“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Particle in a Box by Hand — The node that "traps" the electron

Exercises

Introductory Quantum Mechanics › Unit 2 · Bound states in one dimension › The Particle in a Box by Hand › all problems

Check Problem 10 of 10

The node that "traps" the electron

Problem

A student argues: "The

n = 2

state has a node at

L /2

— zero probability of being found there. An electron on the left side can therefore never reach the right side, since it cannot cross the middle. So once measured on the left, it must stay on the left." Locate every flaw, and describe what actually happens after a position measurement finds the electron on the left half.

Solution

Flaw 1: a node is a zero of the probability density at a point, not a wall — density zero at one point is no obstacle, just as a classical oscillator has zero probability density of being found exactly at maximum displacement with any given velocity sign. Flaw 2: nothing is “crossing” anything — the $n = 2$ state is stationary; its density $∣ ψ_{2} ∣^{2}$ is time-independent and perfectly symmetric between the halves. The picture of a little ball commuting across the box is the classical smuggling the argument runs on.

Flaw 3 is the sharpest: a position measurement that resolves “left half” does not leave the electron in the $n = 2$ state at all. It collapses the state to one localized on the left — a superposition of MANY box eigenstates — which then evolves and sloshes: the right half promptly acquires probability. The premise "it is in $n = 2$ AND known to be on the left" is self-contradictory; no state has both properties.