Quantum Chemistry: Postulates, Model Systems, the Hydrogen Atom and Approximation Methods
1. The postulates: wave functions, operators, commutators and expectation values
The state of a system is described completely by a wave function ψ(r, t). By the Born interpretation, ψ itself has no direct physical meaning but |ψ|² = ψ*ψ is a probability density: |ψ(x)|² dx is the probability of finding the particle between x and x + dx. So an acceptable ψ must be single-valued, continuous, finite and square-integrable, and is normalised so that ∫|ψ|² dτ = 1. In Dirac notation a state is a ket |ψ⟩, its conjugate a bra ⟨ψ|, and ⟨φ|ψ⟩ = ∫φ*ψ dτ is their overlap; normalisation is ⟨ψ|ψ⟩ = 1 and orthogonality ⟨ψₘ|ψₙ⟩ = 0 for m ≠ n.
Every observable corresponds to a linear Hermitian operator: position x̂ = x, momentum p̂ₓ = −iħ ∂/∂x, kinetic energy T̂ = −(ħ²/2m)∇², and the Hamiltonian Ĥ = T̂ + V̂. Hermitian operators have real eigenvalues and orthogonal eigenfunctions for distinct eigenvalues. A measurement can only return an eigenvalue of the operator; for a state that is not an eigenfunction, the average over many measurements is the expectation value ⟨A⟩ = ⟨ψ|Â|ψ⟩/⟨ψ|ψ⟩.
The commutator [Â, B̂] = ÂB̂ − B̂Â decides whether two observables can be known together. The canonical one is [x̂, p̂ₓ] = iħ, which leads to the uncertainty relation Δx·Δpₓ ≥ ħ/2. Operators that commute share a complete set of eigenfunctions, so their observables have simultaneously sharp values: for the hydrogen atom Ĥ, L̂² and L̂_z all commute, which is why n, l and mₗ label one orbital together. The time-dependent Schrödinger equation iħ ∂Ψ/∂t = ĤΨ has, for a time-independent potential, separable solutions Ψ(x, t) = ψ(x)e^(−iEt/ħ), where ψ obeys the time-independent equation Ĥψ = Eψ. These are stationary states: |Ψ|² does not change with time.
2. Particle in a box, finite barriers and tunnelling
For a particle of mass m confined to 0 ≤ x ≤ L by infinite walls, ψ must vanish at both walls. The solutions are ψₙ = √(2/L) sin(nπx/L) with Eₙ = n²h²/(8mL²), n = 1, 2, 3, … The state n has n − 1 interior nodes; the lowest energy is not zero — the zero-point energy h²/(8mL²) is required by the uncertainty principle — and the spacing Eₙ₊₁ − Eₙ = (2n + 1)h²/(8mL²) grows with n and shrinks as the box or the mass grows, which is why a macroscopic object looks classical. In the ground state ⟨x⟩ = L/2 and ⟨pₓ⟩ = 0, while ⟨pₓ²⟩ = h²/(4L²).
In two and three dimensions the Hamiltonian separates: for a rectangular box E = (h²/8m)(nₓ²/a² + n_y²/b² + n_z²/c²). In a square or cubic box different sets of quantum numbers give the same energy — degeneracy that comes from symmetry. In a cube, with ε = h²/(8mL²), the levels are 3ε (1,1,1, non-degenerate), 6ε (three-fold), 9ε (three-fold), 11ε (three-fold), 12ε (2,2,2, non-degenerate) and 14ε (the six permutations of 1,2,3). Stretching one side lifts the degeneracy.
| nₓ² + n_y² + n_z² | Quantum numbers | Degeneracy |
|---|---|---|
| 3 | (1,1,1) | 1 |
| 6 | (2,1,1) ×3 | 3 |
| 9 | (2,2,1) ×3 | 3 |
| 11 | (3,1,1) ×3 | 3 |
| 12 | (2,2,2) | 1 |
| 14 | (3,2,1) ×6 | 6 |
With finite walls of height V₀ the wave function no longer stops at the wall: inside the barrier, where E < V₀, it decays as e^(−κx) with κ = √(2m(V₀ − E))/ħ. There are only a finite number of bound states, each lower in energy than its infinite-well counterpart because the particle has more room. Through a barrier of width a the transmission probability for κa ≫ 1 is roughly T ≈ e^(−2κa): tunnelling falls exponentially with the barrier width and with √m, so it matters for electrons and protons and hardly at all for heavier groups — the basis of the scanning tunnelling microscope, of proton-transfer isotope effects and of the ammonia inversion.
Application — the free-electron model of a conjugated polyene. Treat the N π electrons as particles in a one-dimensional box of length L, two per level. The HOMO is n = N/2 and the LUMO n = N/2 + 1, so the lowest absorption is ΔE = (N + 1)h²/(8mₑL²) and λ = 8mₑL²c/[(N + 1)h]. Lengthening the chain raises L faster than N + 1, so λ moves to longer wavelength — the red shift of longer polyenes and carotenoids.
3. The harmonic oscillator
For V = ½kx² the energies are E_v = (v + ½)hν, v = 0, 1, 2, …, with ν = (1/2π)√(k/μ) and μ the reduced mass m₁m₂/(m₁ + m₂) for a diatomic. The levels are equally spaced by hν and the ground state keeps the zero-point energy ½hν. The wave functions are ψ_v = N_v H_v(y) e^(−y²/2), with y = x√α, α = √(kμ)/ħ, and N_v = (α/π)^¼ (2^v v!)^(−½). The Hermite polynomials are H₀ = 1, H₁ = 2y, H₂ = 4y² − 2, H₃ = 8y³ − 12y, generated by the recursion H_(v+1) = 2yH_v − 2vH_(v−1). Each H_v has v real roots, so ψ_v has v nodes, and parity (−1)^v: even v gives an even function, odd v an odd one — which is what makes ⟨x⟩ = 0 in every state and ⟨v|x|v′⟩ non-zero only for v′ = v ± 1.
For the harmonic potential the virial theorem gives ⟨T⟩ = ⟨V⟩ = E_v/2 in every stationary state. Classically the oscillator never passes its turning points, but ψ₀ extends beyond them: about 16% of the ground-state probability lies in the classically forbidden region. Real bonds are anharmonic — the Morse potential V = Dₑ(1 − e^(−a(r−rₑ)))² is the usual model — and the levels become G(v) = (v + ½)ω̃ₑ − (v + ½)²ω̃ₑxₑ, converging as v rises. The spectroscopic dissociation energy D₀ is measured from v = 0 and is smaller than the well depth Dₑ by the zero-point energy.
4. Rotational motion and angular momentum
The orbital angular momentum operators obey [L̂ₓ, L̂_y] = iħL̂_z and cyclic permutations, so no two components can be sharp together; but L̂² commutes with each component, and the conventional choice is to make L̂² and L̂_z sharp. Their common eigenfunctions are the spherical harmonics Y_l^m(θ, φ) = Θ_lm(θ)e^(imφ), with L̂²Y = l(l + 1)ħ²Y and L̂_zY = mħY, l = 0, 1, 2, … and m = −l, …, +l. The Y_l^m are orthonormal over the sphere, have l angular nodes (nodal planes or cones) and parity (−1)^l. Spin is an intrinsic angular momentum with the same algebra: for an electron s = ½, S² = s(s + 1)ħ² = ¾ħ² and m_s = ±½.
A rigid rotor (a diatomic with fixed bond length r and moment of inertia I = μr²) has energies E_J = J(J + 1)ħ²/(2I), J = 0, 1, 2, …, each (2J + 1)-fold degenerate. In spectroscopic units F(J) = BJ(J + 1) with the rotational constant B = h/(8π²cI) in cm⁻¹, so adjacent levels are 2B(J + 1) apart. A particle confined to a ring (rotation in a plane) has E = m²ħ²/(2I), m = 0, ±1, ±2, …: every level above m = 0 is doubly degenerate, clockwise and anticlockwise.
5. Hydrogen-like atoms and atomic units
For one electron bound to a nucleus of charge Ze the Schrödinger equation separates into ψ_nlm = R_nl(r)Y_l^m(θ, φ). The energy depends only on n: Eₙ = −Z²Eₕ/(2n²) = −13.6 Z²/n² eV, so a level n is n²-fold degenerate (2n² with spin). The orbital ψ_nlm has n − 1 nodes in all: n − l − 1 radial and l angular. The radial distribution function P(r) = r²R_nl² gives the probability of finding the electron in a shell between r and r + dr; for 1s, P(r) ∝ r²e^(−2Zr/a₀) peaks at the most probable radius a₀/Z, while the mean radius is ⟨r⟩ = 3a₀/(2Z). In general ⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)].
Atomic units set ħ = mₑ = e = 4πε₀ = 1. Lengths are then in bohr (a₀ = 52.92 pm) and energies in hartree (Eₕ = 27.211 eV), so the hydrogen ground state is −½ Eₕ and the Hamiltonian of hydrogen is simply −½∇² − 1/r. The angular distribution is |Y_l^m|²: s orbitals are spherical, the p_z orbital ∝ cos θ has a nodal plane at θ = 90°, and real p and d orbitals are linear combinations of the complex Y_l^(±m).
| Orbital | Radial nodes n − l − 1 | Angular nodes l | ⟨r⟩ in units of a₀/Z |
|---|---|---|---|
| 1s | 0 | 0 | 1.5 |
| 2s | 1 | 0 | 6 |
| 2p | 0 | 1 | 5 |
| 3s | 2 | 0 | 13.5 |
| 3p | 1 | 1 | 12.5 |
| 3d | 0 | 2 | 10.5 |
6. Multi-electron atoms, the variational method and perturbation theory
With two or more electrons the repulsion terms e²/(4πε₀r_ij) prevent separation, so the orbital approximation writes the wave function as a product of one-electron orbitals, each electron moving in the average field of the rest (the idea behind the Hartree–Fock method). Electrons are fermions, so the total wave function must be antisymmetric under exchange of any two electrons, space and spin together — the general form of the Pauli exclusion principle. A Slater determinant builds this in: for helium 1s², Ψ = (1/√2)|1sα(1) 1sβ(1); 1sα(2) 1sβ(2)|. Swapping two electrons swaps two rows and changes the sign; putting two electrons in the same spin-orbital makes two columns equal and the determinant vanish.
The variational principle states that for any well-behaved trial function φ, E[φ] = ⟨φ|Ĥ|φ⟩/⟨φ|φ⟩ ≥ E₀, the true ground-state energy. Minimising E with respect to parameters in φ gives the best upper bound of that form. For helium with the trial function a product of two 1s orbitals of effective charge ζ, E(ζ) = ζ² − 27ζ/8 hartree, minimised at ζ = 27/16 = 1.6875, giving E = −(27/16)² = −2.848 Eₕ (−77.5 eV) against the experimental −2.904 Eₕ (−79.0 eV). The effective charge below 2 is the screening of each electron by the other. In a linear variation φ = Σcᵢχᵢ the minimisation leads to the secular equations Σcᵢ(Hᵢⱼ − ESᵢⱼ) = 0, which have non-trivial solutions only when the secular determinant |Hᵢⱼ − ESᵢⱼ| = 0; its roots are upper bounds to successive states.
First-order non-degenerate perturbation theory splits Ĥ = Ĥ⁽⁰⁾ + Ĥ′ with Ĥ⁽⁰⁾ exactly solvable. The first-order energy correction is the expectation value of the perturbation in the unperturbed state, E⁽¹⁾ = ⟨ψ⁽⁰⁾|Ĥ′|ψ⁽⁰⁾⟩, and the first-order wave function mixes in other states with coefficients ⟨ψₖ⁽⁰⁾|Ĥ′|ψₙ⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₖ⁽⁰⁾). The second-order correction to the ground state is always negative. For helium, taking the electron repulsion as Ĥ′, E⁽⁰⁾ = −108.8 eV (two hydrogen-like electrons with Z = 2) and E⁽¹⁾ = (5/8)Z Eₕ = +34.0 eV, giving −74.8 eV — a poorer estimate than the variational −77.5 eV.
Key takeaways
- |ψ|² is the probability density; observables are Hermitian operators, measurements return eigenvalues, and ⟨A⟩ = ⟨ψ|Â|ψ⟩ for a normalised state. Commuting operators have simultaneous eigenfunctions; [x̂, p̂ₓ] = iħ.
- Particle in a box: Eₙ = n²h²/(8mL²), n − 1 nodes; in a cube degeneracy follows from permuting (nₓ, n_y, n_z). Tunnelling through a barrier falls as e^(−2κa).
- Harmonic oscillator: E_v = (v + ½)hν, ν ∝ √(k/μ), Hermite polynomials with parity (−1)^v, ⟨T⟩ = ⟨V⟩ = E/2. Rigid rotor: E_J = BJ(J + 1), degeneracy 2J + 1.
- Hydrogen-like atoms: Eₙ = −13.6Z²/n² eV, n − l − 1 radial and l angular nodes; the 1s radial distribution peaks at a₀/Z and ⟨r⟩ = 3a₀/2Z.
- Slater determinants make many-electron functions antisymmetric. The variational energy is an upper bound (He: ζ = 27/16, −2.848 Eₕ); first-order perturbation adds ⟨ψ⁽⁰⁾|Ĥ′|ψ⁽⁰⁾⟩.
Practice questions (17)
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.
An electron is confined in a one-dimensional box of length 1.0 nm with infinitely high walls. What is the energy of the n = 1 → n = 2 transition, in eV, to two decimal places? (h = 6.626 × 10⁻³⁴ J s, mₑ = 9.109 × 10⁻³¹ kg, 1 eV = 1.602 × 10⁻¹⁹ J)
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Answer: 1.13
E₁ = h²/(8mL²) = (6.626 × 10⁻³⁴)²/(8 × 9.109 × 10⁻³¹ × 10⁻¹⁸) = 6.025 × 10⁻²⁰ J = 0.376 eV. The gap is E₂ − E₁ = (4 − 1)E₁ = 3E₁ = 1.13 eV. Reporting E₂ itself (1.50 eV) or E₁ alone (0.38 eV) is the usual slip.For a particle in a cubic box, what is the degeneracy of the energy level E = 14h²/(8mL²)?
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Answer: 6
nₓ² + n_y² + n_z² = 14 is met only by 1² + 2² + 3², and the three different numbers can be assigned to x, y and z in 3! = 6 ways. The level 12 (2,2,2) is non-degenerate and 11 (3,1,1) is three-fold, which is where answers of 1 or 3 come from.For a particle in the ground state (n = 1) of a one-dimensional box of length L, what is the probability of finding it in the middle third, L/3 ≤ x ≤ 2L/3? Give the answer to two decimal places.
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Answer: 0.61
P = (2/L)∫sin²(πx/L)dx = [x/L − sin(2πx/L)/(2π)] between L/3 and 2L/3 = 1/3 − [sin(4π/3) − sin(2π/3)]/(2π) = 0.3333 + 1.732/6.283 = 0.609, i.e. 0.61. The classical answer 0.33 ignores that |ψ₁|² is largest at the centre.In the free-electron model, the six π electrons of hexatriene are placed in a one-dimensional box of length 0.70 nm. What is the wavelength of the HOMO → LUMO transition, in nm, to the nearest whole number? (h = 6.626 × 10⁻³⁴ J s, mₑ = 9.109 × 10⁻³¹ kg, c = 2.998 × 10⁸ m/s)
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Answer: 231
HOMO n = 3, LUMO n = 4, so ΔE = (16 − 9)h²/(8mL²) = 7h²/(8mL²) and λ = 8mL²c/(7h) = 8 × 9.109 × 10⁻³¹ × (0.70 × 10⁻⁹)² × 2.998 × 10⁸/(7 × 6.626 × 10⁻³⁴) = 2.308 × 10⁻⁷ m = 231 nm. Taking the gap from n = 1 gives a far shorter, wrong wavelength.What is the commutator [x̂, p̂ₓ]?
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Answer: A — iħ
[x̂, p̂ₓ]f = x(−iħ f′) − (−iħ)(f + x f′) = iħ f, so the commutator is iħ. Its non-zero value is the origin of Δx·Δpₓ ≥ ħ/2; −iħ is [p̂ₓ, x̂], the same pair in the other order.Which of the following pairs of operators commute, so that the two observables can have sharp values at the same time?
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Answer: A — L̂² and L̂_z; C — Ĥ and L̂² for the hydrogen atom
L̂² commutes with every component, and for a central potential Ĥ commutes with L̂² and L̂_z — hence the labels n, l, mₗ together. [L̂ₓ, L̂_y] = iħL̂_z and [x̂, p̂ₓ] = iħ are non-zero, so those pairs cannot be sharp together.A diatomic molecule H–X, in which X is very much heavier than hydrogen, has its hydrogen replaced by deuterium. Taking the force constant as unchanged, what is the ratio ν(D–X)/ν(H–X) of the vibrational frequencies, to two decimal places?
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Answer: 0.71
ν = (1/2π)√(k/μ) with k unchanged, and for a very heavy X the reduced mass is essentially the mass of the light atom: μ(H–X) ≈ 1, μ(D–X) ≈ 2 amu. So ν(D)/ν(H) = √(1/2) = 0.707, i.e. 0.71. Answering 0.50 forgets the square root.For a one-dimensional harmonic oscillator in any stationary state v, how are the expectation values of kinetic and potential energy related?
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Answer: A — ⟨T⟩ = ⟨V⟩ = E_v/2
For V ∝ xⁿ the virial theorem gives 2⟨T⟩ = n⟨V⟩; with n = 2 the two are equal, each half of (v + ½)hν. ⟨V⟩ = −2⟨T⟩ is the Coulomb case n = −1, which belongs to the hydrogen atom.Which statements about the harmonic-oscillator wave functions ψ_v = N_v H_v(y)e^(−y²/2) are correct?
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Answer: A — ψ_v has exactly v nodes; B — H₂(y) = 4y² − 2
H_v has v real roots and H₂ = 4y² − 2 from the recursion H₂ = 2yH₁ − 2H₀. ψ₁ ∝ y e^(−y²/2) is odd, parity (−1)¹. About 16% of the ground-state probability lies beyond the turning points — the quantum oscillator penetrates the classically forbidden region.A rigid diatomic rotor has a rotational constant B = 10.0 cm⁻¹. What is the energy of its J = 3 level above J = 0, in cm⁻¹?
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Answer: 120
F(J) = BJ(J + 1) = 10.0 × 3 × 4 = 120 cm⁻¹. The value 60 cm⁻¹ is the J = 2 → 3 spacing 2B(J + 1) with J = 2, not the energy of the level.What is the ground-state energy of the hydrogen-like ion Li²⁺, in eV, to one decimal place? (Take the hydrogen ground state as −13.6 eV.)
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Answer: -122.4
Eₙ = −13.6 Z²/n² eV with Z = 3, n = 1: E = −13.6 × 9 = -122.4 eV. Using Z instead of Z² gives −40.8 eV; the sign must stay negative for a bound state.What is the mean distance ⟨r⟩ of the electron from the nucleus in the 1s orbital of the hydrogen atom, in pm, to one decimal place? (a₀ = 52.92 pm)
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Answer: 79.4
⟨r⟩ = ∫r·(r²R₁₀²)dr = 3a₀/(2Z) = 1.5 × 52.92 = 79.4 pm. The most probable radius, where r²R² peaks, is a₀ = 52.9 pm; the mean lies farther out because the distribution has a long tail.How many radial and angular nodes does a 4d orbital have?
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Answer: A — 1 radial, 2 angular
For n = 4, l = 2: radial nodes n − l − 1 = 1 and angular nodes l = 2, three nodes in all (n − 1). Swapping the two counts gives the 4p answer, 2 radial and 1 angular.A variational calculation on the helium atom uses a trial function that is a product of two hydrogen-like 1s orbitals with effective nuclear charge ζ, for which E(ζ) = ζ² − (27/8)ζ in hartree. What is the minimised energy, in hartree, to three decimal places?
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Answer: -2.848
dE/dζ = 2ζ − 27/8 = 0 gives ζ = 27/16 = 1.6875, and E = (27/16)² − (27/8)(27/16) = −(27/16)² = -2.848 Eₕ. It lies above the true −2.904 Eₕ, as the variational principle requires; putting ζ = 2 gives −2.750 Eₕ, the unscreened result.Why is the many-electron wave function written as a Slater determinant rather than a simple product of spin-orbitals?
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Answer: A — It is antisymmetric under exchange of any two electrons, as the Pauli principle requires
Exchanging two electrons swaps two rows of the determinant and reverses its sign, and two electrons in one spin-orbital give two equal columns and a zero determinant — Pauli exclusion built in. It does not remove the repulsion or make the energy exact; symmetric functions describe bosons, not electrons.An electron tunnels through a rectangular barrier with κa ≫ 1. If the barrier width a is doubled, the transmission probability approximately
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Answer: A — is squared, since T ≈ e^(−2κa) becomes e^(−4κa)
T ≈ e^(−2κa), so doubling a gives e^(−4κa) = (e^(−2κa))²: a small number becomes much smaller. The dependence on width is exponential, not linear, which is why "halves" understates the effect.In first-order perturbation theory applied to helium, the zeroth-order energy is −108.8 eV and the first-order correction from electron repulsion is +34.0 eV. What is the first-order estimate of the ground-state energy, in eV, to one decimal place?
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Answer: -74.8
E ≈ E⁽⁰⁾ + E⁽¹⁾ = −108.8 + 34.0 = -74.8 eV. E⁽⁰⁾ is 2 × (−13.6 × 2²) for two unscreened electrons, and E⁽¹⁾ = ⟨ψ⁽⁰⁾|e²/4πε₀r₁₂|ψ⁽⁰⁾⟩ = (5/8)Z hartree. The experimental −79.0 eV shows the second-order correction is negative.