# Is the set of two-qubit absolutely separable states convex?

Companion question on MathOverflow

Let us order the four nonnegative eigenvalues, summing to 1, of a two-qubit density matrix ($$\rho$$) as $$$$1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0.$$$$ The set ($$S$$) of absolutely separable states (those that can not be "entangled" by global unitary transformations) is defined by the additional inequality (eq. (1) in Halder) $$$$x - z \leq 2 \sqrt{y (1-x-y-z)}.$$$$

Is the set $$S$$, that is, $$$$1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x - z \leq 2 \sqrt{y (1-x-y-z)},$$$$ convex?

If so, I would like to seek to determine the John ellipsoids JohnEllipoids containing and contained within $$S$$ and see if they are simply the same as the circumscribed ($$\mbox{Tr}(\rho^2) \leq \frac{3}{8}$$) and inscribed ($$\mbox{Tr}(\rho^2) \leq \frac{1}{3}$$) sets, respectively Adhikari .

These two sets are determined by the constraints $$$$1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x^2 +y^2 +z^2 +(1-x-y-z)^2 \leq \frac{3}{8}.$$$$ and $$$$1 \geq x \geq y \geq z \geq (1-x-y-z) \geq 0 \land x^2 +y^2 +z^2 +(1-x-y-z)^2 \leq \frac{1}{3}.$$$$ (The latter set corresponds to the separable "maximal ball" inscribed in the set of two-qubit states (sec. 16.7 GeometryQuantumStates).

Further, I am interested in the Hilbert-Schmidt probabilities (relative volumes) Hilbert-Schmidt of these various sets. These probabilities are obtained by integrating over these sets the expression $$$$9081072000 \left(\lambda _1-\lambda _2\right){}^2 \left(\lambda _1-\lambda _3\right){}^2 \left(\lambda _2-\lambda _3\right){}^2 \left(2 \lambda _1+\lambda _2+\lambda _3-1\right){}^2 \left(\lambda _1+2 \lambda _2+\lambda _3-1\right){}^2 \left(\lambda _1+\lambda _2+2 \lambda _3-1\right){}^2,$$$$ where the four eigenvalues are indicated. (This integrates to 1, when only the eigenvalue-ordering constraint--given at the very outset--is imposed.)

In the answer to 4-ball, we report formulas for the Hilbert-Schmidt probabilities (relative volumes) of these inscribed and circumscribed sets, that is, $$$$\frac{35 \pi }{23328 \sqrt{3}} \approx 0.00272132$$$$ and the considerably larger $$$$\frac{35 \sqrt{\frac{1}{3} \left(2692167889921345-919847607929856 \sqrt{6}\right)} \pi}{27518828544} \approx 0.0483353.$$$$ (We also have given an exact--but still quite cumbersome--formula [$$\approx 0.00484591$$] for $$\mbox{Tr}(\rho^2) \leq \frac{17}{50}$$.)

Further, in the answers to AbsSepVol1 and AbsSep2 , the formula for the Hilbert-Schmidt volume (confirming and rexpressing the one given in 2009paper) $$$$\frac{29902415923}{497664}-\frac{50274109}{512 \sqrt{2}}-\frac{3072529845 \pi }{32768 \sqrt{2}}+\frac{1024176615 \cos ^{-1}\left(\frac{1}{3}\right)}{4096 \sqrt{2}} \approx 0.00365826$$$$ of the intermediate absolutely separable set $$S$$ has been given.

As to the total (absolute and non-absolute) separability probability of the 15-dimensional convex set of two-qubit density matrices, compelling evidence of various kinds--though yet no formalized proof--indicate that its value is the considerably larger $$\frac{8}{33} \approx 0.242424$$ MasterLovasAndai . (One can also enquire as to the John ellipsoids for this [known-to-be] convex set JohnEllipsoid2.)

Here is a joint plot of the three sets of central interest here.

ThreeSetPlot

• It's written that $S$ is convex and compact in your first reference. The follow up reference is journals.aps.org/pra/abstract/10.1103/PhysRevA.89.052304 – Danylo Y Nov 12 '20 at 7:26
• Thanks, Danylo Y! I see that the convexity assertion is in the abstract of the follow up reference. – Paul B. Slater Nov 12 '20 at 13:46
• Nathaniel Johnston has given a quite thoughtful answer to the companion question mathoverflow.net/questions/376200/… He distinguishes between the ordered spectra of the absolutely separable states (the focus of my interest) and the (higher-dimensional) absolutely separable states (to which the Danylo comment pertains). – Paul B. Slater Nov 12 '20 at 14:37
• Mark S--the comment above yours references the "companion question". The MO question had the additional "(by definition, $4 \times 4$, Hermitian, nonnegative definite, trace one) "two-qubit density matrix"" intended for non-quantum-oriented folk. – Paul B. Slater Nov 14 '20 at 13:50