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The generalized 2-qubit state is given as: $$ \rho = \frac{1}{4}[ I\otimes I + (m_x\sigma_x + m_y\sigma_y + m_z\sigma_z)\otimes I + I \otimes (n_x\sigma_x + n_y\sigma_y + n_z\sigma_z) + \sum_{ij}t_{ij}\sigma_i\otimes\sigma_j] $$

Then, is there a method to map a given density matrix: $$ \rho_g = \pmatrix{a & b & c & d \\ e &f &g &h \\ i & j & k & l \\ m &n &o &p } $$ to the generalized state in terms of relations between the coefficients, without having to expand and compare terms?

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    $\begingroup$ There is no miraculous way (or divine inspiration). Your "generalized 2-qubit state" $\rho$ is a density matrix too, so just express explicitely the matrices $I \otimes I, \sigma_i \otimes I, I \otimes\sigma_i, \sigma_i \otimes \sigma_j$, and just do $\rho = \rho_g$, and you're done. $\endgroup$
    – Trimok
    Commented Jan 10, 2014 at 9:55

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You can compute the coefficients by virtue of $$ t_{ij} = \mathrm{tr}[\rho_g(\sigma_i\otimes\sigma_j)]\ , $$ $$ m_i = \mathrm{tr}[\rho_g(\sigma_i\otimes I)] $$ and $$ n_i = \mathrm{tr}[\rho_g(I\otimes\sigma_i)]\ . $$ The proof is immediate if you note that all tensor products of Paulis (or of Paulis with the identity) have trace zero, while $\mathrm{tr}[I\otimes I]=4$.

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