Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 2 / a

Solution

Every element of the

n

1

- 1

S_{j}

had

S_{j}^{2} = - 1

, closure of the group would put

- 1

S

, contrary to the definition. Thus

S_{j}^{2} = 1

. Since Pauli operators are unitary,

S_{j}^{- 1} = S_{j}^{†},

and

S_{j}^{- 1} = S_{j}

proves

S_{j} = S_{j}^{†}

If two stabilizers

g, h

anticommute and a nonzero vector

∣ ψ ⟩

belonged to the codespace, then

g h ∣ ψ ⟩ = ∣ ψ ⟩, g h ∣ ψ ⟩ = - h g ∣ ψ ⟩ = - ∣ ψ ⟩,

a contradiction. Hence a nonzero stabilizer code requires

S

C (S) = {P \in P_{n} : P g = g P for every g \in S} .

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