# Applying quantum mechanical logic on simple pendulums

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?

• Your expression for the energy is only valid near the minimum. The unstable position will not produce oscillations in classical or quantum mechanics. – ZeroTheHero Aug 17 '17 at 2:23
• – user191954 Mar 16 at 3:43

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.