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Canonical Transformation of the Hubbard Model

  1. Sep 10, 2011 #1

    Suppose we have a 2 site Hubbard model, with the hopping Hamiltonian given by [itex]H_t[/itex] and the Coulomb interaction Hamiltonian given by [itex]\hat{H}_U[/itex]. In the strong coupling limit (U/t >> 1), we define a canonical transformation of [itex]\hat{H} = \hat{H}_U + \hat{H}_t[/itex], as

    [tex]H' = e^{-t\hat{O}}\hat{H}e^{t\hat{O}} = \hat{H} - t[\hat{O},\hat{H}] + \frac{t^2}{2}[\hat{O},[\hat{O},\hat{H}]] + \ldots[/tex]

    Atland and Simons say (on page 63):

    I don't get this. Even if this choice is made,

    [tex]\hat{H} - t[\hat{O}, \hat{H}] = (\hat{H}_U - t[\hat{O},\hat{H}_t]) + \underbrace{(\hat{H}_t - t[\hat{O},\hat{H}_U])}_{\mbox{0 by choice}} = \hat{H}_U - t[\hat{O},\hat{H}_t][/tex]

    So there's still a t-dependent first order term. What's wrong here?
    Last edited: Sep 11, 2011
  2. jcsd
  3. Sep 11, 2011 #2


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    I would guess that O can be chosen so that it commutes with H_t.
  4. Sep 11, 2011 #3
    upto first order in [itex]t[/itex]:

    [itex] H_U + H_t - t[O,H_U] - t[O,H_t] [/itex] ... (Eq 1)

    Now choosing [itex] H_t - t[O,H_U] =0 \Rightarrow H_t = t[O,H_U] [/itex]
    Put this back to Eq. 1, and you'll have no order [itex]t[/itex] term left.
    Last edited: Sep 11, 2011
  5. Sep 11, 2011 #4
    How is [itex]H_U - t[O, H_t] = 0[/itex]?

    In fact

    [tex]H_U - t[O, H_t] = H_U - t[O, t[O, H_U]] = H_U - t^2[O, O H_U - H_U O][/tex]

    Are you using some projection property of the O, like [itex]O H_U O = 0[/itex] here?
  6. Sep 11, 2011 #5


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    Don t forget that H_t is already first order in t, so t times it s commutator with O is second order.
  7. Sep 11, 2011 #6
    Yes, that's what my last post says DrDu. But the [itex]t^2[/itex] term isn't zero -- or at least I don't see it.

    Shouldn't the expression in the book then say [itex]O(t^2)[/itex] instead of [itex]O(t^3)[/itex]?
  8. Sep 12, 2011 #7


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    As I understood your first post, the transformed hamiltonian in the book is explicitly written down including all second order terms and the third order terms are abbreviated O(t^3). So yes, there is a second order order term in the transformed hamiltonian but not a first order term any more as you originally claimed.
  9. Sep 13, 2011 #8
    You don't want it to vanish. All I meant is, by that particular choice of operator [itex]\hat{O}[/itex] your Hamiltonian becomes [itex]H = H_U + \mathcal{O}(t^2)[/itex], which is what you deduce below.

    When written in this way one can see that the [itex]\mathcal{O}(t^2)[/itex] are perturbations about the Hubbard [itex]U[/itex] term, i.e. your original system is a one with all electrons frozen at their lattice sites and then your introduce hopping by the [itex]t[/itex] terms. In fact the [itex]\mathcal{O}(t^2)[/itex] term is the Heisenberg model and represents super exchange which leads to effective spin flips.
    Last edited: Sep 13, 2011
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