1
$\begingroup$

It is now clearly well-established--though formalized proofs are still largely lacking—that the probability, with respect to Hilbert-Schmidt measure, that a generic two-qubit state is separable/disentangled is $\frac{8}{33} \approx 0.242424$. (See https://arxiv.org/abs/1701.01973, where this and other companion generalized two-qubit results--such as $\frac{29}{64}$ for the two-rebit states--are stated.)

Relatedly, there are the elegant formalized results of Szarek, Bengtsson and Zyczkowski in their paper, “On the structure of the body of states with positive partial transpose” (https://arxiv.org/abs/quant-ph/0509008). In the abstract, they state that we “prove that the probability of finding a random state to be separable equals twice the probability of finding a random boundary state to be separable, provided that the random states are generated uniformly with respect to the Hilbert–Schmidt (Euclidean) measure”. (In the proof of their Proposition 1, they state: “A density matrix lies on the boundary of [the set of density matrices] if and only if it has a zero eigenvalue”.)

Combining these analyses, one should be able to conclude that the probability, with respect to Hilbert-Schmidt measure, that a generic boundary two-qubit state is separable/disentangled is $\frac{1}{2} \cdot \frac{8}{33} =\frac{4}{33} \approx 0.121212$. We may note that Hilbert-Schmidt separability probabilities appear, in a different manner, to be also equally divided between density matrices, the determinants of which are greater that the determinants of their partial transposes, and vice versa (cf. https://arxiv.org/abs/1404.1860).

The question we pose here is whether it is possible to reconcile this $\frac{4}{33}$ assertion with the result reported in Table 2 of the 2017 paper of Khvedelidze and Rogojin, “On the generation of random ensembles of qubits and qutrits: Computing separability probabilities for fixed rank states” https://arxiv.org/abs/1708.07846. The claim there, based on numerical procedures (no sample size is given), is that the Hilbert-Schmidt separability probability of generic ("non-maximal") rank-3 two-qubit density matrices is approximately 0.1652. (Following their [Ginibre-matrix-based] algorithm (eq. (5)), we have obtained an estimate of 0.165237, based on thirteen million [low-discrepancy] quasi-random realizations.) Further, if this is a meaningful result, distinct from the $\frac{4}{33}$ assertion, to what exact value does it presumably correspond?

$\endgroup$
0
$\begingroup$

The reconciliation appears to consist in noting that the Ginibre-matrix-based algorithm presented does not generate rank-3 two-qubit density matrices, as claimed. I have randomly generated 1,000 density matrices, according to the prescription, recording their smallest eigenvalues, and found one with a smallest (of four) eigenvalues as large as 0.07885 (rather than strictly zero). No reference is given for the assertion in the 2017 paper that "it is known that any rank-3 complex $4 \times 4$ matrix admits the following representation...", so it is not clear on what it was based.

Charles Dunkl has suggested to me: "In your rank-3 problem: what if one started with a random upper triangular matrix $C$ (sum of squares of entries =1) with zero at the (4,4) spot. Compute $C*C$ and then find a random $4 \times 4$ permutation matrix $P$ and produce $P^{-1} C* C P$. I don't know what it is but it does have rank 3.

In the 2017 Khevedelidze-Rogojin paper, an estimate of the rank-3 separability probability with respect to the important Bures (minimal monotone) measure of 0.0494 is also given. But since the same Ginibre-based procedure is employed, this estimate would also seem to be in question. Unlike the Hilbert-Schmidt case, exact separability probabilities are not known in the generalized two-qubit Bures case. (In Table X of https://arxiv.org/abs/quant-ph/0410238, an estimate of 1.84676 was given for the ratio of the full-rank two-qubit Bures separability probability to the rank-3 probability, while an estimate of 2.00167 [with the known value as indicated above being exactly 2] was given in the Hilbert-Schmidt case. In Table XI there, corresponding estimated ratios are given in the [$6 \times 6$] qubit-qutrit case.)

In response to the question, K. Zyczkowski wrote:

"I have not a ready program to generate such random states from the boundary of two qubit mixed states.

A simple method to do it which just came to my mind is

  1. Take pure random state | psi> of the bipartite 3 x 3 system.

  2. Find its partial trace rho = Tr_B |psi>

  3. diagonalize rho obtaining three eigenvalues p= p1,p2,p3. They are generated according to the HS measure on the 3-simplex

  4. Extend it by zero writing a diagonal matrix D=diag(p1,p2,p3,0)

  5. draw a random unitary matrix U in U(4) according to the Haar measure.

  6. write sigma := U D U^ Which is or rank three And is generated according to the HS measure.

I hope that this strategy is correct - it would be fine to check that working with this measure one obtains the separability ratio 4/33 as predicted."


It now turns out that the production of rank-4 density matrices--rather than simply rank-3 or lower--was the result of the use (employing the notation [eq. (5)] of the 2017 Khevedelidze-Rogojin paper) of the product form $X A^{-1} Y$, rather than $Y A^{-1} X$. With the correction, only rank-3 or lower rank matrices are, in fact, yielded, as desired. But the associated estimated separability probability--as best we can tell currently--is approximately 0.097 (as opposed to 0.1652, before the corrected product form was used), and not the indicated 0.121212. So, considerable clarification still remains in how to numerically yield this presumed correct value. (An estimated ratio of the separability probabilities of interest, very close to 2 was reported more than a decade ago in Table X of https://arxiv.org/abs/quant-ph/0410238. This result was the starting point for the Szarek, Bengtsson and Zyczkowski JPA analysis, in which the formal proof of the ratio of 2 was given.)

$\endgroup$

Your Answer

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

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