Physic Labs

Quantum mechanics

The wavefunction and its probabilistic interpretation

The wavefunction ψ encodes probability amplitudes; |ψ|² is the position probability density, normalized so total probability is one.

In quantum mechanics, a particle's pure state is described by a complex wavefunction ψ(r,t)\psi(\mathbf r,t). It is neither a trajectory nor a material density; it lets us calculate probabilities of measurement outcomes.

ρ(r,t)=∣ψ(r,t)∣2,P(Ω,t)=∫Ω∣ψ(r,t)∣2 d3r\rho(\mathbf r,t)=|\psi(\mathbf r,t)|^2,\qquad P(\Omega,t)=\int_{\Omega}|\psi(\mathbf r,t)|^2\,d^3r

Definition: Born probability density

∣ψ∣2|\psi|^2 is probability per unit volume. The probability of finding the particle in a region Ω\Omega is the density integrated over that region; in SI, ∣ψ∣2|\psi|^2 has units m−3^{-3}.

Observe ∣ψ∣2|\psi|^2 for a superposition of two box states. Change the energy splitting to see the density oscillate while normalization is preserved.

Normalization and measurement

∫R3∣ψ(r,t)∣2 d3r=1\int_{\mathbb R^3}|\psi(\mathbf r,t)|^2\,d^3r=1

For a particle confined to one dimension, the probability of finding it between aa and bb is P=∫ab∣ψ(x,t)∣2dxP=\int_a^b|\psi(x,t)|^2dx. The density can change with time, but unitary evolution of a closed system preserves its integral over all space at one.

Example: Probability in half a box

A particle in the ground state of an infinite well 0<x<L0<x<L has ψ1=2/Lsin⁡(πx/L)\psi_1=\sqrt{2/L}\sin(\pi x/L). What is the probability of finding it in the left half?

Solution

P=∫0L/2(2/L)sin⁡2(πx/L)dx=1/2P=\int_0^{L/2}(2/L)\sin^2(\pi x/L)dx=1/2. The symmetry of ∣ψ1∣2|\psi_1|^2 about L/2L/2 gives the same result.

For a position measurement, the probability of finding a particle in [a,b][a,b] is P=∫ab∣ψ(x,t)∣2 dxP=\int_a^b |\psi(x,t)|^2\,dx. A state normalized on the full line integrates to one, and probabilities for disjoint regions add. Complex amplitudes can interfere before squaring: for two indistinguishable paths, ∣ψ1+ψ2∣2|\psi_1+\psi_2|^2 contains a cross term that adding separate probabilities would miss. Thus measurement is not merely sampling a pre-existing classical distribution; the chosen observable determines the possible outcomes and their probabilities. In three dimensions replace dxdx by the volume element d3rd^3r.

Quick check

Which quantity gives the probability density for finding a particle at r\mathbf r?

What does normalization of a wavefunction over all space require?

References

  1. David J. Griffiths, Darrell F. Schroeter (2018). Introduction to Quantum Mechanics