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Projection in Hubbard Model

  1. Sep 22, 2014 #1
    This is a question from Altland and Simons book "Condensed Matter Field Theory".

    In the second exercise on page 64, the book claims that if we define [itex] \hat P_s, \hat P_d [/itex] to be the operators that project onto the singly and doubly occupied subspaces respectively, then at half-filling the following equality holds

    [itex] \hat P_s \hat H_t \hat P_s = 0 [/itex] and [itex] \hat P_d \hat H_t \hat P_d =0 [/itex],

    where [itex]\hat H_t[/itex] is the hopping term in the Hubbard model. I am having trouble to see why these two conditions are true. Do I have to write out the projection operator in terms of creation and annihilation operators and directly calculate?

    Thanks in advance.
     
    Last edited: Sep 22, 2014
  2. jcsd
  3. Sep 27, 2014 #2
    You don't need to calculate it directly, though it is a good exercise.

    First, recall that the hopping term looks like [itex]c^\dagger_{i,\sigma} c_{i+1,\sigma}[/itex] (specializing to 1D for simplicity) and assume that we start at half-filling. If we assume (like Altland and Simons do) that [itex]U\gg t[/itex], then effectively every single site is half-occupied.

    In that scenario, whenever the hopping term acts, it moves a particle into an already half-occupied site and leaves an empty site behind. That situation clearly isn't part of the singly occupied subspace, so you get the [itex] \hat P_s \hat H_t \hat P_s = 0 [/itex] statement. On the other hand, if you were to start with a state with two electrons on the same site and let one hop away, you end up with two singly occupied sites, hence [itex] \hat P_d \hat H_t \hat P_d =0 [/itex].

    Effectively, the [itex]\hat H_t[/itex] has no effect to first order in perturbation theory. However, if you let the term act twice, the electron may jump to another site, doubly occupy it for a while and then jump back. That's why the [itex] \hat P_s \hat H_t \hat P_d \hat H_t \hat P_s[/itex] term is important.
     
  4. Oct 13, 2014 #3
    Thank you for your answer. I am having trouble to see why the hopping term only involves two sites. Vaguely I have a feeling that this has to do with the fact that we are only considering the perturbation to second order, but I don't know how to draw the connections.
     
  5. Oct 15, 2014 #4
    Well, terms with an odd number of operators are out immediately if the particle number is conserved. The two-operator term represents direct hopping, the lowest-order process, which presumably is the most important one. You can of course imagine an electron hopping from site 1 to site 2 and then to site 3, e.g. [itex]c_3^\dagger c_2 c_2^\dagger c_1[/itex]. (That's one of the effective terms you end up treating in the 2nd order perturbation theory.) I'm not sure how to interpret this term if it's included in the Hamiltonian from the start, but it can be rewritten as
    [tex]c_3^\dagger c_2 c_2^\dagger c_1 = c_3^\dagger c_1 - c^\dagger_3 c_1 n_2[/tex]
    which looks like the usual direct hopping term plus an interaction with the density at site 2. At half-filling and infinite [itex]U[/itex], [itex]n_2=1[/itex], so the term would actually vanish, at least to first order.
     
  6. Oct 15, 2014 #5
    Thanks again and sorry about the unclear question in my last post.

    My actual question is: just looking at the hopping term in the Hubbard Model on an extended lattice (say a square lattice)
    [tex] \hat H_t = \displaystyle\sum_{\langle i j \rangle} a^{\dagger}_{i \sigma} a_{j \sigma}, [/tex]
    since the summation runs over all neighboring sites, if I start with a configuration where some sites are doubly-occupied (in the doubly-occupied subspace) and apply the hopping term, I don't see that I will always end up with a configuration in the singly-occupied subspace. So one of my explanations is because we are only considering up to second order in perturbation, this is effectively a two-site problem; but I don't know how to go from "second order perturbation" to "a two-site problem".
     
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