University Physics IV · Quantum Potentials · 7.4
Expectation Values in a Box
Once the well's eigenfunctions are in hand, everything measurable about them comes from a handful of integrals. This lesson runs them, turns ⟨x⟩, ⟨x²⟩, ⟨p⟩ and ⟨p²⟩ into two standard deviations, and reads their product for what it says about confinement, zero-point energy, and the classical limit.
Build the model
Connect the measurement to the mechanism.
Born's rule turns a well eigenstate into a distribution, so the honest summary of ψₙ is a set of moments rather than a position. Four of them do all the work: ⟨x⟩ = L/2, which symmetry hands over free; ⟨x²⟩ = L²(1/3 − 1/(2n²π²)), the one integral you actually have to do; ⟨p⟩ = 0, because a real wavefunction vanishing at both walls cannot carry net momentum; and ⟨p²⟩ = 2mEₙ, free of charge because V = 0 inside. Two standard deviations follow, σx = L√(1/12 − 1/(2n²π²)) and σp = nπħ/L, and their product ħ√(n²π²/12 − 1/2) is 1.136 × ħ/2 in the ground state and grows without bound with n.
That one line is where zero-point energy comes from: since ⟨p⟩ = 0, the whole of ⟨p²⟩ is variance, so Eₙ = σp²/2m is variance too, and squeezing L drives it up. The cost is what these numbers refuse to be. They describe an ensemble of identically prepared measurements, never a path — |Ψ|² is time-independent and ⟨x⟩ never moves — and the infinite walls that make the algebra clean kink ψ′ hard enough that ⟨p⁴⟩ diverges.
- Simple definition
- For a well eigenstate, the expectation value of an observable is the probability-weighted mean of the values a measurement could return, and its uncertainty is the standard deviation of that same distribution, σA = √(⟨A²⟩ − ⟨A⟩²).
- Example
- For n = 1 in a 0.50 nm well, ⟨x⟩ = 0.250 nm and ⟨x²⟩ = 0.0707 nm², so σx = √(0.0707 − 0.0625) = 0.0904 nm — 18.1% of the width, not the 28.9% a classical particle bouncing between the walls would give.
Two integrals per observable — the mean, then the mean square. Squaring the mean and stopping gives zero every time.
ψₙ = √(2/L) sin(nπx/L), normalised on 0 ≤ x ≤ L; σA carries A's own unit
⟨x⟩ needs no integral: |ψₙ|² is symmetric about L/2 for every n, so the first moment is the midpoint.
L is the well width in m; n = 1, 2, 3, …; neither moment depends on the particle mass
Caps σx below 0.289L however hard you push n, and reproduces the classical uniform spread in the limit.
n = 1 gives 0.1808L; n = 3 gives 0.2788L; n → ∞ gives L/√12 = 0.2887L
Real ψ forces ⟨p⟩ = 0, and V = 0 inside makes ⟨p²⟩ just 2m times an energy you already have.
σp in kg m s⁻¹; m is the particle mass in kg, ħ = 1.055 × 10⁻³⁴ J s
Proves the bound is satisfied and never saturated, and that excited states sit further from the floor, not closer.
L cancels. n = 1 → 0.568ħ = 1.136 × (ħ/2); n = 3 → 2.63ħ; large n → 0.907nħ
With ⟨p⟩ = 0 all the kinetic energy is variance, so E₁ cannot be removed without widening the box.
Electron in L = 0.50 nm: E₁ = 2.41 × 10⁻¹⁹ J = 1.50 eV; the bound is loose by π² = 9.87
⟨x⟩ = L/2 without doing the integral
Every |ψₙ|² is symmetric about the midpoint: sin(nπ(L − x)/L) = ±sin(nπx/L), and squaring removes the sign, so the density mirrors across x = L/2. A symmetric density has its first moment at the centre, giving ⟨x⟩ = L/2 for every n with no integration at all. Do it the long way as a check: ⟨x⟩ = (2/L)∫₀ᴸ x sin²(nπx/L) dx = (1/L)∫₀ᴸ x[1 − cos(2nπx/L)] dx, and the cosine term integrates to zero over a whole number of periods, leaving (1/L)(L²/2) = L/2. Then read the result carefully. It is the mean of a distribution, not a place the particle prefers: in the n = 2 state |ψ₂(L/2)|² = 0 exactly, so the single point the mean names is the one point a measurement never returns.
⟨x²⟩, and the spread that survives
The second moment needs real work: (2/L)∫₀ᴸ x² sin²(nπx/L) dx = L²(1/3 − 1/(2n²π²)), where 1/3 is what a uniform density would give and the n-dependent term is the correction the nodes impose. Subtract ⟨x⟩² = L²/4 and the position spread is σx = L√(1/12 − 1/(2n²π²)). Put numbers in: 0.1808L at n = 1, 0.2658L at n = 2, 0.2788L at n = 3, and 0.2887L as n → ∞. That limit is exactly L/√12, the standard deviation of a uniform distribution on [0, L] — the classical answer for a particle bouncing at constant speed between the walls, recovered rather than assumed. The ground state is the tightest because its single hump piles density near the middle; each excitation pushes weight outwards, but σx can never exceed 0.2887L.
Momentum without a single integral
Neither momentum moment needs quadrature. For the mean, ⟨p⟩ = −iħ∫₀ᴸ ψₙ ψₙ′ dx = −iħ[ψₙ²/2]₀ᴸ = 0, because ψₙ is real and vanishes at both walls — true of every real bound state, not a quirk of the box. For the mean square, use the equation you already solved: V = 0 inside, so p̂²ψₙ = 2mEₙ ψₙ, giving ⟨p²⟩ = 2mEₙ = (nπħ/L)². Since ⟨p⟩ = 0, that mean square is the entire variance, so σp = nπħ/L. The physical reading is the standing wave: √(2/L) sin(nπx/L) = (1/2i)√(2/L)[e(inπx/L) − e(−inπx/L)], equal weights on +nπħ/L and −nπħ/L, which cancel in the first moment and reinforce in the second. One caution: those exponentials do not satisfy the wall conditions, so the true momentum density is a pair of broadened peaks, not two spikes.
Multiply, and watch the floor recede
Now combine them: σx σp = L√(1/12 − 1/(2n²π²)) × nπħ/L = ħ√(n²π²/12 − 1/2). The width cancels, so the product depends on n alone — a 0.20 nm well and a 2.0 nm well give the same answer. It runs 0.568ħ, 1.670ħ and 2.627ħ for n = 1, 2, 3, and reaches 9.04ħ at n = 10. Three readings follow. The bound holds everywhere, since even the smallest value is 1.136 × ħ/2. The bound is never saturated, because saturation demands a Gaussian and a sine hump clipped at two walls is not one — the ground state misses by 14%. And excited states are worse, not better: σp grows exactly as n while σx is already near its ceiling, so the product climbs towards nπħ/√12 = 0.907nħ, without limit.
Zero-point energy is the bill for confinement
Because ⟨p⟩ = 0, every joule of Eₙ = ⟨p²⟩/2m is momentum variance — and variance is precisely what the uncertainty bound puts a floor under. Anything trapped in [0, L] has σx ≤ L/2, so σp ≥ ħ/(2σx) ≥ ħ/L and therefore E ≥ ħ²/(2mL²). The exact ground state is π² times that, E₁ = π²ħ²/(2mL²): the estimate has the right form and is loose by a factor 9.87, which is what an inequality is for. The scale it sets is the whole point. An electron in L = 0.50 nm gets E₁ = 2.41 × 10⁻¹⁹ J = 1.50 eV, a visible-light energy, which is why a quantum dot's diameter fixes its colour. A 1.0 g bead in a 10 cm box gets E₁ = 5.5 × 10⁻⁶³ J, a speed of 3.3 × 10⁻³⁰ m s⁻¹. Same formula; only one of them is measurable.
What these numbers do not say
|Ψₙ(x, t)|² = |ψₙ(x)|² for all t, so nothing here moves: ⟨x⟩ sits at L/2 forever, σx never changes, and Ehrenfest's d⟨x⟩/dt = ⟨p⟩/m = 0 agrees. There is no bouncing and no orbit to picture — motion appears only once two energies are superposed, which is a different lesson. Nor is σx a blur around one particle: it is the spread of outcomes across many identically prepared measurements, each of which returns a sharp position. It is not a Gaussian σ either — for n = 1 only 65.0% of results land within one σx of L/2, against 68.3% for a normal distribution. Finally the idealisation sends a bill: the wall kinks ψ′, so the momentum density falls only as p⁻⁴. ⟨p²⟩ survives that tail, ⟨p⁴⟩ diverges, and p̂ is not self-adjoint on this domain at all.
Change one variable at a time
Make the relationship visible.
Hold L at 0.50 nm and step n from 1 to 8: the bracket only widens from 0.181 to 0.287 nm while σp climbs eightfold, so the product runs away from the ħ/2 floor. Then shrink L at fixed n — σx falls, σp rises, and the product does not budge.
POSITION SPREAD σx0.090 nm
MOMENTUM SPREAD σp0.66 e-24 kg m/s
σx σp IN UNITS OF ħ/21.136 × ħ/2
ELECTRON ENERGY Eₙ1.50 eV
Live interpretationPOSITION SPREAD σx: 0.090 nm. MOMENTUM SPREAD σp: 0.66 e-24 kg m/s. σx σp IN UNITS OF ħ/2: 1.136 × ħ/2. ELECTRON ENERGY Eₙ: 1.50 eV
Catch the common trap
Explain before calculating.
An electron is in the n = 3 stationary state of an infinite square well of width L. Which statement about its position and momentum spreads is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron occupies the n = 1 state of an infinite square well of width L = 0.50 nm. Find ⟨x⟩, σx, σp, and the product σx σp expressed in units of ħ/2.
- |ψ₁|² = (2/L)sin²(πx/L) is symmetric about the midpoint, so ⟨x⟩ = L/2 = 0.250 nm. ψ₁ is real and vanishes at both walls, so ⟨p⟩ = −iħ[ψ₁²/2]₀ᴸ = 0.
- ⟨x²⟩ = L²(1/3 − 1/(2π²)) = 0.25 nm² × (0.33333 − 0.05066) = 0.070668 nm². Then σx² = 0.070668 − 0.0625 = 0.008168 nm², so σx = 0.0904 nm — 18.1% of the width.
- Inside the well V = 0, so ⟨p²⟩ = 2mE₁ = (πħ/L)². With ⟨p⟩ = 0 that is the whole variance: σp = πħ/L = h/(2L) = 6.626 × 10⁻³⁴ ÷ (1.00 × 10⁻⁹) = 6.63 × 10⁻²⁵ kg m s⁻¹.
- σx σp = (9.04 × 10⁻¹¹ m)(6.63 × 10⁻²⁵ kg m s⁻¹) = 5.99 × 10⁻³⁵ J s. Divide by ħ/2 = 5.27 × 10⁻³⁵ J s to get 1.14.
Answer⟨x⟩ = 0.250 nm, σx = 0.0904 nm, σp = 6.63 × 10⁻²⁵ kg m s⁻¹, and σx σp = 0.568 ħ = 1.14 × (ħ/2) — above the floor, not sitting on it.
MediumThe same electron is excited to n = 3 in the same 0.50 nm well. Find σx, σp and σx σp, then account for the fact that the product grows by a factor 4.63 when n triples.
- ⟨x²⟩ = L²(1/3 − 1/(18π²)) = 0.25(0.333333 − 0.005629) = 0.081926 nm². Subtract ⟨x⟩² = 0.0625 nm² to get σx² = 0.019426 nm², so σx = 0.1394 nm.
- σp = 3πħ/L = 3 × 6.63 × 10⁻²⁵ = 1.988 × 10⁻²⁴ kg m s⁻¹, and E₃ = 9E₁ = 13.5 eV.
- σx σp = (1.394 × 10⁻¹⁰ m)(1.988 × 10⁻²⁴ kg m s⁻¹) = 2.77 × 10⁻³⁴ J s = 2.63 ħ = 5.25 × (ħ/2).
- The two factors do not scale alike. σp is exactly proportional to n, so it triples. σx only rises from 0.1808L to 0.2788L, a factor 1.542, because it is already near its ceiling of L/√12 = 0.2887L. Multiply: 3 × 1.542 = 4.63.
Answerσx = 0.139 nm, σp = 1.99 × 10⁻²⁴ kg m s⁻¹, σx σp = 2.63 ħ = 5.25 × (ħ/2). Exciting the state buys momentum spread at almost no cost in position spread.
HardFor the n = 1 state of the well, what fraction of position measurements falls within one standard deviation of the mean? Compare it with the 68.3% a Gaussian of the same σ would give, and say what the comparison shows about σx.
- σx = 0.180756L, so the interval in units of u = x/L runs from u₁ = 0.5 − 0.180756 = 0.319244 to u₂ = 0.680756.
- ∫(2/L)sin²(πx/L) dx has antiderivative F(u) = u − sin(2πu)/(2π), so the probability is F(u₂) − F(u₁).
- 2πu₁ = 2.00587 rad and 2πu₂ = 4.27732 rad. The two points are mirror images about u = 1/2, so sin(2πu₁) = +0.90684 and sin(2πu₂) = −0.90684.
- P = (0.680756 − 0.319244) − (−0.90684 − 0.90684)/(2π) = 0.361512 + 1.81368/6.28319 = 0.361512 + 0.288657 = 0.65017.
- So 65.0%, about 3.3 percentage points below the Gaussian figure. The single sine hump is flatter across the middle than a bell curve and is cut off hard at the walls, so it puts less weight close to the mean and none at all beyond x = 0 or x = L.
AnswerP = 65.0%, against 68.3% for a normal distribution. σx is a second moment, not a 68% confidence interval; the two coincide only when the density happens to be Gaussian.