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

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?

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.
• 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} Mar 9 at 17:24