Do operators A and Hamiltonian share a set of eigenfunctions if they commute?

[dyn]

If time evolution of a general ket is given by | Ψ > = e^-iHt/ħ | Ψ (0) > where H is the Hamiltonian. If i have a eigenbasis consisting of 2 bases |a> and |b> of a general Hermitian operator A and i write e^{-iHt/ ħ} |a> = e^{-iE_at/ ħ} |a> and e^-iHt/ħ |b> = e^{-iE_bt/ ħ} |b> ; does this mean that operator A and the Hamiltonian share a set of eigenfunctions ? ie they commute ?

And how would the time evolution of a ket be written if its operator did not commute with the Hamiltonian ?

 

[blue_leaf77]

If $E_a \neq E_b$ then $A$ and $H$ share the same eigenvectors. If on the other hand $E_a = E_b$, any linear combination of $|a\rangle$ and $|b\rangle$ is an eigenvector of $H$ but obviously they cannot be an eigenvector of $A$ as well since $a\neq b$.

And how would the time evolution of a ket be written if its operator did not commute with the Hamiltonian ?

Expand the eigenvector of $A$ into the eigenvectors of $H$.

 

[dyn]

If A is a general operator and E_a ≠ E_b then why do A and H share the same eigenvectors ?

 

[blue_leaf77]

If $[A,H]=0$ and the spectrum of $H$ is non-degenerate, then for any pair of eigenvectors $|a\rangle$ and $|b\rangle$ of $H$
$$
\langle a|[A,H]|b\rangle = 0 \\
(E_a-E_b) \langle a|A|b\rangle = 0
$$
This means when $|a\rangle \neq |b\rangle$ thus $E_a -E_b \neq 0$, the scalar $\langle a|A|b\rangle = 0$. That is, the matrix of $A$ is diagonal with respect to the basis of the eigenstates of $H$ and the two operators have the same set of eigenvectors.