- #1
maverick280857
- 1,789
- 5
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):
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?
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?
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