# Why does $\hbar$ appear twice in the axioms of QM?

They're really the same $\hbar$, and they come from the exact same place. To see this, it's best to work in Heisenberg picture, where the Schrodinger equation becomes $$\frac{dA}{dt} = \frac{1}{i \hbar} [A(t), H]$$ for any quantity $A(t)$. Now compare this to the equation of motion in the Hamiltonian formalism of classical mechanics, $$\frac{df}{dt} = \{f, H\}.$$ Similarly, the canonical Poisson brackets and canonical commutators are $$\{x, p\} = 1, \quad [x, p] = i \hbar.$$ Thus the general rule that gives both equations is that a classical equation is turned into a quantum one by replacing the Poisson bracket $\{\cdot, \cdot\}$ with $1/i\hbar$ times a commutator $[\cdot, \cdot]$.

Another way of seeing this is that quantum mechanics specifies a fundamental scale of action, and that can be thought of as an energy times a time (in the Schrodinger equation) or a length times a momentum (in the canonical commutator). However, it's a fair question to ask whether different kinds of particles can have different values of $\hbar$. I asked a question about that here.