Canonical Transformation of the Hubbard Model

[maverick280857]

Hi,

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):

By choosing the operator [itex]\hat{O}[/itex] such that [itex]\hat{H}_t - t[\hat{O}, \hat{H}_U] = 0[/itex], all terms at first order in t can be eliminated from the transformed Hamiltonian. As a result, the effective Hamiltonian is brought to the form

[tex]\hat{H}' = \hat{H}_U + \frac{t}{2}[\hat{H}_t, \hat{O}] + O(t^3)[/tex]

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?

 

[DrDu]

I would guess that O can be chosen so that it commutes with H_t.

 

[vkroom]

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.

 

[maverick280857]

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.

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?

 

[DrDu]

Don t forget that H_t is already first order in t, so t times it s commutator with O is second order.

 

[maverick280857]

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]?

 

[DrDu]

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.

 

[vkroom]

How is [itex]H_U - t[O, H_t] = 0[/itex]?

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.

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]

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.