# Is the set of two-qubit absolutely separable states convex, and if so, what are its John ellipsoids?

Let us order the four nonnegative eigenvalues, summing to 1, of a (by definition, $$4 \times 4$$, Hermitian, nonnegative definite, trace one) "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 \Pi_{j 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

• This could stop at “containing and contained within $S$”, by which time it has asked the two questions in the title. Everything after that looks long and messy. Nov 12 '20 at 6:13
• Thanks, Matt F.! I understand your interest in trim, to-the-point questions. To me the excess is motivation/background. But I do accept the charge of "blowing my own horn". Further, the nice answer of Nathaniel Johnston clarifies the issue with my recently deleted question, to which you objected as to use of the term "ball". What confusion there is arises between simply the ordered spectra (which is the focus of my interest) and the full states themselves, for which the absolutely separable states are known to comprise balls, as discussed in the GeometryQuantumStates link in my question. Nov 12 '20 at 14:31
• Hi! I think this is crossposted on QCSE. Please edit the post to indicate the cross posting. Thanks! Nov 14 '20 at 1:37
• Per Mark S comment--companion post--directed to quantum community-is quantumcomputing.stackexchange.com/questions/14588/… Nov 14 '20 at 13:52

There are two slightly different questions here (even without discussing John ellipsoids, which I do not know the answer to), so I'll try to be careful in my answer.

Question 1: Is the set of absolutely separable states convex? The answer is yes, almost trivially: if $$\rho$$ and $$\sigma$$ are positive semidefinite with trace $$1$$ and are absolutely separable, then so is $$p\rho + (1-p)\sigma$$. This follows immediately from convexity of the set of separable states themselves.

Question 2: Is the set of ordered spectra of absolutely separable states convex? Since absolute separability of a state is determined entirely by the spectrum, this is a reasonable question (and I believe it's the one you're actually interested in). The answer here is also "yes, it is convex", but I don't believe that this follows from any "obvious" argument (in particular, I only know how to prove it in small dimensions, not in all dimensions like with Question 1).

Indeed, convexity follows from Hildebrand's characterization of "absolutely PPT states" from the paper "R. Hildebrand. Positive partial transpose from spectra. Phys. Rev. A, 76:052325, 2007. (arXiv:quant-ph/0502170)", and the fact that the sets of absolutely PPT states and absolutely separable states coincide in the 2-qubit case (and in the qubit-qudit case). He showed in that paper that the spectra of absolutely PPT states are determined by a certain family of linear matrix inequalities (i.e., the set of spectra of absolutely PPT states form a spectrahedron, and is thus convex).

For example, in the particular case of 2-qubit states, the troublesome inequality $$x-z \leq 2\sqrt{y(1-x-y-z)}$$ is equivalent to positive semidefiniteness of the $$2 \times 2$$ matrix $$\begin{bmatrix} 2(1-x-y-z) & z-x \\ z-x & 2y \end{bmatrix}.$$

• Thanks very much N. J. I know that this is certainly a field of expertise of yours. So, if we're considering just the ordered spectra, the circumscribed and inscribed sets--for which I compute Hilbert-Schmidt probabilities--are not "balls", as they are with the two-qubit.....states themselves, and generally thought of. Perhaps they're "sub-simplices"? A couple of days ago, I posted a related question--since deleted--to which Matt. F. raised the objection (correctly, it seemed) that these sets of ordered spectra are not balls, as they are with the states themselves. Nov 12 '20 at 14:18
• As to the "almost trivialness" of "Question 1", doesn't one need to make an argument pertaining to the properties of unitary operators? Nov 12 '20 at 23:51
• The python code "Inner and outer Löwner-John Ellipsoids" docs.mosek.com/9.2/pythonfusion/case-studies-ellipsoids.html indicated by user Dominic in one of his commenst to mathematica.stackexchange.com/questions/234589/… takes as input a matrix P. Does the matrix given at the end of the answer of Nathaniel Johnston serve this purpose? Nov 16 '20 at 15:30
• If we embed the $2 \times 2$ matrix into the upper corner of an originallly $6 \times 6$ null matrix, and insert 1-x, x-y, y-z and z-(1-x-y-z) into the remaining null diagonal positions, the positive-semidefinitenss condition for the $6 \times 6$ matrix would be the constraint for the set S in question. Nov 17 '20 at 13:41

Daniel Huber in his answer CircumscribedConstruction to the companion question "Graph/Construct (John) ellipsoids circumscribing and inscribing a certain 3D convex set" has constructed a circumscribing ellipsoid EllipsoidFit .

The equation of the ellipsoid appears to be most succinctly expressible as $$$$\frac{4}{9} \left(9 \sqrt{2}-2\right) (x+y+2 z-1)^2+4 \left(3 \sqrt{2}-2\right) (x-y)^2+\frac{1}{18} \left(-16 x-16 y+16 z+3 \sqrt{2}+4\right)^2=1.$$$$

I first found its volume to be (the three semiaxes being $$\frac{3}{\sqrt{216 \sqrt{2}-48}},\frac{1}{2 \sqrt{6 \sqrt{2}-4}},\frac{\sqrt{\frac{3}{2}}}{8}$$), $$$$\frac{1}{32} \sqrt{\frac{1}{553} \left(29+12 \sqrt{2}\right)} \pi \approx 0.0283059,$$$$ with a rather amazing, at least at first sight (what happened to the 553?), simplification to $$$$\frac{\pi }{32 \sqrt{29-12 \sqrt{2}}}.$$$$ The volume of the convex set ($$S$$) being circumscribed has been shown to equal (the considerably smaller) $$$$\frac{1}{576} \left(8-6 \sqrt{2}-9 \sqrt{2} \pi +24 \sqrt{2} \cos ^{-1}\left(\frac{1}{3}\right)\right) \approx 0.00227243.$$$$ (We obtain a lesser volume $$\frac{\pi }{864 \sqrt{3}} \approx 0.0020993$$, if we replace the inequality $$x - z \leq 2 \sqrt{y (1-x-y-z)}$$ in the defining constraint for $$S$$ by an inequality $$x^2 + y^2 + (1 - x - y - z)^2 + z^2 \leq 1/3$$ for the sum of the squares of the four eigenvalues comprising the ordered spectra. If the $$\frac{1}{3}$$ is replaced by $$\frac{3}{8}$$ Adhikari, a larger volume $$\frac{\left(14-3 \sqrt{6}\right) \pi }{3456 \sqrt{3}} \approx 0.0034909$$ is found. The former result is for a set contained in $$S$$, and the latter for a set containing $$S$$.)

The ellipsoid was constructed fitting the extremal points of $$S$$ $$$$\left\{\frac{1}{3},\frac{1}{3},\frac{1}{3}\right\},\left\{\frac{1}{4},\frac{1}{4},\frac{1}{4}\right\},\left\{\frac{1}{2},\frac{1}{6},\frac{1}{6}\right\},\left\{\frac{1}{8} \left(2+\sqrt{2}\right),\frac{1}{8} \left(2+\sqrt{2}\right),\frac{1}{8} \left(2-\sqrt{2}\right)\right\}.$$$$

Huber does not specifically address the question of whether or not the ellipsoid constructed is the unique circumscribing (John) ellipsoid of minimal volume--a matter which still seems to be in need of resolution.

The question of the inscribed ellipsoid of maximal volume has been addressed--in somewhat a more purely numerical manner--by Dominic in his answer to CircumscribedConstruction . The resulting plot is InscribedEllipsoid