“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 centered box and parity

Exercises

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

Check Problem 9 of 10

The centered box and parity

Problem

Re-solve the box on

x \in [- L /2, + L /2]

instead of

[0, L]

. Derive the eigenfunctions, show they split into even and odd families, and show the energies are unchanged. Then explain what feature of the potential the even/odd split came from, and what observable consequence it has for dipole transitions

⟨ f ∣ x ∣ i ⟩

Solution

Solutions of $- ψ^{''} = k^{2} ψ$ vanishing at $\pm L /2$ : $cos (k x)$ with $k L = π, 3 π, 5 π, \dots$ (even family) and $sin (k x)$ with $k L = 2 π, 4 π, \dots$ (odd family). Interleaved, they reproduce $k_{n} = nπ / L$ exactly — same $E_{n} = n^{2} h^{2} /8 m L^{2}$ , since shifting the origin cannot change physics.

The split is forced by the symmetry $V (- x) = V (x)$ : the Hamiltonian commutes with parity, so eigenstates can be chosen even or odd, alternating with $n$ . Consequence: $x$ is odd, so $⟨ f ∣ x ∣ i ⟩$ vanishes unless $f$ and $i$ have opposite parity — the selection rule $Δ n$ odd. The $[0, L]$ coordinates hide this structure; the centered ones make it a two-line argument.