An entangled basis with fixed Schmidt number k (EBk) is a set of orthonormal basis states with the same Schmidt number k in a product Hilbert spaceIt is a generalization of both the product basis and the maximally entangled basis. We show here that, for any k ≤ min{d,Consequently, general methods of constructing SEBk (EBk with the same Schmidt coefficients) and EBk (but not SEBk) are proposed. Moreover, we extend the concept of EBk to multipartite case and find out that the multipartite EBk can be constructed similarly.
The unextendible entangled basis with any arbitrarily given Schmidt number k (UEBk) in C d1 ⊗C d2 is proposed in [Phys. Rev. A 90 (2014) which is a set of orthonormal entangled states with Schmidt number k in a d 1 ⊗ d 2 system consisting of fewer than d 1 d 2 vectors which have no additional entangled vectors with Schmidt number k in the complementary space. In this paper, we extend it to multipartite case and a general way of constructing (m + 1)-partite UEBk from m-partite UEBk is proposed (m ≥ 2). Consequently, we show that there are infinitely many UEBks in C d1 ⊗ C d2 ⊗ · · · ⊗ C dN with any dimensions and any N ≥ 3.
Product states are always considered as the states that don't contain quantum correlation. We discuss here when a quantum channel sends the product states to themselves. The exact forms of such channels are proposed. It is shown that such a quantum channel is a local quantum channel, a composition of a local quantum channel and a flip operation, or such that one of the local states is fixed. Both finite-and infinite-dimensional systems are considered.PACS numbers: 03.65. Ud, 03.65.Db, 03.65.Yz. Quantum systems can be correlated in ways inaccessible to classical objects. This quantum feature of correlations not only is the key to our understanding of quantum world, but also is essential for the powerful applications of quantum information and quantum computation. Product state is the state without any quantum correlation [1,2]. It is the only state that has zero mutual information [3] which is interpreted as a measure of total correlations between its two subsystems. It neither contains quantum discord (QD) [4] nor contains the measurementinduced nonlocality (MIN) [5,6]. Recently, it has been shown that the super discord [7] of ρ ab is zero if and only if it is a product state [8].In particular, it is crucial to study the behavior of quantum correlation under the influence of noisy channel [6,[9][10][11][12][13][14][15][16][17][18][19][20][21]. For example, local channel that cannot create QD is investigated in [9,18,19], local channel that preserves the state with vanished MIN is characterized in [6] and local channel that preserves the maximally entangled states is explored in [20]. The goal of this paper is to discuss when a quantum channel preserves the product states.We fix some notations first. Let H, K be separable complex Hilbert spaces, and B(H, K) (B(H) when K = H) be the Banach space of all (bounded linear) operators from We review the definition of the quantum channel. Let T (H), T (K) be the trace classes on H, K respectively. Recall that a quantum channel is described by a tracepreserving completely positive linear map φ : T (H) → T (K). Every quantum channel φ between two systems respectively associated with Hilbert spaces H and K admits the form [22]whereH is the identity operator on H. If dim H = +∞ and dim K = +∞, then there may have infinite X i s in Eq. (1). We call φ is a completely contractive channel if φ(S(H)) is a single state [23], i.e. there exists a fixed state ω 0 ∈ S(H) such thatLet H ab = H a ⊗ H b with dim H a ≤ +∞ and dim H b ≤ +∞ be the state space of the bipartite system A+B. Let S(H ab ) and S P (H ab ) be the set of all quantum states acting on H ab and the set of all product states in S(H ab ) respectively. That is S P (H ab ) = {ρ ⊗ δ : ρ ∈ S(H a ), δ ∈ S(H b )}. Let {|i } and |j ′ be the orthonormal bases of H a and H b respectively. The operatorThe following is the main result of this paper. Theorem 1. Let φ : T (H ab ) → T (H ab ) be a quantum channel. Then φ(S P (H ab )) ⊆ S P (H ab ) if and only if it has one of the following forms.(i) φ = φ a ⊗ φ b , where φ a and ...
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