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Zero-point energy of a linear harmonic oscillator

  1. Oct 18, 2005 #1
    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?
     
  2. jcsd
  3. Oct 18, 2005 #2

    Doc Al

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    Staff: Mentor

    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.
     
  4. Oct 18, 2005 #3
    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?
     
  5. Oct 18, 2005 #4

    Doc Al

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    Staff: Mentor

    That's exactly what I mean.
     
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