How can one integrate over the unit cube, subject to certain (quantum-information-theoretic) constraints?

To begin, we have two constraints $$$$C1=x>0\land z>0\land y>0\land x+2 y+3 z<1$$$$ and $$$$C2=x>0\land y>0\land x+2 y+3 z<1\land x^2+x (3 z-2 y)+(y+3 z)^2<3 z.$$$$ $$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 $$$$b \left(-x \left(a^2-a (b+2)+2 b+1\right)+2 y (a (-a+b+2)+b-1)-3 z (a-b-1) (a+b-1)+(a-1)^2\right)<0,$$$$ with its three parameters $$a,b,c$$ subject to $$$$C4= b>0\land c>0\land 0

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

Alternatively, for $$$$\{a,b,c\} = \left\{\frac{1}{3},\frac{1}{3},\frac{4}{3}\right\},$$$$ the integration of 36 over the unit cube subject to $$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 $$$$\left\{\frac{2}{3} (\cos (\alpha )+1),\frac{2}{3} \left(-\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right),\frac{2}{3} \left(\frac{1}{2} \sqrt{3} \sin (\alpha )-\frac{\cos (\alpha )}{2}+1\right)\right\},$$$$ for $$\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.

• 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)? – LSpice May 17 '19 at 19:44
• 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 $\exists c, C4(a, b, c)$?) – LSpice May 17 '19 at 19:45
• 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 – Paul B. Slater May 17 '19 at 21:47

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 $$$$\frac{-204+56 \sqrt{3} \pi +7 \log (7)-336 \sqrt{3} \csc ^{-1}\left(2 \sqrt{7}\right)}{1134} \approx 0.00325612.$$$$ So, I'll now try to use other values of $$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$$ greatly simplified, yielding $$$$-\frac{(a-2)^3}{9 a^2-30 a+27}.$$$$ For $$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$$, giving the bound-entanglement probability. It took the form $$$$\label{Tessore} -\frac{A+B}{54 (4-3 a)^{3/2} (2 a-3)} ,$$$$ where $$$$A=8 \sqrt{12-9 a} \left(6 a^2-17 a+12\right) \cos ^{-1}\left(\frac{a (3 a-8)+6}{6-4 a}\right)$$$$ and $$$$B=3 \sqrt{a} \left(2 \left(9 a^3-57 a^2+108 a-64\right)+3 (3-2 a) a \log (9-6 a)\right).$$$$