How can I calculate the eigenvalues of $\rho^{T_{B}}$ (PPT) of the following state
$$ \rho =\frac{1}{2}|0\rangle\langle0|\otimes|+\rangle\langle+|+\frac{1}{2}|+\rangle\langle+|\otimes|1\rangle\langle1|. $$
$\begingroup$ $\endgroup$
Add a comment |
1 Answer
$\begingroup$ $\endgroup$
First note that
$$ \begin{align} \rho^{T_B} &= \frac{1}{2}|0\rangle\langle0|\otimes(|+\rangle\langle+|)^T+\frac{1}{2}|+\rangle\langle+|\otimes(|1\rangle\langle1|)^T \\ &= \frac{1}{2}|0\rangle\langle0|\otimes(|+\rangle\langle+|)^\dagger+\frac{1}{2}|+\rangle\langle+|\otimes(|1\rangle\langle1|)^\dagger \\ &= \frac{1}{2}|0\rangle\langle0|\otimes|+\rangle\langle+|+\frac{1}{2}|+\rangle\langle+|\otimes|1\rangle\langle1| \\ &= \rho. \end{align} $$
Now, we can obtain the eigenvalues using a result described in this question which says that the eigenvalues of
$$ \frac{1}{2}(|a\rangle\langle a| + |b\rangle\langle b|) $$
are
$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b \rangle|}{2}. $$
In our case, $|a\rangle = |0\rangle|+\rangle$ and $|b\rangle = |+\rangle|1\rangle$, so
$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle 0|+ \rangle\langle +|1\rangle|}{2} = \frac{1}{2} \pm \frac{1}{4}. $$
Thus, $\lambda_1 = \frac{1}{4}$ and $\lambda_2 = \frac{3}{4}$.