Some questions on planck constant

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SUMMARY

This discussion focuses on the significance of the Planck constant (\hbar) in quantum mechanics and its role in quantization. The commutation relation [\hat{A}, \hat{B}] = i\hbar is emphasized as a fundamental aspect of quantum systems, where \hbar is essential for maintaining the integrity of quantum mechanics. While theoretically, \hbar can be replaced with any small constant, doing so leads to discrepancies with experimental results, indicating that the Planck constant is not arbitrary but rather a fundamental feature of nature. The relationship between \hbar and the Heisenberg uncertainty principle is also explored, highlighting that as \hbar approaches zero, classical mechanics becomes applicable.

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Hi there,
I am reading something on quantization, semi-classical approximation of a quantum system and related topics. There are few statements are pretty confusing. First of all, in many material, whenever they mention the quantization, they refer to a commutator, like

[\hat{A}, \hat{B}] = i\hbar

why's that? Does \hbar really have to be Planck constant? If not, if I make it something small but close to \hbar and will it still work in quantize the system?

Secondly, in some books, they said in theory \hbar could be anything, doesn't really have to be Planck constant. So why the quantum mechanics use the Planck constant instead of the arbitrary number? What's the significant of the Planck constant in nature?

In text, they said if \hbar \to 0, the system approaches to a classical system. Well, if we look at the Heisenberg uncertainty principle

\Delta x\Delta p \ge \hbar/2

Here \hbar is a small but not zero constant, so we cannot simultaneously make precise measurement on x and p. But in classical case, they can. But if you look at the uncertainty principle when \hbar\to 0, \Delta x\Delta p \ge 0 so if x and p are precisely measured at the same time, \Delta x = \Delta p =0, does it mean we can only take the '=' sign in the uncertainty principle?
 
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The commutator of two arbitrary operators A and B need not be related to the Planck constant h; that applies to very specific operators, namely x and p. One can understand that as follows: suppose you have a plane wave

\exp\left(i\frac{px}{\hbar}\right)

The momentum operator looks like

\hat{p} = -i\hbar\frac{\partial}{\partial x}

Acting with it one a plane wave results in the momentum p (which is the eigenvalue of the operator)

\hat{p}\;\exp\left(i\frac{px}{\hbar}\right) = -i\hbar\frac{\partial}{\partial x}\;\exp\left(i\frac{px}{\hbar}\right) = p\;\exp\left(i\frac{px}{\hbar}\right)

It is this hbar in the definition of the operator that results in the hbar in the commutator.

In principle quantization works with any value, so you could try to quantize a system using

\hat{p}_a = -ia\hbar\frac{\partial}{\partial x}

where a is a dimensionless number. Formally the quantization will work, but experimentally the theory will deviate from nature.
 

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