The normalized exponential Coulomb family gives E/|E*|=1/x²−2/x, with x=a/a*. Its exact minimum is −1 at x=1; the interaction and its physical normalization remain inputs.
Try the demonstration
x = a/a* = 1. Kinetic = 1, potential = -2, total E/|E*| = -1. Energy above the minimum: 0. The exact minimum is −1 at x = 1 for the supplied Coulomb trial family.
Exact energy expectation for the normalized exponential family in the supplied nonrelativistic Coulomb Hamiltonian. The graph does not derive the coupling, particle masses, fine structure, or quantum electrodynamic corrections.
Change the size of the trial state
The radius slider chooses x=a/a*, from one quarter to four times the preferred scale. The curve shows the total dimensionless energy, and the highlighted point follows the chosen radius. The live result separates the positive kinetic contribution, negative potential contribution, total energy, and energy above the minimum.
The vertical scale measures energy relative to the magnitude of the bound-state minimum. It is not electron-volts unless the physical Hamiltonian coefficients have been supplied separately. The selected radii are dimensionless comparisons, not measured atomic sizes.
State the interaction before deriving the curve
Use a nonrelativistic relative-coordinate Hamiltonian H=−D∇²−C/r, with D,C>0, and the normalized exponential state ψ(r)=exp(−r/a)/√(πa³). Its expectation value is exactly:
The first term follows from the gradient norm and the second from the expectation of 1/r. This Hamiltonian supplies three-dimensional geometry, an inverse-distance attraction, and a kinetic normalization. Calling the state completed does not derive those ingredients.
Remove the chosen units
Differentiating E(a) gives the minimizing radius a*=2D/C and energy E*=−C²/(4D). Substituting a=xa* and dividing by |E*| yields the displayed universal shape:
E/|E*|+1=(1/x−1)²≥0.
The completed-square identity proves that the minimum is −1, achieved only at x=1. For the supplied Coulomb problem, the optimizing exponential is the exact nonrelativistic ground state. The demonstration examines its radius family rather than integrating a time-dependent wave equation.
Inspect three radii
At x=1, kinetic energy is one and potential energy is minus two, giving total minus one. At x=1/2, the two contributions are four and minus four, so total energy is zero. At x=2, they are one quarter and minus one, giving minus three quarters.
Very small radii incur a rapidly growing kinetic cost. At large radius the energy approaches the separated-particle threshold zero from below. The minimum identifies binding relative to that supplied threshold. A finite-box level spacing alone would not establish the same conclusion.
Binding does not select the coupling
Every positive C in this Hamiltonian produces a bound scale. Doubling C at fixed D halves a* and multiplies |E*| by four. After rescaling, all such choices produce the same dimensionless curve. Its shape therefore cannot determine the electromagnetic coupling or an ideal structural value of alpha.
The SMM/DTT task is to derive geometry, coherent motion, and mediator response from one source with a common clock and action scale. The six inner roles should constrain those coefficients, with outer completion retaining a stable composite. This widget provides an exact benchmark once the Hamiltonian is supplied. Reset restores its minimum, also shown by the initial static plot.
Continue with the articles
- A neutral pair is not yet an atom
Explain why neutrality alone does not establish binding. Introduce the need for a localized state below the separated-particle threshold in the minimal hydrogen model.
- Eliminating a mediator to obtain attraction
Describe the local quadratic mediator and its exact Green-function interaction. Explain how the chosen source signs and response operator enter the effective pair energy.
- A normalizable hydrogen-like ground state
Introduce the lower-bound and bound-state results of the effective model. Explain how the supplied kinetic and potential rules produce a meaningful atomic test case.
- Why binding does not fix the coupling
Introduce the dilation argument showing that a family of positive couplings remains compatible with binding and completion. Identify the extra source-selection law needed to obtain a specific interaction strength.