Lovas and Andai (https://arxiv.org/abs/1610.01410) have recently established that the separability probability (ratio of separable volume to total volume) for the nine-dimensional convex set of two-re[al]bit states (representable by $4 \times 4$ “density matrices” with real entries) is $\frac{29}{64}$. The measure employed was the Hilbert-Schmidt one. Building upon this work of Lovas and Andai, strong evidence has been adduced that the corresponding result for the (standard) fifteen-dimensional convex set of two-qubit states (representable by density matrices with off-diagonal complex entries) is $\frac{8}{33}$ (https://arxiv.org/abs/1701.01973). (A density matrix is Hermitian, positive definite with unit trace.) Further, with respect to a different measure (one of the class of “monotone” ones), the two-qubit separability probability appears, quite strikingly, to be $1-\frac{256}{27 \pi^2}=1-\frac{2^8}{3^3 \pi^2}$. Further, exact values appear to have been found for higher-dimensional sets of states, endowed with Hilbert-Schmidt measure, such as $\frac{26}{323}$ for the 27-dimensional set of two-quater[nionic]bit states.

Now, perhaps the measure upon the quantum states of greatest interest is the Bures (minimal monotone) one (https://arxiv.org/abs/1410.6883). But exact computations pertaining to it appear to be more challenging. Lower-dimensional analyses (having set numbers of entries of the density matrices to zero) have yielded certain exact separability probabilities such as $\frac{1}{4}, \frac{1}{2}, \sqrt{2}-1$ (https://arxiv.org/abs/quant-ph/9911058). Efforts to estimate/determine the (15-dimensional) two-qubit Bures separability probability have been reported in https://arxiv.org/abs/quant-ph/0308037.

Recently (https://arxiv.org/abs/1809.09040, secs. X.B, XI), we have undertaken large-scale numerical simulations—employing both random and quasi-random [low-discrepancy] sequences of points—in this matter. Based on 4,372,000,000 randomly-generated points, we have obtained an estimate of 0.0733181. Further based on ongoing quasi-randomly-generated (sixty-four-dimensional) points, for which convergence should be stronger, we have obtained independent estimates of 0.0733181 and (for a larger sample) 0.0733117.

One approach to the suggestion of possible associated exact formulas, is to feed the estimates into http://www.wolframalpha.com and/or https://isc.carma.newcastle.edu.au (the Inverse Symbolic Calculator) and let it issue candidate expressions.

Certainly, $\frac{8}{11 \pi^2} \approx 0.073688$ would qualify as “elegant”, as well as $\frac{11}{150} \approx 0.073333$, but they do not seem to have the precision required. Also, since in the two cases mentioned above, we have the “entanglement probabilities” of $\frac{25}{33} =1 -\frac{8}{33}$ and $\frac{27}{256 \pi^2} =1-(1-\frac{27}{256 \pi^2})$, it might be insightful to think in such terms.

Bengtsson and Zyczkowski (p. 415 of “Geometry of quantum states: an introduction to quantum entanglement [2017]) ”have observed ``that the Bures volume of the set of mixed states is equal to the volume of an $(N^2-1)$-dimensional hemisphere of radius $R_B=\frac{1}{2}$''. It is also noted there that $R_B$ times the area-volume ratio asymptotically increases with the dimensionality $D=N^2-1$, which is typical for hemispheres. The Bures volume of the $N$-dimensional qubit density matrices is given by $\frac{2^{1-N^2} \pi ^{\frac{N^2}{2}}}{\Gamma \left(\frac{N^2}{2}\right)}$, which for $N=4$, gives $\frac{\pi ^8}{165150720}=\frac{\pi^8}{2^{19} \cdot 3^2 \cdot 5 \cdot 7}$.

Additionaly, we have similarly investigated the two-rebit Bures separability probability question with estimates being obtained of 0.1570934 and (larger sample) 0.1570971. But our level of confidence that some exact simple elegant formula exists for this probability is certainly not as high, based upon the Lovas-Andai two-rebit result for the particular monotone metric they studied.

Strongly related, with a slightly different focus, to this issue is my question Estimate/determine Bures separability probabilities making use of corresponding Hilbert-Schmidt probabilities

  • 2
    What is your question? – Norbert Schuch Oct 11 at 22:53
  • 1
    Well, maybe I kind of answered it in a manner of speaking with the 629/8580 proposal. that I only arrived at in the later stages of formulating the "question", after recalling the availability of the Inverse Symbolic Calculator (previously the Plouffe inverter). But as my ongoing estimation proceeds, other candidates may seem more or less plausible. Hopefully, at some point, some analytic--rather than purely numerical--insights can be gained. The 629/8520 is perhaps not quite as "elegant" as I was envisioning, but the numerator and denominator prime factorizations are somewhat intriguing. – Paul B. Slater Oct 12 at 4:26
  • The original "question" was modified in light of the comment of Norbert Schuch. The original last paragraph--concerning the possible value of 629/8580 was removed, and reinserted in the form of an answer. – Paul B. Slater Oct 12 at 18:10

Use of the Inverse Symbolic Calculator (https://isc.carma.newcastle.edu.au) suggested the possible exact two-qubit separability probability of $\frac{629}{8580} =\frac{17 \cdot 37}{2^2 \cdot 3 \cdot 5 \cdot 11 \cdot 13} \approx 0.07331002$, quite close to the current estimate of ours, based on the largest sample, of 0.0733116 (employing 4,945,000,000 realizations of a quasirandom sequence of interest per the answer of Martin Roberts to https://math.stackexchange.com/questions/2231391/how-can-one-generate-an-open-ended-sequence-of-low-discrepancy-points-in-3d).

As the estimate further evolves, we may be led to modify/expand (if possible) upon this current "suggestion".

Here is a plot of estimates based upon two quasirandom sequences (one using $\alpha_0=0$ and one [the indicated longer one, using $\alpha_0 =\frac{1}{2}$] in the terminology of Roberts) along with the line $\frac{629}{8580}$.

enter image description here

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.