How do I calculate the eigenvalues of the positive partial transpose of this two-qubit state?

Question

How can I calculate the eigenvalues of $\rho^{T_{B}}$ (PPT) of the following state

$$ \rho =\frac{1}{2}|0\rangle\langle0|\otimes|+\rangle\langle+|+\frac{1}{2}|+\rangle\langle+|\otimes|1\rangle\langle1|. $$

Answer 1

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 $\lambda_2 = \frac{3}{4}$.