Quantum Coupling and Strassen Theorem
aa r X i v : . [ qu a n t - ph ] M a r Quantum Coupling and Strassen Theorem
Li Zhou ∗ Shenggang Ying † Nengkun Yu ‡ Mingsheng Ying § September 18, 2018
Abstract
We introduce a quantum generalisation of the notion of coupling inprobability theory. Several interesting examples and basic properties ofquantum couplings are presented. In particular, we prove a quantumextension of Strassen theorem for probabilistic couplings, a fundamentaltheorem in probability theory that can be used to bound the probabilityof an event in a distribution by the probability of an event in anotherdistribution coupled with the first.
Coupling is a powerful technique in probability theory, with which random vari-ables can be linked to or compared with each other. It has been widely used inthe studies of random walks and Markov chains, interacting particle systems anddiffusions, just name a few, in order to establish limit theorems about them, todevelop approximations for them, or to derive correlation inequalities betweenthem [7].Recently, a very successful application of coupling in computer science wasdiscovered by Barthe et al. [4] that it can serve as a solid mathematical foun-dation for defining the semantics of probabilistic relational Hoare logic. Thisdiscovery enables them to develop a series of powerful proof techniques for rea-soning about relational properties of probabilistic computations, in particular,for verification of cryptographic protocols and differential privacy [2, 3, 1, 6]. ∗ Department of Computer Science and Technology, Tsinghua University, Beijing, China;and Centre for Quantum Software and Information, School of Software, Faculty of Engineeringand Information Technology, University of Technology Sydney, NSW, Australia † Centre for Quantum Software and Information, School of Software, Faculty of Engineeringand Information Technology, University of Technology Sydney, NSW, Australia ‡ Centre for Quantum Software and Information, School of Software, Faculty of Engineeringand Information Technology, University of Technology Sydney, NSW, Australia § Centre for Quantum Software and Information, School of Software, Faculty of Engineeringand Information Technology, University of Technology Sydney, NSW, Australia; and Instituteof Software, Chinese Academy of Sciences, Beijing 100190, China; and Department of Com-puter Science and Technology, Tsinghua University, Beijing 100084, China
For convenience of the reader, we first briefly recall the basics of probabilisticcoupling, following [6]. Let A be a finite or countably infinite set. A sub-distribution over A is a mapping µ : A → [0 ,
1] such that P a ∈A µ ( a ) ≤ P a ∈A µ ( a ) = 1, then µ is called a distribution over A . For asub-distribution µ over A , we define:1. The weight of µ is | µ | = P a ∈A µ ( a );2. The support of µ is supp( µ ) = { a ∈ A : µ ( a ) > } ;3. The probability of an event S ⊆ A is µ ( S ) = P a ∈ S µ ( a ) . Moreover, let µ be a joint sub-distribution, i.e. a sub-distribution over Carte-sian product A × A . Then its marginals π ( µ ) , π ( µ ) over A and A are,respectively, defined by π ( µ )( a ) = X a ∈A µ ( a , a ) for every a ∈ A ,π ( µ )( a ) = X a ∈A µ ( a , a ) for every a ∈ A . Now we can define the notion of coupling.
Definition 1 (Probabilistic Coupling) . Let µ , µ be sub-distributions over A , A , respectively. Then a sub-distribution µ over A × A is called a couplingfor ( µ , µ ) if π ( µ ) = µ and π ( µ ) = µ . Here are some simple examples of coupling taken from [6].2 xample 1.
Let
Flip be the uniform distribution over booleans, i.e.
Flip (0) =
Flip (1) = . Then the following are two couplings for ( Flip , Flip ) :1. Identity coupling: µ id ( a , a ) = ( if a = a , .
2. Negation coupling: µ ¬ ( a , a ) = ( if ¬ a = a , . More generally, let
Unif A be the uniform distribution over a finite nonemptyset A , i.e. Unif A ( a ) = |A| for every a ∈ A . Then each bijection f : A → A yields a coupling µ f for ( Unif A , Unif A ) : µ f ( a , a ) = ( |A| if f ( a ) = a , . Example 2.
For any sub-distribution µ over A , the identity coupling for ( µ, µ ) is: µ id ( a , a ) = ( µ ( a ) if a = a = a, . Example 3.
For any distributions µ , µ over A , A , respectively, the inde-pendent or trivial coupling is: µ × ( a , a ) = µ ( a ) · µ ( a ) . Obviously, coupling for a pair of distributions is not unique. Then the notionof lifting can be introduced to choose a desirable coupling.
Definition 2 (Probabilistic Lifting) . Let µ , µ be sub-distributions over A , A ,respectively, and let R ⊆ A × A be a relation. Then a sub-distribution µ over A × A is called a witness for the R -lifting of ( µ , µ ) if:1. µ is a coupling for ( µ , µ ) ;2. supp( µ ) ⊆ R .Whenever a witness exists, we say that µ and µ are related by the R -liftingand write µ R µ . Example 4.
1. Coupling µ f in Example 1 is a witness for the lifting Unif A { ( a , a ) | f ( a ) = a } Unif A .
2. Coupling µ id in Example 2 is a witness for the lifting µ = µ.
3. Coupling µ × in Example 3 is a witness for the lifting µ T µ , where T = A × A . Proposition 1.
1. Let µ , µ be sub-distributions over A , A , respectively.If there exists a coupling for ( µ , µ ) , then | µ | = | µ | .
2. Let µ , µ be sub-distributions over the same A . Then µ = µ if and onlyif µ = µ . .2 Quantum Coupling With the correspondence of probability distributions/density operators (mixedquantum states) and marginal distributions/partial traces mentioned in the In-troduction, we can introduce the notion of quantum coupling. To this end, letus first recall several basic notions from quantum theory; for details, we refer to[8]. Suppose that H is a finite-dimensional Hilbert space. Let Herm( H ) be theset of Hermitian matrices in H . Let Pos( H ) be the set of positive (semidefinite)matrices in H , and D ( H ) ⊂ Pos( H ) is the set of partial density operators, i.e. ,positive (semidefinite) matrices with trace one. A positive operator ρ in H iscalled a partial density operator if its trace tr ( ρ ) = P i h i | ρ | i i ≤
1, where {| i i} is an orthonormal basis of H .We define its support:supp( ρ ) = span { eigenvectors of ρ with nonzero eigenvalues } = span {| ψ i (cid:12)(cid:12) tr( ρ | ψ ih ψ | ) = 0 } ⊥ . If A is an observable, i.e. Hermitian operator, in H , then its expectation in state ρ is h A i ρ = tr ( Aρ ) . Furthermore, let H , H be two Hilbert space. Then partialtrace over H is a mapping tr ( · ) from operators in H ⊗ H to operators in H defined by tr ( | ϕ ih ψ | ⊗ | ϕ ih ψ | ) = h ψ | ϕ i · | ϕ ih ψ | for all | ϕ i , | ψ i ∈ H and | ϕ i , | ψ i ∈ H together with linearity. The partialtrace tr ( · ) over H can be defined dually.Now we are ready to define the concept of coupling. Definition 3 (Quantum Coupling) . Let ρ ∈ D ( H ) and ρ ∈ D ( H ) . Then ρ ∈ D ( H ⊗ H ) is called a coupling for ( ρ , ρ ) if tr ( ρ ) = ρ and tr ( ρ ) = ρ . 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 H be a Hilbert space and B = {| i i} an orthonormal basis of H . Then the uniform density operator on H is Unif H = 1 d X i | i ih i | where d = dim H is the dimension of H . Indeed, the uniform density operatoron H is unique and independent with the choice of orthonormal basis. For eachunitary operator U in H , we write U ( B ) = { U | i i} , which is also an orthonormalbasis of H . Then ρ U = 1 d X i ( | i i U | i i )( h i |h i | U † ) is a coupling for ( Unif H , Unif H ) . In general, for different U and U ′ , ρ U = ρ U ′ ,though they are both the couplings for ( Unif H , Unif H ) . xample 6. Let ρ be a partial density operator in H . Then by the spectraldecomposition theorem, ρ can be written as ρ = P i p i | i ih i | for some orthonormalbasis B = {| i i} and p i ≥ with P i p i ≤ . We define: ρ id( B ) = X i p i | ii ih ii | . Then it is to see that ρ id( B ) is a coupling for ( ρ, ρ ) . A difference between thisexample and Example 2 is that ρ can be decomposed with other orthonormalbases, say D = {| j i} : ρ = P j q j | j ih j | . In general, ρ id( B ) = ρ id( D ) , and we candefine a different coupling: ρ id( D ) = X j q j | jj ih jj | for ( ρ, ρ ) . Example 7.
Let ρ ∈ D ( H ) and ρ ∈ D ( H ) be density operators. Thentensor product ρ ⊗ = ρ ⊗ ρ is a coupling for ( ρ , ρ ) . The notion of lifting can also be easily generalised into the quantum setting.
Definition 4 (Quantum Lifting) . Let ρ ∈ D ( H ) and ρ ∈ D ( H ) , and let X be a subspace of H ⊗ H . Then ρ ∈ D ( H ⊗ H ) is called a witness of the lifting ρ X ρ if:1. ρ is a coupling for ( ρ , ρ ) ;2. supp ( ρ ) ⊆ X . Example 8.
1. The coupling ρ U in Example 5 is a witness for the lifting: Unif H X ( B , U ) Unif H where X ( B , U ) = span {| i i U | i i} is a subspace of H ⊗ H .
2. The coupling ρ id( B ) in Example 6 is a witness of the lifting ρ = B ρ , where = B = span {| ii i} defined by the orthonormal basis B = {| i i} is a subspaceof H ⊗ H . It is interesting to note that the maximal entangled state | Ψ i = √ d P i | ii i is in = B .
3. The coupling ρ ⊗ in Example 7 is a witness of the lifting ρ ( H ⊗ H ) ρ . As a quantum generalisation of Proposition 1, we have:
Proposition 2.
1. Let ρ ∈ D ( H ) and ρ ∈ D ( H ) . If there exists a cou-pling for ( ρ , ρ ) , then tr ( ρ ) = tr ( ρ ) .2. Let ρ , ρ ∈ D ( H ) . Then ρ = ρ if and only if ∃ orthonormal basis B s.t. ρ = B ρ . roof. Part 1 and Part 2 ( ⇒ ) are obvious. Here, we prove Part 2 ( ⇐ ). If ρ = B ρ , then there exists a coupling ρ for ( ρ , ρ ) such that supp ( ρ ) ⊆ span {| ii i} , where B = {| i i} . Then we have: ρ = P j p j | Ψ j ih Ψ j | for some | Ψ j i ∈ span {| ii i} and p j . Furthermore, for each j , we can write: | Ψ j i = P i α ji | ii i . Then it is rou-tine to show that tr ( | Ψ j ih Ψ j | ) = tr ( | Ψ j ih Ψ j | ) = P i | α ji | | i ih i | . Therefore, itholds that ρ = tr ( ρ ) = P j p j tr ( | Ψ j ih Ψ j | ) = P j p j tr ( | Ψ j ih Ψ j | ) = tr ( ρ ) = ρ . As mentioned in the Introduction, a fundamental theorem for probabilistic cou-pling is the following:
Theorem 1 (Strassen Theorem) . Let µ , µ be sub-distributions over A , A ,respectively. Then µ R µ ⇒ ∀ S ⊆ A . µ ( S ) ≤ µ ( R ( S )) (1) where R ( S ) is the image of S under R : R ( S ) = { a ∈ A |∃ a ∈ S s . t . ( a , a ) ∈R} . The converse of (1) holds if | µ | = | µ | . In this section, we prove a quantum generalisation of the above StrassenTheorem. For this purpose, for any subspace X of H ⊗ H , we use P X and P X ⊥ to denote the projections on X and X ⊥ (the ortho-complement of X ),respectively. We use I , I , I to denote the identity matrix of H , H , H ,respectively. h· , ·i is employed to denote the inner product of matrices living inthe same space, h A, B i = tr( A † B )Then a quantum Strassen theorem can be stated as follows: Theorem 2 (Quantum Strassen Theorem) . For any two partial density oper-ators ρ in H and ρ in H with tr( ρ ) = tr( ρ ) , and for any subspace X of H ⊗ H , the following three statements are equivalent:1. ρ X ρ ;2. For all observables (Hermitian operators) Y in H and Y in H satisfying P X ⊥ ≥ Y ⊗ I − I ⊗ Y , it holds that tr( ρ Y ) ≤ tr( ρ Y ) . (2)
3. For all positive observables Y in H and Y in H satisfying P X ⊥ ≥ Y ⊗ I − I ⊗ Y , it holds that tr( ρ Y ) ≤ tr( ρ Y ) . Proof. ( ⇒ ) Suppose ρ is a witness of the lifting ρ X ρ . Then for allobservables (Hermition operators) Y in H and Y in H , if P X ⊥ ≥ Y ⊗ I − ⊗ Y , then we have: tr( ρ Y ) = tr( ρ ( Y ⊗ I )) (3) ≤ tr( ρ ( P X ⊥ + I ⊗ Y )) (4)= tr( ρ ( I ⊗ Y )) (5)= tr( ρ Y ) . (6)Equalities (3) and (6) are derived from the condition that ρ is a coupling for( ρ , ρ ); that is, tr ( ρ ) = ρ and tr ( ρ ) = ρ , (4) is due to the assumption for Y and Y , and (5) is trivial as supp( ρ ) ⊆ X , so tr( ρP X ⊥ ) = 0.( ⇒ ) Let us first define the semidefinite program (Φ , A, B ):Primal problemmaximize: h A, X i subject to: Φ( X ) = B,X ∈ Pos ( H ⊗ H ) . Dual problemminimize: h B, Y i subject to: Φ ∗ ( Y ) ≥ A,Y ∈ Herm ( H ⊕ H ) . where: A = P X , B = (cid:20) ρ ρ (cid:21) , Φ( X ) = (cid:20) tr ( X ) tr ( X ) (cid:21) , Φ ∗ ( Y ) = Φ ∗ (cid:20) Y ·· Y (cid:21) = Y ⊗ I + I ⊗ Y . To show that the above problems are actually primal and dual, respectively, weonly need to check the following equality: ∀ M, N, h Φ( M ) , N i = tr(tr ( M ) N + tr ( M ) N )= tr( M ( N ⊗ I ) + M ( I ⊗ N ))= h M, Φ ∗ ( N ) i . Moreover, the strong duality holds for this semidefinite program as we can checkthat the primal feasible set are not empty and there exists a Hermitian operator Y for which Φ ∗ ( Y ) > A :Primal feasible set A = { X ∈ Pos ( H ⊗ H ) : Φ( X ) = B } ∋ ρ ) ρ ⊗ ρ Choose Y = I ⊕ I ∈ Herm ( H ⊕ H ) , Φ ∗ ( Y ) = 2 I > P X . So, max h P R , X i = min h B, Y i = min {h ρ , Y i + h ρ , Y i} . Now, let us considerthe following condition: 7 A) : For all observable (Hermitian operators) Y in H and Y in H satisfy Y ⊗ I + I ⊗ Y ≥ P X , then h B, Y i = tr( ρ Y + ρ Y ) ≥ tr ρ . If condition (A) holds, then min h B, Y i ≥ tr ρ . Still remember that max h P X , X i ≤ tr X = tr ρ . Due to the strong duality, we have max h P X , X i = tr ρ . So, X max which maximizes h P X , X i must satisfy h P X , X max i = tr ρ = tr X max . Conse-quently, supp X max ⊆ X ; in other words, X max is a witness of ρ X ρ . There-fore, condition (A) ⇒ ρ R ρ . On the other hand, condition (A) is equivalentto statement of the theorem. Indeed, this is not difficult to prove as if wereplace Y ′ = I − Y in condition (A), then Y ∈ Herm( H ) ⇐⇒ Y ′ ∈ Herm( H ) Y ⊗ I + I ⊗ Y ≥ P X ⇐⇒ I ⊗ I − P X ≥ Y ′ ⊗ I − I ⊗ Y ⇐⇒ P ⊥X ≥ Y ′ ⊗ I − I ⊗ Y tr( ρ Y + ρ Y ) ≥ tr ρ ⇐⇒ tr( ρ Y ) ≥ tr( ρ I ) − tr( ρ ( I − Y ′ )) ⇐⇒ tr( ρ Y ) ≥ tr( ρ Y ′ ) . From the above, we can directly derive statement . In summary, we have:statement ⇔ condition (A) ⇒ ρ R ρ . ( ⇒ ) Obvious.( ⇒ ) We only need to show that, for any two observables Y in H and Y in H satisfy P ⊥X ≥ Y ⊗ I − I ⊗ Y , there exist two positive observables Y ′ in H and Y ′ in H such that P ⊥X ≥ Y ′ ⊗ I − I ⊗ Y ′ andtr( ρ Y ) ≤ tr( ρ Y ) ⇐⇒ tr( ρ Y ′ ) ≤ tr( ρ Y ′ ) . Note that Y and Y are Hermitian, so their eigenvalues are real, and we candefine λ = min { eigenvalues of Y and Y } . Choose Y ′ = Y − λI and Y ′ = Y − λI . Obviously, Y ′ and Y ′ are positive observables, and satisfy P ⊥X ≥ Y ⊗ I − I ⊗ Y = Y ⊗ I − λI ⊗ I + λI ⊗ I − I ⊗ Y = Y ′ ⊗ I − I ⊗ Y ′ . Moreover, as tr( ρ ) = tr( ρ ), we havetr( ρ Y ) ≤ tr( ρ Y ) ⇐⇒ tr( ρ Y ) − tr( ρ λI ) ≤ tr( ρ Y ) − tr( ρ λI ) ⇐⇒ tr( ρ Y ′ ) ≤ tr( ρ Y ′ ) . Remark:
In the above proof, it is indeed naturally to employing the methodsof semidefinite programming. In [6], Hsu deliberately constructs a flow network,8nd then using the max-flow min-cut theorem to prove the Strassen theorem inthe finite case. Essentially, the max-flow min-cut theorem is a special case of theduality theorem for linear programs (LP). Considering the fact that quantumstates, quantum operations and so on are all described by matrices, similar toLP, semi-definite programming (SDP) is a powerful and widely used method ofconvex optimization in quantum theory. Indeed, when all matrices appeared ina SDP are diagonal, then the SDP reduces to LP. In the following section, wewill see that in the degenerate case, quantum Strassen theorem also reduces tothe classical 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 µ be a sub-distribution over [ m ] ([ m ] = { i ∈ N (cid:12)(cid:12) ≤ i ≤ m } )and µ over [ n ]. And the corresponding degenerate partial density operators(quantum states) are: ρ = µ (1) µ (2) . . . µ ( m ) , ρ = µ (1) µ (2) . . . µ ( n ) in H = span {| i i : i ∈ [ m ] } and H = span {| j i : j ∈ [ n ] } , respectively. Further-more, let R ⊆ { ( i, j ) (cid:12)(cid:12) i ∈ [ m ] , j ∈ [ n ] } be a classical relation from [ m ] to [ n ].Then the corresponding (quantum relation) subspace of H ⊗ H is defined as X R = span {| i i| j i (cid:12)(cid:12) ( i, j ) ∈ R } . Based on the above definition of the degenerate case, in the rest part of thissection, Proposition 3 shows that the left hand side of Eqn.(1) in Theorem 1is equivalent to the statement in Theorem 2, while Proposition 4 states theequivalence 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 areduction of Theorem 2 (Quantum Strassen Theorem).The following proposition indicates that probabilistic lifting is a special caseof quantum lifting. Proposition 3. µ R µ ⇐⇒ ρ X R ρ . Proof. ( ⇒ ) Suppose that there is a witness µ of the lifting µ R µ . We definethe partial density operator: ρ : h i |h j | ρ | i ′ i| j ′ i = (cid:26) µ ( i, j ) i = i ′ , j = j ′ i = i ′ or j = j ′ .
9t is easy to check: h i | tr ( ρ ) | i ′ i = n X j =1 h i |h j | ρ | i ′ i| j i = (cid:26) P nj =1 µ ( i, j ) = µ ( i ) i = i ′ i = i ′ , h j | tr ( ρ ) | j ′ i = n X i =1 h i |h j | ρ | i i| j ′ i = (cid:26) P mi =1 µ ( i, j ) = µ ( j ) j = j ′ j = j ′ . So, tr ( ρ ) = ρ and tr ( ρ ) = ρ ; that is, ρ is a coupling for ( ρ , ρ ). Furthermore,we have: tr( ρP X R ) = X ( i,j ) ∈ R h i |h j | ρ | i i| j i = X ( i,j ) ∈ R µ ( i, j )= X ( i,j ) ∈ R µ ( i, j ) + X ( i,j ) / ∈ R µ ( i, j )= tr( ρ )Thus, supp( ρ ) ⊆ X R , and ρ is a witness of the quantum lifting ρ X R ρ .( ⇐ ) Suppose there is a witness ρ of the quantum lifting ρ X R ρ . Let usconstruct the joint sub-distribution µ : µ ( i, j ) = h i |h j | ρ | i i| j i for all i, j. It is easy to check: n X j =1 µ ( i, j ) = n X j =1 h i |h j | ρ | i i| j i = h i | ρ | i i = µ ( i ) , m X i =1 µ ( i, j ) = m X i =1 h i |h j | ρ | i i| j i = h j | ρ | j i = µ ( j ) . Also, if ( i, j ) / ∈ R , then | i i| j i ⊥ X R , then µ ( i, j ) = h i |h j | ρ | i i| j i = tr( ρ | i i| j ih i |h j | ) = 0as supp( ρ ) ⊆ X R . Thus, supp( µ ) ⊆ R , and µ is a witness of the lifting µ R µ .The following proposition further shows that in the degenerate case,inequality(2) to (1). Surprisingly, such a reduction can be realized even without the con-dition of lifting. 10 roposition 4. Two statements are equivalent:1. For any S ⊆ [ m ] , µ ( S ) ≤ µ ( R ( S )) ;2. For all positive observables Y in H and Y in H satisfy P ⊥X R ≥ Y ⊗ I − I ⊗ Y , then tr( ρ Y ) ≤ tr( ρ Y ) Proof. As ρ , ρ and P X R are diagonal density operators, so we only need toconsider those Y and Y which are also diagonal. We use the notation Y ,i =( Y ) ii and Y ,j = ( Y ) jj for simplicity. Then it holds that P ⊥X R ≥ Y ⊗ I − I ⊗ Y ⇐⇒ ∀ i, j (cid:26) Y ,j ≥ Y ,i ( i, j ) ∈ RY ,j ≥ Y ,i − i, j ) / ∈ R Now we need a technical lemma:
Lemma 1.
The following two statements are equivalent: ′ . If Z ,i ∈ { , } , Z ,j ∈ { , } , then ∀ i, j (cid:26) Z ,j ≥ Z ,i ( i, j ) ∈ RZ ,j ≥ Z ,i − i, j ) / ∈ R ⇒ m X i =1 µ ( i ) Z ,i ≤ n X j =1 µ ( j ) Z ,j ′ . If Y ,i ≥ , Y ,j ≥ , then ∀ i, j (cid:26) Y ,j ≥ Y ,i ( i, j ) ∈ RY ,j ≥ Y ,i − i, j ) / ∈ R ⇒ m X i =1 µ ( i ) Y ,i ≤ n X j =1 µ ( j ) Y ,j where Z , Z are also diagonal matrices, and Z ,i = ( Z ) ii , Z ,j = ( Z ) jj . For readability, let us first use this lemma to finish the proof of the propo-sition, but postpone the proof of the lemma itself to the end of this section.As tr( ρ Y ) = m X i =1 ( ρ ) ii ( Y ) ii = m X i =1 µ ( i ) Y ,i , tr( ρ Y ) = m X j =1 ( ρ ) jj ( Y ) jj = n X j =1 µ ( j ) Y ,j , it is direct to see that statement 2 of the proposition is equivalent to statement2 ′ of the above lemma. For the statement 1 ′ of the above, we can define the set S = { i ∈ [ m ] (cid:12)(cid:12) Z ,i = 1 } and T = { j ∈ [ n ] (cid:12)(cid:12) Z ,j = 1 } , then it is equivalent to: ∀ S ⊆ [ m ] , T ⊆ [ n ] , R ( S ) ⊆ T ⇒ µ ( S ) ≤ µ ( T ) , which is exactly the statement 1 of the proposition. Therefore, using the abovelemma, we see that statements 1 and 2 in the proposition are equivalent.11ombining Propositions 3 and 4, we see that Theorem 1 (Strassen Theo-rem for probabilistic coupling) is a reduction of Theorem 2 (Quantum StrassenTheorem).To conclude this section, let us present the following: Proof of Lemma 1. (2 ′ ⇒ ′ ) This is trivial as statement 1 ′ is a special case ofstatement 2 ′ .(1 ′ ⇒ ′ ) For any Y , we can construct a decreasing sequence Z > · · · >Z k > Z k +1) > · · · such that: Y = X k λ k Z k , λ k ≥ . We further define S k = { i ∈ [ m ] (cid:12)(cid:12) Z k,i = 1 } and the corresponding T k = R ( S k ).Then, { S k } is a strictly decreasing sequence; that is, S k ⊃ S k +1 for all k , and { T k } is a non-increasing sequence; that is, T k ⊇ T k +1 for all k . Let us alsodefine Z k : Z k,j = (cid:26) j ∈ T k j / ∈ T k and a new operator Y min : Y min = X k λ k Z k . Note that any pair of Z k and Z k satisfy statement ′ . Then m X i =1 µ ( i ) Y ,i = m X i =1 µ ( i ) X k λ k Z k,i = X k λ k m X i =1 µ ( i ) Z k,i ≤ X k λ k n X j =1 µ ( j ) Z k,j = n X j =1 µ ( j ) X k λ k Z k,j = n X j =1 µ ( j ) Y min,j Now it suffices to prove that for any Y satisfying the condition in statement ′ , we have Y ≥ Y min . To show this, let us use I ( · ) to represent the indicationfunction, and consider the following two cases: • Case 1: t / ∈ T . Then of course, ∀ k : t / ∈ T k , so, Y min,t = X k λ k Z k,t = X k λ k I ( t ∈ T k ) = 0 ≤ Y ,t Case 2: ∃ k such that t ∈ T k . Suppose k t = max { k : t ∈ T k } . Then wehave the following two facts: (1) for k ≤ k t , t ∈ T k and for k > k t . t / ∈ T k ;(2) ∃ s ∈ S k t such that ( s, t ) ∈ R , and for k ≤ k t , s ∈ S k . Combining thesetwo facts, wel have: Y ,t ≥ Y ,s = X k λ k Z k,s = X k λ k I ( s ∈ S k ) ≥ k t X k =1 λ k I ( s ∈ S k ) = k t X k =1 λ k I ( t ∈ T k )= X k λ k I ( t ∈ T k ) = X k λ k Z k,t = Y min,t So, for any Y satisfies the condition in statement ′ , we have: n X j =1 µ ( j ) Y ,j ≥ n X j =1 µ ( j ) Y min,j ≥ m X i =1 µ ( i ) Y ,i . In this paper, we defined the notion of quantum coupling and proved a quan-tum generalisation of Strassen theorem for probabilistic coupling. It is well-known that Strassen theorem is true in both the finite-dimensional and infinite-dimensional cases. However, Theorem 2 (quantum Strassen theorem) was provedonly in the finite-dimensional case. So, an open problem is: whether is quantumStrassen theorem still valid in the infinite-dimensional case? Another interestingtopic for further study is to use the coupling techniques to study the behavioursof quantum random walks and quantum Markov chains. As pointed out in theIntroduction, in the future studies, we hope to apply quantum coupling to de-velop quantum relational Hoare logic and then use it in formal verification ofquantum cryptographic protocols and differential privacy.
This work was partly supported by the Australian Research Council (GrantNo: DP160101652, DE180100156) and the Key Research Program of FrontierSciences, Chinese Academy of Sciences (Grant No: QYZDJ-SSW-SYS003).13 eferences [1] G. Barthe, T. Espitau, J. Hsu, T. Sato and P. -Y. Strub, *-liftings for differ-ential privacy, In:
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