Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 342 / 3 / d

Table of contents
- Solution d

Solution

Write

a = γ_{2 M}

and

b = γ_{2 M - 1}

. The ancilla condition

iab ∣ ψ ⟩ = ∣ ψ ⟩

implies

a ∣ ψ ⟩ = ib ∣ ψ ⟩

and lets every occurrence of

a

on the state be replaced by

ib

. Define the two parity projectors

P_{1} = \frac{1 - i γ _{3} b}{2}, P_{2} = \frac{1 + γ _{1} γ _{2} γ _{4} b}{2} .

Their factors square to one, so they are valid fermion-parity measurement projectors. Expanding

P_{1} P_{2}

, using the Majorana anticommutation relations, replacing

a

with

ib

on

∣ ψ ⟩

, and using

e^{π γ_{3} a /4} = \frac{1 + γ _{3} a}{2}, e^{iπ γ_{1} γ_{2} γ_{3} γ_{4} /4} = \frac{1 + i γ _{1} γ _{2} γ _{3} γ _{4}}{2},

gives

e^{iπ γ_{1} γ_{2} γ_{3} γ_{4} /4} ∣ ψ ⟩ \propto e^{π γ_{3} γ_{2 M} /4} \frac{1 - i γ _{3} γ _{2 M - 1}}{2} \frac{1 + γ _{1} γ _{2} γ _{4} γ _{2 M - 1}}{2} ∣ ψ ⟩ .

The proportionality absorbs the probability amplitude for obtaining the two displayed measurement outcomes. This realizes the four-Majorana phase gate using one braid and parity measurements.

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