GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS

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1 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS ARTUR O. LOPES AND MARCOS SEBASTIANI ABSTRACT. We denote by M n the set of n by n complex matrices. Given a fixed density matrix β : C n C n and a fixed unitary operator U : C n C n C n C n, the transformation Φ : M n M n Q ΦQ = Tr Q βu describes the interaction of Q with the external source β. The result of this operation is ΦQ. If Q is a density operator then ΦQ is also a density operator. The main interest is to know what happen when we repeat several times the action of Φ in an initial fixed density operator Q 0. This procedure is known as random repeated quantum iterations and is of course related to the existence of one or more fixed points for Φ. In [3], among other things, the authors show that for a fixed β there exists a set of full probability for the Haar measure such that the unitary operator U satisfies the property that for the associated Φ there is a unique fixed point Q Φ. Moreover, there exists convergence of the iterates Φ n Q 0 Q Φ, when n, for any given initial Q 0 We show here that there is an open and dense set of unitary operators U : C n C n C n C n such that the associated Φ has a unique fixed point. We will also consider a detailed analysis of the case when n =. We will be able to show explicit results. We consider the C 0 topology on the coefficients of U. In this case we will exhibit the explicit expression on the coefficients of U which assures the existence of a unique fixed point for Φ. Moreover, we present the explicit expression of the fixed point Q Φ 1. INTRODUCTION We denote by M n the set of n by n complex matrices. Given a fixed density matrix β : C n C n and a fixed unitary operator U : C n C n C n C n, the transformation Φ : M n M n Q ΦQ = Tr Q βu describes the interaction of Q with the external source β. We assume that all eigenvalues of β are strictly positive. In [3] the model is precisely explained: Q is in the small system and β describes the environment. Then ΦQ gives the output of the action of β over Q for the system which is under the influence of the unitary operator U. Other related papers are [] and [4]. The main question is about the convergence of the iterates Φ n Q 0, when n, for any given Q 0. It is natural to expect that any limit if exists is a fixed point for Φ. Our purpose is to show the following theorem: Theorem 1. Given a fixed density matrix β : C n C n, for an open and dense set of unitary operators U : C n C n C n C n the transformation Φ : M n M n Q ΦQ = Tr Q βu 1

2 ARTUR O. LOPES AND MARCOS SEBASTIANI has a unique fixed point Q Φ. In the case n = we present explicitly the analytic characterization of such family of U and also the explicit formula for Q Φ. This result implies one of the main results in [3] that we mentioned before.. THE GENERAL DIMENSIONAL CASE Suppose V is a complex Hilbert space of dimension n and L V denotes the space of linear transformations of V in itself. Then, Tr : L V V L V satisfies Tr A B = TrBA. There is a canonical way to extend the inner product on V to V V. We fix a density matrix β L V. For each unitary operator U L V V we denote by Φ U : L V L V the linear transformation Φ U A = Tr A βu. We denote by Γ L V the set of density operators. It will be shown that Φ U preserves Γ. As Γ is a convex compact space it has a fixed point. The set of unitary operators is denoted by U. If A is such that Φ U A = A, then it follows that the range of Φ U I is smaller or equal to n 1. We will show that there exists a proper real analytic subset X U such that if U is not in X, then the range Φ U I = n 1. In this case the fixed point is unique. More precisely X = {U U : rangeφ U I < n 1}. This X U is an analytic set because is described by equations given by the determinant of minors equal to zero. It is known that the complement of an analytic set, also known as a Zariski open set, is empty or is open and dense on the analytic manifold see [1]. Therefore, in order to prove our main result we have to present an explicit U such that range of Φ U I is n 1. This will be the purpose of our reasoning which will be described below. The bilinear transformation A,B TrBA from L V L V to L V induces the linear transformation Tr : L V V = [L V L V ] L V. Denote by e 1,e,...,e n an orthonormal basis for V. We also denote L i j L V the transformation such that L i j e j = e i and L i j e k = 0 if k j. The set of L i j provide a basis for L V. If A L V we can write A = i, j a i j L i j and we call [a i j ] 1 i, j n the matrix of A. Note that e i e j, 1 i, j n is an orthonormal basis of V V. Moreover, and L ik L jl e k e l = e i e j, L ik L jl e p e q = 0 if p,q k,l. It is also true that: a L i j L pq = 0 if j p, b L i j L p j = L iq, c Tr L i j = 0 if i j and Tr L ii = 1. One can see that L ik L jl, 1 i,k, j,l n is a basis for L V V.

3 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS 3 Given T L V V denote T = t i, j,k,l L ik L jl. Then, Tr T = t i, j,k, j L ik = ik t i, j,k, j L ik. j In the appendix we give a direct proof that: if A Γ, then Φ U A Γ, for all U U. Now we will express Φ U in coordinates. We choose an orthonormal base e 1,e,..,e n V which diagonalize β. That is β = q λ q L qq, λ q > 0, 1 q n, λ q = 1. q Given r,s, 1 r,s n, we will calculate Φ U L rs. Suppose U = u i, j,k,l L ik L jl, then U = u i, j,k,l L ik L jl and L rs βu = q λ q L rs L qq U = λ j u k,l,s, j L rk L jl. j Now, we write U = u α,β,γ,δ L α γ L β δ. Then, we get Finally, UL rs βu = λ j u α,β,r, j u k,l,s, j L α k L β l. Φ U L rs = λ j u α,l,r, j u k,l,s, j L αk = α,k λ j u α,l,r, j u k,l,s, j λ j L αk. j,l As Γ is convex and compact and ϕ U is continuous then as we said before there exists a fixed point A Γ. In particular the range of ϕ U is smaller or equal to n 1. We will present an explicit U such that range of Φ U I is n 1. This will be described by a certain kind of circulant unitary operator Suppose u 1,u,...,u n are complex number of modulus 1. We define U in the following way Ue 1 e 1 = u 1 e 1 e, Ue 1 e = u e 1 e 3,...,Ue 1 e n = u n e e 1, Ue e 1 = u n+1 e e, Ue e = u n+ e e 3,...,Ue e n = u n e 3 e 1,... Ue n e 1 = u n n+1 e n e, Ue n e = u n n+ e n e 3,...,Ue n e n = u n e 1 e 1, We will show that for some convenient choice of u 1,u,...,u n we will get that the range of Φ U I is n 1. Suppose U = u i, j,k,l L ik L jl, in this case Ue k e l = u i, j,k,l e i e j. i, j By definition of U we get a if l < n, then u i, j,k,l 0, if and only if, i = k, j = l + 1; b if k < n, then u i, j,k,n 0, if and only if, i = k + 1, j = 1; c u i, j,n,n 0, if and only if, i = j = 1.

4 4 ARTUR O. LOPES AND MARCOS SEBASTIANI For fixed r,s such that 1,r,s n we get from a, b and c: 1 r < n, 1 s < n, implies n 1 Φ U L rs = 1 s < n, implies n 1 Φ U L ns = 1 r < n, implies u r, j+1,r, j u s, j+1,s, j λ j L rs + u r+1,1,r,n u s+1,1,s,n λ n L r+1s+1, n 1 Φ U L rn = u n, j+1,n, j u s, j+1,s, j λ j L ns + u 1,1,n,n u s+1,1,s,n λ n L 1s+1, u r, j+1,r, j u n, j+1,n, j λ j L rn + u r+1,1,r,n u 1,1,n,n λ n L r+. In particular for 1 r < n we have Φ U L rr = 1 λ n L rr + λ n L r+1r+1. In order to show that the range of Φ U I is n 1 we will show that the ϕ U L rs L rs are linearly independent for r,s n,n Suppose that c rs ϕ U L rs L rs = 0. r,s n,n The coefficient of L is λ n c, then c = 0. The coefficient of L is λ n c λ n c, then c = The coefficient of L nn is λ n c n 1n 1, then c n 1n 1 = 0. Then, we get that c rs ϕ U L rs L rs = 0. 1 r s We will divide the proof in several different cases. a Case n =. c rs ϕ U L rs L rs = c ϕ U L L + c ϕ U L L. r s By definition of U we have that u 1,,1,1 = u 1, u,1,1, = u, u,,,1 = u 3, u 1,1,, = u 4. Therefore, and ϕ U L L = u 1 u 3 λ 1 1L + u u 4 λ L From 1 it follows that ϕ U L L = u 3 u 1 λ 1 1L + u 4 u λ L. u 1 u 3 λ 1 1c + u 4 u λ c = 0 u u 4 λ c + u 3 u 1 λ 1 1c = 0. Taking U such that u 1 = i, u = u 3 = u 4 = 1 it is easy to see that the determinant of the above system is not equal to zero. Then we get that c = c = 0. Then, we get a U with maximal range.

5 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS 5 b Case n >. We choose u 1,u,...,u n according to Lemma 1 below. The equations we consider before can be written as 1 r < n, 1 s < n, r s, then, Φ U L r s L r s = a r s 1L r s + b r s L r+1s+1, 1 s < n, then, Φ U L ns L ns = a ns 1L ns + b ns L 1s+1, 1 r < n, then, Φ U L r n L r n = a r n 1L r n + b r n L r+. For instance n 1 a rs = u r, j+1,r, j u s, j+1,s, j λ j, and b rs = u r+1,1,r,n u s+1,1,s,n λ n. Note that u r, j+1,r, j u s, j+1,s, j has modulus one and also u r+1,1,r,n u s+1,1,s,n. Moreover, b r s = λ n > 0 and a r s < λ λ n 1. Indeed, note first that the products u r, j+1,r, j u s, j+1,s, j are different by the choice of the u i, j,k,l see Lemma 1. Furthermore, by Lemma we get that a rs can not be equal to λ λ n 1. Therefore, a r s 1 1 a r s > 1 n 1 q=1 λ q = λ n = b i j > 0, for all r,s,i, j and r s, i j. Suppose k n. Remember that the L i j define a linear independent set. The coefficient of L 1k in 1 is The coefficient of L nk 1 in 1 is c 1k a 1k 1 + c nk 1 b nk 1 = 0. c nk 1 a nk c n 1k b n 1k 1 = 0. The coefficient of L n k+1 in 1 is c n k+1 a n k c n k 1n b n k+1n = 0. The coefficient of L n k+1n in 1 is c n k+1n a n k+1n 1 + c n kn 1 b n kn 1 = 0. The coefficient of L n kn 1 in 1 is c n kn 1 a n kn c n k 1n b n k 1n = 0. The coefficient of L k+1 in 1 is... c k+1 a k c 1k b 1k = 0. If c 1k 0, then, from above we get c 1k < c nk 1 <... < c k+1 < c 1k. Then, we get a contradiction. It follows that c 1k = 0. Therefore, c nk 1 = c n 1k =... = c n k+1 = c n k+1n = c n kn 1 =... = c k+1 = 0. From this it follows that c r s = 0 for all r,s, when r s. This shows that for such U we have maximal range equal to n 1. Now we will prove two Lemmas that we used before.

6 6 ARTUR O. LOPES AND MARCOS SEBASTIANI Lemma 1. Given m, there exist complex numbers u 1,...,u m of modulus 1, such that, if 1 i j m, 1 k l m and u i u j = u k u l, then i = k, j = l. Proof: The proof is by induction on m For m =, just take u 1 u not in R. Suppose the claim is true for m and u 1,...,u m the corresponding ones. Consider S = {u i u j 1 i, j m } and T = {u p u q 1 p,q m }. Then, take u m+1 such that u m+1 u p is not in S for all 1 p m, and u m+1 is not in T. Then, u 1,...,u m,u m+1 satisfy the claim. Lemma. Consider λ 1,...,λ m, real positive numbers and z 1,...,z m, complex numbers of modulus 1. Suppose m λ j z j = m λ j, then z 1 = z =... = z m. Proof: The proof is by induction on m. It is obviously true for m = 1. Suppose the claim is true for m 1 and we will show is true for m. Note that m From, this follows that λ j = Then, z 1 = z =... = z m 1 = z. Therefore, m m 1 λ j = z m λ j z j m 1 m 1 λ j z j + λ m m 1 λ j z j = m 1 λ j + z m λ m z λ j. m λ j. λ j + z m λ m = m Given v 1,v complex numbers such that v 1 + v = v 1 + v, then they have the same argument. Then, there exists an s > 0 such that z m 1 λ j = sz m λ m. Now, taking modulus in both sides of the expression above, we get m 1 λ j = z m 1 λ j = sz m λ m = sλ m. λ j. From this follows that z m = z

7 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS 7 3. THE TWO DIMENSIONAL CASE - EXPLICIT RESULTS Our main interest in this section is to present the explicit expression of the U which is the unique fixed point. We restrict ourselves to the two dimensional case. We will consider a two by two density matrix β such that is diagonal in the basis f 1 C, f C. Without lost of generality we can consider that p1 0 β =, 0 p p 1, p > 0. We will describe initially in coordinates some of the definitions which were used before on the paper. If and then and Given R S = R R R = R R S S S = S S,, R S R S R S R S R S R S R S R S R S R S R S R S R S R S R S R S R S Tr R S = + S R S + S R S + S R S + S T = T T T 13 T 14 T T T 3 T 4 T 31 T 3 T 33 T 34 T 41 T 4 T 43 T 44 then, in a consistent way we have T + T Tr T = T 13 + T 4 T 31 + T 4 T 33 + T 44 The action of an operator U on M M in the basis e 1 f 1, e f 1, e 1 f, e f is given by a 4 by 4 matrix U denoted by and U = U = U U U U U U U U U U U U U U U U U U U U U U U U U U U U U U U U.

8 8 ARTUR O. LOPES AND MARCOS SEBASTIANI If U is unitary then U U = I. This relation implies the following set of equations: 1U U 3U 4U U U U U 5U U 6U U 7U U 8U U 9U 10U U U U U U U 13U U 14U U U U U U U U U U U U U U U 15U U 16U U U U = 1, U = 0, U = 0, U = 0, U = 0, U U = 1, U U = 0, U U = 0, U U U U = 0, U = 0, U U U = 1, U U U U U U U = 0, U U = 0. U U = 0. U U = 0. U U = 1. Equation is equivalent to 5, equation is equivalent to 13, equation 8 is equivalent to 15, equation 3 is equivalent to 9, equation 7 is equivalent to 10 and equation 4 is equivalent to 14. Then we have 6 free parameters for the coefficients of U. Using the entries rs we considered above we define U LQ = p 1 i1 U i1 U i1 i=1 U i1 U i1 Q U i U i U i Q p i=1 U i U i U i1 U i1 U i1 U i U i U i +

9 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS 9 We can consider an auxiliary L i j and express LQ = i=1 p 1 i1 Q p 1 U i1 + i=1 L i1 QL i1 + i=1 i=1 L i QL i = p i Q p U i = i. L i j QL i j. From the fact that U U = I it follows after a long computation that LI = I. Note that L preserves the cone of positive matrices. Using the entries Urs i j described above we denote U i1 ˆLQ = p 1 U i1 U i1 i=1 U i1 Q U i1 U i p U i U U i i=1 U i Q i U i U i U i = U i1 U i1 U i1 i. One can also show that ˆLQ = Tr [U Q βu ] see [3]. + L i j QL i j. The first expression is the Kraus decomposition and the second the Stinespring dilation. Moreover ˆL preserve density matrices. This is proved in the appendix but we can present here another way to get that. If Q is a density matrix, then TrˆLQ = Tr Tr i. i. L i j QL i j = i. QL i jl i j = TrQ TrL i j QL i j = i. i. L i jl i j = TrQ = 1 TrQL i jl i j = We denote Then, Q i j = Q Q Q = Q Q. Q Q Q Q i j Q i j Q + i j Q i j Q i j Q i j Q + i j Q i j Q i j Q i j Q + i j Q i j Q i j Q i j Q + i j Q i j Q We have to compute ˆLQ = p 1 [U Q Q ] + p [U Q Q ]. The coordinate a of ˆLQ is p 1 [U Q Q + U p 1 [U Q Q + U p [U Q Q + U Q Q ]+ Q Q ]+ Q Q ]+ =,

10 10 ARTUR O. LOPES AND MARCOS SEBASTIANI The coordinate a is p [U Q Q + U p 1 [U Q Q + U p 1 [U Q Q + U p [U Q Q + U Q Q ]. Q Q ]+ Q Q ]+ Q Q ]+ p [U Q Q + U Q Q ]. 3 We will consider a parametrization of the density matrices taking Q = 1 Q and Q = Q. The variable Q is positive in the real line and smaller than one. Indeed, by positivity of Q, we have 0 Q Q = Q 1 Q = Q Q. Q is in C = R but satisfying Q 1 Q Q Q 0 because we are interested in density matrices which are positive operators. The numbers p 1 and p are fixed. Consider the function G such that GQ,Q = p 1 [U Q Q + U p 1 [U Q Q + U p [U Q Q + U p [U Q Q + U p 1 [U Q Q + U p 1 [U Q Q + U p [U Q Q + U p [U Q Q + U When is there a unique fixed point for G? Q 1 Q ]+ Q 1 Q ]+ Q 1 Q ]+ Q 1 Q ], Q 1 Q ]+ Q 1 Q ]+ Q 1 Q ]+ Q 1 Q ] Example: Suppose U = e iβ σ x σ x =cosβi I + i sinβσ x σ x. In this case cosβ 0 0 i sinβ U = 0 cosβ i sinβ 0 0 i sinβ cosβ 0 i sinβ 0 0 cosβ Therefore, GQ,Q = p 1 p 1 Q + p p Q, p 1 cosβ Q + p 1 sinβ Q + p sinβ Q + p cosβ Q = 1 Q, p 1 cosβ Q + p 1 sinβ Q + p sinβ Q + p cosβ Q One can easily see that given any a R we have that Q = 1/, and Q = a determine a fixed point for G. The fixed point matrix will be positive if 1/ < a < 1/. In this case the fixed point is not unique. Now we return to the general case.

11 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS It is more convenient to express G in terms of the variables Q [0,1], and a,b R, where Q = a + bi. As these parameters describe density matrices there are some restrictions: 1/4 Q 1 Q a + b and 1 Q 0 We denote by Rez the real part of the complex number z and by Imz its imaginary part. In this case we get GQ,a,b = where, Q α 1 + β 1 + a + a a + ia a b, ReQ α + β + a + a a + ia a b, ImQ α + β + a + a a + ia a b. α 1 = p 1 [U U U U U U U ]+ p [U U U U U U U ], β 1 = p 1 [U U U ] + p [U U U ], α = p 1 [U U U U U U U ]+ p [U U U U U U U ], β = p 1 [U U U ] + p [U U U ], a = p 1 [U U U ] + p [U U U ], a = p 1 [U U U ] + p [U U U ], a = p 1 [U U U ] + p [U U U ], a = p 1 [U U U ] + p [U U U ], α 1 is a real number. As Φ takes density matrices to density matrices we have that β 1 is also real. Note that α 1 < 1 and 1 > β 1 > 0. It is easy to see from the above equations that a + a and ia a are both real numbers. We are not able to say the same for a + a a or ia a b. In order to find the fixed point we have to solve which means in matrix form Q α 1 + β 1 + a + a a + ia a b = Q Q α + β + a + a a + ia a b = a + bi, α1 1 a + a ia a α a + a 1 ia a 1 Q a b = β1 β. We are interested in real solutions Q,a,b.

12 ARTUR O. LOPES AND MARCOS SEBASTIANI In the case of the example mentioned above one can show that α 1 = 1 and α 0 = 0 which means that in the expressions above we get a set of two equations in two variables a,b, Remember that we are interested in matrices such that 1/4 Q 1 Q a +b. Notice that 0 Q 1. As Φ takes density matrices to density matrices there is a fixed point for G by the Brower fixed point theorem. The main question is the conditions on U and β such that the fixed point is unique. If there is a solution ˆQ,â, ˆb 0,0,0 in R 3 to the equations ˆQ α a + a â + ia a ˆb = 0 ˆQ α + a + a 1â + ia a 1 ˆb = 0, 4 then, the fixed point is not unique. The condition is necessary and sufficient. A necessary condition for the fixed point to be unique is to be nonull the determinant of the operator a + a K = ia a. a + a 1 ia a 1 Notice that if z 1,z satisfies Kz 1,z = 0,0, then z 1 z is real because a + a and ia a are real. From this follows that there exist a solution a,b R in the kernel of K. In this case 0,a,b is a nontrivial solution of 4. The condition det K 0 is an open and dense property on the unitary matrices U. Indeed, there are six free parameters on the coefficients Urs. i j Consider an initial unitary operator U. One can fix 5 of them and move a little bit the last one. This will change U and will move the determinant of K U in such way that can avoid the value 0 for some small perturbation of the initial U. Suppose U satisfies such property Det U 0. For each real value Q we get a different a Q,b Q which is a solution of Ka,b = Q α 1 1, Q α. In this way we get an infinite number of solutions Q,a Q,b Q R C to 4. α is not real. But, we need solutions on R 3. Denote by S = S U the linear subspace of vectors in C of the form ρα 1 1,α, where ρ is complex. Lemma 3. For an open and dense set of unitary U we get that K 1 S R = {0,0}. For such U, suppose Q,a,b satisfies equation 4, then the non-trivial solutions â, ˆb of Kâ, ˆb = Q α 1 1, Q α are not in R. Proof: Suppose 1 α 1 α = α + βi = z 0 = zu 0. Note that for a generic U we have that α 0. We denote C = a + a, C = ia a, C = a + a 1 and finally C = ia a 1. Suppose Q,a,b R 3 satisfies equation 4. We know that generically on U the value Q is not zero. For each C i j we denote C i j = Ci 1 j +C i j i, where i, j = 1,. If Kâ, ˆb = Q α 1 1, Q α, then In this case C â +C ˆb = z 0 C â +C ˆb = α + βic â +C ˆb. C â +C ˆb = αc 1 â βc â βc 1 ˆb αc ˆb+

13 GENERIC PROPERTIES FOR RANDOM REPEATED QUANTUM ITERATIONS 13 iβc 1 â + αc â + αc 1 ˆb βc ˆb. If â and ˆb are real, then, as C and C are real, then Moreover, If βc 1 + αc â + αc 1 βc ˆb = 0. 5 αc 1 βc C â βc 1 αc C ˆb = 0 6 Det βc 1 + αc αc 1 βc αc 1 βc C βc 1 αc C then just the trivial solution 0,0 satisfies 5 and 6. The above determinant is non zero in an open and dense set of U. Then, the solution Q,a,b R 3 of 4 has to be trivial. 0, Under these two assumptions on U which are open and dense the fixed point for G is unique. Then, it follows that the density matrix Q = Q Φ which is invariant for Φ is unique. Given an initial Q 0 any convergent subsequence Φ n kq 0, κ will converge to the fixed point because is unique. As GQ,a,b = Q α 1 + β 1 + a + a a + ia a b, ReQ α + β + a + a a + ia a b, ImQ α + β + a + a a + ia a b, one can find the explicit solution Q a + bi Q Φ = a bi 1 Q by solving the linear problem GQ,a,b = Q,a,b. 4. APPENDIX Lemma 4. Given A,B L V, then TrA B = TrTr A B. Moreover, TrTr T = TrT, for all T L V V. Proof: Indeed, TrA B = TrATrB = TrTrAB = TrTr A B. Lemma 5. Given T L V V, a if T is selfadjoint, then, Tr is also selfadjoint, b moreover, if T is also positive semidefinite then Tr T is semidefinite.

14 14 ARTUR O. LOPES AND MARCOS SEBASTIANI Proof: a If T is selfadjoint, then, t i jkl = t kli j. This implies that j t i jk j = j t k ji j. Therefore, Tr is selfadjoint. b If T is positive semidefinite, then < T x x, x x > 0, for all x,x V. In particular, < T x e q, x e q > 0, for all x = c 1 e c n e n V and 1 q n. As T x e q = t i jkl L ik x L jl e q = t i jkq c k e i e j., then < T x e q, x e q >= t iqkq c k c i, q = 1,,...,n. i,k From this follows that i,k,q t iqkq c k c i = i,k q t iqkq c k c i 0. Then, < Tr T x,x > 0. Note that the analogous property for positive definite T is also true. Lemma 6. If A Γ, then Φ U A Γ, for all U U. Proof: As A and β are selfadjoint and positive semidefinite the same is true for A β. Then, the same is true for UA βu. From Lemma 5 we get that Φ U A = Tr A βu is selfadjoint. By Lemma 4 TrΦ U A = TrA βu = TrA b = TrAT RB = 1. Instituto de Matematica-UFRGS Av. Bento Gonçalves, CEP , Porto Alegre, Brasil A. O. Lopes partially supported by CNPq, PRONEX Sistemas Dinâmicos, INCT, and beneficiary of CAPES financial support arturoscar.lopes@gmail.com REFERENCES [1] J. Bochnak, M. Coste and M-F Roy, Real Algebraic Geometry, Springer Verlag 1998 [] C. Liu, Unitary conjugation channels with continuous random phases, Quantum Stud.: Math. Found. 014 [3] I. Nechita and C. Pellegrini, Random repeated quantum interactions and random invariant states, Probab. Theory and Relat. Fields, [4] L. Bruneau, A. Joye, and M. Merkli, Repeated interactions in open quantum systems, Journal of Mathematical Physics 55,