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

Question

Physical theories have dimensionful constants. Each constant can be found via measurement, by fitting some equation to data. Mathematically, you would expect each constant to be "defined" in this way by exactly one equation. For example, the gravitational constant $G$ is introduced in exactly one equation of GR, namely in its presence in Einstein's field equation. (Arguably this is more than one equation, but we'll let that pass, as these equations are tied together by the constraint of general covariance).

In quantum mechanics, however, $\hbar$ appears twice in two very different contexts. The first is in Schrodinger's equation

$$i \hbar \frac{\partial}{\partial t} |\Psi\rangle = \hat H |\Psi\rangle$$

and the second is in the canonical commutation relation

$$[\hat x, \hat p] = i \hbar. $$

Is there any reason why these two $\hbar$'s should be the same, morally speaking? Obviously, asking "why" about questions like this is subjective, but I'm still curious if anyone has any good answers.

Answer 1

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 $[\cdot, \cdot]$.

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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.

Answer 2

This question boils down to two questions: (1) why do the Planck-Einstein relations for the energy $E$ and frequency $\omega$ as well as the momentum $\vec p$ and wavevector $\vec k$of a photon hold with the same $\hbar$: $$E=\hbar \omega$$ $$ \vec p=\hbar \vec k$$ and (2) why do these relations also hold for energy and momentum of a free particle (the de Broglie relation). For the photon, the fact that the second equation has the same $\hbar$ as the first equation follows directly from Maxwell's equations (Poynting's theorem). It has also been experimentally confirmed (e.g. Compton effect). For a general free particle, the second equation is the de Broglie relation for the matter wave of a free particle. That this also has the same $\hbar$, as assumed by de Broglie, has been confirmed experimentally, e.g., by the Davisson-Germer electron wave diffraction experiments.