0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – P4PL4
    Jan 7 '21 at 21:49
  • $\begingroup$ Yes, that's correct. $\endgroup$ Jan 7 '21 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.