# Quantum Coupling and Strassen Theorem

We introduce a quantum generalisation of the notion of coupling in probability theory. Several interesting examples and basic properties of quantum couplings are presented. In particular, we prove a quantum extension of Strassen theorem for probabilistic couplings, a fundamental theorem in probability theory that can be used to bound the probability of an event in a distribution by the probability of an event in another distribution coupled with the first.

## Authors

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## 1 Introduction

Coupling is a powerful technique in probability theory, with which random variables can be linked to or compared with each other. It has been widely used in the studies of random walks and Markov chains, interacting particle systems and diffusions, just name a few, in order to establish limit theorems about them, to develop approximations for them, or to derive correlation inequalities between them

[7].

Recently, a very successful application of coupling in computer science was discovered by Barthe et al. [4] that it can serve as a solid mathematical foundation for defining the semantics of probabilistic relational Hoare logic. This discovery enables them to develop a series of powerful proof techniques for reasoning about relational properties of probabilistic computations, in particular, for verification of cryptographic protocols and differential privacy [2, 3, 1, 6].

There is a simple and natural correspondence between probability theory and quantum theory: probability distributions/density operators (mixed quantum states), marginal distributions/partial traces, and more. This correspondence suggests us to explore the possibility of generalising the coupling techniques for reasoning about quantum systems. We expect that these techniques can help us to extend quantum Hoare logic

[10] for proving relational properties between quantum programs and further for verifying quantum cryptographic protocols and differential privacy in quantum computation [11]. But in this paper, we focus on studying quantum couplings themselves.

Strassen theorem [9] is a fundamental theorem in probability theory that can be used to bound the probability of an event in a distribution by the probability of an event in another distribution coupled with the first. The main technical contribution of this paper is proving an elegant (in our opinion) quantum generalisation of Strassen theorem.

## 2 Background and Basic Definitions

### 2.1 Probabilistic Coupling

For convenience of the reader, we first briefly recall the basics of probabilistic coupling, following [6]. Let be a finite or countably infinite set. A sub-distribution over is a mapping such that . In paricular, if , then is called a distribution over . For a sub-distribution over , we define:

1. The weight of is

2. The support of is

3. The probability of an event is

Moreover, let be a joint sub-distribution, i.e. a sub-distribution over Cartesian product . Then its marginals over and are, respectively, defined by

 π1(μ)(a1) =∑a2∈A2μ(a1,a2) for every a1∈A1, π2(μ)(a2) =∑a1∈A1μ(a1,a2) for every a2∈A2.

Now we can define the notion of coupling.

###### Definition 1 (Probabilistic Coupling).

Let be sub-distributions over , respectively. Then a sub-distribution over is called a coupling for if and .

Here are some simple examples of coupling taken from [6].

###### Example 1.

Let

be the uniform distribution over booleans, i.e.

. Then the following are two couplings for :

1. Identity coupling:

2. Negation coupling:

More generally, let be the uniform distribution over a finite nonempty set , i.e. for every . Then each bijection yields a coupling for :

 μf(a1,a2)={1|A|if f(a1)=a2,0otherwise.
###### Example 2.

For any sub-distribution over , the identity coupling for is:

###### Example 3.

For any distributions over , respectively, the independent or trivial coupling is:

Obviously, coupling for a pair of distributions is not unique. Then the notion of lifting can be introduced to choose a desirable coupling.

###### Definition 2 (Probabilistic Lifting).

Let be sub-distributions over , respectively, and let be a relation. Then a sub-distribution over is called a witness for the -lifting of if:

1. is a coupling for ;

2. .

Whenever a witness exists, we say that and are related by the -lifting and write .

###### Example 4.
1. Coupling in Example 1 is a witness for the lifting

2. Coupling in Example 2 is a witness for the lifting

3. Coupling in Example 3 is a witness for the lifting , where

###### Proposition 1.
1. Let be sub-distributions over , respectively. If there exists a coupling for , then

2. Let be sub-distributions over the same . Then if and only if .

### 2.2 Quantum Coupling

With the correspondence of probability distributions/density operators (mixed quantum states) and marginal distributions/partial traces mentioned in the Introduction, we can introduce the notion of quantum coupling. To this end, let us first recall several basic notions from quantum theory; for details, we refer to [8].

Suppose that is a finite-dimensional Hilbert space. Let be the set of Hermitian matrices in . Let be the set of positive (semidefinite) matrices in , and is the set of partial density operators, , positive (semidefinite) matrices with trace one. A positive operator in is called a partial density operator if its trace , where is an orthonormal basis of .

We define its support:

 supp(ρ) =span{eigenvectors of ρ with nonzero eigenvalues} =span{|ψ⟩ ∣∣ tr(ρ|ψ⟩⟨ψ|)=0}⊥.

If is an observable, i.e. Hermitian operator, in , then its expectation in state is Furthermore, let be two Hilbert space. Then partial trace over is a mapping from operators in to operators in defined by

 tr1(|φ1⟩⟨ψ1|⊗|φ2⟩⟨ψ2|)=⟨ψ1|φ1⟩⋅|φ2⟩⟨ψ2|

for all and together with linearity. The partial trace over can be defined dually.

Now we are ready to define the concept of coupling.

###### Definition 3 (Quantum Coupling).

Let and . Then is called a coupling for if and .

This is actually a very special case of the famous quantum marginal problem, see [12, 13, 14, 15] as a very incompleted list for recent development.

###### Example 5.

Let be a Hilbert space and an orthonormal basis of . Then the uniform density operator on is

 UnifH=1d∑i|i⟩⟨i|

where is the dimension of . Indeed, the uniform density operator on is unique and independent with the choice of orthonormal basis. For each unitary operator in , we write , which is also an orthonormal basis of . Then

 ρU=1d∑i(|i⟩U|i⟩)(⟨i|⟨i|U†)

is a coupling for . In general, for different and , , though they are both the couplings for .

###### Example 6.

Let be a partial density operator in . Then by the spectral decomposition theorem, can be written as for some orthonormal basis and with . We define:

 ρid(B)=∑ipi|ii⟩⟨ii|.

Then it is to see that is a coupling for . A difference between this example and Example 2 is that can be decomposed with other orthonormal bases, say : In general, , and we can define a different coupling:

 ρid(D)=∑jqj|jj⟩⟨jj|

for .

###### Example 7.

Let and

be density operators. Then tensor product

is a coupling for .

The notion of lifting can also be easily generalised into the quantum setting.

###### Definition 4 (Quantum Lifting).

Let and , and let be a subspace of . Then is called a witness of the lifting if:

1. is a coupling for ;

2. .

###### Example 8.
1. The coupling in Example 5 is a witness for the lifting:

 UnifHX(B,U)#UnifH

where is a subspace of

2. The coupling in Example 6 is a witness of the lifting , where defined by the orthonormal basis is a subspace of . It is interesting to note that the maximal entangled state is in

3. The coupling in Example 7 is a witness of the lifting .

As a quantum generalisation of Proposition 1, we have:

###### Proposition 2.
1. Let and . If there exists a coupling for , then .

2. Let . Then if and only if orthonormal basis s.t. .

###### Proof.

Part 1 and Part 2 () are obvious. Here, we prove Part 2 (). If , then there exists a coupling for such that where . Then we have: for some and . Furthermore, for each , we can write: Then it is routine to show that Therefore, it holds that

## 3 Quantum Strassen Theorem

As mentioned in the Introduction, a fundamental theorem for probabilistic coupling is the following:

###### Theorem 1 (Strassen Theorem).

Let be sub-distributions over , respectively. Then

 μ1R#μ2⇒∀S⊆A1. μ1(S)≤μ2(R(S)) (1)

where is the image of under : The converse of (1) holds if

In this section, we prove a quantum generalisation of the above Strassen Theorem. For this purpose, for any subspace of , we use and to denote the projections on and (the ortho-complement of ), respectively. We use

to denote the identity matrix of

, respectively. is employed to denote the inner product of matrices living in the same space,

 ⟨A,B⟩=tr(A†B)

Then a quantum Strassen theorem can be stated as follows:

###### Theorem 2 (Quantum Strassen Theorem).

For any two partial density operators in and in with , and for any subspace of , the following three statements are equivalent:

• ;

• For all observables (Hermitian operators) in and in satisfying , it holds that

 tr(ρ1Y1)≤tr(ρ2Y2). (2)
• For all positive observables in and in satisfying , it holds that

###### Proof.

Suppose is a witness of the lifting . Then for all observables (Hermition operators) in and in , if , then we have:

 tr(ρ1Y1) =tr(ρ(Y1⊗I2)) (3) ≤tr(ρ(PX⊥+I1⊗Y2)) (4) =tr(ρ(I1⊗Y2)) (5) =tr(ρ2Y2). (6)

Equalities (3) and (6) are derived from the condition that is a coupling for ; that is, and , (4) is due to the assumption for and , and (5) is trivial as , so .

Let us first define the semidefinite program :

Primal problem
maximize: subject to:
Dual problem
minimize: subject to:

where:

 A=PX,B=[ρ1ρ2], Φ(X)=[tr2(X)tr1(X)], Φ∗(Y)=Φ∗[Y1⋅⋅Y2]=Y1⊗I2+I1⊗Y2.

To show that the above problems are actually primal and dual, respectively, we only need to check the following equality:

 ∀ M, N, ⟨Φ(M),N⟩ =tr(tr2(M)N1+tr1(M)N2) =tr(M(N1⊗I2)+M(I1⊗N2)) =⟨M,Φ∗(N)⟩.

Moreover, the strong duality holds for this semidefinite program as we can check that the primal feasible set are not empty and there exists a Hermitian operator for which :

 Primal feasible set\ A={X∈Pos(H1⊗H2):Φ(X)=B}∋1tr(ρ1)ρ1⊗ρ2 Choose\ Y=I1⊕I2∈Herm(H1⊕H2), Φ∗(Y)=2I12>PX.

So, . Now, let us consider the following condition:

• (A): For all observable (Hermitian operators) in and in satisfy , then

 ⟨B,Y⟩=tr(ρ1Y1+ρ2Y2)≥trρ1.

If condition (A) holds, then . Still remember that . Due to the strong duality, we have . So, which maximizes must satisfy . Consequently, ; in other words, is a witness of . Therefore, On the other hand, condition (A) is equivalent to statement of the theorem. Indeed, this is not difficult to prove as if we replace in condition (A), then

 Y1∈Herm(H1) ⟺Y′1∈Herm(H1) Y1⊗I2+I1⊗Y2≥PX ⟺I1⊗I2−PX≥Y′1⊗I2−I1⊗Y2 ⟺P⊥X≥Y′1⊗I2−I1⊗Y2 tr(ρ1Y1+ρ2Y2)≥trρ1 ⟺tr(ρ2Y2)≥tr(ρ1I1)−tr(ρ1(I1−Y′1)) ⟺tr(ρ2Y2)≥tr(ρ1Y′1).

From the above, we can directly derive statement . In summary, we have:

Obvious.

We only need to show that, for any two observables in and in satisfy , there exist two positive observables in and in such that and

 tr(ρ1Y1)≤tr(ρ2Y2)⟺tr(ρ1Y′1)≤tr(ρ2Y′2).

Note that and

are Hermitian, so their eigenvalues are real, and we can define

. Choose and . Obviously, and are positive observables, and satisfy

 P⊥X ≥Y1⊗I2−I1⊗Y2 =Y1⊗I2−λI1⊗I2+λI1⊗I2−I1⊗Y2 =Y′1⊗I2−I1⊗Y′2.

Moreover, as , we have

 tr(ρ1Y1)≤tr(ρ2Y2) ⟺tr(ρ1Y1)−tr(ρ1λI1)≤tr(ρ2Y2)−tr(ρ2λI2)

Remark:In the above proof, it is indeed naturally to employing the methods of semidefinite programming. In [6]

, Hsu deliberately constructs a flow network, and then using the max-flow min-cut theorem to prove the Strassen theorem in the finite case. Essentially, the max-flow min-cut theorem is a special case of the duality theorem for linear programs (LP). Considering the fact that quantum states, quantum operations and so on are all described by matrices, similar to LP, semi-definite programming (SDP) is a powerful and widely used method of convex optimization in quantum theory. Indeed, when all matrices appeared in a SDP are diagonal, then the SDP reduces to LP. In the following section, we will see that in the degenerate case, quantum Strassen theorem also reduces to the classical Strassen theorem.

## 4 Classical Reduction of Quantum Strassen Theorem

At the first glance, Theorem 1 (Strassen Theorem for Probabilistic Coupling) and Theorem 2 (Quantum Strassen Theorem) are very different. In this section, we show that Theorem 2 is indeed a quantum generalisation of Theorem 1.

To this end, let be a sub-distribution over () and over . And the corresponding degenerate partial density operators (quantum states) are:

 ρ1=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣μ1(1)μ1(2)⋱μ1(m)⎤⎥ ⎥ ⎥ ⎥ ⎥⎦,ρ2=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣μ2(1)μ2(2)⋱μ2(n)⎤⎥ ⎥ ⎥ ⎥ ⎥⎦

in and , respectively. Furthermore, let be a classical relation from to . Then the corresponding (quantum relation) subspace of is defined as

 XR=span{|i⟩|j⟩ ∣∣ (i,j)∈R}.

Based on the above definition of the degenerate case, in the rest part of this section, Proposition 3 shows that the left hand side of Eqn.(1) in Theorem 1 is equivalent to the statement in Theorem 2, while Proposition 4 states the equivalence of the right hand side of Eqn.(1) in Theorem 1 and the statement in Theorem 2, concluding that Theorem 1 (Strassen Theorem) is indeed a reduction of Theorem 2 (Quantum Strassen Theorem).

The following proposition indicates that probabilistic lifting is a special case of quantum lifting.

###### Proof.

() Suppose that there is a witness of the lifting . We define the partial density operator:

 ρ:⟨i|⟨j|ρ|i′⟩|j′⟩={μ(i,j)i=i′, j=j′0i≠i′\ or\ j≠j′.

It is easy to check:

 ⟨i|tr2(ρ)|i′⟩=n∑j=1⟨i|⟨j|ρ|i′⟩|j⟩={∑nj=1μ(i,j)=μ1(i)i=i′0i≠i′, ⟨j|tr2(ρ)|j′⟩=n∑i=1⟨i|⟨j|ρ|i⟩|j′⟩={∑mi=1μ(i,j)=μ2(j)j=j′0j≠j′.

So, and ; that is, is a coupling for . Furthermore, we have:

 tr(ρPXR) =∑(i,j)∈R⟨i|⟨j|ρ|i⟩|j⟩ =∑(i,j)∈Rμ(i,j) =∑(i,j)∈Rμ(i,j)+∑(i,j)∉Rμ(i,j) =tr(ρ)

Thus, , and is a witness of the quantum lifting .

() Suppose there is a witness of the quantum lifting . Let us construct the joint sub-distribution :

 μ(i,j)=⟨i|⟨j|ρ|i⟩|j⟩ for all i,j.

It is easy to check:

 n∑j=1μ(i,j)=n∑j=1⟨i|⟨j|ρ|i⟩|j⟩=⟨i|ρ1|i⟩=μ1(i), m∑i=1μ(i,j)=m∑i=1⟨i|⟨j|ρ|i⟩|j⟩=⟨j|ρ2|j⟩=μ2(j).

Also, if , then , then

 μ(i,j)=⟨i|⟨j|ρ|i⟩|j⟩=tr(ρ|i⟩|j⟩⟨i|⟨j|)=0

as . Thus, , and is a witness of the lifting . ∎

The following proposition further shows that in the degenerate case,inequality (2) to (1). Surprisingly, such a reduction can be realized even without the condition of lifting.

###### Proposition 4.

Two statements are equivalent:

• For any , ;

• For all positive observables in and in satisfy , then

 tr(ρ1Y1)≤tr(ρ2Y2)
###### Proof.

As , and are diagonal density operators, so we only need to consider those and which are also diagonal. We use the notation and for simplicity. Then it holds that

 P⊥XR≥Y1⊗I2−I1⊗Y2⟺∀ i,j {Y2,j≥Y1,i(i,j)∈RY2,j≥Y1,i−1(i,j)∉R

Now we need a technical lemma:

###### Lemma 1.

The following two statements are equivalent:

• If , , then

 ∀ i,j {Z2,j≥Z1,i(i,j)∈RZ2,j≥Z1,i−1(i,j)∉R ⇒ m∑i=1μ1(i)Z1,i≤n∑j=1μ2(j)Z2,j
• If , , then

 ∀ i,j {Y2,j≥Y1,i(i,j)∈RY2,j≥Y1,i−1(i,j)∉R ⇒ m∑i=1μ1(i)Y1,i≤n∑j=1μ2(j)Y2,j

where are also diagonal matrices, and , .

For readability, let us first use this lemma to finish the proof of the proposition, but postpone the proof of the lemma itself to the end of this section. As

 tr(ρ1Y1) =m∑i=1(ρ1)