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2

$\newcommand\dag\dagger$ The Lindblad superoperator is shorthand for a longer operation that generically reads $$\mathcal{L}[\hat{L}]\hat{\rho} = \hat{L} \hat{\rho} \hat{L}^\dag - \tfrac{1}{2}\left(\hat{L}^\dag\hat{L}\hat{\rho} + \hat{\rho}\hat{L}^\dag\hat{L} \right),$$ where the $\dag$ denotes the conjugate transpose (i.e. Hermitian adjoint) of the Lindblad ...


2

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, ...


0

Note that the last formula is incorrect, it should read $$ \rho_{Q} \otimes \rho_X = \sum_{ij} p_i p_j \rho_Q(i) \otimes |j \rangle \langle j| $$ and then you see that in fact the two systems are not correlated.


1

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

$A$ and $B$ are labels for the Hilbert spaces in which each subsystem exists. There is no different physical content between $\mathcal{H}_A\otimes \mathcal{H}_B$ and $\mathcal{H}_B\otimes \mathcal{H}_A$, they are just different ways of bookkeeping. As such, we can immediately trade all of the information about subspaces $A$ and $B$ and write $$|\psi_{BA}\...


2

No, there is no quantum algorithm $\mathcal{D}$ that given a single copy of a quantum state $\rho$ as input determines whether $\rho$ is pure or mixed. Quantum mechanics argument By the principle of deferred measurement, we can assume that $\mathcal{D}$ corresponds to a unitary $U$ followed by a measurement of an observable $M$. Suppose that the eigenvalue $\...


3

(General framework) The general form of this problem is the following. Take a generic Hermitian, unit-trace operator, $X\in\operatorname{Herm}(\mathcal X), \operatorname{Tr}(X)=1$, acting on some $N$-dimensional complex vector space $\mathcal X$. The set of such operators is an affine space of dimension $N^2-1$, embedded in the $N^2$-dimensional real vector ...


2

Since Pauli matrices are traceless and Hermitian, the trace condition $\text{Tr}(\rho)=1$ is satisfied automatically and the numbers $a_k, b_l, E_{kl}$ should be real if we want $\rho$ to be Hermitian. The hard part is to verify the positive semidefinite condition $\rho \ge 0$. The straightforward way is to use Sylvester’s criterion. The positivity condition ...


-1

As a trivial counterexample, let $\mathcal H$ be a Hilbert space with $\dim(\mathcal H)=299792458$, and let $\{|i\rangle\}_{i=1}^{299792458}$ be a basis for it. Then $\rho_s\equiv|1\rangle\!\langle1|\otimes|1\rangle\!\langle1|$ is separable and is already decomposed like you mention, but clearly $|1\rangle$ does not span the full space.


1

In general, no. It can be that the sets of states $\{|\alpha\rangle\}$ and $\{|\beta\rangle\}$ span their respective spaces. However, the size of each set can be larger than the dimension of the space, so the states are not all linearly independent and therefore not a basis. For example, $\rho^{a/b}_1=p_1|0\rangle\langle 0|+(1-p_1)|1\rangle\langle 1|$ and $\...


2

If the density matrices commute then there exists a joint spectral decomposition of both $\rho$ and $\sigma$. I.e. there exists a unitary $U$ and diagonal positive semidefinite matrices $D_{\rho}$ and $D_{\sigma}$ such that $$ \rho = U D_{\rho} U^\dagger \qquad \text{and} \qquad \sigma = U D_{\sigma} U^\dagger. $$ Note that here you are diagonalizing two ...


1

Let $\Phi$ be a channel acting on a state $\rho$ (or more generally, a map acting on a linear operator; we don't actually need restrict to CPTP maps and states for these calculations). Let $J(\Phi)$ be the Choi representation of $\Phi$, i.e. $$J(\Phi)\equiv \sum_{ij} \Phi(E_{ij})\otimes E_{ij},$$ where $E_{ij}\equiv|i\rangle\!\langle j|$. Denote with $J(\Phi)...


2

You want to calculate $$ \rho_{out}=2\text{Tr}_0(\rho_{in}^T\otimes I\cdot\rho^{sys}_{choi}). $$


1

For any map $\Lambda(\cdot)$ with Choi representation $\rho_{C}$ defined as you are doing it (i.e. the channel on the second biparition of the maximally entangled state), the output $\rho_{out} = \Lambda(\rho_{in})$ can be calculated as: $$ \rho_{out} = \mathrm{tr}_{1}\big[\rho_{C}\big(\rho_{in}^{t} \otimes I\big)\big] $$ where the trace is over the ancilla ...


5

Background If $v_1, v_2, \dots, v_n$ is an orthonormal basis in the inner product space $V$, then any vector $u\in V$ can be expressed as a linear combination $$ u = \alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_n v_n.\tag1 $$ Moreover, the coefficients can be computed using $\alpha_k=\langle v_k, u\rangle$, as can be seen by applying $\langle v_k, .\rangle$ ...


5

Of course you can. Take any entanglement measure that can be applied to a system whose overall description is a density matrix, and you can apply that to the density matrix describing your subsystem. (So, you wouldn't use the von Neumann entropy of one qubit because that assumes the overall state is pure.) A particularly good option for a pair of qubits is ...


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