To quickly place our probabilistic (copula) question in its subject matter setting, we note that a fundamental concept in quantum theory is that of entanglement QuantumEntanglement.
The states of certain quantum systems can be considered to be either entangled or, if not, separable.
Now, the probability that a quantum system is separable is an important one SeparableVolume.
With these preliminary remarks, let us examine a specific seven-dimensional situation, in which it has been established (XstatesVolumes Apps. A, B) that the probability of separability is invariant/uniform over each of two (a3, b3 $\in [0,1]$--termed "Bloch radii" BlochSphere) of the seven variables.
Our question here is whether it is suitable--and how--to search in this setting for a bivariate copula having these two marginal uniform distributions.
We begin with a seven-parameter (a3, b3, c11, c12, c21, c22, c33) $4 \times 4$ matrix P forming a zero-nonzero X-pattern \begin{equation} \left( \begin{array}{cccc} P= \frac{1}{4} (\text{a3}+\text{b3}+\text{c12}+\text{c33}+1) & 0 & 0 & \frac{1}{4} (\text{c11}-i \text{c21}-\text{c22}) \\ 0 & \frac{1}{4} (\text{a3}-\text{b3}-\text{c12}-\text{c33}+1) & \frac{1}{4} (\text{c11}+i \text{c21}+\text{c22}) & 0 \\ 0 & \frac{1}{4} (\text{c11}-i \text{c21}+\text{c22}) & \frac{1}{4} (-\text{a3}+\text{b3}-\text{c12}-\text{c33}+1) & 0 \\ \frac{1}{4} (\text{c11}+i \text{c21}-\text{c22}) & 0 & 0 & \frac{1}{4} (-\text{a3}-\text{b3}+\text{c12}+\text{c33}+1) \\ \end{array} \right) \end{equation}
or, in Mathematica notation,
{{1/4 (1 + a3 + b3 + c12 + c33), 0, 0, 1/4 (c11 - I c21 - c22)}, {0, 1/4 (1 + a3 - b3 - c12 - c33), 1/4 (c11 + I c21 + c22), 0}, {0, 1/4 (c11 - I c21 + c22), 1/4 (1 - a3 + b3 - c12 - c33), 0}, {1/4 (c11 + I c21 - c22), 0, 0, 1/4 (1 - a3 - b3 + c12 + c33)}}
Its four eigenvalues--summing to 1--are \begin{equation} \left\{\frac{1}{4} \left(-\sqrt{(\text{a3}-\text{b3})^2+(\text{c11}+\text{c22})^2+\text{c21}^2}-\text{c1 2}-\text{c33}+1\right),\frac{1}{4} \left(\sqrt{(\text{a3}-\text{b3})^2+(\text{c11}+\text{c22})^2+\text{c21}^2}-\text{c12 }-\text{c33}+1\right),\frac{1}{4} \left(-\sqrt{(\text{a3}+\text{b3})^2+(\text{c11}-\text{c22})^2+\text{c21}^2}+\text{c1 2}+\text{c33}+1\right),\frac{1}{4} \left(\sqrt{(\text{a3}+\text{b3})^2+(\text{c11}-\text{c22})^2+\text{c21}^2}+\text{c12 }+\text{c33}+1\right)\right\} \end{equation}
{1/4 (1 - c12 - Sqrt[(a3 - b3)^2 + c21^2 + (c11 + c22)^2] - c33), 1/4 (1 - c12 + Sqrt[(a3 - b3)^2 + c21^2 + (c11 + c22)^2] - c33), 1/4 (1 + c12 - Sqrt[(a3 + b3)^2 + c21^2 + (c11 -c22)^2] + c33), 1/4 (1 + c12 + Sqrt[(a3 + b3)^2 + c21^2 + (c11 - c22)^2] + c33)}
If all four eigenvalues are nonnegative, then P (nonnegative definite--having trace 1) is said to be the "density matrix of a two-qubit [quantum bit] X-state".
In App. A of XstatesVolumes, it was established (integrating over all of the seven parameters, while enforcing the nonnegativity of the four eigenvalues) that the seven-dimensional volume occupied by such matrices is $\frac{\pi^2}{5040}$.
More interestingly still, for our specific question here, it was shown that the intermediate six-fold integral over all but the parameter a3 is $\frac{\pi^2 (1-a3^2)^3}{2304}$.
Now only certain of the $4 \times 4$ X-states correspond to entangled systems. Those that do not are termed "separable" or "classically-correlated". (They are, then, representable as convex sums of Kronecker products of $2 \times 2$ density matrices--in contrast to the entangled systems.)
Further, we would also like to know the seven-dimensional volume (obviously no greater than $\frac{\pi^2}{5040}$) occupied by simply those states that are separable.
To do so, one enforces the "Peres-Horodecki" positive-partial-transposition (PPT) test SeparabilityTest--by taking the "partial transpose" of P. This can be accomplished by transposing in place its four $2 \times 2$ blocks. Doing so for P, simply changes the signs of c12 and c22.
The four eigenvalues of the partial transpose of P are then \begin{equation} \left\{\frac{1}{4} \left(-\sqrt{(\text{a3}-\text{b3})^2+(\text{c11}-\text{c22})^2+\text{c21}^2}+\text{c12 }-\text{c33}+1\right),\frac{1}{4} \left(\sqrt{(\text{a3}-\text{b3})^2+(\text{c11}-\text{c22})^2+\text{c21}^2}+\text{c12} -\text{c33}+1\right),\frac{1}{4} \left(-\sqrt{(\text{a3}+\text{b3})^2+(\text{c11}+\text{c22})^2+\text{c21}^2}-\text{c12 }+\text{c33}+1\right),\frac{1}{4} \left(\sqrt{(\text{a3}+\text{b3})^2+(\text{c11}+\text{c22})^2+\text{c21}^2}-\text{c12} +\text{c33}+1\right)\right\} \end{equation}
{1/4 (1 + c12 - Sqrt[(a3 - b3)^2 + c21^2 + (c11 - c22)^2] - c33), 1/4 (1 + c12 + Sqrt[(a3 - b3)^2 + c21^2 + (c11 - c22)^2] - c33), 1/4 (1 - c12 - Sqrt[(a3 + b3)^2 + c21^2 + (c11 + c22)^2] + c33), 1/4 (1 - c12 + Sqrt[(a3 + b3)^2 + c21^2 + (c11 + c22)^2] + c33)}
Enforcing now both the nonnegativity of P and of its partial transpose--that is, the nonnegativity of both sets of four eigenvalues--the volume (performing the full seven-fold integration) occupied by the separable states ($\frac{\pi ^2}{12600}$) is two-fifths that of the total volume ($\frac{\pi^2}{5040}$). So, the probability that an X-state is separable is simply $\frac{2}{5}$.
Performing a six-fold integration, as before, yields $\frac{\pi^2(1-a3^2)^3}{5760}$ for the volume of separable states.
Now, the ratio of $\frac{\pi^2(1-a3^2)^3}{5760}$ to the earlier-indicated $\frac{\pi^2 (1-a3^2)^3}{2304}$--that is, the separability probability $\frac{2}{5} $--is INDEPENDENT of a3 (a major finding of XstatesVolumes).
By the symmetry as Bloch radii variables between a3 and b3, we know such independence also holds for b3.
So, the probability of separability is uniform over both a3 and b3 $\in [0,1]$.
This finding led us to the question what--if any--is the pertinent bivariate copula having this pair of uniform marginals.
To further proceed, we computed a five-fold integration--while again enforcing the nonnegativity of the four eigenvalues of P--giving us the intermediate total volume result (eq. (7) BlochRadiiRepulsion) \begin{equation} \begin{cases} -\frac{1}{960} \pi ^2 \left(a3-1\right){}^3 \left(a3 \left(a3+3\right)-5 b3^2+1\right) & a3>b3 \\ -\frac{1}{960} \pi ^2 \left(b3-1\right){}^3 \left(-5 a3^2+b3 \left(b3+3\right)+1\right) & a3<b3. \end{cases} \end{equation}
Integrating over b3 $\in [0,1]$ gives us the previously-noted six-fold integration result of $\frac{\pi^2 (1-a3^2)^3}{2304}$.
Similarly, we computed a five-fold integration--while enforcing the nonnegativity of the four eigenvalues of P and the four of its partial transpose--giving us the intermediate separable volume result (eq. (8) BlochRadiiRepulsion) \begin{equation} \begin{cases} -\frac{\pi ^2 (\text{a3}-1)^3 \left(8 \text{a3}^2+5 (\text{a3}+3) \text{b3}^4-10 (3 \text{a3}+1) \text{b3}^2+9 \text{a3}+3\right)}{7680} & \text{a3}>\text{b3} \\ -\frac{\pi ^2 (\text{b3}-1)^3 \left(5 \text{a3}^4 (\text{b3}+3)-10 \text{a3}^2 (3 \text{b3}+1)+8 \text{b3}^2+9 \text{b3}+3\right)}{7680} & \text{b3}>\text{a3} \end{cases} \end{equation}
We note again the presence of the $(\text{a3}-1)^3$ in both of these (total and separable) volume computations.
Unlike the six-fold setting, the ratio 
\begin{equation}
\begin{cases}
\frac{8 \text{a3}^2+5 (\text{a3}+3) \text{b3}^4-10 (3 \text{a3}+1) \text{b3}^2+9
\text{a3}+3}{8 \left(\text{a3} (\text{a3}+3)-5 \text{b3}^2+1\right)} &
\text{a3}>\text{b3} \\
\frac{5 \text{a3}^4 (\text{b3}+3)-10 \text{a3}^2 (3 \text{b3}+1)+\text{b3} (8
\text{b3}+9)+3}{8 \left(-5 \text{a3}^2+\text{b3} (\text{b3}+3)+1\right)} &
\text{a3}<\text{b3}
\end{cases}
\end{equation}
of these five-fold results is not independent of a3 and b3.
As noted in BlochRadiiRepulsion, there appears to be a repulsion between a3 and b3, that is the line a3 = b3 is one of relative minima. (These three figures appear as Figs. 44-46 in BlochRadiiRepulsion, with additional discussion.)
So, can these last two intermediate (total and separable) five-dimensional volume results be employed for the task of constructing a bivariate copula having the demonstrated univariate uniform marginals over the Bloch radii variables a3 and b3? Or if not, would additional calculations be appropriate?
The integral over b3 of the indicated ratio of five-dimensional volumes is \begin{equation} \frac{1}{600} \left(375 \left(\text{a3}^4-6 \text{a3}^2-3\right) \tanh ^{-1}\left(\frac{\text{a3}+4}{9 \text{a3}+6}\right)+\frac{75 \sqrt{5} \left(3 \text{a3}^4+30 \text{a3}^2+7\right) \coth ^{-1}\left(\frac{\sqrt{5} (2 \text{a3}+1)}{\sqrt{4 \text{a3}^2+1}}\right)}{\sqrt{4 \text{a3}^2+1}}-5 \text{a3} (\text{a3} (\text{a3} (8 \text{a3}+33)-60)+99)+\frac{3 \sqrt{5} (\text{a3} (\text{a3} (\text{a3}+11)+20)+8) (\text{a3}-1)^2 \tanh ^{-1}\left(\frac{\sqrt{5} \text{a3}}{\sqrt{\text{a3} (\text{a3}+3)+1}}\right)}{\sqrt{\text{a3} (\text{a3}+3)+1}}+600\right), \end{equation}
1/600 (600 - 5 a3 (99 + a3 (-60 + a3 (33 + 8 a3))) + (75 Sqrt[5] (7 + 30 a3^2 + 3 a3^4) ArcCoth[(Sqrt[5] (1 + 2 a3))/Sqrt[1 + 4 a3^2]])/Sqrt[1 + 4 a3^2] + 375 (-3 - 6 a3^2 + a3^4) ArcTanh[(4 + a3)/(6 + 9 a3)] + (3 Sqrt[5] (-1 + a3)^2 (8 + a3 (20 + a3 (11 + a3))) ArcTanh[(Sqrt[5] a3)/Sqrt[1 + a3 (3 + a3)]])/Sqrt[1 + a3 (3 + a3)]
a plot of which
clearly shows non-uniformity over a3.
This 7-dimensional X-states problem also has a standard 15-dimensional (two-qubit) counterpart in which the entries of the density matrix are not similarly restricted. In that setting, numerous forms of (numerical and analytical) evidence--though not yet a formalized proof--strongly support a separability probability equal to $\frac{8}{33}$--as opposed to $\frac{2}{5}$--as well as again invariance over the Bloch radii a3 and b3 15-DimensionalUnrestrictedProblem CasimirInvariance.
The concept of a quantum copula has recently been introduced QuantumCopula. Its possible relevance to the (classical) question posed here is not at this point clear (to me).
