Quantum Chemistry I: Unit Review

Introduction

This unit began with waves and ended with an approximation method used when exact solutions fail. The particle in a box and the hydrogen atom look different, yet both acquire discrete bound energies because only certain wavefunctions satisfy the Schrödinger equation and physical boundary conditions. The variation principle then tells us what can still be learned when a Hamiltonian cannot be solved exactly.

Core explanation

For a particle in a one-dimensional infinite box of length L, ψ vanishes at both walls. Standing-wave solutions have wavelengths 2L/n and energies Eₙ = n²h²/(8mL²). The positive integers n count distinct stationary states. The ground energy is nonzero, reflecting confinement. Energy grows with n² and falls with L². A finite potential wall changes the condition: wavefunctions can extend into classically forbidden regions, and a finite-width barrier can permit tunnelling. These models teach that the potential and boundary conditions are part of the physical problem, not minor mathematical details.

For hydrogen, the potential is the three-dimensional Coulomb attraction −e²/(4πε₀r). Acceptable bound solutions have principal quantum numbers n = 1, 2, … and gross energies approximately Eₙ = −13.6 eV/n² for ordinary hydrogen. The electron is bound when E < 0 relative to separation at infinity. Each spatial orbital also has l = 0, …, n − 1 and mₗ = −l, …, +l. The n-th shell contains n² spatial states, all sharing one ideal Coulomb energy before finer interactions are considered. Transitions between energy levels give photons with hc/λ equal to the positive energy difference; n low = 1, 2 and 3 generate the Lyman, Balmer and Paschen series.

The box and hydrogen formulas should not be interchanged. Box energies are positive relative to zero potential inside and increase with n²; bound Coulomb energies are negative relative to a free electron and rise toward zero as n increases. Both show quantisation, but their dependence on n, length and physical potential differs. A chemistry question may use the box as a rough model for conjugated π electrons, where a HOMO–LUMO gap estimates an absorption wavelength. That model provides a trend, not a complete prediction of real dye colour.

When exact solutions are unavailable, choose an admissible trial ψ and calculate E trial = ⟨ψ Ĥ ψ⟩/⟨ψ ψ⟩. The variation theorem gives E trial ≥ E₀ for the ground state of the specified Hamiltonian. Optimising an exponent or a set of basis coefficients lowers the estimate within the chosen family. An exponential hydrogen trial reproduces the ideal 1s state exactly because the family contains it; a general molecular trial has no such guarantee. Distinguish this controlled upper bound from perturbation theory, which builds corrections around a solvable reference but does not generally preserve an upper-bound property at truncated order.

Step-by-step reasoning

First identify the potential, spatial dimension and energy zero. Next ask whether an exact model solution and its boundary conditions apply. If calculating a transition, subtract energies in the direction that gives a positive photon energy and then use λ = hc/ΔE. If counting states, enumerate l and mₗ before including spin. If an exact solution is unavailable, build a normalisable trial state, compute a properly normalised expectation and minimise it within the allowed parameter domain.

Visual explanation

Place two energy ladders side by side. The box ladder starts above zero and spreads farther apart as n increases. The hydrogen ladder sits below zero and crowds toward zero from below. Add a photon arrow between two hydrogen levels and a separate box HOMO–LUMO arrow. A third small graph shows variational energy against a trial parameter with its minimum above or touching the exact ground-energy line.

Real-world analogy

Different musical instruments allow different standing-wave patterns because their shapes and boundary conditions differ. Their allowed notes are discrete, but their note frequencies need not follow the same formula. Likewise, the box and hydrogen have distinct energy patterns. Tuning a trial wavefunction resembles adjusting an instrument by ear: it improves a chosen model but cannot create a sound the instrument's design cannot support.

Real-world example

A spectrometer can measure hydrogen's line wavelengths, while an absorption spectrometer can record a dye's visible band. Quantum levels explain why each sample interacts with selected photon energies. Hydrogen's simple spectrum can be predicted accurately from its level formula; a dye's complicated environment requires more than an electron-in-a-box estimate for precise band position and intensity.

Why?

The shared principle is that a physically acceptable wavefunction must satisfy an eigenvalue equation and boundary conditions. Different potentials impose different spectra. Transitions reveal differences between eigenvalues, while variational expectations reveal how a guessed state compares with the lowest exact eigenvalue. These links make quantum chemistry a method for predicting and interpreting measurements rather than just a list of orbital names.

Common misconception

Every formula containing n does not describe the same physics. In particular, inserting hydrogen n values into the box energy expression or treating negative Coulomb energy as a negative photon energy gives nonsense. Nor does a variational minimum prove the trial wavefunction exact: equality is special and must be justified by the Hamiltonian and trial family.

Worked example

Compare a box electron with n = 1 and n = 2 to a hydrogen 2 → 1 transition. In the box, E₂ − E₁ = (4 − 1)h²/(8mL²) = 3h²/(8mL²), so its gap depends strongly on L. For hydrogen, E₁ ≈ −13.6 eV and E₂ ≈ −3.40 eV, so the emitted photon has 10.2 eV. Using hc ≈ 1240 eV nm gives λ ≈ 1240/10.2 = 121.6 nm, a Lyman ultraviolet line. Both gaps are positive, but their energy scales arise from different potentials.

Quick check

1. Why is n = 2 higher in energy than n = 1 in hydrogen even though both energies are negative? Answer: E₂ ≈ −3.40 eV is greater than E₁ ≈ −13.6 eV on the number line. The n = 2 electron is less tightly bound and can release about 10.2 eV by moving to n = 1.

Exam focus

Write the energy reference beside any negative energy. Use the correct n dependence for the specified model. For hydrogen, name a spectral series by its lower level and separate state counting from energy counting. For a variational estimate, state E trial ≥ E₀ and show that the trial obeys normalisation and boundary requirements.

Advanced insight

Quantum chemistry often combines all three viewpoints. Simple exact models build intuition; hydrogen-like functions supply basis functions; variational matrix methods combine those basis functions to approximate molecular states. The computed energies can then predict spectra, but transition strengths require additional wavefunction information. An accurate energy gap and an intense line are related questions, not identical ones.

Summary

The box, finite barriers and hydrogen demonstrate how wavefunctions and potentials determine allowed energies. Box levels scale as n²/L²; hydrogen bound energies scale as −1/n² and produce named spectral series. Quantum numbers count spatial states, and variational minimisation supplies an upper bound when exact solutions are unavailable. Always preserve each model's assumptions and energy reference.

Practice questions

1. A box length doubles without changing the particle mass or n. What happens to that state's energy? Answer: Eₙ is proportional to 1/L², so doubling L reduces its energy to one quarter. The same scaling applies to a fixed pair of box levels and their gap. 2. How many spatial orbitals belong to n = 3 hydrogen in the ideal model, and how many gross energies do they have? Answer: There are n² = 9 spatial orbitals: one s, three p and five d. In ideal field-free Coulomb hydrogen they share one gross n = 3 energy. 3. A valid trial calculation for a Hamiltonian gives E trial = −5 units. What can be concluded about its exact ground energy? Answer: The exact ground energy is at or below −5 units, since E trial ≥ E₀. Without more information one cannot say how close −5 is to the exact value or whether the trial wavefunction has accurate local shape.