Let us order the four eigenvalues ($lambda_1 geq lambda_2 geq lambda_3 geq lambda_4$, $Sigma_{i-1}^4 lambda_i=1$) of a Hermitian, trace-one, positive-definite $4 times 4$ matrix (a “two-qubit density matrix”, in quantum parlance). Those density matrices for which the inequality

$lambda_1-lambda_3 < 2 sqrt{lambda_2 lambda_4}$ holds are termed “absolutely separable” (that is, can not be entangled by any global unitary transformation).

In terms of the Hilbert-Schmidt probability distribution $9081072000 Pi_{j<k}^4 (lambda_j-lambda_k)^2$ on the simplex of ordered eigenvalues (HilbertSchmidt), the volume/probability of the absolutely separable states has been found to be (Evaluate Explicit)

begin{equation}

frac{29902415923}{497664}-frac{50274109}{512 sqrt{2}}-frac{3072529845 pi }{32768

sqrt{2}}+frac{1024176615 cos ^{-1}left(frac{1}{3}right)}{4096 sqrt{2}} approx 0.00365826

end{equation}

while the area/probability of the boundary absolutely separable states (the locus of $lambda_1-lambda_3 =2 sqrt{lambda_2 lambda_4}$ ) has been further shown to be (Confirm)

begin{equation}

-frac{837276448115}{663552}+frac{86031670725 pi }{32768 sqrt{2}}-frac{86031670725

cos ^{-1}left(frac{1}{3}right)}{16384 sqrt{2}} approx 0.15339,

end{equation}

where, notably, $cos ^{-1}left(frac{1}{3}right)$ is the dihedral angle of the regular tetrahedron.

Might these formulas provide any possible insight into the underlying geometry of the two-qubit absolutely separable states?