 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/14a/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The energy fixes

l = 2

m

. Therefore every prepared state is an eigenstate of

\hat{L}^{2}

with

L = l (l + 1) ℏ = 6 ℏ.

An immediate measurement of the total orbital-angular-momentum magnitude cannot yield

ℏ

, so the requested probability is

0 .

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