Quantum Harmonic Oscillator Raising and Lowering operators The commutator of the operators, $[a,a^\dagger] = 1$ is useful in rewriting the Hamiltonian in a neat way in terms of the creation and annihilation operators. 
So my question is, Is there a physical meaning to $[a,a^\dagger] = 1$? for except containing the canonical relationship of $x$ and $p$ and helping in a neat description of the Hamiltonian? 
Like how a commutator of two operators being "$0$" says about common eigenfunctions , is there a meaning to it being $1$? Or is it just to keep further calculations and numbers simple?
 A: *

*One can always remove a positive constant on the RHS of the CCR by normalizing the creation and annihilation operators appropriately. 

*If we define the number operator $$N~:=~a^{\dagger}a,$$ the CCR $$[a,a^{\dagger}]~=~\mathbb{1}$$ leads to 
$$[N,a^{\dagger}]~=~a^{\dagger},$$ 
which states that the creation operator $a^{\dagger}$ raises the number (that $N$ is counting) by $1$ unit, cf. a Fock space. See also this related Phys.SE post.
A: The real meaning of $[x, p] = i\hbar$ can be made clearer when you use it to further derive the Heisenberg's uncertainty principle for these quantities: $\Delta x \Delta p \ge \frac{\hbar}{2}$, which tells us that there's a fundamental limit to the precision with which this pair of physical quantities can be known/measured. 
Unfortunately, $a$ and $a^\dagger$ are not physical properties, and thus, not measurable (even though you could derive similar Heisenberg principle of them). That said I don't think the CCR of $a$ and $a^\dagger$ bears any physical meaning.
In a non-real world, however, imagine you could lower and raise the energy of QHO as you wish, and the CCR just says that this process of lowering and rising energy of the QHO is not commutative.
