T-25 · Wave mechanics

Waves that decide.

Schrödinger's equation, solved three ways. In a box the energies are n²π²/2, and on a grid the box is the chain of springs again. Under a barrier the wave decays but gets through, and what passes and what returns add up, squared, to one. A packet spreads at right angles to its own width: σ(t)² = σ₀² + (t/2σ₀)².

Time t = 0

The box on a grid

On a grid of N points the second derivative becomes tridiag(−1, 2, −1)/h²: the chain of springs from the oscillations sheet. When N + 1 divides 12, its energies are exact in ℚ(√2, √3). They fall short of n²π²/2 by a part in (nπh)²/12: doubling the points quarters the gap.

n	on the grid	exact, n²π²/2
1	144 − 36√2 − 36√6 = 4.90668	4.934802
2	144 − 72√3 = 19.29234	19.739209
3	144 − 72√2 = 42.17662	44.41322

…

Probabilities are squares

A superposition c₁ψ₁ + c₂ψ₂ is found in the first state with probability c₁² and in the second with c₂²; they must add to one, which is Pythagoras for the legs of the state. With c₁ = 3/5 and c₂ = 4/5 that is 9/25 + 16/25.

c₁² + c₂²	9/25 + 16/25 = 1, exactly
⟨x⟩, closed form	0.32707851
⟨x⟩ by quadrature	residual …
one beat, 2π/(E₂ − E₁)	4/(3π) = 0.424413