effective Hamiltonian

1 Formalism of Effective Hamiltonian

Suppose we have a Schrodinger equation such that

H|ψ〉 = E|ψ〉 (1)

Here, we assume all the set of eigenstates spans the eintire Hilbert spaceH. However, in general, we need not have to find all the eigenenergies ofthe Schrodinger equation. we are interested in a physical phenomenona at aspecific energy scale. In many cases of condensed matter physics, low energybehaviour of the system is the main concern. Mathematically speaking, itis sufficient for us to solve the Schrodinger equation in a subspace P of H.All the interesting eigenstates |i〉 are the elements of P. Introducing Q asthe coset of P,

P ⊕ Q = H (2)

A requirement for an proper effective Hamiltonian Heff is that it yieldsthe same spectrum as the original Hamiltionian H does within P. Here,we define two projection operators, P =

∑i∈P |ψi〉〈ψi| and Q = 1 − P .

Obviously, P +Q = 1. The requirement states

HeffP |ψ〉 = EP |ψ〉 (3)

Toward constructing the Heff , we choose H0 so that [H0, P ] = 0, meaningthat H0 is diagonalized in P.

H0|ψi〉 = εi|ψi〉 (4)

Now, H = H0 +V . In addition to this, we sloppily introduce an operator Ωsatisfying ΩP |ψ〉 = |ψ〉.

|ψ〉 = (P +Q)|ψ〉 = ΩP |ψ〉 (5)

Ω retrieves the information tossed by P : therfore, Ω must be a function ofV, at least. Heff can be obtained by the following.

H|ψ〉 = E|ψ〉 (6)

⇒ HΩP |ψ〉 = E|ψ〉 (7)

⇒ PHΩP |ψ〉 = EP |ψ〉 (8)

⇒ PHΩPP |ψ〉 = EP |ψ〉 (P 2 = P ) (9)

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Therefore,

Heff = PHΩP (10)

Now the problem is reduced to find the expression of Ω.

2 Rayleigh-Schrodinger perturbation theory

With these preparation, we will show a scheme so-called Rayleigh-Schrodingerperturbation theory to obtain the effective Hamiltonian Heff . Rewrite theSchrodinger equation.

(E −H0)|ψ〉 = V |ψ〉 (11)

⇒ E|ψ〉 −H0ΩP |ψ〉 = V ΩP |ψ〉 (12)

ΩP (E −H0)|ψ〉 = ΩPV |ψ〉 (13)

⇒ E|ψ〉 − ΩH0P |ψ〉 = ΩPV ΩP |ψ〉 (14)

Subtract both hand sides by the other equation,

[Ω,H0] = (1− ΩP )V ΩP (15)

Note that E has vanished from the equation and Ω has no dependence on E.Now, expand Ω as Ω =

∑n Ωn, where Ωn = O(V n) with Ω0 = 1. Compare

the nth order of V on both hand sides.

[Ωn,H0] = V Ωn−1 −n−1∑m=0

ΩmPV Ωn−m−1 (16)

= QV Ωn−1 −n−1∑m=1

ΩmPV Ωn−m−1 (17)

Each order of Ω can be obtained iteratively. Then, the effective Hamiltonianwill be of the form

Heff = P (H0 + V )ΩP (18)

= P (H0 + V )(Ω0 + Ω1 + Ω2 + · · ·)P (19)

= PH0P + PH0Ω1P + PH0Ω2P + · · ·

+PV Ω0P + PV Ω1P + PV Ω2P + · · · (20)

= H0 +H0PΩ1P +H0PΩ2P + · · ·

+PV Ω0P + PV Ω1P + PV Ω2P + · · · (21)

(22)

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Later we will see the explicit form of Ωn. It will be apparent PΩn containsa term PQ, which is zero. For example,

PΩ1 = P1

εi −H0QV (23)

=1

εi −H0PQV = 0 (24)

So, the effective Hamiltonian is now

Heff = H0 +H1 +H2 +H3 + · · ·

(Hn = PV Ωn−1P ) (25)

We deduce the explicit form of several Ωn’s.Ω1:

[Ω1,H0] = V Ω0 (26)

⇒ (εi −H0)Ω1(εi)|ψi〉 = QV |ψi〉 (27)

⇒ Ω1(εi) =1

εi −H0QV (28)

Ω2:

[Ω2,H0] = V Ω1 − Ω1PV Ω0 (29)

⇒ (εi −H0)Ω2(εi)|ψi〉 =V Ω1(εi)−

∑k∈P

Ω1(εk)|ψk〉〈ψk|V|ψi〉 (30)

⇒ Ω2(εi) =1

εi −H0QV

1εi −H0

QV

−∑k∈P

1εi −H0

1εk −H0

QV |ψk〉〈ψk|V (31)

Let us consider the simplest case, H2. The transition amplitude from |ψi〉 ∈P to |ψj〉 ∈ P due to this second order term is

〈ψj |H2|ψi〉 = 〈ψj |PV Ω1P |ψi〉 (32)

= 〈ψj |PV Ω1(ε(i))P |ψi〉 (33)

= 〈ψj |PV1

εi −H0QV P |ψi〉 (34)

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Then, H3.

〈ψj |H3|ψi〉 = 〈ψj |PV Ω2P |ψi〉 (35)

= 〈ψj |PV Ω2(ε(i))P |ψi〉 (36)

= 〈ψj |PV 1εi −H0

QV1

εi −H0QV

−∑k∈P

1εi −H0

1εk −H0

QV |ψk〉〈ψk|VP |ψi〉 (37)

= 〈ψj |PV1

εi −H0QV

1εi −H0

QV P |ψi〉

−∑k∈P

〈ψj |PV1

εi −H0

1εk −H0

QV |ψk〉〈ψk|V P |ψi〉(38)

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