# Check if operator $A$ is an observable

Given operator $A$ and following relationships: $A|2\rangle=|1\rangle+|3\rangle$ and $A|1\rangle=A|3\rangle=|2\rangle$.

I know that this operator should be self adjoint to correspond to an observable but how exactly do I show that?

• Hint: Presumably, you are supposed to assume that the given basis is orthonormal. – Qmechanic Oct 15 '17 at 13:57
• I've edited your post to put more of it in latex. For future ref, the commands you want are \rangle, or \langle for $\rangle$ and $\langle$, to make nice kets. – CDCM Oct 15 '17 at 13:59
• Before going into elegant proofs, have you written these 3 equations into 3x3 matrix form? is this matrix symmetric? – Cosmas Zachos Oct 15 '17 at 14:07

Assuming $\{\vert 1\rangle, \vert 2\rangle,\vert 3\rangle\}$ form an orthonormal basis, systematically construct $\langle k\vert A\vert m\rangle$ to obtain $$\hat A\mapsto \left(\begin{array}{ccc} 0&1 &0 \\ 1&0& 1\\ 0&1& 0\end{array}\right)$$ and recall that in finite dimensions, there is no difference between hermitian and self-adjoint, and all eigenstates are normalizable.