University Physics V · The Schrödinger Equation · 4.5

Position-Basis Schrödinger Equation

Once H|ψ⟩ = E|ψ⟩ is written out in the position basis it stops being abstract and becomes a second-order ODE you can read like a graph. The habit to build here is getting the shape of an eigenfunction — where it waves, where it dies, where it bends — out of the sign of V − E, before you solve anything.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Take the abstract eigenvalue problem H|ψ⟩ = E|ψ⟩ and resolve it in the position basis by acting with ⟨x| and inserting ∫|x′⟩⟨x′| dx′. Two facts do all the work: ⟨x|p̂|ψ⟩ = −iħ dψ/dx, so the kinetic term becomes −(ħ²/2m)ψ″, and V(x̂) is a function of x̂ alone, hence diagonal here — ⟨x|V(x̂)|x′⟩ = V(x)δ(x − x′) — so it acts by plain multiplication. What was an operator equation on a Hilbert space is now −(ħ²/2m)ψ″(x) + V(x)ψ(x) = Eψ(x): a linear, homogeneous, second-order ODE with E as the eigenvalue, sitting in Sturm–Liouville form with weight 1, which is where real eigenvalues, orthogonality and the node theorem come from.

Rearranged as ψ″/ψ = (2m/ħ²)(V(x) − E) it also becomes readable. Wherever E > V the ratio is negative and ψ curves back toward the axis, oscillating with local wavenumber k(x) = √(2m(E − V))/ħ; wherever E < V it is positive and ψ curves away, dying with κ(x) = √(2m(V − E))/ħ; and where E = V(x) it vanishes, putting an inflection point of ψ exactly at each classical turning point. The cost is locality and dimension.

This picture needs a real V acting as a multiplication operator, one spatial coordinate so the ODE is ordinary, and a spinless particle. Give up any of those — a nonlocal exchange potential, three dimensions, two interacting particles — and the same eigenvalue problem stops being a curve you can sketch.

Simple definition: In the position basis the energy eigenvalue problem becomes the ordinary differential equation −(ħ²/2m)ψ″(x) + V(x)ψ(x) = Eψ(x), in which the second derivative of ψ at each point is fixed by ψ itself, the eigenvalue E, and the local value of V.
Example: For an electron with E − V = 1.0 eV, and ħ²/2mₑ = 0.0381 eV nm², the equation gives ψ″/ψ = −1.0/0.0381 = −26.2 nm⁻², so k = 5.12 nm⁻¹ and the local wavelength is 2π/k = 1.23 nm.

The eigenvalue problem in the position basis−(ħ²/2m) ψ″(x) + V(x) ψ(x) = E ψ(x)

What ⟨x|H|ψ⟩ = E⟨x|ψ⟩ becomes once p̂ → −iħ d/dx and V(x̂) acts by multiplication.

ψ carries units m^(−1/2) in 1D; V and E in J or eV; ħ²/2mₑ = 0.0381 eV nm².

Curvature ratioψ″/ψ = (2m/ħ²) (V(x) − E)

Its sign, not its size, sets the shape: negative bends ψ toward the axis, positive away from it.

Both sides in m⁻² (or nm⁻²); valid at every x where ψ ≠ 0.

Local wavenumber and decay constantk(x) = √(2m(E − V))/ħ · κ(x) = √(2m(V − E))/ħ

Gives the wavelength 2π/k where the state waves, and the penetration depth 1/κ where it dies.

|ψ″/ψ| is k² or κ². For an electron, k or κ = 5.12 √(|E − V| / 1 eV) nm⁻¹.

A turning point is an inflection point of ψE = V(xₜ) ⟹ ψ″(xₜ) = 0

Puts every change of curvature on a hand sketch in the right place without solving for ψ.

xₜ is a classical turning point; needs V continuous there, so a step has none.

Sturm–Liouville form−(d/dx)[p ψ′] + q ψ = λ w ψ, p = ħ²/2m, q = V(x), w = 1, λ = E

This structure, not the physics, guarantees real E, orthogonal eigenfunctions and completeness.

w = 1 is why orthogonality is ∫ψₘ*ψₙ dx = δₘₙ with no weight factor in the integrand.

The natural energy scalex = a u ⟹ −d²ψ/du² + v(u) ψ = ε ψ, with E = ε · ħ²/(2ma²)

ε is a pure number set only by the shape of V, so every level scales as 1/m and as 1/a².

a is the length V supplies. For an electron and a = 1 nm, ħ²/2ma² = 0.0381 eV.

Projecting H|ψ⟩ = E|ψ⟩ onto the position basis

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.

Read the sign of V − E and you have the shape

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.

Local wavelength, and where the amplitude piles up

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.

Turning points, inflections, and how far ψ leaks

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, so no x satisfies V(x) = E, and there is no inflection point at all: ψ and ψ′ stay continuous while ψ″ flips sign discontinuously across the edge.

What the Sturm–Liouville structure guarantees before you solve

Written as −(d/dx)[pψ′] + qψ = λwψ, the equation has p = ħ²/2m > 0, q = V real, w = 1, λ = E. That is a regular Sturm–Liouville problem, and with self-adjoint boundary conditions its 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.

Nondimensionalise before you compute

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 only by the shape of v, so E ∝ 1/m and E ∝ 1/a² without any further calculation. 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 2.08 × 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Energy above the floor E1.0 eV

Step height above E, V₀ − E2.5 eV

Raise E and the ripples on the left get shorter and closer together; raise the step and the tail on the right dies faster. Watch the dot: ψ and ψ′ never break there, but ψ″ changes sign discontinuously, which is why a step has no inflection point.

LOCAL k, E > V5.12 nm⁻¹

WAVELENGTH 2π/k1.226 nm

DECAY κ, E < V8.10 nm⁻¹

PENETRATION 1/κ0.123 nm

Live interpretationLOCAL k, E > V: 5.12 nm⁻¹. WAVELENGTH 2π/k: 1.226 nm. DECAY κ, E < V: 8.10 nm⁻¹. PENETRATION 1/κ: 0.123 nm

Catch the common trap

Explain before calculating.

Misconception check

The rule is toward the axis, not downward. Where E > V, ψ″/ψ is negative — and that makes ψ concave down only while ψ is positive. Cross a node and both ψ and ψ″ change sign, the ratio is unchanged, and the curve is now concave up while still bending toward the axis. Draw every allowed region as concave down and you get a wavefunction that never returns and can have no nodes at all, which quietly destroys the node theorem you are about to use. The mirror error is reading E < V as forbidden, so ψ = 0 there. Nothing in −(ħ²/2m)ψ″ + Vψ = Eψ kills ψ where V exceeds E; the equation only flips the sign of the curvature, and ψ decays as exp(−κx) with a finite penetration depth — 0.195 nm for an electron sitting 1.0 eV below the barrier. What the region forbids is a classical trajectory, not an amplitude.

An electron is in a bound state of energy E = −2.0 eV. At a point x₀ inside the well the potential is V(x₀) = −6.0 eV, and the real eigenfunction there has ψ(x₀) = +0.40 nm^(−1/2). Taking ħ²/2mₑ = 0.0381 eV nm², what is ψ″(x₀), and what does its sign tell you?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron moves through a region of constant potential where E − V = 2.5 eV. Using ħ²/2mₑ = 0.0381 eV nm², find ψ″/ψ, the local wavenumber k, and the wavelength, and say what the sign of ψ″/ψ tells you.

Write the equation as a curvature ratio: ψ″/ψ = (2m/ħ²)(V − E), which for an electron is (V − E)/(0.0381 eV nm²).
Here E − V = +2.5 eV, so V − E = −2.5 eV and ψ″/ψ = −2.5/0.0381 = −65.6 nm⁻². Negative: ψ bends back toward the axis, so the region is classically allowed and ψ oscillates rather than decays.
In an allowed region ψ″ = −k²ψ, so k = √65.6 = 8.10 nm⁻¹.
λ = 2π/k = 6.2832/8.10 = 0.776 nm. Cross-check with the electron shortcut λ = 1.226 nm/√(E − V in eV) = 1.226/1.581 = 0.775 nm; the last digit differs only because ħ²/2m was rounded to 0.0381.

Answerψ″/ψ = −65.6 nm⁻², k = 8.10 nm⁻¹, λ = 0.776 nm. The negative ratio is the whole qualitative answer: oscillation, not decay.

MediumAn electron of energy E = 1.5 eV meets a potential step: V = 0 for x < 0 and V = 4.0 eV for x > 0. Find the wavelength on the left, the decay constant and penetration depth on the right, the ratio of the two curvature magnitudes, and how far into the barrier ψ has fallen to 1% of its value at the edge.

Left of the step E − V = 1.5 eV, so ψ″/ψ = −1.5/0.0381 = −39.4 nm⁻², k = √39.4 = 6.27 nm⁻¹ and λ = 2π/6.27 = 1.00 nm.
Right of the step V − E = 4.0 − 1.5 = 2.5 eV, so ψ″/ψ = +2.5/0.0381 = +65.6 nm⁻², κ = √65.6 = 8.10 nm⁻¹ and the penetration depth is 1/κ = 0.123 nm.
The curvature magnitudes are in the ratio 65.6/39.4 = 2.5/1.5 = 1.67: the common prefactor 2m/ħ² cancels, so ψ bends 1.67 times harder in the barrier than in the well.
Beyond the edge ψ ∝ e^(−κx), so ψ falls to 1% at x = ln(100)/κ = 4.605/8.10 = 0.569 nm.
Note what is missing. V jumps, so no x satisfies V(x) = E and ψ has no inflection point anywhere; ψ and ψ′ are continuous at x = 0 while ψ″ jumps from −39.4 ψ(0) to +65.6 ψ(0).

Answerλ = 1.00 nm on the left; κ = 8.10 nm⁻¹ and 1/κ = 0.123 nm on the right; curvature ratio 1.67; ψ is down to 1% by 0.569 nm into the barrier.

HardFor V(x) = ½mω²x², show that ψ₀(x) = A exp(−x²/2a²) with a = √(ħ/mω) satisfies −(ħ²/2m)ψ″ + Vψ = Eψ, find E, and show that the inflection points of ψ₀ sit exactly at the classical turning points. Then put in numbers for an electron with ħω = 0.50 eV.

Differentiate twice: ψ₀′ = −(x/a²)ψ₀ and ψ₀″ = (x²/a⁴ − 1/a²)ψ₀, so ψ₀″/ψ₀ = x²/a⁴ − 1/a².
The equation demands ψ″/ψ = (2m/ħ²)(½mω²x² − E) = (m²ω²/ħ²)x² − 2mE/ħ². Match the x² terms: 1/a⁴ = m²ω²/ħ², i.e. a² = ħ/mω — exactly the a that was given.
Match the constants: 1/a² = 2mE/ħ², so E = ħ²/(2ma²) = (ħ²/2m)(mω/ħ) = ħω/2. The zero-point energy falls out of the curvature match, with no boundary condition invoked.
Turning points: E = V gives ħω/2 = ½mω²x², so x² = ħ/mω = a² and x = ±a. Inflections: ψ₀″ = 0 when x²/a⁴ = 1/a², again x = ±a. The Gaussian's inflection points are the classical turning points.
Numbers: a² = ħ²/(m⋅ħω) = 2(ħ²/2m)/(ħω) = 2 × 0.0381/0.50 = 0.1524 nm², so a = 0.390 nm, and E = ħω/2 = 0.25 eV.
Check the curvature twice. At x = 0: ψ″/ψ = −1/a² = −6.56 nm⁻², and (V − E)/0.0381 = (0 − 0.25)/0.0381 = −6.56 nm⁻². At x = 2a: V = 4 × 0.25 = 1.00 eV, so (V − E)/0.0381 = 0.75/0.0381 = +19.7 nm⁻², matching 3/a² = 19.7 nm⁻² — deep in the forbidden region, curving away.

AnswerE = ħω/2, and the inflections x = ±a of the Gaussian coincide with the turning points. For ħω = 0.50 eV: a = 0.390 nm, E = 0.25 eV, ψ″/ψ = −6.56 nm⁻² at the centre and +19.7 nm⁻² at x = 2a.