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University Physics IV · Quantum Potentials · 7.1

Piecewise Potentials & Matching

Almost every solvable one-dimensional problem is built the same way: chop the axis wherever V jumps, write down the only two solutions a constant potential allows, then let continuity at the joins do the physics. Learn the bookkeeping once and the well, the step and the barrier all become the same calculation.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

A piecewise-constant potential is a deliberate caricature: replace whatever V(x) really does by a staircase of flat treads, and inside each tread the time-independent Schrödinger equation collapses to ψ″ = (2m/ħ²)(V − E)ψ with a constant coefficient — an equation you met in first-year calculus. Where E lies above the tread the coefficient is negative and ψ oscillates with wavenumber k = √(2m(E − V))/ħ; where E lies below it the coefficient is positive and ψ is a sum of real exponentials with decay constant κ = √(2m(V − E))/ħ. All the physics has been pushed into the joins.

Integrating the equation across a step shows that ψ and ψ′ are both continuous whenever V is finite, and those two conditions per boundary are what stitch the local solutions into one global state — and, by refusing to be satisfied at most energies, are what quantise the spectrum. The cost is honest: a real step is smooth over a finite length, the model is one-dimensional, static and single-particle, and the matching rules themselves break exactly where V stops being finite.

Simple definition
A piecewise-constant potential is flat inside each of a few regions and jumps between them, so the Schrödinger equation is solved separately in every region and the pieces are joined by demanding that ψ and ψ′ agree at each boundary.
Example
For an electron ħ²/2m = 0.0381 eV nm², so 1.0 eV above a flat floor gives k = √(1.0/0.0381) = 5.12 nm⁻¹ and λ = 1.23 nm, while 1.0 eV below it gives κ = 5.12 nm⁻¹ and the wave dies to 1/e in 0.195 nm.
The equation inside one regionψ″ = (2m/ħ²)(V − E) ψ, V constant

The sign of V − E, not its size, decides whether ψ wiggles or grows.

x in m, V and E in J, ψ in m(−1/2); the bracket is one fixed number in the region

Classically allowed region, E > Vψ = A e(ikx) + B e(−ikx), k = √(2m(E − V))/ħ

Two travelling waves, right and left; the real sin/cos pair is the same solution and is handier for bound states.

k in m⁻¹, λ = 2π/k. Electron 1.0 eV above the floor: k = 5.12 nm⁻¹, λ = 1.23 nm.

Classically forbidden region, E < Vψ = C e(κx) + D e(−κx), κ = √(2m(V − E))/ħ

Keep both terms unless the region runs to infinity; only then does one of them fail to be normalisable.

κ in m⁻¹, 1/κ is the penetration depth. Electron 1.0 eV below: 1/κ = 0.195 nm.

Matching at a finite stepψ(x₀⁻) = ψ(x₀⁺) and ψ′(x₀⁻) = ψ′(x₀⁺)

True for any finite jump in V, however large — the height of the step enters only through k and κ.

Two real equations per boundary, in the units of ψ and of ψ per metre

The logarithmic derivative(ψ′/ψ)|₋ = (ψ′/ψ)|₊, both sides in m⁻¹

Divides out the unknown amplitude, turning two equations into one condition on E alone.

The ratio of the two matching equations; invariant under rescaling ψ

Where ψ′ is allowed to jumpψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²) ψ(0)

The jump is the integral of V across the point, and vanishes unless V is unbounded there.

For V = −α δ(x) with α in J m; at an infinite wall ψ = 0 and ψ′ is not matched at all

01

Constant V collapses the equation

Start from −(ħ²/2m)ψ″ + V(x)ψ = Eψ and move everything to one side: ψ″ = (2m/ħ²)(V − E)ψ. Nothing is approximate yet. Now suppose V is constant across some interval. The bracket is a fixed number and the equation is a linear ODE with constant coefficients, with three cases. If E > V, write (2m/ħ²)(V − E) = −k², so ψ″ = −k²ψ and ψ = A e(ikx) + B e(−ikx), or equivalently A′ sin kx + B′ cos kx. If E < V, write it as +κ², so ψ″ = +κ²ψ and ψ = C e(κx) + D e(−κx); the sines are gone. If E = V exactly, ψ″ = 0 and ψ is a straight line, A + Bx. For an electron ħ²/2m = 0.0381 eV nm², so 1.0 eV above the floor gives k = 5.12 nm⁻¹ and λ = 1.23 nm, while 1.0 eV below it gives κ = 5.12 nm⁻¹ and an amplitude that falls by a factor e in 0.195 nm. Two constants per region, always — and that count is the currency the rest of the method spends.

02

The sign of ψ″/ψ is the whole picture

Rearranged once more, ψ″/ψ = (2m/ħ²)(V − E): the curvature of ψ measured relative to ψ itself is fixed by how far E sits from the local floor. Where E > V the ratio is negative, so ψ curves back towards the axis — which is what oscillation is. Where E < V it is positive, so ψ curves away from the axis: a positive ψ that is falling must flatten and turn back up, and a solution that keeps falling has to be a pure decaying exponential. That is why nothing wiggles in a classically forbidden region, and why you can never write a sine there. The same expression gives the local wavelength λ = 2πħ/√(2m(E − V)): as the floor rises towards E the wave stretches out, and at a turning point where E = V the curvature vanishes and ψ has an inflection. Sketch the solution before computing it and most sign errors die on the page. A companion rule from the semiclassical limit runs the other way for amplitude — a slower particle lingers, so |ψ|² tends to be larger where E − V is smaller.

03

Where the two matching conditions come from

The rules are derived, not assumed. Integrate ψ″ = (2m/ħ²)(V − E)ψ from x₀ − ε to x₀ + ε. The left side gives ψ′(x₀ + ε) − ψ′(x₀ − ε); the right side is a bounded function integrated over an interval of width 2ε. If V is finite — however big the jump — the right side vanishes as ε → 0, so ψ′ is continuous. Continuity of ψ itself follows one step earlier: if ψ jumped, ψ′ would contain a delta and ψ″ the derivative of a δ, and no finite (V − E)ψ could match that. So the ordinary rules are ψ continuous everywhere, ψ′ continuous wherever V is bounded. The derivation also tells you when they fail. Against an infinite wall the bound is gone and only ψ = 0 survives, with ψ′ unconstrained. Across V = −α δ(x) the right-hand integral does not vanish; it leaves ψ′(0⁺) − ψ′(0⁻) = −(2mα/ħ²)ψ(0), a fixed kink whose size is the strength of the δ.

04

Divide, don't solve twice

Each boundary hands you two homogeneous equations in the region constants. Solving them as a pair drags the overall normalisation through the algebra for no reason, so divide one by the other and match the logarithmic derivative ψ′/ψ instead. It has dimension 1/length, it is unchanged if you rescale ψ, and it turns two equations into one condition plus a leftover amplitude ratio you can fix at the end. Take an infinite wall at x = 0, a flat floor of width a, and a step of height V₀ beyond it. Then ψI = A sin kx and ψII = B e(−κ(x−a)) are already continuous once B = A sin ka, and equating ψ′/ψ at x = a gives k cot ka = −κ, one transcendental equation in E with no amplitudes in sight. Symmetry buys the same saving twice over: if V(−x) = V(x), every non-degenerate eigenstate is even or odd, so you solve on half the axis and match at one boundary instead of two.

05

Count the constants and you know whether E is free

With N regions you have 2N constants plus one unknown energy. Each of the N − 1 internal boundaries supplies two equations, so 2(N − 1) in all. For a bound state the two outer regions run to infinity and normalisability kills the growing exponential in each, removing two constants and leaving 2N − 2 unknown coefficients against 2N − 2 homogeneous equations. A homogeneous system has only the trivial solution unless its determinant vanishes, and that determinant depends on E: the discrete spectrum is exactly the set of energies where it does, and the one remaining freedom is fixed by normalisation. A scattering state is counted differently. On one side you set the incoming amplitude by hand — you choose how hard to throw the particle — so no eigenvalue condition survives and E is continuous. A three-region barrier has N = 3: six constants, four matching equations, one incoming amplitude set to 1 and no wave arriving from the right, leaving exactly four unknowns for four equations at every E.

06

What the staircase costs

The model is a caricature, and it is worth naming the price. Real potentials are smooth: the image-charge barrier at a metal surface, or the band edge at a semiconductor heterojunction, changes over a length d of a few tenths of a nanometre rather than jumping. Treating it as a sharp step is safe only when d is small compared with the local de Broglie wavelength; for a 1 eV electron λ = 1.23 nm, so d = 0.3 nm is already marginal and the true reflection comes out smaller than the step formula predicts. Push the other way — V varying slowly over many wavelengths — and reflection becomes exponentially weak, which is the WKB limit. The remaining assumptions are structural: one dimension, one particle, a static V with no explicit time dependence, no spin, and non-relativistic kinematics. One rule genuinely changes in a solid. Where the effective mass differs across the junction, matching ψ′ is wrong; continuity of (1/m*)ψ′ is what conserves probability flux.

02

Change one variable at a time

Make the relationship visible.

Interactive model
2.00 eV
6.0 eV

Hold V₀ at 6.0 eV and walk E up from 0.40 eV: the two slope arrows scissor together and the ψ′ mismatch passes through zero near E = 1.10 eV — the one bound state this well holds. Past 1.5 eV, ψ turns over before the step and the decaying tail flips sign.

Interactive physics modelHard wall at x = 0, a flat floor to x = 0.50 nm, then a step to V₀ = 6.0 eV. Inside, E = 2.00 eV sits above V, so ψ oscillates with k = 7.24 nm⁻¹; beyond the step E is below V₀, so ψ decays with κ = 10.25 nm⁻¹. ψ is matched at the step by construction; the two arrows give the slope from each side, and they agree only at a bound state.V₀ = 6.0 eVE = 2.00 eVE > V k = 7.24 nm⁻¹E < V₀ κ = 10.25 nm⁻¹ψ(0) = 0 at the wallψ′ mismatch = −11.16 nm⁻¹

k (E > V region)7.24 nm⁻¹

κ (E < V₀ region)10.25 nm⁻¹

ψ′ MISMATCH AT STEP-11.16 nm⁻¹

PENETRATION 1/κ0.098 nm

Live interpretationk (E > V region): 7.24 nm⁻¹. κ (E < V₀ region): 10.25 nm⁻¹. ψ′ MISMATCH AT STEP: −11.16 nm⁻¹. PENETRATION 1/κ: 0.098 nm

03

Catch the common trap

Explain before calculating.

An electron of energy E tunnels through a rectangular barrier of height V₀ > E and finite width L. Which form should you write for ψ inside the barrier, and why?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron moves through a region where V = 0 with energy 2.0 eV, then meets a flat region where V = 5.0 eV that extends to infinity. Write the form of ψ in each region, and give the wavelength on the left and the penetration depth on the right. Use ħ²/2m = 0.0381 eV nm².
  1. Region I has E − V = 2.0 eV > 0, so ψ″ = −k²ψ and the solution oscillates: ψI = A e(ikx) + B e(−ikx).
  2. k = √((E − V)/0.0381) nm⁻¹ = √52.5 = 7.24 nm⁻¹, so λ = 2π/k = 6.283/7.244 = 0.867 nm.
  3. Region II has V − E = 3.0 eV > 0, so ψ″ = +κ²ψ and the solution is a pair of real exponentials, ψII = C e(κx) + D e(−κx).
  4. κ = √(3.0/0.0381) = √78.7 = 8.87 nm⁻¹. This region runs to infinity, so C = 0 or the state cannot be normalised, and the penetration depth is 1/κ = 0.113 nm.
  5. The two pieces still have to meet: continuity of ψ and of ψ′ at the step fixes D and the ratio B/A. The height 5.0 eV enters nowhere else — only through κ.

AnswerψI = A e(ikx) + B e(−ikx) with k = 7.24 nm⁻¹ and λ = 0.867 nm; ψII = D e(−κx) with κ = 8.87 nm⁻¹, falling to 1/e of its edge value in 0.113 nm.

MediumThe same electron, now at E = 3.0 eV, arrives from the left at a step up to V₀ = 8.0 eV located at x = 0. Write ψI = e(ikx) + r e(−ikx) and ψII = t e(−κx), impose the matching conditions, and find |r|² and |t|².
  1. k = √(3.0/0.0381) = 8.87 nm⁻¹ and κ = √(5.0/0.0381) = 11.45 nm⁻¹. Note k² + κ² = V₀/0.0381 = 210 nm⁻², a fixed number for this step.
  2. Continuity of ψ at x = 0: 1 + r = t.
  3. Continuity of ψ′ at x = 0: ik(1 − r) = −κt.
  4. Substitute t = 1 + r: ik − ikr = −κ − κr, so r(ik − κ) = ik + κ and r = (κ + ik)/(ik − κ).
  5. Numerator and denominator have the same modulus √(k² + κ²), so |r|² = 1 — everything is reflected, as it must be when no flux can escape into a region where ψ merely decays.
  6. t = 1 + r = 2ik/(ik − κ), so |t|² = 4k²/(k² + κ²) = 4E/V₀ = 4(3.0)/8.0 = 1.50.

Answer|r|² = 1 and |t|² = 4E/V₀ = 1.50. The probability density at the wall is half again the incident value, yet the reflected flux is total; the wave leaks 1/κ = 0.087 nm into the step and carries no current there.

HardAn electron is confined between infinite walls at x = 0 and x = 0.80 nm. Inside, V = 0 for 0 < x < 0.40 nm and V = 5.0 eV for 0.40 nm < x < 0.80 nm. Find the ground-state energy to three significant figures and check it against the two rigid-box limits.
  1. Set a = 0.40 nm. The walls force ψ(0) = ψ(2a) = 0, so ψI = A sin kx with k = √(E/0.0381) nm⁻¹, and ψII = B sinh(κ(2a − x)) with κ = √((5.0 − E)/0.0381) nm⁻¹ — the sinh is the combination of e(±κx) that vanishes at the far wall.
  2. Match at x = a. Continuity gives A sin ka = B sinh κa; the derivatives give A k cos ka = −B κ cosh κa. Divide to kill both amplitudes: k cot ka = −κ coth κa.
  3. Bracket the root. At E = 1.00 eV: k = 5.12 nm⁻¹, ka = 2.049, left side −2.66; κ = 10.25 nm⁻¹, right side −10.25, so the left side is the larger. At E = 2.00 eV: ka = 2.898, left side −29.1; right side −8.89, so the ordering has flipped and a root lies between.
  4. Bisect to E = 1.550 eV: k = 6.378 nm⁻¹, ka = 2.5511 rad, cot ka = −1.4925, left side −9.514. κ = 9.515 nm⁻¹, κa = 3.806, coth κa = 1.0010, right side −9.524. Matched to about 0.1%.
  5. Sanity-check the limits. A rigid box of the full 0.80 nm gives E₁ = 0.0381 π²/0.64 = 0.588 eV; one of only 0.40 nm gives 2.35 eV. The answer sits between them: the 5.0 eV step is high enough to push the electron mostly into the left half, yet κa = 3.8 still lets it leak about 1/κ = 0.105 nm past the wall.

AnswerE₁ ≈ 1.55 eV, bracketed by the 0.588 eV of an 0.80 nm rigid box and the 2.35 eV of a 0.40 nm one. Only matched energies exist; at any other E the two pieces meet with a kink and cannot be an eigenstate.