Partial trace

( \mathbb{C}^2 ), for example. Tracing out one subspace results in a 4×4 matrix representation of the resulting operator. Elements in squares connected with lines are added together. In linear algebra and functional analysis, the partial trace is a generalization of the trace. Whereas the trace is a scalar-valued function on operators, the partial trace is an operator-valued function. The partial trace has applications in quantum information and decoherence which is relevant for quantum measurement and thereby to the decoherent approaches to interpretations of quantum mechanics, including consistent histories and the relative state interpretation.

( \mathbb{C}^2 ), for example. Tracing out one subspace results in a 4×4 matrix representation of the resulting operator. Elements in squares connected with lines are added together.]] In linear algebra and functional analysis, the partial trace is a generalization of the trace. Whereas the trace is a scalar-valued function on operators, the partial trace is an operator-valued function. The partial trace has applications in quantum information and decoherence which is relevant for quantum measurement and thereby to the decoherent approaches to interpretations of quantum mechanics, including consistent histories and the relative state interpretation. b_{k, i} = \sum_{j=1}^n a_{k j, i j} This gives a matrix b k,i . The associated linear operator on V is independent of the choice of bases and is by definition the partial trace. Among physicists, this is often called "tracing out" or "tracing over" W to leave only an operator on V in the context where W and V are Hilbert spaces associated with quantum systems (see below). \operatorname{Tr}_W(R \otimes S) = \operatorname{Tr}(S) \, R \quad \forall R \in \operatorname{L}(V) \quad \forall S \in \operatorname{L}(W). To see that the conditions above determine the partial trace uniquely, let v_1, \ldots, v_m form a basis for V , let w_1, \ldots, w_n form a basis for W , let E_{ij} : V \to V be the map that sends v_i to v_j (and all other basis elements to zero), and let F_{kl} \colon W \to W be the map that sends w_k to Since the vectors v_i \otimes w_k form a basis for V \otimes W , the maps E_{ij} \otimes F_{kl} form a basis for From this abstract definition, the following properties follow: \operatorname{Tr}_W (I_{V \otimes W}) = \dim W \ I_{V} \operatorname{Tr}_W (T (I_V \otimes S)) = \operatorname{Tr}_W ((I_V \otimes S) T) \quad \forall S \in \operatorname{L}(W) \quad \forall T \in \operatorname{L}(V \otimes W). \operatorname{Tr}_W\left(\sum_{k,\ell} T^{(k \ell)} \, \otimes \, | k \rangle \langle \ell |\right) = \sum_j T^{(j j)} . The superscripts in parentheses do not represent matrix components, but instead label the matrix itself. The partial trace of ρ with respect to the system B, denoted by \rho ^A , is called the reduced state of ρ on system A. In symbols, \rho^A = \operatorname{Tr}_B \rho. To show that this is indeed a sensible way to assign a state on the A subsystem to ρ, we offer the following justification. Let M be an observable on the subsystem A, then the corresponding observable on the composite system is M \otimes I . However one chooses to define a reduced state \rho^A , there should be consistency of measurement statistics. The expectation value of M after the subsystem A is prepared in \rho ^A and that of M \otimes I when the composite system is prepared in ρ should be the same, i.e. the following equality should hold: \operatorname{Tr}_A ( M \cdot \rho^A) = \operatorname{Tr} ( M \otimes I \cdot \rho). We see that this is satisfied if \rho ^A is as defined above via the partial trace. Furthermore, such operation is unique. Let T(H) be the Banach space of trace-class operators on the Hilbert space H. It can be easily checked that the partial trace, viewed as a map \operatorname{Tr}_B : T(H_A \otimes H_B) \rightarrow T(H_A) is completely positive and trace-preserving. The density matrix ρ is Hermitian, positive semi-definite, and has a trace of 1. It has a spectral decomposition: \rho=\sum_{m}p_m|\Psi_m\rangle\langle \Psi_m|;\ 0\leq p_m\leq 1,\ \sum_{m}p_m=1 Its easy to see that the partial trace \rho ^A also satisfies these conditions. For example, for any pure state |\psi_A\rangle in H_A , we have \langle\psi_A|\rho^A|\psi_A\rangle=\sum_{m}p_m\operatorname{Tr}_B[\langle\psi_A|\Psi_m\rangle\langle \Psi_m|\psi_A\rangle]\geq 0 Note that the term \operatorname{Tr}_B[\langle\psi_A|\Psi_m\rangle\langle \Psi_m|\psi_A\rangle] represents the probability of finding the state |\psi_A\rangle when the composite system is in the state |\Psi_m\rangle . This proves the positive semi-definiteness of \rho ^A . The partial trace map as given above induces a dual map \operatorname{Tr}_B ^* between the C*-algebras of bounded operators on \; H_A and H_A \otimes H_B given by \operatorname{Tr}_B ^* (A) = A \otimes I. \operatorname{Tr}_B ^* maps observables to observables and is the Heisenberg picture representation of . C(X) \otimes C(Y) = C(X \times Y). A state on the composite system is a positive element ρ of the dual of C(X × Y), which by the Riesz–Markov theorem corresponds to a regular Borel measure on X × Y. The corresponding reduced state is obtained by projecting the measure ρ to X. Thus the partial trace is the quantum mechanical equivalent of this operation.

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