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Eigenstate and real space representation

  1. Dec 8, 2013 #1
    Hi All,

    I was going through a paper on quantum simulations. Below is an extract from the paper; I would be obliged if any one can help me to understand it:

    We will use eigenstate representation for transverse direction(HT) and real space for longitudinal direction(HL) Hamiltonians.

    HL= h^2/2m * d^2/dx^2 + U (makes sense because x is longitudinal direction)

    HT= h^2/2m *( d^2/dy^2 + d^2/dz^2) + Uy.z (writing in terms of y,z isn't a real space representation again? This was meant to be eigenstate representaion)

    then,

    Xk(ρ) = exp(i.k.ρ)/area
    where HT.Xk = εk.Xk

    k and ρ are both 2D vectors.

    Now, k is wave vector, Xk(ρ) is function of ρ, what is ρ? What does it physically corresponds to?

    Using finite difference approximation for HL:

    HLψ= -tψn-1 +2t+Un.ψn +....

    here I can interpret ψ is function of n, which is discretized real space. However I am not able to figure out what is ρ?

    Thanks,
    Arya
     
  2. jcsd
  3. Dec 18, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I was able to find the article by Googling. Posting a reference would have been helpful. It seems to be written by someone that has little experience in quantum mechanical numerical simulations and uses a very non-standard notation.

    The Hamiltonian doesn't change because of the representation, the wave function (or rather, its representation) does. The wave function is sampled on a series of grid points ##x_n## along with a finite basis of eigenfunctions ##\chi_\mathbf{k}(y,z)##.

    You truncated the quotation: "both 2D vectors in the y–z plane." I take it that ##\rho = (y,z)##.
     
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