Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
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. –  Trimok Jan 10 at 9:55

1 Answer 1

up vote 3 down vote accepted

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 $\mathrm{tr}[I\otimes I]=4$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.