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.