Zero-point energy of a linear harmonic oscillator

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a linear harmonic oscillator, specifically focusing on the potential energy defined as V = (kx^2)/2 and kinetic energy as KE = 1/2 mv^2. The original poster is tasked with using the Heisenberg Uncertainty principle to determine the minimum energy of the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to combine kinetic and potential energy and what variable to minimize. Some participants suggest using uncertainties in momentum and position to express total energy and then minimizing that expression.

Discussion Status

Participants are exploring the relationship between energy, position, and momentum uncertainties. There is a suggestion to replace variables with their uncertainties and to minimize the energy expression accordingly. The discussion is ongoing, with no explicit consensus reached yet.

Contextual Notes

There is an emphasis on the assumptions regarding the uncertainties in momentum and position, which are considered to be greater than the respective uncertainties themselves. The original poster is navigating how to incorporate these assumptions into their calculations.

mathlete
Messages
148
Reaction score
0
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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
3K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K