Operators and Observables

Introduction

In classical mechanics, momentum and energy are numbers that can be calculated directly from position and velocity. In quantum mechanics the state is a wavefunction, so we need a rule for extracting physical quantities from it. That rule is supplied by operators : each observable property is represented by an operator that acts on the wavefunction. Learning to build and use operators is the practical toolkit of quantum chemistry, and it leads directly to the Schrödinger equation.

Core explanation

What an operator is. An operator is an instruction to do something to a function. The operator d/dx means "differentiate with respect to x"; the operator x means "multiply by x". An operator is written with a circumflex, so Ω̂ψ means "apply Ω̂ to ψ". Operators act on everything to their right, and in general the order matters: x̂(d/dx)ψ is not the same as (d/dx)(x̂ψ).

The postulate. Every observable is represented by an operator. The basic operators are:

- Position: x̂ = x × (multiplication by x). - Linear momentum: p̂ₓ = −iħ d/dx, with ħ = h/2π = 1.055 × 10⁻³⁴ J s.

In three dimensions, p̂ᵧ and p̂ z are defined in the same way with derivatives in y and z.

Building other operators. Any other observable is first written classically in terms of position and momentum, and then x and p are replaced by their operators.

- Kinetic energy. Classically T = pₓ²/2m. Replacing pₓ by −iħ d/dx and applying it twice: T̂ = (1/2m)(−iħ d/dx)(−iħ d/dx) = −(ħ²/2m) d²/dx². In three dimensions T̂ = −(ħ²/2m)∇², where ∇² = ∂²/∂x² + ∂²/∂y² + ∂²/∂z² is the Laplacian. - Potential energy. V depends only on position, so V̂ is simply multiplication by V(x). For an electron and a proton, V̂ = −e²/(4πε₀r). - Total energy. The Hamiltonian operator is Ĥ = T̂ + V̂ = −(ħ²/2m)∇² + V.

The physical meaning of the momentum operator. Kinetic energy is linked to the curvature of the wavefunction: d²ψ/dx² is large where ψ bends sharply. A wavefunction squeezed into a small region must curve steeply, so it has high kinetic energy. This is the operator form of the confinement idea met earlier: shorter wavelengths mean larger momentum.

Real results from complex operators. The momentum operator contains i, yet measured momenta are real. Operators that represent observables are Hermitian , a property that guarantees real measured values. Both x̂ and p̂ₓ are Hermitian, and so are T̂, V̂ and Ĥ.

Commutators. Because operators may not commute, we define [Â, B̂] = ÂB̂ − B̂Â. For position and momentum, acting on any ψ: x̂p̂ₓψ − p̂ₓx̂ψ = −iħx dψ/dx + iħ(ψ + x dψ/dx) = iħψ. Hence [x̂, p̂ₓ] = iħ. A non-zero commutator means the two observables cannot simultaneously have precisely defined values — the formal origin of the uncertainty principle.

Formulae

x̂ = x ×. p̂ₓ = −iħ d/dx. T̂ = −(ħ²/2m) d²/dx² (one dimension), −(ħ²/2m)∇² (three dimensions). V̂ = V(x) ×. Ĥ = T̂ + V̂. [x̂, p̂ₓ] = iħ.

Step-by-step reasoning

To construct a quantum operator for an observable:

1. Write the classical expression in terms of Cartesian positions and linear momenta. 2. Replace each position coordinate by multiplication by that coordinate. 3. Replace each momentum component by −iħ times the derivative with respect to the matching coordinate. 4. Simplify, remembering that i² = −1 and that order of operators matters.

Visual explanation

Draw two wavefunctions of the same height: a gentle, wide hump and a narrow, sharply peaked hump. Mark the curvature at each peak. The narrow hump curves far more steeply, so the kinetic-energy operator returns a much larger value — confinement costs energy.

Real-world analogy

An operator is like a machine in a factory. The raw material (the wavefunction) goes in, and a processed product comes out. A "position machine" stamps each part with its location; a "momentum machine" measures how rapidly the material's pattern changes along the conveyor.

Real-world example

In a quantum chemistry program, the Hamiltonian operator for a molecule is assembled from kinetic-energy terms for every electron, attraction terms between electrons and nuclei, and repulsion terms between electrons. The program then computes how this operator acts on candidate wavefunctions to find molecular energies.

Why?

Why does the momentum operator involve a derivative? Momentum is linked to wavelength through λ = h/p, and a derivative measures how rapidly ψ oscillates in space. Faster oscillation means shorter wavelength and hence greater momentum.

Common misconception

"Applying an operator to ψ always gives a number times ψ." Usually it gives a different function. Only special functions — eigenfunctions — are returned as a constant multiple of themselves, as the next page explains.

Worked example

Question: Apply p̂ₓ to (a) ψ = e^(ikx) and (b) ψ = sin(kx). Comment on the results.

Reasoning: (a) p̂ₓe^(ikx) = −iħ(ik)e^(ikx) = ħk e^(ikx). The result is a constant, ħk, times the original function. (b) p̂ₓ sin(kx) = −iħk cos(kx), which is a different function, not a multiple of sin(kx).

Answer: e^(ikx) has definite momentum +ħk; sin(kx) does not have a single definite momentum, because it is a mixture of waves moving in both directions.

Quick check

1. Write the kinetic-energy operator for a particle of mass m moving in one dimension. Answer: T̂ = −(ħ²/2m) d²/dx², obtained by applying the momentum operator twice and dividing by 2m.

Exam focus

Memorise p̂ₓ = −iħ d/dx and derive T̂ from it rather than memorising T̂ separately. Show working carefully when applying operators: sign errors from i² = −1 are the most common mistake.

Advanced insight

Angular momentum operators are built in the same way from the classical l = r × p. For example l̂ z = x̂p̂ᵧ − ŷp̂ₓ, which in spherical polar coordinates becomes −iħ ∂/∂φ. Different components of angular momentum do not commute, explaining why only one component can be specified with the total.

Summary

Every observable in quantum mechanics is represented by an operator acting on the wavefunction. Position is multiplication by x; momentum is −iħ d/dx. Other operators follow by substituting these into classical expressions, giving the kinetic-energy operator and the Hamiltonian Ĥ = T̂ + V̂. Observables correspond to Hermitian operators, which yield real values. Position and momentum do not commute: [x̂, p̂ₓ] = iħ.

Practice questions

1. Apply the kinetic-energy operator to ψ = sin(kx) and state the result. Answer: T̂ sin(kx) = −(ħ²/2m)(−k²)sin(kx) = (ħ²k²/2m) sin(kx). 2. Write the Hamiltonian for a particle of mass m moving in one dimension under the potential V(x) = ½kₓx². Answer: Ĥ = −(ħ²/2m) d²/dx² + ½kₓx². 3. Evaluate p̂ₓ acting on e^(−ikx) and interpret the result. Answer: p̂ₓe^(−ikx) = −ħk e^(−ikx); the particle has definite momentum ħk directed along −x. 4. Explain why x̂ and p̂ₓ cannot both have precisely defined values in the same state. Answer: Their commutator [x̂, p̂ₓ] = iħ is not zero, so no wavefunction can be an eigenfunction of both, and their uncertainties cannot both be zero.