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There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ The Choi operator of this map is $J(\Lambda)_{BB'} = |B| I_{BB'} - |B| \Phi^+_{BB}$, where $\Phi^+$ is the maximally entangled state. It follows that $J(\Lambda)\... 5 Yes; in fact,$\rho$is both separable and pure. We can start by writing any state$\rho$in its eigenbasis $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|,$$ where$p_i$are probabilities (i.e., positive and sum to unity) and$|\psi_i\rangle$are pure states that may or may not be entangled. If$\rho$is bipartite, each eigenstate$|\psi_i\rangle$is ... 5 In general, the knowledge of the marginals$\rho_A$and$\rho_B$and the operators$A$and$B$is insufficient to compute$\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different density matrices$\rho_{AB}$and$\sigma_{AB}$with the same marginals for which $$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$... 4 TL;DR Tracing out a subsystem corresponds to discarding it. Suppose Alice has subsystem$A$and Bob has subsystem$B$of the composite system$AB$in state$\rho_{AB}$. Tracing out subsystem$B$gives us $$\rho_a=\mathrm{tr}_B\rho_{ab}=\sum_j \langle j_B|\rho_{ab}|j_B \rangle$$ which represents the state of Alice's subsystem in the absence of any ... 4 If$\rho_{AB}, \sigma_{AB} \ge 0$, i.e. they are positive semi-definite then yes (otherwise no). To prove it observe that if$\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$then we can write that $$\sigma_{AB} = \epsilon \cdot \rho_{AB} + \Delta,$$ for some sufficiently small$\epsilon > 0$and some operator$\Delta \ge 0$. Partial trace is ... 4 The equality follows by hitting both sides of $$X_A\mathrm{tr}_B(Y_{AB})=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB})\tag1$$ with the trace and setting$X_A=M$and$Y_{AB}=\rho^{AB}$. We can establish$(1)$for$Y_{AB}$of the form$Y_{AB}=Y_A\otimes Y_Busing $$\mathrm{tr}_B(A\otimes B)=A\mathrm{tr}(B)\tag2$$ as follows \begin{align} X_A\mathrm{tr}_B(Y_{... 3 Sanity check: the statement is indeed true when \rho is a pure state. We can start by finding the singular values of the combination of purified systems, which I will write as |\phi\rangle and |\Psi\rangle. Given the Hermitian matrix M=|\phi\rangle\langle\phi|-|\Psi\rangle\langle\Psi|, we can look for eigenvectors of the form \alpha|\phi\rangle+\... 3 One data point for the general case (that indicates it's not always possible): \rho_1=|0\rangle\langle0|, \rho_2=|1\rangle\langle1|, p_1p_2\neq0 while \sigma_1=p_1\rho_1+p_2\rho_2 and q_1=1. Note that the purifications of the left-hand side are separable, so p_1|\psi_1\rangle\langle\psi_1|+p_2|\psi_2\rangle\langle\psi_2| is separable. Meanwhile, ... 3 Let \newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m) be two arbitrary finite-dimensional quantum states, and let \calU\in \rmU(\CC^n\otimes\CC^m) be a unitary in the total space. We have[U(\rho\otimes\sigma)U^\dagger]_{ij,k\ell} = \sum_{mnpq} U_{... 3 The key thing that it tells you is that the W-state is partially entangled. This is perhaps a little clearer to see if you trace over two qubits: $$\text{Tr}_{AB}(|W\rangle\langle W|)=\frac23|0\rangle\langle 0|+\frac13|1\rangle\langle 1|.$$ The state is mixed, so the overall pure state is entangled, but it's not maximally mixed, so the overall state is not ... 3 For any\theta\in\mathbb{R}$and any operator$T$such that$T^2=Iwe have $$\exp(i\theta T) = I\cos\theta + i T\sin\theta$$ (c.f. exercise 4.2 on p.175 in Nielsen & Chuang). Therefore, $$\exp(i\theta S_{AB}) = I \cos\theta + i S_{AB} \sin\theta$$ and we have \exp(-i\theta S_{AB}) (\rho_A \otimes \sigma_B) \exp(i\theta S_{AB}) = \\ \rho_A\otimes\... 3 Using the fact e^{i\theta S} = \text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S , we calculate \begin{align*} e^{-i\theta S} (\rho \otimes \sigma) e^{i\theta S} & = (\text{cos}(\theta) \cdot I - i \cdot \text{sin}(\theta) \cdot S) \big(\rho \otimes \sigma \big) (\text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S) \\ &... 3 You could try solving this numerically using semidefinite programming. We know the trace norm of an operator X can be formulated as \begin{aligned} \|X\|_1 &= \min_{Y,Z}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\quad \mathrm{s.t.} \quad \begin{pmatrix} Y & X \\ X^* & Z \end{pmatrix} \geq 0. \end{aligned} $$Furthermore, we can write your problem ... 3 No - take \rho one of the four Bell states, which all have the same marginals. Then the trace you give will evaluate to \mathrm{tr}[APBP], with P one of the four Paulis (including I), which is not only a function of A and B (as it depends on the Pauli, which cannot be inferred from the reduced states). 3 In the absence of the constraint on the marginal, it is true that there exists U_{ABC} such that U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle. Indeed, extend |\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle to an orthonormal basis |\rho_{ABC}^{(k)}\rangle and |\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle to an orthonormal basis |\sigma_{ABC}^{... 3 Minimalist formal proof (I'll use \mu_a\equiv \mu(a)): \textrm{(A)}\Rightarrow\textrm{(B)}: Let \Gamma\ge0. Then,$$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$which is a separable decomposition, since \mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0 because it describes ... 3 It's easy to see that it's true for \rho^{AB} = A \otimes B for any matrices A,B (even when \rho^{AB} is not a state, but just a matrix). Any matrix (not only states) is a linear combination of such products, that is \rho^{AB} = \sum_i A_i \otimes B_i, where A_i,B_i are some matrices. Thus tr(M\rho^A)=tr((M\otimes I_B)\rho^{AB}) since both sides ... 3 If you're tracing over two systems, A and B, you can split this into two steps$$ \text{Tr}(Q_{AB})=\text{Tr}\left(\text{Tr}_B(Q_{AB})\right) $$(To see this, let the basis you use for taking the first trace be the standard basis, |ij\rangle. All I'm doing here is separating out the sums over i and j.) So, if you let Q_{AB}=\tilde M\rho^{AB}, ... 2 When we trace out system b, what we are doing is basically reducing the system down to as if we had just measured system a Its as if you had just measured or discarded system b. Otherwise yes, the probability distribution over the computational basis states described by \rho_a on \mathcal{H}_a is precisely the marginal of the distribution described ... 2 That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements \mu_a, and on obtaining outcome a preparing the state \sigma_a. It should be obvious that this breaks any entanglement, since it (destructively) measures the input. (Note that also ... 2 That is indeed some weirdly written exposition with typos, but the result is correct. Let \Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho) and \Phi_k(\rho)=R_k \text{Tr}(F_k\rho). For \Gamma = \rho_1 \otimes \rho_2 we have$$ (I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) =  = \rho_1\text{Tr}(F_k\... 2 Yes, the conjugate is\langle 0|_A\langle 1|_B$. This is also other times written as$\langle 0_A|\langle 1_B|$, or$\langle 0_A|\otimes \langle 1_B|$, or just$\langle 01|$, or similar ways. These are all just notational differences. They are all equivalent as long as one knows what is being discussed. 2 No, this is not the case. Consider the situation where $$\rho_{AB}=\frac12(|0\rangle\langle 0|\otimes |0\rangle\langle 0|+|1\rangle\langle 1|\otimes |1\rangle\langle 1|).$$ So, we have that$\rho_B=\text{Tr}_A(\rho_{AB})=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)$. Now let$\Pi_A=|0\rangle\langle 0|$, which means that$\sigma_{AB}=|00\rangle\langle ...

2

DaftWullie's answer is correct. The key identity they are using is $$\mathrm{tr}_B(\rho_A\otimes\sigma_{BC}) = \rho_A \otimes (\mathrm{tr}_B\sigma_{BC})\tag1$$ which says that we can pull out tensor factors that do not act on the system being traced over. Using $(1)$ and the symbols defined in the linked question and answer, we have \begin{align} \... 2 The two equations are part of the measurement postulate of quantum mechanics which states that probability of the outcome m in a measurement described by operators M_m on a state \rho is p(m) = \mathrm{tr}(M_m^\dagger M_m \rho)\tag1 $$(c.f. (2.159) in Nielsen & Chuang) and the post-measurement state is$$ \frac{M_m\rho M_m^\dagger}{\mathrm{tr}...

2

Computationally, the easiest way to do this is probably as follows: Let your state be $$|\psi\rangle=\sum_{i,j,k,l}c_{ijkl}|ij\rangle_A|kl\rangle_B$$ Rewrite this as a matrix $$C=\sum_{i,j,k,l}c_{ijkl}|ij\rangle\langle kl|$$ Effectively, you just have to reshape your numpy array. Then, you can calculate $$\rho_A=CC^\dagger$$ or $$\rho_B=C^\dagger C.$$...

2

I'm not sure exactly what the question is, but I can expand a bit about these states. The states you mention are sometimes referred to as "one-way quantum-classical correlated states" (eg here and arxiv version) to refect the properties you describe. They differ from "strictly classical-classical states" of the form $\sum_j p_j|j\rangle \... 2 Another interesting example is a single-photon state in a superposition of two different spatial (or any other type of degree of freedom really) modes: $$\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)|0\rangle\equiv \frac{1}{\sqrt2}(|1\rangle+|2\rangle).$$ This type of states is sometimes not considered "entangled", as there is only a single particle ... 2 It absolutely depends on the subdivision of the spaces. Take the 3-qubit system (qubits A, B and C) in a state $$|0\rangle_A(|00\rangle+|11\rangle)_{BC}$$ We can partition these qubits in various different ways. Clearly the$A|BC$partition has no entanglement across the partition, while the$AB|C$partitioning is maximally entangled. Note that if you go ... 2 The question amounts to whether there are nice enough expressions for the partial traces of powers of an operator in a bipartite (finite-dimensional) space. That's where the result with the standard trace comes from: just observe that$\frac{\partial}{\partial X_{ij}}\operatorname{Tr}(X^n)=n (X^{n-1})_{ji}\$ and apply it to the series expansions of the given ...

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