Quantum Mechanics 101 · Unit 3: The Mathematical Machinery

Quantum Academy · Chapter 13: The Quantum Harmonic Oscillator

The universe's favourite spring, and the ladder of evenly spaced energies

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Instructions. Work through every question with the lesson closed, then check yourself. Use ħ = 1.055×10⁻³⁴ J·s, h = 6.626×10⁻³⁴ J·s, c = 2.998×10⁸ m/s, 1 eV = 1.602×10⁻¹⁹ J, 1 atomic mass unit = 1.661×10⁻²⁷ kg and, at room temperature, kT ≈ 0.0257 eV. A handy conversion: a wavenumber of 1 inverse centimetre corresponds to a photon energy of 1.240×10⁻⁴ eV. 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.Harmonic oscillator
2.Spring constant
3.Zero-point energy
4.Classical turning point
5.Hermite polynomials
6.Ladder operators
7.Quantum of excitation
8.Phonon
9.Reduced mass
10.Selection rule
11.Anharmonicity
12.Coherent state

A.A position where a classical oscillator of given energy momentarily stops and reverses, because all its energy is potential.
B.A system whose restoring force is proportional to displacement, F = −kx, giving a parabolic potential and an oscillation frequency that does not depend on amplitude.
C.μ = m₁m₂/(m₁ + m₂), the effective mass of a two-body vibration such as a diatomic molecule.
D.The raising and lowering operators, which move an oscillator state up or down by one rung of energy ħω.
E.A restriction on which transitions light can drive; for an ideal harmonic oscillator, n may change by exactly one.
F.The departure of a real potential from a perfect parabola, which makes the energy rungs unequal, usually closer together higher up.
G.The family of polynomials that, multiplied by a Gaussian, give the oscillator's stationary states; the one of degree n has n zeros.
H.The stiffness k in F = −kx, in newtons per metre; near any potential minimum, the curvature of the potential energy.
I.An oscillator state that swings like a classical one without spreading, with minimum uncertainty and a Poisson spread in its number of quanta.
J.One unit of energy ħω added to an oscillator; the quantum number n counts how many are present.
K.The energy a quantum system keeps above the bottom of its potential even in its lowest state; ½ħω for a harmonic oscillator, present even at absolute zero.
L.A quantum of vibration of a crystal lattice.

Part B Questions

1.
Why does the harmonic oscillator describe the small vibrations of so many different systems?Circle one
1. ABecause most forces in nature are exactly proportional to distance
2. BBecause near the bottom of almost any smooth potential minimum, the potential is very nearly a parabola
3. CBecause quantum mechanics forces every potential to become parabolic
4. DBecause all vibrating systems are literally made of tiny springs
2.
An oscillator's rung spacing is ħω. What is the energy of the state with n = 3?Circle one
1. A3ħω
2. B3.5ħω
3. C4ħω
4. D9ħω
3.
How many nodes does the oscillator wavefunction ψ₄ have, and is it zero at the centre of the valley?Circle one
1. A4 nodes, and it is not zero at the centre
2. B4 nodes, and it is zero at the centre
3. C3 nodes, and it is zero at the centre
4. D5 nodes, and it is not zero at the centre
4.
Cooled to absolute zero, a diatomic molecule stops vibrating completely.Circle one
TrueFalse
5.
For an ideal harmonic oscillator, the energy needed to climb from n = 7 to n = 8 equals the energy needed to climb from n = 0 to n = 1.Circle one
TrueFalse
6.
Applying the lowering operator to the ground state produces a new state with energy −½ħω.Circle one
TrueFalse
7.
The ground-state wavefunction of the harmonic oscillator has the shape of a curve.Fill in the blank
8.
In a crystal, a single quantum of lattice vibration is called a .Fill in the blank
9.
Helium stays liquid at ordinary pressure all the way to absolute zero, yet hydrogen, whose molecules are even lighter, freezes at about 14 K. Explain both facts.Short answer
10.
Nitrogen and oxygen make up about 99 percent of dry air, yet trace carbon dioxide absorbs strongly in the infrared near 15 micrometres while they barely absorb at all. Explain using what you know about molecular vibrations.Short answer
11.
The fundamental vibrational band of hydrogen chloride (hydrogen-1 with chlorine-35, reduced mass 0.9796 atomic mass units) is centred at 2,886 inverse centimetres. Find (a) the wavelength, (b) the rung spacing ħω in eV, (c) the zero-point energy and (d) the bond's force constant k.Show your working
Working
Answer:
12.
A 0.50 kg mass hangs on a laboratory spring with k = 200 N/m. (a) Find ω and the rung spacing ħω in joules. (b) If it oscillates with an amplitude of 1.0 cm, roughly which rung n is it on? (c) Find x₀ and compare it with the size of a proton, about 10⁻¹⁵ m.Show your working
Working
Answer:
13.
The hydrogen molecule H₂ has a harmonic vibration wavenumber of 4,401 inverse centimetres. Replacing both atoms with deuterium multiplies the reduced mass by 2.014/1.008 = 1.998 while leaving the bond's stiffness unchanged. (a) Predict the harmonic wavenumber of D₂. (b) Estimate the difference between the two zero-point energies in eV. (c) Which molecule needs more energy to break apart, and by roughly how much?Show your working
Working
Answer:
14.
A coherent state of an oscillator has a mean of ⟨n⟩ = 4 quanta, and the number of quanta follows a Poisson distribution, P(n) = e^(−⟨n⟩)·⟨n⟩ⁿ/n!. (a) What is the standard deviation of the number of quanta? (b) What is the probability that a measurement finds the oscillator in its ground state? (c) What is its average energy in units of ħω?Show your working
Working
Answer:

Part C Mad scientist challenge bonus

Derive the zero-point energy two ways. (a) Uncertainty route: for a state centred on the valley, ⟨x⟩ = ⟨p⟩ = 0, so the average energy is E = (Δp)²/(2m) + ½mω²(Δx)². Using the smallest momentum spread allowed, Δp = ħ/(2Δx), find the Δx that minimizes E and the minimum energy. (b) Schrödinger route: show that ψ(x) = A·e^(−x²/(2b²)) solves −(ħ²/2m)ψ″ + ½mω²x²ψ = Eψ only if b² = ħ/(m·ω), and find E. (c) Compare the two answers, and evaluate Δx for the carbon monoxide molecule, whose oscillator length is x₀ = 4.79 pm.

Working

Memory aid

Mnemonic Spell SPRING whenever something wobbles. S: Smooth valleys are parabolas at the bottom, so small wobbles are harmonic. P: Plus a half, Eₙ = (n + ½)ħω. R: Rungs are equal, ħω apart, climbed one at a time by the raising and lowering operators. I: Infrared, where molecular springs absorb at their own frequency. N: Nodes, exactly n of them in ψₙ, with tails leaking past the turning points. G: Gaussian ground state, and a Gliding Gaussian, the coherent state, that swings without spreading. To use it, run down the six letters for any oscillator problem: check the valley, set the floor, count the rungs, find the colour, sketch the wave, then ask whether it glides.

In one line Near its bottom, every smooth valley is a spring: its energy climbs equal rungs, (n + ½)ħω, and even the lowest rung never stands still.