Consider the state $$|\psi\rangle=(\cos\theta_A|0\rangle+\sin\theta_A|1\rangle)\otimes(\cos\theta_B|0\rangle+e^{i\phi_B}\sin\theta_B|1\rangle).$$

To calculate the $\rho^{T_B}$ I first calculate the $\rho$ and the change it into a matrix form but It was very complicated to compute the eigenvalues for that matrix.

Is there any easy way to calculate the eigenvalues of $\rho^{T_B}$?


1 Answer 1


In this specific case, absolutely! Note that $$ |\psi\rangle=|\phi_A\rangle|\phi_B\rangle, $$ such that $$ \rho=|\phi_A\rangle\langle\phi_A|\otimes |\phi_B\rangle\langle\phi_B|. $$ Now, it is the case that $|\phi_B\rangle\langle\phi_B|^T=|\phi_C\rangle\langle\phi_C|$ for some state $|\phi_C\rangle=\cos\theta_B|0\rangle+e^{-i\phi_B}\sin\theta_B|1\rangle$.

So, you should now be able to simply observe that $\rho^{T_B}=|\phi_A\rangle\langle\phi_A|\otimes |\phi_C\rangle\langle\phi_C|$ is a rank 1 projector (${\rho^{T_B}}^2=\rho^{T_B}$ and $\text{Tr}(\rho^{T_B})=1$), or that the only non-zero eigenvector is just $|\phi_A\rangle|\phi_C\rangle$, and that it has eigenvalue 1.

Just as a side note, you might wonder how you're supposed to predict this, and therefore know it's a reasonable approach to take for the calculation. The point here is that this is the whole premise of the partial transpose - if you act it on a separable state, then this is just like doing the transpose on pure states, and the transpose of a pure state is always another pure state.


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