How do I calculate Logarithmic Negativity for the given bipartite state?

Question

How can I calculate Logarithmic Negativity for the given state?

$\rho = \frac{1}{2} |0\rangle \langle0| \otimes |+\rangle \langle+| +\frac{1}{2} |+\rangle \langle+| \otimes |1\rangle \langle1| $

Answer 1

If we defined the logarithmic negativity as $E_N(\rho)= \log_2 \|\rho^{\Gamma_A} \|_1$ then given that $$\rho = \frac{1}{2} |0\rangle \langle0| \otimes |+\rangle \langle+| +\frac{1}{2} |+\rangle \langle+| \otimes |1\rangle \langle1| = \begin{pmatrix} 1/4 & 1/4 & 0 & 0\\ 1/4 & 1/2 & 0 & 1/4\\ 0 & 0 & 0 & 0\\ 0 & 1/4 & 0 & 1/4 \end{pmatrix}$$

and since given a matrix $X$ then its partial transpose with respect to the first system $A$, $X^{\Gamma_A} $ is defined as $$X = \begin{pmatrix} x_{11} & x_{12} & x_{13} & x_{14}\\ x_{21} & x_{22} & x_{23} & x_{24}\\ x_{31} & x_{32} & x_{33} & x_{34}\\x_{41} & x_{42} & x_{43} & x_{44} \end{pmatrix} \hspace{0.5cm} X^{\Gamma_A} = \begin{pmatrix} x_{11} & x_{12} & x_{31} & x_{32}\\ x_{21} & x_{22} & x_{41} & x_{42}\\ x_{13} & x_{14} & x_{33} & x_{34}\\x_{23} & x_{24} & x_{43} & x_{44} \end{pmatrix} $$

we have that the partial transpose of $\rho$ with respect to the first system $A$ is itself. That is, $\rho^{\Gamma_A} = \rho$. Hence, $\|\rho^{\Gamma_A}\|_1 = 1$ and thus $ \log_2 \|\rho^{\Gamma_A} \|_1 = \log_2(1) = 0$ where $\|\cdot\|_1$ is the trace norm, that is, the sum of the absolute value of the eigenvalues -- but since $\rho$ is a density matrix, all eigenvalues are positive, so itis just the trace (which equals $1$).