# How to write the eigenvectors of a mixture of two pure states?

Let $$|\psi_1\rangle,|\psi_2\rangle$$ be two pure states. Assume $$\langle\psi_1|\psi_2\rangle\neq0$$, and consider the convex combination $$\rho\equiv p_1 |\psi_1\rangle\!\langle\psi_1| + p_2 |\psi_2\rangle\!\langle\psi_2|,$$ with $$p_1+p_2=1$$.

Suppose we are interested in the pure state corresponding to the largest (smallest) projection probability. In other words, we want a state $$|\phi\rangle$$ that maximises (minimises) $$p_1 |\langle\phi|\psi_1\rangle|^2 + p_2 |\langle\phi|\psi_2\rangle|^2.$$

Algebraically, this can be found by simply computing the eigenvectors of $$\rho$$. Doing so gives $$|\phi_\pm\rangle = N_\pm\left( (A\pm\sqrt{A^2+4z^2})|\psi_1\rangle + 2z |\psi_2\rangle \right),$$ with $$A\equiv p_1 + p_2(2F-1)$$, $$F\equiv |\langle\psi_1|\psi_2\rangle|^2$$, $$z\equiv\langle\psi_1|\psi_2\rangle$$.

Now, because we are effectively operating on the span of two pure states, we can assume without loss of generality that these are qubits, and represent everything in the Bloch sphere.

Is there a nice geometrical characterisation of where the eigenvectors $$|\phi_\pm\rangle$$ end up being represented in the Bloch sphere in relation to $$|\psi_i\rangle$$? By "nice" I mean something like "$$|\phi_\pm\rangle$$ sit in the plane spanned by $$|\psi_i\rangle$$ and at the angle so and so between them".

For $$p_1=p_2$$ I find that $$|\phi_+\rangle$$ sits exactly in between $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$. More precisely, denoting with $$\vec r_\phi$$ the Bloch vector of $$|\phi\rangle$$, we have $$\vec r_{\phi_\pm}=\pm N (\vec r_{\psi_1} + \vec r_{\psi_2}).$$ This stops being the case for $$p_1\neq p_2$$. Is there a similarly nice solution for the general case?

The answer is trivial thinking directly in the Bloch representation of the states. The TL;DR is that if $$\newcommand{\bs}[1]{{\boldsymbol{#1}}}\nu_{\rm op}(\bs r)$$ and $$\nu_{\rm v}(\bs r)$$ denote the operator and $$\mathbb C^2$$ vector corresponding to the Bloch vector $$\bs r$$, respectively, then $$\nu_{\rm v}(\pm\bs r)$$ are the eigenvectors of $$\nu_{\rm op}(\bs r)$$, with eigenvalues $$\frac12(1\pm \|\bs r\|)$$.
If $$\rho_{\bs r}$$ is the state with Bloch representation $$\bs r$$, then $$p\rho_{\bs r}+(1-p)\rho_{\bs s} = \rho_{p\bs r+(1-p)\bs s}.$$ Therefore the question of what are the eigenvectors of $$p|\psi_1\rangle\!\langle\psi_1| + (1-p)|\psi_2\rangle\!\langle\psi_2|$$ reduces to the question of what are the eigenvectors of $$p\rho_1+(1-p)\rho_2$$ where $$\rho_i\equiv |\psi_i\rangle\!\langle\psi_i|$$. And as follows from our above observation, the answer to this is simply that the eigenvectors are $$\pm(p\bs r_1+(1-p)\bs r_2)$$, with $$\bs r_i$$ the Bloch vector corresponding to $$\rho_i$$.
(Eigenvalues and eigenvectors of $$\rho$$) For completeness, I'll include the expressions for eigenvalues and eigenvectors of $$\rho$$ as a function of its matrix and Bloch representations.
The relation between $$\rho$$ and its Bloch representation $$\bs r$$ is $$\rho = \frac{I + \bs r\cdot\bs\sigma}{2} = \frac12\begin{pmatrix}1+z & x-i y \\ x+iy & 1-z\end{pmatrix},$$ with $$\bs\sigma$$ the vector of Pauli matrices, and $$\bs r\equiv(x,y,z)^T$$. The operator $$\rho$$ is a positive semidefinite for $$\bs r\le1$$ and pure for $$\|\bs r\|=1$$.
The eigenvalues of $$\rho$$ are $$\lambda_\pm = \frac{1}{2}\left( 1\pm \sqrt{(\rho_{11}-\rho_{22})^2 + 4\rho_{12}\rho_{21}}\right) = \frac12\left(1 \pm \sqrt{z^2+x^2+y^2}\right) = \frac12(1 \pm \|\bs r\|),$$ as can also be see directly from the expression of $$\rho$$ in terms of $$\bs r$$. The corresponding eigenvectors are $$|\lambda_\pm\rangle = \frac{1}{\sqrt{2r(r\pm z)}}\begin{pmatrix}z\pm r \\ x+i y\end{pmatrix}, \qquad r\equiv\|\bs r\|.$$ The corresponding angles in the Bloch sphere thus read $$\tan(\theta/2) = \frac{\sqrt{x^2+y^2}}{z\pm r} = \frac{\sin\theta}{\cos\theta\pm 1},$$ where in the last step I divided by $$r$$ and used the geometric identities $$\sin\theta=\sqrt{x^2+y^2}/r$$ and $$\cos\theta=z/r$$. It follows that $$\theta_+=\theta$$ and $$\theta_-=\theta+\pi$$. In other words, the angles corresponding to $$|\lambda_\pm\rangle$$ are those corresponding to the same direction of $$\bs r$$.