3
$\begingroup$

On p. 3 of "Separability Probability Formulas and Their Proofs for Generalized Two-Qubit X-Matrices Endowed with Hilbert-Schmidt and Induced Measures" (https://arxiv.org/abs/1501.02289), it is asserted that the Hilbert-Schmidt separability probabilities for the two-rebit, two-qubit and two-quarter[nionic]-bit X-states are, $\frac{16}{3 \pi^2}, \frac{2}{5}$ and $\frac{2}{7}$, respectively. (Of course, by X-states [https://arxiv.org/abs/1407.3021], we mean density matrices having an X-shaped nonzero-zero pattern.)

More generally still, for random induced measure (the case $k=0$, corresponding to the Hilbert-Schmidt instance), for the two-qubit X-states, we have, on the same page, the formula \begin{equation} 1-\frac{2 \Gamma (2 k+4)^2}{\Gamma (k+2) \Gamma (3 k+6)}, \end{equation} giving the indicated $\frac{2}{5}$ for $k=0$, and now $\frac{9}{14}$ for $k=1$,...(Also, 0 for $k=-1$, we note.)

Now, in the abstract and sec. VIII of "Extensions of Generalized Two-Qubit Separability Probability Analyses to Higher Dimensions, Additional Measures and New Methodologies" (https://arxiv.org/abs/1809.09040), it has been asserted that these same probabilities continue to hold in the X-state cases for the higher-dimensional rebit-retrit, two-retrit, qubit-qutrit, two-qutrit,...cases.

Now, on some level this seems counter-intuitive, as the evidence (see previous citation) is that for the (15-dimensional) two-qubit states, in general, the Hilbert-Schmidt separability probability is $\frac{8}{33} \approx 0.242424$ and possibly (certainly much less) $\frac{27}{1000}$ for the qubit-qutrit states,..., also shrinking to zero, asymptotically (as the various works of Aubrun, Szarek, Ye, et al indicate). The numerical evidence is certainly of a similar nature for the rebit/retrit counterparts.

So, it would be nice to have some independent examination of the proposition that for the X-states, separability/PPT-probabilities remain invariant as the dimensions of the spaces in which they are embedded increase.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.