How are the eigenvalues of $rho=frac12(|arangle!langle a| +|brangle!langle b|)$ 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 $|arangle$ and $|brangle$ 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 $rho$.

M. Stern · Accepted Answer

For this it suffices to consider the two-dimensional subspace spanned by $|arangle$ and $|brangle$. Let $|0rangle$ and $|1rangle$ be an orthonormal basis of this subspace. Then
$$begin{align}
|arangle =& a_0 |0rangle + a_1 |1rangle
|brangle =& b_0 |0rangle + b_1 |1rangle
end{align}
$$
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 $frac{1}{2}(a + b pm sqrt{(a-b)^2+ 4 (c^2+d^2)})$.

Adam Zalcman · Answer

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 $rho^2$
$$
detrho = lambda_1lambda_2 = frac{1}{2}left[(lambda_1 + lambda_2)^2 - (lambda_1^2 + lambda_2^2)right] = frac{1}{2}left[(mathrm{tr}rho)^2 - mathrm{tr}rho^2right].
$$
(N.B. this useful relationship underlies the Faddeev-LeVerrier algorithm.) Now, calculate
$$
mathrm{tr}rho^2 = frac{1}{4}mathrm{tr}(|aranglelangle a| + langle a|brangle |aranglelangle b| + langle b|arangle |branglelangle a| + |branglelangle b|) 
= frac{1}{2} + frac{|langle a|brangle|^2}{2}
$$
and so
$$
lambda_1 lambda_2 = frac{1}{2}left[(mathrm{tr}rho)^2 - mathrm{tr}rho^2right] = frac{1}{4} - frac{|langle a|brangle|^2}{4}
$$
from which we see that
$$
lambda_i = frac{1}{2} pm frac{|langle a|brangle|}{2}.
$$

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

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