Excited states, half-wavelengths, and nodes
Half-wavelengths count the state; nodes are not walls.
Past the ground state, the well is read from a picture: count the half-wavelengths or the nodes and you have n, the energy, and how the probability is shared out. This lesson teaches that reading, and corrects the one thing the picture tempts everyone into: treating a node as a wall.
Build the model
Count the lobes and you know the state. A node is a zero of ψ, not a wall.
1The big idea
The nth state of the well, ψₙ = √(2/L) sin(nπx/L), is a standing wave, and almost everything about it can be read from its shape. Inside, ψ″/ψ = −kₙ² at every point, so ψ turns back toward the axis at one steady rate; from the zero forced at x = 0 to the zero forced at x = L its phase advances by kₙL = nπ. That is n half-wavelengths, λₙ = 2L/n, with n − 1 interior zeros at x = jL/n. Each extra half-wavelength steepens the wave, and because ⟨p²⟩ = (h/λₙ)² holds exactly, the energy follows the count: Eₙ = n²E₁. The zeros are nodes of the state, not features of the potential. V is as flat there as anywhere else inside; the two travelling waves e(±ikₙx) that make up the sine simply cancel. Each lobe between nodes holds probability 1/n, and the phase e(−iEₙt/ħ) multiplies every point alike, so the pattern never moves. At large n the lobes grow too fine to resolve and the density averages to the classical 1/L. What a real, finite wall destroys is the exact fit: the wave leaks past each wall, the higher states most, so fewer than n half-wavelengths lie inside and the n² ladder ends at the top of the well. The node count survives: in any one-dimensional well, the nth level has n − 1 nodes.
2In one sentence
An excited state of the infinite well, ψₙ = √(2/L) sin(nπx/L) with n ≥ 2, is a standing wave with n half-wavelengths between the walls and n − 1 interior nodes: points where ψₙ, and so the probability density, is exactly zero.
Example. In a 0.90 nm well, ψ₃ has λ₃ = 2L/3 = 0.600 nm and nodes at 0.30 nm and 0.60 nm; each of its three lobes holds probability 1/3, and an electron in it has E₃ = 9E₁ = 4.18 eV.
3Counting half-wavelengths reads off the energy
The wall conditions gave kₙ = nπ/L (UQMI-06.02). Read it as a picture: the wavelength is λₙ = 2π/kₙ = 2L/n, so ψₙ fits exactly n half-wavelengths between the walls. The sine is itself two travelling waves, sin kₙx = (e(ikₙx) − e(−ikₙx))/(2i), running in opposite directions with momenta ±h/λₙ, which is why de Broglie's relation hands back the energy. For an electron in the n = 3 state of a 1.00 nm well, λ₃ = 0.667 nm, pc = hc/λ₃ = 1239.8 eV nm ÷ 0.667 nm = 1860 eV, and E = (pc)²/(2mc²) = 1860² eV² ÷ 1.022 × 10⁶ eV = 3.38 eV, which is 9E₁ with E₁ = 0.376 eV. This is exact for the energy, because ψₙ″ = −kₙ²ψₙ makes ⟨p²⟩ = ħ²kₙ² = (h/λₙ)² with no approximation. It is not a claim that the momentum is sharp. The wave stops at the walls, so each peak of the momentum distribution is about h/L wide rather than a sharp line; the two peaks approach ±h/λₙ only as n grows, and for n = 1 there is a single peak at p = 0.
4Where the nodes sit, and why there are n − 1
The zeros of sin(nπx/L) fall at x = jL/n. Those at j = 0 and j = n are the walls, imposed by V = ∞; the n − 1 between them are the nodes, spaced L/n = λₙ/2 apart. The count follows from curvature. Inside, ψ″/ψ = −kₙ² at every point, so ψ bends back toward the axis at one steady rate and its phase kₙx advances uniformly; from the zero at one wall to the zero at the other it advances by nπ, passing through π, 2π, …, (n − 1)π on the way, one node at each. Two checks come free. Reflection about the centre gives ψₙ(L − x) = (−1)(n+1) ψₙ(x): odd n is symmetric with an antinode at L/2, even n antisymmetric with a node there. And the nodes interlace: in a 1.00 nm well ψ₄ has nodes at 0.25, 0.50 and 0.75 nm and ψ₅ at 0.20, 0.40, 0.60 and 0.80 nm, one in each of ψ₄'s four lobes. Raising E makes ψ turn faster at every point, so each step up the ladder squeezes in exactly one more node. Made rigorous, that argument is Sturm's oscillation theorem, and it never needs a flat floor, which is why the nth level of any one-dimensional well has n − 1 nodes.
5The shapes of a clamped string, but not its ladder
A string clamped at x = 0 and x = L obeys the same boundary conditions and has the same mode shapes, sin(nπx/L), with n half-wavelengths and n − 1 nodes. The spectra part company because the wave equations differ. The string obeys ∂²y/∂t² = v²∂²y/∂x², so ω = vk and its frequencies form the harmonic series fₙ = nf₁: a string with f₁ = 110 Hz has f₄ = 440 Hz, and every gap is 110 Hz. The Schrödinger equation has one time derivative against two space derivatives, so ω = ħk²/2m and Eₙ = ħωₙ grows as n². In a 1.00 nm well E₄ = 16E₁ = 6.02 eV, not 4E₁ = 1.50 eV, and the gaps widen as (2n + 1)E₁: 1.13 eV from n = 1 to 2, but 2.63 eV from 3 to 4. Two more differences matter. The string's amplitude is free, and its energy grows with the amplitude squared; ψₙ's amplitude is pinned at √(2/L) by normalisation, and n alone sets the energy. And the string's displacement is itself what is observed, while ψ is not, which is the next point.
6Nothing in a stationary state moves, nodes included
With its time dependence, Ψₙ(x, t) = ψₙ(x) e(−iEₙt/ħ), and since ψₙ is real its parts are Re Ψₙ = ψₙ cos(Eₙt/ħ) and Im Ψₙ = −ψₙ sin(Eₙt/ħ) (UQMI-02.05). Each on its own behaves like the vibrating string, swelling and shrinking between fixed nodes, and the two run a quarter-cycle apart. For n = 3 in a 1.00 nm well one cycle takes h/E₃ = 1.22 fs, and at t = 0.306 fs Re Ψ₃ is zero at every x. A string at that instant lies straight; the well is not empty, because Im Ψ₃ is then at full amplitude. The density adds the two in quadrature, |Ψₙ|² = ψₙ²(cos² + sin²) = (2/L) sin²(nπx/L), the same at every t. Both parts vanish at the same points xⱼ at every instant, so the nodes stay put, the lobes do not slosh, and every statistic of position is constant in time. Motion needs two different energies in one state, so that a relative phase can turn; that is UQMI-06.08.
7A node is a zero of ψ, not a wall
Nothing marks a node in the potential: V = 0 at x = L/2 exactly as at x = L/4, and the node of ψ₂ at the centre is where its two travelling waves cancel, not a place the particle is kept out of. Zero density at a point also says less than it seems. Near a node ψₙ is linear in x − xⱼ, so |ψₙ|² ≈ (2/L)(nπ/L)²(x − xⱼ)², and a window of half-width δ centred on it holds P ≈ 4n²π²δ³/(3L³). For n = 2 in a 1.00 nm well with δ = 0.020 nm the formula gives 4.21 × 10⁻⁴ and the exact integral 4.20 × 10⁻⁴, while the same window on the antinode at 0.25 nm holds 0.0796. Small, and shrinking as δ³ rather than δ, but never zero. Nor does anything flow across it: ψₙ is real up to a constant phase, so the probability current j = (ħ/m) Im(ψ* ∂ψ/∂x) is zero at every x (UQMI-02.06), at the antinodes as much as at the node. The state gives position statistics, one half on each side for ψ₂; asking how one particle gets from lobe to lobe asks for a path the theory never supplied.
8At large n the lobes blur into the classical density
A classical particle bouncing between the walls at constant speed spends equal times in equal lengths, so its density is uniform, 1/L. |ψₙ|² is never flat like that: it swings between 0 and 2/L, twice the classical value, n times across the well. Averaged over any stretch wider than the node spacing L/n, though, it gives 1/L. The probability between the left wall and x = aL is a − sin(2nπa)/(2nπ), the classical a plus a remainder no larger than 1/(2nπ). For the left quarter, a = 0.25, it is 0.091 at n = 1, 0.303 at n = 3 and 0.248 at n = 101. How large n gets depends on scale. An electron with 25 meV of kinetic energy in a 10 nm well sits between n = 2 and n = 3, fully quantum; in a 1.0 μm well it is near n = 258, with nodes 3.9 nm apart and every such probability within 6.2 × 10⁻⁴ of the classical value. The node picture has not failed there. It has become finer than any detector can resolve.
Key equations
What each relation says, and the units it is written in.
Half-wavelengths in the nth state
λₙ = 2π/kₙ = 2L/n
With p = h/λₙ it returns Eₙ = n²h²/(8mL²) exactly, because ⟨p²⟩ = (h/λₙ)²; the momentum itself is not sharp, since the wave stops at the walls.
λₙ and L in m, kₙ = nπ/L in m⁻¹; n = 1, 2, 3, … is the number of half-wavelengths between the walls.
Interior nodes of ψₙ
xⱼ = jL/n, j = 1, 2, …, n − 1
n − 1 zeros spaced L/n = λₙ/2 apart; the centre x = L/2 is one of them exactly when n is even.
xⱼ and L in m. The zeros at j = 0 and j = n are the walls, forced by V = ∞, and are not counted as nodes.
Probability in one lobe
∫ |ψₙ|² dx = 1/n, over (j − 1)L/n < x < jL/n
Nodes divide the probability evenly without confining it: a position measurement finds the particle in any chosen lobe with probability 1/n.
Dimensionless, and the same for every lobe j = 1, …, n: each spans one full period of sin².
Probability within δ of a node
P ≈ 4n²π²δ³/(3L³), for δ ≪ L/n
It shrinks as δ³ rather than δ: zero density at one point, but never zero probability in any window around it.
δ is the half-width of a window centred on a node, in m, like L; P is dimensionless. At an antinode the window holds about 4δ/L.
Probability from the wall to x = aL
P(0 < x < aL) = a − sin(2nπa)/(2nπ)
The quantum value differs from the classical one by at most 1/(2nπ), so the lobes blur into the uniform density 1/L as n grows.
a is the dimensionless fraction of the width, 0 ≤ a ≤ 1; a classical particle gives exactly a.
A stationary state's density
|Ψₙ(x, t)|² = (2/L) sin²(nπx/L) at every t
The phase e(−iEₙt/ħ) has modulus 1, so the nodes never move; only a superposition of different n makes the density slosh.
In m⁻¹. Re Ψₙ = ψₙ cos(Eₙt/ħ) and Im Ψₙ = −ψₙ sin(Eₙt/ħ) vanish at the same points xⱼ.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Step n from 1 to 4: each step adds a half-wavelength and a node, and Eₙ grows as n². Sweep the phase: Re Ψ and Im Ψ trade amplitude, yet |Ψ|² and its nodes never move. Drag the strip edge across a node: P levels off there, at exactly j/n, then climbs on.
Live results
- Eₙ, ELECTRON IN 1.00 nm WELL
- 3.384 eV
- WAVELENGTH λₙ = 2L/n
- 0.667 nm
- P(0 < x < aL)
- 0.303
- PHASE PERIOD h/Eₙ
- 1.222 fs
Catch the common trap
The mistake this lesson is built to prevent.
The node looks like a wall inside the well, so students say an electron in ψ₂ is trapped on one side of x = L/2, or must tunnel across, or crosses at infinite speed because the density there is zero. All three treat the node as a property of the potential and ψ as a map of one particle's journey. The potential at L/2 is zero, as at every other interior point, so there is nothing to tunnel through. A node belongs to the state, not to the well: in (ψ₁ + ψ₂)/√2 the density at L/2 is 1/L, and probability does move between the halves. In ψ₂ itself the probability current is zero everywhere, not only at the node, so nothing travels anywhere; the state assigns ½ to each half, and a position measurement on each freshly prepared electron finds it left or right with those odds. A window of half-width δ around the node still holds probability, of order δ³.
Test it: what changes?
- One more half-wavelength and one more node
- Energy rises from n²E₁ to (n + 1)²E₁
- Each lobe's probability drops to 1/(n + 1)
- Phase period h/Eₙ shortens
- Re Ψ and Im Ψ trade amplitude
- |Ψ|² unchanged at every x
- Nodes stay at xⱼ = jL/n
- Strip probability P(0 < x < aL) unchanged
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron is in a stationary state of an infinite well of width L = 1.40 nm. Its probability density is zero at x = 0.35, 0.70 and 1.05 nm and at no other point between the walls. Identify n, find the wavelength and the energy, and find the probability that a position measurement gives a value between 0.35 nm and 1.05 nm.
View solution →
- Three interior nodes, and ψₙ has n − 1 of them, so n = 4. Check the positions: the nodes of ψ₄ sit at jL/4 = 0.35j nm for j = 1, 2, 3, exactly as observed.
- Four half-wavelengths fit in L: λ₄ = 2L/4 = 0.700 nm, twice the node spacing.
- Energy: E₄ = 16 × (ħ²/2m)π²/L² = 16 × 0.0381 eV nm² × 9.870 ÷ 1.96 nm² = 3.07 eV. Units: eV nm² over nm² leaves eV.
- Cross-check with de Broglie: pc = hc/λ₄ = 1239.8 eV nm ÷ 0.700 nm = 1771 eV, and (pc)²/(2mc²) = 1771² eV² ÷ 1.022 × 10⁶ eV = 3.07 eV.
- Probability: 0.35 nm to 1.05 nm spans the second and third lobes, and every lobe holds 1/n = 0.25, so P = 0.500. Normalisation check: four lobes of 0.25 make 1.
Answer. n = 4, λ₄ = 0.700 nm, E₄ = 3.07 eV, and P(0.35 nm < x < 1.05 nm) = 0.500.
MediumExample 2
In an infinite well of width 1.00 nm, consider the point x₀ = 0.40 nm. (a) For which states ψₙ is x₀ a node? Give the three lowest of their energies for an electron. (b) In the n = 3 state, find |Ψ(x₀, t)|² and how it changes over one period h/E₃. (c) Describe what Re Ψ and Im Ψ do at x₀ over that period.
View solution →
- (a) Nodes of ψₙ sit at x = jL/n, so x₀ is one when 0.40 nm = j × 1.00 nm/n, that is when 0.40n is a whole number. That needs n to be a multiple of 5: n = 5, 10, 15, …
- With E₁ = 0.376 eV for an electron in 1.00 nm, E₅ = 25E₁ = 9.40 eV, E₁₀ = 100E₁ = 37.6 eV and E₁₅ = 225E₁ = 84.6 eV. The point is a node of these states, not of the well: for n = 1 to 4 the density there is not zero.
- (b) ψ₃(x₀) = √(2/L) sin(3π × 0.40) = √2 nm(−1/2) × sin 216° = −0.831 nm(−1/2), so |Ψ₃(x₀, t)|² = 0.691 nm⁻¹, a probability per unit length. Beside the classical 1/L = 1.00 nm⁻¹, x₀ lies on the flank of a lobe.
- Period: E₃ = 9E₁ = 3.38 eV, so h/E₃ = 4.136 × 10⁻¹⁵ eV s ÷ 3.38 eV = 1.22 × 10⁻¹⁵ s. The factor e(−iE₃t/ħ) has modulus 1, so over those 1.22 fs the density at x₀ does not change at all.
- (c) Re Ψ₃(x₀, t) = −0.831 cos(E₃t/ħ) and Im Ψ₃(x₀, t) = +0.831 sin(E₃t/ħ), both in nm(−1/2). At t = 0.306 fs, a quarter period, Re Ψ is zero and Im Ψ is 0.831 nm(−1/2); the sum of their squares stays 0.691 nm⁻¹ throughout.
Answer. (a) n = 5, 10, 15, …, at 9.40 eV, 37.6 eV and 84.6 eV; (b) |Ψ₃(x₀)|² = 0.691 nm⁻¹, constant over the 1.22 fs period; (c) Re Ψ and Im Ψ trade an amplitude of 0.831 nm(−1/2), a quarter-cycle apart.
HardExample 3
Ten thousand electrons are prepared one at a time in the same stationary state of a 2.00 nm infinite well, and each position is measured once. In 0.050 nm bins, the histogram is nearly empty in the half-bins against the two walls, where ψ must vanish, and otherwise only in the bins centred on 0.40, 0.80, 1.20 and 1.60 nm. Find n and the energy; predict the counts in 0 < x < 0.40 nm, in the bin centred on 0.80 nm and in the one centred on 1.00 nm; and decide whether the nearly empty bins should read zero.
View solution →
- Four interior nodes spaced 0.40 nm = L/5, so n = 5. E₅ = 25 × 0.0381 eV nm² × π² ÷ (2.00 nm)² = 2.35 eV. Check with λ₅ = 2L/5 = 0.800 nm: (hc/λ₅)²/(2mc²) = (1550 eV)² ÷ 1.022 × 10⁶ eV = 2.35 eV.
- 0 < x < 0.40 nm is the first lobe, which holds 1/n = 0.200 of the probability: 2000 counts expected. Five lobes account for all 10 000, as normalisation requires.
- A bin of half-width δ = 0.025 nm holds (2/L)[δ ∓ (L/(2nπ)) sin(2nπδ/L)], minus when centred on a node and plus when centred on an antinode, because sin² is symmetric about both. Here 2/L = 1.00 nm⁻¹, L/(2nπ) = 0.0637 nm and sin(2nπδ/L) = sin 0.3927 = 0.3827, so the correction term is 0.02436 nm.
- Node bin at 0.80 nm: P = 1.00 nm⁻¹ × (0.02500 − 0.02436) nm = 6.4 × 10⁻⁴, about 6.4 counts; the small-window formula 4n²π²δ³/(3L³) gives 6.43 × 10⁻⁴, so the δ³ law already holds. The bin at 1.00 nm is an antinode, since (j − ½)L/5 = 1.00 nm for j = 3: P = 1.00 × (0.02500 + 0.02436) = 0.0494, about 494 counts.
- Classical check: a uniform density puts 10 000 × 0.050/2.00 = 250 counts in each bin. The antinode bin holds nearly twice that, since |ψ₅|² peaks at 2/L, and the node bin about a fortieth of it; across a whole lobe they balance to the classical 2000.
- The node bins should not read zero. Each expects 6.4 counts with Poisson scatter √6.4 ≈ 2.5, so readings from 2 to 11 are ordinary, and a bin reading zero has probability e(−6.4) ≈ 0.002. A narrower bin would shrink the count as δ³, never to nothing. The half-bins at the walls are each half such a window, so they expect about 3.2 counts, sparse for the same reason.
Answer. n = 5, E₅ = 2.35 eV. About 2000 counts in 0 < x < 0.40 nm, about 6.4 in the node bin at 0.80 nm and about 494 in the antinode bin at 1.00 nm. The node bins, and the half-bins at the walls with about 3.2 each, are sparse, not empty.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
n half-wavelengths, n − 1 nodes, and Eₙ = n²E₁.
Read n from the shape: λₙ = 2L/n and n − 1 interior nodes at jL/n. Each lobe holds probability 1/n and a stationary density never moves. A node is a zero of the state, not a barrier: V is zero there, and a window of half-width δ around it holds probability of order δ³.