Far from a neutral bound system, the slowest-decaying electron-removal amplitude is controlled by the first ionization energy $I=E_{N-1}-E_N>0$. Its asymptotic wavefunction satisfies a free-particle-like equation at negative energy $-I$, giving a decay exponent $\kappa =\sqrt{2m_{e}I}/\hslash$. Density is quadratic in the amplitude, soor $a=2\sqrt{2I}$ in atomic units. Power-law radial factors and angular dependence do not alter this leading exponential coefficient. It is the ionization energy, not the electron affinity, that controls this tail.

For fixed electron-electron interaction and electron number, a nondegenerate ground-state electron density determines the external potential up to an additive constant, and hence determines the Hamiltonian and ground-state observables.

Suppose two potentials $v,v^{\prime }$ differing by more than a constant have different ground states $\Psi ,{\Psi }^{\prime }$ but the same density $\rho$. Strict Rayleigh-Ritz variational principle inequalities giveAdding produces the contradiction $E+E^{\prime }<E+E^{\prime }$. This proves the first Hohenberg-Kohn theorem under the stated nondegeneracy assumptions. A constant potential shift changes energy by $N$ times that constant without changing the density.^[ref]

Define $h_{pq}=⟨p|\hat{h}|q⟩$, the one-electron kinetic plus nuclear-attraction integral, and physicists' two-electron integralsin atomic units. With $E_{\mathrm{NN}}$ the zero-electron nuclear-repulsion term, the Slater determinant expectation isThe occupied sums include spin-orbitals, so the exchange integral vanishes for incompatible spins without needing a separate closed-shell factor.

The first excited unperturbed level is twofold degenerate, spanned by $|1,2⟩$ and $|2,1⟩$, with energy $5h^2/(8m{a}^2)$. The perturbation is $q{\epsilon }_Q(x-a/2)(y-a/2)$. Both diagonal matrix elements vanish, but the off-diagonal element does not:Diagonalizing the $2\times 2$ matrix with zero diagonal gives first-order shiftswith symmetric/antisymmetric combinations $(|1,2⟩\pm |2,1⟩)/\sqrt{2}$. Their order depends on the sign of $q{\epsilon }_Q$; the lower branch has shift $-|W_{12,21}|$. Applying nondegenerate perturbation theory to either product state would miss the splitting.

Let $H^{(0)}|n⟩=E_n^{(0)}|n⟩$ with a nondegenerate reference eigenvalue, and choose intermediate normalization $⟨n|n^{(1)}⟩=0$. The first-order equation isProjection on $|j⟩$, $j\ne n$, gives $⟨j|n^{(1)}⟩=H_{jn}^{(1)}/(E_n^{(0)}-E_j^{(0)})$. The second-order projected equation gives $E_n^{(2)}=⟨n|H^{(1)}|n^{(1)}⟩$, henceThe sum runs over a complete orthonormal reference basis except the reference state; continuum states require integration if present. Degenerate states with zero denominators require degenerate perturbation theory first. These are expansion coefficients in $E_n=E_n^{(0)}+\lambda E_n^{(1)}+{\lambda }^2{E}_n^{(2)}+\cdots$.