The RPA (A, B) Eigenproblem

Everything on the left-hand side is as nice as matrices get: $A$ and $B$ real symmetric, the big block matrix symmetric too. The trouble is the metric $S=\mathrm{diag}(1,-1)$ on the right. To make this an ordinary eigenvalue problem you multiply through by $S^{-1}$ (which is just $S$ again), and the product

is not symmetric. Not even close — the bottom row picked up a global sign. So the workhorse guarantees you lean on everywhere else (real eigenvalues, orthogonal eigenvectors, eigh) are all void. The saving grace is that $SM$ is not a generic nonsymmetric matrix either: it has the block structure of a Hamiltonian (symplectic) operator, and that structure is exactly what the other four framings exploit. An indefinite metric doesn't destroy the symmetry; it reorganizes it.

Write the block equation as two coupled equations:

Now swap the roles of the two halves: feed the vector $(Y,X)$ into the same equations. The first equation asks for $AY+BX$, and the second original equation already says that equals $-\omega Y$. The second asks for $BY+AX$, which the first original equation says is $\omega X$. Both match the pattern of an eigenvector — with eigenvalue $-\omega$:

So the $2n$ eigenvalues are not $2n$ independent numbers: they are $n$ pairs $\pm {\omega }_k$, mirror images through zero, and the $-\omega$ partner carries the same information with $X$ and $Y$ exchanged. In the physics this is time reversal — every excitation has a de-excitation running the same transition backward. In the linear algebra it is the symplectic structure from framing 1, made visible one eigenvector at a time. The validation below finds the pairing holds to $1.2\times 10^{-14}$ on a random stable $(A,B)$.

Add and subtract the two coupled equations and the variables $X+Y$ and $X-Y$ decouple partway: two equations, each mapping one combination onto the other. Eliminate one and you get an $n\times n$ problem with eigenvalue ${\omega }^2$. One more conjugation by the symmetric square root of $A-B$ (which exists whenever $A-B$ is positive-definite) makes it Hermitian:

Every step of that algebra is walked through, move by move, in the Casida Hermitian rewrite derivation — here the point is what it buys. The smallest case makes it concrete: at $n=1$ the matrices are scalars $a$ and $b$, and the rewrite says ${\omega }^2=(a-b)(a+b)=a^2-b^2$. With $a=0.9$, $b=0.3$: $\omega =\sqrt{0.72}=0.848528$ — and the full $2\times 2$ nonsymmetric solve returns exactly $\pm 0.848528$. The coupling $b$ always lowers $\omega$ below the bare $a$: you can read that straight off $a^2-b^2<a^2$.

The payoff is practical, not just aesthetic: one $2n\times 2n$ nonsymmetric eig becomes two $n\times n$ symmetric eigh calls plus matrix products. At $n=500$ that is 7.55 s against 0.91 s below — and eigh hands back guaranteed-real eigenvalues and an orthonormal basis for free.

The rewrite in framing 3 quietly assumed $A-B\succ 0$. That assumption is a physical statement, and the spectrum tells you when it fails. When $A+B$ and $A-B$ are both positive-definite, ${\omega }^2>0$ and every eigenvalue is real. Let $A-B$ lose definiteness — the validation below scales $B\to tB$ and pushes — and an eigenvalue of ${\omega }^2$ crosses zero: a $\pm \omega$ pair slides together, collides at the origin, and re-emerges as a purely imaginary pair $\pm i\gamma$. On real numbers, the smallest pair goes $\pm 0.054$ real just before the crossing and $\pm 0.054i$ just after.

An imaginary frequency means the "ground state" you linearized around is not a minimum at all — there is a direction along which the energy goes down, and the linear response grows exponentially instead of oscillating. In quantum chemistry this is the triplet instability that makes Casida's equation fail loudly at a bad closed-shell reference. So the eigenproblem is doing double duty: its eigenvalues are the excitation spectrum and a certificate that the reference deserves to be expanded around.

Set $B=0$ and the two coupled equations disconnect completely: $AX=\omega X$ and $AY=-\omega Y$. The positive branch is a plain symmetric eigenproblem in $A$ — eigh and done. This is the Tamm-Dancoff approximation (TDA), and it is what most people run first, because it cannot go unstable: no square root of $A-B$ exists to fail.

What you pay: $B$ couples each excitation to the de-excitations, and dropping it pushes every frequency up (in the $1\times 1$ case, from $\sqrt{a^2-b^2}$ to $a$) and breaks the sum rules that make computed intensities gauge-consistent. TDA is the right first move and a slightly wrong final answer — the structured problem on this page is what "not approximating" costs, and the Hermitian rewrite makes that cost small.