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Some questions on planck constant

  1. Sep 18, 2012 #1


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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

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

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

    Secondly, in some books, they said in theory [itex]\hbar[/itex] 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 [itex]\hbar \to 0[/itex], the system approaches to a classical system. Well, if we look at the Heisenberg uncertainty principle

    [tex]\Delta x\Delta p \ge \hbar/2[/tex]

    Here [itex]\hbar[/itex] 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 [itex]\hbar\to 0[/itex], [itex]\Delta x\Delta p \ge 0[/itex] so if x and p are precisely measured at the same time, [itex]\Delta x = \Delta p =0[/itex], does it mean we can only take the '=' sign in the uncertainty principle?
    Last edited: Sep 18, 2012
  2. jcsd
  3. Sep 18, 2012 #2


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    Science Advisor

    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


    The momentum operator looks like

    [tex]\hat{p} = -i\hbar\frac{\partial}{\partial x}[/tex]

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

    [tex]\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)[/tex]

    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

    [tex]\hat{p}_a = -ia\hbar\frac{\partial}{\partial x}[/tex]

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