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Results tagged with Search options user 58382
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Consider the quantum state fidelity $F(\rho,\sigma)$ defined as (I will use the notation used in Nielsen & Chuang here): $$F(\rho,\sigma) \equiv \operatorname{Tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}} = \ … asked Jun 2 '18 by glS 2answers Consider a separable state \rho living in a tensor product space \mathcal H\otimes\mathcal H', with \mathcal H and \mathcal H' of dimensions D and D', respectively. If \rho is separable, … asked Apr 13 '18 by glS The name of the concept you are looking for is probability amplitude. The two states \lvert T\rangle = \frac{1}{\sqrt2}(\lvert 10\rangle + \lvert 01\rangle) and \lvert S\rangle = \frac{1}{\sqrt2}( … answered Nov 6 '16 by glS \newcommand{\bs}{\boldsymbol{#1}}\newcommand{\on}{\operatorname{#1}}Let f:\mathbb R^{d^2-1}\to\mathscr B(\mathscr H) be the mapping from points in \mathbb R^{d^2-1} to bounded operators on … answered Aug 27 '18 by glS 1answer Consider a generic single-qubit state$$\rho=\lambda_1\lvert \lambda_1\rangle\!\langle \lambda_1\rvert+\lambda_2\lvert \lambda_2\rangle\!\langle \lambda_2\rvert\in\mathcal H_S. I am interested in un …
It is in the sense that given any pair of states $\rho$ and $\sigma$, you have \begin{align} \operatorname{Tr}_2(\rho\otimes\sigma)&=\rho,\\ \operatorname{Tr}_1(\rho\otimes\sigma)&=\sigma. \end{align} …
A GHZ in $M=2$ would be $|00\rangle+|11\rangle$. Sure you could still call it a GHZ, but it is not very useful because this state already has a name: it's a Bell state. A standard reference to unders …