On difference in the number of two-qubit stabilizer states that are separable (36) vs those that are maximally entangled (24) and partial entanglement

Question

We have a set of two-qubit stabilizer states. There are 60 of them: 36 separable and 24 maximally entangled (MES).

I was wondering whether we can somehow compare the size of the set of partially entangled states (PES) with those numbers 36 and 24. I understand that there are infinitely many states in between, but let's fix the angle: we take all the possible PES that are half-way between the separable and MES. I.e. if we make Schmidt decomposition of the state it will be of the form $$\frac{\sqrt{2+\sqrt{2}}}{2}|a_0\rangle|b_0\rangle + \frac{\sqrt{2-\sqrt{2}}}{2}|a_1\rangle|b_1\rangle$$ Or in the words of entanglement measures, the Concurrence of such a state should be $\frac{\sqrt2}{2}$. Or in words of rotation it should be a result of state $|a_0\rangle|b_0\rangle$ rotated by $\pi/4$ angle by a matrix that applied by $\pi/2$ angle to the state $|a_0\rangle|b_0\rangle$ gives the state $\frac{1}{\sqrt{2}}(|a_0\rangle|b_0\rangle + |a_0\rangle|b_0\rangle)$

To construct such a PES given separable and MES in the Schmidt decomposition we take the Hamiltonian that is created through tensor of two Schmidt basis Pauli matrices $H = \sigma_y \otimes \sigma_x$. So $$e^{iH\theta} = \begin{pmatrix}cos(\theta)&0&0&-sin(\theta)\\0& cos(\theta)&-sin(\theta)&0\\0&sin(\theta)&cos(\theta)&0 \\sin(\theta)&0&0&cos(\theta)\end{pmatrix} $$

Applying this unitary on $|a_0\rangle|b_0\rangle$ for $\theta = \pi/2$ Gives us the MES, and for $\theta=\pi/4$ the target PES. But in this manner we generally move from separable to a MES.

How I would go about obtaining some measure is this: I would take the set of 15 double-qubit Pauli matrices (all the possible tensors of $I; \sigma_x; \sigma_y; \sigma_z$). And among them I would choose some subset. There are 9 two-Pauli-matrix tensors. I would choose three ($\sigma_x \otimes \sigma_x; \sigma_y \otimes \sigma_y; \sigma_z \otimes \sigma_z$) because the other 6 may be represented as local $I \otimes \sigma$ combined with the three chosen. Then I would apply them as described above to all the MES states, because there's fewer of them. Then I would obtain my set of PES. But I am not sure how do I make sure I don't create same PES for some inputs and how can I be sure I haven't missed some.

Another way (and that's closer to what I want to do) is following. I want to create a measure of the "surface" on which all the infinitely many separable states live; measure of the surface on which all the infinitely many MES live. I guess the surface of separable states will be larger, whatever that means. But I would put a number 36 to that measure, 24 to the measure of MES surface. And then I could also find a "surface" on which the infinitely many half-way-inbetween PES live. Then I could compare the measures of all three surfaces and find a number for the PES surface. But I have totally no idea how to go about measuring those surfaces.

I can think of different reasons why it might be equal to 24, 36, be somewhere in between or even larger than 36 so was wondering whether someone has tried to estimate. Note, let's speak only about pure two-qubit states.

My first intuition tells me that the number should be 24.

Also, does the size of the set change if we vary the angle away from $\pi/4$?

Answer 1

The set of two-qubit maximally-entangled states is isomorphic to the set of single-qubit unitary matrices, because the "type" of entanglement just comes down to how you rotate Alice's measurement results in order to predict Bob's measurements results.

This means the number of maximally entangled stabilizer states is the same as the number of single qubit stabilizer operations. Which is the number of rotational symmetries of a cube: $24$. The number of two-qubit unentangled stabilizer states is the number of single qubit stabilizer states squared, which is $6^2 = 36$.

For similar reasons, the number of maximally-entangled Alice-has-half-the-qubits-and-Bob-has-half-the-qubits $2n$-qubit states should be equal to the number of $n$-qubit stabilizer operations, and the number of unentangled half-and-half $2n$ qubit stabilizer states should be the square of the number of $n$-qubit stabilizer states. Proving these things is probably a good exercise.