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I know that a one-qubit $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, where $\alpha, \beta \in \mathbb C$, can be represented geometrically on a Bloch sphere as $|\psi\rangle = \cos\theta |0\rangle +e^{\text{i}\phi}\sin \theta |1\rangle$.
Now, for a two-qubit state $|\psi\rangle= \alpha |00\rangle + \beta |11\rangle+\gamma|01\rangle+\delta|10\rangle$, where $\alpha, \beta, \gamma, \delta \in \mathbb C$, can we find a geometrical representation analog to that of the Bloch sphere?
A two qubit pure state can be decomposed into 8 "geometric" real parameters:
To compute these values you perform a Schmidt decomposition of the system. The "how entangled is it?" value is the minimum of the two Schmidt coefficients.
To get the entanglement rotation's Euler angles: mutate both Schmidt coefficients to be 0.5, put the state vector back together using the mutated coefficients, then interpret it as a 2x2 unitary matrix (instead of a 1x4 vector) and map that unitary matrix from SU(2) to SO(3) in the usual way.
To get the normalized bloch vectorss: compute bloch vectors in the usual way, normalize them, then convert to latitude and longitude.
Resources: entangled states are like unitary matrices and visualizing 2 qubit entanglement.
