How boundary conditions quantise the energy
Every E solves the equation; boundaries pick the levels
Most students arrive thinking the Schrödinger equation is what makes energy come in steps. It is not: it has solutions at every E. This lesson finds the real source, the conditions at walls and at infinity, turns it into a count you can run on any potential, and uses the same count to say where discrete levels give way to a continuum.
Build the model
The equation accepts every energy. The walls and infinity choose.
1The big idea
The time-independent Schrödinger equation, −(ħ²/2m)ψ″ + Vψ = Eψ, is a second-order linear ODE, and like any such equation it has two independent solutions at every E. Nothing in it forbids an energy. What forbids energies is what the physics demands at the edges: ψ = 0 at a hard wall, and, wherever V stays above E out to infinity, the decaying exponential rather than the growing one, because only then is ∫|ψ|² dx finite. Now count. The solution carries two constants, but the equation and all these conditions are homogeneous, so the overall scale stays free and is spent later on normalisation; one ratio is all that is left to meet the conditions. Two constrained edges against one ratio leave one condition over, and it can only be met by adjusting E, so it becomes an equation for E with isolated roots. For walls at 0 and L it reads sin kL = 0, so kL = nπ and Eₙ = n²π²ħ²/(2mL²). The same count shows where discreteness stops. If E lies above V far out on one side, the solutions there oscillate and stay bounded whatever their mix, so that edge demands nothing, no condition is left over, and every E is allowed: a continuum of scattering states, bounded but not normalisable one by one. Discrete levels can therefore sit only between the minimum of V and the lower of its two asymptotic values. A free particle has none, a harmonic oscillator has nothing else, and hydrogen has both.
2In one sentence
Energy is quantised when the time-independent Schrödinger equation, which has solutions at every E, has solutions that also meet the boundary conditions and stay normalisable only at isolated values Eₙ.
Example. An electron between hard walls 0.50 nm apart: sin kx vanishes at the far wall only when kL = nπ, so E₁ = 1.50 eV and E₂ = 6.02 eV. At 3.00 eV the equation still has a solution, but at the far wall it reaches −0.96 of its amplitude instead of zero.
3The equation solves at every energy
Take the region between two hard walls, where V = 0. The time-independent equation reduces to ψ″ = −k²ψ with k = √(2mE)/ħ, and its general solution A sin kx + B cos kx exists for every E > 0 you care to pick. Read it as an initial-value problem: choose ψ(0) and ψ′(0), integrate, and the equation hands back one smooth curve. For an electron at E = 1.00 eV, k = 5.12 nm⁻¹; start with ψ(0) = 0 and the solution is A sin kx, which at x = 1.00 nm equals A sin(5.12) = −0.92A. Nothing went wrong in the equation. The solution simply fails to vanish at the second wall. E = 0 and E < 0 give solutions too, a straight line A + Bx and a mix of e(±κx). So the list of allowed energies cannot be read off the differential equation. It has to come from extra conditions the equation does not contain.
4Count constants against conditions
The solution has two constants, A and B, and both hard-wall conditions are homogeneous: ψ(0) = 0 reads B = 0, and ψ(L) = 0 reads A sin kL + B cos kL = 0. Written as a matrix acting on (A, B), the rows are (0, 1) and (sin kL, cos kL). A homogeneous linear system has a nonzero solution only when its determinant vanishes, and this determinant is −sin kL, a function of E alone. That is the whole mechanism in one line. Homogeneity also means the walls never fix the overall scale of (A, B); ∫|ψ|² dx = 1 does that later. So the usable freedom is one ratio, B/A, while each edge that constrains ψ costs one condition. One ratio against two conditions leaves one condition over, and the only thing left to adjust is E. At 1.00 eV in a 1.00 nm box the determinant is −sin(5.12) = +0.92, not zero, so that energy is out.
5Solve the box, then check what it claims
With B = 0, ψ(L) = A sin kL = 0 and A ≠ 0 force kL = nπ. Exclude n = 0, which makes ψ ≡ 0 and describes no particle, and drop negative n, which only flips the sign of the same state. Then Eₙ = ħ²kₙ²/2m = n²π²ħ²/(2mL²), with ψₙ = √(2/L) sin(nπx/L). Normalisation: ∫₀ᴸ (2/L) sin²(nπx/L) dx = (2/L)(L/2) = 1. Dimensions: ħ²/(mL²) carries J² s² over kg m², which is J because J = kg m² s⁻². Numbers: for an electron with L = 1.00 nm, E₁ = 0.376 eV, E₂ = 1.50 eV and E₃ = 3.38 eV. The floor: a line A + Bx that vanishes at 0 and at L is identically zero, and so is a sinh, so no level lies at or below V = 0. The lowest level sits above the bottom of the potential, a zero-point energy that a classical particle at rest would not have.
6A decaying tail does the work of a wall
Replace the right wall by a finite step: V = V₀ for x > L, with E < V₀. Beyond L the solutions are e(−κx) and e(+κx) with κ = √(2m(V₀ − E))/ħ, and the equation prefers neither. Normalisability does: ∫|ψ|² dx converges only if the growing piece is absent. That is one condition, the job ψ(L) = 0 used to do, so the count is unchanged: ψ(0) = 0 and no growth, against one free ratio, while continuity of ψ and ψ′ at L only fixes the other constants. Picture it as shooting. Start at the left wall, carry the solution across the well, and at x = L it generally contains some e(κx); only special energies cancel it. The cancellation is delicate: for V₀ − E = 1.00 eV an electron has κ = 5.12 nm⁻¹, so any stray growing term is multiplied by e(5.12) ≈ 167 every nanometre. Let V₀ → ∞ and κ → ∞ with it: the tail shrinks to nothing and the hard wall returns.
7Above the asymptote nothing is left to select E
Now raise E above V₀ in the same step problem. Beyond L the solutions become cos qx and sin qx with q = √(2m(E − V₀))/ħ. Both stay bounded, so neither is worse than the other at infinity, and that edge imposes no condition at all. The count is one condition, ψ(0) = 0, against one ratio, and it is met at every E: those energies form a continuum. A bounded wave that never dies away has an infinite ∫|ψ|² dx, so these states are not normalisable one at a time, but they are acceptable, and superpositions of them make normalisable wave packets. The general rule follows. Discrete levels can lie only where Vₘᵢₙ < E < min[V(−∞), V(+∞)]. Below Vₘᵢₙ, ψ″ has the sign of ψ everywhere, so ψ bends away from the axis and cannot vanish at both ends. A free particle therefore has no discrete levels, a harmonic oscillator, whose V grows without limit, has only discrete ones, and hydrogen has discrete levels below zero and a continuum above.
8Classical correspondence and the node count
Two limits show discreteness fading exactly where it should. Widen the box: Eₙ ∝ 1/L², so as L → ∞ the levels crowd together and the free particle's continuum returns. Climb the ladder: the gap to the next level, relative to the level itself, is (Eₙ₊₁ − Eₙ)/Eₙ = (2n + 1)/n², which is 3 at n = 1, 0.21 at n = 10 and 0.020 at n = 100. A 1.0 g bead sliding at 1.0 cm s⁻¹ between walls 1.0 cm apart has E = 5.0 × 10⁻⁸ J against E₁ = 5.5 × 10⁻⁶¹ J, so n ≈ 3.0 × 10²⁶ and the fractional spacing is 6.6 × 10⁻²⁷: no experiment will ever see the steps, and the bead's continuous classical energy is recovered. The node count sets the order: ψₙ has n − 1 interior nodes, and each extra half-wavelength costs more curvature and so more kinetic energy. That ordering by nodes holds for every one-dimensional well, not only the box.
Key equations
What each relation says, and the units it is written in.
Solutions at any energy in a flat region
ψ(x) = A sin kx + B cos kx, with k = √(2mE)/ħ
A solution exists for every E you name, with two free constants. The equation on its own never restricts E; the restriction has to come from outside it.
V = 0 and E > 0; m in kg, E in J, ħ = 1.055 × 10⁻³⁴ J s, so k is in m⁻¹; A and B in m(−1/2).
Hard walls at x = 0 and x = L
sin kL = 0 ⟹ kₙ = nπ/L, n = 1, 2, 3, …
The second wall finds no free constant left to adjust, so it constrains k, and through k the energy. n counts half-wavelengths across the box: λₙ = 2L/n.
ψ(0) = 0 sets B = 0, then ψ(L) = 0 with A ≠ 0 needs sin kL = 0. kₙ in m⁻¹, L in m; n = 0 gives ψ ≡ 0.
Infinite-well energies
Eₙ = n²π²ħ²/(2mL²)
Levels grow as n² and as 1/L²: halve the box and every level quadruples. E₁ > 0, so the confined particle is never at rest.
Eₙ in J for m in kg and L in m; an electron in a 1.00 nm well has E₁ = 0.376 eV and E₂ = 1.50 eV.
Normalised box eigenfunctions
ψₙ(x) = √(2/L) sin(nπx/L) for 0 ≤ x ≤ L, and 0 outside
Normalisation fixes |A| = √(2/L) and never touches E: the scale is exactly the one constant the walls leave free.
ψₙ in m(−1/2), so |ψₙ|² is a probability per metre; ψₙ has n − 1 nodes between the walls.
A forbidden end acts as a wall
ψ = C e(−κx) + D e(κx), κ = √(2m(V₀ − E))/ħnormalisable only if D = 0
One condition per classically forbidden end, the job ψ = 0 does at a hard wall. Letting V₀ → ∞ sends κ → ∞ and recovers the wall.
Beyond the last turning point, where V₀ > E; κ in m⁻¹, and 1/κ is the penetration depth in m.
Where discrete levels can sit
Vₘᵢₙ < E < min[V(−∞), V(+∞)]
Below Vₘᵢₙ, ψ bends away from the axis everywhere and cannot vanish at both ends; above the lower asymptote one end imposes nothing and E runs continuously.
All in J or eV; Vₘᵢₙ is the lowest value V(x) takes anywhere, the upper bound the lower of its two limits.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At z₀ = 4, even parity, step E down from 0.15 V₀: the tails dive below the axis down to 0.100 and shoot upward from 0.095, so a level sits between (0.098 V₀). Switch to odd and find the flip near 0.38 V₀. Then raise E past V₀: the tails oscillate at every setting.
Live results
- E ÷ E₁ (HARD WALLS AT ±a)
- 0.97
- RIM V₀ ÷ E₁ (HARD WALLS)
- 6.48
- κa OUTSIDE (0 ABOVE RIM)
- 3.69
- BOUND STATES ⌈2z₀/π⌉
- 3
Catch the common trap
The mistake this lesson is built to prevent.
The tempting reading is that the Schrödinger equation only has solutions at the allowed energies, so quantisation is built into the equation. It is not. At any E the equation has two independent solutions, and a numerical integrator started at x = 0 will happily return one at 1.00 eV in a 1.00 nm box; it just reaches −0.92 of its amplitude at the far wall instead of zero. The Eₙ are the energies at which a solution also meets the boundary conditions and stays normalisable. The distinction matters as soon as the walls go. A free particle obeys the same equation and has no discrete levels, and a finite well of depth V₀ has only finitely many, all below V₀, with every energy above V₀ allowed. Change the conditions and the spectrum changes while the equation stays exactly the same.
Test it: what changes?
- Tails beyond ±a oscillate instead of running away
- Solution stays bounded at every E
- κa outside drops to zero
- No edge condition is left to select E
- Every level Eₙ falls as 1/L²
- Levels crowd closer together
- ψₙ keeps its n − 1 nodes
- E₁ stays above zero
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is confined between hard walls 0.80 nm apart, with V = 0 between them. Find its three lowest allowed energies in eV. A classmate proposes that 1.00 eV is also allowed; start the solution at the left wall and show what goes wrong at the right one.
View solution →
- Name the problem before solving it: V = 0 on 0 < x < L with L = 0.80 nm, and ψ(0) = ψ(L) = 0. Inside, ψ = A sin kx + B cos kx, and ψ(0) = 0 gives B = 0.
- ψ(L) = A sin kL = 0 with A ≠ 0 needs kL = nπ, so Eₙ = n²π²ħ²/(2mL²) = n² × π² × (1.055 × 10⁻³⁴ J s)² / (2 × 9.109 × 10⁻³¹ kg × (0.80 × 10⁻⁹ m)²) = n² × 9.42 × 10⁻²⁰ J.
- Divide by 1.602 × 10⁻¹⁹ J per eV: E₁ = 0.588 eV, E₂ = 4E₁ = 2.35 eV, E₃ = 9E₁ = 5.29 eV. Check the scaling against the 1.00 nm box: 0.376 eV ÷ 0.80² ≈ 0.588 eV.
- At 1.00 eV, k = √(2mE)/ħ = 5.12 nm⁻¹ and kL = 4.10 rad. The solution A sin kx exists, but at the far wall it equals A sin(4.10) = −0.82A. Since kL/π = 1.30, the wave puts 1.30 half-wavelengths into the box and cannot vanish at both ends.
Answer. E₁ = 0.588 eV, E₂ = 2.35 eV, E₃ = 5.29 eV. At 1.00 eV the solution reaches −0.82A at x = L instead of zero, so 1.00 eV is not allowed.
MediumExample 2
An electron sees V = ∞ for x < 0, V = 0 for 0 < x < 1.20 nm and V = 0.50 eV for x > 1.20 nm. Take ψ = sin kx inside, since the amplitude is irrelevant. Carry the solution across x = L at E = 0.30 eV and at E = 0.70 eV, and decide in each case whether that energy belongs to the spectrum.
View solution →
- Conditions first: ψ(0) = 0 is built into sin kx; ψ and ψ′ are continuous at L because V is finite there; and for E < V₀ the solution beyond L must not grow. The constant √(2m × 1 eV)/ħ = 5.123 nm⁻¹ turns energies in eV into wavenumbers.
- At 0.30 eV: k = 5.123 × √0.30 = 2.806 nm⁻¹ and κ = 5.123 × √0.20 = 2.291 nm⁻¹. kL = 3.367 rad, so ψ(L) = sin 3.367 = −0.224 and ψ′(L) = k cos kL = 2.806 × (−0.975) = −2.735 nm⁻¹.
- Beyond L write ψ = C e(−κ(x−L)) + D e(κ(x−L)). Matching gives C + D = ψ(L) and κ(D − C) = ψ′(L), so D = ½[ψ(L) + ψ′(L)/κ] = ½(−0.224 − 1.194) = −0.709 and C = +0.485.
- D ≠ 0, and the growing term takes over: each extra nanometre multiplies it by e(2.291) = 9.9. ψ runs off to −∞, ∫|ψ|² dx diverges, and 0.30 eV is rejected.
- At 0.70 eV: k = 4.286 nm⁻¹, and now q = √(2m(E − V₀))/ħ = 2.291 nm⁻¹. kL = 5.144 rad gives ψ(L) = −0.908 and ψ′(L)/q = (k/q) cos kL = 1.871 × 0.418 = 0.782, so beyond L ψ = −0.908 cos q(x−L) + 0.782 sin q(x−L).
- That is an oscillation of amplitude √(0.908² + 0.782²) = 1.20 that never grows. No condition from the right is left to fail, so 0.70 eV is allowed, and so is every E above 0.50 eV: this is the continuum.
Answer. At 0.30 eV the growing coefficient is D = −0.709 per unit inside amplitude, so 0.30 eV is not allowed. At 0.70 eV the outside wave oscillates with amplitude 1.20 and the energy is allowed, as is every E above 0.50 eV.
HardExample 3
Keep a hard wall at x = 0 and a flat floor V = 0 out to x = L, with V = V₀ beyond. (a) Show that a bound state exists only if V₀ > π²ħ²/(8mL²). (b) For an electron with L = 1.00 nm and V₀ = 0.30 eV, find the bound-state energy to three significant figures and show that there is only one.
View solution →
- Inside ψ = sin kx; beyond L only e(−κ(x−L)) survives. Matching ψ′/ψ at L gives k cot kL = −κ. With u = kL and v = κL this reads u cot u = −v, and u² + v² = z₀², where z₀ = L√(2mV₀)/ħ is fixed by the well.
- Since v > 0, a root needs cot u < 0, so u > π/2 on the first branch, while v² = z₀² − u² ≥ 0 caps u at z₀. A root exists only if z₀ > π/2, which is V₀ > π²ħ²/(8mL²), a quarter of the hard-wall E₁: 0.376/4 = 0.094 eV for L = 1.00 nm. As z₀ falls to π/2 the root slides to u = π/2, where κ = 0: ψ is flat beyond L, not normalisable, and the level has merged into the continuum.
- For V₀ = 0.30 eV, z₀ = 5.123 × √0.30 = 2.806, well above π/2 = 1.571. Define f(u) = u cot u + √(z₀² − u²), which vanishes at the level. f(π/2) = +2.33 and f(2.30) = −0.45, so the root lies between.
- Narrow it: f(2.20) = +0.140, f(2.25) = −0.140, f(2.22) = +0.031 and f(2.23) = −0.025. Linear interpolation between the last pair gives u = 2.2255.
- Then E = (u/5.123)² eV = (2.2255/5.123)² eV = 0.189 eV, with v = √(z₀² − u²) = 1.709 and a penetration depth 1/κ = 0.585 nm. Checks: E lies below V₀, and below the hard-wall 0.376 eV, at 0.502 of it, because the tail gives the wave more room.
- A second level needs a root on the next branch, which starts at u = 3π/2 = 4.71. But u cannot exceed z₀ = 2.81, so there is exactly one bound state; every E above 0.30 eV belongs to the continuum.
Answer. Bound states need V₀ > π²ħ²/(8mL²), which is 0.094 eV for L = 1.00 nm. With V₀ = 0.30 eV there is exactly one, at E = 0.189 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Two constrained edges, one free ratio: E must adjust.
Count conditions against one free ratio: ψ = 0 at a hard wall, no growing exponential at a forbidden end. Two such edges leave one condition on E, so levels are isolated. If E lies above V's limit at either end, that edge imposes nothing and E is continuous.
Next: Operators: linear instructions that act on a wavefunction →