Table Of ContentParallel Treatment of
Estimation of SU(2) and Phase Estimation 1
Masahito Hayashi
6 Quantum Computation and Information Project, ERATO, JST
0 5-28-3, Hongo, Bunkyo-ku, Tokyo, 113-0033, Japan
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Abstract
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2 We discuss the accuracy of the estimation of the n identical unknown actions of
SU(2) by using entanglement. This problem has a similar structure with the phase
3
v estimation problem, which was discussed by Buˇzek, Derka, and Massar[1]. The es-
3 timation error asymptotically decreases to zero with an order of 1 at least.
5 n2
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Key words: SU(2), Estimation, Entanglement, Phase estimation
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4 PACS: 03.65.Wj, 03.65.Ud, 02.20.-a
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1 Introduction
u
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i 1.1 Estimation of unitary action
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a
Quantum information processing architecture is constructed with a combina-
tion of various quantum devices. In other words, each quantum device con-
tributes towards obtaining the output quantum state from the input quantum
state. Hence, it is necessary to identify the quantum operation acting on each
device. This problem is referred to as the identification of the quantum opera-
tion; this problem usually entails the estimation of the operation based on the
pair of theinput (initial) stateandthe dataof themeasurement fortheoutput
Email address: [email protected](Masahito Hayashi).
1 Thematerial inthispaperwas presentedinpartatTheNinth QuantumInforma-
tion Technology Symposium (QIT9), Atsugi, Kanagawa, Japan, December 11–12,
2003. It was also presented in part at The Seventh International Conference on
Quantum Communication, Measurement and Computing, Glasgow, UK, July 25–
29,2004.Thefinalsection(Concludingremark)wasaddedafterthesepresentations.
Preprint submitted to Physics Letters A 1 February 2008
state. Generally, such a quantum operation is described by a trace-preserving
completely positive (TP-CP) map. In this paper, we consider a noiseless case
inwhich no noise appears inthe quantum operation.In this case, thequantum
operation is described by a unitary matrix. The adiabatic case is also treated
in a similar manner.
When the reference system is unitarily equivalent to the input system and
the initial state is a maximally entangled state in a joint system between
the original input system and the reference system, the final state in the
joint system is the maximally entangled state described by the unitary matrix
acting on the input system. Hence, by repeating this operation n times, we
can estimate this unitary matrix by performing appropriate measurement for
the total joint system between output system and the reference system. In the
two-dimensional case, Fujiwara [2] showed that a maximally entangled state
is the optimal initial state in the composite system between the single input
system and its reference system. Ballester [3] discussed this problem in the d-
dimensional case. When their method is applied to the case of n applications
of this unknown operation, the estimation error decreases to zero only with
order 1; this is the case even if the output measurement is optimized. This
n
order is obtained from the accuracy of state estimation by using an identical
state preparation. Therefore, it is worthwhile to consider the possibility of
further improvement.
1.2 One-parameter case: Phase estimation
The phase estimation problem is closely related to the problem discussed in
this paper. In the phase estimation problem, we estimate the eigenvalue eiθ of
the unknown unitary matrix acting on a two-level system when we know its
eigenvectors. The phase estimation with the fixed input state was discussed
by Helstrom[4] first. Holevo [5] extended Helstrom’s result to a more general
framework of group covariance. The optimization problem of the input state
forphase estimationwas discussed by Buˇzek et al.[1]inanasymptotic setting.
They proved that the error decreases to zero with speed π2 when we choose
4n2
the optimal input state and optimal measurement. However, the error goes to
zero with the order 1 in the usual statistical parameter estimation. Therefore,
n
this unexpected result indicates the importance of the entangled input state.
1.3 Three-parameter case: Our result
In this study, we discuss whether such a phenomenon occurs in the estimation
of SU(2) unitary action. In this paper, we adopt the error function d(U,Uˆ) d=ef
2
1 − |Tr U−1Uˆ|2 between the true SU(2) matrix U and the estimated matrix
2
Uˆ. Then, we obtain an unexpected relation between our problem and that
discussedbyBuˇzeket al.[1].Thankstothisrelation,wecanshowthattheerror
decreases tozeroinproportionto 1 atleast.Further,thecoefficient isequalto
n2
π2 if the estimator is constructed based on this relation. It is shown that this
bound can be asymptotically attained with no use of the reference system.
Instead of use of the reference system, we regard a part of the composite
system of n input systems as the tensor product of the system of interest and
the reference system. In other words, an effect of “self-entanglement” is used
in this method.
2 Phase estimation: Estimation of eigenvalues with the knowledge
of eigenvectors
In order to demonstrate the interesting relation between phase estimation and
estimation of SU(2) action, we briefly summarize the fact known with regard
to phase estimation. When the eigenvectors of the unknown SU(2) matrix U
are known, the estimation problem can be reduced to the estimation of the
eiθ 0
unknown parameter θ ∈ [0,2π) of the family U d=ef θ ∈ [0,2π)
θ (cid:12)
0 1 (cid:12)(cid:12)
(cid:12)
because the action of U is equivalent to that of cU. (cid:12)
(cid:12)(cid:12)
In our framework, we assume that the tensor product matrix U n acts in the
θ⊗
tensor product space H n and that we can select an arbitrary initial state in
⊗
the tensor product space H n. That is, we can select the input state x ∈ H n
⊗ ⊗
and the estimating POVM Mn(dθˆ) on H n with outcomes in [0,2π).When we
⊗
evaluate the error between the true parameter θ and the estimated parameter
θˆby the function sin2 θ θˆ, the mean error is given by
−
2
2π θ−θˆ
Dn(Mn,x) d=ef sin2 hx|(U n) Mn(dθˆ)U n|xi.
θ 2 θ⊗ † θ⊗
Z0
Thus, when we fix the input state x ∈ H n, our estimation problem can be re-
⊗
duced to the estimation with the state family U n|xihx|(U n) θ ∈ [0,2π) .
θ⊗ θ⊗ †
n (cid:12) o
(cid:12)
In order to treat this problem in a simple manner, we focus o(cid:12)n the phase
estimation problem in the d-dimensional space H spanned by the orthonor-
′
mal basis u ,...,u as follows. We select a vector x = (x )d = d xku
1 d k k=1 k=1 k
satisfying
P
3
d
|x |2 = 1. (1)
k
k=1
X
Suppose that the state to be estimated has the form ρ d=ef U |xihx|U with
θ,x θ θ∗
theunknownparameterθ ∈ [0,2π)andtheunitarymatrixU d=ef d eikθ|u ihu |.
θ k=1 k k
In this case, the error of the estimator M(dθˆ) can be be expressed as
P
2π θ−θˆ
D (M,x) d=ef sin2 TrM(dθˆ)ρ . (2)
θ θ,x
2
Z0
In order to ensure accuracy, it is appropriate to focus on the worst case. In
other words, we minimize the worst error max D (M,x). This problem is
θ θ
refereed to as minimax.
On the other hand, since the state ρ has symmetry
θ,x
ρθ+θ′,x = Uθ′ρθ,xUθ†′, (3)
it is natural to treat the measurement M(θˆ)dθˆwith the same symmetry:
M(θˆ+θˆ′) = Uθˆ′M(θˆ)Uθˆ†′. (4)
A measurement M satisfying (4) is referred to as the covariant measurement
and it has the form
dθˆ
MT(dθˆ) d=ef UθˆTUθˆ†2π, (5)
where the Hermitian matrix T = t |u ihu | satisfies
k,l k,l k l
P
T ≥ 0 and t = 1. (6)
k,k
Holevo proved that the minimum error in the minimax criteria is equal to the
minimum error among covariant measurements, i.e.,
minmaxD (M,x) = min D (M,x).
θ θ
M θ M:covariant
This relation is referred as the quantum Hunt-Stein lemma, and was proved
in a more general covariant setting[5].
By using elementary formulas of trigonometric functions, the equation
4
2π θ −θˆ dθˆ
sin2 2 TrUθ|uk′ihul′|Uθ†Uθˆ|ulihuk|Uθˆ†2π
Z0
1 1 1
= δ δ ( δ − δ − δ ) (7)
k,k′ l,l′ k,l k,l 1 k 1,l
2 4 − 4 −
can be verified. By using this relation, we have
1 d 1 d 1
D (M ,x) = |x |2t − − (x x t +x x t )
θ T k k,k k k+1 k+1,k k+1 k k,k+1
2 4
k=1 k=1
X X
1 d 1 d 1 1 d 1
≥ |x |2 − − (|x x |+|x x |) = 1− − |x ||x | , (8)
k k k+1 k+1 k k k+1
2 4 2 !
k=1 k=1 k=1
X X X
where the inequality follows from |hu |T|u i| ≤ hu |T|u ihu |T|u i =
k k+1 k k k+1 k+1
1. Hence, the equality holds iff q
x x
t = k k+1 . (9)
k,k+1 |x ||x |
k k+1
Since the matrix T d=ef xkxl |u ihu | satisfies the conditions (6) and (9),
we obtain x k,l |xk||xl| k l
P
1 d 1
−
min D (M,x) = 1− |x ||x | . (10)
θ k k+1
M:covariant 2 !
k=1
X
Now, we consider the estimation of the unknown unitary U with its multi-
θ
ple actions. Since the unitary matrix U n has eigenvalues 1,eiθ,...,eniθ, the
θ⊗
application of the relation (10) to the d = n+1 case yields
Dn d=ef min minmaxDn(Mn,x)
opt θ
x ⊗n Mn θ
∈H
1 n n
= min 1− a a a2 = 1,a ≥ 0 .
(ak)nk=0(2 kX=1 k k−1!(cid:12)(cid:12)kX=0 k k )
(cid:12)
(cid:12)
In this problem, the eigenspace H is not one(cid:12)-dimensional for the eigenvalue
k
eikθ. Hence, any initial state can be written as x ek, where ek is a nor-
k k
malized vector in H . Therefore, the minimum error can be expressed in the
k P
above form. Buˇzek, et al.[1] proved that the minimum error is attained by the
coefficient a = √2 sin π(k+1/2) and is almost equal to π2 , i.e.,
k √n+1 n+1 4n2
π2
Dn ∼= . (11)
opt 4n2
5
In this setting, the irreducible representation space of the action of the one-
dimensional circle S1 = [0,2π) is the one-dimensional space spanned by u . In
k
the above discussion, we use n+1 different irreducible representation spaces.
3 Estimation of the unknown SU(2) action
Next, we proceed to the estimation of the unknown SU(2) action g ∈ SU(2)
in the two-dimensional space H. In this problem, the unitary matrix g n acts
⊗
in the tensor product space H n, and a suitable initial state in H n can be
⊗ ⊗
selected for this estimation. On the other hand, Fujiwara[2] focused on the
estimation problem where the initial state is entangled with the reference
system H , in which the unknown SU(2) action g does not act. He proved
R
that this method reduces the estimation error. However, he did not treat
the entanglement between the n-tensor product space H n and its reference
⊗
system.
Now, we consider the amount of the estimation error can be reduced by use
of entanglement among tensor product space. In other words, our framework
has a wider choice. In our problem, we can select an initial state entangled
between H n and the reference space H n that is unitarily equivalent to the
⊗ R⊗
original input system H n. Hence, when we select the initial state x on the
⊗
tensor product space H n ⊗H n and the measurement (POVM) Mn(dgˆ) in
⊗ R⊗
H n ⊗H n with the outcome in SU(2), the error is evaluated as
⊗ R⊗
D (M,x) d=ef d(g,gˆ)hx|(g n ⊗I) M(dgˆ)(g n ⊗I)|xi.
g ⊗ ∗ ⊗
ZSU(2)
Here, we focus on the SU(2) action on the tensor product space H n. Its
⊗
irreducible decomposition is given as follows:
d
H 2d = H ⊗H
⊗ 2k+1 2d,2k+1
k=0
M
d
H 2d 1 = H ⊗H ,
⊗ − 2k 2d 1,2k
−
k=1
M
where H is the k-dimensional irreducible space of the SU(2) action, and H
k n,k
is the corresponding irreducible space of the action of the permutation group,
where SU(2) does not act. Note that the dimension of H is equal to the
n,k
number of representation spaces equivalent to H in the tensor product space
k
H n. Hereafter, we denote the SU(2) action on H by Vk.
⊗ k g
6
For simplicity, first we focus on the estimation of the SU(2) action in a sin-
gle irreducible space H . Now, we select the initial state as the maximally
j
entangled state xj between H and the reference space H that is unitar-
E j j,R
ily equivalent with the space H . The measurement is selected as the POVM
j
M (dgˆ) d=ef j2Vj ⊗ I|xj ihxj |(Vj ⊗ I) µ(dgˆ), where µ is the invariant prob-
j,E gˆ E E gˆ †
ability distribution on SU(2) and the integer j2 is the normalizing factor. By
using Schur’s lemma, we can easily verify that the total integral is constant
times the identity matrix because the state xj is maximally entangled. Since
E
(TrVj) = Tr(Vj) , the average error is calculated as
g ∗ g †
d(g,gˆ)j2hxj |(Vj) Vj|xj ihxj |(Vj) Vj|xj iµ(dgˆ)
E g † gˆ E E gˆ † g E
ZSU(2)
= d(g,gˆ)|TrVj |2µ(dgˆ) = d(I,gˆ)|TrVj |2µ(dgˆ)
gˆ−1g gˆ−1
ZSU(2) ZSU(2)
3 j = 1
= d(I,gˆ)|TrVj|2µ(dgˆ) = 4 (12)
gˆ
ZSU(2) 21 j ≥ 2.
The final equation follows from the elementary calculus of trigonometric func-
tions. This calculation seems to indicate that if we use only one irreducible
space H and even if its dimension is large, the estimation error cannot be
j
reduced2. Hence, in order to reduce the estimation error, it may be needed
to use several irreducible spaces.
Hereafter, we consider the case of n = 2d−1; however, the following discussion
can be applied to the even case. In the odd case, d distinct irreducible spaces
exist. Hence, it is essential to use the correlationbetween them. We investigate
the following subspace of H⊗(2d−1) ⊗HR⊗(2d−1):
m
H ⊗H .
2k 2k,R
k=1
M
Further, we denote the SU(2) representation on this space by U2d 1.
g −
As demonstrated later, this problem can be treated parallel to the phase es-
timation in which the base u corresponds to the maximally entangled state
j
x2k or the vector (2k)x2k. We select a vector ~x = (x )d that satisfies the
E E d k k=1
condition (1), and let the initial state be x2d 1 d=ef d x x2k. In a manner
~xd− k=1 k E
similar to (5), the measurement is selected as
L
2 This fact has been proved by Chiribella et al.[16] after the submission of the
preliminaryversion ofthis paper.Thatis,they proved theoptimality ofthis POVM
in a more general framework.
7
MT (dgˆ)d=ef U2d 1 t |(2k)x2kih(2l)x2l|(U2d 1) µ(dgˆ)
2d 1 gˆ − k,l E E gˆ − †
−
k,l
X
based on a Hermitian matrix T = (t ) satisfying condition (6). Using Schur’s
k,j
lemma and condition (6), we can verify that its total integral is an identity
matrix. In a manner similar to (7), the equations
d(g,gˆ)TrU2d 1|x2k′ihx2l′|(U2d 1) U2d 1|(2k)x2kih(2l)x2l|(U2m 1) µ(dgˆ)
g − E E g − † gˆ − E E gˆ − †
ZSU(2)
1 1 1
=δ δ d(g,gˆ)TrV2k(TrV2l) µ(dgˆ) = δ δ δ − δ − δ
k,k′ l,l′ZSU(2) gˆ gˆ † k,k′ l,l′(cid:18)2 k,l 4 k,l−1 4 k−1,l(cid:19)
(13)
holds, where the final equation is derived in section 4 based on elementary
formulas of trigonometric functions. By using this relation in a manner similar
to (8), we have
d(g,gˆ)TrU2d 1|x2d 1ihx2d 1|(U2d 1) MT (dgˆ)
ZSU(2) g − ~xd− ~xd− g − † 2d−1
1 d 1 d
= |x |2t − (x x t +x x t )
k k,k k 1 k k,k 1 k k 1 k 1,k
2 4 − − − −
k=1 k=1
X X
1 d 1 d 1 1 d 1
≥ |x |2 − − (|x x |+|x x |) = 1− − |x ||x | . (14)
k k k+1 k+1 k k k+1
2 4 2 !
k=1 k=1 k=1
X X X
The equality holds when matrix T = (t ) satisfies (9). Thus, the optimal
k,l
error of this estimation method coincides with Dm 1. That is, our problem is
op−t
reduced tophaseestimation problembyBuˇzek et al.Whenwe select a suitable
initial state and measurement in the case of large n, the estimation error is
equal to π2 asymptotically because Dd 1 ∼= π2 ∼= π2 = π2.
n2 op−t 4(d 1)2 (2d 1)2 n2
− −
In the case n = 2d, the initial state is expressed as x2d d=ef d a x2k+1, where
~a′ k=0 k E
d
the vector ~a = (a )d has no negative element. When we select a measure-
′d k k=0 L
ment similar to the one mentioned above, the estimation error is calculated
to be 12 1− dk−=10akak+1 + 41a0. Hence, the minimum error Dodpt′ satisfies
Dodpt ≤ D(cid:16)odpt′ ≤PDodp−t1. The(cid:17)same conclusion is obtained in the odd case.
Next, we investigate the reference space in the odd-dimensional case. When
we use the above method, the dimension of the reference space H is 2d.
R,2d 1
−
If the dimension of H is greater than that of H , we can use the space
2d 1,2k 2k
−
H as the reference space. For k = d, the dimension of H is 1, i.e.,
2d 1,2k 2d 1,2k
− −
is smaller than that of H . However, for k < d, the dimension of H is
2k 2d 1,2k
2d−1 2d−1 −
− , i.e., it is greater than that of H . Hence, by replacing
d−k d−k −1 2k
(cid:16)therefer(cid:17)enc(cid:16)espaceby(cid:17)thespaceH2d 1,2k,wecanreducetheestimationerrorto
Dd 2 without using the reference sy−stem. Since Dd 2 ∼= π2 ∼= π2 = π2,
op−t op−t 4(d 2)2 (2d 1)2 n2
− −
8
an estimation error of π2 can be achieved without the reference space. This
n2
discussion can also be applied to the case when n is an even number.
4 Technical details
In the following, we investigate the derivations of equations (12) and (13).
Since TrVj and d(I,gˆ) depend only on the eigenvalues of g eiθ/2,e iθ/2, the
gˆ −
invariant distribution µ has the form µ(dgˆ) = 1 sin2 θ dθsinφ dφ dφ in the
π 2 1 1 2
following parameterization:
cosφ sinφ eiφ2 † eiθ/2 0 cosφ sinφ eiφ2
1 1 1 1
gˆ = .
−sinφ e iφ2 cosφ 0 e iθ/2 −sinφ e iφ2 cosφ
1 − 1 − 1 − 1
Hence, the relation
2π f(θ) θ
f(θ)µ(dgˆ) = sin2 dθ (15)
π 2
ZSU(2) Z0
holds. Because d(I,gˆ) = sin2 θ2 and TrVgˆj = jl=1ei(l−j+21)θ and by applying
(15), we have
P
d(I,gˆ)|TrVgˆj|2µ(dgˆ) = 2π π1 sin4 θ2 j ei(l−j+21)θ j ei(l−j+21)θdθ
ZSU(2) Z0 l=1 l=1
X X
2π 1 3 1 j
= −2cosθ+ cos2θ j +2 (j −l)coslθ dθ.
4π 2 2
Z0 (cid:18) (cid:19) l=1
X
When j ≥ 2, this integral is equal to
2π 1 3 1
−2cosθ+ cos2θ (j +2(j −1)cosθ+2(j −2)cos2θ)dθ,
4π 2 2
Z0 (cid:18) (cid:19)
which implies (12). When j = 1, we have
2π 1 3 1 1
−2cosθ+ cos2θ dθ = ,
4π 2 2 2
Z0 (cid:18) (cid:19)
which implies (12).
9
With regard to (13), we can calculate
d(I,gˆ)(TrV2k)(TrV2k′) µ(dgˆ)
gˆ gˆ †
ZSU(2)
= 2π 1 3 −2cosθ+ 1 cos2θ 2k ei(l−2k2+1)θ 2k′ ei(l−2k′2+1)θ dθ.
Z0 4π (cid:18)2 2 (cid:19) l=1 ! l=1 !
X X
In the above integral, the term 2l=k1ei(l−2k2+1)θ l2=k′1ei(l−2k′2+1)θ can be ex-
panded into many terms; howev(cid:16)er, all other ter(cid:17)m(cid:16)s except the con(cid:17)stant term,
P P
2cosθ, and 2cos2θ vanish. That is, the coefficients of only the above three
terms are important.
In the following, we calculate this integral in the three cases: (i) k = k , (ii)
′
|k − k | > 1, and (iii) |k − k | = 1. Case (i) has already been calculated
′ ′
in (12). In case (ii), these three coefficients coincide. Thus, the integral is
equal to 0. Next, we proceed to the case (iii). When k = k + 1, the term
′
2l=k1ei(l−2k2+1)θ 2l=k′1ei(l−2k′2+1)θ can be expanded to k + 2kcosθ + 2(k −
(cid:16)1)cos2θ+···. T(cid:17)h(cid:16)us, the integral(cid:17)is equal to −1. Similarly, we can show that
P P
4
the integral is equal to −1 when k = k −1. Therefore, we obtain (13).
4 ′
5 Concluding remark
In this paper, we derived a remarkable relation between the estimation of the
SU(2) action and phase estimation. By using this relation, we showed that
the optimal estimation error is less than π2. The essence of this relation lies
n2
in the relation between the two similar relations (7) and (13). Indeed, since
sin2θ = 1 − |eiθ/2+e−iθ/2|2, the cost function of phase estimation is equal to
2
1−|χ (θ−θˆ)|2, where χ (θ) is the half of the character of the two-dimensional
2 2
eiθ/2 0
representation eiθ 7→ . Hence, the essence of (7) is the equation
0 e iθ/2
−
2π 1 1 1
(1−|χ (θ)|2)χk(θ)χl(θ)dθ = δ − δ − δ , (16)
2 k,l k,l 1 k 1,l
Z0 2 4 − 4 −
whereχl(θ)isthecharacteroftheone-dimensionalrepresentationofeiθ 7→ eilθ.
Ontheotherhand,whenweletχl(g)assumethecharacterofthel-dimensional
irreducible representation of SU(2) over dimension l, the essence of (13) is
1 1 1
(1−|χ2(g)|2)χ2k(g)χ2l(g)µ(dg) = δ − δ − δ . (17)
k,l k,l 1 k 1,l
ZSU(2) 2 4 − 4 −
10
