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About hamilton operators

  1. Aug 31, 2007 #1
    why i[tex]\hbar[/tex]([tex]\partial[/tex]/[tex]\partial[/tex]t+i[tex]\Omega[/tex])=i[tex]\hbar[/tex]exp(-i[tex]\Omega[/tex]t)[tex]\partial[/tex]/[tex]\partial[/tex]texp(i[tex]\Omega[/tex]t)
     
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  3. Sep 1, 2007 #2

    dextercioby

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    In the RHS you have something that resembles the unitary trasformation of an operator. Where did you get the equation from ?
     
  4. Sep 1, 2007 #3
    [tex]i\hbar(\frac{d}{dt}+ i\Omega) = i\hbar(exp(-i \Omega t) \frac{d}{dt} exp(i \Omega t) [/tex]

    Well if exp(iOt) is your wavefunction, the RHS is just [tex]i\hbar(i \Omega )[/tex]

    are you sure this equation is right? Looks like momentum operator, not hamilton.
     
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