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Harmonic Oscillator

  1. Jul 17, 2008 #1
    I'm reading Griffiths', and I had a question about the harmonic oscillator.

    Griffiths solves the Schrodinger equation using ladder operators, and he then says that there must be a "lowest rung," or [tex]\psi_{0}[/tex], such that a_[tex]\psi_{0}[/tex] = 0. I'm guessing this also means that E = 0 for a_[tex]\psi_{0}[/tex], which is why it wouldn't be allowable.

    Because a_[tex]\psi_{0}[/tex] is (mathematically, at least) a solution to the Schrodinger equation, it corresponds to the energy ([tex]E_{0}[/tex] - [tex]\hbar\omega[/tex]), which equals zero. Doesn't this mean that [tex]E_{0}[/tex] = [tex]\hbar\omega[/tex]?

    But then this conflicts with Griffith's statement that [tex]E_{0}[/tex] = [tex]\frac{1}{2}\hbar\omega[/tex]. HELP!
  2. jcsd
  3. Jul 17, 2008 #2


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    Watch out! The statement [itex]a_- \psi_0 = 0[/itex] just says that when you apply the ladder operator to this state, you get nothing, i.e. the vacuum. Physicists have a tendency to use "natural" notation, and you will often see things like [itex]a_- |0\rangle = 0[/itex]. Note that here the [itex]|0\rangle[/itex] on the left hand side is a state (wavevector) and the 0 on the right hand side is the zero vector, which is not a state. Griffiths is just kind enough to call it [itex]\psi_0[/itex] instead of [itex]|0\rangle[/itex] so you don't get completely confused :smile:

    You say that [itex]a_- \psi_0[/itex] is a solution to the Schrodinger equation: that's right. But it isn't a physical solution. It's just the solution [itex]\psi = 0[/itex]. The energy of [itex]\psi_0[/itex] is [itex]\frac12 \hbar \omega[/itex] (formula 2.61) which can be seen by applying the Hamiltonian in the form (2.56) to this state and using that [itex]a_-[/itex] annihilates it.

    Hope that makes it more clear.
  4. Jul 18, 2008 #3
    Thanks, CompuChip!
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