# Derivation of the low-energy effective Hamiltonian [duplicate]

In the quantum mechanics, the Hamiltonian $H$ satisfies the Schroedinger equation $$H\psi = E\psi.$$ Suppose that $P$ is a projection operator, and $Q=1-P$. The low-energy effective Hamiltonian is $$H_{eff} = H_{PP} + \frac{H_{PQ}H_{QP}}{E-H_{QQ}}.$$ My method to do so is a combination of two algebraic equations. Recently, I found a good paper(L. Petersen $et~~al.$ A simple tight-binding model of spin–orbit splitting of sp-derived surface states), whose authors declared that one can obtain (EQ. 13) $$P\frac{1}{\epsilon-H}P = \frac{1}{\epsilon-H_{eff}}$$ from a standard theorem in linear algebra. Here, I just wonder that "what's the standard theorem"? Please inform me of the omited detail in the above paper.

Say the time evolution for Hamiltonian $H$ is given by $U(t) = \exp(-iHt)$ and the corresponding evolution on the support of $P$ is

$$PU(t)P = P\exp(-iHt)P = \exp(-iH_\text{eff}t) \equiv U_\text{eff}(t)$$

assuming $H_\text{eff}$ exists. The desired identity follows from

$$\lim_{\eta \rightarrow 0} \int_0^\infty dt \; U(t) e^{i (\epsilon + i \eta) t} = \frac{i}{\epsilon - H} \, . \tag{1}$$

Applying $P$ to both sides of Eq. $(1)$ gives

$$\lim_{\eta \rightarrow 0} \int_0^\infty dt \; P U(t) P e^{i (\epsilon + i \eta) t} = P\frac{i}{\epsilon - H}P \, . \tag{2}$$

Then using $PU(t)P = U_\text{eff}(t)$ on the left hand side of Eq. $(2)$ we get

$$\lim_{\eta \rightarrow 0} \int_{0}^{\infty} dt \; U_\text{eff}(t) e^{ i (\epsilon + i \eta) t} = P \frac{i}{\epsilon - H}P \, . \tag{3}$$

Now use Eq. $(1)$ but with the effective instead of full Hamiltonian to replace the left hand side of Eq. $(3)$. The result is

$$\frac{1}{\epsilon - H_\text{eff}} = P\frac{1}{\epsilon - H}P$$ as stated.

• What does $\rightarrow$ mean? Aug 22 '15 at 1:57
• Oh, should've been $\Rightarrow$ = "implies". Thanks for the fixing. However, the way you rearranged the answer might make a beginner wonder about the relation between the 1st and 2nd equalities following the integral over PU(t)P.
– udrv
Aug 22 '15 at 2:48
• You're right, the way I rearranged it wasn't clear. I hope the new version is ok. Aug 22 '15 at 3:14
• @udrv I like the argument, but it ignores the energy dependence of $H_\textrm{eff}$? Jun 7 '17 at 14:38