Chapter 21

Quantum Mechanics: Formalism

College
At a glance
Core ideaA wavefunction ψ evolves by the Schrödinger equation; |ψ|² is a probability density.
Key termCommutator [x̂,p̂]=iℏ — the algebraic root of the uncertainty principle.
You can…Solve the particle in a box and normalize a wavefunction.
Watch outConfinement forces discrete energies and nonzero zero-point energy; measurement collapses ψ.
Theory

The Schrödinger equation and operators

Chapter 17 introduced quantization qualitatively. The full formalism replaces a particle's trajectory with a complex-valued wavefunction ψ(x,t), whose evolution is governed by the time-dependent Schrödinger equation:

iℏ ∂ψ/∂t = Ĥψ = [−ℏ²/2m ∂²/∂x² + V(x)]ψ

Physical quantities become operators acting on ψ: position x̂ = x, momentum p̂ = −iℏ∂/∂x. They do not commute:

[x̂, p̂] = x̂p̂ − p̂x̂ = iℏ

This single algebraic fact is the rigorous origin of the uncertainty principle Chapter 17 stated: for any two operators, ΔAΔB ≥ ½|⟨[Â,B̂]⟩|, and substituting the canonical commutator recovers ΔxΔp ≥ ℏ/2. The Born rule interprets |ψ(x,t)|² as a probability density: the probability of finding the particle in [x, x+dx] is |ψ|²dx, requiring normalization ∫|ψ|²dx = 1.

For a time-independent potential, separable solutions ψ(x,t) = φ(x)e−iEt/ℏ satisfy the time-independent Schrödinger equation Ĥφ = Eφ — an eigenvalue equation whose allowed energies E are often discrete.

Explanation

Why boundaries force quantization

Quantization is not an extra postulate bolted onto waves — it falls straight out of requiring a wave to fit consistently inside boundaries, exactly as a guitar string (Chapter 12) can only vibrate at discrete harmonics fn = nv/2L because it must vanish at both fixed ends. A confined quantum particle's wavefunction must likewise satisfy boundary conditions, and only discrete standing-wave-like solutions survive — hence discrete energy levels. An unconfined (free) particle has no such boundary and its energy spectrum is continuous, matching the classical intuition that quantization is a signature of confinement, not a universal graininess.

The Born rule forces a genuine break from classical thinking: before a measurement, a particle described by a superposition of states does not merely have unknown position — it has no definite position at all. Measurement is described as "collapsing" the wavefunction onto a definite outcome with the corresponding probability. This is not a limitation of our instruments; interference experiments (Chapter 13's double slit, and its single-particle version) show a particle really does explore multiple paths simultaneously until measured.

Connects back. Chapter 17's uncertainty principle ΔxΔp ≥ ℏ/2 is not an isolated rule to memorize — it is a direct, provable consequence of [x̂,p̂]=iℏ, the same way every result in this chapter traces back to treating x and p as non-commuting operators rather than ordinary numbers.
Practical

Worked example — the particle in a box

An electron is confined to an infinite square well of width L = 1.0 nm (V=0 inside, V=∞ at the walls). Find the allowed energy levels.

  1. Inside the well the time-independent Schrödinger equation reduces to −ℏ²/2m φ″ = Eφ, with general solution φ(x) = A sin(kx) + B cos(kx), k = √(2mE)/ℏ.
  2. Boundary condition φ(0) = 0 forces B = 0. Boundary condition φ(L) = 0 forces sin(kL) = 0, so kL = nπ for integer n = 1, 2, 3, … — quantization from the boundary, exactly as in the explanation above.
  3. Solve for energy: k = nπ/L, so En = ℏ²k²/2m = n²π²ℏ²/(2mL²).
  4. Plug in numbers (m = 9.11×10⁻³¹ kg, ℏ = 1.055×10⁻³⁴ J·s, L = 1.0×10⁻⁹ m): E₁ = π²(1.055×10⁻³⁴)²/(2×9.11×10⁻³¹×10⁻¹⁸) = 6.02×10⁻²⁰ J ≈ 0.376 eV.
  5. Higher levels scale as : E₂ = 4E₁ ≈ 1.50 eV, E₃ = 9E₁ ≈ 3.38 eV — sharply discrete, macroscopically-measurable steps, unlike anything in classical mechanics where a confined particle could have any energy at all.
Energy E₁ E₂ = 4E₁ E₃ = 9E₁
Confining a particle to a box only permits discrete energies that grow as n² — not a continuous range.
Q&A
Normalize the ground-state wavefunction φ₁(x) = A sin(πx/L) for 0 ≤ x ≤ L.

Require ∫₀L A²sin²(πx/L) dx = 1. Since ∫₀L sin²(πx/L)dx = L/2, we get A²(L/2) = 1, so A = √(2/L).

Derive Δx·Δp ≥ ℏ/2 from the commutator [x̂,p̂] = iℏ using the general uncertainty relation.

The general relation is ΔAΔB ≥ ½|⟨[Â,B̂]⟩|. With Â=x̂, B̂=p̂, the expectation of the commutator is simply iℏ (a constant), so |⟨[x̂,p̂]⟩| = ℏ. Substituting gives ΔxΔp ≥ ℏ/2 exactly.

Why does the particle-in-a-box ground state have E₁ > 0, unlike a classical particle at rest?

A classical particle can sit motionless with E=0. Quantum mechanically, confining a particle to width L forces some momentum uncertainty Δp ≳ ℏ/L by the uncertainty principle, which forces nonzero kinetic energy — this "zero-point energy" is a direct, testable consequence of quantization and shows up in real confined systems (quantum dots, nuclei).

What physically happens to a superposition state when a measurement is made?

Before measurement, ψ can be a superposition of eigenstates, each with its own probability amplitude. Upon measuring the corresponding observable, the outcome is one of the eigenvalues, selected randomly with probability given by the Born rule (|amplitude|²), and the wavefunction "collapses" to the corresponding eigenstate. Repeated identical measurements on identically-prepared systems reproduce the predicted probability distribution, not a single deterministic value.

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

WavefunctionOperatorsSchrodinger eqEigenvaluesQuantizationProbabilityQuantum Formalism
Infographic

The key facts, visualised

psi
Wavefunction; |psi|^2 is probability density
H*psi = E*psi
Time-independent Schrodinger equation
E_n = n^2*h^2/(8*m*L^2)
Particle-in-a-box energy levels
hbar
Reduced Planck constant, 1.05e-34 J*s
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1Find the ground-state energy of an electron in a 1D box of width L (n=1).

  1. E_n = n^2*h^2/(8*m*L^2)
  2. For n=1: E_1 = h^2/(8*m*L^2)

Example 2What is the ratio of the second to first energy level in a box?

  1. E_n scales as n^2
  2. E_2/E_1 = 2^2 / 1^2
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1What does the wavefunction psi represent physically?
Its squared magnitude |psi|^2 gives the probability density of finding the particle.
2Why does confinement in a box force energy quantization?
Boundary conditions allow only standing waves that fit, giving discrete allowed energies.
3What is an operator in quantum mechanics?
A mathematical action on the wavefunction whose eigenvalues are the measurable values.
4What does the Schrodinger equation determine?
How the wavefunction evolves and which energy eigenstates are allowed.
5Why is measurement probabilistic in quantum theory?
The state is a superposition; measurement yields one eigenvalue with probability from |psi|^2.
6What normalization condition must psi satisfy?
The integral of |psi|^2 over all space must equal 1.