In a randomised-benchmarking sequence

PHYSICS · Multi-part · Fermi University Physics · Quantum Information & Technology · Open quantum systems, decoherence and error correction · Physical realizations and quantum technologies

Quantum Information and Technology, Ch What a qubit must do: requirements, tomography and benchmarking

Problem

In a randomised-benchmarking sequence, each step consists of a uniformly random single-qubit Clifford gate CC (one of the 2424 elements of the single-qubit Clifford group) followed by a fixed error channel E(ρ)=pIρ+px XρX+py YρY+pz ZρZ,pI+px+py+pz=1.\mathcal{E}(\rho) = p_I\rho + p_x\,X\rho X + p_y\,Y\rho Y + p_z\,Z\rho Z, \qquad p_I+p_x+p_y+p_z=1. Because the Clifford group normalises the Pauli group, conjugation by CC sends any fixed nonidentity Pauli to ±X\pm X, ±Y\pm Y or ±Z\pm Z, and as CC ranges uniformly over the 2424 Cliffords each of X,Y,ZX,Y,Z occurs equally often (eight times), independent of the original Pauli. ((a)) Show that the Clifford average of E\mathcal{E}, Eˉ(ρ)=124∑CC† E(CρC†) C\bar{\mathcal{E}}(\rho) = \dfrac{1}{24}\sum_C C^\dagger\,\mathcal{E}(C\rho C^\dagger)\,C, is exactly the depolarising channel Dλ(ρ)=(1−λ)ρ+λI/2\mathcal{D}_\lambda(\rho) = (1-\lambda)\rho + \lambda I/2 with λ=43(px+py+pz)\lambda = \tfrac{4}{3}(p_x+p_y+p_z). ((b)) Show that Dλ\mathcal{D}_\lambda shrinks every Bloch-vector component by the same factor p=1−λp=1-\lambda, i.e. ⟨X⟩→p⟨X⟩\langle X\rangle\to p\langle X\rangle and likewise for Y,ZY,Z. ((c)) A randomised-benchmarking sequence of length mm applies mm such steps, prepares an initial state with some fixed (not necessarily perfect) polarisation, and measures with a fixed (not necessarily perfect) measurement. Show that the surviving polarisation after averaging over all random sequences is shrunk by pmp^m, and hence that the measured sequence fidelity has the form F(m)=Apm+BF(m) = Ap^m + B, with AA and BB independent of mm.

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