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I'm in an introductory quantum mechanics course and we just covered Heisenberg's uncertainty principle. One thing that came to mind is are we allowed to estimate the minimum energy of anything, say a pendulum? If I have a mass and a rigid rod of some fixed length, then I know $\omega$. Can I estimate it's 'lowest' possible energy by using: $$E=\dfrac{\Delta p^2}{2m}+\dfrac{1}{2}m\omega^2\Delta x^2$$ where $$\Delta x\Delta p=\dfrac{\hbar}{2}$$ I did this and I got a really small number (which makes sense I guess). But in classical mechanics, we have two equilibrium points (one pointing directly downwards and one standing perfectly upwards). Since I always have a minimum energy, does this mean the unstable equilibrium solution is impossible (in quantum mechanics)? Am I thinking in the right way? If so can I estimate how long it will take for the unstable equilibrium to fall back to the stable point?

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    $\begingroup$ Your expression for the energy is only valid near the minimum. The unstable position will not produce oscillations in classical or quantum mechanics. $\endgroup$ – ZeroTheHero Aug 17 '17 at 2:23
  • $\begingroup$ Related: physics.stackexchange.com/q/459403 $\endgroup$ – user191954 Mar 16 at 3:43
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There are two things one should say about this:

  1. The unstable equilibrium is already practically impossible in classical mechanics. Or have you ever seen a pendulum standing in upright position and staying there? Unstable equilibrium solutions should be taken as some kind of mathematical curiosity, but they will not occur in the physical world, since the tiniest perturbation by the environment or whatever will destroy them. So if quantum mechanics now would tell you that it may not occur, well, this would not be any surprise at all. The effects that would make a pendulum fall down from the unstable point are thus surely not quantum mechanical but come from very classical effects like the air molecules hitting the pendulum etc. (A different thing is if you build a pendulum out of, say, only a couple of atoms, in that case you might be right and it might be quantum effects driving it away from there.)

  2. The Heisenberg uncertainty relation is about the variance of measured quantities. Not more, not less. (If you ascribe an ontological status to it depends on your interpretation.) It should not be overused. It is true that you can see the following: The value of $h$ is so small that is has no effect in situations where classical physics is valid.

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