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In quantum information theory, negativity is defined as summation of the absolute values of negative eigenvalues of the partial transposed density matrix. The expression of negativity is given as $$ \mathcal{N}\left(\rho_{AB}\right)=\frac{|| \rho_{AB}^{T_B} || -1}{2}. $$ However, I am unable to understand how do we arrive at this expression from the definition. Can anyone please explain?

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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 $$ \sum_i \lambda_i = 1 $$ where $\lambda_i$ are the eigenvalues of $\rho_{AB}^{T_B}$. For a normal matrix $X$ we also have that $\|X\|_1 = \sum_{i} |\lambda_i|$, thus we have $$ \begin{aligned} \|\rho_{AB}^{T_B}\|_1 - 1 &= \sum_i |\lambda_i| - \sum_i \lambda_i \\ &= \sum_{i: \lambda_i \geq 0} \lambda_i + \sum_{i : \lambda_i < 0}(-\lambda_i) - \sum_i \lambda_i \\ &= - 2 \sum_{i:\lambda_i < 0} \lambda_i \end{aligned} $$ where on the second line we split the sum of $|\lambda_i|$ into a sum of the nonnegative eigenvalues and a sum of the negative eigenvalues. Dividing through by $2$ we see we are left with just the total absolute sum of the negative eigenvalues.

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  • $\begingroup$ I have a doubt. We know that transpose map preserves eigenvalues and hence trace. But partial transpose does not always preserve eigenvalues (since a density matrix has all positive eigenvalues but its partial transpose may not). So how can we say that partial transpose will always preserve trace? $\endgroup$ Mar 9 at 16:26
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    $\begingroup$ If $\mathcal{E}$ is a trace-preserving map and $\mathcal{F}$ is a trace-preserving map then $\mathcal{E}\otimes \mathcal{F}$ is a trace-preserving map. To see this note we can write any bipartite matrix $M$ in the form $M = \sum_k X_k \otimes Y_k$ and so $$ \begin{aligned} \mathrm{Tr}[(\mathcal{E}\otimes \mathcal{F})(M)] &= \sum_k \mathrm{Tr}[\mathcal{E}(X_k) \otimes \mathcal{F}(Y_k)]\\ &= \sum_k \mathrm{Tr}[\mathcal{E}(X_k)] \mathrm{Tr}[\mathcal{F}(Y_k)] \\ &= \sum_k \mathrm{Tr}[X_k] \mathrm{Tr}[Y_k] \\ &= \mathrm{Tr}[M] \end{aligned} $$ $\endgroup$
    – Rammus
    Mar 9 at 17:24
  • $\begingroup$ Okay, understood. Thank you! $\endgroup$ Mar 9 at 17:33

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