 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/34b/c/solution

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

Ψ_{k} (x) = n = 0 \sum 2 e^{ikna} ψ (x - na) = e^{ik x} n = 0 \sum 2 ψ (x - na) e^{- ik (x - na)} .

Hence

Ψ_{k} (x) = e^{ik x} u_{k} (x), u_{k} (x) = \sum_{n = 0}^{2} ψ (x - na) e^{- ik (x - na)},

up to the common normalization factor

1/ 3

Under

x \mapsto x + a

, reindexing

m = n - 1

gives the same three terms modulo three sites. The term shifted by three sites is unchanged because

ψ_{m + 3} = ψ_{m}

and

e^{3 ika} = 1

. Therefore

u_{k} (x + a) = u_{k} (x) .

The one-dimensional Bloch theorem states that an eigenfunction in a potential of period

a

can be chosen in the form

Ψ_{k} (x) = e^{ik x} u_{k} (x), u_{k} (x + a) = u_{k} (x) .

It applies because the three-atom ring is invariant under translation by one lattice spacing and obeys periodic boundary conditions.

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