Harmonic Oscillator: Lowest Allowed Energy Not E=0?

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Discussion Overview

The discussion revolves around the concept of the lowest allowed energy in a harmonic oscillator, specifically addressing why this energy is not zero but rather a minimum value denoted as E0. The scope includes theoretical aspects of quantum mechanics, particularly the Time Independent Schrödinger equation and its implications for energy quantization.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why the lowest allowed energy is not zero, suggesting a need for clarification on the concept.
  • Another participant explains that solving the Time Independent Schrödinger equation for the harmonic oscillator leads to quantized energy levels, specifically stating that the lowest energy level is E_n = (n + 1/2) ħω, indicating that E=0 is not permitted.
  • A further contribution presents a hand-wave argument, stating that if E=0, both position (x) and velocity (v) would be zero, which contradicts the principles of Heisenberg's uncertainty principle.

Areas of Agreement / Disagreement

Participants present differing viewpoints on the implications of energy quantization and the relationship to Heisenberg's uncertainty principle. No consensus is reached regarding the interpretation of the lowest allowed energy.

Contextual Notes

The discussion does not resolve the underlying assumptions about the nature of energy levels or the implications of quantum mechanics, leaving open questions regarding the interpretation of the results derived from the Schrödinger equation.

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why is the lowest allowed energy not E=0 but some definite minimum E=E0?
 
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If you solve the Time Independent Schrödinger equation for the Harmonic Oscillator, that is
[tex]-\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} + \frac{1}{2}kx^2 \Psi = E \Psi[/tex]

The quantization of energy comes from the boundary conditions (ie, [itex]\Psi = 0[/itex] when [itex]x= \infty[/itex] or [itex]x = -\infty[/itex]).

The permitted energy levels will be

[tex]E_n = (n+\frac{1}{2}) \hbar \omega[/tex]

So the lowest Energy is not E=0.
 
Last edited:
I could give a hand-wave argument. We have E=1/2mv^2+1/2kx^2.
If E=0 both x and v are zero, which contradicts Heisenberg.
 
thank you very much! :)
 

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