Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I recall faintly from my quantum theory lecture that there was a really neat way to derive Brillouin-Wigner perturbation theory for the special case of two coupled subspaces that involved a geometric series in reverse.

I know that the beginning was to have the Hilbert space split into subspaces 0 and 1 so that the Hamiltonian reads

$$H = H_{00} + H_{01} + H_{01}^\dagger + H_{11}$$ where the two indices indicate what subspace goes "in" and what subspaces comes "out" of each of the components.

If $P$ is a projector onto subspace $0$ and $Q$ a projector onto subspace $1$, this means, for example, that $$H_{00} = PH_{00}P, H_{01} = PH_{01}Q$$ and so on. In matrix notation $$\begin{pmatrix} H_{00} & H_{01} \\ H_{01}^\dagger & H_{11}\end{pmatrix} \begin{pmatrix} \psi_{0} \\ \psi_1\end{pmatrix} = E \begin{pmatrix} \psi_0 \\ \psi_1\end{pmatrix}$$

Now we can formally solve the Schrödinger equation for $\psi_1$ only: $$|\psi_1\rangle = \frac{1}{E - H_{11}} H_{01}^\dagger |\psi_0\rangle$$ and insert that back into the SG for $\psi_0$ to obtain

$$H_{00} |\psi_0\rangle + H_{01} \frac{1}{E-H_{11}} H_{01}^\dagger |\psi_0\rangle = E|\psi_0\rangle$$

Now I know that some cool trick with geometric series and an inverted matrix was going on to immediately write down the expansion of $|\psi\rangle$. Usually in the literature one first shows the iterative formula and THEN notes that this is a geometric series, but here it was done in the reverse, somehow an operator of the type $1/(A-B)$ with $A$ and $B$ matrices, was found and then expanded in the geometric series, but no matter how I massage the equations, I cannot seem to make it work, because there are so many different points where one could substitute one form into the other etc.

Any ideas on how this works? I'm also relatively sure that the work was done using operators, not for "individual" matrix elements. Personally, I find this more elegant because it makes it clearer what's going on on a more abstract level instead of dealing with the low-level matrix elements.

It could also be that at this point already the approximation $E \approx E_0$ was made where $E_0$ is a typical energy of subspace $0$, assuming that this subspace is near-degenerate and well-separated from subspace 1.

share|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.