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3 votes
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How is the expression $\frac{\|\rho^{T_B}\|-1}{2}$ obtained from the definition of negativity?

Firstly $$ \mathrm{Tr}[\rho_{AB}^{T_B}] = 1 $$ as the transpose map is trace-preserving. As the trace of $\rho_{AB}^{T_B}$ is equal to the sum of its eigenvalues we have $$ \sum_i \lambda_i = 1 $$ ...
Rammus's user avatar
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3 votes
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If $\rho_{AB}$ is a separable then the partial transpose w.r.t to A is PSD

The main result that you need to complete all the steps that you mention is that if $\rho$ is a density matrix, then $\rho^T$ is also a density matrix. So, what are the key properties of a density ...
DaftWullie's user avatar
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