University Physics V · Particle in a Box · 5.6
Probability Density, Nodes & Intervals
Once ψₙ is in hand the physics is still one step away. Here you turn the eigenvector into something a detector could check: normalise the density, read the node pattern straight off n, and write “is it in [a, b]?” as a projector whose expectation value is the answer.
Build the model
Connect the measurement to the mechanism.
The Born rule says the eigenvector ⟨x|n⟩ = √(2/L) sin(nπx/L) is not itself the observable; what is observable is |⟨x|n⟩|², a probability density of dimension 1/length that means nothing at a point and everything over an interval. Written properly, “the particle is in [a, b]” is the projector Π̂ = ∫ₐᵇ|x⟩⟨x|dx — Hermitian, idempotent, spectrum {0, 1} — and the prediction is ⟨n|Π̂|n⟩, which for the box integrates in closed form to the classical share (b−a)/L minus an oscillatory term of size 1/(2πn). Everything else the density knows follows from the boundary condition: Dirichlet walls select sin, sin puts zeros at x = jL/n, and Sturm oscillation fixes the count of interior zeros at exactly n − 1, so the node pattern labels the level before any energy is computed.
What the model costs is any picture of motion. The density of a stationary state does not move — the phase e(−iEₙt/ħ) cancels in the modulus, and the probability current j = (ħ/m) Im(ψ*ψ′) is identically zero because ψₙ is real — so it predicts the histogram of position measurements on identically prepared boxes and says nothing about how a single particle reached where it was found. The classical 1/L is never recovered pointwise at any n; it appears only after averaging over a window at least one period L/n wide.
- Simple definition
- The probability density of a stationary state is |⟨x|ψ⟩|², a per-unit-length quantity whose integral over an interval — not its value at a point — is the probability that a position measurement lands there.
- Example
- In the n = 1 box state |ψ₁|² = (2/L)sin²(πx/L) peaks at 2/L in the middle and integrates over the middle half [L/4, 3L/4] to ½ + 1/π = 0.818, against a classical bouncer's 0.500.
The 2/L, not 1/L, is what normalisation demands: sin² averages to ½ over the n whole half-periods the walls enclose.
ρ in m⁻¹, x in [0, L] and identically zero outside; ∫₀ᴸ ρₙ dx = 1 for every n
Turns “where is it?” into a yes/no observable. Disjoint intervals add because Π̂₁Π̂₂ = 0, and the whole box gives Π̂ = 1̂, so P = 1.
Π̂ = ∫ₐᵇ|x⟩⟨x|dx is Hermitian and idempotent with spectrum {0, 1}; a and b in metres, P dimensionless
Sturm oscillation makes the interior node count identify the level, so counting sign changes sorts a numerical spectrum without eigenvalues.
n − 1 interior nodes spaced by L/n; n antinodes, each reaching the same peak ρ = 2/L
A node suppresses the count by roughly (nε/L)² instead of forbidding it: about 340 times fewer hits than the same window on an antinode at n = 3, ε = 0.02L.
valid for ε ≪ L/n; ψ crosses zero linearly there, so ρ vanishes quadratically and its integral cubically
Continuity ∂ρ/∂t + ∂j/∂x = 0 then reads 0 = 0 — nothing flows, so a node is not something anything has to cross.
j = (ħ/m) Im(ψ*ψ′) in m⁻¹ s⁻¹; it vanishes because ψₙ is real up to one global phase
Exactly 1/L whenever Δ is a whole number of periods L/n. Pointwise ρₙ still swings from 0 to 2/L at every n, so convergence is never pointwise.
Δ is the window width about centre x; the correction is at most L/(nπΔ) of 1/L
From the ket to a density, with units attached
⟨x|n⟩ is not the observable. The Born rule says the observable object is ρₙ(x) = |⟨x|n⟩|², whose dimension in one dimension is 1/length, so a value of ρ is never a probability — only ρ dx is. Normalisation is the statement ⟨n|n⟩ = 1 read in the position basis: ∫₀ᴸ (2/L)sin²(nπx/L) dx = 1, which works for every n because the interval holds exactly n half-periods of sin² and sin² averages to ½ over any whole number of them. Outside [0, L] the density is identically zero, and that is not an approximation but the domain: H is defined on H²([0, L]) with ψ(0) = ψ(L) = 0, so the walls are imposed before any integral is attempted. Note what the modulus discards. ψ₂ changes sign at L/2; ρ₂ does not. Every fact about relative phase — the thing that makes superpositions interfere — is invisible in the density of a single eigenstate.
“Is it in [a, b]?” is a projector, not a phrase
“The particle is in [a, b]” is a yes/no observable, and its operator is the projector Π̂ = ∫ₐᵇ|x⟩⟨x|dx. It is Hermitian, satisfies Π̂² = Π̂, and has spectrum {0, 1}, so its expectation value in any normalised state lies between 0 and 1 — exactly what a probability must do. In the position basis ⟨n|Π̂|n⟩ collapses to ∫ₐᵇ ρₙ dx, and with sin²θ = (1 − cos 2θ)/2 the integral is elementary: P = (b−a)/L − [sin(2πnb/L) − sin(2πna/L)]/(2πn). Two properties come free. Disjoint intervals add, because Π̂₁Π̂₂ = 0; and the whole box gives Π̂ = 1̂, so P = 1 without further work. For n = 1 and [0, L/4] the formula returns 0.25 − 1/(2π) = 0.0908 — nine per cent of the hits where a classical bouncer would spend twenty-five per cent of its time, because the ground-state density is still climbing out of the wall there.
Counting nodes labels the level
Zeros of ρₙ are zeros of sin(nπx/L), so they sit at x = jL/n. Of the values j = 0…n, two are the walls, which are boundary conditions rather than nodes; the n − 1 interior ones are the nodes, evenly spaced by L/n. Between them sit n antinodes at (2j−1)L/(2n), each rising to the same 2/L, so no lobe is favoured over another within a level. The count is not an accident of the box: Sturm's oscillation theorem says that for a real Sturm–Liouville problem with Dirichlet conditions, the eigenfunctions ordered by increasing eigenvalue carry 0, 1, 2, … interior zeros. That makes the node count a label. Given eigenvectors from numpy.linalg.eigh on a grid, counting sign changes identifies each level without comparing eigenvalues, and it keeps working when V is not flat and no closed form exists. Physically nodes are curvature, and curvature is kinetic energy: ⟨p̂²⟩ = (nπħ/L)² rises as the square of the number of half-waves squeezed between the walls.
A node suppresses the count; it does not forbid it
Because ρ is a density, the probability of any single point is zero — at a node and at an antinode alike. What separates them is how fast a shrinking window loses probability. At a generic x₀ the density is locally flat, so P ≈ ρ(x₀)ε for a window of width ε. At a node ψ crosses zero linearly with slope √(2/L)(nπ/L), so ρ ≈ (2/L)(nπ/L)²(x − x₀)², and integrating gives P ≈ (nπ)²ε³/(6L³): a cubic vanishing, not a linear one. Put numbers on it with n = 3 and ε = 0.02L. On the node at L/3 the exact formula returns 1.18 × 10⁻⁴; on the antinode at L/6 it returns 3.99 × 10⁻², a factor of about 340. The node is a deep suppression of the count, not a wall inside the box.
Static density, zero current, one-shot statistics
Attach the time factor and nothing happens: Ψ(x, t) = ⟨x|n⟩e(−iEₙt/ħ) has |Ψ|² = ρₙ(x), because the phase has modulus one. The stronger statement is that nothing is flowing either. The probability current j = (ħ/m) Im(ψ*∂ψ/∂x) vanishes identically for ψₙ, since ψₙ is real up to a constant phase, and continuity ∂ρ/∂t + ∂j/∂x = 0 then reads 0 = 0. So ρₙ is the frequency histogram of position measurements made on many identically prepared boxes, not the trace of one particle's motion. And that histogram is a one-shot object: a measurement that finds the particle in [a, b] leaves the state Π̂|n⟩ / ‖Π̂|n⟩‖, which is no longer ψₙ and no longer an energy eigenstate, so a second measurement on the same box is drawn from a different density altogether.
The classical 1/L survives only an average
A classical particle bouncing at constant speed spends equal time in equal lengths, giving a flat 1/L. ρₙ never approaches that pointwise: at every n, however large, it still swings between 0 and 2/L. What converges is the local average. Over a window of width Δ centred on x, ⟨ρₙ⟩ = 1/L − sin(nπΔ/L)cos(2πnx/L)/(nπΔ), a correction bounded by L/(nπΔ) relative to 1/L, and exactly zero whenever Δ is a whole number of periods L/n. Take L = 5.00 nm and n = 20, so the period is 0.250 nm. A window of 0.600 nm centred at 1.00 nm averages 0.1748 nm⁻¹, 13% below 1/L = 0.200 nm⁻¹, because 0.600 nm is 2.4 periods. Change it to 0.250 nm — a narrower window, not a wider one — and the average is exactly 0.200 nm⁻¹. Correspondence here is a statement about resolution matched to L/n, not about ħ becoming small.
Change one variable at a time
Make the relationship visible.
Set n = 4 and the window width to 0.25 L — one whole period of the density — and the probability equals the classical share exactly, wherever you slide the left edge. Any other width leaves a residue that shrinks as 1/n, while the curve itself still runs from 0 to 2/L at every n.
P IN WINDOW0.335
CLASSICAL SHARE Δ/L0.350
MEAN DENSITY × L0.958
INTERIOR NODES2
Live interpretationP IN WINDOW: 0.335. CLASSICAL SHARE Δ/L: 0.350. MEAN DENSITY × L: 0.958. INTERIOR NODES: 2
Catch the common trap
Explain before calculating.
A particle is in the n = 3 stationary state of an infinite well on [0, L]. What is the probability that a position measurement finds it in the middle third, [L/3, 2L/3]?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle is in the ground state (n = 1) of an infinite well of width L = 1.00 nm. Find the probability that a position measurement finds it in the middle half, [L/4, 3L/4], compare that with the classical bouncer's answer, and state how many interior nodes ψ₁ has.
- The density is ρ₁(x) = (2/L)sin²(πx/L). Use sin²θ = (1 − cos 2θ)/2, so the antiderivative is x/L − sin(2πx/L)/(2π).
- Evaluate between a = L/4 and b = 3L/4: P = (3/4 − 1/4) − [sin(3π/2) − sin(π/2)]/(2π) = 0.5 − (−1 − 1)/(2π).
- So P = 0.5 + 1/π = 0.5 + 0.3183 = 0.818. A classical particle at constant speed would spend exactly half its time there, P = 0.500.
- Nodes: sin(πx/L) vanishes only at x = 0 and x = L, and those are the walls, imposed as boundary conditions. So there are n − 1 = 0 interior nodes, as Sturm oscillation requires of any ground state.
AnswerP = ½ + 1/π = 0.818 against 0.500 classically — the ground state is 64% more likely to be found in the middle half than a bouncing bead. ψ₁ has no interior nodes.
MediumFor n = 3 in a well of width L = 1.20 nm: (a) locate the interior nodes and antinodes; (b) find the probability in the left lobe [0, L/3]; (c) compare the probability in a 0.0240 nm window centred on the node at L/3 with the same window centred on the antinode at L/6.
- (a) Nodes at x = jL/3 for j = 1, 2, i.e. 0.400 nm and 0.800 nm; the walls at 0 and 1.20 nm are boundary conditions, not nodes. Antinodes at (2j−1)L/6, i.e. 0.200, 0.600 and 1.000 nm, each with ρ = 2/L = 1.667 nm⁻¹.
- (b) Use P = (b−a)/L − [sin(2πnb/L) − sin(2πna/L)]/(2πn) with a = 0, b = L/3, n = 3. The sine arguments are 0 and 2π, both giving zero, so P = 1/3 exactly — each of the three identical lobes carries one third.
- (c) Node window: a = 0.388 nm, b = 0.412 nm, so (b−a)/L = 0.0200. The sine arguments are 2π ± 0.06π, giving sines ±0.18738; the correction is 0.37476/(6π) = 0.019882.
- Pnode = 0.0200 − 0.019882 = 1.18 × 10⁻⁴, matching the cubic estimate (nπ)²ε³/(6L³) = (3π)²(0.02L)³/(6L³) = 1.18 × 10⁻⁴.
- Antinode window at 0.200 nm: the arguments are π ± 0.06π, so the sines are ∓0.18738 and the correction changes sign: Pₐₙₜᵢ = 0.0200 + 0.019882 = 0.0399, close to the flat estimate ρₘₐₓ ε = 1.667 × 0.0240 = 0.0400.
AnswerNodes at 0.400 and 0.800 nm; antinodes at 0.200, 0.600 and 1.000 nm. P(left lobe) = 1/3 exactly. Pnode = 1.18 × 10⁻⁴ against Pantinode = 0.0399, a ratio of about 340.
HardA well of width L = 5.00 nm holds a particle in n = 20. A detector integrates the density over a window of width Δ = 0.600 nm centred at x = 1.00 nm. Find the probability it records and the mean density over that window. Then find a window width narrower than 0.600 nm that returns the classical value exactly, and say why the pointwise density never does.
- Window edges: a = 0.700 nm, b = 1.300 nm. The classical share is Δ/L = 0.600/5.00 = 0.120.
- Sine arguments: 2πnb/L = 2π(20)(1.300)/5.00 = 10.4π and 2πna/L = 5.6π. So sin(10.4π) = sin(0.4π) = 0.95106 and sin(5.6π) = sin(1.6π) = −0.95106.
- Correction: [0.95106 − (−0.95106)]/(2π × 20) = 1.90211/125.66 = 0.015137. Hence P = 0.120 − 0.01514 = 0.10486, and ⟨ρ⟩ = P/Δ = 0.10486/0.600 = 0.1748 nm⁻¹ against 1/L = 0.200 nm⁻¹ — 13% low.
- The averaged form ⟨ρₙ⟩ = 1/L − sin(nπΔ/L)cos(2πnx/L)/(nπΔ) shows why: the correction dies only when nπΔ/L is a multiple of π, i.e. Δ = kL/n = k(0.250 nm). Take Δ = 0.250 nm: edges 0.875 and 1.125 nm give arguments 7π and 9π, both sines zero, so P = 0.250/5.00 = 0.0500 = Δ/L exactly and ⟨ρ⟩ = 0.200 nm⁻¹ — at any centre.
- Pointwise nothing has improved: ρ₂₀ = (2/L)sin²(20πx/L) still runs from 0 at each of the 19 interior nodes to 0.400 nm⁻¹ at each of the 20 antinodes. Correspondence is about a window matched to the period L/n, not about the density flattening.
AnswerP = 0.1049 over the 0.600 nm window and ⟨ρ⟩ = 0.1748 nm⁻¹, 13% below 1/L. A window of exactly 0.250 nm = L/n gives P = 0.0500 = Δ/L and ⟨ρ⟩ = 0.200 nm⁻¹ at any centre, while the density itself still swings from 0 to 2/L.