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How to apply ladder operators?

  1. Aug 20, 2013 #1
    The total energy of a particle in a harmonic oscillator is found to be 5/2
    ~!. To change the energy,
    if i applied the lowering operator 4 times and then the raising operator 1 times successively. What
    will be the new total energy?
    i want the calculation plz
     
  2. jcsd
  3. Aug 20, 2013 #2

    CompuChip

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    The energy is the eigenvalue of the Hamiltonian operator. So you have some state ##\phi## satisfying ##\hat H \phi = E \phi## with ##E = \tfrac52##.
    For brevity let me write the ladder operators as ##\hat A## (so ##\hat A = a^\dagger a^\dagger \cdots a##); then you can work out what ##\hat H(\hat A \phi)## is by commuting it to the front, i.e. rewriting it as
    $$\hat H \hat A \phi = \hat A \hat H \phi + [\hat H, \hat A] \phi = \tfrac52 \phi' + \Delta E \phi'$$
    where ##\phi' = A \phi## is your new state. You can then read off your new energy ##E = \tfrac52 + \Delta E##.
    The trick, of course, is calculating ##[\hat H, \hat A] ##.
     
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