Master Lovas–Andai and equivalent formulas verifying the $\frac{8}{33}$ two-qubit Hilbert–Schmidt separability probability and companion rational-valued conjectures

Abstract

We begin by investigating relationships between two forms of Hilbert–Schmidt two-rebit and two-qubit “separability functions”—those recently advanced by Lovas and Andai (J Phys A Math Theor 50(29):295303, 2017), and those earlier presented by Slater (J Phys A 40(47):14279, 2007). In the Lovas–Andai framework, the independent variable $\varepsilon \in [0,1]$ is the ratio $\sigma (V)$ of the singular values of the $2 \times 2$ matrix $V=D_2^{1/2} D_1^{-1/2}$ formed from the two $2 \times 2$ diagonal blocks ($D_1, D_2$) of a $4 \times 4$ density matrix $D= \left||\rho _{ij}\right||$. In the Slater setting, the independent variable $\mu$ is the diagonal-entry ratio $\sqrt{\frac{\rho _{11} \rho _ {44}}{\rho _ {22} \rho _ {33}}}$—with, of central importance, $\mu =\varepsilon$ or $\mu =\frac{1}{\varepsilon }$ when both $D_1$ and $D_2$ are themselves diagonal. Lovas and Andai established that their two-rebit “separability function” $\tilde{\chi }_1 (\varepsilon )$ ($\approx \varepsilon$) yields the previously conjectured Hilbert–Schmidt separability probability of $\frac{29}{64}$. We are able, in the Slater framework (using cylindrical algebraic decompositions [CAD] to enforce positivity constraints), to reproduce this result. Further, we newly find its two-qubit, two-quater[nionic]-bit and “two-octo[nionic]-bit” counterparts, $\tilde{\chi _2}(\varepsilon ) =\frac{1}{3} \varepsilon ^2 \left( 4-\varepsilon ^2\right)$, $\tilde{\chi _4}(\varepsilon ) =\frac{1}{35} \varepsilon ^4 \left( 15 \varepsilon ^4-64 \varepsilon ^2+84\right)$ and $\tilde{\chi _8} (\varepsilon )= \frac{1}{1287}\varepsilon ^8 \left( 1155 \varepsilon ^8-7680 \varepsilon ^6+20160 \varepsilon ^4-25088 \varepsilon ^2+12740\right)$. These immediately lead to predictions of Hilbert–Schmidt separability/PPT-probabilities of $\frac{8}{33}$, $\frac{26}{323}$ and $\frac{44482}{4091349}$, in full agreement with those of the “concise formula” (Slater in J Phys A 46:445302, 2013), and, additionally, of a “specialized induced measure” formula. Then, we find a Lovas–Andai “master formula,” $\tilde{\chi _d}(\varepsilon )= \frac{\varepsilon ^d \Gamma (d+1)^3 \, _3\tilde{F}_2\left( -\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;\varepsilon ^2\right) }{\Gamma \left( \frac{d}{2}+1\right) ^2}$, encompassing both even and odd values of d. Remarkably, we are able to obtain the $\tilde{\chi _d}(\varepsilon )$ formulas, $d=1,2,4$, applicable to full (9-, 15-, 27-) dimensional sets of density matrices, by analyzing (6-, 9, 15-) dimensional sets, with not only diagonal $D_1$ and $D_2$, but also an additional pair of nullified entries. Nullification of a further pair still leads to X-matrices, for which a distinctly different, simple Dyson-index phenomenon is noted. C. Koutschan, then, using his HolonomicFunctions program, develops an order-4 recurrence satisfied by the predictions of the several formulas, establishing their equivalence. A two-qubit separability probability of $1-\frac{256}{27 \pi ^2}$ is obtained based on the operator monotone function $\sqrt{x}$, with the use of $\tilde{\chi _2}(\varepsilon )$.

Appendices

Appendix A: Absolute separability probabilities

Those separable states that cannot be entangled through unitary operations have been designated as absolutely separable [54, p. 392].

In [55], we reported exact (but now decidedly not rational-valued, and much smaller-valued) formulas for the Hilbert–Schmidt absolute separability probabilities for the two-rebit, two-qubit and two-quaterbit states. For the convenience and interest of the reader, we present them here, while simplifying the forms of the last two.

The two-rebit absolute separability probability is expressible as [55, eq. (32)]

$$\begin{aligned} \frac{6928-2205 \pi }{16 \sqrt{2}} \approx 0.0348338, \end{aligned}$$

(A1)

the two-qubit as [55, eq. (34)]

$$\begin{aligned}&1-\frac{3217542976-5120883075 \pi +16386825840 \tan ^{-1}\left( \sqrt{2}\right) }{32768 \sqrt{2}}-\frac{29901918259}{497664} \nonumber \\&\quad \approx 0.00365826 \end{aligned}$$

(A2)

and the two-quaterbit as [55, eq. (36)]

$$\begin{aligned}&\frac{13}{3043362286338048} \times \Bigg (806338156306739134839776\nonumber \\&\quad -658857590468226345222144 \sqrt{2}+1048604423167357891775325 \sqrt{2} \pi \nonumber \\&\quad -3355534154135545253681040 \sqrt{2} \tan ^{-1}\left( \sqrt{2}\right) \Bigg ) \approx 0.0000401326. \end{aligned}$$

(A3)

Let us note that the integer components of the denominators appearing above are either simply powers of 2, or involve high powers of 2. Also, $\tan ^{-1}\left( \sqrt{2}\right) ) \approx 0.955317$—the angle between the space diagonal of a cube and any of its three connecting angles—has been termed the “magic angle” (see the eponymous Wikipedia article). We have also been able to obtain absolute separability probabilities in the two-rebit case, again featuring this particular angle prominently, when the Hilbert–Schmidt ($k=0$) measure is replaced by random induced measures [53] for $k=1,2,3$.

It appears to be a substantial challenge—using eq. (4) of [56] to find the $6 \times 6$ counterparts to these three formulas for $4 \times 4$ systems.

Appendix B: Rebit-retrit and qubit-qutrit analyses

Let us now attempt to extend the two-rebit and two-qubit line of analysis above to rebit-retrit and qubit-qutrit settings—now, of course, passing from consideration of $4 \times 4$ density matrices to $6 \times 6$ ones. Lovas and Andai, in their quite recent study, had not yet addressed such issues. In [2], candidate (Slater-type) separability functions had been proposed. Two dependent variables (cf. the use of $\mu =\sqrt{\frac{\rho _ {11} \rho _ {44}}{\rho _ {22} \rho _ {33}}}$ in the lower-dimensional setting above) had been employed [2, eq. (44)]. Let us now refer to these two variables as $\tau _1=\sqrt{\frac{\rho _ {11} \rho _{55}}{\rho _ {22} \rho _ {44}}}$ and $\tau _2 =\sqrt{\frac{\rho _ {22} \rho _{66}}{\rho _ {33} d _{55}}}$. But, interestingly, it was argued that only a single dependent variable $\tau = \tau _1 \tau _2 = \sqrt{\frac{\rho _ {11} \rho _{66}}{\rho _ {33} \rho _ {44}}}$ sufficed for modeling the corresponding separability functions. The separability function in the rebit-retrit case was proposed to be simply proportional to $\tau$ [2, eq.(98)].

In our effort to extend the Lovas–Andai analyses [1] to this setting, we now took $D_1$ and $D_2$ to equal the upper and lower diagonal $3 \times 3$ blocks of the $6 \times 6$ density matrix in question. Then, we computed the three singular values ($s_1 \ge s_2 \ge s_3$) of $D_2^{1/2} D_1^{-1/2}$, and took the ratio variables $\varepsilon _1 = \frac{s_2}{s_1}$ and $\varepsilon _2 = \frac{s_3}{s_2}$ as the dependent ones in question. (An issue of possible concern is that, unlike the $4 \times 4$ case [32], positivity of the determinant of the partial transpose of a $6 \times 6$ density matrix is only a necessary, but not sufficient condition for separability.) Also, in the case of diagonal D, again the two variables in the $\mu$ framework are equal to those in the $\varepsilon$ setting, or to their reciprocals.

Then, we generated 3436 million rebit-retrit and 2379 million qubit-qutrit density matrices, randomly with respect to Hilbert–Schmidt measure. (These sizes are much larger than those employed in 2007—for similar purposes—in [2].) We appraised the separability of the density matrices D by testing whether the partial transpose, using the four $3 \times 3$ blocks, had all its six eigenvalues positive. The separability probability estimates were $0.13180011 \pm 0.0000113109$ and $0.02785302 \pm 6.6124281 \cdot 10^{-6}$, respectively. (We can reject the qubit-qutrit conjecture of $\frac{32}{1199} \approx 0.0266889$ advanced in [2, sec. 10.2]. A possible alternative candidate is $\frac{72}{2585} =\frac{5 \cdot 11 \cdot 47}{2^3 \cdot 3^2} \approx 0.027853$, while in the rebit-retrit case, we have $\frac{298}{2261} =\frac{7 \cdot 17 \cdot 19}{2 \cdot 149} \approx 0.1318001$.) Further, our estimates of the probabilities that D had two (the most possible [57]) negative eigenvalues, and hence a positive determinant, although being entangled, were $0.0334197 \pm 0.0000409506$ in the rebit-retrit case and $0.0103211 \pm 0.000031321$ in the qubit-qutrit instance.)

In the two-variable settings, we partition the square $[0,1]^2$ of possible separability probability results into an $80 \times 80$ grid, and in the one-variable setting, use a partitioning (as in the two-rebit and two-qubit analyses above) into 200 subintervals of [0,1]. In Fig. 25 we show the ratio of the square of the rebit-retrit separability probability to the qubit-qutrit separability probability as a function of $\tau$, while in Fig. 26, we show a two-dimensional version. Figure 27 is the analog of this last plot using the singular-value ratios $\varepsilon _1$ and $\varepsilon _2$. As in Figs. 17, 18 and 19, we observe a gradual increase in these Dyson-index-oriented analyses. In Figs. 28 and 29, we show the highly linear (“diagonal”) rebit-retrit and qubit-qutrit separability probabilities, holding $\tau _1=\tau _2$.

Fig. 25

The ratio of the square of the rebit-retrit separability probability to the qubit-qutrit separability probability as a function of $\tau = \tau _1 \tau _2 = \sqrt{\frac{\rho _ {11} \rho _{66}}{\rho _ {33} \rho _ {44}}}$

Fig. 26

The ratio of the square of the rebit-retrit separability probability to the qubit-qutrit separability probability as a function of $\tau _1=\sqrt{\frac{\rho _ {11} \rho _{55}}{\rho _ {22} \rho _ {44}}}$ and $\tau _2 =\sqrt{\frac{\rho _ {22} d _{66}}{\rho _ {33} d _{55}}}$

Fig. 27

The ratio of the square of the rebit-retrit separability probability to the qubit-qutrit separability probability as a function of the singular-value ratios $\varepsilon _1$ and $\varepsilon _2$

Fig. 28

Rebit-retrit separability probabilities for $\tau _1=\tau _2$

Fig. 29

Qubit-qutrit separability probabilities for $\tau _1=\tau _2$

Let us note that Mendonça, and Marchiolli, and Hedemann have recently shown [58, App. A] that for qubit-qutrit X-states, the partial transposes can—in contrast to more general such $6 \times 6$ systems—have no more than one negative eigenvalue. Therefore, positivity of the determinant of the partial transpose is both necessary and sufficient for separability, in this case. Nevertheless, Dunkl has been able to conclude that the Hilbert–Schmidt separability probabilities reported in [44] for two-qubit X-states, continue to hold in these higher-dimensional qubit-qutrit X-state systems.

Appendix C: Comparison of Ginibre and Cholesky methods for quaternion positive-definite matrices—by C. F. Dunkl

The calculations depend on integrating monomials over the unit sphere in $\mathbb {R}^{N}$. We use the Pochhammer symbol $\left( a\right) _{n} :=\prod _{i=1}^{n}\left( a+i-1\right)$. If $a\ne 0,-1,-2,\ldots$ then $\Gamma \left( a+n\right) /\Gamma \left( a\right) =\left( a\right) _{n}$.

Lemma C.1

Let $S_{N-1}$ be the unit sphere in $\mathbb {R}^{N}$ with the inherited rotation-invariant measure $\hbox {d}\omega$ and let $n_{1},n_{2} ,\ldots ,n_{N}\in \mathbb {N}_{0}$ ($\left\{ 0,1,2,3\ldots \right\}$) then

$$\begin{aligned} \int _{S_{N-1}}\prod _{i=1}^{N}\left| x_{i}\right| ^{n_{i}} \hbox {d}\omega \left( x\right) =2\prod _{i=1}^{N}\Gamma \left( \frac{n_{i}+1}{2}\right) \Bigg /\Gamma \left( \frac{1}{2}\left( \sum _{i=1}^{N}n_{i}+N\right) \right) . \end{aligned}$$

Proof

Let $n=\sum _{i=1}^{N}n_{i}$ and $f\left( x\right) =\prod _{i=1}^{N}\left| x_{i}\right| ^{n_{i}}$. In spherical polar coordinates

$$\begin{aligned} \int _{\mathbb {R}^{N}}f\left( x\right) \exp \left( -\frac{\left| x\right| ^{2}}{2}\right) \hbox {d}x=\int _{0}^{\infty }r^{n}\exp \left( -\frac{r^{2}}{2}\right) r^{N-1}\hbox {d}r\int _{S_{N-1}}f\left( x\right) \hbox {d}\omega \left( x\right) . \end{aligned}$$

The left-hand side equals (by the substitution $x_{i}^{2}=2t$)

$$\begin{aligned} \prod _{i=1}^{N}\int _{-\infty }^{\infty }\left| x_{i}\right| ^{n_{i} }{\text {e}}^{-\left| x_{i}\right| ^{2}/2}\hbox {d}x_{i}=\prod _{i=1}^{N}2^{\left( n_{i}+1\right) /2}\Gamma \left( \frac{n_{i}+1}{2}\right) , \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{\infty }r^{n}\exp \left( -\frac{r^{2}}{2}\right) r^{N-1} \hbox {d}r=2^{\left( N+n\right) /2-1}\Gamma \left( \frac{n+N}{2}\right) . \end{aligned}$$

Divide the left side by this to obtain $\int _{S}f\left( x\right) \hbox {d}\omega \left( x\right)$. $\square$

Denote the right-hand side by $I\left[ n_{1},n_{2},\ldots ,n_{N}\right]$. We use $0\$n$ to denote 0 listed n times. Thus the surface measure of $S_{N-1}$ is $I[0\$N]$. We need another lemma for integrating powers of sums of squares.

Lemma C.2

Suppose $1\le A\le B$ and $n=1,2,3,\ldots$ then the normalized integral

$$\begin{aligned} \int _{S_{B-1}}\left( \sum _{i=1}^{A}y_{i}^{2}\right) ^{n}\hbox {d}\omega \left( y\right) \left\{ \int _{S_{B-1}}\hbox {d}\omega \left( y\right) \right\} ^{-1}=\dfrac{\left( A/2\right) _{n}}{\left( B/2\right) _{n}}. \end{aligned}$$

Proof

The argument is similar to the proof of Lemma C.1. An alternative approach would rely on Dirichlet integrals $\square$

The Cholesky method begins with a random point on $S_{27}\subset \mathbb {R}^{28}$ to form an upper triangular matrix A such that $A_{11},A_{22},A_{33},A_{44}\ge 0$ and $A_{ij}\in \mathbb {H}$ for $1\le i<j\le 4$. Then $Q:=A^{*}A$ is positive-definite and $trQ=1$. In particular $Q_{11}=A_{11}^{2}$ and $\det Q=\left( A_{11}A_{22}A_{33}A_{44}\right) ^{2}$. The Jacobian is $A_{11}^{13}A_{22}^{9}A_{33}^{5}A_{44}$. For the measure $\left( \det Q\right) ^{k}$ the $n^{th}$ moment of $Q_{11}$ (that is $\mathcal {E}\left( Q_{11}^{n}\right)$) is given by

$$\begin{aligned} \mu _{n}&=\frac{I\left[ 2n+2k+13,2k+9,2k+5,2k+1,0\$24\right] }{I[2k+13,2k+9,2k+5,2k+1,0\$24]}\\&=\frac{\Gamma \left( k+7+n\right) \Gamma \left( 4k+28\right) }{\Gamma \left( k+7\right) \Gamma \left( 4k+28+n\right) }=\frac{\left( k+7\right) _{n}}{\left( 4k+28\right) _{n}},\\ \mu _{1}&=\frac{1}{4},~\mu _{2}=\frac{k+8}{4\left( 4k+29\right) },\\ \mu _{3}&=\frac{\left( k+8\right) \left( k+9\right) }{8\left( 4k+29\right) \left( 2k+15\right) }. \end{aligned}$$

The Ginibre method for $M\times 4$ begins with a random point on $S_{16M-1}\subset \mathbb {R}^{16M}$ to form an $M\times 4$ matrix H with $H_{ij} \in \mathbb {H}$ for $1\le i\le M,1\le j\le 4$ and $\sum _{i=1}^{M}\sum _{j=1}^{4}\left| H_{ij}\right| ^{2}=1$. (For a quaternion $q=x_{1}+x_{2}\varvec{i}+x_{3}\varvec{j}+x_{4}\varvec{k}$ define $\overline{q}=x_{1}-x_{2}\varvec{i}-x_{3}\varvec{j}-x_{4} \varvec{k}$ then $\left| q\right| ^{2}=\overline{q}q=\sum _{i=1}^{4}x_{i}^{2}$.) Then $Q:=H^{*}H$ is positive-definite and $trQ=1$. In particular $Q_{11}=\sum _{i=1}^{M}\left( A^{*}\right) _{1i}A_{i1} =\sum _{i=1}^{M}\overline{A_{i1}}A_{i1}=\sum _{i=1}^{M}\left| A_{i1} \right| ^{2}$. In real terms each $\left| A_{i1}\right| ^{2}$ is a sum of four squared real variables so rewrite $Q_{11}=\sum _{j=1}^{4M}x_{j} ^{2}$ where $\left( x_{j}\right) _{j=1}^{16M}$ is a random point on $S_{16M-1}$. Apply Lemma C.2 with $A=4M$ and $B=16M$ to obtain

$$\begin{aligned} \nu _{n}=\mathcal {E}\left( Q_{11}^{n}\right) =\mathcal {E}\left( \sum _{j=1}^{4M}x_{j}^{2}\right) ^{n}=\frac{\left( 2M\right) _{n}}{\left( 8M\right) _{n}} \end{aligned}$$

for $n=1,2,3,\ldots$. In particular $\nu _{1}=\frac{1}{4}$ and $\nu _{2} =\dfrac{2M+1}{4\left( 8M+1\right) }$.

Thus $\nu _{n}=\dfrac{\left( 2M\right) _{n}}{\left( 8M\right) _{n}}$ and $\mu _{n}=\dfrac{\left( k+7\right) _{n}}{\left( 4k+28\right) _{n}}$ are equal for all n exactly when $k+7=2M$. In itself this is not a proof that the Ginibre method produces $\left( \det Q\right) ^{2M-7}$ times the HS measure. This statement is a consequence of equation (4.6) in [53].

In particular the Ginibre method does not lead to the HS measure for any M since k is necessarily odd.

The above calculations can be adapted to other values of the parameter $\alpha$ (with $\alpha =\frac{1}{2}$ for $\mathbb {R}$, $\alpha =1$ for $\mathbb {C}$, and $\alpha =2$ for $\mathbb {H}$). The Cholesky method starts with the sphere in $\mathbb {R}^{4+12\alpha }$ (4 on diagonal, $12\alpha$ off-diagonal), and the Jacobian is $A_{11}^{1+6\alpha }A_{22}^{1+4\alpha } A_{33}^{1+2\alpha }A_{44}$. The nth moment of $Q_{11}$ is

$$\begin{aligned} \frac{I\left[ 2n+1+6\alpha +2k,1+4\alpha +2k,1+2\alpha +2k,1+2k,0\$12\alpha \right] }{I\left[ 1+6\alpha +2k,1+4\alpha +2k,1+2\alpha +2k,1+2k,0\$12\alpha \right] }=\frac{\left( k+3\alpha +1\right) _{n}}{\left( 4k+12\alpha +4\right) _{n}}. \end{aligned}$$

Similarly to above the Ginibre matrix size $M\times 4$ comes from a random point on $S_{8\alpha M-1}\subset \mathbb {R}^{8\alpha M}$ and $Q_{11}$ is the sum of $2\alpha M$ squares $x_{i}^{2}$ and the nth moment of $Q_{11}$ is

$$\begin{aligned} \frac{\left( \alpha M\right) _{n}}{\left( 4\alpha M\right) _{n}}, \end{aligned}$$

which agrees with the $\left( \det Q\right) ^{k}\times HS$ when $k=\alpha M-3\alpha -1=\left( M-3\right) \alpha -1$. To apply the formula:

$$\begin{aligned} \alpha&=\frac{1}{2},k=\frac{1}{2}(M-5),\\ \alpha&=1,k=M-4,\\ \alpha&=2,k=2M-7. \end{aligned}$$

Thus the Ginibre method does produce HS random density matrices with $M=5$ for $\mathbb {R}$ and $M=4$ for $\mathbb {C}$.

Appendix D: A further formula for the Lovas–Andai integral—by Charles Dunkl

We deal here with the evaluation of the integral for even d.

$$\begin{aligned} \int _{0}^{1}t^{-3\left( 1+d\right) }~_{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d}{4+6d} ;1-\frac{1}{t^{2}}\right) \left( 1-t^{2}\right) ^{d}~_{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{-\frac{d}{2},\frac{d}{2},d}{1+\frac{d}{2},1+\frac{3d}{2}} ;t^{2}\right) \mathrm {d}t \end{aligned}$$

(D1)

The first step is to rewrite the $_{2}F_{1}$ series as a power series in $t^{2}$. By the use of the identity

$$\begin{aligned} _{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{a,b}{c} ;x\right) =\left( 1-x\right) ^{-a}~_{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{a,c-b}{c} ;\frac{x}{x-1}\right) \end{aligned}$$

we obtain

$$\begin{aligned} _{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d}{4+6d} ;1-\frac{1}{t^{2}}\right) =t^{2\left( 2+3d\right) }~_{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d}{4+6d} ;1-t^{2}\right) \end{aligned}$$

Thus the desired integral equals

$$\begin{aligned} \int _{0}^{1}t^{1+3d}~_{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d}{4+6d} ;1-t^{2}\right) \left( 1-t^{2}\right) ^{d}~_{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{-\frac{d}{2},\frac{d}{2},d}{1+\frac{d}{2},1+\frac{3d}{2}} ;t^{2}\right) \mathrm {d}t \end{aligned}$$

By expanding the series we can integrate term by term. The typical term is

$$\begin{aligned} \int _{0}^{1}t^{1+3d+2j}\left( 1-t^{2}\right) ^{d+n}\mathrm {d}t&=\frac{1}{2}\int _{0}^{1}s^{j+3d/2}\left( 1-s\right) ^{d+n}\mathrm {d}s\\&=\frac{1}{2}B\left( j+\frac{3d}{2}+1,d+n+1\right) \\&=\frac{1}{2} \frac{\Gamma \left( j+\frac{3d}{2}+1\right) \Gamma \left( d+n+1\right) }{\Gamma \left( j+\frac{5d}{2}+2+n\right) }\\&=\frac{1}{2}\frac{\Gamma \left( \frac{3d}{2}+1\right) \Gamma \left( d+1\right) }{\Gamma \left( \frac{5d}{2}+2\right) }\frac{\left( d+1\right) _{n}}{\left( j+\frac{5d}{2}+2\right) _{n}}\frac{\left( \frac{3d}{2}+1\right) _{j}}{\left( \frac{5d}{2}+2\right) _{j}}, \end{aligned}$$

by the use of the change of variable $s=t^{2}$ and the beta function. The result for integral (D1) is (with d even)

$$\begin{aligned} \frac{1}{2}\frac{\Gamma \left( \frac{3d}{2}+1\right) \Gamma \left( d+1\right) }{\Gamma \left( \frac{5d}{2}+2\right) }\sum _{j=0}^{d/2} \frac{\left( -\frac{d}{2}\right) _{j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}\left( 2+\frac{5d}{2}\right) _{j}j!}~_{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d,d+1}{4+6d,j+\frac{5d}{2}+2} ;1\right) . \end{aligned}$$

(D2)

By the use of the Gauss sum and contiguous hypergeometric series we can produce finite expressions for the integral. We state the Gauss sum (with $\dot{c}>a+b$) and define utility functions.

$$\begin{aligned} S\left( a,b,c\right)&:=\frac{\Gamma \left( c-a-b\right) \Gamma \left( c\right) }{\Gamma \left( c-a\right) \Gamma \left( c-b\right) },\\ _{2}F_{1}\left( \genfrac{}{}{0.0pt}{}{a,b}{c} ;1\right)&=S\left( a,b,c\right) . \end{aligned}$$

For $k=0,1,2\ldots$ and $k<\min \left( c-a-b,g\right)$ define

$$\begin{aligned} S_{32}\left( a,b,c;g,k\right) :=~_{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{a,b,g}{c,g-k} ;1\right) . \end{aligned}$$

Proposition D.1

For $k=0,1,2\ldots$ and $k<\min \left( c-a-b,g\right)$

$$\begin{aligned} S_{32}\left( a,b,c;g,k\right) =\frac{\Gamma \left( c-a-b\right) \Gamma \left( c\right) }{\Gamma \left( c-a\right) \Gamma \left( c-b\right) }\sum _{j=0}^{k}\frac{\left( -k\right) _{j}\left( a\right) _{j}\left( b\right) _{j}}{j!\left( 1+a+b-c\right) _{j}\left( g-k\right) _{j}}. \end{aligned}$$

Proof

For any n there is the formula (easy to verify, by finite differences, or the Chu–Vandermonde sum)

$$\begin{aligned} \frac{\left( g+n-k\right) _{k}}{\left( g-k\right) _{k}}=\sum _{j=0} ^{k}\left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{n\left( n-1\right) \left( n-2\right) \cdots \left( n-j+1\right) }{\left( g-k\right) _{j}}. \end{aligned}$$

Then

$$\begin{aligned} ~_{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{a,b,g}{c,g-k} ;1\right) =\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( b\right) _{n}}{\left( c\right) _{n}n!}\frac{\left( g\right) _{n}}{\left( g-k\right) _{n}} \end{aligned}$$

and

$$\begin{aligned} \frac{\left( g\right) _{n}}{\left( g-k\right) _{n}}=\frac{\left( g\right) _{n-k}\left( g+n-k\right) _{k}}{\left( g-k\right) _{k}\left( g\right) _{n-k}}=\frac{\left( g+n-k\right) _{k}}{\left( g-k\right) _{k}}. \end{aligned}$$

Also

$$\begin{aligned}&\sum _{n=0}^{\infty }\frac{\left( a\right) _{n}\left( b\right) _{n}}{\left( c\right) _{n}n!}n\left( n-1\right) \left( n-2\right) \cdots \left( n-j+1\right) =\sum _{n=j}^{\infty }\frac{\left( a\right) _{n}\left( b\right) _{n}}{\left( c\right) _{n}\left( n-j\right) !}\\&\quad =\frac{\left( a\right) _{j}\left( b\right) _{j}}{\left( c\right) _{j} }\sum _{m=0}^{\infty }\frac{\left( a+j\right) _{m}\left( b+j\right) _{m} }{\left( c+j\right) _{m}m!}=\frac{\left( a\right) _{j}\left( b\right) _{j}}{\left( c\right) _{j}}S\left( a+j,b+j,c+j\right) . \end{aligned}$$

making the change of index $m=n-j$ (observe that $n\cdots \left( n-j+1\right) =0$ for $0\le n\le j-1$). The S-term equals

$$\begin{aligned} \frac{\Gamma \left( c-a-b-j\right) \Gamma \left( c+j\right) }{\Gamma \left( c-a\right) \Gamma \left( c-b\right) }=\frac{\left( c\right) _{j}}{\left( c-a-b-j\right) _{j}}\frac{\Gamma \left( c-a-b\right) \Gamma \left( c\right) }{\Gamma \left( c-a\right) \Gamma \left( c-b\right) } \end{aligned}$$

by the use of the relation $\Gamma \left( t\right) \left( t\right) _{j} =\Gamma \left( t+j\right)$; also by reversal $\left( c-a-b-j\right) _{j}=\left( -1\right) ^{j}\left( 1+a+b-c\right) _{j}$. The binomial coefficient $\left( {\begin{array}{c}k\\ j\end{array}}\right) =\left( -1\right) ^{j}\frac{\left( -k\right) _{j}}{j!}$. Combine the ingredients and this proves the formula. $\square$

Consider the typical term in (D2)

$$\begin{aligned} _{3}F_{2}\left( \genfrac{}{}{0.0pt}{}{2+3d,2+3d,d+1}{4+6d,j+\frac{5d}{2}+2} ;1\right) . \end{aligned}$$

Set $a=2+3d,b=1+d,c=4+6d,g=2+3d$, $k=\frac{d}{2}-j$. Thus desired integral (D1) equals

$$\begin{aligned}&\frac{1}{2}\frac{\Gamma \left( \frac{3d}{2}+1\right) \Gamma \left( d+1\right) }{\Gamma \left( \frac{5d}{2}+2\right) }\\&\quad \times \sum _{j=0}^{d/2} \frac{\left( -\frac{d}{2}\right) _{j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}\left( 2+\frac{5d}{2}\right) _{j}j!}S_{32}\left( 2+3d,1+d,4+6d;2+3d,\frac{d}{2}-j\right) \\&=\frac{1}{2}\frac{\Gamma \left( \frac{3d}{2}+1\right) \Gamma \left( d+1\right) }{\Gamma \left( \frac{5d}{2}+2\right) }\frac{\Gamma \left( 4+6d\right) \Gamma \left( 1+2d\right) }{\Gamma \left( 2+3d\right) \Gamma \left( 3+5d\right) }\\&\quad \times \sum _{j=0}^{d/2}\frac{\left( -\frac{d}{2}\right) _{j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}\left( 2+\frac{5d}{2}\right) _{j}j!}\sum _{i=0}^{d/2-j}\frac{\left( j-\frac{d}{2}\right) _{i}\left( 2+3d\right) _{i}\left( 1+d\right) _{i} }{i!\left( -2d\right) _{i}\left( 2+\frac{5d}{2}+j\right) _{i}}. \end{aligned}$$

The double sum in the last line can be rewritten as

$$\begin{aligned} \sum _{i\ge 0,j\ge 0}^{i+j\le d/2}\frac{\left( -\frac{d}{2}\right) _{i+j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}\left( 2+3d\right) _{i}\left( 1+d\right) _{i}}{\left( 2+\frac{5d}{2}\right) _{i+j}\left( 1+\frac{d}{2}\right) _{j}i!j!\left( -2d\right) _{i}}. \end{aligned}$$

Collect the (independent of t) prefactors [from the product of (82) and (83)]

$$\begin{aligned} 18\frac{\Gamma \left( \frac{1}{2}+\frac{d}{2}\right) ^{3}\Gamma \left( \frac{7}{6}+\frac{d}{2}\right) ^{2}\Gamma \left( \frac{5}{6}+\frac{d}{2}\right) ^{2}\Gamma \left( 1+2d\right) }{\pi ^{7/2}\Gamma \left( 1+\frac{d}{2}\right) \Gamma \left( 3+5d\right) }3456^{d}. \end{aligned}$$

Evaluate for even d, $\Gamma \left( \frac{1}{2}+\frac{d}{2}\right) =\Gamma \left( \frac{1}{2}\right) \left( \frac{1}{2}\right) _{d/2}$ (and $\Gamma \left( \frac{1}{2}\right) =\sqrt{\pi }$);

$$\begin{aligned} \Gamma \left( \frac{7}{6}+\frac{d}{2}\right) \Gamma \left( \frac{5}{6} +\frac{d}{2}\right)&=\left( \frac{7}{6}\right) _{d/2}\left( \frac{5}{6}\right) _{d/2}\Gamma \left( \frac{7}{6}\right) \Gamma \left( \frac{5}{6}\right) \\&=\left( \frac{7}{6}\right) _{d/2}\left( \frac{5}{6}\right) _{d/2} \frac{1}{6}\Gamma \left( \frac{1}{6}\right) \Gamma \left( \frac{5}{6}\right) , \end{aligned}$$

and $\Gamma \left( \frac{1}{6}\right) \Gamma \left( \frac{5}{6}\right) =\dfrac{\pi }{\sin \frac{\pi }{6}}=2\pi$ (recall $\Gamma \left( t\right) \Gamma \left( 1-t\right) =\dfrac{\pi }{\sin \pi t}$). Put it all together (even d)

$$\begin{aligned}&\mathcal {P}_{\mathrm {sep/PPT}}\left( d\right) \\&\quad =3456^{d}\frac{\left( \frac{1}{2}\right) _{d/2} ^{3}\left( \frac{7}{6}\right) _{d/2}^{2}\left( \frac{5}{6}\right) _{d/2}^{2}\left( 2d\right) !}{\left( \frac{d}{2}\right) !\left( 3\right) _{5d}}\sum _{i\ge 0,j\ge 0}^{i+j\le d/2}\frac{\left( -\frac{d}{2}\right) _{i+j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}\left( 2+3d\right) _{i}\left( 1+d\right) _{i}}{\left( 2+\frac{5d}{2}\right) _{i+j}\left( 1+\frac{d}{2}\right) _{j}i!j!\left( -2d\right) _{i}}. \end{aligned}$$

Appendix E: Remark on Lovas–Andai paper

It certainly appears that the work of Lovas and Andai [1]—inspired by that of Milz and Strunz [24]—is highly innovative and successful in finding the two-rebit separability function $\tilde{\chi }_1(\varepsilon )$, and verifying the conjecture that the two-rebit Hilbert–Schmidt separability probability is $\frac{29}{64}$. However, in our study of the Lovas–Andai paper, we remain unconvinced by the chain of arguments on page 13 leading to the result $\frac{1}{4}$ and have posted a stack exchange question (https://mathematica.stackexchange.com/q/144277/29989) in this regard.