Are nearly all pure two-qubit state entangled?

Question

I am using the code below, utilizing QETLAB's RandomStateVector(4) and IsPPT, to generate a random state and to judge whether the state is entangled or separable:

clear all; clc;
for i = 1:30000
    psi = RandomStateVector(4);
    rho = psi*psi';
    if IsPPT(rho)
        'yes'
    end
end

But why does the program nearly never get inside the if block? I didn't see one single yes output. But the program works well, and if I use RandomDensityMatrix(4) instead, it's easy to see the output yes.

So, does it mean that for pure two qubits state, nearly all state are entangled?

Answer 1

Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled.

Let $\mathcal{F}$ denote the set of all pure states of two qubits and $\mathcal{S}$ denote the set of all pure product states of two qubits. Note that $\mathcal{S}$ can be thought of as the Cartesian product of two Bloch spheres. Therefore, $\mathcal{S}$ is a four-dimensional real manifold. On the other hand, one needs six real numbers to describe an element of $\mathcal{F}$. Thus, $\mathcal{S}$ is a low dimensional submanifold of $\mathcal{F}$.

Consequently, a pure two-qubit state selected uniformly$^1$ at random is entangled almost surely, i.e. with probability one. Finally, two qubit state is entangled if and only if its partial transpose is not positive. Therefore, the partial transpose of a pure two-qubit state selected uniformly at random fails to be positive with probability one.

---

^{$^1$ The same applies to every probability measure which is absolutely continuous with respect to the uniform probability measure.}