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Quantum Mechanical Simple Harmonics Question

  1. Nov 14, 2004 #1
    So let us assume we have the following system. A harmonic oscillator consists of a mass 1g on a spring. Its frequency is 1 Hz and the masses passes through the equilibrium position with a velocity of 10cm/s.

    I have already calculated that the magnitude of the quantum number (n = 7.6 x 10^27)

    Now, what is meant by the average spacing between zeros of an eigenstate with such a quantum number? How would I calculate this, and what is this representing physically.

    Thanks to all who have given this a look.
  2. jcsd
  3. Nov 15, 2004 #2


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    With the spacing between two zero's of an eigenstate they most probably mean the distance between two neighbouring nodes (points where the wavefunction is zero). That's where you have zero probabability of finding the particle.

    By the way. Since you have a mass of 1 g (!), are you dealing with a system of billions and billions of particles??
    How can you say it passes the equilibrium position with a certain velocity. That's a violation of Heisenbergs principle.
  4. Nov 15, 2004 #3
    I think because n is so great the quantum mechanical model approaches that of classical physics where Heisenberg's Uncertainty principle can be ignored???
  5. Nov 15, 2004 #4


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    No,The Uncertainty Principle has nothing to do when speaking of bodies whose size and hence mass are macroscopical...The quantum number "n" can be as big as u like,the microscopic/macroscopic scale of the problem is the issue.You can't approach by quantum mechanical methods a problem involving macroscopical bodies,because the results give no realistic meaning...
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