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

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

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

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