Quantum Mechanics 101 · Unit 3: The Mathematical Machinery

Quantum Academy · Chapter 11: Particle in a Box

Trap a wave between two walls and quantization appears on its own

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Instructions. Try every question with the lesson closed, then check your work. Use h = 6.626×10⁻³⁴ J·s, electron mass 9.109×10⁻³¹ kg, proton mass 1.673×10⁻²⁷ kg, 1 eV = 1.602×10⁻¹⁹ J and hc = 1239.8 eV·nm. Once you have derived it, you may use the shortcut that an electron's E₁ is 0.376 eV divided by the square of the box width in nanometres. Show every step in the calculation questions. One point for each vocabulary match and each question; the challenge is a bonus.

Part A Vocabulary matching

Write the letter of the matching definition in the blank beside each term.

1.Infinite square well
2.Boundary condition
3.Stationary state
4.Quantum number
5.Node
6.Ground state
7.Zero-point energy
8.Level spacing
9.Quantum beat
10.Quantum dot
11.Conjugated molecule
12.Correspondence principle

A.The oscillation at frequency (E₂ − E₁)/h that appears when two energy states are superposed, making the probability density slosh.
B.A potential that is zero inside a region of width L and infinite outside it, so a particle is perfectly confined; also called the particle in a box.
C.The minimum energy a confined quantum system keeps even at absolute zero; E₁ = h²/(8·m·L²) for the box.
D.The lowest-energy state of a system; n = 1 for the particle in a box.
E.The requirement, in its modern textbook form, that quantum predictions reproduce classical physics in the limit of large quantum numbers.
F.A semiconductor nanocrystal a few nanometres across that confines electrons in all three directions, so its colour is set by its size.
G.A state of definite energy whose probability density does not change in time; also called an energy eigenstate.
H.A point where the wavefunction is zero, so the probability of finding the particle exactly there is zero.
I.A molecule with alternating single and double bonds whose π electrons spread along the chain as if confined in a one-dimensional box.
J.An integer label, here n = 1, 2, 3 and so on, that counts the allowed states and fixes their energies.
K.A requirement a solution must satisfy at the edges of its region; for the box, ψ = 0 at both walls.
L.The energy gap between neighbouring levels; in the box it is (2n + 1)E₁ and grows as the box shrinks.

Part B Questions

1.
An electron is trapped in an infinite square well. Which single change makes its ground-state energy four times larger?Circle one
1. AHalving the box width
2. BDoubling the box width
3. CDoubling the electron's mass
4. DRaising the temperature fourfold
2.
How many interior nodes does the wavefunction ψ₅ have?Circle one
1. A5
2. B6
3. C4
4. D0
3.
For which of these states is the probability density exactly zero at the centre of the box?Circle one
1. An = 1
2. Bn = 2
3. Cn = 3
4. Dn = 5
4.
The ground-state energy of a particle in an infinite square well is zero.Circle one
TrueFalse
5.
In the particle in a box, the energy gap between neighbouring levels grows as you climb the ladder.Circle one
TrueFalse
6.
In the n = 1 stationary state, the probability density slowly sloshes from one side of the box to the other.Circle one
TrueFalse
7.
Requirements that a wavefunction must meet at the edges of a region, such as ψ = 0 at both walls of the box, are called conditions.Fill in the blank
8.
Making a quantum dot smaller shifts the light it emits toward the end of the spectrum.Fill in the blank
9.
In two or three sentences, explain why a particle in a box has quantized energy while a free particle does not.Short answer
10.
Why does an equal superposition of n = 1 and n = 2 slosh back and forth, while either state on its own has a probability density that never changes?Short answer
11.
An electron is confined to an infinite well 0.50 nm wide. Find its ground-state energy in eV and the wavelength of the photon emitted in the jump from n = 2 to n = 1. In which part of the spectrum does it lie?Show your working
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12.
Model a proton inside a small nucleus as a particle in a one-dimensional box of width 5.0 fm (5.0×10⁻¹⁵ m). Estimate its ground-state energy in MeV.Show your working
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Answer:
13.
For the ground state ψ₁ = √(2/L) sin(πx/L), find the probability of finding the particle in the middle third of the box, between L/3 and 2L/3. Compare with a classical particle bouncing at constant speed.Show your working
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14.
An electron in a 2.0 nm box is placed in an equal superposition of n = 1 and n = 2. How long does one full slosh take?Show your working
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Part C Mad scientist challenge bonus

Prove that the box obeys Heisenberg. For ψₙ you are given ⟨x⟩ = L/2, ⟨x²⟩ = L²(1/3 − 1/(2n²π²)), ⟨p⟩ = 0 and ⟨p²⟩ = 2mEₙ. (a) Find Δx and Δp, then the product Δx·Δp for n = 1 and n = 2, and check each against Δx·Δp ≥ ħ/2. (b) Show that as n grows, Δx approaches the value for a classical particle spread evenly across the box. (c) Explain in one sentence how part (a) guarantees zero-point energy.

Working