University Physics V · Particle in a Box · 5.7
Expectation Values & the σₓ σₚ Product
Once the eigenfunctions are in hand, every measurable prediction is one sandwich away. This is where you learn which moments symmetry hands you free, which single integral has to be done by force, and why the ground state's momentum spread is the zero-point energy in disguise.
Build the model
Connect the measurement to the mechanism.
A stationary state is a probability distribution, and everything measurable about it is a moment of that distribution taken with the right operator. Which operator matters: |ψₙ|² alone delivers every moment of position, but momentum needs p̂ = −iħ d/dx acting on the state itself, and it returns numbers the density never hints at. Parity about x = L/2 settles the two first moments without an integral — ⟨x⟩ = L/2 and ⟨p⟩ = 0 for every n — so all the physics sits in the second moments.
One genuine integral gives ⟨x²⟩ = L²(1/3 − 1/(2n²π²)); the Schrödinger equation gives ⟨p²⟩ = 2mEₙ = (nπħ/L)² for free, because V = 0 inside. Their standard deviations multiply to σₓ σₚ = (ħ/2)√(n²π²/3 − 2), which is 0.568ħ in the ground state: above Heisenberg's floor, never on it, and growing without bound as n rises. What the model costs is any picture of motion — σₚ is the half-separation of a two-peaked momentum distribution, not a speed, and σₓ is the width of a density that does not move at all.
What it buys is E₁ = σₚ²/(2m) = π²ħ²/(2mL²), a zero-point energy charged by the boundary condition alone, which no amount of cooling can remove.
- Simple definition
- An expectation value is the sandwich ⟨A⟩ = ⟨ψ|Â|ψ⟩ — the mean of many measurements of A on identically prepared copies — and its uncertainty is the standard deviation σA = √(⟨²⟩ − ⟨A⟩²) of that same distribution.
- Example
- In the box ground state ⟨x⟩ = L/2 and ⟨x²⟩ = 0.28267L², so σₓ = 0.18076L. For L = 1.00 nm that is a spread of 0.181 nm — well under a fifth of the box, not half of it.
Both are ensemble statistics. No single measurement returns ⟨A⟩, and σA is not an instrument error.
 Hermitian, ⟨ψ|ψ⟩ = 1; σA carries A's own unit
σₓ = L√(1/12 − 1/(2n²π²)): 0.1808L at n = 1, 0.2658L at n = 2, rising to 0.2887L.
L in m, n = 1, 2, 3…; ⟨x⟩ is the same for every n
σₚ = nπħ/L with no integral done — and it equals the classical |p| at the same energy exactly.
p in kg m s⁻¹; ⟨p²⟩ comes from H, since V = 0 inside
0.5679ħ at n = 1, 1.6703ħ at n = 2, 5.3953ħ at n = 6, approaching 0.9069nħ. L cancels.
the bracket is dimensionless; ħ = 1.0546 × 10⁻³⁴ J s
That condition integrates to a Gaussian, which is nowhere zero — so no box state can ever sit on the floor.
equality needs (x̂ − ⟨x⟩)|ψ⟩ = iλ(p̂ − ⟨p⟩)|ψ⟩ with λ real
Halve L and E₁ quadruples. It scales with confinement, not with temperature or with any material property.
m in kg, L in m; an electron in L = 1.00 nm gives 0.376 eV
Every number is the same sandwich
For a normalised state the mean of a long run of A measurements is ⟨A⟩ = ⟨ψ|Â|ψ⟩, and the spread of that run is σA = √(⟨²⟩ − ⟨A⟩²). In a stationary state the time dependence is the global phase e(−iEₙ t/ħ), which cancels between bra and ket, so every moment computed here is a constant: nothing in |n⟩ evolves. Now watch which object each moment needs. Position moments read straight off the density, ⟨xk⟩ = ∫₀L xk |ψₙ(x)|² dx, so |ψₙ|² is sufficient. Momentum moments are not like that: ⟨p²⟩ = ∫₀L ψₙ*(−ħ²ψₙ″) dx requires the operator acting on the state, and |ψₙ|² is identical for ψₙ and for ψₙ e(iθ(x)), two states with completely different momentum content. So the working habit is to name the operator and its domain before integrating anything — here p̂ = −iħ d/dx and Ĥ = p̂²/2m, both on square-integrable functions with square-integrable second derivative satisfying ψ(0) = ψ(L) = 0 — and only then to reach for a table of integrals.
Parity about the midpoint does two integrals for free
Let Π be reflection about the centre, (Πf)(x) = f(L − x). It commutes with H because the box is symmetric, and it maps the eigenbasis onto itself: Πψₙ = (−1)(n+1) ψₙ, so ψ₁ is even about L/2, ψ₂ is odd, and so on. Split the position operator as x̂ = L/2 + (x̂ − L/2). The second piece is odd under Π while |ψₙ|² is even, so its integral vanishes and ⟨x⟩ = L/2 for every n, with no integration performed. The same argument kills ⟨p⟩: p̂ is odd under Π, so ⟨n|p̂|n⟩ = ⟨n|Π†p̂Π|n⟩ = −⟨n|p̂|n⟩, and the only number equal to minus itself is zero. There is a blunter second route worth knowing, because it generalises to any real bound state: ψₙ is real up to a constant phase, so ⟨p⟩ = −iħ∫₀L ψₙ ψₙ′ dx = (−iħ/2)[ψₙ²]₀L = 0, killed by the boundary condition itself. Both routes say a real standing wave carries no net momentum. Neither says the particle is at rest.
The one integral you actually do: ⟨x²⟩
Write ⟨x²⟩ = (2/L)∫₀L x² sin²(nπx/L) dx and split sin² = (1 − cos(2nπx/L))/2. The first half gives (2/L)(L³/6) = L²/3. The second needs ∫₀L x² cos(2nπx/L) dx; two integrations by parts leave only the term in cos(2nπ) = 1, giving L³/(2n²π²). So ⟨x²⟩ = L²/3 − L²/(2n²π²), and σₓ² = ⟨x²⟩ − (L/2)² = L²(1/12 − 1/(2n²π²)). Recognise the two pieces. L²/3 and L²/12 are exactly the second moment and the variance of a uniform density 1/L on [0, L] — the classical bouncer. All the quantum content is the 1/(2n²π²) deficit, which removes 60.8% of the classical variance in the ground state, leaving σₓ = 0.1808L, then shrinks fast: 0.2658L at n = 2, 0.2788L at n = 3, 0.2862L at n = 6, against the classical 0.2887L. Note what converges. σₓ is an average over the whole box and it does converge; |ψₙ|² itself still swings between 0 and 2/L at every n and never converges pointwise to 1/L.
Momentum: the second moment comes from H, not from an integral
Inside the box V = 0, so H = p̂²/2m and ⟨p²⟩ = 2m⟨H⟩ = 2mEₙ = (nπħ/L)². No integral required. With ⟨p⟩ = 0 that makes σₚ = nπħ/L — exactly |p| for a classical particle of the same energy bouncing between the walls, whose momentum distribution is two spikes at ±nπħ/L with mean zero and standard deviation nπħ/L. The box's momentum spread is already classical; only its position spread is not. The quantum version smears those spikes: |φ(p)|² ∝ n²[1 − (−1)ⁿ cos(pL/ħ)] / [(nπ)² − (pL/ħ)²]², two peaks of width of order ħ/L, unresolved at n = 1 and clearly separate by n = 4. Their tails fall only as p⁻⁴, and that is a warning. ⟨p²⟩ converges, and Parseval applied to ψₙ′ confirms ħ²(nπ/L)². But ∫p⁴|φ(p)|² dp diverges, while squaring 2mH returns a finite ħ⁴(nπ/L)⁴. The two disagree because ψₙ has a kink at each wall: ψₙ lies in the domain of p̂, but p̂ψₙ does not, so p̂ cannot legally be applied twice. Topic 5.1's remark that p̂ is symmetric but not self-adjoint here surfaces as an actual number.
Why the product is 0.568ħ and never ħ/2
Multiply the two spreads: σₓ σₚ = L√(1/12 − 1/(2n²π²)) × nπħ/L. The L cancels — the product cannot depend on the box size, which is why it is the clean invariant of this problem — and tidying gives (ħ/2)√(n²π²/3 − 2). At n = 1 that is 0.5679ħ, 13.6% above the Robertson floor; at n = 2, 1.6703ħ; at n = 6, 5.3953ħ; asymptotically nπħ/√12 = 0.9069nħ. Two expectations students bring are false. First, the ground state does not saturate the bound. Saturation requires (x̂ − ⟨x⟩)|ψ⟩ = iλ(p̂ − ⟨p⟩)|ψ⟩ with λ real, a first-order differential equation whose only solution is a Gaussian; a Gaussian is nowhere zero, so no state obeying ψ(0) = ψ(L) = 0 can ever reach ħ/2. Second, high n does not creep toward the bound — it runs away, because σₚ grows linearly in n while σₓ is capped below L/√12. One loose end: Schrödinger's sharper inequality adds a covariance term ½⟨(Δx̂, Δp̂)⟩, but ψₙ is real, so x̂ψₙ is real and p̂ψₙ is purely imaginary, that term vanishes, and Robertson's bound is the whole story here.
Zero-point energy, and the moments as matrix sums
E₁ = σₚ²/(2m) = π²ħ²/(2mL²) is not an approximation and not a thermal residue: it is the price of the Dirichlet condition. An electron in L = 1.00 nm has E₁ = h²/(8mL²) = 0.376 eV, about fifteen times kBT at 300 K (0.0259 eV), and cooling the sample does not touch it. Halve the box and E₁ quadruples while σₚ doubles and σₓ halves. The same moments make a sharp numerical check in the energy basis, where completeness gives ⟨n|x̂²|n⟩ = Σₘ |xₘₙ|². With xₙₙ = L/2 and xₘₙ = −8Lmn/(π²(m² − n²)²) for odd m − n, truncating at N = 4 for n = 1 gives (0.250000 + 0.032445 + 0.000208)L² = 0.282653L² against the exact 0.282673L² — 0.007% low, and low from below, since every omitted term is positive. Do the same for p and the identical truncation gives (8/3)² + (16/15)² = 8.249 (ħ/L)² against π² = 9.870, 16.4% low. The tail is the reason: |x₁m|² falls as m⁻⁶ while |p₁m|² falls only as m⁻², so ⟨p²⟩ needs of order eighty basis states for 1%. Truncation is not a uniform approximation; it punishes whichever operator has the slower matrix elements.
Change one variable at a time
Make the relationship visible.
Raise n: the two momentum dots march outward in proportion to n while the σₓ bar only creeps from 0.181 L to 0.286 L, so the product climbs as 0.907 nħ. Then hold n and stretch L — the bar widens, the dots close in, and the product does not move at all.
σₓ / L0.1808
σₚ3.14 ħ/nm
σₓ σₚ0.568 ħ
Eₙ, electron0.38 eV
Live interpretationσₓ / L: 0.1808. σₚ: 3.14 ħ/nm. σₓ σₚ: 0.568 ħ. Eₙ, electron: 0.38 eV
Catch the common trap
Explain before calculating.
An electron occupies the n = 3 state of a one-dimensional infinite square well of width L. Which statement about the product σₓ σₚ is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is confined to a one-dimensional infinite square well of width L = 0.500 nm, in its ground state. Find σₓ, σₚ, their product in units of ħ, and E₁ in eV. Take ħ = 1.0546 × 10⁻³⁴ J s and m = 9.109 × 10⁻³¹ kg.
- σₓ = L√(1/12 − 1/(2π²)) = L√(0.083333 − 0.050661) = L√0.032673 = 0.18076L = 9.038 × 10⁻¹¹ m.
- σₚ = nπħ/L with n = 1: π(1.0546 × 10⁻³⁴)/(5.00 × 10⁻¹⁰) = 6.626 × 10⁻²⁵ kg m s⁻¹.
- σₓ σₚ = (9.038 × 10⁻¹¹)(6.626 × 10⁻²⁵) = 5.989 × 10⁻³⁵ J s. Divide by ħ: 0.568. The floor ħ/2 = 5.273 × 10⁻³⁵ J s is cleared by 13.6%.
- Since ⟨p⟩ = 0, ⟨p²⟩ = σₚ², so E₁ = σₚ²/(2m) = (6.626 × 10⁻²⁵)²/(1.8219 × 10⁻³⁰) = 2.410 × 10⁻¹⁹ J = 1.50 eV. The check h²/(8mL²) returns the same 1.504 eV.
Answerσₓ = 0.0904 nm, σₚ = 6.63 × 10⁻²⁵ kg m s⁻¹, σₓ σₚ = 0.568ħ, E₁ = 1.50 eV.
MediumShow that σₓ for the box state |n⟩ rises toward the classical value L/√12 strictly from below, find the smallest n for which it sits within 1% of that value, and evaluate σₓ σₚ there.
- σₓ(n) = L√(1/12 − 1/(2n²π²)). Divide by the classical L/√12, noting that (1/(2n²π²)) ÷ (1/12) = 6/(n²π²): the ratio is √(1 − 6/(n²π²)).
- That ratio increases with n and is strictly less than 1 for every finite n, so σₓ approaches L/√12 = 0.28868L from below and never reaches it.
- Within 1% means √(1 − 6/(n²π²)) ≥ 0.99, so 6/(n²π²) ≤ 1 − 0.9801 = 0.0199, so n² ≥ 6/(0.0199 × 9.8696) = 30.55 and n ≥ 5.53.
- Test the integers: n = 5 gives ratio 0.9878, which is 1.22% low; n = 6 gives 0.9915, which is 0.85% low. So n = 6, with σₓ = 0.9915 × 0.28868L = 0.28623L.
- At n = 6, σₓ σₚ = (ħ/2)√(36π²/3 − 2) = (ħ/2)√116.435 = 5.395ħ. The position spread is within 1% of classical while the product sits almost eleven times the Heisenberg floor — the two limits have nothing to do with each other.
Answern = 6, where σₓ = 0.2862L, 0.85% below the classical L/√12 = 0.2887L, and σₓ σₚ = 5.40ħ.
HardIn the energy basis xₙₙ = L/2 and xₘₙ = −8Lmn/(π²(m² − n²)²) for odd m − n, while pₙₙ = 0 and pₘₙ = −4iħmn/(L(m² − n²)) for odd m − n. Using completeness, evaluate ⟨1|x̂²|1⟩ and ⟨1|p̂²|1⟩ from a 4 × 4 truncation, compare each with the exact value, and say why the two truncations are not equally good.
- Inserting Σₘ |m⟩⟨m| gives ⟨n|²|n⟩ = Σₘ |Aₘₙ|² for Hermitian  — a sum of non-negative terms, so any truncation is an underestimate, never an overestimate.
- For x with n = 1 the surviving m ≤ 4 are m = 1, 2, 4: x₁₁ = L/2 → 0.250000L²; x₂₁ = −16L/(9π²) = −0.180127L → 0.032446L²; x₄₁ = −32L/(225π²) = −0.014410L → 0.000208L². Sum = 0.282653L².
- Exact ⟨x²⟩ = L²(1/3 − 1/(2π²)) = 0.282673L². The truncation is low by 6.9 × 10⁻⁵ in relative terms — 0.007%.
- For p with n = 1: p₂₁ = −8iħ/(3L) → (8/3)² = 7.1111 (ħ/L)²; p₄₁ = −16iħ/(15L) → (16/15)² = 1.1378 (ħ/L)². Sum = 8.2489 (ħ/L)² against the exact ⟨p²⟩ = π²ħ²/L² = 9.8696 (ħ/L)² — 16.4% low.
- The tails decide it. |x₁m|² ∝ m⁻⁶, so its omitted tail is negligible, while |p₁m|² ≈ 16/m², whose omitted tail is about 8/N. Setting 8/N = 0.01 × π² needs N ≈ 80 basis states for 1% on ⟨p²⟩.
Answer⟨x²⟩ ≈ 0.28265L², 0.007% low; ⟨p²⟩ ≈ 8.249ħ²/L², 16.4% low. Both err downward, but p² converges far more slowly because its matrix elements decay as 1/m rather than 1/m³.