What is the Physical Significance of The Coefficients of Angular Momentum Ladder Operators? For example, in a spin 1 system in a state $|-1\rangle$, lets raise it twice, and the state becomes $2\hbar^2|+1\rangle$, now lets lower it twice, we get $4\hbar^4|-1\rangle$. Keep going and the $\hbar$ factors keep piling on. What is the physical significance of this factor? Why don't we get back to $|-1\rangle$?
 A: The angular momentum operators have units of $\hbar$ so their action contains $\hbar$; these accumulate just in the same way that $E_n$ accumulates when you write 
\begin{align}
\hat H\vert n\rangle &=E_n\vert n\rangle\, ,\\
(\hat H)^k\vert n\rangle &=(E_n)^k\vert n\rangle\, .
\end{align}
The angular version would be
$$
\hat L_z\vert \ell,m\rangle=m\hbar \vert \ell,m\rangle\, ,
\quad (\hat L_z)^k\vert \ell,m\rangle=(m\hbar)^k \vert \ell,m\rangle.
$$
Such repeated actions of operators are useful in generating finite transformation, such as the time evolution 
$$
U(t)=e^{-i t \hat H/\hbar}= 1+\sum_{k=1}^{\infty} 
\frac{1}{k!}\left(\frac{-i t\hat H}{\hbar}\right)^k
$$
or a finite rotation
$$
R_z(\theta)=e^{-i\theta \hat L_z/\hbar}=1+
\sum_{k=1}^\infty\frac{1}{k!} 
\left(\frac{-i \theta\hat L_z}{\hbar}\right)^k\, .
$$
A: There is some flexibility in how one can define these operators. Ladder operators don't need to carry these constants. For instance, for the number states, one can define the ladder operator such that they simply give
$$ a^{\dagger} |n\rangle = |n+1\rangle \sqrt{1+n} , \\ a |n\rangle = |n-1\rangle \sqrt{n} . $$
The same applies for spin operators. Often one can get rid of the Planck constant through normalization. 
It is a good idea to remove the Planck constant as far as possible to avoid some confusion the comes with the association between the Planck constant and quantum effects.
For instance, there does not need to be a Planck in the commutation relations of spin operators. The reason is that these spin operators are just the generators for rotation, which is not intrinsically a quantum process.
The situations where one cannot get rid of the Planck constant are those situations that involve interactions. These are situations that represent intrinsic quantum effects.
