We derive a formula for the time-energy costs of general quantum channels proposed in [Phys. Rev. A 88, 012307 (2013)]. This formula allows us to numerically find the time-energy cost of any quantum channel using positive semidefinite programming. We also derive a lower bound to the time-energy cost for any channels and the exact the time-energy cost for a class of channels which includes the qudit depolarizing channels and projector channels as special cases.postprin

Chau, HF

Fung, CHF

Li, CK

Sze, NS

Quantum Information and Computation

We derive a formula for the time-energy costs of general quantum channels proposed in [Phys. Rev. A 88, 012307 (2013)]. This formula allows us to numerically find the time-energy cost of any quantum channel using positive semidefinite programming. We also derive a lower bound to the time-energy cost for any channels and the exact the time-energy cost for a class of channels which includes the qudit depolarizing channels and projector channels as special cases.Department of Applied Mathematic

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Solution to time-energy costs of quantum channels

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Title Solution to Time-energy Costs of Quantum ChannelsAuthor(s) Fung, CHF; Chau, HF; Li, CK; Sze, NSCitation Quantum Information & Computation, 2015, v. 15 n. 7-8, p. 685-693Issued Date 2015URL http://hdl.handle.net/10722/208704Rights Creative Commons: Attribution 3.0 Hong Kong LicensearXiv:1404.4134v2  [quant-ph]  16 Dec 2014 SOLUTION TO TIME-ENERGY COSTS OF QUANTUM CHANNELSCHI-HANG FRED FUNGaDepartment of Physics and Center of Theoretical and Computational Physics,University of Hong Kong, Pokfulam Road, Hong KongH. F. CHAUDepartment of Physics and Center of Theoretical and Computational Physics,University of Hong Kong, Pokfulam Road, Hong KongCHI-KWONG LIDepartment of Mathematics, College of William & Mary, Williamsburg, Virginia 23187-8795, USANUNG-SING SZEDepartment of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Hong KongWe derive a formula for the time-energy costs of general quantum channels proposedin [Phys. Rev. A 88, 012307 (2013)]. This formula allows us to numerically find thetime-energy cost of any quantum channel using positive semidefinite programming. Wealso derive a lower bound to the time-energy cost for any channels and the exact thetime-energy cost for a class of channels which includes the qudit depolarizing channelsand projector channels as special cases.Keywords: Time-energy cost, quantum channel, fidelityCommunicated by : to be filled by the Editorial1 IntroductionA time-energy cost of a unitary matrix U ∈ U(r) is defined as [1]‖U‖max= max1≤j≤r|θj | (1)where U has eigenvalues exp(iθj) for j = 1, . . . , r. Here, we denote by U(r) the group ofr× r unitary matrices, and we take the convention that θj ∈ (−π, π]. This definition of time-energy cost was motivated [1, 2] from time-energy uncertainty relations [3, 4]. Essentially,this time-energy cost captures the idea that time and energy are a trade-off against each otherand may be used as an indicator for the resource used by a quantum system. In particular,aPresent address: Huawei Noah’s Ark Lab, Hong Kong Science Park, Shatin, Hong Kong. E-mail: chf-fung.app@gmail.com12 Solution to time-energy costs of quantum channelsa closed quantum system with a time-independent Hamiltonian H evolves from the initialstate |ψi〉 to the final state |ψf〉 according to the Schro¨dinger equation: |ψf〉 = U |ψi〉 whereU = exp(−iHt/~) and t is the evolution time. The eigenvalues of the Hamiltonian H arethe energies and thus the eigenvalues of logU correspond to the time-energy products, theabsolute maximum of which is the time-energy cost ‖U‖maxdefined above. Note that toimplement the same information processing task characterized by U , one may use a highenergy H run for a short time or a low energy H run for a long time. The time-energyproducts in both cases are the same.The definition for ‖U‖maxin Eq. (1) is for unitary quantum channels. The time-energycost has been extended to cover general quantum channels [2]. A quantum channel mappingn-dimensional density matrices to n-dimensional density matrices can be written asK(ρ) =d∑j=1KjρK†j , (2)where Kj ∈ Cn×n are the Kraus operators and∑dj=1K†jKj = In. In this paper, we onlyconsider finite dimensional systems. The time-energy cost for quantum channel K is definedas the time-energy cost of the most efficient unitary extension that implements K [2]:‖K‖max≡ minU‖U‖max(3)s.t. K(ρ) = TrB[UBA(|0〉B〈0| ⊗ ρA)U†BA] ∀ρ,where the channel K acts on quantum state ρ in system A and the unitary extension UBAincludes system B prepared in a standard state.The time-energy cost has an interesting informational meaning. The cosine of this costfor a general quantum channel is exactly the worst-case entanglement fidelity of the chan-nel [5], establishing a connection between the physical aspect (the time-energy cost) and theinformation aspect (the fidelity) of quantum channels. Fidelity is a popular quantity oftenused to characterize the performance of information processing tasks including quantum keydistribution (as a security measure [6, 7]) and state discrimination (as the inconclusive prob-ability [8, 9, 10]). Thus the study of the time-energy cost is important from a quantuminformation theoretical perspective. To be specific, the result of Ref. [5] shows that for anyquantum channel K, the worst-case entanglement fidelity Fmin(K) of the channel is related tothe time-energy cost bybFmin(K) = cos ‖K‖max . (4)Here, the worst-case entanglement fidelity Fmin(K) is defined asFmin(K) ≡ min|Ψ〉F(|Ψ〉AC〈Ψ|, (KA ⊗ IC)(|Ψ〉AC〈Ψ|)), (5)where the channel acts on system A and the fidelity is taken between the channel input state(allowed to be entangled in systems A and C) and the corresponding output state. Here,F (ρ, ρ′) ≡ Tr√ρ1/2ρ′ρ1/2 is the fidelity between two mixed quantum states ρ and ρ′ [11, 12].bNote that Ref. [5] originally shows that Fmin(K) = max(cos ‖K‖max , 0). However, we should always con-sider taking the freedom of including an all-zero Kraus operator in the channel representation. In this case,cos ‖K‖maxis never negative. See Theorem 1 and its proof.C.-H. F. Fung, H. F. Chau, C.-K. Li, and N.-S. Sze 3This paper derives a formula for the time-energy cost ‖K‖maxdefined in Eq. (3) andprovides a numerical solution method via semidefinite programming. This in turn allows usto compute the the worst-case entanglement fidelity using Eq. (4). The difficulty in solvingfor ‖K‖maxstems from the freedom in the unitary extension. All the freedom we have forchoosing different U without changing the channel consists of the following operations:1. Change the last (d+ 1)n− n columns of U .2. Apply V ⊗ In to U on the left, where V ∈ U(d+ 1).It turns out that one can apply an abstract mathematical result in unitary dilation theory [13]to solve the problem. One can then determine the optimal solution using semidefinite pro-gramming. Thus, we have a theoretical optimal solution that can be determined by numericalmethod. This is one of the best scenarios in solving an optimization problem if there is aclosed form for the optimal solution of the given problem.The organization of this paper is as follows. We solve problem (3) for ‖K‖maxin Sec. 2,and we derive a lower bound to the time-energy cost for any channels and compute the exacttime-energy costs for special channels in Sec. 3. We formulate in Sec. 4 the problem of findingthe time-energy cost as a semidefinite program (SDP) which can be solved numerically andefficiently. We give some mathematical remarks in Sec. 5 and conclude in Sec. 62 Main resultTheorem 1‖K‖max= cos−1[maxv12λmin(Kv +K†v)](6)where v ∈ Cd has ℓ2-norm ‖v‖ ≤ 1, Kv =∑dj=1 vjKj, λmin(·) denotes the minimum eigen-value of its argument, and we take the convention that cos−1 returns an angle in the range[0, π].Proof: The most general form of U in Eq. (3) isU = (V ⊗ In)K1 ∗ ∗ · · · ∗K2 ∗ ∗ · · · ∗......Kd ∗ ∗ · · · ∗Kd+1 ∗ ∗ · · · ∗︸ ︷︷ ︸U ′(7)where V ∈ U(d+1) and only the first n columns of U ′ are fixed. Here, we append an all-zeroKraus operator Kd+1 = 0 in order to make U the most general unitary implementing thechannel K. Certainly, both {K1, . . . ,Kd} and {K1, . . . ,Kd+1} are valid representations of K.As we shall see, there is no need to add more than one extra all-zero operator.We first consider the freedom in U ′. Let d′ = d+ 1. We want to choose the last d′n − ncolumns of U ′ so that its norm is the smallest. This is described as an optimization problem4 Solution to time-energy costs of quantum channelsas follows:ϕ ≡ minU ′‖U ′‖maxs.t. U ′i1 = Ki for all i = 1, . . . , d′,with U ′ ∈ U(d′n) (8)where U ′ij denotes the (i, j) block of size n× n.By the result in Ref. [13], we know that there is a unitary matrix U˜ = (U˜rs)1≤r,s≤2 ∈ U(2n)with eigenvalues e±iθj for j = 1, . . . , n, such that U˜11 = K1 and U˜21 =√In −K†1K1 whereπ ≥ θ1 ≥ · · · ≥ θn ≥ 0 and cos(θ1) = λmin(K1+K†1)/2. Note that there existsW ∈ U(d′n−n)such that (In ⊕W )(U˜ ⊕ Id′n−2n)(In ⊕W )† satisfies the constraints in Eq. (8) and thusϕ ≤∥∥∥U˜∥∥∥max= cos−1[12λmin(K1 +K†1)]. (9)Next, we lower bound ϕ. Consider U ′ satisfying the constraints in Eq. (8). By theinterlacing inequalities (see, e.g., Ref. [14]), because (K1+K†1)/2 is the principal submatrix of(U ′+U ′†)/2, the eigenvalues a1 ≥ · · · ≥ ad′n of (U′+U ′†)/2 and the eigenvalues b1 ≥ · · · ≥ bnof (K1 +K†1)/2 satisfyad′n ≤ bn ≤ an,and socos−1(ad′n) ≥ cos−1(bn).If U ′ has eigenvalues exp(iθj), where j = 1, . . . , d′n and θj ∈ (−π, π], then ad′n = cos(maxj |θj |),givingmaxj|θj | ≥ cos−1[12λmin(K1 +K†1)].Thus, (8) is bounded byϕ ≥ cos−1[12λmin(K1 +K†1)]. (10)Combining with Eq. (9) givesϕ = cos−1[12λmin(K1 +K†1)]. (11)Finally, we optimize V in Eq. (7) to obtain ‖K‖max. Note that ϕ which corresponds to theoptimal solution of U ′ after adjusting the last d′n− n columns depends only on the principalsubmatrix of U ′. Thus,‖K‖max= cos−1[maxv: ‖v‖=112λmin(Kv +K†v)](12)where v ∈ Cd+1 is the first row of V . Here, Kv =∑d+1j=1 vjKj represents the principalsubmatrix of U , where v = [v1, . . . , vd+1]. Taking into account Kd+1 = 0 gives the claim ofthe theorem.We remark that cos ‖K‖max≥ 0.C.-H. F. Fung, H. F. Chau, C.-K. Li, and N.-S. Sze 53 Time-energy costs for special channelsIn this section, we use Theorem 1 to compute the time-energy costs for a class of channelswhich includes the qudit depolarizing channels and projector channels as special cases.Lemma 1 Any channel K can be described by an equivalent form with the Kraus operators{Kj ∈ Cn×n : j = 1, . . . , d} satisfyingTr(Kj) = 0, j = 2, . . . , d.Proof: Two sets of Kraus operators {K1, . . . ,Kd} and {K˜1, . . . , K˜d} describe the same quan-tum channel if and only ifKi =d∑j=1wijK˜j , for i = 1, . . . , d (13)and for some unitary matrix W ≡ [wij ] of dimension d (see, e.g., Theorem 8.2 of Ref. [15]).By taking the trace of Eq. (13), we see that there must exist W that can bring d − 1 termsto zero. In particular, we haveK1 = d∑j=1|Tr(K˜j)|2−12 d∑j=1Tr†(K˜j)K˜j. (14)(If d = 1, we can pad the channel with K2 = 0 to make Lemma 1 automatically hold.)Lemma 2 For any channel K that can be described by Kraus operators {Kj ∈ Cn×n : j =1, . . . , d} of the formTr(Kj) = 0, j = 2, . . . , d,we havecos−1[1n|Tr (K1)|]≤ ‖K‖max. (15)Proof: We consider the middle term of Eq. (6):12λmin(Kv +K†v)≤12nn∑i=1λi(Kv +K†v)=12nTr(Kv +K†v)=1nRe [Tr (Kv)]=1nRe [v1Tr (K1)]where the first line is because the minimum is no greater than the average and λi denotes theith eigenvalue. Maximizing over v gives the claim.6 Solution to time-energy costs of quantum channelsTheorem 2 (Time-energy lower bound) For any channel K described by Kraus op-erators {Kj ∈ Cn×n : j = 1, . . . , d}, we havecos−1 1n√√√√ d∑j=1|Tr (Kj)|2 ≤ ‖K‖max. (16)Proof: This follows from Lemma 1 and Lemma 2.Theorem 3 (Time-energy for special channels) For any channel K that can be de-scribed by Kraus operators {Kj ∈ Cn×n : j = 1, . . . , d} of the formK1 = αI where α ∈ CTr(Kj) = 0, j = 2, . . . , d,(17)its time-energy cost is‖K‖max= cos−1|α|. (18)Proof: From Eq. (15), we have cos−1 |α| ≤ ‖K‖max.On the other hand, by choosing a particular v,maxv12λmin(Kv +K†v)≥maxθ112λmin(eiθ1K1 + e−iθ1K†1)=|α|.Therefore, ‖K‖max≤ cos−1 |α| and the claim is proved.Note that this theorem is slightly more general than Eq. (52) of Ref. [2] in which α is realand positive. As noted in Ref. [2], channels satisfying Eq. (17) include the qudit depolarizingchannels. In the following, we show that projector channels also satisfy Eq. (17).In general, given a channel, we can find an equivalent form according to Lemma 1 andcompute the new K1 using Eq. (14). If this new K1 satisfies Eq. (17), then the time-energycost of the channel is immediately given by Theorem 3. Otherwise, we can lower bound itusing Theorem 2.Theorem 4 (Projector channels) For any channel K that can be described by Krausoperators {Kj ∈ Cn×n : j = 1, . . . , d} of the form Kj = sjPj with Pj = P2j = P†j being aprojector of rank r and sj ∈ C, we have‖K‖max= cos−1(√rn). (19)Proof: Note that Tr(Kj) = sjr for all j. Using Lemma 1 and Eq. (14), an equivalentdescription of K satisfiesK ′1 =1√∑di=1 |si|2I,Tr(K ′j) = 0, j = 2, . . . , d.Next, note that the trace-preserving constraint of quantum channels implies that In =∑dj=1K†jKj =∑dj=1 |sj |2Pj and taking the trace of it gives n/r =∑dj=1 |sj |2. Then byTheorem 3, the claim is proved.C.-H. F. Fung, H. F. Chau, C.-K. Li, and N.-S. Sze 74 Efficient numerical solution using semidefinite programmingOur main result (6) in Theorem 1 can be formulated as an SDP. We can write Kj = Aj+ iBj,where Aj , Bj ∈ Cn×n are Hermitian, and also write vj = aj − ibj with aj , bj ∈ R for j =1, . . . , d. Then the problem is equivalent tomax λmin(d∑i=1(ajAj + bjBj))s.t.d∑j=1(a2j + b2j) ≤ 1(20)where the maximization is over a1, b1, . . . , ad, bd ∈ R. We show that this problem can be castas a complex SDP which has the following form:min gTxs.t. x1G1 + · · ·+ xmGm +H  0(21)where the minimization is over x ∈ Rm. Here, g ∈ Rm, and G1, . . . , Gm, H are complexHermitian matrices. Note that a complex SDP can always be cast as a real SDP in whichG1, . . . , Gm, H are real symmetric matrices.Note that we can rewrite the objective function as follows:min − λs.t.d∑j=1(a2j + b2j) ≤ 1d∑i=1(ajAj + bjBj)  λI(22)where the maximization is over a1, b1, . . . , ad, bd, λ ∈ R. Next, we convert this inequalityconstraint to a positive semidefinite constraint. Let c =√∑dj=1(a2j + b2j). Consider thematrixC =[1 cc 1]which has eigenvalues 1± c. Thus, the constraint c ≤ 1 is equivalent to the constraint C  0.Note that C ⊕ I2d−1 is unitarily similar toa1F1 + . . . adFd + b1Fd+1 + · · ·+ bdF2d + I2d+1where Fj = Ej,2d+1 +E2d+1,j and Ei,j is an (2d+ 1)× (2d+ 1) matrix with one at the (i, j)position. Then, the problem becomesmin − λs.t. a1F1 + . . . adFd + b1Fd+1 + · · ·+ bdF2d + I2d+1  0d∑i=1(ajAj + bjBj)− λI  0(23)8 Solution to time-energy costs of quantum channelswhere the maximization is over a1, b1, . . . , ad, bd, λ ∈ R. This is in the SDP form (21). Thus,one can apply standard positive semidefinite programming to determine the time-energy costof a general quantum channel given in Eq. (6).5 Mathematical remarks• We may replace K1 by eiθ1K1 without affecting the quantum channel. Thus, we canselect θ1 ∈ [0, 2π) to maximize the smallest eigenvalue of eiθ1K1+e−iθ1K†1 . To this end,we can use the numerical range of K1 defined asW (K1) = {〈x|K1|x〉 : |x〉 ∈ Cn, 〈x|x〉 = 1}.This is a compact convex set in C, and can be obtained as the intersection of the halfspacesQθ1 ={µ ∈ C : eiθ1µ+ e−iθ1 µ¯ ≥λmin(eiθ1K1 + e−iθ1K†1)}, θ1 ∈ [0, 2π).So, maximizing the smallest eigenvalue of eiθ1K1 + e−iθ1K†1 corresponds to finding thehalf space Qθ1 whose intersection with the unit disk has the smallest area.• A heuristic approach to upper bound Eq. (6) is as follows. We separately considervjKj, j = 1, . . . , d and let vj = cj exp(iθj) where cj ∈ R+. Choose θj ∈ [0, 2π) tomaximize the smallest eigenvalue σj of eiθjKj + e−iθjK†j . This is equivalent to rotatingthe numerical range W (Kj) so that the left support line is as close to the right side aspossible. Then choose a nonnegative unit vector (c1, . . . , cd) to maximize∑dj=1 cjσj .If Kv =∑dj=1 cj exp(iθj)Kj , then λmin(Kv +K†v)≥∑dj=1 cjσj . Thus, ‖K‖max ≤cos−1(∑dj=1 cjσj/2).6 ConclusionsThe physical meaning of the time-energy cost is its relation with the channel fidelity [5]. Inthis paper, we show that the time-energy cost of any general quantum channel is given byEq. (6). It has closed formulas for special channels. For general channels, the problem offinding the time-energy cost can be formulated as an SDP which can be solved efficiently oncomputers.AcknowledgmentsLi would like to thank Mikio Nakahara for some inspiring discussion. Chau and Fung werepartially supported by the Hong Kong RGC grant No. 700712P. Sze was partially supportedby the Hong Kong RGC grant PolyU 502512. Li was supported by a USA NSF grant anda Hong Kong RGC grant. He was visiting the Hong Kong Polytechnic University, the Uni-versity of Hong Kong, the Shanghai University, and the Institute for Quantum Computingat the University of Waterloo in the Spring of 2014. 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Solution to Time-energy Costs of Quantum Channels

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Solution to Time-energy Costs of Quantum Channels

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