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Negative energy eigenvalues of Hamiltonian

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data

    If I have a Hamiltonian matrix, [itex]\mathcal{H}[/itex], that only depends on a kinetic energy operator, do the energy eigenvalues have to be non-negative? I have an [itex]\mathcal{H}[/itex] like this, and some of its eigenvalues are negative, so I was wondering if they have any physical significance, or if I should just reject them and their associates eigenstates. Also, one of the eigenstates was just the zero vector, which I think can be ignored since it's a trivial solution.

    2. Relevant equations

    [itex]\mathcal{H} = \mathcal{K} + \mathcal{V} = \mathcal{K}[/itex]

    3. The attempt at a solution

    I don't see any way that the system could have negative energy if the potential is zero everywhere, but I also don't feel entirely comfortable just ignoring eigenstates like that.
     
  2. jcsd
  3. Nov 24, 2012 #2

    Redbelly98

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    FYI, I have moved your thread to "Advanced Physics".

    Your suspicions seem right to me. The energy eigenvalues are the allowed energies for the system, which really should be positive or zero if there can only be kinetic energy.

    Can you post your actual Hamiltonian and describe the system?
     
  4. Nov 24, 2012 #3
    Thanks for the reply, I wasn't completely sure if this belonged in the advanced section or not. The system is an ##X_{6}## molecule with a single electron able to move between the different ##X## ions. The basis ket ##|e_{j}\rangle## represents the electron occupying the ##j^{\text{th}}## ion with ##|e_{j}\rangle = |e_{j\pm6}\rangle##. We're to take the kinetic energy operator to be:

    $$
    \mathcal{K}=-\sum_{i=1}^{6}\left( |e_{j}\rangle\langle e_{j+1}| +|e_{j+1}\rangle\langle e_{j}| \right)
    $$

    And the potential operator to be zero. The Hamiltonian is then (assuming I did this right):

    $$
    \mathcal{H} = \mathcal{K}=
    \begin{pmatrix}
    0& -1 & 0 & 0 & 0 & -1\\
    -1& 0& -1& 0& 0& 0\\
    0& -1& 0& -1& 0& 0\\
    0& 0& -1& 0& -1& 0\\
    0& 0& 0& -1& 0& -1\\
    -1& 0& 0& 0& -1&0
    \end{pmatrix}
    $$

    With this eigensystem. As you can see, some of the eigenvalues are negative!
     
  5. Nov 25, 2012 #4

    Redbelly98

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    Hmm. I agree that that Hamiltonian has negative eigenvalues, but am puzzled as to why the kinetic energy operator would have the form that you are told to use. I'm used to seeing KE in the form p2/(2m).

    I'm going to see if any of the Homework Helpers have an idea here.
     
  6. Nov 25, 2012 #5

    Mute

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    The Hamiltonian looks like it is basically a tight-binding model for one election. You might look up that model in some textbooks to see what they get (here is the relevant section of the wikipedia article in second-quantized notation). I suspect that calling this term the "kinetic energy" is somewhat of a convention and so perhaps one can encounter negative energy eigenvalues. If that's the case then perhaps you can just add a constant energy shift to make the ground state energy zero.
     
  7. Nov 25, 2012 #6
    Hmm yes, that does seem to be of the same form. Honestly though, the material in the article seems way too advanced for the level that we're at (ie. the "do this because it works" stage) so I don't know if I'm even supposed to be worrying about this. You're probably right in that calling it the "kinetic energy" operator is just more of a jargon thing. The only reason I even considered this an issue is because I couldn't solve for the state vector for the given initial condition unless I rejected the zero vector solution and instead chose a different eigenvector that also had eigenvalue 1 (which I had to find manually.) It sort of begs the question as to why Mathematica would report the zero vector as a solution unless it has no other choice, but there's probably some sort of linear algebra argument behind that. So then I started to wonder whether or not I should also reject the negative energies, but I couldn't solve for the initial state that way.
     
  8. Nov 26, 2012 #7

    Redbelly98

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    One of the Homework Helpers told me -- and I hope I'm not oversimplifying what he said -- that this kinetic energy Hamiltonian is pretty much the change or difference in KE relative to some reference level for an electron that is bound to an atom and does not jump to another atom.

    For more details, you can look at Feynman's lectures, in the first section of chapter 13 in Vol III which deals with an electron living on a 1D lattice. See especially Equation (13.3).

    Hope that helps.

    p.s.: I am about 17 years removed from grad school, and really do not deal with material like this nowadays.
     
  9. Nov 26, 2012 #8
    Ah, that does make a lot more sense. Thanks again everyone!
     
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