Operator with 3 degenerate orthonormal eigenstates

[Celso]

Homework Statement

I've been given an operator $\hat{A}$ that has 3 orthonormal vectors as degenerate eigenstates corresponding to the same eigenvalue $a$.

I know how the hamiltonian acts on these vectors and I want to use this information to check whether $\hat{H}$ and $\hat{A}$ commute or not.

Relevant Equations

$\hat{A} |1> a|1>, \hat{A} |2> a|2>, \hat{A} |3> a|3>$
$\hat{H} = \begin{bmatrix} \sigma & 0 & \sigma \\ 0 & \sigma & \delta \\ \sigma & \delta & \sigma \end{bmatrix}$

With this information I concluded that the diagonal elements of $\hat{A}$ are equal to the eigenvalue $a$, so $\hat{A} = \begin{bmatrix} a & A_{12} & A_{13} \\ A_{21}& a & A_{23}\\A_{31} & A_{32} & a \end{bmatrix}$ but I can't see how to go from this to the commuting relation, since I don't know the other terms.

 

[kuruman]

Can you show that any normalized linear combination of the three orthonormal vectors is also an eigenstate of $\hat A$ with eigenvalue $a$? If you can, what does this have to do with the question being asked here?

 

[vela]

Relevant Equations:: $\hat{A} |1> a|1>, \hat{A} |2> a|2>, \hat{A} |3> a|3>$

With this information I concluded that the diagonal elements of $\hat{A}$ are equal to the eigenvalue $a$, so $\hat{A} = \begin{bmatrix} a & A_{12} & A_{13} \\ A_{21}& a & A_{23}\\A_{31} & A_{32} & a \end{bmatrix}$ but I can't see how to go from this to the commuting relation, since I don't know the other terms.

Note that your relevant equations say $\hat{A} |n\rangle = a|n\rangle$, not $\langle n \lvert \hat{A} \rvert n \rangle = a$.