Canonical Transformation of the Hubbard Model

[maverick280857]

Hi,

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

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

Atland and Simons say (on page 63):

By choosing the operator \hat{O} such that \hat{H}_t - t[\hat{O}, \hat{H}_U] = 0, 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

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

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

\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]

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 t:

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

Now choosing H_t - t[O,H_U] =0 \Rightarrow H_t = t[O,H_U]
Put this back to Eq. 1, and you'll have no order t term left.

 

[maverick280857]

upto first order in t:

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

Now choosing H_t - t[O,H_U] =0 \Rightarrow H_t = t[O,H_U]
Put this back to Eq. 1, and you'll have no order t term left.

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

In fact

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]

Are you using some projection property of the O, like O H_U O = 0 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 t^2 term isn't zero -- or at least I don't see it.

Shouldn't the expression in the book then say O(t^2) instead of O(t^3)?

 

[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 H_U - t[O, H_t] = 0?

You don't want it to vanish. All I meant is, by that particular choice of operator \hat{O} your Hamiltonian becomes H = H_U + \mathcal{O}(t^2), which is what you deduce below.

In fact

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]

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