# Canonical Transformation of the Hubbard Model

1. Sep 10, 2011

### 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):

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?

Last edited: Sep 11, 2011
2. Sep 11, 2011

### DrDu

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

3. Sep 11, 2011

### 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.

Last edited: Sep 11, 2011
4. Sep 11, 2011

### maverick280857

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?

5. Sep 11, 2011

### DrDu

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

6. Sep 11, 2011

### 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 $O(t^3)$?

7. Sep 12, 2011

### 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.

8. Sep 13, 2011

### vkroom

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.

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.

Last edited: Sep 13, 2011