4
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ I suspect you're just going to have to brute force it. $\endgroup$
    – DaftWullie
    Sep 13 at 13:54
  • $\begingroup$ 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? $\endgroup$
    – Tom
    Sep 13 at 15:12
  • $\begingroup$ Yes - linear combinations of the eigenvectors will span the eigenbasis of the density matrix $\endgroup$ Sep 13 at 15:14
  • 1
    $\begingroup$ You could write a diagonal matrix with canonical eigestates and do a generic unitary transformation $\endgroup$
    – Mauricio
    Sep 13 at 15:25
  • 2
    $\begingroup$ Correcting my comment: linear combinations of the eigenvectors will span the range of the density matrix - there's also the null space $\endgroup$ Sep 14 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.