Quantum Mechanics II: Central Potentials, Hydrogen-like Atoms, Angular Momentum and Its Addition, Variational and Perturbation Methods, and Born Scattering
1. Central potentials and hydrogen-like atoms
For V(r), ψ = R(r)Yₗᵐ(θ, φ). The angular part is the same for every central potential; with u = rR the radial equation is one-dimensional, −(ħ²/2μ)u″ + [V + l(l + 1)ħ²/(2μr²)]u = Eu, the second term the centrifugal barrier. For the Coulomb potential −Ze²/(4πε₀r), Eₙ = −13.6 Z²/n² eV, depending on n alone: the degeneracy is n² (2n² with spin), l = 0 … n − 1, and R_nl has n − l − 1 radial nodes. The Bohr radius is a₀ = 0.529 Å and the scale is a₀/Z.
- Transitions: hν = 13.6 Z²(1/n₁² − 1/n₂²) eV. H: 3 → 2 (Balmer Hα) gives 1.89 eV; 2 → 1 (Lyman α) gives 10.2 eV, λ = 1240/10.2 = 121.6 nm. He⁺ (Z = 2) has an ionisation energy of 54.4 eV.
- Ground state ψ₁₀₀ = e−r/a₀/√(πa₀³): the most probable radius is a₀ (the peak of r²|ψ|²), but ⟨r⟩ = 3a₀/2, and ⟨1/r⟩ = 1/a₀ gives ⟨V⟩ = 2E₁ and ⟨T⟩ = −E₁, the virial theorem.
- Reduced mass μ replaces m, so deuterium’s lines sit slightly to the blue of hydrogen’s, and positronium (μ = m/2) has half hydrogen’s binding energy.
2. Orbital and spin angular momentum
[Lₓ, L_y] = iħL_z and cyclically, while each component commutes with L². Simultaneous eigenstates |l, m⟩ have L² = l(l + 1)ħ² and L_z = mħ, with m = −l … l. The ladder operators L± = Lₓ ± iL_y give L±|l, m⟩ = ħ√(l(l + 1) − m(m ± 1)) |l, m ± 1⟩; for l = 2, L₊|2, 1⟩ = ħ√(6 − 2)|2, 2⟩ = 2ħ|2, 2⟩. Orbital l is an integer, because Yₗᵐ must be single-valued; the same algebra also allows half-integers, which is where spin lives.
An electron has spin ½: S = (ħ/2)σ with the Pauli matrices σₓ = [[0, 1], [1, 0]], σ_y = [[0, −i], [i, 0]], σ_z = [[1, 0], [0, −1]], each of eigenvalues ±1, squaring to 1 and anticommuting with the others. S² = (3/4)ħ² for every spin-½ state. A spin in a field B along z precesses at the Larmor frequency, and a Stern–Gerlach magnet sorts atoms into 2s + 1 beams — two for spin ½, which is how half-integer angular momentum was first seen.
3. Addition of angular momenta
Combining j₁ and j₂ gives total J = j₁ + j₂ with j = |j₁ − j₂|, |j₁ − j₂| + 1, …, j₁ + j₂, and the count of states is conserved: Σ(2j + 1) = (2j₁ + 1)(2j₂ + 1). The coupled states |j, m⟩ are combinations of product states |m₁, m₂⟩ with m = m₁ + m₂; the coefficients are the Clebsch–Gordan coefficients.
- Two spin-½ particles: 2 × 2 = 4 states = a triplet (s = 1: |↑↑⟩, (|↑↓⟩ + |↓↑⟩)/√2, |↓↓⟩), symmetric under exchange, plus a singlet (s = 0: (|↑↓⟩ − |↓↑⟩)/√2), antisymmetric.
- l = 1 with s = ½: j = 3/2 (4 states) and j = 1/2 (2 states), 6 in all, as 3 × 2 requires.
- L·S = [J² − L² − S²]/2 = (ħ²/2)[j(j + 1) − l(l + 1) − s(s + 1)]: for l = 1, it is +½ħ² at j = 3/2 and −ħ² at j = 1/2, a separation of (3/2)ħ² — the spin–orbit splitting the next section’s chapter uses.
4. The variational method and time-independent perturbation theory
Variational principle: for any normalised trial state, ⟨ψ|H|ψ⟩ ≥ E₀. Minimising over a parameter gives an upper bound on the ground-state energy — exact if the true state is in the family (a Gaussian for the oscillator), too high otherwise (a Gaussian for hydrogen gives −(8/3π) × 13.6 = −11.5 eV against −13.6 eV). The energy is accurate to second order in the error of the wavefunction, which is why a crude trial state gives a good energy.
Perturbation theory, H = H₀ + H′. First order: Eₙ⁽¹⁾ = ⟨n|H′|n⟩. Second order: Eₙ⁽²⁾ = Σm≠n |⟨m|H′|n⟩|²/(Eₙ − Eₘ), which is always negative for the ground state. Worked: a box 0 < x < L with H′ = V₀ on the left half shifts every level by V₀/2 at first order (by symmetry of |ψₙ|²). An oscillator with its spring stiffened by 21%, H′ = ½(0.21k)x², shifts E₀ by 0.21k⟨x²⟩/2 = 0.0525ħω, so E₀′/E₀ ≈ 1.105 against the exact √1.21 = 1.100.
5. Elementary scattering theory and the Born approximation
Far from the target the wavefunction is eikz + f(θ)eikr/r, and the differential cross-section is dσ/dΩ = |f(θ)|². In the first Born approximation, valid when the potential is weak compared with the incident energy, f is the Fourier transform of the potential at the momentum transfer q = 2k sin(θ/2): for a spherically symmetric V, f(θ) = −(2m/ħ²q)∫₀^∞ rV(r) sin(qr) dr. The Yukawa potential V₀e−μr/r gives f = −2mV₀/[ħ²(μ² + q²)]; as μ → 0 it becomes the Coulomb potential and reproduces the Rutherford formula, dσ/dΩ ∝ 1/sin⁴(θ/2).
The partial-wave view complements it: at low energy (ka ≪ 1) only l = 0 scatters, and the scattering is isotropic. A hard sphere of radius a then has σ = 4πa² — four times its geometric cross-section, a wave effect; for a = 1 fm, σ = 12.57 fm². The optical theorem ties the total cross-section to the forward amplitude, σ = (4π/k) Im f(0).
Key takeaways
- Hydrogen-like: Eₙ = −13.6Z²/n² eV, degeneracy n² (2n² with spin); ground state most probable at a₀ but ⟨r⟩ = 3a₀/2.
- L² = l(l + 1)ħ², L_z = mħ, L±|l, m⟩ = ħ√(l(l + 1) − m(m ± 1))|l, m ± 1⟩; spin ½ is (ħ/2)σ.
- j runs from |j₁ − j₂| to j₁ + j₂ and the states total (2j₁ + 1)(2j₂ + 1); two spins ½ give a triplet and a singlet.
- Variational energies are upper bounds; E⁽¹⁾ = ⟨n|H′|n⟩, E⁽²⁾ < 0 for the ground state; diagonalise H′ first in a degenerate level.
- Born: f is the Fourier transform of V at q = 2k sin(θ/2); Yukawa → Rutherford as μ → 0; a hard sphere at low energy has σ = 4πa².
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Using Eₙ = −13.6/n² eV, the energy of the photon emitted in the hydrogen transition n = 3 → n = 2, in eV, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.89
ΔE = 13.6(1/4 − 1/9) = 13.6 × 5/36 = 1.89 eV (the red Hα line). 13.6(1/2 − 1/3) = 2.27 eV uses 1/n instead of 1/n²; 13.6/9 = 1.51 eV is only the binding energy of n = 3.Using Eₙ = −13.6/n² eV and hc = 1240 eV nm, the wavelength of the Lyman-α line of hydrogen (n = 2 → 1), in nm, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 121.6
ΔE = 13.6(1 − 1/4) = 10.2 eV and λ = hc/ΔE = 1240/10.2 = 121.6 nm, in the ultraviolet. 1240/13.6 = 91.2 nm is the series limit, not Lyman-α.For the hydrogen ground state ψ = e−r/a₀/√(πa₀³), the expectation value ⟨r⟩ in units of a₀ is ____.
Numerical answer — type the value.
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Answer: 1.5
⟨r⟩ = (4/a₀³)∫₀^∞ r³e−2r/a₀ dr = (4/a₀³) × 3!(a₀/2)⁴ = 3a₀/2 = 1.5a₀. The most probable radius, where r²|ψ|² peaks, is a₀ — the tempting wrong answer, because the distribution has a long tail.Including spin, the number of electron states in the n = 3 shell of hydrogen is ____.
Numerical answer — type the value.
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Answer: 18
l = 0, 1, 2 give 1 + 3 + 5 = 9 = n² orbital states, and two spin states each give 2n² = 18. Without spin the answer is 9.L₊|l = 2, m = 1⟩ = cħ|l = 2, m = 2⟩. The value of c is ____.
Numerical answer — type the value.
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Answer: 2
c = √(l(l + 1) − m(m + 1)) = √(6 − 2) = 2. Using m(m − 1), the lowering form, gives √6 = 2.45; using l − m gives 1.For an electron with l = 1, the difference in ⟨L·S⟩ between the j = 3/2 and j = 1/2 levels, in units of ħ², is ____.
Numerical answer — type the value.
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Answer: 1.5
L·S = ½[j(j + 1) − l(l + 1) − s(s + 1)]ħ². j = 3/2: ½(15/4 − 2 − 3/4) = ½. j = 1/2: ½(3/4 − 2 − 3/4) = −1. Difference = 1.5ħ². Omitting the ½ gives 3.Two spin-½ particles are combined. The total spin states are:
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Answer: A — A triplet (S = 1) and a singlet (S = 0)
½ ⊗ ½ gives S = 0 and 1; 1 + 3 = 4 = 2 × 2 states. The triplet is symmetric under exchange and the singlet antisymmetric. A quartet needs three spin-½ particles.An orbital angular momentum l = 1 is combined with a spin s = ½. Which statements are correct?
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Answer: A — j = 3/2 occurs; B — j = 1/2 occurs; D — The total number of states is 6
j runs from |1 − ½| = ½ to 1 + ½ = 3/2 in integer steps, so only ½ and 3/2 occur, with 2 + 4 = 6 = 3 × 2 states. 5/2 would need l = 2 or a larger spin.A trial wavefunction is used in the variational method for the ground state. The energy it gives is:
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Answer: A — Greater than or equal to the true ground-state energy
Expanding the trial state in eigenstates, ⟨H⟩ = Σ|cₙ|²Eₙ ≥ E₀Σ|cₙ|² = E₀: an upper bound, reached only if the trial state is the ground state. That is why lowering ⟨H⟩ over a parameter is always an improvement.A particle in an infinite well 0 < x < L is perturbed by H′ = V₀ for 0 < x < L/2 and zero elsewhere. The first-order energy shift of any level, in units of V₀, is ____.
Numerical answer — type the value.
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Answer: 0.5
E⁽¹⁾ = V₀∫₀L/2(2/L)sin²(nπx/L) dx = V₀/2 = 0.5V₀, because |ψₙ|² is symmetric about L/2 and each half holds probability ½. Answering V₀ assumes the perturbation covers the whole well.The spring constant of a harmonic oscillator is increased from k to 1.21k. Treating H′ = ½(0.21k)x² as a perturbation, the first-order estimate of the ratio of the new ground-state energy to the old, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.105
⟨0|x²|0⟩ = ħ/(2mω), so E⁽¹⁾ = ½(0.21k)(ħ/2mω) = 0.21 × ħω/4 = 0.0525ħω, and (0.5 + 0.0525)/0.5 = 1.105. The exact ratio is √1.21 = 1.100; the difference is the (negative) second-order term.In non-degenerate perturbation theory, the second-order correction to the ground-state energy is:
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Answer: A — Never positive
E₀⁽²⁾ = Σm≠0|⟨m|H′|0⟩|²/(E₀ − Eₘ): every numerator is non-negative and every denominator negative for the ground state, so the sum is ≤ 0. For an excited state the terms have both signs.At very low energy, the total scattering cross-section of a hard sphere of radius 1.0 fm, in fm², to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 12.57
Only the s-wave scatters, with phase shift −ka, so σ = 4πa² = 4π × 1.0 = 12.57 fm² — four times the geometric πa² = 3.14 fm², a wave effect. The classical answer is πa².In the first Born approximation for a spherically symmetric potential, the scattering amplitude depends on the angle θ only through:
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Answer: A — q = 2k sin(θ/2), the momentum transfer
f(θ) is the Fourier transform of V evaluated at |k′ − k| = 2k sin(θ/2), so all angular dependence enters through q. The impact parameter belongs to classical trajectories, and partial waves are a different expansion.