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Harmonic oscillator

  1. Dec 4, 2005 #1
    why is the lowest allowed energy not E=0 but some definite minimum E=E0?
  2. jcsd
  3. Dec 4, 2005 #2


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    If you solve the Time Independent Schrodinger 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: Dec 4, 2005
  4. Dec 4, 2005 #3


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    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.
  5. Dec 4, 2005 #4
    thank you very much!!! :)
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