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.

  • $\begingroup$ What does $\rightarrow$ mean? $\endgroup$
    – DanielSank
    Aug 22 '15 at 1:57
  • $\begingroup$ 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. $\endgroup$
    – udrv
    Aug 22 '15 at 2:48
  • $\begingroup$ You're right, the way I rearranged it wasn't clear. I hope the new version is ok. $\endgroup$
    – DanielSank
    Aug 22 '15 at 3:14
  • 2
    $\begingroup$ @udrv I like the argument, but it ignores the energy dependence of $H_\textrm{eff}$? $\endgroup$ Jun 7 '17 at 14:38

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