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Negative sign in hopping amplitude

  1. Jun 22, 2011 #1
    Might be a naive question.

    In many places, specially in the Hubbard model, people use a minus sign before the tight-binding hopping amplitude. What does the negative sign signify for? Any special intention?


    Thanks.
     
  2. jcsd
  3. Jun 22, 2011 #2

    DrDu

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    The hopping amplitude is something like [tex] \langle\phi_j| T_\mathrm{kin}|\phi_i\rangle [/tex], hence it's sign is well defined although there is a certain arbitrariness in the choice of the phase of the phi_i.
     
  4. Jun 22, 2011 #3
    That's true. But there must be a meaning for keeping this minus sign. I have seen Hubbard's original paper. Surprisingly that paper does not have this minus sign!

    If the sign is just a matter of convention, then why unnecessarily an extra minus was chosen as a convention?


    Thanks.
     
  5. Jun 22, 2011 #4

    DrDu

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    No, it is not just a convention, it makes a real physical difference, at least in lattices where you cannot consistently number neighbouring atoms as 1 or 2. A simple example is azulene in the Hueckel model (which is basically the Hubbard model in quantum chemistry)
     
  6. Jun 22, 2011 #5
    Sorry. Still not clear to me. What if I write

    $t_{ij}$ instead of $-t_{ij}$ in the kinetic part of the Hubbard Hamiltonian?

    Does it make any difference? If yes, then what difference?
     
  7. Jun 22, 2011 #6
    @DrDu

    Sorry, the latex didn't work! I guess, you could follow the notation.

    And I think, the Huckel model and the tight-binding model are equivalent. Hopping from one site to other site is similar to hopping from one orbital to another orbital.
     
  8. Jun 25, 2011 #7
    Consider the infinite U limit. Here the electrons are frozen at the lattice sites because the U prohibits double occupancy, even as a virtual state. Now if you tune U to a finite value then electrons start hopping in the process getting delocalized, thus lowering its energy.

    The sign of t captures this lowering of energy.

    P.S. : I can't seem to construct an argument as to why delocalization would lower the energy :(
     
  9. Jun 27, 2011 #8

    DrDu

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    At large U, t can be taken into account perturbationally. As t is non-diagonal, it can make a contribution beginning with second order which will be proportional to t^2/(-U). Hence it is negative (stabilizing) and does not depend on the sign of t.
    However, when going around a ring with an odd number of atoms, the sign does matter in higher orders, e.g. for a ring of three atoms the you get a contribution t_12*t_23*t_31/(-U)^2. For simple orbitals, like hydrogen 1s, t is negative, typically, whence these terms lead to a stabilization.
     
  10. Jun 27, 2011 #9

    M@2

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    The sign of t is significant. For example in tight binding model it determines where is the minimum energy orbital: at k=0 or at k=Pi/2

    Let us for example consider polymer of atoms with s atomic valent orbital. It is evident,that k=0 (LCAO) orbital has minimum energy.
    If we take polymer of atoms with p atomic orbital we get minimum energy orbital with |k|=Pi/2

    LCAO: Linear Combination of Atomic Orbitals

    I can recommend the book of Nobel Prize winner in chemistry:
    Roald Hoffmann, Solids and surfaces: A Chemist's View of bonding in extended structures, 1988
    pages: 4-9
     
    Last edited by a moderator: Jun 27, 2011
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