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Sakurai problem 1.9.

  1. Oct 11, 2006 #1
    Of Modern Quantum Mechanics. This starts with a Hamiltonian

    H = a(|1\rangle\langle 1| - |2\rangle\langle 2| + |1\rangle\langle 2| + |2\rangle\langle 1|)

    This has eigenvalues [itex]\pm a\sqrt{2}[/itex]. Shouldn't a Hamiltonian have only non-negative eigenvalues? If the sign in front of the [itex]|2\rangle\langle 2|[/itex] is [itex]+[/itex] instead of a [itex]-[/itex] you get eigenvalues [itex]0[/itex] and [itex]2a[/itex], which makes more sense (assuming a is real and positive). So might this be a typo, or am I wrong in general about the eigenvalues of a Hamiltonian? Or am I taking this toy "Hamiltonian" too seriously?
    Last edited: Oct 11, 2006
  2. jcsd
  3. Oct 11, 2006 #2

    George Jones

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    In non-relativistic quantum mecanics, the signs of energy eigenvalues depend on where the zero for energy is (arbitrarily) chosen. For example, in the harmonic oscillator, the zero is chosen at the equilibrium position, and the energies are all positve, while in the hydrogen atom, the zero is chosen at r = infinity. For hydrogen, bound states have negative energies, while scattering states have positive energies because they can "make it to infinity".

    Both of these zeros are arbitary and chosen for convenience.

    What is necessary, is that the spectrum of energies has a lower bound, e.g., -13.7 ev for hydrogen. If no lower bound exists, then the system can spiral down to lower and lower energies, and, in the process, release an infinite amount of energy.
  4. Oct 11, 2006 #3
    Well, duh, how did I forget bound states? Well, I did. That's not even QM, the bound "states" in Newtonian mechanics have negative energy. Thanks for jogging my fuzzy memory. Guess I have a lot of review to do.
    Last edited: Oct 11, 2006
  5. Oct 13, 2006 #4
    Nevermind. I figured it out.
    Last edited: Oct 13, 2006
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