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Ladder operator for harmonic oscillator, I don't get a mathematical

  1. Jun 6, 2013 #1

    fluidistic

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    If the ladder operator ##a=\sqrt {\frac{m\omega}{2\hbar}}x+\frac{ip}{\sqrt{2m\hbar \omega}}## and ##a^\dagger=\sqrt {\frac{m\omega}{2\hbar}}x-\frac{ip}{\sqrt{2m\hbar \omega}}## then I get that the number operator N, defined as ##a^\dagger a## is worth ##\frac{m \omega x^2}{2\hbar}+\frac{p^2}{2m\hbar \omega}##.
    In wikipedia and http://quantummechanics.ucsd.edu/ph130a/130_notes/node167.html, as well as in books, one can read that for the harmonic oscillator ##\hat H=\hbar \omega \left ( a^\dagger a +\frac{1}{2} \right )##. When I calculate the right hand side of this equation using my calculated value for N, I reach that ##\hbar \omega \left ( a^\dagger a +\frac{1}{2} \right )=\frac{\hbar \omega }{2}+\frac{m\omega ^2 x^2}{2}+\frac{p^2}{2m}##.
    However for the harmonic oscillator the hamiltonian operator is ##\hat H=\frac{\hat {\vec p ^2}}{2m}+\frac{m\omega ^2 \hat x ^2}{2}##.
    So that I do not reach that the left hand side is equal to the right hand side. Instead, I get that the left hand side is worth the right hand side + ##E_0##, the ground state energy. Am I making an algebra mistake?! If not, what the heck is going on?
     
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  3. Jun 6, 2013 #2

    kith

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    This is not true because x and p don't commute, so the mixed terms don't vanish. xp is not equal to px, but to px + [x,p].
     
  4. Jun 6, 2013 #3

    fluidistic

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    My bad, you are right. For a moment I forgot that x and p were operators. Thank you very much!
     
  5. Jun 6, 2013 #4

    kith

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    You're welcome ;-)
     
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