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About role of Planck constant in classical physics

  1. May 23, 2015 #1

    KFC

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    Hi there,
    I am confusing on the statement that we have classical physics when Planck constant approaching zero. I search the similar topic in Physics Forums and I saw that most of the answers refer to the size effect. It argues that when we measure something in the scale of meter comparing to Planck constant (of ##10^{-34}##), we could ignore that so it becomes classical but in microscopic, the measuring scale is comparable with the Planck constant so quantum effect is significant. Frankly, I don't understand this very well from the physical point of view. I know that quantum effect observed in microscopic scale but how do you related it to the Planck constant?

    I am thinking this from the point of commutation relation on momentum and position operator
    ##[\hat{x}, \hat{p}] = i\hbar##

    As we know in classical world, the order of measuring position and momentum does not matter, which is achieved when ##\hbar \to 0##.

    I have two questions. First, ##\hbar## is that small, why can't we approximate the commutation relation to zero? What happens if Planck constant is much bigger, how does it change our world?
     
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  3. May 23, 2015 #2

    KFC

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    By the way, I am reading something on so-called quasi-classical theory (e.g. http://link.springer.com/chapter/10.1007/978-3-540-70626-7_182), I am not quite following the detail explanation but from the brief introduction, I think what the theory about is if we have some way to rescale the equation such that ##\hbar \to \hbar/N## and when ##N## becomes big, ##\hbar/N \to 0## so it becomes classical. They call ##\tilde{\hbar}=\hbar/N## scale Plank constant. So does it mean we can scale the system in some way so we can tweak the system between classical and quantum?

    What's the term 'quasi' really mean? When we say 'quasi-classical' does it mean the system is still quantum mechanical but very close to the classical behavior?
     
  4. May 23, 2015 #3
    Let me try to answer your first question first.
    The essential thing, as you said, is to compare Planck's constant to other variables in your system. If other variables in your system are 'small enough', Planck's constant becomes larger in comparison.
    For example, let's look at the scale of the product ## x p##, where ## x## is the position (operator) and ## p## is the momentum of the particle.
    Quantum effects are visible in a Bose-Einstein condensate. Each atom has roughly a kinetic energy (they're cold, so non-relativistic) of ## k_B T= p^2/ 2 m##, so that ## p= \sqrt{2 m k_B T} ##. Let's say the condensate is realised with Rubidium atoms, ## m= 1.42 \cdot 10^{-25} \mathrm{kg}## per atom. The condensation temperature is around ##200 \cdot 10^{-9} \mathrm{K}##. Thus ## p = 8.85 \cdot 10^{-28} \mathrm{kg \, m/s^2} ##. If you have, say, ##10^{18}## atoms in a cubic meter, each atom 'takes a volume' of ##10^{-18}## cubic meters, which is ##x= 10^{-6}## m to each direction of space. With these values the aforementioned product ## x p \sim 8.85 \cdot 10^{-34} \mathrm{J s}##, which is the same order as ## \hbar##. Thus, in this case, one wouldn't expect ## x## and ## p## to commute. When they don't commute, if you know an atom's momentum precisely (error is much smaller than the actual value), it's position spreads out (measuring the same momentum gives substantially different position at different measurements), or vice versa.
    Note that the above calculations are extremely crude. I tried to look at the needed numbers from some actual experiments so that they'd be realistic.

    Did this answer your first question?

    It's probably not easy to answer that. Basically quantum effects would be visible in a larger scale than they are now, and one would have to go further up the scale for the classical limit. To be honest, I don't know is there any motivation to the value (i.e. a theoretical way to predict it from a model).
    I think you've understood right. That's how I understand it, too. At least, in that context. There are also cases where part of the system is treated quantum mechanically and part of it classically. For example, when studying an atom making a transition from one quantum state to another when in contact with an external radiation field, sometimes only the atom is modelled quantum mechanically, while the radiation field is taken to be classical. But I'm not sure if this approach goes by the same name.
     
    Last edited: May 23, 2015
  5. May 23, 2015 #4

    anorlunda

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    My favorite definition of the domain where quantum mechanics becomes necessary is when we consider things so small that they can not be observed without disturbing the thing being observed.

    That definition doesn't refer to Plank's constant at all.

    For example, bouncing a photon off the thing is often the gentlest thing we can do to observe. But a photon emparts a tiny bit of energy and momentum to the thing. If that energy and momentum is negligably small compared to the thing, we have classical mechanics, if not negligable, then QM applies.
     
  6. May 23, 2015 #5

    KFC

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    I appreciate your detail example. I think your explanation definitely help me to understand more in QM but I need more time to think about it.

    I agree. A big question in my mind why Planck constant is a universal constant, why it take that value instead of any other value.

    Your last example remind me something in the text, where they call it semi-classical. I don't know if semi-classical is same as quasi-classical. To me, it sounds like semi-classical treat the problem partial QM and partial classical (e.g. as you mentioned the field is classical while the state is quantum). And quasi-classical sounds like that the quantum problem could be treated as classical one or using classical model to study the system and it manifests the quantum behavior from classical point of view. I may be wrong but after reading other references online, that's how I understand so far.
     
  7. May 23, 2015 #6

    KFC

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    Thanks. I agree with that and that's the first aspects when I learn quantum mechanics at the very beginning. It is tough for me to understand why we need to introduce a constant called Planck constant to describe quantum effects. I know how it originally from but don't know why there will a universal constant and why it takes that value.
     
  8. May 24, 2015 #7

    anorlunda

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    There are three important constants that appear frequently in physics. h = planks constant, c = speed of light, G = Newton's gravitation constant. All three appear to be universal and to have specific values. That is determined by observation. None of the three, as far as I know, have an underlying theory about why they have the values that they do. We can imangine (but not observe) other universes in which physical constants have different values.
     
  9. May 24, 2015 #8
    I agree, now that you said it, my last example probably goes by the name 'semiclassical'.
    And one more note about your first question. The intuition I tried to give is more of a guideline, I think, to determine whether or not one should be aware of the commutation relations. I didn't really comment on the actual difference.
    Let's assume you measure two variables, one right after another (not letting the system evolve in between). For operators that commute, the order of measurement doesn't matter, and for operators that don't, the result will be different (statistically) depending on which variable was measured first. You can also pick a state, or an amplitude in the position representation and check that depending on whether you operate it with ## x## or ##-i \hbar \partial_x ## first, you will get a different state as a result. The difference will be ## x p \psi - p x \psi= i \hbar \psi ##.
     
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