Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 342 / 3 / c

Table of contents
- Solution c

Solution

At

t = Δ

and

μ = 0

, direct substitution of

a_{j} = \frac{1}{2} (c_{2 j - 1} + i c_{2 j})

into the bond Hamiltonian makes the hopping and pairing terms cancel except for

h_{j} = i t c_{2 j} c_{2 j + 1} .

Thus

H_{0} = i t j \sum c_{2 j} c_{2 j + 1} .

The terms pair disjoint Majoranas, so this is already Majorana diagonal, with nonzero single-particle values

ϵ_{j} = 2 t

. For periodic boundaries the final pair closes around the chain; for open boundaries

c_{1}

and

c_{2 M}

remain unpaired.

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