates belong to three of the classes , , ,
AB AC BC
K K K
then there exist two gates belongs to the same class. Due to the symmetric
property of the Fredkin gate, we need to consider the following two cases:
Case1: Twogatesbelongto ,thenatleastoneofthemliesinthefront
AB
K
or the end of the circuit. then we conclude that is impossible by apply Lemma
4.1 and
V(0,θ θ ) =W( θ ) V(θ ,θ ) =V(θ ,θ ) W( θ ) .
2 1 ABC 1 AB 1 2 ABC 1 2 ABC 1 AB
− − −
Case 2: Two gates belong to . The four gates are U , U
BC AB AB AC
K ∈K ∈
and U ,V . According to symmetric properties of the Fredkin
AC BC BC BC
K ∈K
gate, the three possible circuits are
Subcase 1: U U V U = V(θ ,θ ) , this is impossible since U
BC AC BC AB 1 2 ABC AB
lies in the end.
8
Subcase 2: U U V U = F . this circuit can be reduced to
BC AB BC AC ABC
Lemma 4.2 by noticing that S U ,S V and S U S
BC BC BC BC BC BC AC BC
∈ K ∈
and S F S is the controlled controlled gate with control on sys-
AB BC ABC BC
K
tems A and C,
(S U )U (S V )(S U S )=S F S .
BC BC AB BC BC BC AC BC BC ABC BC
Subcase 3: U U U V = V(θ ,θ ) . Observe that U U is a
BC AB AC BC 1 2 ABC AB AC
controlunitary with controlon A, we can conclude that R share eigenvalues
BC
with a local unitary by invoking Proposition 2. That is impossible.
5 Optimal simulation of Fredkin gate
The Fredkin gate is the three-qubit gate that swaps the last two-qubits if the
first-qubit is 1 . The matrix form of Fredkin gate is given as
| i
1 0 0 0 0 0 0 0
 0 1 0 0 0 0 0 0 
0 0 1 0 0 0 0 0
 
 0 0 0 1 0 0 0 0 
F = .
 0 0 0 0 1 0 0 0 
 
 0 0 0 0 0 0 1 0 
 
 0 0 0 0 0 1 0 0 
 
 0 0 0 0 0 0 0 1 
 
ItisshowedthattheFredkingatecanbesimulatedbyemployingfivetwo-qubit
gate in the following circuit [20] by letting V2 =X with X being the pauli flip
matrix,
• • •
• •
V V V†
• •
In this section, we will show that
Theorem 5.1. Four two-qubit gates are not sufficient for implementing F .
ABC
The proof of Theorem 1 heavily depends on the discussion of the possible
circuit structures. The following symmetric properties of the Fredkin gate are
quite helpful to decrease the number of cases: the Fredkin gate is invariant
underthepermutationofB andC anditissymmetric,i.e.,F =F and
ABC ACB
F =FT with F =S F S .
ABC ABC ACB BC ABC BC
F canberegardedasatripartiteunitaryofHilbertspace .
ABC A B C
H ⊗H ⊗H
We can easily verify F is a control unitary with control on A by noticing
ABC
F = 0 0 I + 1 1 S ,
ABC BC BC
| ih |⊗ | ih |⊗
where I and S stand for an identity operator and the swap gate, respectively,
theindexBC meansthesystemitappliesonis ,aqubit-quditunitary
B C
H ⊗H
9
U is called a control unitary if there exist unitaries U and U such that U =
0 1
0 0 U + 1 1 U .
0 1
| ih |⊗ | ih |⊗
In order to explain our idea and key technique of showing four gates are
not enough, we first demonstrate that a Fredkin gate can not be decomposed
into two two-qubit gates by dividing the problem into two cases: Case 1: The
two gates belong to classes , , then U must be a control unitary
AB BC AB
K K
with control on A, A direct calculation leads to the confliction; Case 2: The
two gates belong to two of the classes , , we assume the circuit is
AB AC
K K
U U = F , invoking Proposition 2, we can assert that S is a local
AB AC ABC BC
unitary by figuring directly out the form of control unitary. That is again
impossible. Therefore,
Lemma 5.1. Two two-qubit gates are not sufficient for implementing F .
ABC
In the rest, we showthat four nonlocaltwo-qubitgates arenot sufficient for
implementing F . We first study two special kinds of circuits.
ABC
Lemma 5.2. There is no U ,V and U ,V such that
AB AB AB BC BC BC
∈ K ∈ K
the Fredkin gate can be implemented in the following circuit,
A
V U
AB AB
B
V U
BC BC
C
.
Proof—: The circuit is U U V V = F , then U U V is a
AB BC AB BC ABC AB BC AB
controlunitarywithcontrolonA. Moreover,foranyinputstate 0 ψ ,the
| iA| iBC
A’s part state of U U V i ψ is 0 . We assume V 0 0 =
AB BC AB | iA| iBC | iA AB| iA| iB
0 0 after moving the local unitaries by invoking Proposition 1. Thus, the
| iA| iB
A’s part’s state of the following state is 0 ,
| iA
U U V 0 0 z =U U 0 0 z .
AB BC AB| iA| iB| iC AB BC| iA| iB| iC
There are three cases about the states U 0 y :
BC| iB| iC
Case 1: There is some z such that U 0 z becomes entangled.
| 0iC BC| iB| 0iC
Assume there is 0<λ<1 such that
U 0 z =√λ0 α +√1 λ1 α⊥ .
BC| iB| 0iC | iB| iC − | iB C
(cid:12) (cid:11)
Define χ =U U 0 0 z , we know that (cid:12)
| iABC AB BC| iA| iB| 0iC
χ =√λΦ α +√1 λΨ α⊥ ,
| iABC | iAB| iC − | iAB C
χ =λΦ +(1 λ)Ψ Φ =Ψ =(cid:12) 0(cid:11)0,
A A A A A (cid:12)
⇒ − ⇒ | ih |
where Φ = U 00 and Ψ = U 01 . Therefore, U is a control unitary
AB AB AB
| i | i | i | i
with control on system A, so is V . We can assume that
AB
U = 0 0 I + 1 1 v ,
AB | ih |⊗ B | ih |⊗ B1
V = 0 0 I + 1 1 v .
AB | ih |⊗ B | ih |⊗ B2
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