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Harmonic Oscillator problem

  1. Aug 3, 2011 #1
    The book derives the wavefunction for the ground state of a harmonic oscillator. It's found to be a Gaussian with dispersion [itex]l = \sqrt{\frac{\hbar}{2m\omega}}[/itex]. The probability distribution for momentum is found to be Gaussian as well with dispersion [itex]\sigma_{p} = \frac{\hbar}{2l}[/itex]. The following is the line of the book that I'm having trouble understanding:

    "By inserting [itex]x = l[/itex] and [itex]p = \sigma_{p}[/itex] in the Hamiltonian, we obtain estimates of the typical kinetic and potential energies of the particle when it’s in its ground state."

    where the Hamiltonian of course is [itex]\frac{1}{2m}\{ p^{2} + \left( m \omega x \right)^{2} \} [/itex].

    Problem is: Why does inserting dispersion or variance produce estimates for typical values? Shouldn't this happen when we insert expectation or mean values?

    The book goes on to say that minimizing the Hamiltonian after substituting the expressions for the dispersion values, as described above, under the constraint that their product will remain less than or equal to [itex]\frac{\hbar}{2}[/itex] (uncertainty principle), will produce the appropriate value of energy for ground-state. Why does this happen? Why is inserting dispersion values producing these correct solutions? But I think this confusion emerges from same question as before.
     
  2. jcsd
  3. Aug 3, 2011 #2
    variance IS by definition the expectation value of x^2
    so by inserting variance you are indeed getting what you should get
     
  4. Aug 3, 2011 #3
    Because the H of SHO, as you point out consists of [itex]p^2[/itex] and [itex]x^2[/itex], so [itex]\langle H \rangle[/itex] consists of [itex]\langle p^2[/itex] and [itex]\langle x^2[/itex]. Given that it is clear that [itex]\langle p \rangle 0[/itex] and [itex]\langle x \rangle = 0[/itex], the expectation of H can be computed from the variant.
     
  5. Aug 3, 2011 #4
    Oh, yes, right.... thanks alot!
     
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