Physical meaning of commutation

Question

I was reading the solution to quantum harmonic oscillator by J.J. Sakurai. He uses the annihilation and creation operators and there's a key step (I think) which is $$[a,a^{\dagger}]=1$$
I know we can interpret the commutation operator as if it were saying if we can measure two observable at the same time (commutation relation between position and momentum operator, for example), but in this case, how should I take this result?

Answer 1

The result $[a, a^{\dagger}] = 1$ means that there is no common set of eigenfunctions and hence are not simultaneously diagonalizable. When two operators A & B commute, i.e. $[A,B] = AB - BA = 0$, then they can be simultaneously diagonalized (this is the reason it is said they can be observed together). Normal operators commute with their adjoint (a normal operator has a representation like $\alpha = \beta + i\delta$ , where $\beta,\delta$ are self-adjoint and commute).

Answer 2

For your information: the operator $a$ may have eigenstates - the so called coherent states, and they are observable. They are states with uncertain energy - they are superpositions of different energy eigenstates, so the Hamiltonian $H\propto a^{\dagger}a$ does not commute with $a$.

The operator $a^{\dagger}$ does not have any eigenstate at all. It commutes with nothing.