The entanglement negativity, introduced in (Vidal and Werner 2002), is defined as $$\mathcal N(\rho) \equiv \frac{\|\rho^{T_B}\|_1-1}{2}.$$ It is mentioned there that this equals the sum of the absolute values of the negative values of the eigenvalues of the partial transpose $\rho^{T_B}$: $\mathcal N(\rho)=\sum_{\lambda<0} \lvert\lambda\rvert$.

How does one show the equivalence of these two quantities?


1 Answer 1


This is also discussed in the paper linked above. The trace norm of $X$ is defined as the sum of the absolute values of the eigenvalues of $X$: $\|X\|_1=\sum_i \lvert\lambda_i\rvert$. $\newcommand{\tr}{\operatorname{tr}}$Given a state $\rho$, the normalisation condition amounts to $$\tr(\rho) = \sum_i \lambda_i = \sum_{\lambda\in\sigma_+}\lambda + \sum_{\lambda\in\sigma_-}\lambda = 1,$$ where $\sigma_\pm$ is the set of positive (negative) eigenvalues. Given $\rho=\rho_{ij,k\ell}|ij\rangle\!\langle k\ell|$, the partial transpose $\rho^{T_B}$ reads $\rho^{T_B}=\sum_{ijk\ell}\rho_{ij,k\ell}|i\ell\rangle\!\langle kj|$, which means that $$\tr(\rho^{T_B})=\sum_{ij} \rho_{ij,ij} = \tr(\rho) = 1.$$

It follows that $$\|\rho^{T_B}\|_1 = \sum_i \lvert\lambda_i\rvert = \sum_{\lambda\in\sigma_+}\lambda - \sum_{\lambda\in\sigma_-}\lambda = 1 - 2 \sum_{\lambda\in\sigma_-}\lambda. $$ And therefore $$\mathcal N(\rho) = \frac{\|\rho^{T_B}\|_1-1}{2} = -\sum_{\lambda\in\sigma_-}\lambda = \sum_{\lambda\in\sigma_-}\lvert\lambda\rvert.$$


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