The questions
Q1. What are simple ways to think mathematically about the physical meanings of the Planck constant?
Q2. How does the Planck constant appear in mathematics of quantum mechanics? In particular, quantization is an important notion in mathematical physics and there are various forms of quantization for classical Hamiltonian systems. What is the role of the Planck constant in mathematical quantization?
Q3. How does the Planck constant relate to the uncertainty principle and to mathematical formulations of the uncertainty principle?
Q4. What is the mathematical and physical meaning of letting the Planck constant tend to zero? (Or to infinity, if this ever happens.)
Motivation:
One purpose of this question is for me to try to get better early intuition towards a seminar we are running in the fall. Another purpose is that the Planck constant plays almost no role (and, in fact, is hardly mentioned) in the literature on quantum computation and quantum information, and I am curious about it.
Related MO questions: Does quantum mechanics ever really quantize classical mechanics? ;
Best Answer
Let's give it a try. Of course, the precise mathematical meaning is perhaps absent, so the answers are sort of heuristic. But if I understand correctly, you want to gain intuition ;)
The first observation is that Planck's constant has units, it is not a numerical constant but carries physical dimension. In some sense, this makes all the difference: in your favorite unit system, $\hbar$ has the numerical value $1$. Period. Nothing more to say...
Now what is the impact? If you think of e.g. Fourier transform, the phase in your integral is $e^{ikx}$ and for a mathematician, everything is fine with this. Now for a physicist (depending on the application), $x$ has a unit of length. So there is no way to exponentiate a length per se, you can only plug in dimension-less quantities in a transcendental function. This means that the $k$ has a physical dimension of $1/\text{length}$. The physical interpretation is that $k$ is an inverse wave length. Now in quantum physics, there comes the time that you want your wave functions in the momentum representation. So you have to replace $k$ by the physical momentum $p$ which has a different unit, namely that of momentum. This requires dividing the product $px$ in your phase by a quantity of dimension momentum times length which is action. So the observation in physics is that there is a universal constant providing a scale for doing exactly that, $\hbar$. The familiar uncertainty relation you know from Fourier transform becomes thereby scaled by $\hbar$ as well.
Another, perhaps more important observation is that in quantization in the Schrödinger approach you consider wave functions on configuration space, depending on the position $x$ of dimension length. Now the relevant operators are the momentum operators encoded by a derivative. But derivatives have dimension $1/\text{length}$ instead of momentum. Thus you have to rescale the derivative by a physical constant of dimension action to get a quantity of dimension momentum. Again, $\hbar$ is the one doing the job.
From what I said (and much more can be said) it should be clear that $\hbar$ does not tend to zero at all (it's $1$, right?). So what should these statements $\hbar \to 0$ then really mean from a physical point of view? This is in fact a quite subtle and, I guess, ultimately not well understood point. The observation in daily life is that quantum effects do not play any role, the world behaves classically. So classical physics is at least a perfect approximation in many situations. Quantum physics only enters the picture if we perform very precise measurements etc. Now the idea is that the more fundamental theory (quantum) has a certain less fundamental theory (classical) as limit. The interpretation of this limit is subtle. But in many situations, the limit relates to parameters of the system which carry dimensions, typically the dimension of action. Now it is the ratio of this system-dependent parameter (say a combination of masses, lengths etc) and $\hbar$ which tells us whether the classical theory is a good approximation or not. The main point is that we need parameters $\alpha$ of the same dimension as $\hbar$ to have a dimension-less ratio $\hbar/\alpha$. Only such a dimension-less parameter can be considered to be "small" or "large". The classical limit is thus better understood as $\hbar/\alpha \to 0$, meaning that we look at many different systems with different values of $\alpha$ (whatever its concrete physical interpretation may be) and get the classical limit as limiting scenario if these parameters assume values much larger than $\hbar$. We can compare them since they have both the same physical dimension.
Now why don't we see these arguments in math? The (perhaps pretty obscure) observation is that in the explicit situations we can handle, the mathematical limit $\hbar/\alpha \to 0$ can also be understood as $\hbar \to 0$ while $\alpha$ is fixed. Mathematically this is not a big deal, but physically an absurd interpretation: $\hbar$ is a fundamental constant of nature and we have no $\hbar$-wheel where we can adjust its value.
Hmm, lot of blabla, but I hope that this clarifies the physical side of the story a bit. Mathematically, one has several ways to incorporate "dimensional" constants. One nice way is to look at graded algebras where the grading refers to the power of the dimension you are looking at. Then the grading helps to keep track of the correct dimensions. In particular in quantization theory this turned out to be a very useful tool.