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5

I know the question is already answered, but there was some question on my comment and I wanted to elaborate on that. First, let us consider one system only. The $\mathbb{R}$-span of all states $\rho$ is the space of Hermitian operators. Indeed, by the spectral decomposition, already the set of pure states is enough. This also implies that the $\mathbb{C}$-...


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If all $c_{ij}$ are non-negative, this is the definition of a separable state. But you can argue this by giving a protocol that lets you construct, using LOCC, the state $\rho_{AB}$. You probably could use this description to cover entangled states if you let the $c_{ij}$ be negative. Here's a simple argument for bipartite states of qubits: the Pauli ...


1

In quantum computing, we compute the tensor product exactly as you specify. This includes the case of taking a tensor product with two vectors. Now, it is also true that $|\psi\rangle\langle\phi|$ can be thought of as a tensor product between $|\psi\rangle$ and $\langle\phi|$. But in QC we would never define the tensor product between $|\psi\rangle$ and $|\...


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The tensor product of two objects with $m$ and $n$ components is an object with $mn$ components that consists of the pairwise products of the components of the inputs. The Kronecker product and the $v w^T$ product (vector outer product) are both tensor products by that definition. The reason for first defining a tensor product by matrix multiplication in the ...


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