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

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 $$\rho$$.

• please try to use titles that actually describe what is being asked. This makes the question of greater reusability and easier to find in the future
– glS
Nov 30 '20 at 11:22

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 $$\frac{1}{2}(a + b \pm \sqrt{(a-b)^2+ 4 (c^2+d^2)})$$.

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

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

• Nice. I like this! Feb 10 '21 at 7:46
• is there a shortcut that you used to evaluate $\text{Tr} \rho^2 = \frac{1}{4} \text{Tr} (|a\rangle\langle a| + \dots )$ instead of substituting $|a\rangle,|b\rangle$ written in terms of orthonormal basis like $|a\rangle = a_0 |0\rangle + a_1 |1\rangle$ ? Feb 10 '21 at 20:19
• No shortcut. I computed $\rho^2$ under the trace getting $\frac{1}{4}(|a\rangle\langle a|a\rangle\langle a| + |a\rangle\langle a|b\rangle\langle b| + |b\rangle\langle b|a\rangle\langle a| + |b\rangle\langle b|b\rangle\langle b|)$. Then, I computed the inner products, e.g. $\langle a|a\rangle=1$ and $\langle a|b\rangle$ (which I left unchanged). Next, I applied trace to each term, e.g. $\mathrm{tr}(|a\rangle\langle a|) = \langle a|a\rangle = 1$ and $\mathrm{tr}(|a\rangle\langle b|) = \langle b|a\rangle$. Finally, I collected all terms into $\frac{1}{2} + \frac{|\langle a|b\rangle|^2}{2}$. Feb 10 '21 at 20:50
• I intentionally tried to avoid expanding $|a\rangle$ and $|b\rangle$ in a basis since both the density matrix and the formula for eigenvalues were given using Dirac notation in terms of kets and bras like $|a\rangle$. I thought to myself that we should be able to go from one to the other entirely using Dirac notation and the two kets and bras. Happily, it turned out that it works out :-) Feb 10 '21 at 20:58