Is there a way to write down the eigenstates of this two-qubit density matrix?

I am considering the density matrix which represents an arbitrary state for a pair of qubits. When written out in terms of the Pauli operators, this is as follows (certain terms vanish for another reason so there are fewer than $$16$$ terms):

$$\rho = \frac{1}{4} \bigg( I \otimes I + t_{01}I \otimes \sigma_x + t_{10} \sigma_x \otimes I + t_{02} I \otimes \sigma_y + t_{20} \sigma_y \otimes I + t_{11} \sigma_x \otimes \sigma_x + t_{22}\sigma_y \otimes \sigma_y + t_{33} \sigma_z \otimes \sigma_z + t_{12} \sigma_x \otimes \sigma_y + t_{21}\sigma_y \otimes \sigma_x \bigg),$$

where the coefficients take values between $$-1$$ and $$1$$.

It is possible to have eigenstates for a density matrix but this is quite a long and complicated expression, so I am just curious to how would one would write down the possible eigenvectors for this matrix? Do the eigenvectors here relate to the usual eigenvectors for the Pauli matrices themselves?

• I suspect you're just going to have to brute force it. Sep 13, 2021 at 13:54
• Will this have to be done numerically? Is it possible to have a general form that the eigenvector takes as a linear combination of the eigenvectors for each individual term in the brackets?
– Tom
Sep 13, 2021 at 15:12
• Yes - linear combinations of the eigenvectors will span the eigenbasis of the density matrix Sep 13, 2021 at 15:14
• You could write a diagonal matrix with canonical eigestates and do a generic unitary transformation Sep 13, 2021 at 15:25
• Correcting my comment: linear combinations of the eigenvectors will span the range of the density matrix - there's also the null space Sep 14, 2021 at 13:58