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To begin, we have two constraints \begin{equation} C1=x>0\land z>0\land y>0\land x+2 y+3 z<1 \end{equation} and \begin{equation} C2=x>0\land y>0\land x+2 y+3 z<1\land x^2+x (3 z-2 y)+(y+3 z)^2<3 z. \end{equation} $C1$ ensures the nonnegative-definiteness of a class of $9 \times 9$ ("two-qutrit") density matrices ($\rho$).

$C2$ also ensures this, as well as the nonnegative-definiteness of the "partial transpose" ($\rho^{PT}$) of $\rho$.

The integration--subject to $C1$--of the value 36 over the unit cube $\{x,y,z\} \in [0,1]^3$ yields 1.

The integration--subject to $C2$--of the value 36 over the unit cube yields (the Hilbert-Schmidt positive partial transpose probability) $\frac{8 \pi}{27 \sqrt{3}} \approx 0.537422$.

Now, we are interested in similarly enforcing both $C1 \land C3$ (yielding an "entanglement probability") and $C2 \land C3$ (yielding a "bound-entanglement probability"), where, the entanglement constraint $C3$ is \begin{equation} b \left(-x \left(a^2-a (b+2)+2 b+1\right)+2 y (a (-a+b+2)+b-1)-3 z (a-b-1) (a+b-1)+(a-1)^2\right)<0, \end{equation} with its three parameters $a,b,c$ subject to \begin{equation} C4= b>0\land c>0\land 0<a<1\land a+b+c=2\land (a-1)^2=b c. \end{equation}

Now, I suspect these last two problems are too difficult to resolve in their full generality (leaving $a,b,c$ unspecified).

But, to begin, if we take \begin{equation} \{a,b,c\}=\left\{\frac{1}{4} \left(3-\sqrt{5}\right),\frac{1}{2},\frac{1}{4} \left(3+\sqrt{5}\right)\right\}, \end{equation} the integration of 36 over the unit cube subject to $C1 \land C3$ yields $\frac{5}{132} \left(5+\sqrt{5}\right) \approx 0.274093$.

Alternatively, for \begin{equation} \{a,b,c\} = \left\{\frac{1}{3},\frac{1}{3},\frac{4}{3}\right\}, \end{equation} the integration of 36 over the unit cube subject to $C1 \land C3$ yields $\frac{125}{486} \approx 0.257202$.

However, I have not been so far able to obtain the counterparts for these last two results for $C2 \land C3$. (Using numerical integration, we get the much lower values of 0.001497721920258410 and 0.003256122941383665, respectively.)

A set of values of $a,b,c$ which satisfy $C4$ are \begin{equation} \left\{\frac{2}{3} (\cos (\alpha )+1),\frac{2}{3} \left(-\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right),\frac{2}{3} \left(\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right)\right\}, \end{equation} for $\frac{\pi}{3} \leq \alpha \leq \frac{5 \pi}{3}$.

So, I would like to obtain results of integration of the value 36 over $[0,1]^3$ of $C1 \land C3$ and $C2 \land C3$ for either specific values of $a,b,c$, satisfying $C4$, or even without particular values being specified.

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    $\begingroup$ Presumably "The integration—subject to $C(x, y, z)$—over the unit cube" means "The integration over $\{(x, y, z) \in [0, 1]^3 : C(x, y, z)\}$"? In which case, you are just looking for volumes (multiplied by 36, for some reason)? $\endgroup$ – LSpice May 17 '19 at 19:44
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    $\begingroup$ And, further, you are looking for an answer in terms of $a$, $b$, $c$, but only requiring the validity of that formula when $C4(a, b, c)$? (However, your $C3(a, b, c, x, y, z)$ doesn't seem actually to depend on $c$; so do you mean just to require that $\exists c, C4(a, b, c)$?) $\endgroup$ – LSpice May 17 '19 at 19:45
  • $\begingroup$ The 36 is simply a needed normalization constant. Also, the $a,b,c$'s can not be independent variables--constrained by both the $a+b+c==2$ and $b c==(1-a)^2$ requirements, given in equation (1) of arxiv.org/pdf/1107.2720.pdf $\endgroup$ – Paul B. Slater May 17 '19 at 21:47
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Well, maybe the second comment of LSpice made me fully realize that the constraint $C4$ means that there is only one degree of freedom between $a,b,c$. So, as a start I took $a=\frac{1}{3}$. Then, using the Mathematica GenericCylindricalDecomposition command and choosing the order of integration of $x,y,z$ to select the simplest (as measured by LeafCount) output, I was able to obtain an entanglement probability of $\frac{125}{486}=\frac{5^3}{2 \cdot 3^5} \approx 0.257202$.

Additionally, using the same approach, the bound-entanglement probability proved to be \begin{equation} \frac{-204+56 \sqrt{3} \pi +7 \log (7)-336 \sqrt{3} \csc ^{-1}\left(2 \sqrt{7}\right)}{1134} \approx 0.00325612. \end{equation} So, I'll now try to use other values of $a$--rather than $\frac{1}{3}$ (and extend this answer, if I so succeed, and the results are of interest).

I thought that the problem of getting results for general $0 \leq a \leq 1$ would be much more formidable, as the resulting forms that $C1 \land C3$ and $C2 \land C3$ take seemed quite involved.

However, much to my surprise, the calculation for $C1 \land C3$ for the entanglement probability as a function of $a$ greatly simplified, yielding \begin{equation} -\frac{(a-2)^3}{9 a^2-30 a+27}. \end{equation} For $a=\frac{1}{3}$, the function gives the above-reported $\frac{125}{486}$.

Further, Nicholas Tessore was able to find the formula for $C2 \land C3$, giving the bound-entanglement probability. It took the form \begin{equation} \label{Tessore} -\frac{A+B}{54 (4-3 a)^{3/2} (2 a-3)} , \end{equation} where \begin{equation} A=8 \sqrt{12-9 a} \left(6 a^2-17 a+12\right) \cos ^{-1}\left(\frac{a (3 a-8)+6}{6-4 a}\right) \end{equation} and \begin{equation} B=3 \sqrt{a} \left(2 \left(9 a^3-57 a^2+108 a-64\right)+3 (3-2 a) a \log (9-6 a)\right). \end{equation}

So, a complete answer has now been successfully given to the question put.

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