Energy quantisation: the condition for a nontrivial solution
Only special energies let a nonzero ψ fit both walls.
This is where the ladder of levels is earned rather than quoted. Take E below zero, at zero and above it, apply the two walls one at a time, and watch every energy fail except a discrete set. The same bookkeeping says why the count starts at n = 1, and why n and −n are one state when the free particle had two, ±k, at every energy.
Build the model
Zero fits both walls at every energy. A nonzero ψ fits them only at Eₙ.
1The big idea
Between the walls the time-independent equation is ψ″ = −(2mE/ħ²)ψ, and the walls add ψ(0) = ψ(L) = 0. Equation and conditions are linear and homogeneous, so ψ ≡ 0 satisfies them at every energy; it has ∫|ψ|² dx = 0 and describes no particle. Quantisation is therefore the question of when a second, nonzero solution exists, just as (H − E)v = 0 has a nonzero v only where det(H − E) = 0. Take the energies in order. For E < 0 the solutions are sinh κx and cosh κx, which bend away from the axis, so one that leaves zero never returns. At E = 0 the solution is a straight line, and a line through two zeros is the axis itself. For E > 0 the first wall removes cos kx and leaves A sin kx at every energy; the second wall then has nothing to adjust but the scale A, which cannot move a zero, so it falls on E and demands sin kL = 0. The survivors are kₙ = nπ/L and Eₙ = n²π²ħ²/(2mL²) with n = 1, 2, 3, …, once n = 0 (no function at all) and negative n (the same function times −1) are set aside. The result is exact for the model and only as good as it: the walls must be high enough that ψ(L) = 0 holds to the accuracy wanted (UQMI-06.01), and the speed low enough for E = p²/2m. And the steps fade where they should: the gap relative to the level, (2n + 1)/n², falls toward zero as n grows, every gap closes as 1/L² as the box widens, and the classical continuum returns.
2In one sentence
An allowed energy of the infinite well is a value of E at which ψ″ = −(2mE/ħ²)ψ has a solution that vanishes at both walls without vanishing everywhere; these are Eₙ = n²π²ħ²/(2mL²) with n = 1, 2, 3, …
Example. An electron at E = 1.20 eV in a 1.00 nm well has kL = 5.61 rad, so the solution leaving ψ(0) = 0 reaches the far wall at sin(5.61) = −0.62 of its peak. Only 0.376, 1.50, 3.38, 6.02 eV, … bring it back to zero.
3Zero always fits; the demand is a nonzero ψ
Inside the well V = 0 and the equation is ψ″ = −(2mE/ħ²)ψ, with the walls adding ψ(0) = 0 and ψ(L) = 0. All three are linear and homogeneous, so ψ ≡ 0 satisfies them at every energy. That solution has ∫|ψ|² dx = 0, cannot be normalised and describes no particle, so the allowed energies are the values of E at which a second, nonzero solution exists: the question (H − E)v = 0 asks of a matrix. A string clamped at both ends obeys y″ = −(ω/v)²y with the same two conditions and gets the same answer, a discrete set of wavenumbers. What makes the well quantum is what the wavenumber means. A string's energy is set by its amplitude, so a single mode can hold any energy. The amplitude of ψ is fixed by normalisation and carries no energy at all, while E = ħ²k²/(2m) ties the energy to k alone. Discrete k therefore means discrete E.
4Below the floor, a curve that cannot turn back
Try E < 0 first. Then ψ″ = +κ²ψ with κ = √(2m|E|)/ħ, and the general solution is A sinh κx + B cosh κx. The left wall gives ψ(0) = B = 0; the right wall gives A sinh κL = 0, and sinh κL > 0 for every κL > 0, so A = 0 and nothing survives. The geometry says why: wherever ψ > 0, ψ″ > 0 too, so a curve that leaves the axis bends further away and can never come back. For an electron at E = −0.500 eV in a 1.00 nm well, κ = 3.62 nm⁻¹ and ψ(L) = A sinh 3.62 = 18.7A. E = 0 needs its own look, because at k = 0 sin kx and cos kx collapse to 0 and 1 and one independent solution goes missing. The true general solution there is A + Bx: ψ(0) = 0 gives A = 0, and ψ(L) = BL = 0 gives B = 0. So n = 0 is not a state, but the reason is this straight line, not sin 0 = 0.
5An integral that forbids E ≤ 0 for any shape
The same verdict follows without solving anything. Multiply ψ″ = −(2mE/ħ²)ψ by ψ* and integrate across the well. By parts, ∫ψ*ψ″ dx equals the boundary term [ψ*ψ′] from 0 to L minus ∫|ψ′|² dx, and the boundary term is zero because ψ vanishes at both walls. What remains is E ∫|ψ|² dx = (ħ²/2m) ∫|ψ′|² dx. The right side is an integral of a square: never negative, and zero only if ψ′ = 0 everywhere, that is only for a constant ψ, which the walls make zero. So every nonzero solution has E real and strictly positive. The identity says the energy inside is all kinetic, and that a ψ pinned to zero at both ends must have slope somewhere. It also gives the scale before the solution arrives: ψ must rise from zero and fall back within L, so |ψ′| is of order |ψ|/L and E of order ħ²/(2mL²). The exact ground level turns out to be π² times that.
6For E > 0, one wall fixes the shape and the other the energy
Now take E above zero, where k is real and ψ = A sin kx + B cos kx. Apply the walls one at a time. ψ(0) = B = 0 removes the cosine, and it does so at every energy: after the first wall a solution A sin kx exists for any E you name. That wall fixed the shape, not the energy. The second wall asks A sin kL = 0, and the only constant left is A, which cannot help: the condition is homogeneous, so scaling A scales ψ(L) and never moves its zeros. Either A = 0, the trivial solution again, or sin kL = 0, a condition on k and through k on E. Normalisation later fixes |A| = √(2/L) and never touches E. For an electron in a 1.00 nm well at E = 1.20 eV, k = 5.61 nm⁻¹ and the solution reaches the far wall at sin 5.61 = −0.62 of its peak. Slide E and that wall value passes through zero only at kL = π, 2π, 3π, …, the energies 0.376, 1.50 and 3.38 eV. The figure runs this scan.
7Counting the roots: n = 0 and negative n
sin kL = 0 holds for kL = nπ with n any integer, and each root must be checked for being a new state. n = 0 means k = 0 and E = 0, already shown to leave nothing. Replacing n by −n gives sin(−nπx/L) = −sin(nπx/L): the same function times −1 = e(iπ), a global phase that changes no probability and no expectation value, at the same energy because E depends on k². So n = −2 is n = 2 again, and n = 1, 2, 3, … exhausts the states, one per level. That is a real change from the free particle, whose e(ikx) and e(−ikx) are independent states at every E > 0 (UQMI-05.04). In the box sin kx = (e(ikx) − e(−ikx))/(2i): the wall fixes the amplitudes of the two travelling waves in the ratio −1, reflecting each into the other. The ladder Eₙ = n²E₁ has gaps (2n + 1)E₁ that widen as it climbs: 1.13, 1.88 and 2.63 eV for the first three in a 1.00 nm well.
8Checking the ladder: units, limits and the classical bounce
Dimensions: ħ²/(mL²) is (J s)² ÷ (kg m²), and since kg m² = J s², that is J. Scale: an electron in a 1.00 nm well has E₁ = 0.376 eV, while a proton in a 5.0 fm box, a crude nucleus, has E₁ = 8.2 MeV; one formula spans chemistry and nuclear physics through 1/(mL²). Limit: as L → ∞ every gap closes as 1/L² and the free particle's continuum returns. Classical correspondence, in Bohr's form: a classical particle with energy Eₙ moves at v = √(2Eₙ/m) = nπħ/(mL) and bounces between the walls at angular frequency πv/L = nπ²ħ/(mL²). The quantum jump from n + 1 to n emits (Eₙ₊₁ − Eₙ)/ħ = (n + ½)π²ħ/(mL²). Their ratio, (2n + 1)/(2n), is 1.5 at n = 1, 1.05 at n = 10 and 1.005 at n = 100: at large n the emitted light carries the frequency of the classical motion.
Key equations
What each relation says, and the units it is written in.
Inside the well, and the wavenumber
ψ″ = −k²ψ, with k = √(2mE)/ħ
Real k needs E > 0. For E < 0 the same equation reads ψ″ = +κ²ψ with κ = √(2m|E|)/ħ, and its solutions curve away from the axis.
For 0 < x < L, where V = 0. k in m⁻¹, E in J, m in kg, ħ in J s. For an electron ħ²/2m = 0.0381 eV nm².
Positivity identity
E ∫|ψ|² dx = (ħ²/2m) ∫|ψ′|² dx, both integrals over 0 to L
The right side is never negative and is zero only for a constant ψ, which the walls force to zero, so every state has E > 0 whatever its shape.
ψ in m(−1/2), ψ′ in m(−3/2); both sides in J. The boundary term ψ*ψ′ vanishes because ψ(0) = ψ(L) = 0.
Wall value once the first wall is met
ψ(L) = A sin kL for E > 0, AL at E = 0, A sinh κL for E < 0
L and sinh κL are positive, so E ≤ 0 needs A = 0. Rescaling A never moves a zero, so the second wall constrains E, not A.
A is the one constant ψ(0) = 0 leaves, in m(−1/2) (m(−3/2) at E = 0); κ = √(2m|E|)/ħ in m⁻¹; L in m.
Condition for a nontrivial solution
sin kL = 0 with A ≠ 0, so kₙ = nπ/L
One state per root: the wall at x = 0 locks e(ikx) and e(−ikx) in the ratio −1, so the free particle's ±k pair becomes a single standing wave.
kₙ in m⁻¹ and L in m, for n = 1, 2, 3, … only: n = 0 gives ψ ≡ 0, and −n gives −ψₙ, the same state.
The energy ladder
Eₙ = n²π²ħ²/(2mL²), n = 1, 2, 3, …
Scales as 1/(mL²): halve L and every level quadruples. The gaps Eₙ₊₁ − Eₙ = (2n + 1)E₁ widen as n climbs.
Eₙ in J with m in kg, L in m, ħ in J s. Electron, L = 1.00 nm: 0.376, 1.50, 3.38 and 6.02 eV.
Levels at or below an energy E
N(E) = ⌊kL/π⌋ = ⌊√(E/E₁)⌋
For large N, N ≈ kL/π = 2pL/h: the area 2pL enclosed by the classical bounce in phase space, divided by h. One state per h of area.
⌊ ⌋ takes the integer part; N is a pure number, E and E₁ in J. Electron, L = 1.00 nm, E = 5.00 eV: N = 3.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag E up from −1 eV. Below zero, and up to E₁/4, the dot at the wall stays pinned at the top; at L = 1.00 nm it first reaches the axis near 0.38 eV, then 1.50, 3.38 and 6.02 eV. Then hold E at 1.50 eV and drag L: the wall meets a node only at 1.00 nm and 1.50 nm.
Live results
- kL/π
- 1.786
- WALL VALUE ψ(L) ÷ PEAK
- -0.622
- LEVELS AT OR BELOW E
- 1
- GROUND LEVEL E₁
- 0.376 eV
Catch the common trap
The mistake this lesson is built to prevent.
Fresh from the free particle, where e(ikx) and e(−ikx) are independent states at every energy, students carry the pairing into the box and count n and −n as two states, so every level looks twofold degenerate. It is not. ψ₋ₙ = −ψₙ: the two differ by the constant factor −1, so they are linearly dependent, and multiplying a wavefunction by a nonzero constant never makes a new state. Here the factor is a pure phase, so even |ψ|² is untouched, along with the energy and every expectation value. The wall at x = 0 is exactly what removes the free particle's second solution: it forces e(ikx) and e(−ikx) into the one combination sin kx, leaving one state per allowed k. The error does not stay a counting error. Put 20 non-interacting electrons, two per spatial state for spin, into a 5.00 nm well: the top occupied level is n = 10 at 1.50 eV, but counting ±n separately stops at n = 5 and 0.376 eV, a factor of four too low.
Test it: what changes?
- kL/π grows as √E for E > 0
- Wall value ψ(L) vanishes only at each Eₙ
- Level count steps up by one at each Eₙ
- Ground level E₁ unchanged
- Every level Eₙ quadruples
- Every gap (2n + 1)E₁ quadruples
- Ratios Eₙ/E₁ = n² unchanged
- kL/π at a given E halves
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is confined to an infinite well of width L = 0.500 nm. (a) Test the trial energy E = −0.500 eV: write the interior solution that satisfies ψ(0) = 0 and find ψ(L) in terms of its constant. (b) Find the lowest energy that satisfies both walls.
View solution →
- For E < 0 the interior equation is ψ″ = +κ²ψ with κ = √(2m|E|)/ħ = √(|E|/(ħ²/2m)). With ħ²/2m = 0.0381 eV nm² for an electron, κ = √(0.500/0.0381) nm⁻¹ = 3.623 nm⁻¹.
- The general solution is A sinh κx + B cosh κx. The left wall gives ψ(0) = B = 0, leaving ψ = A sinh κx.
- At the far wall κL = 3.623 × 0.500 = 1.811, so ψ(L) = A sinh 1.811 = 2.98A. That is zero only for A = 0, which is ψ ≡ 0: E = −0.500 eV admits no state, and no E < 0 does, since sinh is positive for every positive argument.
- For E > 0 the walls demand sin kL = 0 with A ≠ 0. The smallest positive root is k₁ = π/L = π/0.500 nm = 6.28 nm⁻¹.
- E₁ = (ħ²/2m)k₁² = 0.0381 × 6.283² eV = 1.50 eV. Check against the 1.00 nm well: halving L should multiply E₁ by four, and 4 × 0.376 eV = 1.50 eV.
Answer. At E = −0.500 eV, ψ(L) = 2.98A, which vanishes only for A = 0, so there is no state; the lowest allowed energy is E₁ = 1.50 eV.
MediumExample 2
An electron is held at E = 1.50 eV while the width L of an infinite well is varied. (a) Write the wall value ψ(L)/A of the solution A sin kx that meets ψ(0) = 0, and find every width from 0.600 nm to 2.20 nm at which 1.50 eV is an allowed energy, naming its level n each time. (b) At L = 1.25 nm, find ψ(L)/A and the two allowed levels either side of 1.50 eV. (c) Say how far apart the allowed widths are, and why.
View solution →
- The first wall leaves ψ = A sin kx, and k is set by E alone: k = √(E/(ħ²/2m)) = √(1.50/0.0381) nm⁻¹ = 6.275 nm⁻¹. So ψ(L)/A = sin kL, which at fixed E is a function of the width only.
- The second wall needs sin kL = 0 with A ≠ 0, so kL = nπ, now read as a condition on L: Lₙ = nπ/k = n × 0.5007 nm. From 0.600 nm to 2.20 nm that leaves n = 2, 3 and 4, at L = 1.00, 1.50 and 2.00 nm; n = 1 (0.501 nm) and n = 5 (2.50 nm) fall outside.
- Check each with Eₙ = n² × 0.0381π²/L² eV, L in nm: 4 × 0.376 eV at 1.00 nm, 9 × 0.167 eV at 1.50 nm and 16 × 0.0940 eV at 2.00 nm, all 1.50 eV. The same energy is the second, third or fourth level depending on the box.
- At L = 1.25 nm, kL = 6.275 × 1.25 = 7.84 = 2.50π, so ψ(L)/A = sin 7.84 = 1.00: the wall value is as far from zero as it can be. There E₁ = 0.0381 × (π/1.25)² eV = 0.241 eV, so E₂ = 0.963 eV and E₃ = 2.17 eV bracket 1.50 eV.
- Neighbouring widths differ by π/k = 0.501 nm, half of the de Broglie wavelength 2π/k = 1.00 nm: at fixed k the wall value returns to zero every half-wavelength, so each extra half-wave that fits between the walls raises n by one.
Answer. 1.50 eV is allowed at L = 1.00, 1.50 and 2.00 nm, as n = 2, 3 and 4; at L = 1.25 nm, ψ(L)/A = 1.00 and 1.50 eV falls between E₂ = 0.963 eV and E₃ = 2.17 eV; the allowed widths are spaced by half a wavelength, 0.501 nm.
HardExample 3
Write the interior solution of a well of width L as ψ = A e(ikx) + B e(−ikx), the free particle's two travelling waves. (a) Impose ψ(0) = ψ(L) = 0 as two equations for (A, B) and find when a nonzero pair exists. (b) Your condition also holds at k = 0; decide whether that is a state. (c) Say what becomes of the free particle's twofold degeneracy. (d) Give the first three levels of an electron in a 1.20 nm well.
View solution →
- The walls give A + B = 0 at x = 0 and A e(ikL) + B e(−ikL) = 0 at x = L. A homogeneous pair of equations has a nonzero solution only if its determinant vanishes: e(−ikL) − e(ikL) = −2i sin kL = 0, so kL = nπ, exactly as in the sine and cosine basis.
- At k = 0 the determinant is zero and (A, B) = (1, −1) is a nonzero pair, but both basis functions equal 1 there, so it builds ψ = 1 − 1 = 0. A zero determinant guarantees a nonzero coefficient pair, not a nonzero function. At E = 0 the proper general solution is C + Dx, which the two walls force to zero: n = 0 is not a state.
- At kL = nπ with n ≠ 0 the first equation gives B = −A, so ψ = A(e(ikx) − e(−ikx)) = 2iA sin(nπx/L). The factor 2iA is a constant whose size normalisation fixes later and whose phase is unobservable. Replacing n by −n swaps the two exponentials and gives −2iA sin(nπx/L), the same state.
- A free electron at energy E has two independent states, e(ikx) and e(−ikx). In the box the first wall fixes B/A = −1, so each allowed k carries one state: the wall reflects each travelling wave into the other and fuses them into a standing wave, which carries zero probability current.
- E₁ = (ħ²/2m)(π/L)² = 0.0381 × (π/1.20)² eV = 0.261 eV, then E₂ = 4E₁ = 1.04 eV and E₃ = 9E₁ = 2.35 eV. Check: (1.00/1.20)² × 0.376 eV = 0.261 eV.
Answer. −2i sin kL = 0 gives kL = nπ; k = 0 builds ψ ≡ 0 and is no state; the ±k pair fuses into one standing wave per level; E₁ = 0.261 eV, E₂ = 1.04 eV, E₃ = 2.35 eV.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
The second wall fixes the energy, not the amplitude.
For E > 0 the first wall leaves A sin kx; the second needs sin kL = 0 with A ≠ 0, so Eₙ = n²π²ħ²/(2mL²). Use n = 1, 2, 3, … only: n = 0 gives ψ ≡ 0 and −n repeats a state. Needs high walls and E = p²/2m.
Next: Normalisation, the ground state, and zero-point energy →