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A Three-Factor Solovay–Kitaev Error Budget

Statement

Suppose a fixed three-factor unitary block is U=G3G2G1, where each Gj=eiϕjSj with ϕj∈R and Sj∈SU(2). If unitary gate words Wj satisfy ∥Sj−Wj∥≤ε/3, then the word product C=W3W2W1 satisfies min⁡∣λ∣=1∥U−λC∥≤ε.

Example

Take three one-qubit factors in a fixed decomposition of a block. Each Solovay–Kitaev call is assigned one third of the total error tolerance. The scalar phases are recorded as one phase for the whole unconditional block.

Verification

Given: Gj=eiϕjSj and unitary words Wj with ∥Sj−Wj∥≤ε/3 for j=1,2,3.

[F2] The operator norm is the maximum over unit vectors (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||), and the inner-product vector norm satisfies the triangle inequality (The inner-product norm is definite, homogeneous, and satisfies the triangle inequality).

1.1givenalgebra

Insert one factor at a time: S3S2S1−W3W2W1=(S3−W3)S2S1+W3(S2−W2)S1+W3W2(S1−W1).

2.1F1F2step 1.1

Each Sj and Wj is unitary by the Given. For any unitary Q, ∥Qv∥=∥v∥ for every unit vector v, so [F2] gives ∥Q∥=1. The submultiplicativity in [F1] and the triangle inequality in [F2] therefore bound the norm of the right side by ∥S3−W3∥+∥S2−W2∥+∥S1−W1∥≤ε.

3.1givenstep 2.1algebra∎

By Given, the scalar factors in G3G2G1 multiply to λ=ei(ϕ1+ϕ2+ϕ3). Therefore ∥U−λC∥=∥S3S2S1−W3W2W1∥≤ε by step 2.1. This is an unconditional-block guarantee; it does not license discarding λ inside a later control.

Depends on

Used by

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Sources