Expectation Values for the Particle in a Box

Introduction

A single measurement of the position of a particle in a box can give any value between the walls. What quantum mechanics predicts precisely is the average of a large number of measurements on identically prepared systems: the expectation value. Calculating ⟨x⟩, ⟨x²⟩, ⟨p⟩ and ⟨p²⟩ for the box is a classic exercise that brings together operators, integration and the uncertainty principle, and it shows how kinetic energy and confinement are linked.

Core explanation

The recipe. For a normalised wavefunction, the expectation value of an observable with operator Ω̂ is

⟨Ω⟩ = ∫ ψ Ω̂ ψ dx

For ψₙ = (2/L)^½ sin(nπx/L), the functions are real, so ψ = ψ.

Mean position. The position operator is simply multiplication by x:

⟨x⟩ = (2/L) ∫₀ᴸ x sin²(nπx/L) dx = L/2

This result holds for every n. It follows from symmetry: ψₙ² is symmetric about the centre, so positions to the left and right of L/2 are equally likely and average to the midpoint.

Mean-square position. A longer integration (by parts, twice) gives

⟨x²⟩ = L² [1/3 − 1/(2n²π²)]

so the spread in position is

Δx = (⟨x²⟩ − ⟨x⟩²)^½ = L [1/12 − 1/(2n²π²)]^½

For n = 1 this is 0.181L. As n increases, Δx approaches L/√12 = 0.289L, the value for a uniform distribution.

Mean momentum. The momentum operator is p̂ = −iħ d/dx. Acting on ψₙ gives −iħ(nπ/L)(2/L)^½ cos(nπx/L), so

⟨p⟩ = −iħ(2/L)(nπ/L) ∫₀ᴸ sin(nπx/L) cos(nπx/L) dx = 0

because sin θ cos θ = ½ sin 2θ integrates to zero over whole half-periods. Physically, each box state is a standing wave, an equal mixture of waves travelling to the right (momentum +nh/2L) and to the left (−nh/2L), so the average momentum is zero.

Mean-square momentum. Applying p̂ twice gives p̂² = −ħ² d²/dx², and d²ψₙ/dx² = −(nπ/L)²ψₙ. Therefore

⟨p²⟩ = ħ²(nπ/L)² ∫ψₙ² dx = n²π²ħ²/L² = n²h²/4L²

Since the potential energy inside the box is zero, the energy is all kinetic: Eₙ = ⟨p²⟩/2m = n²h²/8mL², which reproduces the familiar energy formula. Because ⟨p⟩ = 0, the momentum spread is Δp = ⟨p²⟩^½ = nh/2L.

Uncertainty check. For n = 1, ΔxΔp = 0.181L × h/2L = 0.568ħ, which is greater than the minimum ħ/2 = 0.5ħ. The Heisenberg relation is satisfied, and the product grows with n.

Formulae

⟨x⟩ = L/2; ⟨x²⟩ = L²[1/3 − 1/(2n²π²)]; ⟨p⟩ = 0; ⟨p²⟩ = n²h²/4L²; Eₙ = ⟨p²⟩/2m; Δp = nh/2L.

Step-by-step reasoning

To calculate an expectation value for the box:

1. Confirm the wavefunction is normalised. 2. Write the operator: x for position, −iħ d/dx for momentum, −ħ² d²/dx² for p². 3. Let the operator act on ψ first, then multiply by ψ . 4. Integrate from 0 to L, using standard integrals or symmetry. 5. Interpret the result as the mean of many measurements.

Visual explanation

Picture ψ₁² as a dome centred at L/2: its balance point is at the middle, which is ⟨x⟩. Now picture the standing wave as two arrows of equal length pointing left and right: they cancel to give ⟨p⟩ = 0, but their squared lengths add, giving a non-zero ⟨p²⟩.

Real-world analogy

A pendulum swinging back and forth has an average velocity of zero over a full swing, yet it clearly has kinetic energy. The average of the square of velocity is not zero. The particle in a box behaves the same way: ⟨p⟩ vanishes but ⟨p²⟩ does not.

Real-world example

In conjugated dyes, the π electrons confined along the chain have a kinetic energy set by ⟨p²⟩ = n²h²/4L². Lengthening the chain lowers ⟨p²⟩ and the spacing of the energy levels, which is why longer conjugated molecules absorb at longer wavelengths.

Why?

Why is ⟨p²⟩ never zero, even in the ground state? Confining a particle to a length L forces its wavefunction to curve, and curvature means kinetic energy. The narrower the box, the sharper the curvature and the larger ⟨p²⟩, exactly as the uncertainty principle requires.

Common misconception

"If ⟨p⟩ = 0, the particle is stationary." A zero average momentum only means that motion to the left and right is equally likely. The non-zero ⟨p²⟩ shows that the particle always has kinetic energy; a measurement of momentum would give ±nh/2L (spread somewhat by the finite box), never exactly zero.

Worked example

Question: An electron is in the n = 1 state of a box of length 1.00 nm. Calculate Δp and the root-mean-square speed.

Reasoning: Δp = h/2L = 6.626 × 10⁻³⁴ / (2 × 1.00 × 10⁻⁹) = 3.31 × 10⁻²⁵ kg m s⁻¹. The rms speed is ⟨p²⟩^½/m = 3.31 × 10⁻²⁵ / 9.109 × 10⁻³¹ = 3.64 × 10⁵ m s⁻¹.

Answer: Δp ≈ 3.3 × 10⁻²⁵ kg m s⁻¹; rms speed ≈ 3.6 × 10⁵ m s⁻¹.

Quick check

1. Why is ⟨x⟩ equal to L/2 for every state of the particle in a box? Answer: Because every ψₙ² is symmetric about the centre, so equal probabilities on each side average to the midpoint.

Exam focus

Examiners expect you to show the operator acting before integrating, especially for momentum. State that ⟨p⟩ = 0 because the state is a standing wave, and link ⟨p²⟩ to the energy through E = ⟨p²⟩/2m. A final uncertainty check (ΔxΔp ≥ ħ/2) earns extra marks.

Advanced insight

The box wavefunctions are eigenfunctions of p̂² but not of p̂, because applying −iħ d/dx turns a sine into a cosine. So kinetic energy has a definite value in each state while momentum does not. Only operators that commute share eigenfunctions; p̂ and Ĥ commute for a free particle, but the walls of the box spoil that relationship.

Summary

For a particle in a box, ⟨x⟩ = L/2 by symmetry and ⟨x²⟩ = L²[1/3 − 1/(2n²π²)]. The average momentum is zero because each state is a standing wave, while ⟨p²⟩ = n²h²/4L² gives the kinetic energy n²h²/8mL². The spreads Δx and Δp satisfy the uncertainty principle, with ΔxΔp ≈ 0.57ħ in the ground state.

Practice questions

1. Show that ⟨p²⟩/2m reproduces the particle-in-a-box energy. Answer: ⟨p²⟩/2m = (n²h²/4L²)/2m = n²h²/8mL², which is Eₙ. 2. Calculate Δx for the n = 2 state in terms of L. Answer: Δx = L[1/12 − 1/(8π²)]^½ = L(0.0833 − 0.0127)^½ = 0.266L. 3. Why is the box wavefunction not an eigenfunction of p̂? Answer: Because −iħ d/dx turns sin(nπx/L) into a multiple of cos(nπx/L), which is not a constant times the original function. 4. How does ⟨p²⟩ change if the box length is halved? Explain physically. Answer: It increases by a factor of four, since ⟨p²⟩ ∝ 1/L². Tighter confinement forces greater curvature of the wavefunction and so greater kinetic energy.