The well, the boundary conditions, and what they cost
Infinite walls pin ψ to zero but free its slope.
Every result of the next two weeks, from the quantised energies to the sine basis and its phases, is squeezed out of two boundary conditions. This lesson earns them: where ψ(0) = ψ(L) = 0 comes from, which rule about ψ′ the infinite wall suspends, and how tall a real wall must be before the idealisation deserves trust.
Build the model
The walls, not the equation, do all of the confining.
1The big idea
The infinite square well sets V = 0 for 0 < x < L and V = ∞ everywhere else. Between the walls the equation is the free one, ψ″ = −k²ψ, which on its own accepts every positive energy, so all of the confinement has to enter through conditions at x = 0 and x = L. The honest way to find them is to give the walls a finite height V₀ and take the limit. Beyond a finite wall ψ decays as e(−κ(x−L)), with κ = √(2m(V₀ − E))/ħ, and matching ψ and ψ′ there gives ψ(L) = −ψ′(L⁻)/κ: at a fixed inside slope the wall value falls as (V₀ − E)(−1/2), and the probability beyond it faster still. In the limit ψ is zero outside, continuity carries that zero to the walls, and ψ(0) = ψ(L) = 0. What the limit costs is the continuity of ψ′. The slope goes from ψ′(L⁻) to zero across a tail whose length has shrunk to nothing, because (2m/ħ²)Vψ, infinity times zero, still integrates to the finite κψ(L) = −ψ′(L⁻). That kink belongs to the idealisation, not to any material. Real walls are finite, of order 10 eV at a metal surface and a few tenths of an eV in a semiconductor quantum well, and the model is only as good as the tail length 1/κ is small beside L: for an electron in a 1.00 nm well behind 1.0 eV walls, the true ground energy is about half the infinite-well value.
2In one sentence
The infinite square well is the potential V = 0 for 0 < x < L and V = ∞ elsewhere; its states obey ψ(0) = ψ(L) = 0, with ψ continuous at each wall and ψ′ free to jump there.
Example. Where V₀ − E = 5.00 eV, an electron's ψ reaches past a wall over δ = ħ/√(2m(V₀ − E)) = 0.0873 nm; where V₀ − E = 500 eV, over 0.00873 nm. The condition ψ(L) = 0 is the limit in which that length reaches zero.
3Inside the well, the equation alone permits every energy
Between the walls V = 0, so the time-independent equation is the free one of UQMI-05: −(ħ²/2m)ψ″ = Eψ, or ψ″ = −k²ψ with k = √(2mE)/ħ. Its general solution A sin kx + B cos kx exists at every E > 0, and nothing in the region 0 < x < L tells it how wide the region is: an electron with k = 4.00 nm⁻¹, at E = 0.610 eV, satisfies the equation inside a 1.00 nm well exactly as well as inside a 1.00 m one. Everything that will distinguish the well from free space, the discrete ladder of levels, the zero-point energy, the sine basis, must therefore enter through what happens at x = 0 and x = L. A second-order equation carries two free constants and needs two conditions to pin them, and the two walls are where they come from. The work of this lesson is to derive those conditions rather than assert them, and to name the rule about ψ that they suspend.
4Beyond an infinite wall, ψ has to vanish
Give the wall a finite height V₀ > E first. For x > L the equation reads ψ″ = κ²ψ, with κ = √(2m(V₀ − E))/ħ; of its solutions e(±κx) the growing one cannot be normalised, so ψ = ψ(L) e(−κ(x−L)), a tail of length δ = 1/κ. The region is classically forbidden, yet ψ is not zero there: forbidden means decaying, not absent. What makes ψ vanish is V₀ → ∞, which sends δ to zero. An energy argument reaches the same place without solving anything. The potential energy stored beyond the wall is at least V₀ times the probability found there, so if even 1% of the probability sat behind a 1.00 MeV wall, ⟨V⟩ would be at least 10.0 keV; with V₀ infinite, any nonzero share makes the energy infinite. A state of finite energy puts zero probability where V is infinite, and a continuous ψ whose |ψ|² integrates to zero over a region is zero at every point of it. The classical correspondence holds as well: δ scales as ħ/√m, so 1.00 eV below the top of a wall an electron's tail is 0.195 nm long, a proton's 4.56 pm and a 1.0 μg grain's 6 × 10⁻²¹ m.
5Continuity carries the zero to the wall
ψ must be continuous everywhere, walls included. A jump would put a delta function in ψ′ and its derivative in ψ″, which nothing in the Schrödinger equation, not even an infinite step times ψ, can balance (UQMI-01.09). So the zero outside meets the inside solution at each wall, and the conditions are ψ(0) = 0 and ψ(L) = 0. The finite wall shows how the zero arrives. There V is still finite, so ψ and ψ′ both match at x = L, and since the tail has ψ′ = −κψ, the inside slope obeys ψ′(L⁻) = −κψ(L), or ψ(L) = −ψ′(L⁻)/κ: the wall value is minus the inside slope times the tail length. With ψ′(L⁻) = −2.00 nm(−3/2) and V₀ − E = 3.00 eV, κ = 8.87 nm⁻¹ and ψ(L) = +0.225 nm(−1/2). Raise the wall at the same slope and ψ(L) falls as (V₀ − E)(−1/2), while the probability beyond it, |ψ(L)|²/(2κ), carries one more factor of 1/κ. ψ(L) = 0 is the V₀ → ∞ limit of this relation between value and slope, not a separate postulate. The figure runs the limit backwards: choose how nearly the inside wave fits half a wavelength into L, and it reports the wall that makes that the ground state.
6The slope is released, and its jump has a definite size
Integrate the equation across the wall, from L − ε to L + ε: ψ′(L + ε) − ψ′(L − ε) = (2m/ħ²)∫(V − E)ψ dx. Where V is finite the integrand is bounded, the integral vanishes as ε → 0, and ψ′ is continuous; the finite well of UQMI-07 relies on exactly this. At an infinite wall the integrand is V, growing without limit, times ψ, shrinking to zero, and the product need not vanish. Evaluate it on the finite wall's tail: (2m/ħ²)(V₀ − E) × ψ(L)/κ = κψ(L) = −ψ′(L⁻), the same finite number at every wall height. In the limit the slope goes from ψ′(L⁻) to zero across a distance δ → 0: a corner, with a delta function in ψ″. The ground state of UQMI-06.03, √(2/L) sin(πx/L), reaches x = L with slope −(π/L)√(2/L), −4.44 nm(−3/2) for L = 1.00 nm, and leaves it with slope zero. The classical counterpart is a string clamped to rigid posts: its displacement is pinned at each end while the slope there is whatever the vibration makes it, and a post that yields slightly lets the end move, as a finite wall lets ψ(L) differ from zero.
7A nonzero value at the wall would cost unlimited energy
Turn the question round: could a state keep a fixed value c at the wall while V₀ grows? Its tail must still decay, say as c e(−q(x−L)), and it then carries kinetic plus potential energy ∫[(ħ²/2m)|ψ′|² + V₀|ψ|²]dx = |c|²[(ħ²/2m)(q/2) + V₀/(2q)], the kinetic part written in the |ψ′|² form that ⟨p²⟩/2m takes after integrating by parts. The best choice of q, worked through in the Hard example, leaves |c|²√((ħ²/2m)V₀), which grows as √V₀ without limit: for |c|² = 1.00 nm⁻¹, the uniform density of a 1.00 nm well, that is 0.617 eV behind a 10.0 eV wall and 6.17 eV behind a 1.00 keV one, against a ground energy of 0.376 eV. So ψ(L) = 0 is not a convention; it is the price of a finite energy. The true solution pays it by letting c fall as 1/κ, which makes the same tail energy vanish. The argument also sorts trial functions: a flat ψ = 1/√L across the well, the density of a classical particle bouncing between the walls, cannot be prepared in an infinite well at any finite energy. UQMI-06.06 will still expand it on the sine basis, but as mathematics, not as a state.
8No material supplies the infinity
The exact zeros, the kink and a ladder of levels that never ends all belong to the infinite wall, and nature supplies none. A metal surface holds its conduction electrons behind a barrier of order 10 eV above the band bottom; a semiconductor quantum well offers a few tenths of an eV. The idealisation is trustworthy when the tail is short beside the well, δ ≪ L, the same statement as V₀ − E ≫ ħ²/(2mL²), and its errors are then first order in δ/L. For an electron in a 1.00 nm well that first order is not small. Behind 5.0 eV walls δ = 0.090 nm, 0.83% of the probability lies outside, and the ground energy is 0.272 eV rather than the infinite-well 0.376 eV, 28% lower; even 100 eV walls leave it 7% low, and UQMI-07.08 traces the shift to a well that behaves as if wider by about δ at each wall. Two failures remain at any finite height. A real well holds a finite number of bound levels, ending below V₀, while the model's ladder climbs forever; and a real wall has finite thickness, so a particle near its top can tunnel through it or pass over it.
Key equations
What each relation says, and the units it is written in.
Boundary conditions of the infinite well
ψ(0) = 0 and ψ(L) = 0
One condition from each wall, two in all, as a second-order equation needs; UQMI-06.02 shows they permit a nonzero ψ only at special energies.
ψ in m(−1/2); L, the well width, in m. The slopes ψ′(0) and ψ′(L) are left free.
Tail beyond a finite wall
ψ(x) = ψ(L) e(−κ(x − L)) for x > L, with κ = √(2m(V₀ − E))/ħ
The finite step that the infinite wall idealises. The growing exponential is dropped because no multiple of it can be normalised.
κ in m⁻¹, V₀ and E in J, m in kg; δ = 1/κ is the tail length. For an electron ħ²/2m = 0.0381 eV nm².
Wall value set by matching ψ and ψ′
ψ(L) = −ψ′(L⁻)/κ
At fixed inside slope the wall value falls as (V₀ − E)(−1/2); ψ(L) = 0 is its V₀ → ∞ limit, not a separate postulate.
ψ′(L⁻) is the slope just inside the wall, in m(−3/2); dividing by κ in m⁻¹ leaves m(−1/2), the unit of ψ.
Slope change across a wall
ψ′(L + ε) − ψ′(L − ε) = (2m/ħ²) ∫ (V − E) ψ dx
Why ψ′ is continuous wherever V is finite, and why an infinite wall can change it by a finite amount out of infinity times zero.
The integral runs from L − ε to L + ε as ε → 0. Finite V: it vanishes. Infinite wall: it tends to κψ(L) = −ψ′(L⁻).
Cheapest tail for a fixed wall value c
Eₜₐᵢₗ ≥ |c|² √((ħ²/2m) V₀)
Grows as √V₀ without limit, so a state that keeps c ≠ 0 at an infinite wall would need infinite energy.
c = ψ(L) in m(−1/2), V₀ and Eₜₐᵢₗ in J. Equality holds for the tail c e(−κ₀(x − L)), κ₀ = √(2mV₀)/ħ.
When a wall may be treated as infinite
δ = ħ/√(2m(V₀ − E)) ≪ L
The model's errors are first order in δ/L: in that well the ground energy already sits 3.4% below the infinite-well value.
δ and L in m; equivalent to V₀ − E ≫ ħ²/(2mL²). Electron, 10.0 nm well, 5.0 eV walls: δ = 0.087 nm, δ/L = 0.0087.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag k from 0.85 π/L to 0.95 and 0.99 π/L and read the wall each needs in a 1.00 nm well: about 5 eV, 55 eV, then 1.5 keV, while the tails and circles shrink onto the dashed curve's zeros. Then halve L: the same picture needs walls four times higher.
Live results
- WALL HEIGHT V₀ NEEDED
- 4.99 eV
- TAIL LENGTH δ = 1/κ
- 0.090 nm
- WALL VALUE ψ(L) ÷ ψ(L/2)
- 0.233
- PROBABILITY OUTSIDE
- 0.8306 %
Catch the common trap
The mistake this lesson is built to prevent.
The rule from finite potentials, ψ and ψ′ both continuous, gets carried to the infinite wall, and since both are zero outside, students impose ψ(0) = 0 and ψ′(0) = 0 together. That leaves nothing: a second-order equation whose value and slope are both fixed at one point has exactly one solution, and with both zero it is ψ = 0 everywhere. The slope rule came from integrating the equation across a point, where a bounded V times ψ contributes nothing over a shrinking interval. At an infinite wall V is unbounded, the same integral tends to κψ(L) = −ψ′(L⁻), which is finite, and ψ′ is free to jump. Only ψ is continuous there. The kink this leaves does not make the infinite-well states wrong, only idealised, to exactly the degree that the infinite wall is.
Test it: what changes?
- Wall value ψ(L) falls as (V₀ − E)(−1/2)
- Tail length δ = 1/κ shrinks
- Probability beyond the wall falls as (V₀ − E)(−3/2)
- Slope integral κψ(L) stays equal to −ψ′(L⁻)
- Wall height needed rises fourfold
- Tail length δ halves
- Wall value ratio ψ(L) ÷ ψ(L/2) unchanged
- Probability outside the well unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is confined to an infinite well of width L = 0.600 nm. Three candidates each solve the Schrödinger equation inside: (i) A sin(2πx/L), (ii) A cos(πx/L) and (iii) A sin(πx/b) with b = 0.800 nm. Find the energy the inside equation assigns to each, then decide which are allowed states of the well.
View solution →
- Each is sin kx or cos kx, so ψ″ = −k²ψ and E = ħ²k²/2m, with ħ²/2m = 0.0381 eV nm² for an electron.
- (i): k = 2π/0.600 nm = 10.47 nm⁻¹, E = 4.18 eV. (ii): k = π/0.600 nm = 5.236 nm⁻¹, E = 1.04 eV. (iii): k = π/0.800 nm = 3.927 nm⁻¹, E = 0.588 eV. The inside equation accepts all three; only the walls can choose.
- (i): ψ(0) = A sin 0 = 0 and ψ(L) = A sin 2π = 0, so both conditions hold. Its slope at x = 0, 2πA/L = 10.5A nm⁻¹, is not zero, and at an infinite wall it need not be.
- (ii): ψ(0) = A cos 0 = A and ψ(L) = A cos π = −A. It fails at both walls: joined to ψ = 0 outside, it would jump at each one, and continuity forbids that.
- (iii): ψ(0) = 0, but ψ(L) = A sin(π × 0.600/0.800) = A sin(3π/4) = 0.707A. It fails at the right wall only; it would fit a well 0.800 nm wide.
Answer. Only (i) is a state of the 0.600 nm well, at 4.18 eV. (ii) fails at both walls and (iii) leaves 0.707A at x = L, although the inside equation accepts them at 1.04 eV and 0.588 eV.
MediumExample 2
An electron with E = 0.30 eV meets the right wall of a well, a step of height V₀, with slope ψ′(L⁻) = −4.44 nm(−3/2) just inside; treat that slope as fixed. For V₀ = 2.30 eV and V₀ = 200.3 eV find ψ(L), the probability Pₒᵤₜ beyond the wall, the potential energy stored there, V₀Pₒᵤₜ, and the slope integral (2m/ħ²)(V₀ − E)∫ψ dx over the tail. Which of these products of a large V₀ and a small ψ survive the limit?
View solution →
- κ = √((V₀ − E)/(ħ²/2m)) with ħ²/2m = 0.0381 eV nm². V₀ − E = 2.00 eV gives κ = 7.245 nm⁻¹ (δ = 0.138 nm); V₀ − E = 200 eV gives κ = 72.45 nm⁻¹ (δ = 0.0138 nm).
- Matching ψ and ψ′: ψ(L) = −ψ′(L⁻)/κ = 4.44/7.245 = 0.613 nm(−1/2) for the low wall and 0.0613 nm(−1/2) for the high one. Units: m(−3/2) over m⁻¹ is m(−1/2).
- Probability beyond: Pₒᵤₜ = |ψ(L)|²/(2κ) = 0.3755/14.49 = 2.59 × 10⁻² for the low wall and 2.59 × 10⁻⁵ for the high one.
- Stored potential energy: V₀Pₒᵤₜ = 2.30 eV × 2.59 × 10⁻² = 0.0596 eV, then 200.3 eV × 2.59 × 10⁻⁵ = 5.19 × 10⁻³ eV. V₀ rose 87-fold and the product fell about 11.5-fold: for V₀ ≫ E it tends to (ħ²/2m)|ψ′(L⁻)|²/(2κ), which goes to zero as 1/κ.
- Slope integral: (2m/ħ²)(V₀ − E) × ψ(L)/κ = κψ(L) = 7.245 × 0.613 = 4.44 nm(−3/2), and 72.45 × 0.0613 = 4.44 nm(−3/2) again: always equal to −ψ′(L⁻), whatever the wall height.
- The two infinity-times-zero products part company. The tail's energy vanishes, which is why the infinite wall costs nothing in energy; the slope integral stays at 4.44 nm(−3/2), which is exactly the jump in ψ′ that the kink records.
Answer. ψ(L) = 0.613 then 0.0613 nm(−1/2); Pₒᵤₜ = 2.59 × 10⁻² then 2.59 × 10⁻⁵; V₀Pₒᵤₜ = 0.0596 eV then 5.19 × 10⁻³ eV, heading to zero; the slope integral is 4.44 nm(−3/2) at both heights, the finite kink of the infinite wall.
HardExample 3
A classical particle bouncing between walls has uniform density 1/L. Suppose an electron in a 2.00 nm well kept that density up to a wall, |c|² = |ψ(L)|² = 0.500 nm⁻¹, then continued past a wall of height V₀ as c e(−q(x − L)). Find the least energy this tail can carry at V₀ = 100 eV, 10.0 keV and 1.00 MeV. Compare a state meeting the wall with slope |ψ′(L⁻)| = 1.57 nm(−3/2) and the matched value c = |ψ′(L⁻)|/κ₀, with E negligible beside V₀.
View solution →
- The tail's energy is ∫[(ħ²/2m)|ψ′|² + V₀|ψ|²]dx over x > L, the kinetic part in its integrated-by-parts form. With ∫|ψ|²dx = |c|²/(2q) and ∫|ψ′|²dx = q|c|²/2, Eₜₐᵢₗ(q) = |c|²[(ħ²/2m)(q/2) + V₀/(2q)].
- Minimise: dEₜₐᵢₗ/dq = |c|²[(ħ²/2m)/2 − V₀/(2q²)] = 0 at q = κ₀ = √(2mV₀)/ħ. Substituting gives Eₘᵢₙ = |c|²(ħ²/2m)κ₀ = |c|²√((ħ²/2m)V₀). Dimensions: m⁻¹ × √(J m² × J) = J.
- Numbers, with ħ²/2m = 0.0381 eV nm²: √(0.0381 eV nm² × 100 eV) = 1.952 eV nm, so Eₘᵢₙ = 0.500 nm⁻¹ × 1.952 eV nm = 0.976 eV at 100 eV; likewise 9.76 eV at 10.0 keV and 97.6 eV at 1.00 MeV. Each hundredfold rise in V₀ multiplies the cost by ten.
- Normalisation is not the culprit: at 100 eV, κ₀ = 51.2 nm⁻¹ and the tail holds |c|²/(2κ₀) = 0.0049 of the probability, less at higher walls. Yet the cost at 100 eV is already ten times the 0.0940 eV ground energy of a 2.00 nm infinite well.
- Matched state: c = 1.57/51.23 = 0.0306 nm(−1/2) at 100 eV, and Eₜₐᵢₗ = (ħ²/2m)κ₀c² = (ħ²/2m)|ψ′(L⁻)|²/κ₀ = 0.0381 × 2.465/51.23 = 1.83 × 10⁻³ eV; at 10.0 keV, 1.83 × 10⁻⁴ eV. It falls as V₀(−1/2).
- A second wall doubles every figure. A wall value held fixed costs energy growing as √V₀; one falling as 1/κ₀ costs energy shrinking as 1/√V₀. Only ψ(L) → 0 survives V₀ → ∞ at finite energy.
Answer. Eₘᵢₙ = |c|²√((ħ²/2m)V₀) = 0.976 eV, 9.76 eV and 97.6 eV per wall, without limit; the matched state's tail costs 1.83 meV at 100 eV and 0.183 meV at 10.0 keV, falling toward zero.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Impose ψ(0) = ψ(L) = 0 and leave the slope free.
Only ψ is continuous at an infinite wall: ψ(L) = 0 is the V₀ → ∞ limit of ψ(L) = −ψ′(L⁻)/κ, and ψ′ may jump. Trust the model only while the tail δ = ħ/√(2m(V₀ − E)) is short beside L; its errors are first order in δ/L.
Next: Energy quantisation: the condition for a nontrivial solution →