How to combine three quantum states - arXiv
How to combine three quantum states Maris Ozols
arXiv:1508.00860v1 [quant-ph] 4 Aug 2015
Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge, CB3 0WA, U.K. August 4, 2015 Abstract We devise a ternary operation for combining three quantum states: it consists of permuting the input systems in a continuous fashion and then discarding all but one of them. This generalizes a binary operation recently studied by Audenaert et al. [arXiv:1503.04213] in the context of entropy power inequalities. In fact, our ternary operation continuously interpolates between all such nested binary operations. Our construction is based on a unitary version of Cayley’s theorem: we use representation theory to show that any finite group can be naturally embedded in a continuous subgroup of the unitary group. Formally, this amounts to characterizing when a linear combination of certain permutations is unitary.
1
Introduction
A basic result in group theory known as Cayley’s theorem states that every finite group G is isomorphic to a subgroup of the symmetric group. In other words, the elements of G can be faithfully represented by permutation matrices of size | G | × | G |. This gives a natural embedding of G in the symmetric group Sn on n = | G | elements, also known as the regular representation of G. Since permutation matrices are unitary, one might ask whether the resulting subgroup of S|G| can be further extended to a continuous subgroup of the unitary group U(| G |). Such extension would allow to treat the otherwise discrete group G as continuous (e.g., it would allow to perturb the elements of G and continuously interpolate between them in a meaningful sense). To illustrate this, consider the simplest non-trivial example, namely, the finite group Z2 . This group has two elements which according to Cayley’s theorem can be represented by 2 × 2 permuta 1 0 tion matrices as I := 0 1 and X := 01 10 . Interestingly, the following complex linear combination cos α i sin α iϕ iϕ U ( ϕ, α) := e cos α I + i sin α X = e (1) i sin α cos α of I and X is unitary for any ϕ, α ∈ [0, 2π ). Note that U (0, 0) = I and U (3π/2, π/2) = X, so by changing ϕ and α we can continuously interpolate between the two original matrices representing Z2 . In fact, U ( ϕ, α)U ( ϕ0 , α0 ) = U ( ϕ + ϕ0 , α + α0 ), so the matrices U ( ϕ, α) themselves form a group—a two-parameter subgroup of U(2). One of our contributions is to generalize this idea to any finite group G. Namely, we use representation theory to characterize complex linear combinations of matrices from the regular representation of G that are unitary (see Theorem 7). As a consequence, any finite group G can be naturally embedded in a continuous subgroup of the unitary group U(| G |). We also show that 1
this subgroup is isomorphic to a direct sum of smaller unitary groups—one copy of U(dτ ) for each irreducible representation τ of G where dτ is the dimension of the representation τ.
1.1
The partial swap operation
Ability to make discrete groups continuous has an interesting application to quantum information. If we apply our construction to the symmetric group Sn , we get a unitary extension of Sn that continuously interpolates between different permutations. This then can be further extended to permuting n quantum systems in a continuous fashion. To illustrate this, let us focus again on the simplest non-trivial instance, S2 . This group can act on two d-dimensional quantum systems (qudits) either by leaving them alone or by swapping them. In other words, we can represent the elements of S2 by d2 × d2 permutation matrices I and S, where I is the identity matrix and S|i i| ji := | ji|i i for all i, j ∈ {1, . . . , d}. Making this representation continuous would allow to interpolate between these two operations and thus swap two quantum systems in a continuous fashion. This was exactly the starting point of [ADO15]. Following [ADO15], we define the partial swap operation Uλ ∈ U(d2 ) for λ ∈ [0, 1] as the following linear combination of the identity I and the two qudit swap S: √ √ Uλ := λ I + i 1 − λ S (2) This can easily be verified to be unitary. In fact, Uλ is very similar to U ( ϕ, α) in eq. (1), except we ignore the global phase eiϕ and take only one sector of the unit circle corresponding to α ∈ [0, π/2]. Note also that U1 = iS rather than S; however, this global phase mismatch will be unimportant. Given two qudit states ρ1 , ρ2 ∈ D(d), we can combine them using the partial swap as follows: h i ρ1 λ ρ2 := Tr2 Uλ (ρ1 ⊗ ρ2 )Uλ† (3) where λ ∈ [0, 1]. This operation (in the present context) was introduced in [ADO15], however it has independently appeared earlier in [LMR14]. With some tricks (see Section 2.3), eq. (3) can be expanded as q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 +
λ (1 − λ ) i [ ρ2 , ρ1 ].
(4)
Note that ρ1 0 ρ2 = ρ2 and ρ1 1 ρ2 = ρ1 . Also, for commuting states, ρ1 λ ρ2 = λρ1 + (1 − λ)ρ2 is just a convex combination of the two states. This operation obeys some interesting properties—most notably, an analogue of the entropy power inequality [ADO15]. Infull generality, consider a function f : D(d) → R. We say that f is concave if f λρ + (1 − λ)σ ≥ λ f (ρ) + (1 − λ) f (σ) for any λ ∈ [0, 1] and ρ, σ ∈ D(d), and symmetric if f (ρ) depends only on the eigenvalues of ρ and is symmetric in them. Theorem 1 ([ADO15]). Let ρ, σ ∈ D(d), λ ∈ [0, 1], and ρ λ σ be as in eq. (4). If f is concave and symmetric then f ( ρ λ σ ) ≥ λ f ( ρ ) + (1 − λ ) f ( σ ). (5) Motivated by this result, our goal is to generalize the partial swap operation to any number of systems. The rest of this paper is devoted to obtaining such generalization and using it to derive an analogue of eq. (4) for three states. Proving an analogue of Theorem 1 for this generalization is left as an open problem.
2
1.2
A nested generalization
Before we proceed with our general approach, it is instructive first to consider a simple method of combining states in an iterative manner. This will highlight some of the issues that appear when dealing with three or more states, hence providing further motivation for our approach. This will also serve as a reference point to come back to once we know the answer. A straightforward way of generalizing eq. (4) to more systems is by feeding the output state into another operation of the same kind. For example, for any a, a0 ∈ [0, 1], q ρ1 a (ρ2 a0 ρ3 ) = aρ1 + (1 − a)(ρ2 a0 ρ3 ) + a(1 − a) i [ρ2 a0 ρ3 , ρ1 ] (6) q = aρ1 + (1 − a) a0 ρ2 + (1 − a0 )ρ3 + a0 (1 − a0 ) i [ρ3 , ρ2 ] (7) q q h i + a (1 − a ) i a 0 ρ2 + (1 − a 0 ) ρ3 + a 0 (1 − a 0 ) i [ ρ3 , ρ2 ], ρ1 (8)
= aρ1 + (1 − a) a0 ρ2 + (1 − a)(1 − a0 )ρ3 q + (1 − a ) a 0 (1 − a 0 ) i [ ρ3 , ρ2 ] q + a (1 − a ) a 0 i [ ρ2 , ρ1 ] q + a (1 − a ) (1 − a 0 ) i [ ρ3 , ρ1 ] q q + a (1 − a ) a 0 (1 − a 0 ) i [ i [ ρ3 , ρ2 ], ρ1 ].
(9) (10) (11) (12) (13)
It then follows trivially from Theorem 1 that f ρ1 a ( ρ2 a 0 ρ3 ) ≥ a f ( ρ1 ) + (1 − a ) f ρ2 a 0 ρ3 0
(14) 0
≥ a f (ρ1 ) + (1 − a) a f (ρ2 ) + (1 − a)(1 − a ) f (ρ3 )
(15)
for any symmetric and concave function f : D(d) → R. However, there are other nested ways of combining three states, such as
(ρ1 b ρ2 ) b0 ρ3 = b0 bρ1 + b0 (1 − b)ρ2 + (1 − b0 )ρ3 + · · ·
(16)
for any b, b0 ∈ [0, 1]. The omitted terms are similar to eqs. (10) to (13), except for the double commutator which is i [ρ3 , i [ρ2 , ρ1 ]] in this case, thus making the expression different from the one above. Since there are several other nested ways of combining three states, it is not clear which of them, if any, should be preferred. Considering this, it is more desirable to have a ternary operation that treats each of the input states in the same way and combines them all at once. Note that the coefficients a, (1 − a) a0 , (1 − a)(1 − a0 ) and b0 b, b0 (1 − b), 1 − b0 (17) in front of the states ρ1 , ρ2 , ρ3 in eqs. (9) and (16) are probability distributions. It is thus natural to demand the combined state to be of the form p1 ρ1 + p2 ρ2 + p3 ρ3 + · · ·
(18)
for some probability distribution ( p1 , p2 , p3 ), where, unlike in (13), all three states are treated symmetrically in the omitted higher order terms (in particular, the third order). 3
1.3
The most general way of combining two states
Before we attempt to generalize eq. (4), it is worthwhile first checking whether Uλ in eq. (2) is actually the most general unitary matrix that can be expressed as a linear combination of I and S. As it turns out, up to an overall global phase and the sign of i, this is indeed the case. Proposition 2. Let z1√ , z2 ∈ C and U := √ z1 I + z2 S where S swaps two qudits (d ≥ 2). Then U is unitary if and only if z1 = eiϕ λ and z2 = ±ieiϕ 1 − λ for some λ ∈ [0, 1] and ϕ ∈ [0, 2π ). In other words, √ √ (19) U = eiϕ λ I ± i 1 − λ S . Proof. If U := z1 I + z2 S for some z1 , z2 ∈ C then UU † = (z1 z¯1 + z2 z¯2 ) I + (z1 z¯2 + z¯1 z2 )S.
(20)
Since I and S are linearly independent and we want UU † = I, we get (21) | z1 |2 + | z2 |2 = 1 and z1 z¯2 = −z1 z¯2 . √ √ From the first equation, z1 = eiϕ1 λ and z2 = eiϕ2 1 − λ for some λ ∈ [0, 1] and ϕ1 , ϕ2 ∈ [0, 2π ). The second equation says that z1 z¯2 is purely imaginary. Thus ei( ϕ1 − ϕ2 ) = ±i, which is equivalent to eiϕ2 = ±ieiϕ1 . In other words, we can take eiϕ1 := eiϕ and eiϕ2 := ±ieiϕ for some ϕ ∈ [0, 2π ). Since the map in eq. (3) involves conjugation, the global phase of U in eq. (19) is irrelevant and we only have the freedom of choosing the sign in front of i. Up to the sign, eq. (2) indeed provides the most general unitary for our purpose. To account for the two possible signs, we define q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 + λ(1 − λ) i [ρ2 , ρ1 ], (22) q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 − λ(1 − λ) i [ρ2 , ρ1 ]. (23) These two operations are related as follows: ρ1 λ ρ2 = ρ2 1−λ ρ1 .
2
(24)
Generalization to three states
Now that we fully understand the case of two states, let us investigate how to combine three states. We first look in more detail at the nested combinations suggested in Section 1.2, and then consider a more general ternary operation that combines all three states at once.
2.1
Twelve nested ways of combining three states
We can use eqs. (22) and (23) to work out all possible nested ways of combining three states. Because of the relation (24), there are 12 different nested ways of combining three states: ρ1 a ( ρ2 a0 ρ3 )
(25)
ρ 2 b ( ρ 3 b 0 ρ 1 )
(26)
ρ 3 c ( ρ 1 c 0 ρ 2 )
(27)
4
where each can independently be replaced by either or . In general, all twelve expressions will be different. However, if we choose the parameters so that ( p1 , p2 , p3 ) = a, a0 (1 − a), (1 − a0 )(1 − a) (28) 0 0 = (1 − b )(1 − b), b, b (1 − b) (29) 0 0 = c (1 − c), (1 − c )(1 − c), c (30) for some probability distribution ( p1 , p2 , p3 ), then all twelve expressions will begin with the same convex combination p1 ρ1 + p2 ρ2 + p3 ρ3 + · · · . Our goal is to obtain a more general ternary operation that continuously interpolates between these twelve expressions.
2.2
Combining n states at once
Recall from eq. (2) that Uλ is a linear combination of two-qudit permutation matrices. One way of generalizing this to more systems is as follows: h i (ρ1 , . . . , ρn ) 7→ Tr2,...,n U (ρ1 ⊗ · · · ⊗ ρn )U † , (31) where ρ1 , . . . , ρn ∈ D(d) are n qudit states and the unitary matrix U ∈ U(dn ) is a linear combination of all n! permutations that act on n qudits. The goal of this paper is to develop a better understanding of this special type of unitaries and the corresponding map resulting from eq. (31).
2.3
Graphical notation for evaluating partial traces
Let us work out the details of eq. (31) for n = 3. Let ρ1 , ρ2 , ρ3 ∈ D(d) and define ρ := Tr2,3 U (ρ1 ⊗ ρ2 ⊗ ρ3 )U †
(32)
for some unitary matrix U ∈ U(d3 ) such that 6
U :=
∑ zi Qi ,
(33)
i =1
where zi ∈ C and each Qi ∈ U(d3 ) is one of the six permutations that act on three qudits. We can represent them using the following graphical notation (see [WBC15] for more details): Q1 : =
Q2 : =
Q3 : =
Q4 : =
Q5 : =
Q6 : =
(34)
Note that one can easily recover the matrix representation of each Qi from this pictorial notation. For example, 2
1
Q2 |ψ1 i ⊗ |ψ2 i ⊗ |ψ3 i =
2
=
3
3
= |ψ2 i ⊗ |ψ3 i ⊗ |ψ1 i
(35)
1
for any |ψ1 i, |ψ2 i, |ψ3 i ∈ Cd , which is enough to determine all d3 × d3 entries of the matrix Q2 . The diagrams of matrices Qi can be composed in a natural way, and this operation is compatible with the matrix multiplication (in fact, the diagrams as well as the matrices form a group). For example, Q4 Q5 =
= 5
= Q2 .
(36)
Matrix inverses can be found by reversing the diagram: †
Q2† =
=
= Q3 .
(37)
Except for Q2 and Q3 which are inverses of each other, all other Qi s are self-inverse, i.e., Qi† = Qi when i ∈ / {2, 3}. Just like in eq. (35), subsystems of a mixed product state are permuted under conjugation by Qi . For example, 2
1
Q2 (ρ1 ⊗ ρ2 ⊗ ρ3 ) Q2† =
2 3
=
3
= ρ2 ⊗ ρ3 ⊗ ρ1 .
(38)
1
Using this pictorial representation, we would like to expand U on both sides of eq. (32) and obtain an explicit formula for the output state ρ in terms of the input states ρk and coefficients zi . This requires evaluating ξ i,j := Tr2,3 zi Qi (ρ1 ⊗ ρ2 ⊗ ρ3 )z¯ j QTj (39) for each combination of i, j ∈ {1, . . . , 6}. This daunting task becomes much more straightforward using the graphical notation for the partial trace [WBC15]. For example, " # 1 1
ξ 1,1 = z1 z¯1 Tr2,3
= z1 z¯1
2
3
3
" ξ 4,1 = z4 z¯1 Tr2,3
1
= z4 z¯1
ξ 5,1 = z5 z¯1 Tr2,3
= z5 z¯1
ξ 2,1 = z2 z¯1 Tr2,3
1 2 3
1 3
3
3
"
2
= z5 z¯1
2
3
2
1
#
2
3
1
= z4 z¯1
2 3
3 1
1
2
1
#
2
"
= z1 z¯1
2
= z1 z¯1 ρ1 Tr ρ2 Tr ρ3 ,
(40)
= z4 z¯1 ρ1 Tr(ρ2 ρ3 ),
(41)
= z5 z¯1 ρ2 ρ1 Tr ρ3 ,
(42)
= z2 z¯1 ρ2 ρ3 ρ1 .
(43)
1
#
= z2 z¯1
= z2 z¯1
2
2
3
1
3
The remaining cases are similar and are summarized in Fig. 1. Since ρ = ∑6i,j=1 ξ i,j , we can find the resulting state by putting all 36 terms together: 2 2 ρ = |z1 | + |z4 | + 2 Re(z1 z¯4 ) Tr(ρ2 ρ3 ) ρ1 + |z2 |2 + |z5 |2 + 2 Re(z2 z¯5 ) Tr(ρ3 ρ1 ) ρ2 + |z3 |2 + |z6 |2 + 2 Re(z3 z¯6 ) Tr(ρ1 ρ2 ) ρ3 + z1 z¯5 + z4 z¯2 ρ1 ρ2 + c.t. + z2 z¯6 + z5 z¯3 ρ2 ρ3 + c.t. (44) + z3 z¯4 + z6 z¯1 ρ3 ρ1 + c.t. + z2 z¯1 + z5 z¯4 ρ2 ρ3 ρ1 + c.t. + z3 z¯2 + z6 z¯5 ρ3 ρ1 ρ2 + c.t. + z1 z¯3 + z4 z¯6 ρ1 ρ2 ρ3 + c.t. where c.t. stands for the conjugate transpose of the previous term. 6
1 2 3 1 3
2 1
1
2
2
3
3
3
2
3
1
2 3
1
1
1
1
2
2
2
3
3
3
2
1
3
2
1
3 2
1
1
1
1
1
2
2
2
2
3
3
3
3
1
2
1 3
3
1
1
3
2
1
1
1
1
1
2
2
2
2
2
3
3
3
3
3
2
2
2
1 3
3
3
2 1
3
2
2
1
3
1
2 3
1
1
1
1
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
3
1 2
3
3
2 1
1
3
2
2
1
3
1
2
3 1
2
Figure 1: Tensor contraction diagrams for computing ξ i,j := Tr2,3 Qi (ρ1 ⊗ ρ2 ⊗ ρ3 ) QTj for each pair of three qudit permutations Qi and QTj (i ≥ j). By combining all 36 terms we obtain eq. (44).
7
2.4
Extra constraints
Before we can call eq. (44) a generalization of eq. (4), we need to answer the following two questions: Q1. Under what constraints on coefficients zi ∈ C is the matrix U in eq. (33) unitary? Q2. Are additional constraints needed to make the Tr(ρi ρ j ) terms in eq. (44) go away? We demand unitarity since we want ρ to be a valid quantum state—this is very natural considering eq. (32). Note that for n = 2 the answer to Q1 is already provided by Proposition 2. Using representation theory, we will answer Q1 in full generality (in Theorem 9 we characterize when U := ∑π ∈Sn zπ Qπ is unitary for any n ≥ 1). In addition to constraints from Q1, we would also like that Re(z1 z¯4 ) = Re(z2 z¯5 ) = Re(z3 z¯6 ) = 0.
(45)
Then the first order terms in eq. (44) would form a linear (in fact, convex) combination of ρi with coefficients depending only on zi and not the states themselves. We will be able to answer Q2 for n = 3 where the answer is “yes” (see Proposition 10 for more details). In fact, we will see in Section 4.3 that constraint (45) can be imposed in a fairly natural way and it leads to some further nice structure in eq. (44). Note that for n = 2 the answer to Q2 is trivially “no”, since no terms of the form Tr(ρi ρ j ) appear.
3
When is a linear combination of permutations unitary?
We begin by first answering Q1 which will require some representation theory. Particularly relevant to us is reference [Chi13] that provides a very concise introduction to representation theory and Fourier analysis of non-abelian groups. For more background on representation theory of finite groups, see the standard reference [Ser12].
3.1
Background on representation theory
Let U(d) denote the set of all d × d unitary matrices. A d-dimensional representation of a finite group G is a map τ : G → U(d) such that τ ( gh) = τ ( g)τ (h) for all g, h ∈ G. We will always use e to denote the identity element of G. Note that τ (e) = Id , the d × d identity matrix. We call d the dimension of representation τ. Let Sn denote the symmetric group consisting of all n! permutations acting on n elements. We write π (i ) = j to mean that permutation π ∈ Sn maps i to j, and we write π −1 to denote the inverse permutation of π. The following are four different representations of the symmetric group—two of them (Qπ and Lπ ) will play an important role later in this paper. Example (Representations of Sn ). The symmetric group Sn can be represented by permutation matrices in several different ways. In each case we write down how the matrix associated to permutation π ∈ Sn acts in the standard basis: • natural representation Pπ ∈ U(n): Pπ : |i i 7→ |π (i )i,
∀i ∈ {1, . . . , n};
(46)
• tensor representation Qπ ∈ U(dn ): Q π : | i 1 i ⊗ · · · ⊗ | i n i 7 → | i π −1 (1 ) i ⊗ · · · ⊗ | i π −1 ( n ) i , 8
∀i1 , . . . , in ∈ {1, . . . , d};
(47)
• left regular and right regular representations Lπ , Rπ ∈ U(|Sn |): Lπ : |σi 7→ |πσ i,
∀ σ ∈ Sn ,
(48)
Rπ : |σi 7→ |σπ −1 i,
∀ σ ∈ Sn .
(49)
We call d the local dimension of the tensor representation (we will typically required that d ≥ n). Note that for the regular representations, the standard basis of the underlying space is labeled by permutations themselves, so the space has n! dimensions: CSn ∼ = Cn! . Finally, note that the regular representations can be defined for any finite group G in a similar manner. Since Qπ will play an important role later, let us verify that it is indeed1 a representation of Sn . First, note that Qπ
n O
|ψi i =
i =1
n O
|ψπ −1 (i) i =
i =1
n O
|φi i
(50)
i =1
where |φi i := |ψπ −1 (i) i. Following the same rule, we see that Qσ
n O
|φi i =
i =1
n O
|φσ−1 (i) i =
i =1
n O
|ψπ −1 (σ−1 (i)) i =
i =1
n O
|ψ(σπ )−1 (i) i.
(51)
i =1
In other words, Qσ Qπ acts in exactly the same way as Qσπ .
3.2
From permutations to unitaries—importance of the left regular representation
Consider the following linear combination: U :=
∑
zπ Qπ ,
(52)
π ∈Sn
where zπ ∈ C and Qπ are the matrices from the tensor representation of Sn . Can we parametrize in some simple way all coefficient tuples z := (zπ : π ∈ Sn ) ∈ CSn such that U is unitary for all local dimensions d ≥ 1 simultaneously? We will provide a nice answer to this question using representation theory. Let us start by first reducing this problem from the tensor representation Qπ to the left regular representation Lπ . As a first step, we need to show that the matrices Lπ are linearly independent (this holds more generally, i.e., not just for Sn but for any finite group G). Proposition 3. For any finite group G, the matrices { L g : g ∈ G } have disjoint supports and thus are linearly independent. Proof. For any x, y, g ∈ G: (
h x | L g |yi = h x | gyi =
1 0
if g = xy−1 , otherwise.
(53)
Thus, for any fixed x and y there is exactly one matrix, namely L xy−1 , with a non-zero entry in row x and column y. Hence, the matrices L g have disjoint supports and thus are linearly independent. 1 It
would not be a representation if we would use π instead of π −1 on the RHS of eq. (47).
9
Remark. The matrices L g form a (non-commutative) association scheme [Ban93]. Their linear span is known as the Bose–Mesner algebra of this scheme. Our Theorem 7 (see below) thus characterizes all unitary matrices within this particular algebra. Example (G = S3 ). Because of Proposition 3, any linear combination of the matrices L g has a particularly nice structure. For example, if G = S3 and we order the permutations according to eq. (34) then 1 3 2 4 5 6 2 1 3 6 4 5 6 3 2 1 5 6 4 (54) ∑ kLk = 4 6 5 1 2 3 . k =1 5 4 6 3 1 2 6 5 4 2 3 1 One can easily read off the matrix representation of each Li from this. Let us now show that the matrices Qπ are also linearly independent when d is sufficiently large. Lemma 4. If d ≥ n then Qπ ∼ = Lπ ⊕ τ (π ) and thus the matrices { Qπ : π ∈ Sn } are linearly independent. Proof. It suffices to show that Qπ are linearly independent even when restricted to some invariant N Nn n −1 subspace of Cd . Let |Ψσ i := Qσ in=1 |i i = 1 | σ (i )i (we needNd ≥ n for this to make Nn Ni= n − 1 sense). Note that Qπ |Ψσ i = Qπ i=1 |σ (i )i = i=1 |σ−1 (π −1 (i ))i = in=1 |(πσ)−1 (i )i = |Ψπσ i, n so L := span{|Ψπ i : π ∈ Sn } is an invariant subspace of Cd under the action of the tensor representation. In fact, the vectors |Ψπ i form an orthonormal basis of the subspace L, and the tensor representation Qπ acts as the left regular representation Lπ in this subspace. In other words, Qπ ∼ = Lπ ⊕ τ (π )
(55)
for some representation τ acting on the orthogonal complement of L. According to Proposition 3, the matrices { Lπ : π ∈ Sn } are linearly independent so the result follows. Lemma 5. Let zπ ∈ C for each π ∈ Sn . If d ≥ n then ∑π ∈Sn zπ Qπ is unitary if and only if ∑π ∈Sn zπ Lπ is unitary (the reverse implication holds for any d ≥ 1). n
Proof. As we noted in the proof of Lemma 4, for d ≥ n there is a subspace L of Cd where Qπ acts as Lπ , see eq. (55); this immediately gives the forward implication. For the reverse implication, note that we can decompose any representation of a finite group into a direct sum of its irreducible representations or irreps [Ser12]. If we obtain such decomposition for the left regular representation, a standard result from representation theory says that each irrep will appear at least once in this decomposition [Ser12]. Thus, the unitarity of ∑π ∈Sn zπ Lπ is equivalent to the simultaneous unitarity of ∑π ∈Sn zπ τ (π ) for all irreps τ of Sn . Since for any d ≥ 1 the tensor representation Qπ can also be decomposed as a direct sum of irreps, we conclude that ∑π ∈Sn zπ Qπ must therefore be unitary. According to Lemma 5, our problem now reduces to analyzing the left regular representation of Sn and characterizing when U := ∑π ∈Sn zπ Lπ is unitary. We will analyze this in the Fourier basis where the left regular representation decomposes as a direct sum of irreducible representations.
10
3.3
Fourier transform over finite groups
Recall from [Chi13] that the Fourier transform over a finite group G is the unitary matrix s d τ dτ τ ( g) j,k |(τ, j, k )ih g|, F := ∑ ∑ | G | j,k∑ ˆ g∈ G =1
(56)
τ ∈G
where Gˆ denotes the set of irreducible representations of G, dτ is the dimension of irrep τ, and τ ( g) j,k are the matrix elements of τ ( g). The output space of F is labeled by triples (τ, j, k ) and has the same dimension as the input space—indeed, it is a standard result in representation theory that
∑ d2τ = |G|.
(57)
τ ∈ Gˆ
One can easily check that F is unitary using the orthogonality of characters [Chi13]. We will need the following standard result from representation theory [Chi13, Ser12]. Let Lˆ g := FL g F † denote L g in the Fourier basis. Then Lˆ g =
M
τ ( g) ⊗ Idτ
(58)
τ ∈ Gˆ
where Idτ is the dτ × dτ identity matrix. In other words, Lˆ g is block-diagonal and contains each irrep τ the number of times equal to its dimension dτ .
3.4
Characterization of unitary linear combinations
Let us first show a simple preliminary fact. Recall from Proposition 3 that { L g : g ∈ G } is a linearly independent set. In fact, when properly normalized, these matrices are orthonormal. p Proposition 6. For any finite group G, the matrices { L g / | G | : g ∈ G } are orthonormal with respect to the Hilbert–Schmidt inner product h A, Bi := Tr( A† B). Proof. First, note from eq. (53) that for any g ∈ G, Tr L g =
∑ hh| Lg |hi = ∑ δg,e = |G|δg,e .
h∈ G
(59)
h∈ G
Thus, for any a, b ∈ G, the corresponding Hilbert–Schmidt inner product is 1 1 1 h L a , Lb i = Tr L†a Lb = Tr L a−1 b = δa,b . |G| |G| |G| p Hence the normalized matrices L g / | G | are orthonormal.
(60)
Theorem 7. Let G be a finite group, L g be its left regular representation, and z g ∈ C for each g ∈ G. Then ∑ g∈G z g L g is unitary if and only if zg =
∑
τ ∈ Gˆ
dτ Tr τ ( g)† Uτ |G|
for some choice of Uτ ∈ U(dτ ), one for each irrep τ of G. 11
Proof. Clearly, ∑ g∈G z g L g is unitary if and only if ∑ g∈G z g Lˆ g is unitary where M Lˆ g = FL g F † = τ ( g) ⊗ Idτ
(61)
τ ∈ Gˆ
according to eq. (58). We prefer to work in the Fourier basis, since then all Lˆ g become simultaneously block diagonal and we can write M ˆ (62) ∑ zg Lg = ∑ zg τ ( g) ⊗ Idτ . g∈ G
g∈ G
τ ∈ Gˆ
This matrix is unitary if and only if each of its blocks is unitary, i.e., M Uτ ⊗ Idτ =: U ∑ zg Lˆ g = g∈ G
(63)
τ ∈ Gˆ
p for some set of unitaries Uτ ∈ U(dτ ). Since B := { L g / | G | : g ∈ G } is an orthonormal set, see p ˆ = | G | = ∑ ˆ d2τ Proposition 6, so is Bˆ := { FBF † : B ∈ B} = { Lˆ g / | G | : g ∈ G }. Since |B| τ ∈G according to eq. (57), Bˆ is in fact an orthonormal basis for the set of all block matrices that have the same block structure as U in eq. (63). Thus, we can obtain the coefficients z g in the expansion U p = |G|
Lˆ g
∑ z g p| G |
(64)
g∈ G
simply by projecting on the corresponding basis vector: ! ˆ i Mh Lg U 1 zg = p (τ ( g)† Uτ ) ⊗ Idτ ,p = Tr = |G| |G| |G| ˆ τ ∈G
∑
τ ∈ Gˆ
dτ Tr(τ ( g)† Uτ ). |G|
(65)
where we substituted eqs. (61) and (63). The reverse implication follows by applying all steps in reverse order. As a byproduct of our proof, we observe that all unitary linear combinations ∑ g∈G z g L g form a L group of their own. Indeed, we see from eq. (63) that this group is isomorphic to τ ∈Gˆ U(dτ ) and it contains G as a subgroup (as represented by the matrices L g ). Indeed, if we take any g ∈ G and set Uτ := τ ( g) for all irreps τ, then z g = 1 while all other coefficients vanish according to Proposition 6, hence reducing the linear combination to L g . Considering this, we can intuitively think of n o (66) ∑ zg Lg ∈ U(|G|) : zg ∈ C g∈ G
as a natural continuous extension of the discrete finite group G. Corollary 8 (Unitary version of Cayley’s theorem). Every finite group G can be naturally extended to a continuous subgroup of the unitary group U(| G |). If we specialize Theorem 7 to G = Sn and apply Lemma 5, we get the following result. Theorem 9. Let zπ ∈ C for each π ∈ Sn . If d ≥ n then ∑π ∈Sn zπ Qπ is unitary if and only if zπ =
∑
τ ∈Sˆ n
dτ Tr τ (π )† Uτ n!
(67)
for some choice of Uτ ∈ U(dτ ), one for each irrep τ of Sn (the reverse implication holds for any d ≥ 1). 12
π τ1 (π ) τ2 (π )
1 1
1 0 0 1
1 0 0 1
τ3 (π ) τ30 (π )
1 1
1 2
√ − 1 − 3 √ 3 −1 ω 0 0 ω2
1 1 −√1 1 2 − 3 2 ω 0
1 √ −1 3 1 0 0 −1 −1 0 0 1 ω 1 0
1 −1 √ −√1 − 3 1 2 − 3 1 0 ω2 ω 0
1 −1 √ −1 3 1 √ 2 3 1 0 ω ω2 0
Table 1: All irreducible representations of S3 . The matrix elements of τ3 are not unique but depend on the choice of basis—we provide two simple choices (irreps τ3 and τ30 are isomorphic).
4
How to combine three quantum states?
We will now use Theorem 9 to parametrize all tuples of complex coefficients (zπ : π ∈ S3 ) such that ∑π ∈S3 zπ Qπ is a unitary matrix for all d ≥ 1. Recall that Qπ permutes three qudits according to permutation π, see eq. (34). First, we need to work out all irreps of S3 .
4.1
Irreducible representations of S3
The symmetric group S3 has three irreducible representations [Ser12]: two 1-dimensional representations (the trivial representation τ1 and the sign representation τ2 ) and a 2-dimensional representation τ3 . Recall that S3 ∼ = D3 (the dihedral group), so geometrically τ3 corresponds to rotations and reflections in 2D that preserve an equilateral triangle centered at the origin and pointing along the x axis. These representations are written out explicitly in Table 1. One can easily verify that τk (πσ) = τk (π )τk (σ) for all π, σ ∈ S3 and k ∈ {1, 2, 3}. For example, √ √ 1 −1 − 3 1 −1 − 3 1 0 √ √ τ3 = τ3 τ3 = · . = = τ3 0 −1 2 − 3 2 1 3 −1
4.2
General solution for S3
According to Theorem 9, we need to assign one unitary matrix Uk ∈ U(dτk ) to each irrep τk of S3 . Since the global phase of ∑6i=1 zi Qi has no effect, we can without loss of generality assume one of the unitaries Uk to be in the special unitary group. We take U1 , U2 ∈ U(1) and U3 ∈ SU(2), and parametrize these unitaries as follows: a c iϕ iϕ 2 1 : : : U1 = e , U2 = e , U3 = (68) −c¯ a¯
13
2
2
where ϕ1 , ϕ2 ∈ [0, 2π ) and a, c ∈ C are such that | a| + |c| = 1. We can now use eq. (67) and irreps τ1 , τ2 , τ3 from Table 1 to compute the coefficients z1 , . . . , z6 : z1 = z2 = z3 = z4 = z5 = z6 =
1 iϕ1 e + eiϕ2 + 4 Re( a) , 6 √ 1 iϕ1 e + eiϕ2 − 2 Re( a + 3c) , 6 √ 1 iϕ1 e + eiϕ2 − 2 Re( a − 3c) , 6 1 iϕ1 e − eiϕ2 + 4i Im( a) , 6 √ 1 iϕ1 e − eiϕ2 − 2i Im( a + 3c) , 6 √ 1 iϕ1 e − eiϕ2 − 2i Im( a − 3c) . 6
(69) (70) (71) (72) (73) (74)
Up to an overall global phase, this parametrizes precisely the set of coefficients for which the following matrix, cf. eq. (54), is unitary:
z1 z2 6 z3 ∑ zi Li = z4 i =1 z5 z6
z3 z1 z2 z6 z4 z5
z2 z3 z1 z5 z6 z4
z4 z6 z5 z1 z3 z2
z5 z4 z6 z2 z1 z3
z6 z5 z4 . z3 z2
(75)
z1
√ Remark. If we insist, in addition to eqs. (69) to (74), that |z1 | = · · · = |z6 | = 1/ 6, we get a 6 × 6 ˙ flat unitary—this is also known as a complex Hadamard matrix [TZ06]. Such matrices are relevant + to the MUB problem in six dimensions [BBE 07]. Using computer, we can find 72 discrete flat solutions (z1 , . . . , z6 ). Unfortunately, the corresponding unitaries appear to be equivalent to the 6 × 6 Fourier matrix. Nevertheless, this method can in principle be used to find flat unitaries of size | G | × | G | for any finite group G. It would be interesting to know whether this construction can ˙ yield anything beyond what is already known [TZ06].
4.3
Imposing extra constraints
Recall that eq. (44) involves terms with coefficients Re(z1 z¯4 ), Re(z2 z¯5 ), and Re(z3 z¯6 ), and we would like to understand when these terms vanish (see Q2 in Section 2.3). Proposition 10. Re(z1 z¯4 ) = Re(z2 z¯5 ) = Re(z3 z¯6 ) = 0 if ϕ1 = − ϕ2 . Proof. Note from eqs. (69) to (74) that (up to an overall constant) each coefficient zi has one of the following two forms: u := eiϕ1 + eiϕ2 + r cos α, v := e
iϕ1
−e
iϕ2
+ ir sin α
(76) (77)
for some ϕ1 , ϕ2 , α ∈ [0, 2π ) and r ≥ 0. Furthermore, each u-type coefficient is paired up with a
14
corresponding v-type coefficient. A straightforward calculation gives Re(uv¯ ) = Re(u) Re(v) + Im(u) Im(v)
(78)
= (cos ϕ1 + cos ϕ2 + r cos α)(cos ϕ1 − cos ϕ2 ) + (sin ϕ1 + sin ϕ2 )(sin ϕ1 − sin ϕ2 + r sin α)
(79)
= (cos ϕ1 )2 − (cos ϕ2 )2 + (cos ϕ1 − cos ϕ2 )r cos α
(80)
+ (sin ϕ1 )2 − (sin ϕ2 )2 + (sin ϕ1 + sin ϕ2 )r sin α = (cos ϕ1 − cos ϕ2 )r cos α + (sin ϕ1 + sin ϕ2 )r sin α = r cos(α − ϕ1 ) − cos(α + ϕ2 ) .
(81) (82)
We can guarantee that Re(uv¯ ) = 0 irrespectively of the values of r and α by choosing − ϕ1 = ϕ2 . This makes all three terms vanish simultaneously. Taking Proposition 10 into account, we define ϕ := ϕ1 = − ϕ2 . Then eqs. (69) to (74) become: z1 = z2 = z3 = z4 = z5 = z6 =
1 cos ϕ + 2 Re( a) , 3 √ 1 cos ϕ − Re( a + 3c) , 3 √ 1 cos ϕ − Re( a − 3c) , 3 i sin ϕ + 2 Im( a) , 3 √ i sin ϕ − Im( a + 3c) , 3 √ i sin ϕ − Im( a − 3c) . 3 2
(83) (84) (85) (86) (87) (88) 2
Here we can choose any ϕ ∈ [0, 2π ) and a, c ∈ C such that | a| + |c| = 1, so in total we have four degrees of freedom. The output state can be obtained by substituting eqs. (83) to (88) in eq. (44).
4.4
Alternative parametrizations
Unfortunately the parametrization in eqs. (83) to (88) is somewhat cumbersome. In this section we derive two alternative parametrizations that are much simpler and more insightful. We will derive them from eqs. (83) to (88). However, with an educated guess, they can also be derived from scratch without invoking the irreps of S3 (see Appendix A). 4.4.1
Parametrization by C3
It is natural to pair up the coefficients zi as follows: 1 iϕ (e + 2a), 3 √ 1 q2 := z2 + z5 = (eiϕ − a − 3c), 3 √ 1 q3 := z3 + z6 = (eiϕ − a + 3c). 3 q1 : = z1 + z4 =
15
(89)
The coefficients zi in terms of the new parameters q1 , q2 , q3 ∈ C are expressed as follows:
(z1 , z2 , z3 , z4 , z5 , z6 ) := (Re q1 , Re q2 , Re q3 , i Im q1 , i Im q2 , i Im q3 ).
(90)
With this parametrization, the output state ρ from eq. (44) looks as follows: ρ = | q1 |2 ρ1 + | q2 |2 ρ2 + | q3 |2 ρ3
(91)
+ Im(q1 q¯2 ) i [ρ1 , ρ2 ] + Im(q2 q¯3 ) i [ρ2 , ρ3 ] + Im(q3 q¯1 ) i [ρ3 , ρ1 ]
(92)
+ Re(q1 q¯2 )(ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) + Re(q2 q¯3 )(ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) + Re(q3 q¯1 )(ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 )
(93)
where q1 , q2 , q3 ∈ C are subject to the following constraints:
|q1 |2 + |q2 |2 + |q3 |2 = 1,
|q1 + q2 + q3 |2 = 1.
To derive these constraints, we solve eq. (89) for the original parameters: √ 1 3 iϕ e = q1 + q2 + q3 , a = q1 − ( q2 + q3 ), c= ( q3 − q2 ). 2 2
(94)
(95)
From the first equation we immediately get the second constraint: 1 = |q1 + q2 + q3 |2 = |q1 |2 + |q2 |2 + |q3 |2 + 2 Re(q1 q¯2 ) + 2 Re(q2 q¯3 ) + 2 Re(q3 q¯1 ).
(96)
From the next two equations we get 1 1 |q2 |2 + |q3 |2 − Re(q1 q¯2 ) − Re(q3 q¯1 ) + Re(q2 q¯3 ), 4 2 3 3 |c|2 = |q2 |2 + |q3 |2 − Re(q2 q¯3 ). 4 2
| a |2 = | q1 |2 +
(97) (98)
Putting these together gives another constraint: 1 = | a|2 + |c|2 = |q1 |2 + |q2 |2 + |q3 |2 − Re(q1 q¯2 ) − Re(q2 q¯3 ) − Re(q3 q¯1 ).
(99)
From eqs. (96) and (99) we conclude that Re(q1 q¯2 ) + Re(q2 q¯3 ) + Re(q3 q¯1 ) = 0.
(100)
Consequently, constraints (96) and (99) are equivalent to eq. (94). These constraints are in fact also equivalent to the original constraints guaranteeing unitarity of the matrices in eq. (68). Indeed, if we substitute eq. (89) in eq. (94), we recover the original constraints |eiϕ |2 = 1 and | a|2 + |c|2 = 1. Another way of writing constraints (94) is as follows. If we let q1 1 1 | q i : = q2 , | u i : = √ 1 (101) 3 1 q3 16
then eq. (94) is equivalent to
|hq|qi|2 = 1,
|hq|ui|2 = 1/3,
(102)
i.e., |qi ∈ C3 is a unit vector that is mutually unbiased to |ui. One advantage of the new parametrization is that it makes it more apparent that neither the constraints (102) nor the expression of ρ in eqs. (91) to (93) depend on the global phase of |qi; this feature was not so obvious from the original parametrization in eqs. (83) to (88). We can make two further observations: • Due to the first constraint in eq. (94), the first order terms in eq. (91) form a convex combination of ρ1 , ρ2 , ρ3 . This is analogous to eq. (4) for n = 2. • According to eq. (100), the coefficients of the third order terms (93) sum to zero. This allows writing the third order terms as a linear combination of double commutators (more details on this are provided in Section 5.1). 4.4.2
Parametrization by a probability distribution and phases
As we just noted, (|q1 |2 , |q2 |2 , |q3 |2 ) is a probability distribution according to eq. (94). We can highlight this aspect by choosing √ qk := eiφk pk (103) for some probability distribution ( p1 , p2 , p3 ) and some phases φ1 , φ2 , φ3 ∈ [0, 2π ). Note that p Re(qi q¯ j ) = pi p j cos(φi − φj ), (104) p Im(qi q¯ j ) = pi p j sin(φi − φj ). (105) If we further denote the differences between consecutive phases by δ12 := φ1 − φ2 , δ23 := φ2 − φ3 , δ31 := φ3 − φ1 ,
(106) (107) (108)
we can rewrite eqs. (91) to (93) as follows: ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 √ + p1 p2 sin δ12 i [ρ1 , ρ2 ] √ + p2 p3 sin δ23 i [ρ2 , ρ3 ] √ + p3 p1 sin δ31 i [ρ3 , ρ1 ] √ + p1 p2 cos δ12 (ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) √ + p2 p3 cos δ23 (ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) √ + p3 p1 cos δ31 (ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 ).
(109)
(110)
(111)
Parameters pi and δij are subject to the following constraints. For any distribution ( p1 , p2 , p3 ), the angles δij must satisfy δ12 + δ23 + δ31 = 0,
√
p1 p2 cos δ12 + 17
√
p2 p3 cos δ23 +
√
p3 p1 cos δ31 = 0.
(112)
The first constraint is apparent from eqs. (106) to (108), while the second constraint is equivalent to eq. (100). Except for deterministic distributions, these constraints yield a one-parameter family of angles (δ12 , δ23 , δ31 ). This is qualitatively different from the n = 2 case, where the coefficients of the first order terms are completely determined by λ, see eq. (4). Once λ is fixed, only a discrete degree of freedom remains corresponding to the sign in front of the commutator, see Section 1.3.
5
Further observations
In this section we highlight some further features of our operation. In particular, we show that the third order terms can be expressed as a linear combination of double commutators. We also show that the twelve nested compositions discussed in Section 2.1 are special cases of our operation.
5.1
Double commutators
Let us elaborate more on the meaning of the constraint (100). As mentioned earlier, it has to do with double commutators, i.e., expressions of the form [1, [2, 3]].2 In what follows, we do not specify the states ρi but rather treat them as abstract non-commutative variables. With this convention, for example, ρ1 and ρ2 are always considered to be linearly independent. Note that there are 6 ways of ordering three states and 2 ways of putting brackets, so there are twelve double commutators in total. However, many of them are identical (such as [1, [2, 3]] and [[3, 2], 1]) or differ only by a sign (such as [1, [2, 3]] and [1, [3, 2]]). Furthermore, we know from the Jacobi identity that [1, [2, 3]] + [2, [3, 1]] + [3, [1, 2]] = 0. (113) This leaves us with only two linearly independent double commutators. Somewhat arbitrarily, we can choose them as [1, [2, 3]] and [[1, 2], 3]. After expanding both of them, we get the following coefficients in front of the six different products of the matrices 1, 2, 3:
[1, [2, 3]] [[1, 2], 3] x i [1, i [2, 3]] + y i [i [1, 2], 3]
123 132 213 231 312 321 1 −1 0 −1 0 1 1 0 −1 0 −1 1 z x y x y z
(114)
where the last row is a linear combination of the first two rows and z := − x − y. In other words, x (231 + 132) + y(312 + 213) + z(123 + 321) = x i [1, i [2, 3]] + y i [i [1, 2], 3]
(115)
whenever x + y + z = 0. We can use this to simply the third order terms in eqs. (93) and (111). Indeed, since the coefficients in (93) sum to zero, see eq. (100), we can rewrite (93) as a linear combination of two double commutators: Re(q1 q¯2 ) i [ρ1 , i [ρ2 , ρ3 ]] (116) + Re(q2 q¯3 ) i [i [ρ1 , ρ2 ], ρ3 ]. Similarly, because of identity (112), expression (111) is equal to
√ p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]] √ + p2 p3 cos δ23 i [i [ρ1 , ρ2 ], ρ3 ]. 2 We
write i instead of ρi for brevity.
18
(117)
5.2
Relation to nested expressions
It is interesting to know whether our ternary operation can produce the 12 nested expressions discussed in Section 2.1 as special cases. More precisely, we would like to know whether, for fixed distribution ( p1 , p2 , p3 ), the one-parameter family of states described by eqs. (109) to (111) under constraints (112) contains all 12 nested expressions with first order terms p1 ρ1 + p2 ρ2 + p3 ρ3 . For example, consider nested expressions of the form ρ1 a (ρ2 a0 ρ3 ) where each is either or , see eqs. (22) and (23). Recall from eqs. (7) and (8) that q 0 ρ1 a (ρ2 a0 ρ3 ) = aρ1 + (1 − a) a0 ρ2 + (1 − a0 )ρ3 + (−1)s a0 (1 − a0 ) i [ρ3 , ρ2 ] (118) q q h i 0 +(−1)s a(1 − a) i a0 ρ2 + (1 − a0 )ρ3 + (−1)s a0 (1 − a0 ) i [ρ3 , ρ2 ], ρ1 , (119) where s, s0 ∈ {0, 1} depend on the signs of the two operations (0 stands for while 1 stands for ). Importantly, this expression has only one double commutator, namely i [ρ1 , i [ρ2 , ρ3 ]], while a general expression involves two. Indeed, recall from eq. (117) that for the ternary operation the terms of order three can be written as √ p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]] (120) √ + p2 p3 cos δ23 i [i [ρ1 , ρ2 ], ρ3 ]. To get only one double commutator, we can set cos δij = 0 for some ij ∈ {12, 23, 31}. However, we must also be able to solve eq. (112) for the remaining two phases. For example, to get only i [ρ1 , i [ρ2 , ρ3 ]], we can set cos δ23 = 0 and solve eq. (112) for δ12 and δ31 . Similarly, cos δ31 = 0 would yield i [ρ2 , i [ρ3 , ρ1 ]] via eq. (112) and the Jacoby identity. The following lemma establishes that we can indeed get all 12 nested expressions in this way. Recall that throughout this section we treat ρi as abstract non-commutative variables. Lemma 11. Let ρ be given by eqs. (109) to (111) and assume p1 , p2 , p3 6= 0. Then ρ admits a nested expression if and only if cos δij = 0 for some ij ∈ {12, 23, 31}. Proof. We have already proved the forward implication. Indeed, nested expressions have only one double commutator, so we get cos δij = 0 for some ij ∈ {12, 23, 31}. For the reverse implication, we assume cos δ23 = 0 (the other cases follow similarly due to symmetry). Under this assumption, let us argue that eq. (112) has only four discrete solutions, and that these solutions correspond to the four nested expressions ρ 1 p 1 ρ 2 p2 ρ 3 (121) p2 + p3
with arbitrary signs s, s0 ∈ {0, 1} for the two operations. First, we can write 0 cos δ23 = 0 and sin δ23 = −(−1)s
(122)
s0
for some s0 ∈ {0, 1}. In other words, δ23 := −(−1) π/2. From the first half of eq. (112), δ12 = 0 −δ23 − δ31 = (−1)s π/2 − δ31 . Hence 0
0
cos δ12 = (−1)s sin δ31 and sin δ12 = (−1)s cos δ31 . (123) √ √ From the second half of eq. (112), p2 cos δ12 + p3 cos δ31 = 0. If we substitute eq. (123), this √ √ 0 becomes p2 (−1)s sin δ31 + p3 cos δ31 = 0 and we get r r p3 p2 s s+s0 sin δ31 = (−1) and cos δ31 = −(−1) (124) p2 + p3 p2 + p3 19
for some s ∈ {0, 1}. We can then find δ12 by substituting eq. (124) back in eq. (123): r r p2 p3 s s+s0 sin δ12 = −(−1) and cos δ12 = (−1) . p2 + p3 p2 + p3
(125)
Equations (122), (124) and (125) with s, s0 ∈ {0, 1} give us the four solutions. It remains to argue that these solutions produce the four states in eq. (121). We begin by restating eqs. (109) to (111) when cos δ23 = 0: ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 √ + p1 p2 sin δ12 i [ρ1 , ρ2 ] √ + p2 p3 sin δ23 i [ρ2 , ρ3 ] √ + p3 p1 sin δ31 i [ρ3 , ρ1 ] √ + p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]]. Let us group the terms together to make this appear more similar to eqs. (118) and (119): √ ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 − p2 p3 sin δ23 i [ρ3 , ρ2 ] √ √ √ + i − p1 p2 sin δ12 ρ2 + p3 p1 sin δ31 ρ3 + p1 p2 cos δ12 i [ρ3 , ρ2 ], ρ1 . If we pull out the desired prefactors, we get √ p2 p3 p2 p3 ρ = p1 ρ1 + (1 − p1 ) ρ2 + ρ3 − sin δ23 i [ρ3 , ρ2 ] p2 + p3 p2 + p3 p2 + p3 √ √ √ q p2 sin δ12 p3 sin δ31 p2 cos δ12 ρ2 + √ ρ3 + √ i [ ρ3 , ρ2 ], ρ1 . + p1 (1 − p1 ) i − √ p2 + p3 p2 + p3 p2 + p3
(126)
(127)
(128)
(129) (130)
(131) (132)
This becomes one of the four states in eq. (121) once we substitute eqs. (122), (124) and (125). Example. Let ρ( x, y, z) := 12 ( I + xσx + yσy + zσz ) where σx , σy , σz are the Pauli matrices. Take ρ1 := ρ(1, 0, 0), ρ2 := ρ(0, 1, 0), ρ3 := ρ(0, 0, 1) and let p1 = p2 = p3 := 1/3. (Note that ρ1 = |+ih+|, ρ2 = |+i ih+i |, ρ3 = |0ih0|, i.e., pure states pointing along the three axis of the Bloch sphere.) Then the combined state given by eqs. (109) to (111) is 1 1 1 (1 − sin δ23 ), (1 − sin δ31 ), (1 − sin δ12 ) , (133) ρ 3 3 3 where the angles δij are subject to the following relations, see eq. (112): δ12 + δ23 + δ31 = 0,
cos δ12 + cos δ23 + cos δ31 = 0.
(134)
This describes a one-parameter orbit shown in Fig. 2. As proved by Lemma 11, this orbit contains all twelve nested combinations which are also illustrated in Fig. 2.
20
ρ3
ρ2 ( ρ3 ρ1 )
ρ1 ( ρ2 ρ3 )
ρ3 ( ρ1 ρ2 )
ρ3 ( ρ1 ρ2 )
ρ1 ( ρ2 ρ3 )
ρ2 ( ρ3 ρ1 )
ρ2 ( ρ3 ρ1 )
ρ1 ( ρ2 ρ3 )
ρ1
ρ3 ( ρ1 ρ2 )
ρ3 ( ρ1 ρ2 )
ρ1 ( ρ2 ρ3 )
ρ2
ρ2 ( ρ3 ρ1 )
Figure 2: One-parameter orbit obtained by combining three mutually unbiased pure qubit states ρ1 , ρ2 , ρ3 with equal weights p1 = p2 = p3 = 1/3. This orbit contains all twelve nested combinations of ρ1 , ρ2 , ρ3 as special cases (the two operations have parameters 1/3 and 1/2, respectively).
21
6
Open problems
The most obvious open problem is generalizing Theorem 1 (originally from [ADO15]) to three states. Here is a formal statement of this conjecture. Conjecture (EPI for three states). For any concave and symmetric function f : D(d) → R, any states ρ1 , ρ2 , ρ3 ∈ D(d), and any probability distribution ( p1 , p2 , p3 ), f ( ρ ) ≥ p1 f ( ρ1 ) + p2 f ( ρ2 ) + p3 f ( ρ3 )
(135)
where ρ is given by eqs. (109) to (111) with pi and δij subject to eq. (112). It would also be interesting to understand how an arbitrary number of states can be combined. Towards this goal, the two main steps are: 1. Finding a generalization of eq. (44). Since for general n the final expression of ρ would consist of (n!)2 terms, we need a more efficient way of contracting tensor diagrams. 2. Answering Q2 for any n. While for n = 3 it was sufficient to adjust the global phases of the unitaries Uτ , for general n it is not clear at all how to turn the first order terms of ρ into a convex combination of ρi , with coefficients depending only on the parameters zπ but not the input states ρi themselves. In particular, it is worthwhile investigating when higher order terms of ρ can be written as a linear combination of nested commutators. Perhaps, as suggested by the n = 3 case, better understanding of the second problem might make it possible to deal with both problems simultaneously, since many terms are likely to drop out simultaneously.
Acknowledgements I would like to thank Toby Cubitt, Nilanjana Datta, and Will Matthews for very helpful discussions. This work was supported by the European Union under project QALGO (Grant Agreement No. 600700). In loving memory of L.O.V. Coffeebean.
References [ADO15] Koenraad Audenaert, Nilanjana Datta, and Maris Ozols. Entropy power inequalities for qudits. 2015. arXiv:1503.04213. [Ban93]
Eiichi Bannai. Association schemes and fusion algebras (an introduction). Journal of Algebraic Combinatorics, 2(4):327–344, 1993. doi:10.1023/A:1022489416433.
˚ Ericsson, Jan-Ake ˚ [BBE+ 07] Ingemar Bengtsson, Wojciech Bruzda, Asa Larsson, Wojciech Tadej, ˙ and Karol Zyczkowski. Mutually unbiased bases and Hadamard matrices of order six. Journal of Mathematical Physics, 48(5):052106, 2007. arXiv:quant-ph/0610161, doi: 10.1063/1.2716990. [Chi13]
Andrew M. Childs. Fourier analysis in nonabelian groups. Lecture notes at University of Waterloo, 2013. URL: http://www.cs.umd.edu/~amchilds/teaching/w13/l06.pdf.
22
[LMR14] Seth Lloyd, Masoud Mohseni, and Patrick Rebentrost. Quantum principal component analysis. Nature Physics, 10(9):631–633, 2014. arXiv:1307.0401, doi:10.1038/ nphys3029. [Ser12]
Jean-Pierre Serre. Linear Representations of Finite Groups. Springer, 2012.
˙ [TZ06]
˙ Wojciech Tadej and Karol Zyczkowski. A concise guide to complex Hadamard matrices. Open Systems & Information Dynamics, 13(02):133–177, 2006. arXiv:quant-ph/0512154, doi:10.1007/s11080-006-8220-2.
[WBC15] Christopher J. Wood, Jacob D. Biamonte, and David G. Cory. Tensor networks and graphical calculus for open quantum systems. Quantum Information and Computation, 15(9&10):0759–0811, July 1 2015. URL: http://www.rintonpress.com/journals/ qiconline.html#v15n910, arXiv:1111.6950.
A
Deriving the parametrization from scratch
In this appendix we derive from scratch (namely, without using the irreps of S3 ) the parametrization of ρ obtained in Section 4.4.1. The only assumption that goes into our derivation is that z1 , z2 , z3 are real while z4 , z5 , z6 are imaginary—this is something we observed in Section 4.3, eqs. (83) to (88). This assumption is in fact sufficient for deriving eqs. (91) to (93) and the constraints in eq. (94). Following eq. (90), we choose the coefficients zi as
(z1 , z2 , z3 , z4 , z5 , z6 ) := ( a1 , a2 , a3 , ib1 , ib2 , ib3 )
(136)
for some a1 , a2 , a3 , b1 , b2 , b3 ∈ R. Without any additional constraints, this can potentially capture only more than eqs. (83) to (88). At the same time, zi chosen according to eq. (136) still satisfy the constraints imposed in Proposition 10. In other words, this choice automatically takes care of Q2. Let us rewrite the output state ρ from eq. (44) in terms of the new parameters: ρ = a21 + b12 ρ1 + a22 + b22 ρ2 + a23 + b32 ρ3
(137)
+ a1 b2 − a2 b1 i [ρ2 , ρ1 ] + a2 b3 − a3 b2 i [ρ3 , ρ2 ] + a3 b1 − a1 b3 i [ρ1 , ρ3 ] + a1 a2 + b1 b2 (ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) + a2 a3 + b2 b3 (ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) + a3 a1 + b3 b1 (ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 ).
(138)
(139)
Clearly, this is identical to eqs. (91) to (93) if we let qk := ak + ibk . Let us now work backwards to find what extra constraints should be imposed on the coefficients ai and bi to satisfy the unitarity requirement Q1. Recall from Lemma 5 that we want the following
23
matrix, see eq. (75), to be unitary: a1 a3 a2 ib1 ib2 ib3 a2 a1 a3 ib3 ib1 ib2 6 a3 a2 a1 ib2 ib3 ib1 A iBT ∑ zk Lk = ib1 ib3 ib2 a1 a2 a3 = iB AT =: U k =1 ib2 ib1 ib3 a3 a1 a2 ib3 ib2 ib1 a2 a3 a1
(140)
where
a1 a3 a2 A : = a2 a1 a3 , a3 a2 a1
b1 b3 b2 B := b2 b1 b3 . b3 b2 b1
We can write the unitarity condition as T A iBT AAT + BT B i [ BT , A ] A −iBT † UU = = = I. −iB A iB AT i [ B, AT ] AT A + BBT
(141)
(142)
Note that [ BT , A] = 0 holds automatically, so the remaining constraints follow solely from the diagonal blocks: AAT + BT B = AT A + BBT = I. These constraints are: a21 + a22 + a23 + b12 + b22 + b32 = 1,
(143)
a1 a2 + a2 a3 + a3 a1 + b1 b2 + b2 b3 + b3 b1 = 0.
(144)
It is not hard to see that these constraints are equivalent to eq. (94).
24
arXiv:1508.00860v1 [quant-ph] 4 Aug 2015
Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge, CB3 0WA, U.K. August 4, 2015 Abstract We devise a ternary operation for combining three quantum states: it consists of permuting the input systems in a continuous fashion and then discarding all but one of them. This generalizes a binary operation recently studied by Audenaert et al. [arXiv:1503.04213] in the context of entropy power inequalities. In fact, our ternary operation continuously interpolates between all such nested binary operations. Our construction is based on a unitary version of Cayley’s theorem: we use representation theory to show that any finite group can be naturally embedded in a continuous subgroup of the unitary group. Formally, this amounts to characterizing when a linear combination of certain permutations is unitary.
1
Introduction
A basic result in group theory known as Cayley’s theorem states that every finite group G is isomorphic to a subgroup of the symmetric group. In other words, the elements of G can be faithfully represented by permutation matrices of size | G | × | G |. This gives a natural embedding of G in the symmetric group Sn on n = | G | elements, also known as the regular representation of G. Since permutation matrices are unitary, one might ask whether the resulting subgroup of S|G| can be further extended to a continuous subgroup of the unitary group U(| G |). Such extension would allow to treat the otherwise discrete group G as continuous (e.g., it would allow to perturb the elements of G and continuously interpolate between them in a meaningful sense). To illustrate this, consider the simplest non-trivial example, namely, the finite group Z2 . This group has two elements which according to Cayley’s theorem can be represented by 2 × 2 permuta 1 0 tion matrices as I := 0 1 and X := 01 10 . Interestingly, the following complex linear combination cos α i sin α iϕ iϕ U ( ϕ, α) := e cos α I + i sin α X = e (1) i sin α cos α of I and X is unitary for any ϕ, α ∈ [0, 2π ). Note that U (0, 0) = I and U (3π/2, π/2) = X, so by changing ϕ and α we can continuously interpolate between the two original matrices representing Z2 . In fact, U ( ϕ, α)U ( ϕ0 , α0 ) = U ( ϕ + ϕ0 , α + α0 ), so the matrices U ( ϕ, α) themselves form a group—a two-parameter subgroup of U(2). One of our contributions is to generalize this idea to any finite group G. Namely, we use representation theory to characterize complex linear combinations of matrices from the regular representation of G that are unitary (see Theorem 7). As a consequence, any finite group G can be naturally embedded in a continuous subgroup of the unitary group U(| G |). We also show that 1
this subgroup is isomorphic to a direct sum of smaller unitary groups—one copy of U(dτ ) for each irreducible representation τ of G where dτ is the dimension of the representation τ.
1.1
The partial swap operation
Ability to make discrete groups continuous has an interesting application to quantum information. If we apply our construction to the symmetric group Sn , we get a unitary extension of Sn that continuously interpolates between different permutations. This then can be further extended to permuting n quantum systems in a continuous fashion. To illustrate this, let us focus again on the simplest non-trivial instance, S2 . This group can act on two d-dimensional quantum systems (qudits) either by leaving them alone or by swapping them. In other words, we can represent the elements of S2 by d2 × d2 permutation matrices I and S, where I is the identity matrix and S|i i| ji := | ji|i i for all i, j ∈ {1, . . . , d}. Making this representation continuous would allow to interpolate between these two operations and thus swap two quantum systems in a continuous fashion. This was exactly the starting point of [ADO15]. Following [ADO15], we define the partial swap operation Uλ ∈ U(d2 ) for λ ∈ [0, 1] as the following linear combination of the identity I and the two qudit swap S: √ √ Uλ := λ I + i 1 − λ S (2) This can easily be verified to be unitary. In fact, Uλ is very similar to U ( ϕ, α) in eq. (1), except we ignore the global phase eiϕ and take only one sector of the unit circle corresponding to α ∈ [0, π/2]. Note also that U1 = iS rather than S; however, this global phase mismatch will be unimportant. Given two qudit states ρ1 , ρ2 ∈ D(d), we can combine them using the partial swap as follows: h i ρ1 λ ρ2 := Tr2 Uλ (ρ1 ⊗ ρ2 )Uλ† (3) where λ ∈ [0, 1]. This operation (in the present context) was introduced in [ADO15], however it has independently appeared earlier in [LMR14]. With some tricks (see Section 2.3), eq. (3) can be expanded as q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 +
λ (1 − λ ) i [ ρ2 , ρ1 ].
(4)
Note that ρ1 0 ρ2 = ρ2 and ρ1 1 ρ2 = ρ1 . Also, for commuting states, ρ1 λ ρ2 = λρ1 + (1 − λ)ρ2 is just a convex combination of the two states. This operation obeys some interesting properties—most notably, an analogue of the entropy power inequality [ADO15]. Infull generality, consider a function f : D(d) → R. We say that f is concave if f λρ + (1 − λ)σ ≥ λ f (ρ) + (1 − λ) f (σ) for any λ ∈ [0, 1] and ρ, σ ∈ D(d), and symmetric if f (ρ) depends only on the eigenvalues of ρ and is symmetric in them. Theorem 1 ([ADO15]). Let ρ, σ ∈ D(d), λ ∈ [0, 1], and ρ λ σ be as in eq. (4). If f is concave and symmetric then f ( ρ λ σ ) ≥ λ f ( ρ ) + (1 − λ ) f ( σ ). (5) Motivated by this result, our goal is to generalize the partial swap operation to any number of systems. The rest of this paper is devoted to obtaining such generalization and using it to derive an analogue of eq. (4) for three states. Proving an analogue of Theorem 1 for this generalization is left as an open problem.
2
1.2
A nested generalization
Before we proceed with our general approach, it is instructive first to consider a simple method of combining states in an iterative manner. This will highlight some of the issues that appear when dealing with three or more states, hence providing further motivation for our approach. This will also serve as a reference point to come back to once we know the answer. A straightforward way of generalizing eq. (4) to more systems is by feeding the output state into another operation of the same kind. For example, for any a, a0 ∈ [0, 1], q ρ1 a (ρ2 a0 ρ3 ) = aρ1 + (1 − a)(ρ2 a0 ρ3 ) + a(1 − a) i [ρ2 a0 ρ3 , ρ1 ] (6) q = aρ1 + (1 − a) a0 ρ2 + (1 − a0 )ρ3 + a0 (1 − a0 ) i [ρ3 , ρ2 ] (7) q q h i + a (1 − a ) i a 0 ρ2 + (1 − a 0 ) ρ3 + a 0 (1 − a 0 ) i [ ρ3 , ρ2 ], ρ1 (8)
= aρ1 + (1 − a) a0 ρ2 + (1 − a)(1 − a0 )ρ3 q + (1 − a ) a 0 (1 − a 0 ) i [ ρ3 , ρ2 ] q + a (1 − a ) a 0 i [ ρ2 , ρ1 ] q + a (1 − a ) (1 − a 0 ) i [ ρ3 , ρ1 ] q q + a (1 − a ) a 0 (1 − a 0 ) i [ i [ ρ3 , ρ2 ], ρ1 ].
(9) (10) (11) (12) (13)
It then follows trivially from Theorem 1 that f ρ1 a ( ρ2 a 0 ρ3 ) ≥ a f ( ρ1 ) + (1 − a ) f ρ2 a 0 ρ3 0
(14) 0
≥ a f (ρ1 ) + (1 − a) a f (ρ2 ) + (1 − a)(1 − a ) f (ρ3 )
(15)
for any symmetric and concave function f : D(d) → R. However, there are other nested ways of combining three states, such as
(ρ1 b ρ2 ) b0 ρ3 = b0 bρ1 + b0 (1 − b)ρ2 + (1 − b0 )ρ3 + · · ·
(16)
for any b, b0 ∈ [0, 1]. The omitted terms are similar to eqs. (10) to (13), except for the double commutator which is i [ρ3 , i [ρ2 , ρ1 ]] in this case, thus making the expression different from the one above. Since there are several other nested ways of combining three states, it is not clear which of them, if any, should be preferred. Considering this, it is more desirable to have a ternary operation that treats each of the input states in the same way and combines them all at once. Note that the coefficients a, (1 − a) a0 , (1 − a)(1 − a0 ) and b0 b, b0 (1 − b), 1 − b0 (17) in front of the states ρ1 , ρ2 , ρ3 in eqs. (9) and (16) are probability distributions. It is thus natural to demand the combined state to be of the form p1 ρ1 + p2 ρ2 + p3 ρ3 + · · ·
(18)
for some probability distribution ( p1 , p2 , p3 ), where, unlike in (13), all three states are treated symmetrically in the omitted higher order terms (in particular, the third order). 3
1.3
The most general way of combining two states
Before we attempt to generalize eq. (4), it is worthwhile first checking whether Uλ in eq. (2) is actually the most general unitary matrix that can be expressed as a linear combination of I and S. As it turns out, up to an overall global phase and the sign of i, this is indeed the case. Proposition 2. Let z1√ , z2 ∈ C and U := √ z1 I + z2 S where S swaps two qudits (d ≥ 2). Then U is unitary if and only if z1 = eiϕ λ and z2 = ±ieiϕ 1 − λ for some λ ∈ [0, 1] and ϕ ∈ [0, 2π ). In other words, √ √ (19) U = eiϕ λ I ± i 1 − λ S . Proof. If U := z1 I + z2 S for some z1 , z2 ∈ C then UU † = (z1 z¯1 + z2 z¯2 ) I + (z1 z¯2 + z¯1 z2 )S.
(20)
Since I and S are linearly independent and we want UU † = I, we get (21) | z1 |2 + | z2 |2 = 1 and z1 z¯2 = −z1 z¯2 . √ √ From the first equation, z1 = eiϕ1 λ and z2 = eiϕ2 1 − λ for some λ ∈ [0, 1] and ϕ1 , ϕ2 ∈ [0, 2π ). The second equation says that z1 z¯2 is purely imaginary. Thus ei( ϕ1 − ϕ2 ) = ±i, which is equivalent to eiϕ2 = ±ieiϕ1 . In other words, we can take eiϕ1 := eiϕ and eiϕ2 := ±ieiϕ for some ϕ ∈ [0, 2π ). Since the map in eq. (3) involves conjugation, the global phase of U in eq. (19) is irrelevant and we only have the freedom of choosing the sign in front of i. Up to the sign, eq. (2) indeed provides the most general unitary for our purpose. To account for the two possible signs, we define q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 + λ(1 − λ) i [ρ2 , ρ1 ], (22) q ρ1 λ ρ2 := λρ1 + (1 − λ)ρ2 − λ(1 − λ) i [ρ2 , ρ1 ]. (23) These two operations are related as follows: ρ1 λ ρ2 = ρ2 1−λ ρ1 .
2
(24)
Generalization to three states
Now that we fully understand the case of two states, let us investigate how to combine three states. We first look in more detail at the nested combinations suggested in Section 1.2, and then consider a more general ternary operation that combines all three states at once.
2.1
Twelve nested ways of combining three states
We can use eqs. (22) and (23) to work out all possible nested ways of combining three states. Because of the relation (24), there are 12 different nested ways of combining three states: ρ1 a ( ρ2 a0 ρ3 )
(25)
ρ 2 b ( ρ 3 b 0 ρ 1 )
(26)
ρ 3 c ( ρ 1 c 0 ρ 2 )
(27)
4
where each can independently be replaced by either or . In general, all twelve expressions will be different. However, if we choose the parameters so that ( p1 , p2 , p3 ) = a, a0 (1 − a), (1 − a0 )(1 − a) (28) 0 0 = (1 − b )(1 − b), b, b (1 − b) (29) 0 0 = c (1 − c), (1 − c )(1 − c), c (30) for some probability distribution ( p1 , p2 , p3 ), then all twelve expressions will begin with the same convex combination p1 ρ1 + p2 ρ2 + p3 ρ3 + · · · . Our goal is to obtain a more general ternary operation that continuously interpolates between these twelve expressions.
2.2
Combining n states at once
Recall from eq. (2) that Uλ is a linear combination of two-qudit permutation matrices. One way of generalizing this to more systems is as follows: h i (ρ1 , . . . , ρn ) 7→ Tr2,...,n U (ρ1 ⊗ · · · ⊗ ρn )U † , (31) where ρ1 , . . . , ρn ∈ D(d) are n qudit states and the unitary matrix U ∈ U(dn ) is a linear combination of all n! permutations that act on n qudits. The goal of this paper is to develop a better understanding of this special type of unitaries and the corresponding map resulting from eq. (31).
2.3
Graphical notation for evaluating partial traces
Let us work out the details of eq. (31) for n = 3. Let ρ1 , ρ2 , ρ3 ∈ D(d) and define ρ := Tr2,3 U (ρ1 ⊗ ρ2 ⊗ ρ3 )U †
(32)
for some unitary matrix U ∈ U(d3 ) such that 6
U :=
∑ zi Qi ,
(33)
i =1
where zi ∈ C and each Qi ∈ U(d3 ) is one of the six permutations that act on three qudits. We can represent them using the following graphical notation (see [WBC15] for more details): Q1 : =
Q2 : =
Q3 : =
Q4 : =
Q5 : =
Q6 : =
(34)
Note that one can easily recover the matrix representation of each Qi from this pictorial notation. For example, 2
1
Q2 |ψ1 i ⊗ |ψ2 i ⊗ |ψ3 i =
2
=
3
3
= |ψ2 i ⊗ |ψ3 i ⊗ |ψ1 i
(35)
1
for any |ψ1 i, |ψ2 i, |ψ3 i ∈ Cd , which is enough to determine all d3 × d3 entries of the matrix Q2 . The diagrams of matrices Qi can be composed in a natural way, and this operation is compatible with the matrix multiplication (in fact, the diagrams as well as the matrices form a group). For example, Q4 Q5 =
= 5
= Q2 .
(36)
Matrix inverses can be found by reversing the diagram: †
Q2† =
=
= Q3 .
(37)
Except for Q2 and Q3 which are inverses of each other, all other Qi s are self-inverse, i.e., Qi† = Qi when i ∈ / {2, 3}. Just like in eq. (35), subsystems of a mixed product state are permuted under conjugation by Qi . For example, 2
1
Q2 (ρ1 ⊗ ρ2 ⊗ ρ3 ) Q2† =
2 3
=
3
= ρ2 ⊗ ρ3 ⊗ ρ1 .
(38)
1
Using this pictorial representation, we would like to expand U on both sides of eq. (32) and obtain an explicit formula for the output state ρ in terms of the input states ρk and coefficients zi . This requires evaluating ξ i,j := Tr2,3 zi Qi (ρ1 ⊗ ρ2 ⊗ ρ3 )z¯ j QTj (39) for each combination of i, j ∈ {1, . . . , 6}. This daunting task becomes much more straightforward using the graphical notation for the partial trace [WBC15]. For example, " # 1 1
ξ 1,1 = z1 z¯1 Tr2,3
= z1 z¯1
2
3
3
" ξ 4,1 = z4 z¯1 Tr2,3
1
= z4 z¯1
ξ 5,1 = z5 z¯1 Tr2,3
= z5 z¯1
ξ 2,1 = z2 z¯1 Tr2,3
1 2 3
1 3
3
3
"
2
= z5 z¯1
2
3
2
1
#
2
3
1
= z4 z¯1
2 3
3 1
1
2
1
#
2
"
= z1 z¯1
2
= z1 z¯1 ρ1 Tr ρ2 Tr ρ3 ,
(40)
= z4 z¯1 ρ1 Tr(ρ2 ρ3 ),
(41)
= z5 z¯1 ρ2 ρ1 Tr ρ3 ,
(42)
= z2 z¯1 ρ2 ρ3 ρ1 .
(43)
1
#
= z2 z¯1
= z2 z¯1
2
2
3
1
3
The remaining cases are similar and are summarized in Fig. 1. Since ρ = ∑6i,j=1 ξ i,j , we can find the resulting state by putting all 36 terms together: 2 2 ρ = |z1 | + |z4 | + 2 Re(z1 z¯4 ) Tr(ρ2 ρ3 ) ρ1 + |z2 |2 + |z5 |2 + 2 Re(z2 z¯5 ) Tr(ρ3 ρ1 ) ρ2 + |z3 |2 + |z6 |2 + 2 Re(z3 z¯6 ) Tr(ρ1 ρ2 ) ρ3 + z1 z¯5 + z4 z¯2 ρ1 ρ2 + c.t. + z2 z¯6 + z5 z¯3 ρ2 ρ3 + c.t. (44) + z3 z¯4 + z6 z¯1 ρ3 ρ1 + c.t. + z2 z¯1 + z5 z¯4 ρ2 ρ3 ρ1 + c.t. + z3 z¯2 + z6 z¯5 ρ3 ρ1 ρ2 + c.t. + z1 z¯3 + z4 z¯6 ρ1 ρ2 ρ3 + c.t. where c.t. stands for the conjugate transpose of the previous term. 6
1 2 3 1 3
2 1
1
2
2
3
3
3
2
3
1
2 3
1
1
1
1
2
2
2
3
3
3
2
1
3
2
1
3 2
1
1
1
1
1
2
2
2
2
3
3
3
3
1
2
1 3
3
1
1
3
2
1
1
1
1
1
2
2
2
2
2
3
3
3
3
3
2
2
2
1 3
3
3
2 1
3
2
2
1
3
1
2 3
1
1
1
1
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
3
1 2
3
3
2 1
1
3
2
2
1
3
1
2
3 1
2
Figure 1: Tensor contraction diagrams for computing ξ i,j := Tr2,3 Qi (ρ1 ⊗ ρ2 ⊗ ρ3 ) QTj for each pair of three qudit permutations Qi and QTj (i ≥ j). By combining all 36 terms we obtain eq. (44).
7
2.4
Extra constraints
Before we can call eq. (44) a generalization of eq. (4), we need to answer the following two questions: Q1. Under what constraints on coefficients zi ∈ C is the matrix U in eq. (33) unitary? Q2. Are additional constraints needed to make the Tr(ρi ρ j ) terms in eq. (44) go away? We demand unitarity since we want ρ to be a valid quantum state—this is very natural considering eq. (32). Note that for n = 2 the answer to Q1 is already provided by Proposition 2. Using representation theory, we will answer Q1 in full generality (in Theorem 9 we characterize when U := ∑π ∈Sn zπ Qπ is unitary for any n ≥ 1). In addition to constraints from Q1, we would also like that Re(z1 z¯4 ) = Re(z2 z¯5 ) = Re(z3 z¯6 ) = 0.
(45)
Then the first order terms in eq. (44) would form a linear (in fact, convex) combination of ρi with coefficients depending only on zi and not the states themselves. We will be able to answer Q2 for n = 3 where the answer is “yes” (see Proposition 10 for more details). In fact, we will see in Section 4.3 that constraint (45) can be imposed in a fairly natural way and it leads to some further nice structure in eq. (44). Note that for n = 2 the answer to Q2 is trivially “no”, since no terms of the form Tr(ρi ρ j ) appear.
3
When is a linear combination of permutations unitary?
We begin by first answering Q1 which will require some representation theory. Particularly relevant to us is reference [Chi13] that provides a very concise introduction to representation theory and Fourier analysis of non-abelian groups. For more background on representation theory of finite groups, see the standard reference [Ser12].
3.1
Background on representation theory
Let U(d) denote the set of all d × d unitary matrices. A d-dimensional representation of a finite group G is a map τ : G → U(d) such that τ ( gh) = τ ( g)τ (h) for all g, h ∈ G. We will always use e to denote the identity element of G. Note that τ (e) = Id , the d × d identity matrix. We call d the dimension of representation τ. Let Sn denote the symmetric group consisting of all n! permutations acting on n elements. We write π (i ) = j to mean that permutation π ∈ Sn maps i to j, and we write π −1 to denote the inverse permutation of π. The following are four different representations of the symmetric group—two of them (Qπ and Lπ ) will play an important role later in this paper. Example (Representations of Sn ). The symmetric group Sn can be represented by permutation matrices in several different ways. In each case we write down how the matrix associated to permutation π ∈ Sn acts in the standard basis: • natural representation Pπ ∈ U(n): Pπ : |i i 7→ |π (i )i,
∀i ∈ {1, . . . , n};
(46)
• tensor representation Qπ ∈ U(dn ): Q π : | i 1 i ⊗ · · · ⊗ | i n i 7 → | i π −1 (1 ) i ⊗ · · · ⊗ | i π −1 ( n ) i , 8
∀i1 , . . . , in ∈ {1, . . . , d};
(47)
• left regular and right regular representations Lπ , Rπ ∈ U(|Sn |): Lπ : |σi 7→ |πσ i,
∀ σ ∈ Sn ,
(48)
Rπ : |σi 7→ |σπ −1 i,
∀ σ ∈ Sn .
(49)
We call d the local dimension of the tensor representation (we will typically required that d ≥ n). Note that for the regular representations, the standard basis of the underlying space is labeled by permutations themselves, so the space has n! dimensions: CSn ∼ = Cn! . Finally, note that the regular representations can be defined for any finite group G in a similar manner. Since Qπ will play an important role later, let us verify that it is indeed1 a representation of Sn . First, note that Qπ
n O
|ψi i =
i =1
n O
|ψπ −1 (i) i =
i =1
n O
|φi i
(50)
i =1
where |φi i := |ψπ −1 (i) i. Following the same rule, we see that Qσ
n O
|φi i =
i =1
n O
|φσ−1 (i) i =
i =1
n O
|ψπ −1 (σ−1 (i)) i =
i =1
n O
|ψ(σπ )−1 (i) i.
(51)
i =1
In other words, Qσ Qπ acts in exactly the same way as Qσπ .
3.2
From permutations to unitaries—importance of the left regular representation
Consider the following linear combination: U :=
∑
zπ Qπ ,
(52)
π ∈Sn
where zπ ∈ C and Qπ are the matrices from the tensor representation of Sn . Can we parametrize in some simple way all coefficient tuples z := (zπ : π ∈ Sn ) ∈ CSn such that U is unitary for all local dimensions d ≥ 1 simultaneously? We will provide a nice answer to this question using representation theory. Let us start by first reducing this problem from the tensor representation Qπ to the left regular representation Lπ . As a first step, we need to show that the matrices Lπ are linearly independent (this holds more generally, i.e., not just for Sn but for any finite group G). Proposition 3. For any finite group G, the matrices { L g : g ∈ G } have disjoint supports and thus are linearly independent. Proof. For any x, y, g ∈ G: (
h x | L g |yi = h x | gyi =
1 0
if g = xy−1 , otherwise.
(53)
Thus, for any fixed x and y there is exactly one matrix, namely L xy−1 , with a non-zero entry in row x and column y. Hence, the matrices L g have disjoint supports and thus are linearly independent. 1 It
would not be a representation if we would use π instead of π −1 on the RHS of eq. (47).
9
Remark. The matrices L g form a (non-commutative) association scheme [Ban93]. Their linear span is known as the Bose–Mesner algebra of this scheme. Our Theorem 7 (see below) thus characterizes all unitary matrices within this particular algebra. Example (G = S3 ). Because of Proposition 3, any linear combination of the matrices L g has a particularly nice structure. For example, if G = S3 and we order the permutations according to eq. (34) then 1 3 2 4 5 6 2 1 3 6 4 5 6 3 2 1 5 6 4 (54) ∑ kLk = 4 6 5 1 2 3 . k =1 5 4 6 3 1 2 6 5 4 2 3 1 One can easily read off the matrix representation of each Li from this. Let us now show that the matrices Qπ are also linearly independent when d is sufficiently large. Lemma 4. If d ≥ n then Qπ ∼ = Lπ ⊕ τ (π ) and thus the matrices { Qπ : π ∈ Sn } are linearly independent. Proof. It suffices to show that Qπ are linearly independent even when restricted to some invariant N Nn n −1 subspace of Cd . Let |Ψσ i := Qσ in=1 |i i = 1 | σ (i )i (we needNd ≥ n for this to make Nn Ni= n − 1 sense). Note that Qπ |Ψσ i = Qπ i=1 |σ (i )i = i=1 |σ−1 (π −1 (i ))i = in=1 |(πσ)−1 (i )i = |Ψπσ i, n so L := span{|Ψπ i : π ∈ Sn } is an invariant subspace of Cd under the action of the tensor representation. In fact, the vectors |Ψπ i form an orthonormal basis of the subspace L, and the tensor representation Qπ acts as the left regular representation Lπ in this subspace. In other words, Qπ ∼ = Lπ ⊕ τ (π )
(55)
for some representation τ acting on the orthogonal complement of L. According to Proposition 3, the matrices { Lπ : π ∈ Sn } are linearly independent so the result follows. Lemma 5. Let zπ ∈ C for each π ∈ Sn . If d ≥ n then ∑π ∈Sn zπ Qπ is unitary if and only if ∑π ∈Sn zπ Lπ is unitary (the reverse implication holds for any d ≥ 1). n
Proof. As we noted in the proof of Lemma 4, for d ≥ n there is a subspace L of Cd where Qπ acts as Lπ , see eq. (55); this immediately gives the forward implication. For the reverse implication, note that we can decompose any representation of a finite group into a direct sum of its irreducible representations or irreps [Ser12]. If we obtain such decomposition for the left regular representation, a standard result from representation theory says that each irrep will appear at least once in this decomposition [Ser12]. Thus, the unitarity of ∑π ∈Sn zπ Lπ is equivalent to the simultaneous unitarity of ∑π ∈Sn zπ τ (π ) for all irreps τ of Sn . Since for any d ≥ 1 the tensor representation Qπ can also be decomposed as a direct sum of irreps, we conclude that ∑π ∈Sn zπ Qπ must therefore be unitary. According to Lemma 5, our problem now reduces to analyzing the left regular representation of Sn and characterizing when U := ∑π ∈Sn zπ Lπ is unitary. We will analyze this in the Fourier basis where the left regular representation decomposes as a direct sum of irreducible representations.
10
3.3
Fourier transform over finite groups
Recall from [Chi13] that the Fourier transform over a finite group G is the unitary matrix s d τ dτ τ ( g) j,k |(τ, j, k )ih g|, F := ∑ ∑ | G | j,k∑ ˆ g∈ G =1
(56)
τ ∈G
where Gˆ denotes the set of irreducible representations of G, dτ is the dimension of irrep τ, and τ ( g) j,k are the matrix elements of τ ( g). The output space of F is labeled by triples (τ, j, k ) and has the same dimension as the input space—indeed, it is a standard result in representation theory that
∑ d2τ = |G|.
(57)
τ ∈ Gˆ
One can easily check that F is unitary using the orthogonality of characters [Chi13]. We will need the following standard result from representation theory [Chi13, Ser12]. Let Lˆ g := FL g F † denote L g in the Fourier basis. Then Lˆ g =
M
τ ( g) ⊗ Idτ
(58)
τ ∈ Gˆ
where Idτ is the dτ × dτ identity matrix. In other words, Lˆ g is block-diagonal and contains each irrep τ the number of times equal to its dimension dτ .
3.4
Characterization of unitary linear combinations
Let us first show a simple preliminary fact. Recall from Proposition 3 that { L g : g ∈ G } is a linearly independent set. In fact, when properly normalized, these matrices are orthonormal. p Proposition 6. For any finite group G, the matrices { L g / | G | : g ∈ G } are orthonormal with respect to the Hilbert–Schmidt inner product h A, Bi := Tr( A† B). Proof. First, note from eq. (53) that for any g ∈ G, Tr L g =
∑ hh| Lg |hi = ∑ δg,e = |G|δg,e .
h∈ G
(59)
h∈ G
Thus, for any a, b ∈ G, the corresponding Hilbert–Schmidt inner product is 1 1 1 h L a , Lb i = Tr L†a Lb = Tr L a−1 b = δa,b . |G| |G| |G| p Hence the normalized matrices L g / | G | are orthonormal.
(60)
Theorem 7. Let G be a finite group, L g be its left regular representation, and z g ∈ C for each g ∈ G. Then ∑ g∈G z g L g is unitary if and only if zg =
∑
τ ∈ Gˆ
dτ Tr τ ( g)† Uτ |G|
for some choice of Uτ ∈ U(dτ ), one for each irrep τ of G. 11
Proof. Clearly, ∑ g∈G z g L g is unitary if and only if ∑ g∈G z g Lˆ g is unitary where M Lˆ g = FL g F † = τ ( g) ⊗ Idτ
(61)
τ ∈ Gˆ
according to eq. (58). We prefer to work in the Fourier basis, since then all Lˆ g become simultaneously block diagonal and we can write M ˆ (62) ∑ zg Lg = ∑ zg τ ( g) ⊗ Idτ . g∈ G
g∈ G
τ ∈ Gˆ
This matrix is unitary if and only if each of its blocks is unitary, i.e., M Uτ ⊗ Idτ =: U ∑ zg Lˆ g = g∈ G
(63)
τ ∈ Gˆ
p for some set of unitaries Uτ ∈ U(dτ ). Since B := { L g / | G | : g ∈ G } is an orthonormal set, see p ˆ = | G | = ∑ ˆ d2τ Proposition 6, so is Bˆ := { FBF † : B ∈ B} = { Lˆ g / | G | : g ∈ G }. Since |B| τ ∈G according to eq. (57), Bˆ is in fact an orthonormal basis for the set of all block matrices that have the same block structure as U in eq. (63). Thus, we can obtain the coefficients z g in the expansion U p = |G|
Lˆ g
∑ z g p| G |
(64)
g∈ G
simply by projecting on the corresponding basis vector: ! ˆ i Mh Lg U 1 zg = p (τ ( g)† Uτ ) ⊗ Idτ ,p = Tr = |G| |G| |G| ˆ τ ∈G
∑
τ ∈ Gˆ
dτ Tr(τ ( g)† Uτ ). |G|
(65)
where we substituted eqs. (61) and (63). The reverse implication follows by applying all steps in reverse order. As a byproduct of our proof, we observe that all unitary linear combinations ∑ g∈G z g L g form a L group of their own. Indeed, we see from eq. (63) that this group is isomorphic to τ ∈Gˆ U(dτ ) and it contains G as a subgroup (as represented by the matrices L g ). Indeed, if we take any g ∈ G and set Uτ := τ ( g) for all irreps τ, then z g = 1 while all other coefficients vanish according to Proposition 6, hence reducing the linear combination to L g . Considering this, we can intuitively think of n o (66) ∑ zg Lg ∈ U(|G|) : zg ∈ C g∈ G
as a natural continuous extension of the discrete finite group G. Corollary 8 (Unitary version of Cayley’s theorem). Every finite group G can be naturally extended to a continuous subgroup of the unitary group U(| G |). If we specialize Theorem 7 to G = Sn and apply Lemma 5, we get the following result. Theorem 9. Let zπ ∈ C for each π ∈ Sn . If d ≥ n then ∑π ∈Sn zπ Qπ is unitary if and only if zπ =
∑
τ ∈Sˆ n
dτ Tr τ (π )† Uτ n!
(67)
for some choice of Uτ ∈ U(dτ ), one for each irrep τ of Sn (the reverse implication holds for any d ≥ 1). 12
π τ1 (π ) τ2 (π )
1 1
1 0 0 1
1 0 0 1
τ3 (π ) τ30 (π )
1 1
1 2
√ − 1 − 3 √ 3 −1 ω 0 0 ω2
1 1 −√1 1 2 − 3 2 ω 0
1 √ −1 3 1 0 0 −1 −1 0 0 1 ω 1 0
1 −1 √ −√1 − 3 1 2 − 3 1 0 ω2 ω 0
1 −1 √ −1 3 1 √ 2 3 1 0 ω ω2 0
Table 1: All irreducible representations of S3 . The matrix elements of τ3 are not unique but depend on the choice of basis—we provide two simple choices (irreps τ3 and τ30 are isomorphic).
4
How to combine three quantum states?
We will now use Theorem 9 to parametrize all tuples of complex coefficients (zπ : π ∈ S3 ) such that ∑π ∈S3 zπ Qπ is a unitary matrix for all d ≥ 1. Recall that Qπ permutes three qudits according to permutation π, see eq. (34). First, we need to work out all irreps of S3 .
4.1
Irreducible representations of S3
The symmetric group S3 has three irreducible representations [Ser12]: two 1-dimensional representations (the trivial representation τ1 and the sign representation τ2 ) and a 2-dimensional representation τ3 . Recall that S3 ∼ = D3 (the dihedral group), so geometrically τ3 corresponds to rotations and reflections in 2D that preserve an equilateral triangle centered at the origin and pointing along the x axis. These representations are written out explicitly in Table 1. One can easily verify that τk (πσ) = τk (π )τk (σ) for all π, σ ∈ S3 and k ∈ {1, 2, 3}. For example, √ √ 1 −1 − 3 1 −1 − 3 1 0 √ √ τ3 = τ3 τ3 = · . = = τ3 0 −1 2 − 3 2 1 3 −1
4.2
General solution for S3
According to Theorem 9, we need to assign one unitary matrix Uk ∈ U(dτk ) to each irrep τk of S3 . Since the global phase of ∑6i=1 zi Qi has no effect, we can without loss of generality assume one of the unitaries Uk to be in the special unitary group. We take U1 , U2 ∈ U(1) and U3 ∈ SU(2), and parametrize these unitaries as follows: a c iϕ iϕ 2 1 : : : U1 = e , U2 = e , U3 = (68) −c¯ a¯
13
2
2
where ϕ1 , ϕ2 ∈ [0, 2π ) and a, c ∈ C are such that | a| + |c| = 1. We can now use eq. (67) and irreps τ1 , τ2 , τ3 from Table 1 to compute the coefficients z1 , . . . , z6 : z1 = z2 = z3 = z4 = z5 = z6 =
1 iϕ1 e + eiϕ2 + 4 Re( a) , 6 √ 1 iϕ1 e + eiϕ2 − 2 Re( a + 3c) , 6 √ 1 iϕ1 e + eiϕ2 − 2 Re( a − 3c) , 6 1 iϕ1 e − eiϕ2 + 4i Im( a) , 6 √ 1 iϕ1 e − eiϕ2 − 2i Im( a + 3c) , 6 √ 1 iϕ1 e − eiϕ2 − 2i Im( a − 3c) . 6
(69) (70) (71) (72) (73) (74)
Up to an overall global phase, this parametrizes precisely the set of coefficients for which the following matrix, cf. eq. (54), is unitary:
z1 z2 6 z3 ∑ zi Li = z4 i =1 z5 z6
z3 z1 z2 z6 z4 z5
z2 z3 z1 z5 z6 z4
z4 z6 z5 z1 z3 z2
z5 z4 z6 z2 z1 z3
z6 z5 z4 . z3 z2
(75)
z1
√ Remark. If we insist, in addition to eqs. (69) to (74), that |z1 | = · · · = |z6 | = 1/ 6, we get a 6 × 6 ˙ flat unitary—this is also known as a complex Hadamard matrix [TZ06]. Such matrices are relevant + to the MUB problem in six dimensions [BBE 07]. Using computer, we can find 72 discrete flat solutions (z1 , . . . , z6 ). Unfortunately, the corresponding unitaries appear to be equivalent to the 6 × 6 Fourier matrix. Nevertheless, this method can in principle be used to find flat unitaries of size | G | × | G | for any finite group G. It would be interesting to know whether this construction can ˙ yield anything beyond what is already known [TZ06].
4.3
Imposing extra constraints
Recall that eq. (44) involves terms with coefficients Re(z1 z¯4 ), Re(z2 z¯5 ), and Re(z3 z¯6 ), and we would like to understand when these terms vanish (see Q2 in Section 2.3). Proposition 10. Re(z1 z¯4 ) = Re(z2 z¯5 ) = Re(z3 z¯6 ) = 0 if ϕ1 = − ϕ2 . Proof. Note from eqs. (69) to (74) that (up to an overall constant) each coefficient zi has one of the following two forms: u := eiϕ1 + eiϕ2 + r cos α, v := e
iϕ1
−e
iϕ2
+ ir sin α
(76) (77)
for some ϕ1 , ϕ2 , α ∈ [0, 2π ) and r ≥ 0. Furthermore, each u-type coefficient is paired up with a
14
corresponding v-type coefficient. A straightforward calculation gives Re(uv¯ ) = Re(u) Re(v) + Im(u) Im(v)
(78)
= (cos ϕ1 + cos ϕ2 + r cos α)(cos ϕ1 − cos ϕ2 ) + (sin ϕ1 + sin ϕ2 )(sin ϕ1 − sin ϕ2 + r sin α)
(79)
= (cos ϕ1 )2 − (cos ϕ2 )2 + (cos ϕ1 − cos ϕ2 )r cos α
(80)
+ (sin ϕ1 )2 − (sin ϕ2 )2 + (sin ϕ1 + sin ϕ2 )r sin α = (cos ϕ1 − cos ϕ2 )r cos α + (sin ϕ1 + sin ϕ2 )r sin α = r cos(α − ϕ1 ) − cos(α + ϕ2 ) .
(81) (82)
We can guarantee that Re(uv¯ ) = 0 irrespectively of the values of r and α by choosing − ϕ1 = ϕ2 . This makes all three terms vanish simultaneously. Taking Proposition 10 into account, we define ϕ := ϕ1 = − ϕ2 . Then eqs. (69) to (74) become: z1 = z2 = z3 = z4 = z5 = z6 =
1 cos ϕ + 2 Re( a) , 3 √ 1 cos ϕ − Re( a + 3c) , 3 √ 1 cos ϕ − Re( a − 3c) , 3 i sin ϕ + 2 Im( a) , 3 √ i sin ϕ − Im( a + 3c) , 3 √ i sin ϕ − Im( a − 3c) . 3 2
(83) (84) (85) (86) (87) (88) 2
Here we can choose any ϕ ∈ [0, 2π ) and a, c ∈ C such that | a| + |c| = 1, so in total we have four degrees of freedom. The output state can be obtained by substituting eqs. (83) to (88) in eq. (44).
4.4
Alternative parametrizations
Unfortunately the parametrization in eqs. (83) to (88) is somewhat cumbersome. In this section we derive two alternative parametrizations that are much simpler and more insightful. We will derive them from eqs. (83) to (88). However, with an educated guess, they can also be derived from scratch without invoking the irreps of S3 (see Appendix A). 4.4.1
Parametrization by C3
It is natural to pair up the coefficients zi as follows: 1 iϕ (e + 2a), 3 √ 1 q2 := z2 + z5 = (eiϕ − a − 3c), 3 √ 1 q3 := z3 + z6 = (eiϕ − a + 3c). 3 q1 : = z1 + z4 =
15
(89)
The coefficients zi in terms of the new parameters q1 , q2 , q3 ∈ C are expressed as follows:
(z1 , z2 , z3 , z4 , z5 , z6 ) := (Re q1 , Re q2 , Re q3 , i Im q1 , i Im q2 , i Im q3 ).
(90)
With this parametrization, the output state ρ from eq. (44) looks as follows: ρ = | q1 |2 ρ1 + | q2 |2 ρ2 + | q3 |2 ρ3
(91)
+ Im(q1 q¯2 ) i [ρ1 , ρ2 ] + Im(q2 q¯3 ) i [ρ2 , ρ3 ] + Im(q3 q¯1 ) i [ρ3 , ρ1 ]
(92)
+ Re(q1 q¯2 )(ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) + Re(q2 q¯3 )(ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) + Re(q3 q¯1 )(ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 )
(93)
where q1 , q2 , q3 ∈ C are subject to the following constraints:
|q1 |2 + |q2 |2 + |q3 |2 = 1,
|q1 + q2 + q3 |2 = 1.
To derive these constraints, we solve eq. (89) for the original parameters: √ 1 3 iϕ e = q1 + q2 + q3 , a = q1 − ( q2 + q3 ), c= ( q3 − q2 ). 2 2
(94)
(95)
From the first equation we immediately get the second constraint: 1 = |q1 + q2 + q3 |2 = |q1 |2 + |q2 |2 + |q3 |2 + 2 Re(q1 q¯2 ) + 2 Re(q2 q¯3 ) + 2 Re(q3 q¯1 ).
(96)
From the next two equations we get 1 1 |q2 |2 + |q3 |2 − Re(q1 q¯2 ) − Re(q3 q¯1 ) + Re(q2 q¯3 ), 4 2 3 3 |c|2 = |q2 |2 + |q3 |2 − Re(q2 q¯3 ). 4 2
| a |2 = | q1 |2 +
(97) (98)
Putting these together gives another constraint: 1 = | a|2 + |c|2 = |q1 |2 + |q2 |2 + |q3 |2 − Re(q1 q¯2 ) − Re(q2 q¯3 ) − Re(q3 q¯1 ).
(99)
From eqs. (96) and (99) we conclude that Re(q1 q¯2 ) + Re(q2 q¯3 ) + Re(q3 q¯1 ) = 0.
(100)
Consequently, constraints (96) and (99) are equivalent to eq. (94). These constraints are in fact also equivalent to the original constraints guaranteeing unitarity of the matrices in eq. (68). Indeed, if we substitute eq. (89) in eq. (94), we recover the original constraints |eiϕ |2 = 1 and | a|2 + |c|2 = 1. Another way of writing constraints (94) is as follows. If we let q1 1 1 | q i : = q2 , | u i : = √ 1 (101) 3 1 q3 16
then eq. (94) is equivalent to
|hq|qi|2 = 1,
|hq|ui|2 = 1/3,
(102)
i.e., |qi ∈ C3 is a unit vector that is mutually unbiased to |ui. One advantage of the new parametrization is that it makes it more apparent that neither the constraints (102) nor the expression of ρ in eqs. (91) to (93) depend on the global phase of |qi; this feature was not so obvious from the original parametrization in eqs. (83) to (88). We can make two further observations: • Due to the first constraint in eq. (94), the first order terms in eq. (91) form a convex combination of ρ1 , ρ2 , ρ3 . This is analogous to eq. (4) for n = 2. • According to eq. (100), the coefficients of the third order terms (93) sum to zero. This allows writing the third order terms as a linear combination of double commutators (more details on this are provided in Section 5.1). 4.4.2
Parametrization by a probability distribution and phases
As we just noted, (|q1 |2 , |q2 |2 , |q3 |2 ) is a probability distribution according to eq. (94). We can highlight this aspect by choosing √ qk := eiφk pk (103) for some probability distribution ( p1 , p2 , p3 ) and some phases φ1 , φ2 , φ3 ∈ [0, 2π ). Note that p Re(qi q¯ j ) = pi p j cos(φi − φj ), (104) p Im(qi q¯ j ) = pi p j sin(φi − φj ). (105) If we further denote the differences between consecutive phases by δ12 := φ1 − φ2 , δ23 := φ2 − φ3 , δ31 := φ3 − φ1 ,
(106) (107) (108)
we can rewrite eqs. (91) to (93) as follows: ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 √ + p1 p2 sin δ12 i [ρ1 , ρ2 ] √ + p2 p3 sin δ23 i [ρ2 , ρ3 ] √ + p3 p1 sin δ31 i [ρ3 , ρ1 ] √ + p1 p2 cos δ12 (ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) √ + p2 p3 cos δ23 (ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) √ + p3 p1 cos δ31 (ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 ).
(109)
(110)
(111)
Parameters pi and δij are subject to the following constraints. For any distribution ( p1 , p2 , p3 ), the angles δij must satisfy δ12 + δ23 + δ31 = 0,
√
p1 p2 cos δ12 + 17
√
p2 p3 cos δ23 +
√
p3 p1 cos δ31 = 0.
(112)
The first constraint is apparent from eqs. (106) to (108), while the second constraint is equivalent to eq. (100). Except for deterministic distributions, these constraints yield a one-parameter family of angles (δ12 , δ23 , δ31 ). This is qualitatively different from the n = 2 case, where the coefficients of the first order terms are completely determined by λ, see eq. (4). Once λ is fixed, only a discrete degree of freedom remains corresponding to the sign in front of the commutator, see Section 1.3.
5
Further observations
In this section we highlight some further features of our operation. In particular, we show that the third order terms can be expressed as a linear combination of double commutators. We also show that the twelve nested compositions discussed in Section 2.1 are special cases of our operation.
5.1
Double commutators
Let us elaborate more on the meaning of the constraint (100). As mentioned earlier, it has to do with double commutators, i.e., expressions of the form [1, [2, 3]].2 In what follows, we do not specify the states ρi but rather treat them as abstract non-commutative variables. With this convention, for example, ρ1 and ρ2 are always considered to be linearly independent. Note that there are 6 ways of ordering three states and 2 ways of putting brackets, so there are twelve double commutators in total. However, many of them are identical (such as [1, [2, 3]] and [[3, 2], 1]) or differ only by a sign (such as [1, [2, 3]] and [1, [3, 2]]). Furthermore, we know from the Jacobi identity that [1, [2, 3]] + [2, [3, 1]] + [3, [1, 2]] = 0. (113) This leaves us with only two linearly independent double commutators. Somewhat arbitrarily, we can choose them as [1, [2, 3]] and [[1, 2], 3]. After expanding both of them, we get the following coefficients in front of the six different products of the matrices 1, 2, 3:
[1, [2, 3]] [[1, 2], 3] x i [1, i [2, 3]] + y i [i [1, 2], 3]
123 132 213 231 312 321 1 −1 0 −1 0 1 1 0 −1 0 −1 1 z x y x y z
(114)
where the last row is a linear combination of the first two rows and z := − x − y. In other words, x (231 + 132) + y(312 + 213) + z(123 + 321) = x i [1, i [2, 3]] + y i [i [1, 2], 3]
(115)
whenever x + y + z = 0. We can use this to simply the third order terms in eqs. (93) and (111). Indeed, since the coefficients in (93) sum to zero, see eq. (100), we can rewrite (93) as a linear combination of two double commutators: Re(q1 q¯2 ) i [ρ1 , i [ρ2 , ρ3 ]] (116) + Re(q2 q¯3 ) i [i [ρ1 , ρ2 ], ρ3 ]. Similarly, because of identity (112), expression (111) is equal to
√ p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]] √ + p2 p3 cos δ23 i [i [ρ1 , ρ2 ], ρ3 ]. 2 We
write i instead of ρi for brevity.
18
(117)
5.2
Relation to nested expressions
It is interesting to know whether our ternary operation can produce the 12 nested expressions discussed in Section 2.1 as special cases. More precisely, we would like to know whether, for fixed distribution ( p1 , p2 , p3 ), the one-parameter family of states described by eqs. (109) to (111) under constraints (112) contains all 12 nested expressions with first order terms p1 ρ1 + p2 ρ2 + p3 ρ3 . For example, consider nested expressions of the form ρ1 a (ρ2 a0 ρ3 ) where each is either or , see eqs. (22) and (23). Recall from eqs. (7) and (8) that q 0 ρ1 a (ρ2 a0 ρ3 ) = aρ1 + (1 − a) a0 ρ2 + (1 − a0 )ρ3 + (−1)s a0 (1 − a0 ) i [ρ3 , ρ2 ] (118) q q h i 0 +(−1)s a(1 − a) i a0 ρ2 + (1 − a0 )ρ3 + (−1)s a0 (1 − a0 ) i [ρ3 , ρ2 ], ρ1 , (119) where s, s0 ∈ {0, 1} depend on the signs of the two operations (0 stands for while 1 stands for ). Importantly, this expression has only one double commutator, namely i [ρ1 , i [ρ2 , ρ3 ]], while a general expression involves two. Indeed, recall from eq. (117) that for the ternary operation the terms of order three can be written as √ p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]] (120) √ + p2 p3 cos δ23 i [i [ρ1 , ρ2 ], ρ3 ]. To get only one double commutator, we can set cos δij = 0 for some ij ∈ {12, 23, 31}. However, we must also be able to solve eq. (112) for the remaining two phases. For example, to get only i [ρ1 , i [ρ2 , ρ3 ]], we can set cos δ23 = 0 and solve eq. (112) for δ12 and δ31 . Similarly, cos δ31 = 0 would yield i [ρ2 , i [ρ3 , ρ1 ]] via eq. (112) and the Jacoby identity. The following lemma establishes that we can indeed get all 12 nested expressions in this way. Recall that throughout this section we treat ρi as abstract non-commutative variables. Lemma 11. Let ρ be given by eqs. (109) to (111) and assume p1 , p2 , p3 6= 0. Then ρ admits a nested expression if and only if cos δij = 0 for some ij ∈ {12, 23, 31}. Proof. We have already proved the forward implication. Indeed, nested expressions have only one double commutator, so we get cos δij = 0 for some ij ∈ {12, 23, 31}. For the reverse implication, we assume cos δ23 = 0 (the other cases follow similarly due to symmetry). Under this assumption, let us argue that eq. (112) has only four discrete solutions, and that these solutions correspond to the four nested expressions ρ 1 p 1 ρ 2 p2 ρ 3 (121) p2 + p3
with arbitrary signs s, s0 ∈ {0, 1} for the two operations. First, we can write 0 cos δ23 = 0 and sin δ23 = −(−1)s
(122)
s0
for some s0 ∈ {0, 1}. In other words, δ23 := −(−1) π/2. From the first half of eq. (112), δ12 = 0 −δ23 − δ31 = (−1)s π/2 − δ31 . Hence 0
0
cos δ12 = (−1)s sin δ31 and sin δ12 = (−1)s cos δ31 . (123) √ √ From the second half of eq. (112), p2 cos δ12 + p3 cos δ31 = 0. If we substitute eq. (123), this √ √ 0 becomes p2 (−1)s sin δ31 + p3 cos δ31 = 0 and we get r r p3 p2 s s+s0 sin δ31 = (−1) and cos δ31 = −(−1) (124) p2 + p3 p2 + p3 19
for some s ∈ {0, 1}. We can then find δ12 by substituting eq. (124) back in eq. (123): r r p2 p3 s s+s0 sin δ12 = −(−1) and cos δ12 = (−1) . p2 + p3 p2 + p3
(125)
Equations (122), (124) and (125) with s, s0 ∈ {0, 1} give us the four solutions. It remains to argue that these solutions produce the four states in eq. (121). We begin by restating eqs. (109) to (111) when cos δ23 = 0: ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 √ + p1 p2 sin δ12 i [ρ1 , ρ2 ] √ + p2 p3 sin δ23 i [ρ2 , ρ3 ] √ + p3 p1 sin δ31 i [ρ3 , ρ1 ] √ + p1 p2 cos δ12 i [ρ1 , i [ρ2 , ρ3 ]]. Let us group the terms together to make this appear more similar to eqs. (118) and (119): √ ρ = p1 ρ1 + p2 ρ2 + p3 ρ3 − p2 p3 sin δ23 i [ρ3 , ρ2 ] √ √ √ + i − p1 p2 sin δ12 ρ2 + p3 p1 sin δ31 ρ3 + p1 p2 cos δ12 i [ρ3 , ρ2 ], ρ1 . If we pull out the desired prefactors, we get √ p2 p3 p2 p3 ρ = p1 ρ1 + (1 − p1 ) ρ2 + ρ3 − sin δ23 i [ρ3 , ρ2 ] p2 + p3 p2 + p3 p2 + p3 √ √ √ q p2 sin δ12 p3 sin δ31 p2 cos δ12 ρ2 + √ ρ3 + √ i [ ρ3 , ρ2 ], ρ1 . + p1 (1 − p1 ) i − √ p2 + p3 p2 + p3 p2 + p3
(126)
(127)
(128)
(129) (130)
(131) (132)
This becomes one of the four states in eq. (121) once we substitute eqs. (122), (124) and (125). Example. Let ρ( x, y, z) := 12 ( I + xσx + yσy + zσz ) where σx , σy , σz are the Pauli matrices. Take ρ1 := ρ(1, 0, 0), ρ2 := ρ(0, 1, 0), ρ3 := ρ(0, 0, 1) and let p1 = p2 = p3 := 1/3. (Note that ρ1 = |+ih+|, ρ2 = |+i ih+i |, ρ3 = |0ih0|, i.e., pure states pointing along the three axis of the Bloch sphere.) Then the combined state given by eqs. (109) to (111) is 1 1 1 (1 − sin δ23 ), (1 − sin δ31 ), (1 − sin δ12 ) , (133) ρ 3 3 3 where the angles δij are subject to the following relations, see eq. (112): δ12 + δ23 + δ31 = 0,
cos δ12 + cos δ23 + cos δ31 = 0.
(134)
This describes a one-parameter orbit shown in Fig. 2. As proved by Lemma 11, this orbit contains all twelve nested combinations which are also illustrated in Fig. 2.
20
ρ3
ρ2 ( ρ3 ρ1 )
ρ1 ( ρ2 ρ3 )
ρ3 ( ρ1 ρ2 )
ρ3 ( ρ1 ρ2 )
ρ1 ( ρ2 ρ3 )
ρ2 ( ρ3 ρ1 )
ρ2 ( ρ3 ρ1 )
ρ1 ( ρ2 ρ3 )
ρ1
ρ3 ( ρ1 ρ2 )
ρ3 ( ρ1 ρ2 )
ρ1 ( ρ2 ρ3 )
ρ2
ρ2 ( ρ3 ρ1 )
Figure 2: One-parameter orbit obtained by combining three mutually unbiased pure qubit states ρ1 , ρ2 , ρ3 with equal weights p1 = p2 = p3 = 1/3. This orbit contains all twelve nested combinations of ρ1 , ρ2 , ρ3 as special cases (the two operations have parameters 1/3 and 1/2, respectively).
21
6
Open problems
The most obvious open problem is generalizing Theorem 1 (originally from [ADO15]) to three states. Here is a formal statement of this conjecture. Conjecture (EPI for three states). For any concave and symmetric function f : D(d) → R, any states ρ1 , ρ2 , ρ3 ∈ D(d), and any probability distribution ( p1 , p2 , p3 ), f ( ρ ) ≥ p1 f ( ρ1 ) + p2 f ( ρ2 ) + p3 f ( ρ3 )
(135)
where ρ is given by eqs. (109) to (111) with pi and δij subject to eq. (112). It would also be interesting to understand how an arbitrary number of states can be combined. Towards this goal, the two main steps are: 1. Finding a generalization of eq. (44). Since for general n the final expression of ρ would consist of (n!)2 terms, we need a more efficient way of contracting tensor diagrams. 2. Answering Q2 for any n. While for n = 3 it was sufficient to adjust the global phases of the unitaries Uτ , for general n it is not clear at all how to turn the first order terms of ρ into a convex combination of ρi , with coefficients depending only on the parameters zπ but not the input states ρi themselves. In particular, it is worthwhile investigating when higher order terms of ρ can be written as a linear combination of nested commutators. Perhaps, as suggested by the n = 3 case, better understanding of the second problem might make it possible to deal with both problems simultaneously, since many terms are likely to drop out simultaneously.
Acknowledgements I would like to thank Toby Cubitt, Nilanjana Datta, and Will Matthews for very helpful discussions. This work was supported by the European Union under project QALGO (Grant Agreement No. 600700). In loving memory of L.O.V. Coffeebean.
References [ADO15] Koenraad Audenaert, Nilanjana Datta, and Maris Ozols. Entropy power inequalities for qudits. 2015. arXiv:1503.04213. [Ban93]
Eiichi Bannai. Association schemes and fusion algebras (an introduction). Journal of Algebraic Combinatorics, 2(4):327–344, 1993. doi:10.1023/A:1022489416433.
˚ Ericsson, Jan-Ake ˚ [BBE+ 07] Ingemar Bengtsson, Wojciech Bruzda, Asa Larsson, Wojciech Tadej, ˙ and Karol Zyczkowski. Mutually unbiased bases and Hadamard matrices of order six. Journal of Mathematical Physics, 48(5):052106, 2007. arXiv:quant-ph/0610161, doi: 10.1063/1.2716990. [Chi13]
Andrew M. Childs. Fourier analysis in nonabelian groups. Lecture notes at University of Waterloo, 2013. URL: http://www.cs.umd.edu/~amchilds/teaching/w13/l06.pdf.
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[LMR14] Seth Lloyd, Masoud Mohseni, and Patrick Rebentrost. Quantum principal component analysis. Nature Physics, 10(9):631–633, 2014. arXiv:1307.0401, doi:10.1038/ nphys3029. [Ser12]
Jean-Pierre Serre. Linear Representations of Finite Groups. Springer, 2012.
˙ [TZ06]
˙ Wojciech Tadej and Karol Zyczkowski. A concise guide to complex Hadamard matrices. Open Systems & Information Dynamics, 13(02):133–177, 2006. arXiv:quant-ph/0512154, doi:10.1007/s11080-006-8220-2.
[WBC15] Christopher J. Wood, Jacob D. Biamonte, and David G. Cory. Tensor networks and graphical calculus for open quantum systems. Quantum Information and Computation, 15(9&10):0759–0811, July 1 2015. URL: http://www.rintonpress.com/journals/ qiconline.html#v15n910, arXiv:1111.6950.
A
Deriving the parametrization from scratch
In this appendix we derive from scratch (namely, without using the irreps of S3 ) the parametrization of ρ obtained in Section 4.4.1. The only assumption that goes into our derivation is that z1 , z2 , z3 are real while z4 , z5 , z6 are imaginary—this is something we observed in Section 4.3, eqs. (83) to (88). This assumption is in fact sufficient for deriving eqs. (91) to (93) and the constraints in eq. (94). Following eq. (90), we choose the coefficients zi as
(z1 , z2 , z3 , z4 , z5 , z6 ) := ( a1 , a2 , a3 , ib1 , ib2 , ib3 )
(136)
for some a1 , a2 , a3 , b1 , b2 , b3 ∈ R. Without any additional constraints, this can potentially capture only more than eqs. (83) to (88). At the same time, zi chosen according to eq. (136) still satisfy the constraints imposed in Proposition 10. In other words, this choice automatically takes care of Q2. Let us rewrite the output state ρ from eq. (44) in terms of the new parameters: ρ = a21 + b12 ρ1 + a22 + b22 ρ2 + a23 + b32 ρ3
(137)
+ a1 b2 − a2 b1 i [ρ2 , ρ1 ] + a2 b3 − a3 b2 i [ρ3 , ρ2 ] + a3 b1 − a1 b3 i [ρ1 , ρ3 ] + a1 a2 + b1 b2 (ρ2 ρ3 ρ1 + ρ1 ρ3 ρ2 ) + a2 a3 + b2 b3 (ρ3 ρ1 ρ2 + ρ2 ρ1 ρ3 ) + a3 a1 + b3 b1 (ρ1 ρ2 ρ3 + ρ3 ρ2 ρ1 ).
(138)
(139)
Clearly, this is identical to eqs. (91) to (93) if we let qk := ak + ibk . Let us now work backwards to find what extra constraints should be imposed on the coefficients ai and bi to satisfy the unitarity requirement Q1. Recall from Lemma 5 that we want the following
23
matrix, see eq. (75), to be unitary: a1 a3 a2 ib1 ib2 ib3 a2 a1 a3 ib3 ib1 ib2 6 a3 a2 a1 ib2 ib3 ib1 A iBT ∑ zk Lk = ib1 ib3 ib2 a1 a2 a3 = iB AT =: U k =1 ib2 ib1 ib3 a3 a1 a2 ib3 ib2 ib1 a2 a3 a1
(140)
where
a1 a3 a2 A : = a2 a1 a3 , a3 a2 a1
b1 b3 b2 B := b2 b1 b3 . b3 b2 b1
We can write the unitarity condition as T A iBT AAT + BT B i [ BT , A ] A −iBT † UU = = = I. −iB A iB AT i [ B, AT ] AT A + BBT
(141)
(142)
Note that [ BT , A] = 0 holds automatically, so the remaining constraints follow solely from the diagonal blocks: AAT + BT B = AT A + BBT = I. These constraints are: a21 + a22 + a23 + b12 + b22 + b32 = 1,
(143)
a1 a2 + a2 a3 + a3 a1 + b1 b2 + b2 b3 + b3 b1 = 0.
(144)
It is not hard to see that these constraints are equivalent to eq. (94).
24
