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

How can I calculate Logarithmic Negativity for the given state?

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

• The state is separable hence its logarithmic negativity is zero. Feb 6 at 16:04

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 $$1$$).
• The $\|\cdot\|_1$ norm is the trace norm, which is the sum of the singular values and is different from the column 1-norm. Otherwise, this is correct. Note that we can simplify the calculation by observing that if $\rho$ is a separable state then $\rho^\Gamma$ is a state and therefore $\|\rho^\Gamma\|_1=1$. Consequently, $E_N(\rho)=0$ for all separable states. Feb 6 at 6:31