Possible measurements of an observable described by an operator composed of two identical states

Question

Consider we have an operator $\hat{B}$ which is composed of two states i.e. $$ \hat{B}=|\theta_1\rangle\langle\theta_1| $$ The state $\theta_1$ is in turn composed of basis vectors $$ |\theta_1\rangle=\alpha_1|\phi_1\rangle+\alpha_2|\phi_2\rangle $$ where $\alpha_1,\alpha_2\in\mathbb{C}$ and $\{\phi_1,\phi_2\}$ form an orthonormal basis for the Hilbert space. If we want to find the possible measurements for this observable, we need to solve the eigenvalue equation $$ \hat{B}|\phi_{i}\rangle=\epsilon_{i}|\phi_{i}\rangle \;\;\;\;\;\; i=1,2 $$ I tried approaching this in the manner described in jinawees answer. However, my interpretation of how it applied to this question would not work. In this case we get (for $i=1$) \begin{align*} \hat{B}|\phi_1\rangle&=|\theta_1\rangle\langle\theta_1|\phi_1\rangle \\ &=(\alpha_1 |\phi_1\rangle+\alpha_2 |\phi_2\rangle)(\alpha_1^*\langle\phi_1|\phi_1\rangle)\\ &=\alpha_1^{*}(\alpha_1 |\phi_1\rangle+\alpha_2 |\phi_2\rangle) \end{align*} which does not give us our required eigenvalue equation as both orthonormal bases are still present. Not really sure how to solve this, any help would be greatly appreciated.

Answer 1

The $\mid\phi_i\rangle$ in the eigenvalue equation are states to be determined, not some given basis states. The equation should look like this:

$$B\vert\psi\rangle=\varepsilon\vert\psi\rangle,$$

and you need to determine what the possible combinations of states $\vert\psi\rangle$ and eigenvalues $\varepsilon$ are which solve this equation. In this case you don't have to do all that much algebra, and instead notice that $\vert\theta_1\rangle$ itself is an obvious eigenstate of $B$ with eigenvalue $1$, and the state orthogonal to $\vert\theta_1\rangle$ is also an eigenstate with eigenvalue $0$. Since your Hilbert space is 2-dimensional, that's all the eigenstates you'll find.