Find the ratio of two volumes of 6 x 6 positive-definite symmetric matrices

Consider the class $$A$$ of $$6 \times 6$$ positive definite matrices with real entries and unit trace (that is, the sum of the six diagonal entries is 1). (In quantum information-theoretic parlance, this is the class of "rebit-retrit density matrices".)

What is the twenty-dimensional Euclidean volume comprised by such matrices? This is the denominator of the ratio in which we are interested.

As to the numerator of the ratio, we want the twenty-dimensional Euclidean volume comprised by a certain subset $$B$$ of $$A$$. In addition to satisfying the conditions for membership in $$A$$, the "partial transposes" of such matrices must also be positive definite. To construct the partial transposes, one transposes in place the four $$3 \times 3$$ blocks of the rebit-retrit density matrices.

If one analogously considers the (9-dimensional) set of $$4 \times 4$$ "two-rebit density matrices" and the partial transposes of their $$2 \times 2$$ blocks, the desired ratio has been shown

to equal $$$$\frac{29}{64} =\frac{29}{2^6} \approx 0.453125.$$$$

As to the presently-desired ratio in the $$6 \times 6$$ instance, the conjecture--based on numerical computations (RebitRetritConjecture)--of $$$$\frac{860}{6561} = \frac{2^2 \cdot 5 \cdot 43}{3^8} \approx 0.131078$$$$ has been given.

These several ratios comprise what is termed "Hilbert-Schmidt separability probabilities".

As to the particular use of Mathematica in approaching this problem, I have considered the use of the GenericCylindricalDecomposition command (so far, with no successful outcomes). I have also considered the use of the positivity of the leading minors as a test for positive-definiteness. Perhaps the transformation of variables to such leading minors might be helpful.

Also, certain "simplified" forms of this daunting problem can be of interest--such as the 14-dimensional scenario in which the off-diagonal entries of the two diagonal (or two off-diagonal) $$3 \times 3$$ blocks are set to 0.

If one allows the off-diagonal entries of the matrices in question to be complex-valued, one moves to the realm of "two-qubit" and "qubit-qutrit" density matrices. In the 15-dimensional two-qubit case, the ratio (though not yet formally demonstrated to be such) is $$\frac{8}{33}$$. In the qubit-qutrit scenario, a conjecture of $$\frac{27}{1000}$$ has been stated.

According to eq. (7.7) $$$$V^{(1)}_N = \frac{2^{\frac{1}{4} (N-1) N+N} \sqrt{N} \pi ^{\frac{1}{4} (N-1) N-\frac{1}{2}} \Gamma \left(\frac{N+1}{2}\right) \prod _{k=1}^N \Gamma \left(\frac{k}{2}+1\right)}{N! \Gamma \left(\frac{1}{2} N (N+1)\right)}.$$$$ in Hilbert–Schmidt volume, with $$N= 6$$, the (Hilbert-Schmidt-valued) denominator of the ratio we seek should be $$$$\frac{\pi ^9}{2252687044608000 \sqrt{3}} \approx \text{7.63989457197784\grave{ }*{}^{\wedge}-12}.$$$$ Alternatively, based on Lebesgue measure, the volume of the denominator should be (Theorem 1 in LebesgueVolume)

V[k_] :=  Pi^(k^2) (2 k)! Product[(2 i)!, {i, 1,
k - 1}]/(2^(k^2 + k) k! (2 k^2 + k - 1)!)


with $$k=3$$, that is, $$$$\frac{\pi ^9}{1730063650258944000} \approx \text{1.7230059326999088\grave{ }*{}^{\wedge}-14}.$$$$

The ratio of the latter (Lebesgue) volume to the former (Hilbert-Schmidt) one is $$$$\frac{1}{256 \sqrt{3}}.$$$$

• Is it true that each volume individually is diverging? Jul 11 at 14:55
• No, yarchik, the volumes are not diverging arxiv.org/abs/math-ph/0604032 Jul 11 at 15:00
• yarchik--The "Hilbert-Schmidt metric" is defined by the line element squared--$\mbox{d}s^2_{HS}=(1/2) \mbox{Tr[}(d \rho)^2]$, where $\rho$ is a density matrix. This formula is (14.29) in the 2006 edition of "Geometry of Quantum States" by Zyczkowski and Bengtsson. I'm not sure if it is equivalently a "Haar measure". I just treat the problem in question as one in Euclidean/flat (Frobenius?) space. Jul 11 at 15:17
• There are other possible measures of strong interest--in particular, the "Bures" (see the "Geometry of Quantum States" book--or the literature, in general). These are still more challenging to compute, it would seem. Jul 11 at 15:26