- #1
hokhani
- 552
- 15
- TL;DR Summary
- I can not go one step further in this expansion.
In the youtube lecture “electron interaction and the Hubbard model” at the time 2:23:00, we have the following self-consistent equation with energy appearing at both sides:
$$(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P) |\phi \rangle_{g.s.} =E |\phi \rangle_{g.s.}$$
Where ##H_0## is the unperturbed Hamiltonian, ##H_1## the perturbation, ##|\phi \rangle_{g.s.}## is the ground state ket of the full Hamiltonian ##(H_0+H_1)##, and ##\hat P (\hat Q)## is the projection operator on the ground (excited) states of ##H_0##.
By defining the effective Hamiltonian as:
$$H_{eff}=(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P)$$
The self-consistent equation is as:
$$H_{eff}|\phi \rangle_{g.s.}=E|\phi \rangle_{g.s.}$$
So, my question:
How does the solution of this effective Hamiltonian, recursively, give the following equation?
$$H_{eff}=\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P +$$
Where the first, second and third lines are respectively zero, second and third order terms.
I would be grateful if anyone could please provide any help with that.
$$(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P) |\phi \rangle_{g.s.} =E |\phi \rangle_{g.s.}$$
Where ##H_0## is the unperturbed Hamiltonian, ##H_1## the perturbation, ##|\phi \rangle_{g.s.}## is the ground state ket of the full Hamiltonian ##(H_0+H_1)##, and ##\hat P (\hat Q)## is the projection operator on the ground (excited) states of ##H_0##.
By defining the effective Hamiltonian as:
$$H_{eff}=(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P)$$
The self-consistent equation is as:
$$H_{eff}|\phi \rangle_{g.s.}=E|\phi \rangle_{g.s.}$$
So, my question:
How does the solution of this effective Hamiltonian, recursively, give the following equation?
$$H_{eff}=\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P +$$
Where the first, second and third lines are respectively zero, second and third order terms.
I would be grateful if anyone could please provide any help with that.
