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arXiv:0711.4021v1 [quant-ph] 26 Nov 2007

How many CNOT gates does it take to generate a three-qubit state ?

MarkoˇZnidariˇ c1, Olivier Giraud2and Bertrand Georgeot2

1Department of Physics, Faculty of Mathematics and Physics,

University of Ljubljana, SI-1000 Ljubljana, Slovenia

2Laboratoire de Physique Th´ eorique, Universit´ e de Toulouse, CNRS, 31062 Toulouse, France

(Dated: November 26, 2007)

The number of two-qubit gates required to transform deterministically a three-qubit pure quantum

state into another is discussed. We show that any state can be prepared from a product state

using at most three CNOT gates, and that, starting from the GHZ state, only two suffice. As a

consequence, any three-qubit state can be transformed into any other using at most four CNOT

gates. Generalizations to other two-qubit gates are also discussed.

PACS numbers: 03.67.-a, 03.67.Ac, 03.67.Bg

Quantum information and computation (see e.g. [1]) is

usually described using qubits as elementary units of in-

formation which are manipulated through quantum oper-

ators. In most practical implementations, such operators

have to be realized as sequences of local transformations

acting on a few qubits at a time. Whereas one-qubit gates

alone cannot create entanglement, it has been shown that

together with two-qubit gates they can form universal

sets, from which the set of all unitary transformations of

any number of qubits can be generated [2]. The complex-

ity of a quantum algorithm is usually measured by assess-

ing the number of elementary gates needed to perform

the computation. The Controlled-Not (CNOT) gate, a

two-qubit gate whose action can be written |00? → |00?,

|01? → |01?, |10? → |11?, |11? → |10?, is one of the most

widely used both for theory and implementations. It can

be shown that the CNOT gate together with one-qubit

gates is a universal set [2]. Experimental implementa-

tions of a CNOT gate (or the equivalent controlled phase-

flip) have been recently reported using e.g. atom-photon

interaction in cavities [3], linear optics [4], superconduct-

ing qubits [5, 6] or ion traps [7, 8].

quantum computers are still far away, small platforms of

a few qubits exist or can be envisioned in the framework

of these existing experimental techniques. In most such

implementations, two-qubit gates such as the CNOT are

much more demanding that one-qubit gates.

While large size

Theoretical quantum computation has been usually fo-

cused on assessing the number of elementary gates to

build a given unitary operator performing a given com-

putation. Some works have tried to focus on two-qubit

gates and to minimize their number in order to build a

given unitary transformation for several qubits [9, 10].

Still, unitary transformations in many applications are

a tool to transform an initial state to a given state. It

seems therefore natural to try and assess how costly this

process is in itself. In this paper, we thus study the min-

imal number of two-qubit gates needed to change a given

quantum state to obtain another one. Of course, this

number is necessarily upper bounded by the number re-

quired for a general unitary transformation. We focus

on the case of two and three qubits. For two qubits, we

generalize the result of [11] and show that one CNOT

is enough to go from any given pure state to any other

one. For three qubits, we show that three CNOTs are

enough to go from |000? to any other pure state, and

that two CNOTs suffice if one starts from the GHZ state

(|000?+|111?)/√2. A corollary of the latter is thus that

four CNOTs are enough to go from any pure state to any

other pure state. The number of CNOT gates required to

go from a state to another defines a discrete distance on

the Hilbert space. Given any fixed state |ψ?, the Hilbert

space can be partitioned according to the distance to |ψ?.

It is known that if stochastic one-qubit operations are

used, entanglement of three [12] and four [13] qubits fall

into respectively two and nine different classes. Our clas-

sification according to the number of CNOTs is different,

although there are some relations. Our results generalize

to other universal two-qubit gates, in particular to the

iSWAP gate which has been shown to be implementable

for superconducting qubits [14].

Weconsider purestates

dimensional Hilbert space C2n. The space of normal-

ized quantum states is the sphere S2n+1−1.

cost of one-qubit gates is negligible, we are interested

only in equivalence classes of states modulo local uni-

tary transformations (LU). We thus consider the sets

En= S2n+1−1/U(2)nof states nonequivalent under LU.

In the case of two and three qubits the dimension of

E2 and E3 was determined in [15, 16] and their topol-

ogy has been described in [17].

per we will make use of the one-qubit LU operations

R(k)

jj

matrices acting on qubit k.

ation R(k)

y (ξ) corresponds to a rotation of the qubit

cos(ϕ)|0? + sin(ϕ)|1? ?→ cos(ϕ + ξ)|0? + sin(ϕ + ξ)|1?.

Two-qubit states. Let us first consider the two-qubit

case. It has been shown in [11] that one can transform

an arbitrary two-qubit state |ψ? to |00? by using only

one CNOT. Here we prove that the same holds for two

general two-qubit states |ψ? and |ψ′?.

Proof: Since E2 is homeomorphic to [0,1], only one

belonging tothe2n-

As the

Throughout the pa-

(ξ) = exp(−iξσ(k)

) where the σ(k)

In particular, the oper-

j

are the Pauli

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parameter (e.g., one Schmidt coefficient) characterizes

a state up to LU. More precisely, by LU each two-

qubit state can be brought to the canonical form |ψ? =

cosϕ|00? + sinϕ|11?, which is just Schmidt decomposi-

tion. We want to transform state |ψ? with parameter ϕ

to state |ψ′? with parameter ϕ′. When ϕ′?= ϕ, we need

at least one CNOT. It turns out that one CNOT is in

fact sufficient, as can be easily seen by checking that the

relation R(1)

convention, qubit 1 is the leftmost one).

In contrast, one needs three CNOTs in general to con-

struct a specific two-qubit unitary transformation [9].

Transforming one state to another is thus clearly easier.

We now turn to the three-qubit case.

Classification with respect to |000?. We start with

the case where we want to prepare a state |ψ? from |000?.

The distance (in number of CNOTs) from |000? to |ψ? is

a criterion for the difficulty to prepare |ψ?. We will show

that this distance partitions the Hilbert space into four

classes, and that any state can be prepared from |000?

using three CNOT gates or less. We will examine each

of these four classes in turn.

Class 0: One needs zero CNOT gates to transform |ψ?

to |000? iff the state is of the product form |ψ? = |αβγ?,

where |α?,|β?,|γ? are normalized single qubit states (this

is trivial, since only LU are used).

Class 1: One needs one CNOT iff the state is of the

form |ψ? = |α?1|χ?23(i.e., it is bi-separable), where |χ?23

is an arbitrary entangled state of the last two qubits [18].

Proof: By LU |ψ? can be transformed into canonical

form |ψ?LU

to this canonical form we obtain |0?(cosϕ|00?+sinϕ|10?),

i.e., state |0?(cosϕ|0? + sinϕ|1?)|0? which is in class 0.

We can therefore reach |000? in a single CNOT step.

Conversely, applying one CNOT gate on a state from

class 0 the state reached is bi-separable and therefore all

states that need 1 step to get from |000? are of the above

form.

In their canonical form states from class 1 can be

parametrized by a single real parameter ϕ.

Class 2: One needs two CNOT gates iff the state

is of the form |ψ? = cosϕ|αβγ? + sinϕ|α⊥β′γ′?, with

?α|α⊥? = 0, |?β|β′?| < 1 and |?γ|γ′?| < 1 (if |?β|β′?| or

|?γ|γ′?| are equal to 1 then |ψ? belongs to class 1).

Proof: We first bring the state by LU to the canon-

ical form |ψ?

are absorbed into the definition of local bases, |γ?

can be written as cosξ|0? + sinξ|1?.

R(3)

y

?

(cosϕ|00?+sinϕ|1β?)|γ′?, which is a state of class 1 from

which we can reach state |000? in a single step.

To prove the converse, we have to show that by using

two CNOT gates one can reach only states of the form

|ψ? = cosϕ|αβγ? + sinϕ|α⊥β′γ′?, or states in class 0 or

class 1. Starting from class 0, we are in class 1 after one

y (−ϕ)CNOT12R(1)

y (ϕ′)|ψ? = |ψ′? holds (by

= |0?(cosϕ|00?+sinϕ|11?). Applying CNOT23

LU

= cosϕ|000? + sinϕ|1βγ?. If the phases

The rotation

π

4−ξ

2

?

followed by a CNOT13 yields a state

step. Any class 1 state can be written as |α?(cosϕ|0γ?+

eiξsinϕ|1γ′?). Applying CNOT23 or CNOT32 we get a

state in class 0 or class 1. Applying CNOT21 on the

other hand we get cosϕ|α0γ? + eiξsinϕ|α1γ′?, where

|α? = σx|α?, which is indeed of the canonical form of

class 2 states. Last possibility is applying CNOT12. In

this case it is better to write our state in the basis |±? =

(|0? ± |1?)/√2 for the second qubit. That is, any class 1

state can be written as |α?(cosϕ| − γ? + eiξsinϕ| + γ′?).

Writing |α? = cosϕ′|0? + eiξ′sinϕ′|1?, we then get after

applying CNOT12state cosϕ(cosϕ′|0?−eiξ′sinϕ′|1?)|−

γ?+sinϕeiξ(cosϕ′|0?+eiξ′sinϕ′|1?)|+γ′?, which is again

of the canonical form expected. For CNOT13or CNOT31

the argument is similar.

One needs 3 real parameters to describe states of class

2 in their canonical form. Note that class 2 states con-

stitute a subset of GHZ type states which are of (unnor-

malized) form |αβγ? + |α′β′γ′? [12].

Class 3: One needs three CNOT gates iff a state is

not in class 0, 1 or 2.

Proof: States not in the previous classes are of two

types:(i) W-like states (according to the classifica-

tion in [12]) for which the range of the reduced den-

sity matrix of qubits 2 and 3 contains only one prod-

uct state. Such states are of the (unnormalized) form

|ψ? = |αβγ?+|α′?|χ?23, where |χ?23is entangled and or-

thogonal to |βγ?. Under LU they can be written in the

following canonical form [12]

|ψ?LU

with |α? = cosξ|0? + sinξ|1? and, (ii) GHZ-like states

with the canonical form

= cosϕ|000? + sinϕ|α?(cosϕ′|10? + sinϕ′|01?), (1)

|ψ?LU

= a|000? + eiξb|αβγ?, (2)

where |α?, |β? and |γ? are real single qubit states

parametrized by one parameter each and a,b are real

parameters, one of which is fixed by normalization [19].

To exclude class 2 states we must demand that none of

|α?, |β? and |γ? be equal to |1?. To exclude class 1 and 0

states in (1),(2) ρ23must be of rank 2. W-like states (1)

require 3 parameters. GHZ-like (2) states need 5, and

thus are the generic states.

First we show that by using single CNOT one can

transform class 3 states to class 2. For W-like states (1)

we just have to apply CNOT23to the canonical form (1)

and we immediately get cosϕ|000?+ sinϕ|α?(cosϕ′|1? +

sinϕ′|0?)|1? which is of class 2 (states on the third qubit

are orthogonal). For GHZ-like states (2) it is a bit more

work. Note that GHZ-type states can be, by rearranging

terms and after LU, written as cosϕ|000?+ sinϕ|1?|χ?23

(expanding |0? on the first qubit in (2) into |α? and |α⊥?

and adding |α? part to the second term), where |χ?23can

in turn be expanded as |χ?23 = cosϕ′|0δ? + sinϕ′|1δ′?)

with |?δ|δ′?| < 1 (otherwise state would be in class

2).Finally, rotating third qubit brings the state to

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cosϕ|00γ′′?+sinϕ|1?(cosϕ′|00?+eiξ′sinϕ′|1γ′?) with real

|γ′? = cosg′|0? + sing′|1?. After application of the ro-

tation R(3)

y

?

cosϕ|00˜ γ′′? + sinϕ|1?(cosϕ′|0? + eiξ′sinϕ′|1?)|˜ γ′? which

is of class 2.

Since the canonical forms of classes 0, 1, 2 and 3 span

the whole Hilbert space and since the forms of classes 0,

1 and 2 are the only ones that can be reached in 2 steps,

it immediately follows that the states of canonical forms

(1) and (2) require exactly three steps.

Thus any state is at a distance less than or equal to 3

from |000?.

Classification with respect to GHZ state. Let

us now examine the distance to the GHZ state |GHZ? =

(|000? + |111?)/√2. It turns out that any state is at a

distance less than or equal to 2 from GHZ. We use the fact

that any state in E3can be characterized by a set of six

polynomial invariants [21, 22]. Following [22], we denote

the first three invariants by Ii= trρ2

ρi is the reduced density matrix of the ith qubit. Any

three-qubit state is equivalent under LU to its canonical

form [22]

π

4−g′

2

?

followed by CNOT23 we arrive at

i, 1 ≤ i ≤ 3, where

λ0|000? + λ1eiφ|100? + λ2|101? + λ3|110? + λ4|111?, (3)

where the six parameters λ0,...,λ4,ϕ label the state in

E3and?λ2

in the canonical form (3) are unique.

Class 0: States at distance 0 from GHZ are states

whose canonical form (3) is GHZ (this is trivial).

Class 1: States at distance 1 from GHZ are states

whose canonical form (3) is (up to relabeling)

i= 1. If we take 0 ≤ ϕ ≤ π the parameters

1

√2|000? + λ2|101? + λ3|110? + λ4|111?,

(with λ4?= 1/√2) i.e., states with I1=1

Proof: It is straightforward to check that the state

(4) has the same invariants (i.e.

to LU) as CNOT23R(2)

1

2arcsin(λ2

distance 1 from GHZ. Conversely, if a state |ψ? is at dis-

tance 1 from GHZ, then one of its invariants I1,I2,I3has

(4)

2.

is the same up

y (θ2)R(3)

2arccos(λ3

y (θ3)|GHZ? with θ2 =

√2), and thus is at

√2) and θ3=1

to be the same as for GHZ. Since GHZ is symmetric un-

der permutation of the qubits, one can consider that the

CNOT gate applied is CNOT23(up to relabeling of the

qubits before applying the CNOT). In this case the in-

variant I1=1

of the first qubit of |ψ? therefore verifies trρ2

in turn (since it is a 2 × 2 density matrix) implies that

ρ1 =

2

as |ψ? =

canonical form of such states is precisely (4).

2is conserved. The reduced density matrix

1=1

2which

1

?. All states with this property can be written

1

√2|0?|χ? +

1

√2|1?|χ⊥?, with ?χ|χ⊥? = 0. The

Class 2: All other states are at distance 2 from GHZ.

Proof: We show that all other states are at distance

1 from states of canonical form (4). States of canonical

form (4) are characterized by I1=1

to prove that any state |ψ? not in class 0 or 1 can be trans-

formed to a state verifying I1=1

gate. The reduced density matrix of the first qubit is a

?

B∗1 − A

thus given by I1 = A2+ (1 − A)2+ 2|B|2. It is equal

to

2iff A =

yields three equations. One-parameter rotations of the

qubits before applying CNOT yield free parameters. It

turns out that a solution to the three equations always

exists. Indeed, the state |ψ? can be reduced either to

the canonical form (1) or to the canonical form (2). If

|ψ? is in canonical form (1), one can check that the state

CNOT12R(1)

2. Therefore we have

2with only one CNOT

2 × 2-matrix ρ1=

AB

?

. The first invariant is

1

1

2and B = 0, which (since B is complex)

y (θ1)R(2)

y (θ2)|ψ?, with

tan2θ1 =

cot2ϕ + cos2ξ

sin2ξ

(5)

tan2θ2 =

sin(ξ + 2θ1)cosϕ′sin2ϕ

sin(2ξ + 2θ1)cos2ϕ′sin2ϕ − cos2ϕsin2θ1

,

is such that I1=

canonical form (2). The one-qubit states |α?, |β?, |γ? can

be written in the form cosϕ|0?+sinϕ|1? with parameters

respectively ϕ1, ϕ2and ϕ3. Normalization imposes that

a2+b2+2abcosξ cosϕ1cosϕ2cosϕ3= 1. One can check

that the state CNOT12R(1)

parameters of the rotations given by

1

2. Let us now suppose that |ψ? is in

x (χ)R(1)

y (θ1)R(2)

y (θ2)|ψ?, with

tan2θ1 =

(b2− b4)cos2ϕ1− a2?a2− 1 − 2b2cos2ϕ1cos2ϕ2+ 2b4cos2ϕ2sin2(2ϕ1)sin2ϕ3

b2sin(2ϕ1)?1 − a2− b2+ 2a2cos2ϕ2(1 − a2sin2ϕ3+ b2cos(2ϕ1)sin2ϕ3)?

tan2θ2 = −

a2sin2θ1+ b2sin(2ϕ1+ 2θ1)cos2ϕ2+ 2abcosξ sin(ϕ1+ 2θ1)cosϕ2cosϕ3

tan2χ = −2absinξ sinϕ1sinϕ2cosϕ3(a2sin2θ1− b2sin(2ϕ1+ 2θ1))

?

b2sin(2ϕ1+ 2θ1)sin2ϕ2+ 2abcosξ sin(ϕ1+ 2θ1)sinϕ2cosϕ3

(6)

2asinϕ1sinϕ2(2ab2cosϕ1cosϕ2+ b(a2+ b2)cosξ cosϕ3),

verifies I1=1

2.

The results above show that the number of CNOTs

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needed to transform a state to another state is much less

than the number to produce all unitary transformations.

Indeed, according to [10], one needs at least 14 CNOT

gates to produce any three-qubit unitary transformation.

We also note that the best available algorithm actually

does not saturate the bound, needing 20 CNOTs [10].

Our procedure is explicit, and selects a specific unitary

transformation leading from one state to the other using

less CNOTs than a general unitary transformation.

Let us now discuss the applicability of these results to

physical systems. In an experimental context, our re-

sults can be used to construct any desired state from the

initial state which is easiest to produce with a given sys-

tem. In some experimental set-ups, two-qubit gates are

built from a nearest neighbor interaction (for example

in [5, 6]). In this case, the common procedure is to use

additional SWAP gates to transfer the states of two dis-

tant qubits to nearest neighbors before performing the

CNOTs. However, in our case, due to the symmetry of

GHZ state, one can go from any state to the GHZ state

in two CNOTs without the need of any quantum SWAP.

Thus even if only nearest-neighbors CNOTs are available

for three qubits on a line, still only 4 CNOTs are enough

to go from any state to any other state. If one starts

from |000? and one allows for relabeling of qubits in the

final state, 3 CNOTs are still enough to go to any state,

except for the GHZ-type class 3 states, Eq. (2), which

need an additional CNOT in this architecture.

Any two-qubit gate can be expressed in terms of

CNOTs and one-qubit gates. Thus our result will im-

ply a bound in the number of two-qubit gates needed to

go from one three-qubit state to another, for any other

choice of universal two-qubit gate. We note that another

popular two-qubit gate is the iSWAP which is natural

for implementations corresponding to a XY-interaction.

As the iSWAP can be expressed in terms of one CNOT

and one SWAP gate plus one-qubit gates [23], our re-

sults apply directly to this particular gate provided the

SWAP can be made classically: the number of iSWAPs

needed to transform three qubits is then the same as for

CNOTs. This in particular arises when the physical im-

plementation allows for a coupling between any pair of

qubits, as swapping two qubits is equivalent to relabeling

the qubits for all subsequent gates by interchanging the

role of the qubits. An important example is the case of

superconducting qubits coupled to each other via cavity

bus [14], one of the most promising recent developments,

where the resonance can be tuned to couple any pair.

In conclusion, we have shown that one needs only

three CNOTs plus additional one-qubit gates to trans-

form |000? to any pure three-qubit states. If one starts

from the GHZ state, only two CNOTs are enough, and

thus one needs only four CNOTs plus additional one-

qubit gates to transform any initial pure three-qubit state

to any other pure three-qubit states. An interesting open

question is to find out whether four is the maximal dis-

tance between two three-qubit states (it should be at

least three since |000? is at distance 3 from class 3 states).

It would be also interesting to know how these results

translate to mixed states, or to pure states using SLOCC

instead of LU. At last, generalizations to higher numbers

of qubits may pave the way to better optimization of

quantum algorithms, which are usually described as uni-

tary operators but are sometimes just transformations of

a given state into another one.

We thank the French ANR (project INFOSYSQQ) and

the IST-FET program of the EC (project EUROSQIP)

for funding. MˇZ would like to acknowledge support by

Slovenian Research Agency, grant J1-7437, and hospi-

tality of Laboratoire de Physique Th´ eorique, Toulouse,

where this work has been started.

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