How are the eigenvalues of $\rho=\frac12(|a\rangle\!\langle a| +|b\rangle\!\langle b|)$ derived?

Question

Let's say I have a density matrix of the following form:

$$ \rho := \frac{1}{2} (|a \rangle \langle a| + |b \rangle \langle b|), $$ where $|a\rangle$ and $|b\rangle$ are quantum states. I saw that the eigenvalues of this matrix are: $$ \frac{1}{2} \pm \frac{|\langle a | b \rangle|}{2}. $$ I was just wondering how this is derived. It seems logical, i.e if $|\langle a | b \rangle| = 1$ then the eigenvalues are $0$ and $1$, otherwise if $|\langle a | b \rangle| = 0$ then they are half and half. This means that the entropy of the system would either be $0$ or $1$. But I was just wondering how to calculate the eigenvalues from $\rho$.

Answer 1

For this it suffices to consider the two-dimensional subspace spanned by $|a\rangle$ and $|b\rangle$. Let $|0\rangle$ and $|1\rangle$ be an orthonormal basis of this subspace. Then $$\begin{align} |a\rangle =& a_0 |0\rangle + a_1 |1\rangle\\ |b\rangle =& b_0 |0\rangle + b_1 |1\rangle \end{align} $$ and $$\rho = \frac{1}{2}\left(\begin{array}{cc} a_0 a_0^*+b_0b_0^* & a_0a_1^* + b_0 b_1^*\\ a_1 a_0^*+ b_1b_0^* & a_1a_1^*+b_1b_1^* \end{array}\right).$$ That is, now you have a 2x2 Hermitian matrix and calculate its eigenvalues as usual. Hint: A Hermitian matrix $$\left(\begin{array}{cc} a & c + d i\\ c - d i & b \end{array}\right)$$ has eigenvalues $\frac{1}{2}(a + b \pm \sqrt{(a-b)^2+ 4 (c^2+d^2)})$.

Answer 2

Let $\lambda_1$ and $\lambda_2$ denote the eigenvalues of $\rho$. Then $\lambda_1 + \lambda_2 = \mathrm{tr}\rho = 1$ and $\lambda_1 \lambda_2 = \det \rho$. We can compute the determinant using trace of $\rho$ and $\rho^2$

$$ \det\rho = \lambda_1\lambda_2 = \frac{1}{2}\left[(\lambda_1 + \lambda_2)^2 - (\lambda_1^2 + \lambda_2^2)\right] = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right]. $$

(N.B. this useful relationship underlies the Faddeev-LeVerrier algorithm.) Now, calculate

$$ \mathrm{tr}\rho^2 = \frac{1}{4}\mathrm{tr}(|a\rangle\langle a| + \langle a|b\rangle |a\rangle\langle b| + \langle b|a\rangle |b\rangle\langle a| + |b\rangle\langle b|) \\ = \frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2} $$

and so

$$ \lambda_1 \lambda_2 = \frac{1}{2}\left[(\mathrm{tr}\rho)^2 - \mathrm{tr}\rho^2\right] = \frac{1}{4} - \frac{|\langle a|b\rangle|^2}{4} $$

from which we see that

$$ \lambda_i = \frac{1}{2} \pm \frac{|\langle a|b\rangle|}{2}. $$