Exact maximum purity of absolutely separable qubit–qudit states
Establish the exact maximum purity of absolutely separable states ρ ∈ M_2 ⊗ M_n for n ≥ 3 by proving that the maximum equals 2/(3n) when n is even and equals (6n+4)/( (3n+1)^2 ) when n is odd, and demonstrate that the maximizers are precisely those states whose spectra satisfy λ1 = … = λ_{n/2} = 1/n and λ_{n/2+1} = … = λ_{2n} = 1/(3n) for even n, and λ1 = … = λ_{(n+1)/2} = 3/(3n+1) and λ_{(n+3)/2} = … = λ_{2n} = 1/(3n+1) for odd n.
References
On the basis of these numerical estimates, we propose the following conjecture regarding the exact maximum purity. Conjecture \ref{conj1} The maximum purity of absolutely separable states \rho \in M_2\otimes M_n, n\ge 3, is given by \frac{2}{3n} if n is even, and \frac{6n+4}{(3n+1)2} if n is odd. Furthermore, this maximum value is reached when \lambda_1= \cdots =\lambda_{\frac{n}{2}}=\frac{1}{n}, \lambda_{\frac{n}{2}+1}=\cdots=\lambda_{2n}=\frac{1}{3n} if n is even, and \lambda_1= \cdots =\lambda_{\frac{n+1}{2}}=\frac{3}{3n+1}, \lambda_{\frac{n+3}{2}}=\cdots=\lambda_{2n}=\frac{1}{3n+1} if n is odd.