Advanced Topics in Quantum Information Theorywatrous/CS798/Slides/01.handout.pdf · Advanced Topics...

CS 798/QIC 890 Spring 2020

Advanced Topics in

Quantum Information Theory

John Watrous

Institute for Quantum Computing

and

Cheriton School of Computer Science

Lecture 1

Conic Programming

Conic programming preliminaries

We will begin this course with a generalization of semidefinite programming (which youwill recall from CS 766/QIC 820) known as conic programming.

Let V be a finite-dimensional inner-product space over the real numbers. (Note: it isimportant that our ground field is R and not C.)

A subset C ⊆ V is convex if, for all u, v ∈ C and λ ∈ [0, 1], we have

λu+ (1− λ)v ∈ C.

A subset K ⊆ V is a cone if, for all v ∈ K and λ > 0, we have

λv ∈ K.

For any subset A ⊆ V, the dual cone to A is defined as

A∗ ={v ∈ V : 〈u, v〉 > 0 for every u ∈ A

}.

Example

A highly relevant example is given by the cone of positive semidefinite operatorsacting on a complex Euclidean space.

In detail, let X be a complex Euclidean space, and consider the real inner-product spaceV = Herm(X), with the inner-product being our usual inner-product for operators:

〈A,B〉 = Tr(A∗B) = Tr(AB),

with the second equality reflecting the assumption that A is Hermitian.

The set Pos(X) ⊆ Herm(X) is a closed, convex cone that happens to be its own dual(i.e., it is self-dual):

Pos(X)∗ ={H ∈ Herm(X) : 〈P,H〉 > 0 for every P ∈ Pos(X)

}= Pos(X).

Definition of conic programs

Now suppose that V and W are real inner-product spaces, K ⊆ V is a closed, convexcone, φ ∈ L(V,W) is a linear map, and a ∈ V and b ∈W are vectors.

With such a choice of objects, we define a conic program, which is associated withthese two optimization problems:

Primal problem

maximize: 〈a, x〉subject to: φ(x) = b

x ∈ K

Dual problem

minimize: 〈b,y〉subject to: φ∗(y) − a ∈ K∗

y ∈W

In the dual problem, φ∗ ∈ L(W,V) is the adjoint map to φ, which is the (unique) linearmap satisfying

〈φ∗(y), x〉 = 〈y,φ(x)〉

for all x ∈ V and y ∈W.

Semidefinite programs as conic programsRecall that a semidefinite program is specified by a Hermitian- preserving mapΦ ∈ T(X,Y) and Hermitian operators A ∈ Herm(X) and B ∈ Herm(Y), for X and Y

complex Euclidean spaces.

Primal problemmaximize: 〈A,X〉subject to: Φ(X) = B

X ∈ Pos(X)

Dual problemminimize: 〈B, Y〉

subject to: Φ∗(Y) > A

Y ∈ Herm(Y)

Note that the dual constraint may alternatively be expressed as

Φ∗(Y) −A ∈ Pos(X) = Pos(X)∗.

This is a conic program for which V = Herm(X), W = Herm(Y), and K = Pos(X), andwhere we view Φ as a linear map of the form

Φ : Herm(X)→ Herm(Y).

Motivation for conic programming

1. The level of abstraction offered by conic programs is useful, and offers a directconnection to interesting notions in convex analysis.

2. Some important properties of semidefinite programming are true for conicprogramming, while others are not. Examples:

• Slater’s theorem applies not only to semidefinite programming, but to conicprogramming in general.

• While a computer can be used to efficiently approximate the optimal value of asemidefinite program (under mild restrictions), this is not true of conicprogramming. For example, it is NP-hard to optimize a linear function over theconvex cone Sep(X : Y) of separable operators.

An understanding of conic programming therefore helps to illuminate the role thatthe cone of positive semidefinite operators plays in the context of semidefiniteprogramming.

Notation and terminology for conic programs

Let us return to the primal and dual problems of a conic program:

Primal problemmaximize: 〈a, x〉subject to: φ(x) = b

x ∈ K

Dual problemminimize: 〈b,y〉

subject to: φ∗(y) − a ∈ K∗

y ∈W

The primal feasible and dual feasible vectors are given by

A ={x ∈ K : φ(x) = b

}B =

{y ∈W : φ∗(y) − a ∈ K∗}

and the primal optimum and dual optimum are given by

α = sup{〈a, x〉 : x ∈ A

}β = inf

{〈b,y〉 : y ∈ B

}

Weak duality

Conic programs possess the property of weak duality:

α = sup{〈a, x〉 : x ∈ A

}6 β = inf

{〈b,y〉 : y ∈ B

}

duality gapβ

α

minimizationin the dual

maximizationin the primal

If A = ∅, then α = −∞, and if B = ∅, then β =∞, so weak duality is trivially satisfied ineither case.

Weak duality


α = sup{〈a, x〉 : x ∈ A

}6 β = inf

{〈b,y〉 : y ∈ B

}

duality gapβ

α



Otherwise, if x ∈ A and y ∈ B, then x ∈ K and φ∗(y) − a ∈ K∗, so⟨x,φ∗(y) − a

⟩> 0.

Therefore,〈a, x〉 6 〈x,φ∗(y)〉 = 〈φ(x),y〉 = 〈b,y〉.

Weak duality


α = sup{〈a, x〉 : x ∈ A

}6 β = inf

{〈b,y〉 : y ∈ B

}

duality gapβ

α



Finally, observe that if we ever find a primal feasible x ∈ A and a dual feasible y ∈ B

such that〈a, x〉 = 〈b,y〉,

then we know for certain that these are the optimal values, and moreover we have α = β(a condition known as strong duality).

Minimizing versus maximizing

An alternative form of a conic program where the primal is a minimization problem andthe dual is a maximization problem:

Primal problemminimize: 〈a, x〉

subject to: φ(x) = b

x ∈ K

Dual problemmaximize: 〈b,y〉subject to: a− φ∗(y) ∈ K∗

y ∈W

These problems are equivalent, up to the negation of the optimal values, to an instance ofthe standard formulation, substituting (φ,a,b)→ (−φ,−a,−b).

Primal problemmaximize: 〈−a, x〉subject to: − φ(x) = −b

x ∈ K

Dual problemminimize: 〈−b,y〉

subject to: − φ∗(y) − (−a) ∈ K∗

y ∈W

The relative interior of a convex set

In a moment we will be stating and proving Slater’s theorem for conic programs, but firstwe require the concept of the relative interior of a set.

This is simply the interior of the set, assuming that we restrict our attention to thesmallest affine subspace that contains the set.

For any convex set C, there is a simple-to-state definition:

relint(C) ={x ∈ C : (∀y ∈ C)(∃ε > 0)

((1+ ε)x− εy ∈ C

)}.

λx+ (1− λ)y(1+ ε)x− εy

y

x

C

Statement of Slater’s theorem

We are now prepared to state Slater’s theorem for conic programs, which provides anoften easy-to-check condition for strong duality.

Theorem (Slater’s theorem for conic programs)

Let V and W be real inner-product spaces, let K ⊆ V be a closed, convex cone, and letφ ∈ L(V,W), a ∈ V, and b ∈W be given.

With respect to the notations defined previously, the following two statements are true:

1. If B is nonempty and there exists x ∈ relint(K) such that φ(x) = b, then there mustexist y ∈ B such that 〈b,y〉 = α.

2. If A is nonempty and there exists y ∈W for which φ∗(y) − a ∈ relint(K∗), thenthere must exist x ∈ A such that 〈a, x〉 = β.

Both statements imply the equality α = β.

Proof of Slater’s theorem

We will prove the first statement. Let us recall once again the primal and dual problems,along with the statement to be proved:

Primal problemmaximize: 〈a, x〉subject to: φ(x) = b

x ∈ K

Dual problemminimize: 〈b,y〉

subject to: φ∗(y) − a ∈ K∗

y ∈W

Statement: If B is nonempty and there exists x ∈ relint(K) such that φ(x) = b, thenthere must exist y ∈ B such that 〈b,y〉 = α.

We can make these assumptions without loss of generality:

V = span(K) and W = im(φ).

(Consider the two assumptions, one at a time, and see if you can convince yourself thatindeed no loss of generality is incurred.)


Statement: If B is nonempty and there exists x ∈ relint(K) such that φ(x) = b, thenthere must exist y ∈ B such that 〈b,y〉 = α.

Define two convex sets:

C ={(b− φ(x), z, 〈a, x〉) : x, z ∈ V, x− z ∈ K

},

D ={(0, 0,η) : η > α

}.

They are disjoint, so they are separated by a hyperplane: there must exist a nonzerovector (y,u, λ) ∈W⊕ V⊕ R such that

〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη

for all x, z ∈ V for which x− z ∈ K and all η > α. By scaling this vector, we may assumeλ ∈ {−1, 0, 1}.

We will draw various conclusions from the inequality. Note that the vector y will be the(optimal) dual solution we seek.

Proof of Slater’s theoremThere exists a nonzero vector (y,u, λ) ∈W⊕ V⊕ R such that

〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη

for all x, z ∈ V for which x− z ∈ K and all η > α.

This inequality must be true when x = 0 and z = 0, and therefore

〈y,b〉 6 λη

for all η > α.

This implies that λ > 0, for otherwise the right-hand side of the inequality tends to −∞as η becomes large, while the right-hand side remains fixed.

Conclusion: λ > 0.


〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη


For any choice of x ∈ A and z ∈ −K, it is the case that x− z ∈ K. Substituting thesevectors into the inequality and rearranging yields

〈u, z〉 6 λ(η− 〈a, x〉)

for every η > α.

This implies u ∈ K∗, for otherwise the left-hand side of the above inequality can be madeto approach∞ while the right-hand side remains bounded.

Conclusion: u ∈ K∗.


〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη


Fix any vector x ∈ A ∩ relint(K) and assume toward contradiction that λ = 0. Theinequality simplifies to

〈u, z〉 6 0

for every choice of z ∈ V for which x− z ∈ K.

Setting z = x, we have that x− z ∈ K, and therefore

〈u, x〉 6 0.

As x ∈ K and u ∈ K∗, we conclude that 〈u, x〉 = 0.


〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη


As x ∈ relint(K), for any vector v ∈ K, there exists ε > 0 such that

x− ε(v− x) = (1+ ε)x− εv ∈ K.

Setting z = ε(v− x), we conclude that x− z ∈ K, and so

ε〈u, v〉 = ε〈u, v〉− ε〈u, x〉 = 〈u, z〉 6 0.

As it was was for x, we find that 〈u, v〉 = 0.

This is true for all v ∈ K, and we have u ∈ V = span(K), so u = 0.


〈y,b− φ(x)〉+ 〈u, z〉+ λ〈a, x〉 6 λη


Finally, if λ = 0 and u = 0, then we may free the vector x to range over all of V and setz = x to conclude that

〈y,b− φ(x)〉 6 0

for every x ∈ V. Bearing in mind the assumption that W = im(φ), we conclude thaty = 0.

The assumption λ = 0 has implied that (y,u, λ) = (0, 0, 0), contradicting the fact that thisvector is nonzero.

Conclusion: λ 6= 0, and therefore λ = 1.


The argument just presented has allowed us to conclude that there exist vectors y ∈W

and u ∈ K∗ such that〈y,b− φ(x)〉+ 〈u, z〉 6 η− 〈a, x〉


Setting z = 0 and flipping sign, we find that

〈φ∗(y) − a, x〉 > 〈y,b〉− η

for all x ∈ K and η > α. This implies

φ∗(y) − a ∈ K∗,

for otherwise the left-hand side of the previous inequality can be made to approach −∞while the right-hand side remains fixed. The vector y is therefore a dual-feasible point:y ∈ B.


The argument just presented has allowed us to conclude that there exist vectors y ∈W

and u ∈ K∗ such that〈y,b− φ(x)〉+ 〈u, z〉 6 η− 〈a, x〉


Finally, again considering the possibility that x = 0 and z = 0, we concude that

〈b,y〉 6 η

for all η > α. It is therefore the case that 〈b,y〉 6 α, and hence 〈b,y〉 = α by weakduality.

We have therefore proved the statement we aimed to prove: If B is nonempty and thereexists x ∈ relint(K) such that φ(x) = b, then there must exist y ∈ B such that〈b,y〉 = α.

Example: optimal measurements

Suppose X is a complex Euclidean space and K ⊆ Herm(X) is a closed, convex cone.

Suppose further that H1, . . . ,Hn ∈ Herm(X), and consider this optimization problem:

maximize: 〈H1,X1〉+ · · ·+ 〈Hn,Xn〉

subject to: X1 + · · ·+ Xn = 1X

X1, . . . ,Xn ∈ K

This is an optimal measurement problem, where we demand that the operatorsX1, . . . ,Xn are drawn from K.

The cone K = Pos(X) yields the ordinary optimal measurement problem, but we canconsider any closed, convex cone K we choose. For example, we may take K to be thecone of separable operators when X = Y⊗ Z is a bipartite tensor product space.


We can set this problem up as a conic program as follows.

First, observe that Kn is a closed, convex cone. The problem may therefore beexpressed as the primal problem of a conic program:

maximize:⟨(H1, . . . ,Hn), (X1, . . . ,Xn)

⟩subject to: φ(X1, . . . ,Xn)

�= X1 + · · ·+ Xn = 1X

(X1, . . . ,Xn) ∈ Kn

Here is the dual problem:

minimize:⟨1X, Y

⟩subject to: φ∗(Y) − (H1, . . . ,Hn) ∈

(Kn

)∗Y ∈ Herm(X)

We can simplify this by observing that φ∗(Y) = (Y, . . . ,Y) and (Kn)∗ = (K∗)n.


We can set this problem up as a conic program as follows.

First, observe that Kn is a closed, convex cone. The problem may therefore beexpressed as the primal problem of a conic program:

maximize:⟨(H1, . . . ,Hn), (X1, . . . ,Xn)

⟩subject to: φ(X1, . . . ,Xn)

�= X1 + · · ·+ Xn = 1X

(X1, . . . ,Xn) ∈ Kn

Simplified dual problem:minimize: Tr(Y)

subject to: Y −H1 ∈ K∗

...

Y −Hn ∈ K∗

Y ∈ Herm(X)


In summary, here are the (simplified) primal and dual problems:

Primal problemmaximize: 〈H1,X1〉+ · · ·+ 〈Hn,Xn〉

subject to: X1 + · · ·+ Xn = 1X

X1, . . . ,Xn ∈ K

Dual problemminimize: Tr(Y)

subject to: Y −H1 ∈ K∗

...

Y −Hn ∈ K∗

Y ∈ Herm(X)

Example: optimal separable measurements

Consider the 4 pure states |ψ1〉, . . . , |ψ4〉 ∈ C4 ⊗ C4 defined as

|ψ1〉 =1

2

(|0〉|0〉+ |1〉|1〉+ |2〉|2〉+ |3〉|3〉

),

|ψ2〉 =1

2

(|0〉|3〉+ |1〉|2〉+ |2〉|1〉+ |3〉|0〉

),

|ψ3〉 =1

2

(|0〉|3〉+ |1〉|2〉− |2〉|1〉− |3〉|0〉

),

|ψ4〉 =1

2

(|0〉|1〉+ |1〉|0〉− |2〉|3〉− |3〉|2〉

),

Yu, Duan, and Ying (2012) identified these states and proved that they cannot beperfectly discriminated by a PPT measurement.

Cosentino (2013) showed that the optimal PPT discrimination probability is 7/8 (bysolving the associated semidefinite program).


We will prove that no separable measurement can discriminate these states withprobability greater than 3/4 (easily achievable by coarse-graining a measurement withrespect to the standard basis).

Primal problem

maximize:1

4〈ψ1 |X1|ψ1〉+ · · ·+

1

4〈ψ4 |X4|ψ4〉

subject to: X1 + · · ·+ X4 = 14 ⊗ 14

X1, . . . ,X4 ∈ Sep(C4 : C4)

Dual problem

minimize: Tr(Y)

subject to: Y −1

4|ψk〉〈ψk | ∈ Sep(C4 : C4)∗ (1 6 k 6 4)

Y ∈ Herm(C4 ⊗ C4)


Our goal will be to describe a dual-feasible solution having objective value 3/4, for thiswill then be an upper-bound on the probability of a correct discrimination by weak duality.

To do this, it will be helpful to observe that the pure states in question may be expressedas follows:

|ψ1〉 =1

2

(|0〉|0〉+ |1〉|1〉+ |2〉|2〉+ |3〉|3〉

)=

1

2vec(1⊗ 1),

|ψ2〉 =1

2

(|0〉|3〉+ |1〉|2〉+ |2〉|1〉+ |3〉|0〉

)=

1

2vec(σx ⊗ σx)

|ψ3〉 =1

2

(|0〉|3〉+ |1〉|2〉− |2〉|1〉− |3〉|0〉

)=i

2vec(σy ⊗ σx)

|ψ4〉 =1

2

(|0〉|1〉+ |1〉|0〉− |2〉|3〉− |3〉|2〉

)=

1

2vec(σz ⊗ σx)


Let us incorporate these expressions into the dual problem:

minimize: Tr(Y)

subject to: Y −1

4|ψk〉〈ψk | ∈ Sep(C4 : C4)∗ (1 6 k 6 4)

Y ∈ Herm(C4 ⊗ C4)

The dual-feasibility of a given Y ∈ Herm(C4 ⊗ C4) is equivalent to

Y −1

16vec(Uk) vec(Uk)

∗ ∈ Sep(C4 : C4)∗

forU1 = 1⊗ 1, U2 = σx ⊗ σx, U3 = iσy ⊗ σx, U4 = σz ⊗ σx.

The Breuer–Hall Lemma

Lemma (Breuer–Hall). Let U,V ∈ U(Cn) be unitary operators such that VTU isanti-symmetric: (VTU)T = −VTU. The operator

Z = 1n ⊗ 1n − vec(U) vec(U)∗ − (T⊗1)(vec(V) vec(V)∗)

is contained in Sep(Cn : Cn)∗.

Proof. For any unit vector z ∈ Cn, we find that

(1⊗ z)∗Z(1⊗ z) = 1−UzzTU∗ − Vzz∗VT > 0;

the vectors Uz and Vz must be orthogonal unit vectors:⟨Vz,Uz

⟩= z∗VTUz =

⟨zzT,VTU

⟩=

⟨(zzT)T, (VTU)T

⟩= −

⟨zzT,VTU

⟩= 0

The Breuer–Hall Lemma

Lemma (Breuer–Hall). Let U,V ∈ U(Cn) be unitary operators such that VTU isanti-symmetric: (VTU)T = −VTU. The operator

Z = 1n ⊗ 1n − vec(U) vec(U)∗ − (T⊗1)(vec(V) vec(V)∗)

is contained in Sep(Cn : Cn)∗.

Proof. For any unit vector z ∈ Cn, we find that

(1⊗ z)∗Z(1⊗ z) = 1−UzzTU∗ − Vzz∗VT > 0;

the vectors Uz and Vz must be orthogonal unit vectors.

It follows that〈yy∗ ⊗ zz∗,Z〉 = (y⊗ z)∗Z(y⊗ z) > 0

for every y ∈ Cn, implying the required containment.


Let us now return to the dual problem, which is the minimization of Tr(Y) over allY ∈ Herm(C4 ⊗ C4) for which

Y −1

16vec(Uk) vec(Uk)

∗ ∈ Sep(C4 : C4)∗

forU1 = 1⊗ 1, U2 = σx ⊗ σx, U3 = iσy ⊗ σx, U4 = σz ⊗ σx.

Let V = iσy ⊗ σz, and verify that VTU1, VTU2, VTU3, and VTU4 are all anti-symmetric.

By the Breuer–Hall lemma, the operator

Y =1

16

(14 ⊗ 14 − (T⊗1)(vec(V) vec(V)∗)

)is dual-feasible, and we have Tr(Y) = (16− 4)/16 = 3/4.

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