Position-basis Schrödinger equation

Act on both sides with ⟨x|. The kinetic term is easy: ⟨x|p̂²|ψ⟩ = (−iħ d/dx)²ψ(x) = −ħ²ψ″(x), because p̂ generates translations and is represented by −iħ d/dx on wavefunctions. The potential term is the interesting one. V(x̂) is a function of the position operator alone, so it is diagonal in this basis, ⟨x|V(x̂)|x′⟩ = V(x)δ(x − x′), and the ∫dx′ collapses to a single product V(x)ψ(x). Put them together and −(ħ²/2m)ψ″ + Vψ = Eψ. Notice what bought the ODE: locality. A nonlocal potential — the exchange term of Hartree–Fock, an optical potential — has a kernel that does not collapse to a δ, and the same eigenvalue problem is an integro-differential equation instead. Change basis and it changes character again: in the momentum representation p̂² is the diagonal one and V becomes a convolution. The differential equation is a property of the representation you chose, not new physics.

Divide the equation by ψ at any point where ψ ≠ 0 and it rearranges to ψ″/ψ = (2m/ħ²)(V(x) − E). Where E > V the right-hand side is negative, so ψ and ψ″ carry opposite signs: where ψ is positive it is concave down, where ψ is negative it is concave up, and in both cases it is bending back toward the axis. That is what forces it to cross and cross again — oscillation. Where E < V the ratio is positive, ψ and ψ″ share a sign, the curve bends away from the axis and, once launched outward, generically runs off to ±∞. Numbers make the two regimes concrete for an electron: E − V = 1.0 eV gives ψ″/ψ = −26.2 nm⁻², while V − E = 4.0 eV gives +105 nm⁻². The magnitude matters too, since |ψ″/ψ| is k² in the first case and κ² in the second, but it is the sign that decides whether you are drawing a ripple or a tail.

In a classically allowed region the equation reads ψ″ = −k²ψ with k(x) = √(2m(E − V))/ħ. If V is flat, that is solved exactly by A cos(kx + φ). If V varies slowly on the scale of one wavelength, the same form survives locally with an x-dependent k — the semiclassical reading. For an electron the shortcut is k = 5.12 √(E − V in eV) nm⁻¹: at E − V = 1.0 eV, k = 5.12 nm⁻¹ and λ = 1.23 nm; at 4.0 eV, k = 10.2 nm⁻¹ and λ = 0.61 nm. So the ripple gets finer the deeper you sit in the well, and a plotted eigenfunction is visibly more crowded near the bottom of a potential than near its edges. The envelope runs the other way. The WKB amplitude goes as 1/√k, largest where the classical particle is slowest, which is how |ψ|² recovers the classical density ∝ 1/v in the large-n limit. That is a statement about the envelope only; |ψ|² itself still drops to zero at every node.

Where E = V(xₜ) the right-hand side vanishes and ψ″(xₜ) = 0 exactly. That inflection is the most reliable landmark on any hand-drawn eigenfunction: inside the turning points the curve is concave toward the axis, outside them convex away from it, and the changeover is at the classical turning point and nowhere else. Beyond it the solution behaves as exp(−∫κ dx) with κ(x) = √(2m(V − E))/ħ, so the leak is a pure exponential only where V is flat; as the wall rises κ grows and the tail steepens. Two numbers for an electron: 1.0 eV below the barrier top the penetration depth 1/κ is 0.195 nm, and 4.0 eV below it is 0.098 nm. A potential step is the exception that proves the rule. V jumps past E, so no x satisfies V(x) = E and ψ″ does not pass through zero at the edge. ψ and ψ′ stay continuous while ψ″ flips sign discontinuously across it, which makes the edge itself the place where the curve switches from bending toward the axis to bending away.

Written as −(d/dx)[pψ′] + qψ = λwψ, the equation has p = ħ²/2m > 0, q = V real, w = 1, λ = E. On a finite interval, such as a box, this is a regular Sturm–Liouville problem; on the whole line it is singular, and what follows then holds for its bound states. With self-adjoint boundary conditions the consequences arrive free of charge: eigenvalues are real and bounded below and can be ordered E₁ < E₂ < …; eigenfunctions belonging to different eigenvalues are orthogonal, with weight 1 so the integral is plain ∫ψₘ*ψₙ dx; and the oscillation theorem fixes that ψₙ has exactly n − 1 interior nodes, which lets you name a level off a plot without computing an energy. Non-degeneracy in one dimension comes from the Wronskian: with no first-derivative term, W = ψ₁ψ₂′ − ψ₂ψ₁′ is constant, and it must vanish at infinity for bound states, so W ≡ 0 and any two solutions at the same E are proportional. The w = 1 is worth flagging — in spherical coordinates the radial measure is r² dr, and the substitution u = rR is exactly what restores w = 1 so this whole toolkit can be reused.

The equation contains ħ, m, and whatever length a the potential supplies, and those three build exactly one energy: ħ²/2ma². Set x = au and divide through by it, and you get −d²ψ/du² + v(u)ψ = εψ with v = V/(ħ²/2ma²) and ε = E/(ħ²/2ma²). Every level is that unit times a pure number fixed by v alone. Where V brings no energy scale of its own, as in the infinite well, v does not change with m or a, so E ∝ 1/m and E ∝ 1/a² without any further calculation; a finite well of depth V₀ instead carries the strength V₀/(ħ²/2ma²) into v, and ε moves with it. For an electron with a = 1 nm the unit is 0.0381 eV, and the infinite-well ground state, ε = π², gives E₁ = 0.376 eV. Put a proton in the same box and the unit falls to 2.07 × 10⁻⁵ eV and E₁ to 2.05 × 10⁻⁴ eV. This is also the step that makes the numerical work behave: solving in u keeps every matrix entry of order one, instead of carrying ħ²/2m = 6.1 × 10⁻³⁹ J m² through a finite-difference Hamiltonian and hoping the eigensolver copes.