Zero-point energy of a linear harmonic oscillator

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SUMMARY

The discussion centers on calculating the zero-point energy of a linear harmonic oscillator defined by the potential V = (kx^2)/2 and kinetic energy KE = 1/2 mv^2. Participants emphasize the application of the Heisenberg Uncertainty Principle, specifically using the relationships Δx and Δp to express total energy (E = KE + PE) in terms of uncertainties. The goal is to minimize this energy expression to determine the lowest allowable energy state of the system, which is a fundamental concept in quantum mechanics.

PREREQUISITES
  • Understanding of harmonic oscillators and their potential energy functions.
  • Familiarity with kinetic energy equations and their application in physics.
  • Knowledge of the Heisenberg Uncertainty Principle and its implications in quantum mechanics.
  • Ability to perform minimization techniques in calculus.
NEXT STEPS
  • Study the derivation of the zero-point energy for quantum harmonic oscillators.
  • Learn how to apply the Heisenberg Uncertainty Principle in various quantum systems.
  • Explore the mathematical techniques for minimizing functions in physics.
  • Investigate the implications of zero-point energy in quantum field theory.
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, as well as researchers interested in the foundational principles of energy quantization in harmonic oscillators.

mathlete
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Hi. I'm given a problem with a harmonic oscillator where the potential is V= (kx^2)/2 with a mass m (KE = 1/2 mv^2). I have to use the Heisenberg Uncertainty principle to show what the minimum energy is, but I'm not sure where to start... I think I have to combine KE + V and minimize that, but with respect to what? And how do I fit in the uncertainty principle?
 
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Make the (handwaving, but usual) assumption that the momentum must be greater than the uncertainty in momentum, and that the position must be greater than the uncertainty in position. Start by writing the total energy (KE + PE) in terms of those uncertainties. Minimize that to find the lowest allowable energy.
 
Doc Al said:
Make the (handwaving, but usual) assumption that the momentum must be greater than the uncertainty in momentum, and that the position must be greater than the uncertainty in position. Start by writing the total energy (KE + PE) in terms of those uncertainties. Minimize that to find the lowest allowable energy.

Do you mean something like replace x with Δx and p with Δp and then replace one of those with h-bar/Δx (or h-bar/Δp) and then minimize E with respect to Δx/Δp?
 
That's exactly what I mean.
 

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