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Ground state of wavefunction

  1. Feb 24, 2006 #1
    A particle has mass 'm' and potential [itex]V(x) = x^2[/itex]. Find the ground state of this wavefunction.

    Should I use the Hamiltonian operator here?
     
  2. jcsd
  3. Feb 24, 2006 #2
    Yes. You should get something very similiar to the harmonic oscillator.
     
  4. Feb 25, 2006 #3
    Thanks, inha! I compared it to the PE function of a harmonic oscillator.

    [tex]V(x) = {1\over 2} m \omega^2 x^2[/tex]

    Comparing it with [itex]V(x) = x^2[/tex],

    [tex]{1\over 2} m \omega^2 = 1[/tex]

    [tex]\omega = \sqrt{2\over m}[/tex]

    And I substitute this value in the formula:
    [tex]E_n = \hbar \omega \left(n + {1\over 2}\right)[/tex]

    Am I going right? :confused:
     
  5. Feb 25, 2006 #4
    It looks good (Of course, n=0 for the ground state, don't forget that!), but make sure you know how to do the problem from scratch. I'm guessing that's what your professor wanted you to do.

    -Dan
     
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