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The two-qubit eigenvalue ($\lambda_i$ >= 0, $i=1,\ldots,4$, $\lambda_4=1-\lambda_1-\lambda_2-\lambda_3$) condition of Verstraete, Audenaert, de Bie and de Moor

AbsoluteSeparability

(p. 6) for absolute separability is

\begin{equation} \lambda_1-\lambda_3 < 2 \sqrt{\lambda_2 \lambda_4},\hspace{.5in}\lambda_1 >\lambda_2 >\lambda_3>\lambda_4. \end{equation} Implementing this condition,

2009paper

I presented an explicit formula \begin{equation} 1-\frac{3217542976-5120883075 \pi +16386825840 \tan ^{-1}\left(\sqrt{2}\right)}{32768 \sqrt{2}}-\frac{29901918259}{497664} \approx 0.00365826 \end{equation}

for the two-qubit absolute separability probability in terms of the Hilbert-Schmidt measure. (``[C]opious use was made of trigonometric identities involving the tetrahedral dihedral angle $\phi=\cos ^{-1}\left(\frac{1}{3}\right)$, assisted by V. Jovovic.")

A "typo" in this formula was corrected, and a somewhat reexpressed pair of formulas, \begin{equation} \label{HSabs} \frac{29902415923}{497664}+\frac{-3217542976+5120883075 \pi -16386825840 \tan ^{-1}\left(\sqrt{2}\right)}{32768 \sqrt{2}} = \end{equation} \begin{equation} \frac{32(29902415923 - 24433216974 \sqrt{2})+248874917445 \sqrt{2}(5 \pi - 16 \tan ^{-1}\left(\sqrt{2}\right))}{2^{16} \cdot 3^5} \approx 0.00365826 \end{equation}

were given in eq.(A2) in

"Quasirandom estimations of two-qubit operator-monotone-based separability probabilities"

Quasirandom

(We, incidentally observe here that \begin{equation} 5 \pi -16 \tan ^{-1}\left(\sqrt{2}\right)=\cos ^{-1}\left(\frac{5983}{3^8}\right).) \end{equation}

Relatedly, in Table 2 (based on eq. (75) there) of

ON THE GEOMETRIC PROBABILITY OF ENTANGLED MIXED STATES

KhvedelidzeRogojin

a fully consistent numerical value of $0.365826 \%$ was given for the Hilbert-Schmidt two-qubit absolute separability probability.

In regard to this result, A. Khvedelidze recently wrote in an email:

"As far I remember we got this result using two numerical methods:

  1. Evaluating integrals over the absolute separability area ;
  2. Generating HS random states and testing them on the absolute separability

Our attempts to evaluate all integrals analytically fell down."

So, let me pose the problem of giving an explicit analytic derivation of the two-qubit absolute Hilbert-Schmidt separability probability, as Khvedelidze and Rogojin were not able to do. ( Possibly, a somewhat different, but equivalent formula could emerge.)

The 2009 analytic derivation of mine was lengthy, "messy", with no full record of the step-by-step derivation given.

A Mathematica formulation of the problem, incorporating the Hilbert-Schmidt measure, proportional to (eq. (15.35) in GeometryOfQuantumStates) \begin{equation} \Pi_{j<k}^4 (\lambda_j-\lambda_k)^2 \end{equation} is

MathematicaFormulation

I've so far been able to perform the first two integrations (with a large outcome), but not the full three.

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A highly skillful analysis in Mathematica of the problem has been given by the user "JimB" in his answer to

MathematicaFormulation .

That analysis was based on a transformation by Nicolas Tessore of the original 3D constrained integration problem to an unconstrained one.

The answer given by JimB \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}} \end{equation} or in Mathematica notation,

29902415923/497664 - 50274109/(512 Sqrt[2]) - (3072529845 π)/(32768 Sqrt[2]) +(1024176615 ArcCos[1/3])/(4096 Sqrt[2])

fully agrees with the earlier set of results-shown in the question statement.

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