Maximal tree size of few-qubit states

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by  Huy Nguyen Le

Maximal tree size of few-qubit states
Huy Nguyen Le,1 Yu Cai,1 Xingyao Wu,1 Rafael Rabelo,2 and Valerio Scarani1,3
1Centre for Quantum Technologies, National University of Singapore, 3 Science Drive 2, Singapore 117543, Singapore
2Departamento de Matema´tica, Universidade Federal de Minas Gerais,
Caixa Postal 702, Belo Horizonte, Minas Gerais 30123-970, Brazil
3Department of Physics, National University of Singapore, 2 Science Drive 3, Singapore 117542, Singapore
Tree size (TS) is an interesting measure of complexity for multiqubit states: not only is it in
principle computable, but one can obtain lower bounds for it. In this way, it has been possible to
identifyfamiliesofstateswhosecomplexityscalessuperpolynomiallyinthenumberofqubits. With
thegoalofprogressinginthesystematicstudyofthemathematicalpropertyofTS,inthisworkwe
characterizethetreesizeofpurestatesforthecasewherethenumberofqubitsissmall,namely,3or
4. Thestudyofthreequbitsdoesnotholdgreatsurprises,insofarasthestructureofentanglement
israthersimple;themaximalTSisfoundtobe8,reachedforinstancebythe|W(cid:105)state. Thestudy
4 offourqubitsyieldsseveralinsights: inparticular,themosteconomicdescriptionofastateisfound
1 nottoberecursive. ThemaximalTSisfoundtobe16,reachedforinstancebyastatecalled|Ψ(4)(cid:105)
0 which was already discussed in the context of four-photon down-conversion experiments. We also
2 find that the states with maximal tree size form a set of zero measure: a smoothed version of tree
size over a neighborhood of a state ((cid:15)−TS) reduces the maximal values to 6 and 14, respectively.
l
u Finally,weintroduceanotionoftreesizeformixedstatesanddiscussitforaone-parameterfamily
J of states.

0
1
I. INTRODUCTION muststressthatbra-ketnotationisnottheonlypossible

] classical description of states; another well known one is
h the matrix product representation (MPS) in which the
p It is likely that the origin of the “speed up” quantum
cost is associated with the size of the matrices [9]. In
- computers offer over classical computers lie in the quan-
t Ref.[3], weobtainedarelationbetweentreesizeandthe
n tum states. There are many evidences that if the state
sizeofthematricesinaMPS,whichshowsthattreesize
a in a quantum computation is simple, it can be simulated
u efficiently with classical computers. Examples are quan- is only a polynomial in the number of qubits when the
q matrix size in a MPS is bounded.
tum circuits where the states at every step have poly-
[ While lower bounds on tree size can be obtained by
nomial Schmidt rank [1], and measurement-based quan-
utilizing Raz’s theorem on multilinear formulas [7, 8],
2 tumcomputationonresourcestateswithlogarithmically
finding the exact tree size of a given state remains to be
v bounded Schmidt-rank width [2], or polynomial tree size
0 investigated. In this paper, we use an exhaustive proce-
[3]. States that are useful for quantum computing must
9 dure to compute the tree size and find the most complex
alsoberealizablebyaquantumcircuitofpolynomialsize.
9 state for three and four qubits. From now on when we
So, for quantum computation to have an advantage over
0 mention the “most complex” state we mean the state
. itsclassicalcounterpart,astatemustbesufficientlycom-
1 withmaximal tree size. Ourapproachreliesonentangle-
plex in some sense but can be prepared efficiently. Thus,
0 mentclassificationbystochasticlocaloperationandclas-
studyingthecomplexityofstatesisimportantbecauseit
4 sical communication (SLOCC) for three and four qubits
1 is not only a fundamental concept of nature but also a [10–12].
: relevant aspect in quantum computing.
v We fist give a brief description of tree-size complex-
i Among the different measures of complexity, quantum ity in Sec. II before reviewing relevant results in en-
X
Kolmogorov complexity is the attempt to quantify com- tanglement classification in Sec. IIIA. We then show in
r plexity of states in the most general way [4–6]; however, Sec. IIIB that the most complex class of three-qubit
a this measure suffers from the setback that it is not com- states is the W class with tree size 8. However, the tree
putable and only upper bounds can be given. Therefore size of these states is not “stable” in the sense that an
arbitrarilysmallperturbationinthestatesleadstoade-
itispossibletocertifythatastateisnotcomplex, butit
crease in the tree size to 6. The TS of mixed states,
is impossible to certify that a state is complex. Things
particularly that of the generalized Werner states, is
arebetterdefinedwhenwerestrictconsiderationtosome
also considered. The case of four qubits is discussed in
particularrepresentations. Ifwerestrictourattentionto
Sec.IV:Wefindthatthemostcomplexclassisanentan-
themostcommondescriptionofquantumstates, Dirac’s glement class which had been overlooked in the previous
bra-ket notation, it is possible to prove superpolynomial worksoninductiveclassificationofentanglement[11,12].
lower bounds [7, 8]. The cost of expressing a state with One example of this class is the state
bra-ket notation gives rise to the definition of tree-size (cid:114) (cid:20)
1 1
complexity. A state with a very long bra-ket representa- |Ψ(4)(cid:105)= (|0110(cid:105)+|0101(cid:105)+|1001(cid:105)+|1010(cid:105))
3 2
tion is hard to generate with classical computers, so tree
(cid:21)
sizeisa“classicalcomplexity”measureasitindicatesthe
−|0011(cid:105)−|1100(cid:105) . (1)
difficulty of simulating a state using classical means. We
2
The reader may refer to Ref. [3] for a proof of this
proposition. This is a very useful observation because if
we know the tree size of a given state, we know the tree
size of all states in its class. Also, entanglement classifi-
cation by SLOCC has been studied with great detail in
the literature.
ForagivennumberofqubitsnandasizeS,thenumber
oftreeswithsizeatmostSisfinite. Forexample, allthe
treesofthreequbitswithatmosteightleavesarelistedin
Fig. 2 according to their depth. A reader who is familiar
withentanglementclassificationmayimmediatelyrecog-
FIG. 1. Rooted trees of two-qubit entangled states.
nizethatthesetreescorrespondtotheproduct,bisepara-
ble,Greenberger-Horne-Zeilinger(GHZ),andWfamilies
of states, respectively.
II. TREE-SIZE COMPLEXITY
Our procedure to find the most complex pure states
is as follows: First, we find a decomposition that can
The most obvious way to write down a state is using
describeaparticularsetofn-qubitstatesanddenotethe
bra-ketnotation. Anymultiqubitstatewritteninitsbra-
size of this decomposition S. Next, we show that these
ket form can be described by a rooted tree of ⊗ and +
states cannot be described by the trees with size smaller
gates; each leaf vertex is labeled with a single-qubit su-
thanS,buttheotherstatesthatarenotinthisparticular
perposition α|0(cid:105)+β|1(cid:105) (this state need not be normal-
√ set can be. It follows that the maximal tree size is S and
ized) [7]. For example, the Bell state (|00(cid:105)+|11(cid:105))/ 2
the set of states mentioned above are the most complex
can be described by the rooted tree T of entangled
E states.
two-qubitstatesinFig.1byassigningappropriatesingle
Tree size can also be extended to mixed states. A
qubit states to each leaf. The size of a rooted tree is de-
mixedstatecanbedecomposedintoanensembleofpure
fined as the number of leaves. A quantum state may be
states |ψ (cid:105) as
representedbydifferentrootedtreeseachwithadifferent i
size. For example, the state (|00(cid:105)+|01(cid:105)+|10(cid:105)+|11(cid:105))/2
(cid:88)
whose size is 8 can also be written as |+(cid:105)|+(cid:105) with size ρ= pi|ψi(cid:105)(cid:104)ψi|. (4)
2. The tree size of a state is taken as the minimum size i
over allpossible trees. Itcanbeunderstoodasthelength
Following the approach of Ref. [14], we define the TS of
of the shortest bra-ket representation of a state.
mixed states as
Going to the example of three qubits, any pure state
can be written as [13]
TS(ρ)=min{max[TS(ψ)]}, (5)
i
i
|Ψ(cid:105)=cosθ|000(cid:105)+sinθ|1(cid:105)(cosω|0(cid:48)0(cid:48)(cid:48)(cid:105)+sinω|1(cid:48)1(cid:48)(cid:48)(cid:105)),
(2) where the minimization is taken over all possible decom-
positions of ρ. The TS of a mixed states is equal to
where the prime and double prime indicate different the TS of the most complex pure state in its decomposi-
bases. Thus, tree size is at most 8 for three qubits. We tion. The motivation behind this definition is that, if in
see below that there are indeed states with tree size 8, apreparationprocedureapurestate|ψ (cid:105)isrealizedwith
i
thatis,thesestatesdonothaveasimplerdecomposition. nonzeroprobability,thenthepreparedstateisatleastas
A more physical measure of complexity which allows complex as this pure state.
for deviations over a neighborhood is the (cid:15)-approximate
tree size of a state. Given a positive (cid:15) < 1, the
(cid:15)-approximate tree size TS (|Ψ(cid:105)) of the state |Ψ(cid:105) is
(cid:15) III. THREE QUBITS
the minimum tree size over all states |ϕ(cid:105) such that
|(cid:104)Ψ|ϕ(cid:105)|2 ≥ 1−(cid:15) [7]. Under a distance error of (cid:15), the
state |Ψ(cid:105) is not distinguishable from |ϕ(cid:105) and hence can A. Tool: SLOCC classification of three qubits
be approximated by the latter.
Animportantpropertyoftreesizewhichweuseexten- Given the central role of Proposition 1 in this work,
sively in this work is that it is invariant under SLOCC. we start by reviewing the SLOCC entanglement classifi-
More specifically, we have the following propositions: cation of three qubits. These are mostly known results,
but we recast them in a form useful for this work.
Proposition 1. [3]Ifthereexistinvertiblelocaloperators
(ILOs) A such that Theinequivalentclassesofastatecanbeinferredfrom
i
its coefficient matrix. Let us first start with a two-qubit
|ψ(cid:105)=A ⊗···⊗A |φ(cid:105), (3) state
1 n
then TS(|ψ(cid:105))=TS(|φ(cid:105)). |Ψ(cid:105)=c |00(cid:105)+c |01(cid:105)+c |10(cid:105)+c |11(cid:105) (6)
00 01 10 11
3
FIG. 2. Rooted trees with size at most 8 for three qubits.
with the coefficient matrix such that |Ψ(cid:105)=A ⊗A ⊗A |W(cid:105). The product, bisep-
1 2 3
arable, and GHZ classes are defined similarly.
(cid:18) (cid:19)
C = c00 c01 . Again, one can determine which class a state belongs
c10 c11 to by studying its coefficient matrix. The state is first
expressed in a basis expansion
It is straightforward to check that this state is a product
state if and only if (iff) det(C) = 0, that is, c00c11 = |Ψ(cid:105)=c000|000(cid:105)+c001|001(cid:105)+c010|010(cid:105)+c011|011(cid:105)
c c . And it is entangled iff det(C)(cid:54)=0. These are the
01 10 +c |100(cid:105)+c |101(cid:105)+c |110(cid:105)+c |111(cid:105)
100 101 110 111
only two inequivalent SLOCC classes of two qubits.
=|0(cid:105)|φ (cid:105)+|1(cid:105)|φ (cid:105), (8)
For the case of three qubits, there are six differ- 0 1
ent classes: product, biseparable, GHZ and W classes
where the two-qubit states are
[10, 11]. Permutation of qubits lead to three inequiva-
lent biseparable classes. The most common examples of |φ (cid:105)=c |00(cid:105)+c |01(cid:105)+c |10(cid:105)+c |11(cid:105),
0 000 001 010 011
states in these classes are
|φ (cid:105)=c |00(cid:105)+c |01(cid:105)+c |10(cid:105)+c |11(cid:105) (9)
1 100 101 110 111
|P(cid:105)=|000(cid:105),
with the coefficient matrices
1
|B(cid:105)=√ |0(cid:105)(|01(cid:105)+|10(cid:105)), (cid:18)c c (cid:19) (cid:18)c c (cid:19)
2 C1|23 = 000 001 , C1|23 = 100 101 , (10)
0 c c 1 c c
1 010 011 110 111
|GHZ(cid:105)=√ |000(cid:105)+|111(cid:105),
2 where the superscript 1|23 indicates the partition of
1 qubits. Since it will be clear from the context which
|W(cid:105)=√ (|001(cid:105)+|010(cid:105)+|100(cid:105)). (7)
3 partition is considered, this superscript will be dropped
fromthetextbelow. Inthisworkwearemoreconcerned
Astate|Ψ(cid:105)issaidtobeintheWclass,forexample,ifand with the states in the GHZ and W classes because they
only if there exist invertible local operators A ,A ,A have larger tree size. We now state the conditions that
1 2 3
4
the matrices C and C must satisfy for |Ψ(cid:105) to be in the B. Maximal tree size
0 1
GHZ or W class, which is derived by Lamata et al. in
Ref.[11]. Thetheoremisstatedinaslightlymodifiedbut It follows from Eq. (2) that the tree size of a three-
equivalent form which we think is easier to work with. qubit state cannot exceed 8. It turns out that all the
Proposition 2. [11] Let |Ψ(cid:105) be a three-qubit pure state, states in the W class have tree size of exactly 8.
then
Proposition 3. The most complex states of three qubits
(1)|Ψ(cid:105)isaGHZstateiffoneofthefollowingconditions
are the states in the W class, and they have tree size 8.
holds:
(a)Thereisapartitioni|jk forwhichC0 andC1 are Proof. This is done by ruling out the smaller trees in
linearly independent, det(C0) (cid:54)= 0, and C0−1C1 has two Fig.2aspossiblerepresentationsofthe|W(cid:105)state. First,
distinct eigenvalues. we observe that the tree T is a special case of T . In-
P B
(b) The same as (a) but with C0 and C1 inter- deed, if one branch of the + gate in TB vanishes we get
changed. T . Similarly,T isaspecialcaseofT becauseifwe
P B GHZ
(c) For all partitions i|jk C0 and C1 are linearly expand the + gate in TB we obtain the form of TGHZ.
independent,andthereisapartitionsuchthatdet(C0)= Let us denote by |T(cid:105) the set of states describable by the
det(C1)=0. tree T, so |TGHZ(cid:105) can be a GHZ, biseparable or product
(2) |Ψ(cid:105) is a W state iff one of the following conditions state. We can parametrize |T (cid:105) as
GHZ
holds:
(a)Thereisapartitioni|jk forwhichC andC are 3 3
0 1 (cid:79) (cid:79)
linearly independent, det(C0) (cid:54)= 0, and C0−1C1 has only |TGHZ(cid:105)= (xi|0(cid:105)+yi|1(cid:105))+ (x(cid:48)i|0(cid:105)+yi(cid:48)|1(cid:105))
one eigenvalue. i=1 i=1
(b) The same as (a) but with C and C inter- (15)
0 1
changed.
with complex coefficients x ,y ,x(cid:48) and y(cid:48). If |W(cid:105) =
Note that the eigenvalue equation of a 2×2 matrix M i i i i
|T (cid:105),byequatingcoefficientsofbothsidesweobtaina
is GHZ
system of equations for the variables x ,y ,x(cid:48),y(cid:48). These
i i i i
λ2−tr(M)λ+det(M)=0; (11) equationscanbeeasilyshowntohavenosolution. There-
fore, the tree T cannot describe the |W(cid:105) state. Now
GHZ
hence it has only one eigenvalue when the discriminant
that all the smaller trees have been ruled out, we con-
vanishes, which yields [tr(M)]2 = 4det(M); otherwise it
cludethatthe|W(cid:105)statehastreesize8whichismaximal
has two distinct eigenvalues.
forthree-qubitstates. Proposition1thenimpliesthatall
Agenericstateofthreequbitsalmostalwaysobey(1a)
states in the W class have tree size 8.
or(1b)ofProposition2. Theotherconstraintsareequa-
tionsthatonlyaspecificsetofmatrixcoefficientssatisfy. Whatisthetreesizeofthesecondmostcomplexclass?
Thus, most of the states in the Hilbert space of three One can also show in a similar way that the GHZ state
qubits are in the GHZ class. cannot be described by the tree T . A reader familiar
B
The states in the W class have a special property that with entanglement classification would find this obvious
isimportantforfindingtheir(cid:15)-approximatetreesize: For because there is no genuine three-qubit entanglement in
every state in the W class, there is a GHZ state that is T . Note that T is a special case of T so need not be
B P B
arbitrarily close to it [15]. It is straightforward to verify considered. Therefore, the GHZ state, as well as all the
that the coefficient matrix of the |W(cid:105) state satisfies con- states in its class, has tree size 6.
dition (2a) of Proposition 2. But if we introduce a small Since states are determined only up to a finite preci-
fluctuation sion, it is necessary to consider the more physical defini-
tion of the tree size — the (cid:15)-approximate tree size. We
|W(cid:105)→|W(cid:105)+µ|111(cid:105), (12)
see in Sec. IIIA that adding an arbitrarily small pertur-
then the coefficient matrix satisfies condition (1a) of bation to |W(cid:105) results in a state in the GHZ class with
Proposition 2 and hence the perturbed state belongs to tree size 6. Thus, TS(cid:15)(|W(cid:105)) is at most 6 for any posi-
the GHZ class. Not only |W(cid:105) but every state in its class tive (cid:15). This is also true for all the states in the W class,
is “unstable” under an arbitrarily small fluctuation. For which means that the maximal tree size of three-qubit
a state |φ (cid:105) in this class, there exist invertible local op- states under nonvanishing perturbation is 6.
W
erators A ,A ,A such that How large can (cid:15) be before |W(cid:105) can be approximated
1 2 3
by the tree T with size 5? Any normalized state that
B
|φ (cid:105)=A ⊗A ⊗A |W(cid:105). (13)
W 1 2 3 is described by T must adopt the following form, up to
B
permutation of qubits:
We have
|φ (cid:105)+µA ⊗A ⊗A |111(cid:105) |T (cid:105)=|u(cid:105)(α|0(cid:105)|v(cid:105)+β|1(cid:105)|v(cid:48)(cid:105)), (16)
W 1 2 3 B
=A ⊗A ⊗A (|W(cid:105)+µ|111(cid:105)), (14)
1 2 3 where|u(cid:105),|v(cid:105),and|v(cid:48)(cid:105)arenormalizedsingle-qubitstates,
which is a state in the GHZ class for arbitrarily small µ. and|α|2+|β|2 =1. Thisisingeneralabiseparablestate,
5
but it is a product state when |v(cid:105) = |v(cid:48)(cid:105) up to a phase.
The |W(cid:105) state is particularly easy to work with since it
is invariant under permutation of qubits, so we do not
need to consider the other two partitions of qubits, 2|13
and 3|12, in |T (cid:105). It is known that the squared overlap
B
|(cid:104)W|T (cid:105)|2 obeys the inequality
B
2
|(cid:104)W|T (cid:105)|2 ≤ . (17)
B 3
The2/3upperboundisstrictbecausetheequalityholds
when,forinstance,|T (cid:105)=|B(cid:105)giveninEq.(7). Fromthe
B FIG.3. EachSLOCCclassofmixedthree-qubitstatesforms
definition of the (cid:15)-approximate tree size and the above aconvexsetwithextremalpointsrepresentingthepurestates
inequality one concludes that T(cid:15)(|W(cid:105)) = 6 for 0 < (cid:15) < intheclass. Theboundariesofsmallersetsareinfinitesimally
1/3. closetothoseofthebiggersets,becauseforeverypurestatein
theformerthereisoneinthelatterthatisarbitrarilycloseto
it. Theverticalvectorshowsqualitativelyhowthegeneralized
Werner state transfoms from one set to the next when the
parameter p is increased.
C. Mixed states
Computing the TS of mixed states of two qubits is IV. FOUR QUBITS
straightforward. TheTSofaseparablestateis2andthe
TSofanentangledstate,whichhasatleastoneentangled
Inthissectionweusethesameapproachasinthepre-
pure state in its decomposition, is 4. One can determine
vious section to find the most complex four-qubit states.
whetheramixedstateisseparableorentangledusingthe
We find that every state can be be described by a tree
positive partial transpose criterion [16].
with at most 16 leaves only if the state is written in the
The case of three qubits is more difficult. The TS form (37). This form seems to preclude a recursive con-
of mixed states of three qubits can be computed based struction of the most economic description in terms of
on the entanglement classification of three qubits intro- tree size. Forms that look recursive, like (19), do require
ducedbyAcin et. al. [17]: Amixedstateofthreequbits 18 leaves for some states. In the process, we find that
belongs to the class S of separable states if it can be ex- the most complex four-qubit state belong to a SLOCC
pressed as a convex sum of separable pure states, class class not described in previous classifications, which we
B of biseparable states if it can be expressed as a con- call “states with irreducible A—BCD form”. Finally, as
vex sum of separable and biseparable pure states, class wedidforthreequbits,wedescribetheapproximatetree
W if it can be expressed as a convex sum of separable, size.
biseparable, and W pure states, and class GHZ if it can
be expressed as a convex sum of all possible states. It
follows that S ⊂ B ⊂ W ⊂ GHZ (see Fig. 3). From the A. A|BCD form
definition of the TS for mixed states we see that TS=3
for the set S, TS=5 for the set B\S, TS=8 for the set We begin our search by the observation that any four-
W\B, and TS=6 for the set GHZ\W. qubit state |Ψ(cid:105) can be written as
However, unlike in the case of two qubits, there is no
systematic way to determine the entanglement class of |Ψ(cid:105)=|0(cid:105)|φ0(cid:105)+|1(cid:105)|φ1(cid:105) (19)
an arbitrary mixed state. But it is possible to identify
with some three-qubit states |φ (cid:105) and |φ (cid:105). We will refer
the class of some states with the help of entanglement 0 1
toitastheA|BCD form intherestofthetext. Thesize
witnesses [17]. Let us consider the example of the gener-
oftheabovedecompositionisatmost18andreachesthe
alized Werner state of three qubits [18]
maximal value when both |φ (cid:105) and |φ (cid:105) are states in the
0 1
W class. We now define the irreducible A|BCD form as
1−p
ρ (p)=p|GHZ(cid:105)(cid:104)GHZ|+ I , (18) follows:
WS 8 8
Definition 1. Astate|Ψ(cid:105)hastheirreducibleA|BCDform
ifforallA∈{1,2,3,4}andallILOsA theA|BCDform
where the parameter p ranges from 0 to 1. It is shown 1
of A |Ψ(cid:105) always has |φ (cid:105) and |φ (cid:105) in the W class.
thatρ (p)belongstothesetSwhenp≤1/5,B\Swhen 1 0 1
WS
1/5 < p ≤ 3/7, W\B when 3/7 < p ≤ p , and GHZ\W In other words, both the three-qubit states |φ (cid:105) and
W 0
when p <p≤1 where p ≈0.6955427 [18]. Thus, the |φ (cid:105) in the A|BCD form cannot be brought out of the W
W W 1
generalized Werner state has the maximal TS = 8 when class by switching to a different partition and applying
3/7<p≤p . ILOs. Note that permuting qubits in the BCD part and
W
6
applyingILOsonthesequbitsdonotbring|φ (cid:105)and|φ (cid:105) one zero regardless of the values of c , c , and c . So
0 1 1 3 4
out of the W class. Therefore, we come to a stronger we try (2b): We have detC = (c +λ)c −c c , which
1 5 8 6 7
statementthatifastatehastheirreducibleA|BCDform, is not zero for all λ if and only if c = 0 and c c (cid:54)=
8 6 7
then for all permutations (A,B,C,D) of (1,2,3,4) and 0; the equation (cid:2)tr(C−1C )(cid:3)2 = 4det(C−1C ) yields a
1 0 1 0
ILOs A1,A2,A3,A4 the A|BCD form of A1⊗A2⊗A3⊗ quadratic equation,
A |Ψ(cid:105) has |φ (cid:105) and |φ (cid:105) in the W class. These states
4 0 1
were overlooked in the original work on the inductive a λ2+a λ+a =0, (26)
2 1 0
classification of entanglement for four qubits [12].
wherea ,a ,a arefunctionsofc ,c ,...,c . Thisequa-
It is shown in Sec. IVB that the states with irre- 0 1 2 1 3 7
tion is satisfied for all λ if all a ,a ,a vanish, which
ducible A|BCD form are indeed the most complex four- 0 1 2
yields three constraints:
qubitstates,butletusfirstidentifythesetofthesestates.
When both |φ0(cid:105) and |φ1(cid:105) are in the W class, first we use (c +c −c )2 =4c c ,
6 7 4 6 7
ILOs on the last three qubits to transform |φ (cid:105) to the
1
(c c −c c )(c +c −c )=2c c c ,
|W(cid:105) state. We denote the state after this transformation 3 6 4 5 6 7 4 3 6 7
by (c c −c c )2 =−4c c c c . (27)
3 6 4 5 1 4 6 7
|Ψ(cid:105)=|0(cid:105)|φ (cid:105)+|1(cid:105)|W(cid:105), (20) Andwiththeseconstraints,itisnotdifficulttoshowthat
w
C and C are linearly independent for all λ if and only
0 1
where |φ (cid:105) is a state in the W class with the coefficient if c (cid:54)= 0. In Ref. [12], the possibility of a ,a ,a being
w 4 0 1 2
matrices all zero was not noticed; as a consequence, the authors
missedthefamilyofstateswithirreducibleA|BCDform.
(cid:18) (cid:19) (cid:18) (cid:19)
c c c c
C = 1 2 , C = 5 6 . (21) Indeed, as soon as one of the coefficients is not zero, the
0 c c 1 c c
3 4 7 8 quadratic equation has at most two solutions: Then one
can choose any λ not in the set of solutions to transform
Wecanassumethatc =0becauseifc (cid:54)=0wecanapply
2 2 at least one of |φ (cid:105) and |φ (cid:105) out of the W class.
A on the first qubit such that A |0(cid:105) = |0(cid:105),A |1(cid:105) = 0 1
1 1 1 We need to carry out the same analysis for other par-
−c |0(cid:105)+|1(cid:105). After applying A the new state |φ (cid:105) has
2 1 W titions A = 2,3,4. The algebra is lengthy but straight-
c =0.
2 forward. As it turns out, all additional equations and
By applying a local invertible operator
inequalities can be derived from the already known con-
(cid:18) (cid:19) straints. So the state |Ψ(cid:105) in Eq. (20) has the irreducible
a a
A1 = a11 a12 , (22) A|BCD form if the coefficients of the three-qubit state
21 22
|φ (cid:105) obey the following set of constraints:
w
with
c ,c ,c (cid:54)=0,
4 6 7
det(A1)=a11a22−a12a21 (cid:54)=0, (23) c2 =c8 =0,
(c +c −c )2 =4c c ,
on the first qubit, we have 6 7 4 6 7
(c c −c c )(c +c −c )=2c c c ,
3 6 4 5 6 7 4 3 6 7
A1|Ψ(cid:105)=|0(cid:105)(a11|φw(cid:105)+a12|W(cid:105)) (c3c6−c4c5)2 =−4c1c4c6c7. (28)
+|1(cid:105)(a |φ (cid:105)+a |W(cid:105)). (24)
21 w 22
These constraints can be greatly simplified when we
Thus, we must find |φw(cid:105) such that (1) a11|φw(cid:105)+a12|W(cid:105) consider two possible cases: c1 =0 and c1 (cid:54)=0. If c1 =0
remains in the W class, and (2) a |φ (cid:105)+a |W(cid:105) re- the above constraints become
21 w 22
mains in the W class, for all a obeying the invertibility
ij c ,c ,c (cid:54)=0,
condition. This constraint makes sure that a and a 4 6 7
11 12
arenotbothzero,andneitherarea21 anda22. Ifa11 =0 c1 =c2 =c3 =c5 =c8 =0,
√ √
then a12 (cid:54)=0, and thus (1) is always satisfied. If a11 (cid:54)=0 c4 =( c6± c7)2, (29)
let λ = a /a and (1) becomes finding |φ (cid:105) such that
12 11 w √
|φ (cid:105)+λ|W(cid:105)forallλ. Similarargumentsforthepaira where z denotes the principal square root of the com-
w 21
and a result in the same requirement. plex number z. If c (cid:54)= 0, by applying A on the first
22 1 1
Recall that we can choose c = 0, the coefficient ma- qubit such that A |0(cid:105) = −|0(cid:105)/c ,A |1(cid:105) = |1(cid:105) we get a
2 1 1 1
trices of |φ (cid:105)+λ|W(cid:105) is new |φ (cid:105) with c =−1. Substituting this into the set of
w w 1
equations we obtain
(cid:18) (cid:19) (cid:18) (cid:19)
c λ c +λ c
C0 = c3+1 λ c4 , C1 = 5c7 c68 . (25) c4,c6,c7 (cid:54)=0,
c =−1, c =c =0,
1 2 8
Nowwerefertocondition(2)ofProposition2. Condition
(cid:16)c (cid:17)2 (cid:16)c (cid:17)2 (cid:18)c −c (cid:19)2
(a2lla)λ, wbhecicahusreeqdueitreCs de=tC−0λ(cid:54)=2 0−, ccaλnn+otcbce sahtaissfiaetdlweaitsht c4 = 23 , c6 = 25 , c7 = 3 2 5 . (30)
0 3 1 4
7
The simplest example of the first case is c = 4,c = Proposition 4. Afour-qubitstate|Ψ(cid:105)withtheirreducible
4 6
c =1, which yields the state A|BCD form has the decomposition
7
|Ψ(cid:105)=|0(cid:105)(4|011(cid:105)+|101(cid:105)+|110(cid:105)) |Ψ(cid:105)=|φ (cid:105)|ϕ (cid:105)+|φ(cid:48) (cid:105)|ϕ(cid:48) (cid:105), (37)
12 34 13 24
+|1(cid:105)(|001(cid:105)+|010(cid:105)+|100(cid:105)). (31)
where |φ(cid:105), |ϕ(cid:105), |φ(cid:48)(cid:105), |ϕ(cid:48)(cid:105) are two-qubit entangled states,
It is not difficult to find ILOs to transform |Ψ(cid:105) to the
and the subscripts indicate the qubits assigned to each
following more symmetric state:
partition.
1
|Ψ(4)(cid:105)=√ (|0(cid:105)|W0(cid:105)+|1(cid:105)|W1(cid:105)), (32) Here |φ12(cid:105) is an entangled state of the first qubit and
2 the second qubit and so on. Since T is a tensor product
8
where of two two-qubit entangled states (see Fig. 4), it is clear
that the above decomposition can be described by the
1
|W0(cid:105)= √6(|110(cid:105)+|101(cid:105)−2|011(cid:105)), tree T8+T8. Note that the orders of qubits assigned to
each branch are not the same. Hence, this decomposi-
1
|W (cid:105)= √ (|001(cid:105)+|010(cid:105)−2|100)(cid:105)). (33) tion is not similar to those usually seen in entanglement
1
6 theory. This “crossing” of qubits is required to obtain
Or, explicitly, the minimal tree for the states with irreducible A|BCD
form.
(cid:114) (cid:20)
1 1
|Ψ(4)(cid:105)= (|0110(cid:105)+|0101(cid:105)+|1001(cid:105)+|1010(cid:105))
3 2 Proof. We show this by explicit constructions. For the
(cid:21) firstcasewherethecoefficientsaregiveninEq.(29),one
−|0011(cid:105)−|1100(cid:105) . (34) can verify that the explicit form is
√ √
Coincidentally, this state was already realized in exper- |φ12(cid:105)=(|10(cid:105)∓ c6 c7|01(cid:105)),
√
iments by using photons from a down-conversion source c
6
|ϕ (cid:105)=∓√ |01(cid:105)+|10(cid:105),
[19, 20]; and the genuine four photon entanglement was 34 c
7
confirmed by measuring a witness [19]. The state |Ψ(4)(cid:105) √ √ √
|φ(cid:48) (cid:105)= c ( c ± c )|01(cid:105)+|10(cid:105),
issimilartotheDickestatewithtwoexcitations,thatis, 13 7 7 6
√ √
c ± c
1 (cid:0) |ϕ(cid:48) (cid:105)= 7√ 6 |01(cid:105)+|10(cid:105). (38)
|D2(cid:105)=√6 |0011(cid:105)+|0101(cid:105)+|0110(cid:105) 24 c7
(cid:1)
+|1001(cid:105)+|1010(cid:105)+|1100(cid:105) , (35) The above state is always well defined because the con-
straints of Eq. (29) require that c (cid:54)=0.
exceptforthefactorsof−2. Despitethissimilarity, |D (cid:105) 7
2 For the second case of Eq. (30) the explicit form is
doesnothaveirreducibleA|BCDform,asonecanverify
more complicated:
that applying the ILO
A =(cid:18)1 1 (cid:19) (36) |φ (cid:105)= 4 |10(cid:105)+|0(cid:105)(cid:20) 2c3 |0(cid:105)+|1(cid:105)(cid:21),
1 1 −1 12 c (c −c ) c (c −c )
5 5 3 5 5 3
on the first qubit gives the state |0(cid:105)|GHZ(cid:48)(cid:105)+|1(cid:105)|GHZ(cid:48)(cid:48)(cid:105) |ϕ (cid:105)= c25 |01(cid:105)+(cid:20)c5 |0(cid:105)+ c5(c5−c3)|1(cid:105)(cid:21)|0(cid:105),
where |GHZ(cid:48)(cid:105) and |GHZ(cid:48)(cid:48)(cid:105) belong to the GHZ class. 34 4 2 4
Inthenextsectionweshowthatafour-qubitstatehas (cid:20) c c (cid:21) 2
maximal tree size if and only if it has irreducible A|BCD |φ(cid:48)13(cid:105)=|0(cid:105) c −5c |0(cid:105)+ 23 |1(cid:105) + c −c |10(cid:105),
3 5 3 5
form. Moreover, while the maximal size of the A|BCD
c −c (cid:104) c (cid:105)
formis18,themaximaltreesizeisonly16,whichmeans |ϕ(cid:48) (cid:105)= 3 5 |10(cid:105)+|0(cid:105) |0(cid:105)+ 3 |1(cid:105) , (39)
24 2 2
that the A|BCD form is not the optimal decomposition
for the most complex four-qubit states. which is again always well defined since the constraints
of Eq. (30) make sure that c (cid:54)=c .
3 5
B. Maximal tree size The decomposition of Eq. (37) turns out to be opti-
mal for all states with irreducible A|BCD form. In other
First, we list all the four-qubit trees with a ⊗ gate at words, these states do not possess decompositions with
the root as shown in Fig. 4. The subscript of each tree size smaller than 16.
indicates the number of its leaves. All the other trees Proposition 5. If |Ψ(cid:105) is a state with irreducible A|BCD
(with a + gate at the root) are combinations of these form, its minimal tree is T +T . Thus, the tree size of
8 8
trees. Then we proceed to prove that the tree size of these states is 16.
the states with irreducible A|BCD form is 16, which will
later be shown to be the maximal tree size of four-qubit Proof. First we need to draw the trees with 15 leaves or
states. less. There are a lot of them, but most are special cases
8
FIG. 4. Rooted trees for four qubits with a ⊗ gate at the root. The subscript of each tree indicates the number of its leaves.
The trees T , T , and T are used to label branches.
E GHZ W
state can be then expressed as α|u (cid:105)+β|u (cid:105) where |u (cid:105)
0 1 0
isthenormalizedstatethatformsabasiswith|u (cid:105). The
1
four-qubit state described by T +T is
6 9
|ϕ(cid:105)=(α|u (cid:105)+β|u (cid:105))|T (cid:105)+γ|u (cid:105)|T (cid:105)
0 1 B 1 W
=α|u (cid:105)|T (cid:105)+|u (cid:105)(β|T (cid:105)+γ|T (cid:105)). (40)
0 B 1 B W
Note that the tree states with letters as subscripts are
three-qubit states (see Fig. 2). The size of the three-
qubit state β|T (cid:105)+γ|T (cid:105) is at most 8, so the size of
B W
the above A|BCD form is at most 15 and hence cannot
be maximal.
In the second case, the qubit assigned to the leaf at
therootofT isassignedtooneoftheleavesinthetwo-
9
qubit entangled subtree T of T . Then, we can express
FIG. 5. Rooted trees of four qubits with 15 leaves. All the E 6
the two-qubit state of T as α|u (cid:105)|v (cid:105)+β|u (cid:105)|v (cid:105). Af-
treeswithfewerthan15leavesarespecialcasesofthesetrees. E 0 0 1 1
ter writing down |ϕ(cid:105) and grouping the terms with |u (cid:105)
1
together, one sees that it has the form of Eq. (40) with
|T (cid:105) instead of |T (cid:105). The size of this A|BCD form is at
of others. Let us first consider the set of trees shown in P B
most 13 and is again not maximal.
Fig. 4. We see that T is a special case of T because
4 6 FortheT +T tree, denotebyγ|u (cid:105)thesingle-qubit
the two-qubit product state is a special case of T . We 7 8 1
E state assigned to the leaf at the root of T . Using the
denote this relation as T ⊂ T . After examining the 7
4 6 same procedure one can show that a state described by
structuresofthesetrees,onecanbeconvincedthatT ⊂
4 this tree has an A|BCD form with size at most 15. Sim-
T ⊂T ⊂T , T ⊂T and T ⊂T +T . From these
6 7 9 6 8 7 4 4 ilarly, a state described by T +T +T has an A|BCD
relations we have T +T ⊂ T +T ⊂ T +T +T 4 4 7
6 6 7 7 4 4 7 formwithsizeatmost16. Recallthatthemaximalsizeof
and so on. After listing the trees with at most 15 leaves
theA|BCDformis18;allthetreeswith15leavescannot
weseethatallofthemarespecialcasesofthesetoftrees
describe a state that has irreducible A|BCD form. Since
with exactly 15 leaves. This set is shown in Fig. 5.
the trees with fewer than 15 leaves are special cases of
WenowproveProposition5byshowingthatifastate
the trees with 15 leaves, we conclude that all the trees
is described by a tree with at most 15 leaves, it cannot
with 15 leaves or fewer cannot describe the states with
have the irreducible A|BCD form. Only T +T is con-
6 9 irreducible A|BCD form. Therefore, the T +T tree is
8 8
sidered as the arguments for the other trees are similar.
theoptimaldecompositionforthesestatesandtheirtree
For better clarity, we draw T +T explicitly in Fig. 6.
6 9 size is 16.
Let us denote by γ|u (cid:105) the single-qubit state assigned
1
to the leaf at the root of T with |u (cid:105) normalized. The NowthetreesizeofthestateswithirreducibleA|BCD
9 1
samequbitmaybeassignedtoanyleafontheT branch. form has been found, one still needs to prove that these
6
There are two inequivalent cases. In the first, this qubit states are the most complex, that is, the other states
is assigned to one of the two leaves at the root of T . Its (withnoirreducibleA|BCDform)havesmallertreesize.
6
9
FIG. 6. Explicit drawing of T +T with its two main branches T and T .
6 9 6 9
Proposition 6. If a four-qubit state |Ψ(cid:105) does not have an size of four-qubit states is therefore 16. Note that for
irreducible A|BCD form, it can be described by a tree 2≤n≤4 the maximal TS is 2n, which is the dimension
with at most 15 leaves. of the Hilbert space. Whether this relation holds for all
n≥2 remains an open question.
Proof. If |Ψ(cid:105) does not have the irreducible A|BCD form,
thereexistsapartitionA|BCDandanILOA suchthat
1
after the application of this ILO we have
C. Approximate tree size
|Ψ(cid:105)=|0(cid:105)|φ (cid:105)+|1(cid:105)|φ (cid:105), (41)
0 1
What is the maximal (cid:15)-approximate tree size of four
where at least one of the three-qubit states, say |φ1(cid:105), is qubit states? Even in the worst case scenario when |φ0(cid:105)
not in the W class. If |φ1(cid:105) is a biseparable state, it is and |φ1(cid:105) in Eq. (19) are in the W class, we know from
clear that |Ψ(cid:105) can be described by the tree T9+T6 with Sec. IIIA that they can be approximated with arbitrary
15leaves. If|φ1(cid:105)isintheGHZclass,weuseILOsonthe precision by two states in the GHZ class. More con-
last three qubits to transform |Ψ(cid:105) to cretely, for (cid:15) arbitrarily close to 0 there exists a state of
the form
|Ψ(cid:105)=|0(cid:105)|φ (cid:105)+|1(cid:105)|GHZ(cid:105). (42)
w
|ϕ(cid:105)=|0(cid:105)|GHZ(cid:48)(cid:105)+|1(cid:105)|GHZ(cid:48)(cid:48)(cid:105) (45)
Consider the state |φ (cid:105) + λ|GHZ(cid:105), a necessary condi-
w
tion for this state to be in the W class is that C C−1
0 1 such that |(cid:104)Ψ|ϕ(cid:105)|2 ≥ 1−(cid:15). Here |GHZ(cid:48)(cid:105) and |GHZ(cid:48)(cid:48)(cid:105)
or C−1C has only one eigenvalue. Both cases yield the
1 0 are the two states in the GHZ class. Because the tree
same equation
size of |ϕ(cid:105) is at most 14, we conclude that the TS (|Ψ(cid:105))
(cid:15)
λ4+a λ3+a λ2+a λ+a =0, (43) is at most 14 for every (cid:15) > 0. Thus, if fluctuation over
3 2 1 0
aneighborhoodis allowed, themaximal treesize offour-
wherea ,a ,a ,a arefunctionsofthecoefficientsofφ . qubit states is at most 14.
0 1 2 3 w
This equation has at most four distinct solutions. Thus, Proposition 7. The(cid:15)-approximatetreesizeof|Ψ(4)(cid:105)is14
it is always possible to find a value λ∗ (cid:54)= 0 that is not for 0<(cid:15)< 1 .
in the set of solutions. Then, |φ (cid:105)+λ∗|GHZ(cid:105) is not in 12
w
the W class and hence can be described by T . Next,
GHZ Proof. We need to show that if |ϕ(cid:105) is a state described
we apply on the first qubit an ILO such that A1|0(cid:105) = by a tree with fewer than 14 leaves, then |(cid:104)ϕ|Ψ(4)(cid:105)|2 ≤
|0(cid:105),A |1(cid:105)=λ∗|0(cid:105)+|1(cid:105), whichisinvertiblesinceλ∗ (cid:54)=0.
1 1−1/12=11/12. Inotherwords,thestateswithsmaller
After that the state becomes size are a finite distance away from |Ψ(4)(cid:105). For this pur-
pose we employ the same elimination procedure used in
|Ψ(cid:105)=|0(cid:105)(|φ (cid:105)+λ∗|GHZ(cid:105))+|1(cid:105)|GHZ(cid:105), (44)
w Sec. IIIB to find the most complex three-qubit state.
First, we draw all the trees with 13 leaves or fewer. We
which can be described by T +T with 14 leaves.
7 7
observe that all of the trees in this set are special cases
A direct corollary of Propositions 5 and 6 is that the of the four particular trees listed in Fig. 7. Thus, Propo-
states with irreducible A|BCD form are the states with sition7holdsifwecanshowthatthestatesdescribedby
the maximal tree size, and vice versa. The maximal tree these four trees are a finite distance away from |Ψ(4)(cid:105).
10
immediately that |(cid:104)W |φ(cid:105)|2+|(cid:104)W |φ(cid:105)|2 ≤1. This and
0 1
Eqs. (47), (48), and (49) yields
|(cid:104)Ψ(4)|ϕ(cid:105)|2 ≤ 1(cid:0)|(cid:104)W |T (cid:105)|2+|(cid:104)W |T (cid:105)|2+1(cid:1).
2 0 B 1 B
(50)
Thus, we need to maximize
f =|(cid:104)W |T (cid:105)|2+|(cid:104)W |T (cid:105)|2 (51)
0 B 1 B
to find the largest overlap between |Ψ(4)(cid:105) and |ϕ(cid:105).
As a biseparable state |T (cid:105) has the following form:
B
|T (cid:105)=(a|0(cid:105)+b|1(cid:105))
FIG. 7. Rooted trees of four qubits with sizes 12 and 13. B
⊗(c |00(cid:105)+c |01(cid:105)+c |10(cid:105)+c |11(cid:105)) (52)
00 01 10 11
with the constraints
Eliminating T +T , T +T , and T +T — Using
4 8 6 7 4 9
the argument in the proof of Proposition 6 that leads to
|a|2+|b|2 =|c |2+|c |2+|c2 +|c |2 =1. (53)
Eq. (40), one can show that a state described by one of 00 01 10 11
T +T , T +T , and T +T has the form
4 8 6 7 4 9 Substituting this in f we obtain
|ϕ(cid:105)=α|u0(cid:105)|TB(cid:105)+β|u1(cid:105)|φ(cid:105), (46) f = 1(cid:0)|b(c +c )−2ac |2+|a(c +c )−2bc |2(cid:1).
6 01 10 11 01 10 00
where |α|2+|β|2 =1, |u (cid:105) and |u (cid:105) are two orthonormal (54)
0 1
single-qubit states, |φ(cid:105) is a three-qubit state, and |T (cid:105)
B
Maximizing f with respect to the constraints of Eq. (53)
is a biseparable state (product states are treated as a
gives us f =2/3.
special case of biseparable states). In addition, |φ(cid:105) and max
Let us now consider the situations when instead of
|T (cid:105) are both normalized. One sees later that it is the
B
qubit 1, qubits 2,3,4 are assigned to part A of the bipar-
biseparable component |T (cid:105) of |ϕ(cid:105) that keeps it away
B
from |Ψ(4)(cid:105). titecutA|BCD. Sinceweareconcernedwiththeoverlap
|(cid:104)Ψ(4)|ϕ(cid:105)|, this is equivalent to keeping |ϕ(cid:105) unchanged
Thestate|ϕ(cid:105)hastheformofthebipartitecutA|BCD
while permuting the qubits in |Ψ(4)(cid:105). Under permuta-
where any of the four qubits can be assigned to part A.
tion |Ψ(4)(cid:105) still has the same form as Eq. (32) but the
Wefirstconsiderthecasewhentheyarethestatesofthe
factor2changesitsplacewithin|W (cid:105)and|W (cid:105). Thanks
first qubit. Complications due to permutation of qubits 0 1
tothehighsymmetryintheformof|Ψ(4)(cid:105),thereareonly
will be dealt with later. As in the case for three qubits,
two different cases:
we look at the overlap
|(cid:104)Ψ(4)|ϕ(cid:105)|2 =|α(cid:104)Ψ(4)|u0TB(cid:105)+β(cid:104)Ψ(4)|u1φ(cid:105)|2 |Ψ(14)(cid:105)=√112(cid:2)|0(cid:105)(|110(cid:105)−2|101(cid:105)+|011(cid:105))
≤|(cid:104)Ψ(4)|u0TB(cid:105)|2+|(cid:104)Ψ(4)|u1φ(cid:105)|2, (47) +|1(cid:105)(|001(cid:105)−2|010(cid:105)+|100(cid:105))(cid:3),
wherethelastlinefollowsfromanapplicationofCauchy- |Ψ(4)(cid:105)=√1 (cid:2)|0(cid:105)(−2|110(cid:105)+|101(cid:105)+|011(cid:105))
2 12
Schwarz inequality. Moreover, we have
(cid:3)
+|1(cid:105)(−2|001(cid:105)+|010(cid:105)+|100(cid:105)) . (55)
1
|(cid:104)Ψ(4)|u T (cid:105)|2 = |(cid:104)0|u (cid:105)(cid:104)W |T (cid:105)+(cid:104)1|u (cid:105)(cid:104)W |T (cid:105)|2
0 B 2 0 0 B 0 1 B Following the same argument as for |Ψ(4)(cid:105), we arrive at
≤ 1(cid:0)|(cid:104)W |T (cid:105)|2+|(cid:104)W |T (cid:105)|2(cid:1), (48) the same inequality as in Eq. (50) for |Ψ(14)(cid:105) and |Ψ(24)(cid:105)
2 0 B 1 B with
and f = 1(cid:2)|ac +b(c −2c )|2+|bc +a(c −2c )|2(cid:3)
1 6 11 10 01 00 01 10
1 (56)
|(cid:104)Ψ(4)|u φ(cid:105)|2 = |(cid:104)0|u (cid:105)(cid:104)W |φ(cid:105)+(cid:104)1|u (cid:105)(cid:104)W |φ(cid:105)|2
1 2 1 0 1 1 for |Ψ(4)(cid:105), and
1
≤ 12(cid:0)|(cid:104)W0|φ(cid:105)|2+|(cid:104)W1|φ(cid:105)|2(cid:1). (49) f = 1(cid:2)|ac +b(c −2c )|2+|bc +a(c −2c )|2(cid:3)
2 6 11 01 10 00 10 01
Since |W (cid:105) and |W (cid:105) are two orthonormal states, we (57)
0 1
can expand |φ(cid:105) in a basis in which |W (cid:105) and |W (cid:105) are for|Ψ(4)(cid:105). Sincef becomesf afterinterchangingc ↔
0 1 2 1 2 01
basis vectors. From this observation we can conclude c , they have the same maximum. And maximizing f
10 1