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

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.

## 1 Answer

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.

• Thanks! That makes sense. So basically a system does not inherently have a basis associated with it, rather the choice of observable determines what the useful basis vectors are? Jan 7 '21 at 21:49
• Yes, that's correct. Jan 7 '21 at 21:58