# Tag Info

Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them. Suppose that $$A=B=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}.$$ Then $$A^{T_A}=B^{T_A}=\begin{... 4 There is no good definition of what is an "amount of entanglement". We have some requirements, such as saying that a measure of entanglement must be convex and cannot increase under local operations, but beyond that it is really a matter of taste. There is a nice interpretation of entanglement negativity, though, in the case that \rho^{T_A} only ... 2 Hint 1: Logarithmic negativity is easily computed once we know the eigenvalues of \rho_w^\Gamma - where ^\Gamma denotes the partial transpose - since then it becomes a matter of substitution into the formulas in the Wikipedia article. Hint 2: Finding the eigenvalues becomes easier once we observe that if |v\rangle is an eigenvector of \rho_w^\Gamma ... 2 If we defined the logarithmic negativity as E_N(\rho)= \log_2 \|\rho^{\Gamma_A} \|_1 then given that$$\rho = \frac{1}{2} |0\rangle \langle0| \otimes |+\rangle \langle+| +\frac{1}{2} |+\rangle \langle+| \otimes |1\rangle \langle1| = \begin{pmatrix} 1/4 & 1/4 & 0 & 0\\ 1/4 & 1/2 & 0 & 1/4\\ 0 & 0 & 0 & 0\\ 0 & 1/4 &...
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_{\...