= Solution
Assume $H$ is <self-adjoint> on its specified <operator domain>, and the admissible variations preserve that domain and any <boundary conditions>, so no boundary terms spoil self-adjointness. Let $N=\langle\phi,\phi\rangle>0$. The first variation of the <Rayleigh quotient> is
$$
\delta F=\frac{\langle\delta\phi,(H-F[\phi])\phi\rangle+\langle(H-F[\phi])\phi,\delta\phi\rangle}{N}=\frac{2\operatorname{Re}\langle\delta\phi,(H-F[\phi])\phi\rangle}{N}.
$$
At $\phi=\psi_n$, the residual vanishes, so $\boxed{\delta F[\psi_n]=0}$. More explicitly, for a normalized $\psi_n$ and $\phi=\psi_n+\eta$,
$$
F[\psi_n+\eta]-E_n=\frac{\langle\eta,(H-E_n)\eta\rangle}{1+2\operatorname{Re}\langle\psi_n,\eta\rangle+\|\eta\|^2}.
$$
For a fixed admissible direction scaled by a small parameter, the change is second order. This is stationarity, and is not necessarily a minimum for an excited state.
Conversely, vanishing first variation for every admissible complex direction yields the stationary equation
$$
\boxed{(H-F[\phi])\phi=0\quad\text{at a stationary }\phi.}
$$
At the original stationary point, this is $(H-F[\psi_n])\psi_n=0$. \b[The printed equation with an arbitrary perturbed $\phi$ in the final factor is not generally true.] Taking $\phi=\psi_n+\epsilon\psi_m$ with $E_m\neq E_n$ gives $(H-E_n)\phi=\epsilon(E_m-E_n)\psi_m\neq0$, even though the first-order change in $F$ is zero. Thus the valid claim is the stationary equation at the extremal state, not an exact <eigenvalue> equation for every variation.
Back to article page
