For $H=p^2/(2m)-g\delta (x)$ with $g>0$, put $\kappa =mg/{\hslash }^2$ and $T>0$. The bound state in a delta potential has squared amplitude $\kappa$ at the origin. The even normalized continuum states have squared amplitude $k^2/[\pi (k^2+{\kappa }^2)]$ for $k>0$, and the odd states vanish there. The spectral resolution of a quantum propagator gives the displayed result. At long times, the continuum stationary point at $k=0$ is suppressed by its quadratic spectral weight: expanding that weight gives a continuum term $e^{-3i\pi /4}(2m/\hslash T{)}^{3/2}/(4\sqrt{\pi }{\kappa }^2)+O(T^{-5/2})$. The bound state term persists with constant modulus. This stationary phase expansion requires $\hslash {\kappa }^{2}T/(2m)\gg 1$.

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.

Neglect radiation in cosmology and take pressureless matter plus a separately conserved dark energy component with $p_{\mathrm{DE}}=w{\rho }_{\mathrm{DE}}{c}^2$, where $\rho$ here denotes mass-equivalent energy density. The Friedmann acceleration equation givesThus $w<-1/3$ for a positive-density component is necessary for that component to accelerate expansion, but the total must also overcome pressureless matter. For a cosmological constant, $w=-1$ and $q_0={\Omega }_{m,0}/2-{\Omega }_{\Lambda ,0}$, so acceleration requires ${\Omega }_{\Lambda ,0}>{\Omega }_{m,0}/2$. Spatial curvature of an FLRW universe enters the distance relation but has no separate term in this acceleration criterion. A Hubble diagram alone does not identify a microscopic dark energy mechanism.