1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Hint: Presumably, you are supposed to assume that the given basis is orthonormal. $\endgroup$ – Qmechanic Oct 15 '17 at 13:57
  • 1
    $\begingroup$ 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. $\endgroup$ – CDCM Oct 15 '17 at 13:59
  • 2
    $\begingroup$ Before going into elegant proofs, have you written these 3 equations into 3x3 matrix form? is this matrix symmetric? $\endgroup$ – Cosmas Zachos Oct 15 '17 at 14:07
4
$\begingroup$

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.

$\endgroup$

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.