I am interested in the state:

$\frac{1}{\sqrt{2}} (\left|11\right> + \left|22\right>)$

If I find the density matrix of this, I find the $9 \times 9$ matrix $\rho$.

If I want to find the reduced density matrix of Alice's qubit, I obtain:

$$\rho_{A} = \left|1\right>\left<1\right| + \left|2\right>\left<2\right|$$ this gives us a $3 \times 3$ matrix with 1's in $\rho_{A_{2,2}}$ and $\rho_{A_{3,3}}$

However, this would imply that the eigenvalues of the system are $0,1,1$. How is this possible?


1 Answer 1


Maybe you forgot about the coefficient $\frac{1}{\sqrt{2}}$. The correct reduced density matrix is $$\frac12 (\left|1\right>\left<1\right| + \left|2\right>\left<2\right|)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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