Consider a Hamiltonian $H$ on a Hilbert space $\mathcal H_A \oplus\mathcal H_B$ and let $P$ (and $Q$) denote the projection operator onto $A$ (and $B$). There are two common definitions of an effective Hamiltonian $H_\textrm{eff}(E)$ on subspace $A$:

  1. The Schroedinger equation can be written as $$ \left( \begin{array}{cc} PHP & PHQ\\ QHP & QHQ \end{array} \right) \left( \begin{array}{c} \psi_A \\ \psi_B\end{array} \right) = E \left( \begin{array}{c} \psi_A \\ \psi_B\end{array} \right)$$ One can straightforwardly eliminate $\psi_B$ to obtain $$ H_\textrm{eff}(E) |\psi_A\rangle = E |\psi_A\rangle \qquad \textrm{with } \; \boxed{H_\textrm{eff}(E) := P \left( H + H Q \frac{1}{E-QHQ} Q H \right) P}$$
  2. In the Green's function/resolvent approach, one simply defines $$ \boxed{\frac{1}{E-H_\textrm{eff}(E)} := P \frac{1}{E-H} P }$$ (The motivation for this has to do with the fact that the left-hand side has the same poles as the right-hand side would have even without the $P$ projectors, hence $H_\textrm{eff}(E)$ has the same eigenvalues as $H$ for eigenvectors which have a non-zero support on $A$.)

My question is simply: how can I see whether these two definitions are equivalent?

(Note that this is not a duplicate! A very similar question was asked before, but there the answer that was accepted in fact did not answer the exact question that I am asking. There is also another question which has to do with deriving the effective Hamiltonian, which is not what I am asking here.)


I'll prove $(2\implies 1)$; the reverse implication can be found just by reading the proof bottom to top. To get equation (1) from equation (2), we need to figure out how to invert $P\frac{1}{E-H}P$. Once we've figured that out, the rest of the proof follows straightforwardly.

For any operator of the form $$ \hat{O}=\left(\begin{array}{cc}\hat{A} & \hat{B}\\\hat{C}&\hat{D}\end{array}\right) $$ it is a theorem from linear algebra that $$ \hat{O}^{-1}=\left(\begin{array}{cc}(\hat{A}-\hat{B}\hat{D}^{-1}\hat{C})^{-1} & \square\\\square&\square\end{array}\right) $$ where the $\square$s denote elements we don't care about.

For the specific case of $\hat{O}=E-H=\left(\begin{array}{cc}E-PHP&-PHQ\\-QHP&E-QHQ\end{array}\right)$, we have that $$ \hat{O}^{-1}=\frac{1}{E-H}=\left(\begin{array}{cc}([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1} & \square\\\square&\square\end{array}\right) $$ Thus, we have $$ P\frac{1}{E-H}P=([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1} $$ therefore, $$ \frac{1}{E-H_{eff}}=([E-PHP]-PHQ[E-QHQ]^{-1}QHP)^{-1} $$ We can now easily take the inverse of both sides! As promised, we very quickly get to the final answer:

\begin{array}{rcl} E-H_{eff}&=&[E-PHP]-PHQ[E-QHQ]^{-1}QHP\\ H_{eff}&=&PHP+PHQ[E-QHQ]^{-1}QHP\\ \end{array}


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