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Momentum space representation for finite lattices

  1. Mar 19, 2012 #1
    Hi all,
    I have a question. For sure the momentum representation used in solid state physics works for infinite lattices or periodic ones.

    But when it comes to finite lattice, i.e. 100 sites, can the momentum representation be used? What are the errors? Where does this fail?

    Thanks for discussions!
  2. jcsd
  3. Mar 19, 2012 #2


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    Science Advisor

    The representation can still be used. Particularly if you have finite lattices with periodic boundary conditions. The situation is: This is not an approximation. What you are doing is to re-express the basis of lattice sites |r> of real space in another space, which, for example, can be |k> = \sum_r exp(i k r) |r>. As long as the set of |k>s spans the same space as the |i>s before, this is an completely equivalent basis.

    If you are using periodic boundary conditions (in a finite lattice), then in fact you still have translational symmetry (which means that the Hamiltonian commutes with all translations T: [H,T]=0). Consequently, the eigenstates of the Hamiltonian can be made to transform according to irreducible representations of the translation symmetry groups. And these happen to be labelled by exactly the same k-vectors as in the infinite case -- it is just that you only have a finite number of them. This is most easily seen when you set up a tight-binding hopping Hamiltonian (-t, this emulates the kinetic energy operator) with periodic boudary conditions:
    \mathbb{t} \equiv
    0 & -t & 0 & 0 & \ldots & 0 & -t
    \\ -t & 0 & -t & 0 & \ldots & 0 & 0
    \\ 0 & -t & 0 & -t & \ldots & 0 & 0
    \\ 0 & 0 & -t & 0 & \ldots & 0 & 0
    \\ \vdots & & & & \ddots &
    \\ 0 & 0 & 0 & 0 & -t & 0 & -t
    \\ -t & 0 & 0 & 0 & 0 & -t & 0
    In that case the eigenfunctions of that operator are still plane waves (i.e., the operator is diagonal in the plane wave basis), just as in the infinite case.

    If you have a finite lattice but *no* periodic boundary conditions, then translational symmetry is gone, and the eigenstates do no longer exactly transform according to irreps of the translation group. You can still define the plane wave basis, however (as described above), and you will see that for example the kinetic energy operator still is very diagonal dominant in this basis (but not exactly diagonal anymore).
  4. Mar 19, 2012 #3
    Ok, I got the point. thanks for replying!
    It's just a change of basis that under boundary condition diagonalize the Hamiltonian. But then a subtle point:

    In order for k-representation to be a good basis change (i.e. orthogonality and completeness properties) I guess that one has to choose the k values AS if there were periodic boundary conditions (i.e. k=2pi*n/L). Right?

    Then another last point:
    how do you really model a finite size chain? A finite size chain has zero boundary condition. In this case I use the k representation with k chosen as if there were periodic boundary condition and just then reconstruct the physical wavefunction by making a wavepacket that is zero on the borders?
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