Volume of the set of locally diagonalizable bipartite states
arXiv:1611.09586v1 [quant-ph] 29 Nov 2016
Lin Zhang1 ,∗ Seunghun Hong2†
1 Institute
of Mathematics, Hangzhou Dianzi University, Hangzhou 310018, PR China
2 Northwestern
College, 101 7th St SW, Orange City, 51041 IA, USA
Abstract
The purpose of this article is to investigate the geometry of the set of locally diagonalizable bipartite
quantum states. Instead of characterizing which bipartite quantum states may be locally diagonalizable
except the two-qubit situation, we calculate the Hilbert-Schmidt volume of all locally diagonalizable bipartite quantum states. Besides, we partition the set of all locally diagonalizable states as local unitary
orbits (or co-adjoint orbits) of diagonal forms. It is well-known that the Riemannian volume of a co-adjoint
orbit for a regular point in a specified Weyl chamber can be calculated exactly by Harish-Chandra’s volume formula. By modifying Harish-Chandra’s volume formula, we firstly give a specific formula for the
Riemannian volume of a co-adjoint local unitary orbit of a regular point in a specified Weyl chamber.
Several open questions are presented as well.
Keywords: Euclid volume; Hilbert-Schmidt measure; Harish-Chandra’s volume formula; local unitary
orbit
1 Introduction
Qubits and qubit quantum channels are the simplest building blocks for quantum information processing
and quantum computations. A qubit is the quantum analog of the classical bit; a qubit quantum channel
is just the quantum analog of the transition probability matrix. Recently, Lovas and Andai [11] analyze the
structure of these qubit channels by duality between quantum maps and quantum states, i.e., via ChoiJamiłkowski correspondence [19]. They give a calculation about the (Euclid) volume of general and unital
qubit channels (real and complex case) with respect to the Lebesgue measure. For unital qubit channels,
they are essential equivalent to two-qubit states with the same marginal states–completely mixed states
via Choi-Jamiłkowski representation.
In the recent decades, the geometric separability probability of bipartite systems, i.e., the ratio of
volumes of the set of separable bipartite states to the set of all bipartite states on the same tensor space
of two Hilbert spaces, has been extensively studied. In 1998, Życzkowski et al. raised this question and
∗ E-mail:
† E-mail:
[email protected]; [email protected]
[email protected]
1
gave a detailed discussion about it [18]. In order to solve this problem, as suggested by the definition of
separability probability, we must calculate two volumes: (1) the volume of the set of all states acting on the
same Hilbert space; (2) the volume of the set of all separable states acting on the bipartite Hilbert spaces.
Luckily, the volume of the set of all states with respect to the Hilbert-Schmidt measure was computed
by Życzkowski [21] and Andai [1]. Later a detailed reasoning for such volume formula was reviewed by
Zhang [17]. One of the simplest but very difficult problem is to compute the separability probability of
two-qubit quantum states with respect to the Hilbert-Schmidt measure. Numerical simulations leads an
intriguing formula for separability probability, presented in 2013 by Slater [16]: the separability probability
sep
sep
volHS (D(C2 ⊗C2 ))
vol (D(R2 ⊗R2 ))
8
for real two-qubit state is volHS (D(R2 ⊗R2 )) = 29
64 and for complex two-qubit is vol (D(C2 ⊗C2 )) = 33 , where
HS
HS
D (K m ⊗ K n ) stands for the set of all bipartite density matrices acting on K m ⊗ K n . Here K = R or C.
29
64 has been proved by Lovas
8
is still open at present.
being 33
Now the separability probability for real two-qubit state being
[12]. But the separability probability for complex two-qubit
and Andai
The purpose of this article is to infer some information about the set of bipartite quantum states
which may be locally diagonalizable. Instead of characterizing which bipartite qudits may be locally
diagonalizable, we calculate the volume of all bipartite quantum states that are locally unitary equivalent
to some diagonal matrix. It is well-known that the volume of a co-adjoint orbit for a regular point in a
specified Weyl chamber can be calculated exactly by Harish-Chandra’s volume formula. By modifying
Harish-Chandra’s volume formula, we firstly give a specific formula for the volume of a co-adjoint local
unitary orbit of a regular point in a specified Weyl chamber. As an application, we calculate the volume of
the set of all bipartite quantum states that are locally unitary equivalent to some diagonal quantum states.
The paper is organized as follows. After introducing basic notions used in the present paper, in Sect. 2,
we give a derivation of joint probability distribution density of all eigenvalues of all locally diagonalizable bipartite states. At this section, we present a necessary and sufficient condition for two-qubit being
locally unitary (LU) equivalent to a given diagonal form. Sequentially, we prove one of our main results
(Theorem 2.6). In Sect. 3, we present demonstration of Harish-Chandra’s volume formula in the explicit
case such as unitary group and the tensor product of two unitary groups. First main result (Theorem 3.5)
in this section lead to the analytical formula of the volume of tensor product of two unitary groups. Subsequently, such formula is applied to study the local unitary orbit of a diagonal form. The second main
result (Theorem 3.8) in this section leads to a conclusion that the volume of locally co-adjoint orbits of
tensor product is exactly the product of volumes. We conclude Sect. 4 with discussion and several open
problems.
2 Volume of the set of all locally diagonalizable states
Suppose that we are given two finite-dimensional Hilbert spaces H and K. Specifically, let H = C m and
n
m
K = C n . Chose the standard basis {|i i}m
C n , respectively. A qudit is represented
i =1 and {| j i} j =1 for C and
by a positive semi-definite matrices of unit trace. Denote by D C k the set of all k × k density matrices.
By this, a bipartite qudit is represented by a density matrices in D (C m ⊗ C n ). Throughout this paper, we
do not distinguish the meaning of a state and a density matrix. That is, we view both as the same thing.
A bipartite state ρ AB is separable if it can be written as a probabilistic mixture of product states ρ µA ⊗ ρµB ,
2
where ρµA ∈ D (C m ) and ρµB ∈ D (C n ), i.e.,
ρ AB =
∑ pµ ρµA ⊗ ρµB .
(2.1)
µ
Here { pµ }µ is a probability distribution. A bipartite state is called entangled if it cannot be written as a
probabilistic mixture of product states. It is well-known that each Hermitian matrix can be diagonalized
by the adjoint action of a unitary matrix via Spectral Decomposition Theorem. In view of this, we see that
a bipartite state ρ AB ∈ D (C m ⊗ C n ) can be diagonalized by some global unitary matrix U ∈ U(mn), the
unitary group consisting of all unitary matrices on C m ⊗ C n . That is,
ρ AB = UΛU †,
m
Λ=
n
∑ ∑ λij |ijihij|,
(2.2)
i =1 j =1
where † denotes the adjoint. Note that all eigenvalues λij of ρ AB are indexed by two indices.
A bipartite state ρ AB is said to be locally diagonalizable if it can be brought to be the diagonal form
ρ AB = (U A ⊗ UB )Λ(U A ⊗ UB )† ,
(U A ∈ U(m), UB ∈ U(n)).
(2.3)
Here we can require that U A ∈ SU(m), UB ∈ SU(n). Denote the set of all locally diagonalizable bipartite
states from D (C m ⊗ C n ) by the following notation
DLU (C m ⊗ C n ) := {ρ ∈ D (C m ⊗ C n ) : ρ is locally diagonalizable} .
(2.4)
Whether a bipartite state is locally diagonalizable is intimately related to the question of the so-called local
unitary equivalence. Let ρ and ρ′ be two bipartite states in D (C m ⊗ C n ). They are called locally unitary
equivalent (abbr. LU) if
ρ ′ = (U ⊗ V ) ρ (U ⊗ V ) †
(2.5)
for some unitary matrices U ∈ SU(m) and V ∈ SU(n). Thus in (2.3), ρ AB is LU equivalent to the diagonal
form Λ. Clearly, not every bipartite state can be locally diagonalizable, even so for separable states. A
simple dimension counting may account for the reason [2]. It is easily seen that dim (D (C m ⊗ C n )) =
(mn)2 − 1. We can identify the submanifold DLU (C m ⊗ C n ) with (U(m) ⊗ U(n))/( T(m) ⊗ T(n) ) × ∆mn−1 .
That is,
DLU (C m ⊗ C n )
where ∆k−1 =
n
≃ (U(m) ⊗ U(n))/( T(m) ⊗ T(n) ) × ∆mn−1
(2.6)
≃ (U(m)/T(m)) ⊗ (U(n)/T(n)) × ∆mn−1 ,
(2.7)
( p1 , . . . , pk ) ∈ R k+ : ∑kj=1 p j = 1
o
denotes the (k − 1)-dimensional probability simplex.
Also, T( m) and T( n) denote the maximal tori of Lie groups U(m) and U(n), respectively. More on this in
Sect. 3. Therefore
dim (DLU (C m ⊗ C n )) = (m2 − 1) + (n2 − 1) + (m − 1)(n − 1).
(2.8)
Besides, the set of all product mixed states forms a (m2 + n2 − 2)-dimensional subset of the whole state
space.
3
Lemma 2.1. The joint probability density of eigenvalues of all locally diagonalizable bipartite states from DLU (C m ⊗
C n ) is given by
"
P(Λ) ∝
∏′
#
∑ (λij − λi′ j )2 
n
16 i < i 6 m j = 1
m

∏′ ∑ (λij − λij′ )2  [dΛ],
16 j < j 6 n i = 1
(2.9)
n
where [dΛ] := ∏m
i =1 ∏ j =1 dλij is the Lebesgue volume element for the diagonal matrix Λ mentioned in (2.2).
Proof. Let ρ ∈ DLU (C m ⊗ C n ). Then it must be of the form: ρ = (U ⊗ V )Λ(U ⊗ V )† for U ∈ U(m)/T( m)
and V ∈ U(n)/T( n) . Thus
dρ = (U ⊗ V ) (dΛ + [dG, Λ]) (U ⊗ V )† .
(2.10)
Here dG = (U ⊗ V )† d(U ⊗ V ) = dG1 ⊗ 1 n + 1 m ⊗ dG2 , where dG1 = U † dU and dG2 = V † dV. It suffices
to identify the volume element generated by [dG, Λ]. Clearly
hdρ, dρi = hdΛ + [dG, Λ] , dΛ + [dG, Λ]i
= hdΛ, dΛi + hdΛ, [dG, Λ]i + h[dG, Λ] , dΛi + h[dG, Λ] , [dG, Λ]i
= Tr dΛ2 + Tr [dG, Λ]† [dG, Λ] ,
where
Tr [dG, Λ]† [dG, Λ] = 2 Tr (ΛdGΛdG ) − 2 Tr Λ2 dG2 .
n
If Λ = ∑m
i =1 ∑ j =1 λij |ij ihij |, we then have
E
D Tr [dG, Λ]† [dG, Λ] = 2 ∑ ∑ λij λi′ j′ ij |dG | i ′ j′ i ′ j′ |dG | ij − 2 ∑ λ2ij ij dG2 ij
i,j i ′ ,j ′
E
D 2
= −2 ∑ ∑ λij λi′ j′ ij |dG | i ′ j′ − 2 ∑ λ2ij ij dG2 ij
i,j i ′ ,j ′
= 2∑∑
i,j i ′ ,j ′
λ2ij
i,j
i,j
ij |dG | i ′ j′ 2 − 2 ∑ ∑ λij λi′ j′ ij |dG | i ′ j′ 2
i,j i ′ ,j ′
2
= 2 ∑ ∑ λij (λij − λi′ j′ ) ij |dG | i ′ j′ .
i,j i ′ ,j ′
Now
ij |dG | i ′ j′ 2
Thus
=
=
i |dG1 | i ′ h j| j′ i + j |dG2 | j′ hi |i ′ i hi |dG1 | i ′ i h j| j′ i + h j |dG2 | j′ i hi |i ′ i
i |dG1 | i ′ 2 δjj′ + j |dG2 | j′ 2 δii′
+ i |dG1 | i ′ h j |dG2 | j′ iδii′ δjj′ + hi |dG1 | i ′ i j |dG2 | j′ δii′ δjj′ .
Tr [dG, Λ]† [dG, Λ]
2
= 2 ∑ ∑ λij (λij − λi′ j′ ) i |dG1 | i ′ δjj′
i,j i ′ ,j ′
2
+2 ∑ ∑ λij (λij − λi′ j′ ) j |dG2 | j′ δii′
i,j i ′ ,j ′
+2 ∑ ∑ λij (λij − λi′ j′ ) i |dG1 | i ′ h j |dG2 | j′ iδii′ δjj′
i,j i ′ ,j ′
+2 ∑ ∑ λij (λij − λi′ j′ )hi |dG1 | i ′ i j |dG2 | j′ δii′ δjj′ .
i,j i ′ ,j ′
4
That is,
Tr [dG, Λ]† [dG, Λ]
2
= 2 ∑ λij (λij − λi′ j ) i |dG1 | i ′ i ′ ,i,j
+2
∑′ λij (λij − λij′ ) i,j,j
2
2
2
j |dG2 | j′ .
Note that |hi |dG1 | i ′ i| = |hi ′ |dG1 | i i| (both vanishes if i = i ′ because dG1 is skew-Hermitian), we have
"
#
2
∑ ∑ λij (λij − λi′ j ) i |dG1 | i′ i′ 6=i
=
j
∑′
i<i
=
"
#
2
∑ λij (λij − λi′ j ) i |dG1 | i′ +
j
"
j
∑′ ∑(λij − λi′ j )2
i<i
j
#
Therefore, we get
hdρ, dρi = ∑ dλ2ij + 2
i,j
=∑
i,j
dλ2ij
+2 ∑
∑
′
i 6 = i,j
i>i
j
i |dG1 | i ′ 2 .
j
2
!
2
"
(m2 )+(n2 )
j
"
j< j
#
i |dG1 | i ′ 2
i,j 6 = j
∑′ ∑(λij − λij′ )
j< j
∑ ′ λij (λij − λij′ ) i
∏ 2 ∑(λij − λi′ j ) ∏′ 2 ∑(λij − λij′ )
i<i′
=2
j
i |dG1 | i ′ 2
i |dG1 | i ′ 2 + 2
The Hilbert-Schmidt volume element is given by
#
"
[dρ] =
#
2
λij (λij − λi′ j ) i |dG1 | i ′ + 2
∑(λij − λi′ j )
i<i′
"
∑′ ∑ λij (λij − λi′ j )
∑′ ∑ λij (λij − λi′ j ) + ∑ λi′ j (λi′ j − λij )
i<i
=
"
2
i
#
2
!
2
j |dG2 | j′ j |dG2 | j′ 2 .
[dΛ][dG1 ][dG2 ]

#
∏ ∑(λij − λi′ j )2  ∏ ∑(λij − λij′ )2  [dΛ][dG1 ][dG2 ].
i<i′ j
j< j′ i
By integrating out [dG1 ][dG2 ], we get the desired result.
Remark 2.2. We make some remarks here: the Lebesgue volume element [dG1 ] is defined on the flag
manifold U(m)/T( m) , where we endow the quotient space with the quotient measure. Similarly, [dG2 ] is
defined on the flag manifold U(n)/T( n). They can be described in terms of Haar measure defined over Lie
groups U(m) and U(n). Specifically,
[dG1 ] =
vol(U(m))
dµHaar (U ),
vol( T( m) )
[dG2 ] =
vol(U(n))
dµHaar (V ),
vol( T( n) )
U ∈ U( m ), V ∈ U( n ).
By such interpretation, we see that
m
[dρ] = vol(U(m))vol(U(n))
×
"
∏′ ∑(λij − λi′ j )2
i<i
j
n
2( 2 )+( 2 )
(2π )m+n
#

 ∏ ∑(λij − λij′ )2  [dΛ]dµHaar (U )dµHaar (V ).
j< j′ i
5
(2.11)
In most cases of this paper, we will focus on two-qubit situation since relevant analytical computations
to two-qubit case can be performed. Let us recall the notion of the Bloch sphere representation. In
quantum mechanics, the Bloch sphere is a geometrical representation of the pure state space of a two-level
quantum mechanical system (qubit). Any qubit state can be represented by Pauli matrices:
ρ=
1
(1 + r( ρ ) · σ ),
2
r( ρ ) ∈ R3 .
(2.12)
Here the vector r(ρ) is called the Bloch vector of ρ and is restricted to have k r(ρ)k :=
Here σ = (σx , σy , σz ) is the vector of Pauli’s matrices where
!
!
0 1
0 −i
σx =
, σy =
, σz =
1 0
i 0
1
0
0
−1
q
r2x + r2y + r2z 6 1.
!
(2.13)
with properties



σ σ

 x y
σy σz



σ σ
= −σy σx = iσz ,
= −σx σz = iσy .
z x
It also holds that communication relations



[σ , σ ]

 x y
[σy , σz ]



[σ , σ ]
z
and anti-communication relations
(2.14)
= −σz σy = iσx ,
x
= 2iσz ,
(2.15)
= 2iσx ,
= 2iσy ,
{σi , σj } = σi σj + σj σi = 0 (i 6= j).
(2.16)
Moreover σk2 = 1 for each k = x, y, z. For U ∈ SU(2), there corresponds O ∈ SO(3) such that
r(UρU †) = Or(ρ).
(2.17)
This means that the adjoint action of a unitary matrix U ∈ SU(2) on a qubit state ρ amounts to a rotation
of the corresponding Bloch vector of ρ. With this Bloch sphere representation of a qubit, we have that any
two-qubit state can be written as
ρ AB
1
=
4
1⊗1+r·σ⊗1+1⊗s·σ+
∑
i,j = x,y,z
tij σi ⊗ σj
!
.
(2.18)
From this, any two-qubit is given by the 3-dimensional real column vectors r, s ∈ R3 , and the real 3 × 3
matrix T = (tij ) ∈ R3×3 . Now both two-qubit states ρ AB and ρ′AB are locally unitary (LU) equivalent, i.e.,
ρ′AB = (U A ⊗ UB )ρ AB (U A ⊗ UB )† for some U A , UB ∈ SU(2), if and only if there are O A , OB ∈ SO(3) such
that
r′ = O A r,
s′ = OB s,
(2.19)
T ′ = O A TOTB .
(2.20)
6
Theorem 2.3 ([10]). Two generic two-qubit states are LU equivalent if and only if they have the same values for the
following 12 invariants: for k = 0, 1, 2,
D D k E
k E
s TT T s ,
r TT T r ,
D k E
k +1 r TT T T s , Tr TT T
.
(2.21)
(2.22)
In order to answer which two-qubits are locally diagonalizable, we need the notion of X-states: the
density matrix of a two-qubit X-state is of the general form:

ρ11

 0
ρ X := 
 0

ρ41
0
0
ρ22
ρ23
ρ32
ρ33
0
0
ρ14


0 
.
0 

ρ44
(2.23)
This expression describes a quantum state, ρ X ∈ D C2 ⊗ C2 , provided the unit trace and positivity
conditions:
(i) ∑4j=1 ρ jj = 1
(ii) ρ22 ρ33 > | ρ23 |2 and ρ11 ρ44 > | ρ14 |2
We denote DX (C2 ⊗ C2 ) the set of all two-qubit X-states. Apparently DX (C2 ⊗ C2 ) is the 7-dimensional
submanifold of D C2 ⊗ C2 . So is DLU (C2 ⊗ C2 ). Note that the Hilbert-Schmidt volume of the set of all
π2 1
two-qubit X-states is calculated in [14]: volHS DX (C2 ⊗ C2 ) = 630
.
Proposition 2.4. If a two-qubit state ρ AB is an X-state (2.23), then it can be written as
ρ AB
=
1
(1 ⊗ 1 + az σz ⊗ 1 + bz 1 ⊗ σz + r xx σx ⊗ σx
4
+ r xy σx ⊗ σy + ryx σy ⊗ σx + ryy σy ⊗ σy + rzz σz ⊗ σz ),
(2.24)
where


a


 z
bz



r
zz


r xx





r xy
= ρ11 − ρ22 + ρ33 − ρ44 ,
= ρ11 + ρ22 − ρ33 − ρ44 ,
and


ryx




r
yy
= ρ11 − ρ22 − ρ33 + ρ44 ,
= ρ14 + ρ23 + ρ32 + ρ41 ,
= i(ρ14 + ρ23 − ρ32 − ρ41 ),
= i(ρ14 − ρ23 + ρ32 − ρ41 ),
= −ρ14 + ρ23 + ρ32 − ρ41 .
Moreover all eigenvalues of such state are given by
q
1
2
2
2
λ1,2 (ρ X ) =
(1 + rzz ) ± ( az + bz ) + (r xx − ryy ) + (r xy + ryx ) ,
4
q
1
2
2
2
(1 − rzz ) ± ( az − bz ) + (r xx + ryy ) + (r xy − ryx ) .
λ3,4 (ρ X ) =
4
1 This
(2.25)
(2.26)
fraction is a correction to the result of Milz and Strunz [14]; they misused a factor leading to an incorrect Hilbert-Schmidt
volume of
π2
5040 .
But their calculation of the Euclid volume of X-states is correct.
7
By using this result, we have that, for Λ = diag(λ1 , λ2 , λ3 , λ4 ),
Λ=
1
(1 ⊗ 1 + az σz ⊗ 1 + bz 1 ⊗ σz + rzz σz ⊗ σz ),
4
(2.27)
where



a = λ1 − λ2 + λ3 − λ4

 z
bz = λ 1 + λ 2 − λ 3 − λ 4



r = λ − λ − λ + λ
zz
2
3
1
4
.
(2.28)
Then r(Λ) = az |3i, s(Λ) = bz |3i, and T (Λ) = rzz |3ih3|, where |3i := (0, 0, 1)T. If ρ is LU equivalent to the
diagonal form Λ, then we have the following statement.
Theorem 2.5. A generic two-qubit state ρ is locally diagonalizable if and only if the following conditions are satisfied:
r = a z O A |3 i ,
s = bz O B | 3 i ,
T = rzz O A |3ih3|OTB,
(2.29)
where O A , OB ∈ SO(3); and the triple ( az , bz , rzz ) is given by (2.28).
Theorem 2.6. The Hilbert-Schmidt volume of DLU (C2 ⊗ C2 ) is given by
(2π )2
volHS DLU (C2 ⊗ C2 ) =
.
105
(2.30)
ρ = (U ⊗ V ) Λ (U ⊗ V ) † ,
(2.31)
Proof. For any generic two-qubit density matrices ρ ∈ DLU (C2 ⊗ C2 ), it can be parameterized as follows:
where Λ = diag(λ1 , λ2 , λ3 , λ4 ) with λ j ’s being pairwise different and ∑ j λ j = 1. Thus
δ(1 − Tr (ρ))[dρ]
= 4δ 1 − ∑ λ j
j
!
(2.32)
h
ih
i
× (λ1 − λ2 )2 + (λ3 − λ4 )2 (λ1 − λ3 )2 + (λ2 − λ4 )2 [dΛ]
×
[vol(U(2))]2
dµHaar (U )dµHaar (V ).
(2π )4
(2.33)
(2.34)
Thus
Z
DLU (C2 ⊗C2 )
δ(1 − Tr (ρ))[dρ]
4[vol(U(2))]2
=
(2π )4
Z
δ 1 − ∑ λj
j
!
h
( λ1 − λ2 ) 2 + ( λ3 − λ4 ) 2
ih
i
(λ1 − λ3 )2 + (λ2 − λ4 )2 [dΛ].
Note that [17]
vol(U(k)) =
2k π (
k +1
2 )
∏kj=1 Γ( j)
8
.
(2.35)
Then vol(U(2)) = 4π 3 . Therefore
Z
volHS DLU (C2 ⊗ C2 ) =
δ(1 − Tr (ρ))[dρ]
DLU (C2 ⊗C2 )
!
Z
i 4
ih
h
= 4π 2 δ 1 − ∑ λ j (λ1 − λ2 )2 + (λ3 − λ4 )2 (λ1 − λ3 )2 + (λ2 − λ4 )2 ∏ dλ j .
(2.36)
(2.37)
j =1
j
Next we will calculate the above last integral. Clearly we have
i
ih
h
( λ1 − λ2 ) 2 + ( λ3 − λ4 ) 2 ( λ1 − λ3 ) 2 + ( λ2 − λ4 ) 2
= ( λ1 − λ2 ) 2 ( λ1 − λ3 ) 2 + ( λ3 − λ1 ) 2 ( λ3 − λ4 ) 2 + ( λ2 − λ1 ) 2 ( λ2 − λ4 ) 2 + ( λ4 − λ2 ) 2 ( λ4 − λ3 ) 2 .
Then the following four integrals are equal:
!
Z
δ 1 − ∑ λj
j
!
Z
δ 1 − ∑ λj
Z
δ 1 − ∑ λj
Z
δ 1 − ∑ λj
j
j
!
j
4
(λ1 − λ2 )2 (λ1 − λ3 )2 ∏ dλ j ,
j =1
4
(λ3 − λ1 )2 (λ3 − λ4 )2 ∏ dλ j ,
j =1
4
(λ2 − λ1 )2 (λ2 − λ4 )2 ∏ dλ j ,
!
j =1
Denote
f (t) =
Z
δ t − ∑ λj
j
4
(λ4 − λ2 )2 (λ4 − λ3 )2 ∏ dλ j .
j =1
!
4
(λ1 − λ2 )2 (λ1 − λ3 )2 ∏ dλ j .
(2.38)
j =1
Performing Laplace transform (t → s) to f (t), we get fe(s)(s > 0):
fe(s)
=
=
Z ∞
Z ∞Z ∞Z ∞Z ∞
f (t)e−st dt
"Z
Z ∞Z ∞Z ∞Z ∞
exp −s ∑ λ j
= L ( f )(s) =
0
0
0
0
0
0
0
0
0
∞
0
dte−st δ
j
t − ∑ λj
!
j
!#
(2.39)
4
(λ1 − λ2 )2 (λ1 − λ3 )2 ∏ dλ j .
(2.40)
j =1
4
(λ1 − λ2 )2 (λ1 − λ3 )2 ∏ dλ j .
(2.41)
j =1
That is,
fe(s)
= s −8
= s
−8
Z ∞Z ∞Z ∞Z ∞
0
0
0
Z ∞Z ∞Z ∞
0
0
0
0
4
exp − ∑ x j
j =1
3
exp − ∑ x j
j =1
!
!
4
( x1 − x2 )2 ( x1 − x3 )2 ∏ dx j
3
( x1 − x2 )2 ( x1 − x3 )2 ∏ dx j .
j =1
Since
( x1 − x2 )2 ( x1 − x3 )2 = x14 − 2( x2 + x3 ) x13 + ( x22 + x32 + 4x2 x3 ) x12 − 2x2 x3 ( x2 + x3 ) x1 + x22 x32
9
(2.42)
j =1
(2.43)
and
Γ( n) =
Z ∞
0
x n−1 e− x dx,
n ∈ N,
it follows that
Z ∞
0
dx1 e− x1 ( x1 − x2 )2 ( x1 − x3 )2
= Γ(5) − 2( x2 + x3 )Γ(4) + ( x22 + x32 + 4x2 x3 )Γ(3) − 2x2 x3 ( x2 + x3 )Γ(2) + x22 x32 .
Hence
Z ∞Z ∞Z ∞
0
0
0
dx1 dx2 dx3 e− x1 e− x2 e− x3 ( x1 − x2 )2 ( x1 − x3 )2
= Γ(5) − 2[Γ(2) + Γ(2)]Γ(4) + [Γ(3) + Γ(3) + 4Γ(2)Γ(2)]Γ(3)
− 2[Γ(3)Γ(2) + Γ(2)Γ(3)]Γ(2) + Γ(3)Γ(3)
= 12.
Therefore,
fe(s) = 12 · s−8 .
Based on this fact, we get that
(2.44)
t7
1 7
f (t) = L −1 ( fe)(t) = 12 ·
=
t .
7!
420
(2.45)
Finally, we can draw our conclusion as follows:
This completes the proof.
(2π )2
volHS DLU (C2 ⊗ C2 ) = 4π 2 · 4 f (1) =
.
105
(2.46)
Remark 2.7. By the reasoning method used in the proof of Theorem 2.6, we can generalize the volume
formula for two-qubit case to arbitrary bipartite state case. Specifically, we have
volHS (DLU (C m ⊗ C n )) =
m
n
(2π )( 2 )+(2 )
= m
∏ i=1 Γ(i ) ∏nj=1 Γ( j)
×
∏′
n
Z
Z
DLU (C m ⊗C n )
∑ (λij − λi′ j )2
16 i < i 6 m j = 1
m
δ 1− ∑
n

(2.47)
∑ λij
i =1 j =1
!
δ(1 − Tr (ρ))[dρ]
!
m
(2.48)

m
n
∏′ ∑ (λij − λij′ )2  ∏ ∏ dλij.
16 j < j 6 n i = 1
(2.49)
i =1 j =1
Theoretically, we can write down the volume formula. But, however, such expression for this volume
formula will be rather complicated.
10
3 Harish-Chandra’s volume formula
Let U(m) be the unitary group acting on C m with Lie algebra u(m). Denote by T( m) the maximal torus of
√
∼
as the set of diagonal
U(m), and t
= −1R m is its Lie algebra. Without loss of generality, we take t
( m)
( m)
matrices with purely imaginary diagonal entries. Let K = U(m) ⊗ U(n). Then its Lie algebra k may be
described as
k = u( m ) ⊗ 1 n + 1 m ⊗ u( n ) .
(3.1)
Moreover, the Lie algebra t of the maximal torus T = T( m) ⊗ T( n) of K may also described as, in terms of
t( m) and t( n) ,
t = t( m ) ⊗ 1 n + 1 m ⊗ t( n ) .
(3.2)
It is a routine exercise to see that dim(t) = dim(t( m) ) + dim(t( n) ) − 1 = m + n − 1. We shall see that
(Proposition 3.4) dim(k) = dim(u(m)) + dim(u(n)) − 1 = m2 + n2 − 1. We also have
[U(m) ⊗ U(n)]/[ T(m) ⊗ T(n) ] ≃ [U(m)/T(m)] ⊗ [U(n)/T(n) ].
(3.3)
It goes without saying that, for volumes of quotient spaces, we shall use quotient measures.
Proposition 3.1 (Harish-Chandra’s volume formula, [3, 6]). Let K be a compact, connected Lie group. Let T be
the maximal torus of K. The Riemannian volume of the flag manifold K/T is given by
2π
∏+ hα, ̟ i ,
vol(K/T ) =
(3.4)
α∈ Φk
where Φk+ is the set of all positive roots for k and ̟ :=
1
2
positive roots of k.
∑α∈Φ+ α, which is Weyl vector, i.e., a half the sum of all
k
Remark 3.2. From Harish-Chandra’s volume formula, one may calculate the volume vol(K ) for a compact
connected Lie group K whenever one finds the volume vol( T ) of the maximal torus T of K and all positive
roots for k, assuming that K/T is endowed with the quotient measure. Indeed, we have
vol(K ) = vol( T )
∏
α∈ Φk+
2π
.
hα, ̟ i
(3.5)
Example 3.3. In the present example, we want to calculate the volume of the flag manifold U(n)/T( n)
where T( n) ∼
= U(1)×n is the maximal torus of unitary group U(n). It suffices to find all positive roots for
u(n). The set of all positive roots of u(n) is given by
n
o
Φu+( n) = αij ∈ t∗( n) : αij ( X ) = xi − x j for any X = diag( x1 , . . . , x n ), i < j .
We can view a diagonal matrix X as a real vector X = ( x1 , . . . , x n ). In turn, we may view αij as the real
vector
i
j
z}|{
z}|{
αij = (· · · 1 · · · −1 · · · )
11
(1 6 i < j 6 n),
where · stands for zeroes, so that we have αij ( X ) = αij , X . Then
̟=
1
1
αij = (n − 1, n − 3, . . . , 3 − n, 1 − n).
∑
2 i< j
2
1
−
j
ej.
Let {e j : j = 1, . . . , n} be the standard orthonormal basis for R n . Then αij = ei − e j and ̟ = ∑ j n+
2
1
n +1
So αij , ̟ = ̟i − ̟ j = n+
2 −i −
2 − j = j − i. We also have α ij , λ = λi − λ j . Then, by Eq. (3.4),
n
vol U(n)/T( n) = ∏
i< j
Since
2π
(2π )(2 )
2π
=∏
= n
.
j−i
αij , ̟
∏ j =1 Γ ( j )
i< j
vol( T( n) ) = vol(U(1))n = (2π )n ,
(3.6)
(3.7)
we have
n ( n +1 )
(2π ) 2
volHS (U(n)) = n
.
∏ j =1 Γ ( j )
(3.8)
Concerning the volume of unitary groups and orthogonal groups, we have given a comprehensive treatment in more elementary approach in [17].
Let Φu( m) denote the set of roots of a Lie algebra u(m). For each α( m) ∈ Φu( m) , denote its associated
root space by u(m)α(m) . Let τm (·) :=
Tr (·) be the normalized trace form on the m × m matrices so that
∈ Φu(m) , the tensor product α(m) ⊗ τn is a purely imaginary-valued R-linear
map t( m) ⊗ 1 n → C. Extend, by zero, the domain of α( m) ⊗ τn to t = t( m) ⊗ 1 n + 1 m ⊗ t( n) (recall that
√
the intersection (t( m) ⊗ 1 n ) ∩ (1 m ⊗ t( n) ) = −1R · 1 m ⊗ 1 n , and on such intersection, α( m) ⊗ τn yields the
value 0 because α( m) depends only on the differences in the diagonal entries). We denote this extension by
τm (1 m ) = 1. For any
α( m)
1
m
α̃( m) = α( m) ⊗ τn . Symmetrically, for each root α( n) of u(n), we denote by α̃( n) the purely imaginary-valued
R-linear map t → C obtained by extending τm ⊗ α( n) by zero.
o
o
n
n
e u( n) = α̃( n) | α( n) ∈ Φu( n) . Their disjoint
e u( m) = α̃( m) | α( m) ∈ Φu( m) and Φ
Proposition 3.4. Denote Φ
union yields the set of roots for k = u(m) ⊗ 1 n + 1 m ⊗ u(n):
e u( m )
Φk = Φ
G
e u( n ) .
Φ
(3.9)
e u( m) is ũ(m) := u(m) ⊗ 1 n . The root space associated with α̃( n) ∈ Φ
e u( n) is
The root space associated with α̃( m) ∈ Φ
ũ(n) := 1 m ⊗ u(n).
Proof. Take the root space decompositions
u( m ) C
u( n ) C
= t(m),C ⊕
m
M
u( m )
= t(n),C ⊕
n
M
u( n )
12
i6= j
k6=l
(m)
αij ,C
(n)
αkl ,C
.
,
(3.10)
(3.11)
Here, u( p)C (p = m or n) denotes the complexification of u( p). Then,
kC = u(m)C ⊗ 1 n + 1 m ⊗ u(n)C

m
M
= t(m),C ⊗ 1 n + 1 m ⊗ t(n),C +  u(m)
We need to show that
Lm
i 6 = j u( m ) α ( m )
ij
⊗ 1 n ’s and
vector Z ∈ t( m),C ⊗ 1 n + 1 m ⊗ t( n),C ; then
Ln
i6= j
k6=l
(m)
αij
1 m ⊗ u( n )


⊗ 1n  + 
( n ) ’s
αkl
n
M
k6=l
1 m ⊗ u( n )
(n)
αkl
(3.12)

.
(3.13)
are root spaces for k. Take an arbitrary
Z = X ⊗ 1n + 1m ⊗ Y
for some X ∈ u(m)C and Y ∈ u(n)C . Observe that
[ Z, u(m)
So u(m)
(m)
αij
(m)
αij
⊗ 1 n ] = [ X, u(m)
(m)
αij
( m)
] ⊗ 1 n = αij ( X )(u(m)
(m)
αij
⊗ 1 n ).
(3.14)
⊗ 1 n is indeed a root space for k. Similar argument proves that each 1 m ⊗ u(n)
(n)
αkl
is a root
space for k.
e u( m ) ∪ Φ
e u( n) . To show that this is a disjoint union, we claim
We have demonstrated so far that Φk = Φ
that, for any pair (α( m) , α( n) ) ∈ Φu( m) × Φu( n) , the two root spaces u(m)α(m) ⊗ 1 n and 1 m ⊗ u(n)α(n) are
distinct. Since root spaces are 1-dimensional, our claim is equivalent to saying that
u ( m ) α ( m ) ⊗ 1 n ∩ 1 m ⊗ u ( n ) α ( n ) = {0 }.
(3.15)
The argument goes as follows: Say X ∈ u(m)α(m) and Y ∈ u(n)α(n) . As root vectors, the nonzero entries of
X and Y are all off-diagonal. Now suppose X ⊗ 1 n = 1 m ⊗ Y. Because all diagonal entries of both X and
Y are zero, then it is easy to see that both X and Y are zero matrices. This completes the proof.
With the above Proposition 3.4, one may calculate the volume vol(K ) theoretically for K = U(m) ⊗ U(n)
using Harish-Chandra’s volume formula:
vol(U(m) ⊗ U(n)) = vol( T( m) ⊗ T( n) )
= (2π )m+n−1
∏
α(m ) ∈ Φu+(m )
D
2π
∏
α∈ Φk+
2π
hα, ̟ i
e u( m ) + ̟
e u( n )
α̃( m), ̟
E
∏
α(n ) ∈ Φu+(n )
(3.16)
D
2π
e u( m ) + ̟
e u( n )
α̃( n) , ̟
where Φk+ is the set of all positive roots for k = u(m) ⊗ 1 n + 1 m ⊗ u(n). Note that
E
vol( T( m) ⊗ T( n) ) = (2π )m+n−1
(3.17)
(3.18)
since dim(t( m) ⊗ 1 n + 1 m ⊗ t( n) ) = m + n − 1. Now
E
E D
E D
D
( m)
e u( m ) + ̟
e u( n) = α(ijm) ⊗ τn , ̟u( m) ⊗ τn + α(ijm) ⊗ τn , τm ⊗ ̟u( n)
α̃ij , ̟
E
E
ED
D
D
( m)
( m)
= αij , ̟u(m) hτn , τn i + αij , τm τn , ̟u(n) ,
that is,
D
E
j−i
( m)
e u( m ) + ̟
e u( n ) =
α̃ij , ̟
.
n
13
(3.19)
Similarly, we have
D
Furthermore,
∏
α(m ) ∈ Φu+(m )
D
E
l−k
(n)
e u( m ) + ̟
e u( n ) =
α̃kl , ̟
.
m
2π
e
α̃( m) , ̟
u( m )
e u( n )
+̟
E
∏
=
16 i < j 6 m
∏
=
16 i < j 6 m
D
(3.20)
2π
( m)
e u( m ) + ̟
e u( n )
α̃ij , ̟
E
m
2nπ
= n( 2 ) vol(U(m)/T(m) )
j−i
(3.21)
(3.22)
and
∏
α(n ) ∈ Φu+(n )
D
2π
e u( m ) + ̟
e u( n )
α̃( n) , ̟
E
∏
=
16 k < l 6 n
=
D
2π
(n)
e u( m ) + ̟
e u( n )
α̃kl , ̟
E
n
2mπ
= m(2 ) vol(U(n)/T(n) ).
l−k
16 k < l 6 n
∏
(3.23)
(3.24)
Therefore we can draw the following conclusion:
Theorem 3.5. For m, n > 2, it holds that
vol U(m) ⊗ U(n)/T( m) ⊗ T( n)
n
m
= m(2 ) n( 2 ) vol U(m)/T(m) vol U(n)/T(n)
n
m
= m ( 2 ) n( 2 )
(2π )
(m2 )+(n2 )
.
n
∏m
i =1 Γ ( i ) ∏ j =1 Γ ( j )
(3.25)
(3.26)
As a consequence, with
vol( T( m) ⊗ T( n) ) = (2π )m+n−1,
(3.27)
we have
m +1
n
m
vol (U(m) ⊗ U(n)) = m(2 ) n( 2 )
n +1
(2π )( 2 )+( 2 )−1
.
n
∏m
i =1 Γ ( i ) ∏ j =1 Γ ( j )
(3.28)
In particular, for (m, n) = (2, 2), we have
vol (U(2) ⊗ U(2)) = 128π 5 .
(3.29)
Proposition 3.6 ([3]). Let K be a compact, connected Lie group of which T is a maximal torus. Let λ be a regular
√
point in −1t∗>0 . OK,λ is a coadjoint orbit through λ. Then
vol(OK,λ ) =
∏
α∈ Φk+
Here ̟ =
1
2
hλ, αi
.
h̟, αi
(3.30)
∑α∈Φ+ α.
k
Example 3.7. Consider the unitary group U(n) and λ ∈ R n with λ1 > · · · > λn . The Hilbert-Schmidt
inner product gives the identification of iu(n)∗ and iu(n). Adjoint and coadjoint orbits of compact Lie
14
groups U(n) can be identified, using any Ad(U(n))-invariant inner product on the Lie algebra u(n). In
√
the following, we calculate the volume of an adjoint orbit Ad(U(n))Λ for Λ ∈ −1t where t is the
maximal torus of diagonal matrices from iu(n), i.e.,
OΛ = Ad(U(n))Λ = {UΛU ∗ : Λ = diag(λ1 , . . . , λn ) with λ ∈ R n and λ1 > · · · > λn } .
Continuing with the description of the positive roots of u(n) given in Ex. 3.3, we have, by Eq. (3.30),
αij , λ
λi − λ j
∏ i < j ( λi − λ j )
=∏
vol (Ad(U(n))Λ) = ∏ =
.
(3.31)
j−i
1! · · · (n − 1)!
i < j α ij , ̟
i< j
That is,
vol(OΛ ) = vol (Ad(U(n))Λ) =
∏i < j ( λi − λ j )
| ∆(λ)|
= n
,
∏nj=1 Γ( j)
∏ j =1 Γ ( j )
(3.32)
where ∆(λ) = ∏i< j (λ j − λi ).
Theorem 3.8. Let K = U(m) ⊗ U(n). Denote its Lie algebra k by
k = u( m ) ⊗ 1 n + 1 m ⊗ u( n ) .
(3.33)
Moreover, the Lie algebra t of the maximal torus T = T( m) ⊗ T( n) of K is denoted by
t = t( m ) ⊗ 1 n + 1 m ⊗ t( n ) .
(3.34)
Denote Λ = diag(λ11 , . . . , λm1 , . . . , λ1n , . . . , λmn ) where λ11 > . . . > λmn . It is a regular point in a maximal
√
√
√
torus algebra −1t( mn) of −1u(mn). Now let λ ∈ −1t>0 be a regular point derived from Λ, i.e., λ =
Trn (Λ) ⊗ 1 n + 1 m ⊗ Trm (Λ), then it holds that
i h
i
h
λ
−
λ
(
)
∏ 16 i < j 6 n ∑ m
∏16i< j6m ∑nk=1 λik − λ jk
ik
il
i =1
.
vol (OK,λ ) =
n
Γ
(
i
)
Γ
(
k
)
∏m
∏
i =1
k =1
Proof. In fact, using the obtained positive roots for k, we have
D
E
α̃( m) , λ
E
vol (OK,λ ) =
∏ D ( m)
∏
e u( m ) + ̟
e u( n ) α ( n ) ∈ Φ +
α̃ , ̟
α(m ) ∈ Φ +
u( m )
u( n )
That is,
vol (OK,λ ) =
∏
16 i < j 6 m
Since
and
D
D
D
( m)
D
( m)
α̃ij , λ
E
e u( m ) + ̟
e u( n )
α̃ij , ̟
E
D
D
α̃( n), λ
E
e u( m ) + ̟
e u( n )
α̃( n) , ̟
∏
16 i < j 6 n
D
(n)
D
(n)
α̃kl , λ
E
(3.35)
E.
e u( m ) + ̟
e u( n )
α̃kl , ̟
(3.36)
E.
(3.37)
E D
E
1
1
( m)
( m)
α̃ij , λ = αij ⊗ τn , λ = Tr (( Eii − Ejj ) ⊗ 1 n )Λ = Tr ( Eii − Ejj ) Trn (Λ)
n
n
(3.38)
E D
E
1
1
(n)
(n)
α̃kl , λ = τm ⊗ αkl , λ =
Tr ((1 m ⊗ ( Ekk − Ell ))Λ) =
Tr (( Ekk − Ell ) Trm (Λ))
m
m
(3.39)
15
it follows that
D
and
E
1
( m)
α̃ij , λ =
n
D
Therefore
vol (OK,λ )
E
1
(n)
α̃kl , λ =
m
∏
=
16 i < j 6 m
∏
=
16 i < j 6 m
1
n
(3.40)
∑ (λik − λil ) .
(3.41)
n
∑
k =1
λik − λ jk
m
i =1
∑nk=1 λik − λ jk
1
n ( j − i)
∑nk=1 λik − λ jk
j−i
∏
16 i < j 6 n
1
m
∑m
i =1 ( λik − λil )
1
m (l − k)
∑m
i =1 ( λik − λil )
.
l−k
16 i < j 6 n
∏
(3.42)
(3.43)
The proof is done.
Essentially, Theorem 3.8 give a formula for the volume of local unitary orbit of product form, i.e.,
√
√
vol Oµ⊗ν = vol Oµ vol (Oν ), where µ ∈ −1t( m) and ν ∈ −1t( n) .
Corollary 3.9. Assume that the local unitary orbit, of a generic probability distribution Λ = diag(λ1 , λ2 , λ3 , λ4 )
where λ1 > λ2 > λ3 > λ4 > 0 with ∑4j=1 λ j = 1, is denoted by
n
o
LU
OΛ
:= (U ⊗ V )Λ(U −1 ⊗ V −1 ) : U, V ∈ U(2) .
Then the Riemannian volume of such local unitary orbit is given by
vol OλLU = (λ1 + λ2 − λ3 − λ4 )(λ1 + λ3 − λ2 − λ4 ).
(3.44)
Remark 3.10. With the notations in Corollary 3.9, denote
n
o
GU
OΛ
:= WΛW −1 : W ∈ U(4) .
(3.45)
1
GU
vol OΛ
=
( λi − λ j ).
12 16∏
i < j64
(3.46)
Proof. It is a direct consequence of Theorem 3.8.
From Eq. (3.32), we have
Meanwhile, with the restriction λ1 > λ2 > λ3 > λ4 > 0 with ∑4j=1 λ j = 1, we have
vol OλLU = [(λ1 − λ3 ) + (λ2 − λ4 )][(λ1 − λ2 ) + (λ3 − λ4 )]
q
> 4 (λ1 − λ3 )(λ2 − λ4 )(λ1 − λ2 )(λ3 − λ4 )
> 4(λ1 − λ3 )(λ2 − λ4 )(λ1 − λ2 )(λ3 − λ4 )
GU
GU
> 4 ∏ (λi − λ j ) > 48vol OΛ
> vol OΛ
.
16i < j64
So we see that vol
GU
OΛ
LU is a submanifold of O GU since U(2) ⊗ U(2)
< vol OλLU , and apparently OΛ
Λ
LU and O GU have no a
is a Lie subgroup of U(4). This is not a contradiction because the measures for OΛ
Λ
LU is a set of zero-measure in O GU since
priori relation, so one cannot directly compare the two. In fact, OΛ
Λ
LU ) < dim(O GU ).
dim(OΛ
Λ
16
4 Discussion and concluding remarks
We know that there may exist many Λ’s corresponds to single one λ (see Theorem 3.8). Indeed, a relevant
famous problem is the so-called quantum marginal problem; and for the two-qubit system there is a nice
solution for it [4]. Specifically, mixed two-qubit state ρ AB with spectrum λ1 > λ2 > λ3 > λ4 > 0 and
margins ρ A and ρ B exists if and only if minimal eigenvalues λ A , λ B of the margins satisfy inequalities



min(λ A , λ B ) > λ3 + λ4 ,

(4.1)
λ A + λ B > λ2 + 2λ3 + λ4 ,



| λ − λ | 6 min(λ − λ , λ − λ ).
1
3 2
4
A
B
Here we present a specific example showing this property. Let ρ AB be any two-qudit in D C d ⊗ C d .
Then there exists a global unitary V ∈ U(d2 ) such that ρ′AB = Vρ AB V † with two marginal states as
ρ′A = ρ′B = 1 d /d. Indeed, by the Spectral Decomposition Theorem, we have the following decomposition:
denoting [k] := {1, . . . , k} for any positive integer k,
d2
ρ AB =
|Ψ j i ∈ C d ⊗ C d ,
∑ λ j |Ψ j ihΨ j |,
j =1
(4.2)
where λ j > 0 for each j ∈ [d2 ] and |Ψ j i : j ∈ [d2 ] are the eigenvectors corresponding to eigenvalues λ j .
There exists a collection of unitary matrices, called discrete Weyl unitary matrices, Wj ∈ U(d)( j = 1, . . . , d2 ),
that form a unitary matrix basis for Md (C ), the set of all d × d complex matrices. Now vec(Wj ) : j ∈ [d2 ]
is a maximally entangled basis for C n ⊗ C n (see also for its generalization in [7]). Note that the notation
vec( M ) is the vectorization of a complex rectangular matrix M, which is defined by vec( M ) = ∑i,j Mij |iji,
where M = ∑i,j Mij |i ih j|. Thus exists a global unitary matrix V ∈ U(d2 ) such that
n
since |Ψ j i : j ∈ [d2 ] and √1
d
This means that
ρ′AB
1
(4.3)
V |Ψ j i = √ vec(Wj ) , j = 1, . . . , d2
d
o
vec(Wj ) : j ∈ [d2 ] are two orthonormal basis for the same space C d ⊗ C d .
d2
d2
1
= Vρ AB V = ∑ λ j V |Ψ j ihΨ j |V =
d
j =1
†
†
∑ λ j vec(Wj ) vec(Wj )† .
(4.4)
j =1
Thus
ρ′A
1
=
d
d2
∑
j =1
λ j Wj Wj†
1
= d
d
d2
∑ λj =
j =1
1
1d
Tr (ρ AB ) = d
d
d
(4.5)
and
ρ′B
1
=
d
d2
∑
j =1
λ j (Wj† Wj )T
1
= d
d
d2
∑ λj =
j =1
1
1d
Tr (ρ AB ) = d .
d
d
(4.6)
Therefore ρ′A = ρ′B = 1 d /d. This example also indicates that the maximum of mutual information along a
global unitary orbit of a given bipartite state with the prescribed spectrum Λ is 2 ln(d) − S(Λ) [8, 9], where
17
S(Λ) is the von Neumann entropy. Let ρ A and ρ B be fixed. Denote by C(ρ A , ρ B ) the set of all bipartite
states ρ AB with fixed marginal states ρ A and ρ B , respectively. It is known that C(ρ A , ρ B ) is a compact
convex set. Moreover, a necessary and sufficient criterion is presented by Pathasarathy [15] for an element
ρ AB in C(ρ A , ρ B ) to be an extreme point. In the two-qubit case, the extreme points of C(1/2, 1/2) are
characterized as well, namely, a two-qubit state ρ AB ∈ C(1/2, 1/2) is extreme if and only if it is maximally
entangled state.
There are several open questions which are presented below:
1. Assume that both generic bipartite states ρ and ̺ are not LU equivalent, where ρ is locally diagonalizable and ̺ is not. As we see previously, the volume of OρLU can be calculated, the question is how to
calculate the volume of O̺LU ?
2. It is easily seen that DLU (C m ⊗ C n ) can be partitioned as local unitary orbits. That is,
DLU (C m ⊗ C n ) =
[
Λ ∈ ∆ mn −1
LU
OΛ
.
(4.7)
LU ) of local unitary orbits
How do we get the volume vol (DLU (C m ⊗ C n )) from the volumes vol(OΛ
LU ? This question is motivated by the fact that the whole set of the state space D (C n ) can be
OΛ
partitioned as unitary orbits:
D (C n ) =
[
Λ ∈ ∆ n −1
OΛ .
(4.8)
We can get the volume vol(D (C n )) through the volumes vol(OΛ ) in terms of Duistermaat-Heckman
measure [5].
3. Can we establish some kind of "canonical form" for any bipartite state just like establishing Singular
Value Decomposition (SVD) for any complex matrix and/or Spectral Decomposition for any normal
matrix, especially for any two-qubit states? And how?
4. Because C(ρ A , ρ B ) is a compact convex set, its volume can be calculated in theoretical. To the best of
our knowledge, analytical expression for its volume is never found out. But, recently, we find that a
formula is obtained for C(1/2, 1/2). That is, vol(C(1/2, 1/2)) =
2π 4
315
[11]. Here we identify C(1/2, 1/2)
with the set of all unital qubit quantum channels via Choi-Jamiłkowski isomorphism. The problem
about the calculation of the volume of C(1/2, 1/2) is previously raised in [17].
In summary, we make an analysis of the geometry of locally diagonalizable states, and calculate its
Hilbert-Schmidt volume. Theoretically, although we work out analytical formula for it, a specific value
is just presented in the two-qubit situation. In addition, we have obtained the necessary and sufficient
condition for two-qubit state being LU equivalent to a given diagonal form. But, however, the problem in
a high dimensional space is still open. After introducing Harish-Chandra’s volume formula for co-adjoint
orbit, we turn to the geometry of local unitary orbits. We find that Harish-Chandra’s volume formula can
be applied to calculate the volume of local unitary orbits. As by-product, we get the volume formula of
tensor product U(m) ⊗ U(n) as a compact Lie subgroup of the compact Lie group U(mn). Although this is
a direct consequence of Harish-Chandra’s volume formula, it is never given explicitly before. We believe
these results obtained and questions raised in the paper can shed new lights and spur relevant researches
in quantum information theory.
18
Acknowledgements
LZ is supported by Natural Science Foundation of Zhejiang Province of China (LY17A010027) and by
National Natural Science Foundation of China (No.11301124&No.61673145). SH was partially supported
under Northwestern Scholarship Grant.
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