Error decomposition proof

5.7. Error decomposition proof 125

call a nonempty subsetSofEacandidate level-0 error ifS consists of either phase flip errors or XXXX measurement errors, but not both, and if it is contained in a spacetime box [x, x+ 1]×[y, y+ 1]×[t, t]. S will be an actual level-0 error if S andE\S are (a, a, b)-separated, whereaand bare fixed positive integers. The union of all actual level-0 errors is called level-0 noise, which we denote by E₀.

Now we proceed to define candidate and actual level-n errors inductively. For n >0, suppose that candidate level-kerrors, actual level-kerrors, and level-knoise Ek have been defined for k < n.

A nonempty set S⊆E\En−1 is a candidate level-n error if (i) S is contained in a spacetime box of sizeQⁿ×Qⁿ×Uⁿ, and (ii) S contains at least two disjoint candidate level-(n-1) errors that are

(aQⁿ⁻¹, aQⁿ⁻¹, bUⁿ⁻¹)-linked.

S will be an actual level-n error if, additionally,

(iii) S does not contain two candidate level-nerrors that are (4(a+ 2)Qⁿ⁻¹,4(a+ 2)Qⁿ⁻¹,4(b+ 2)Uⁿ⁻¹)-separated, and (iv) S andE\(S∪En−1) are (aQⁿ, aQⁿ, bUⁿ)-separated.

Level-nnoiseEn is defined as the union ofEn−1 and all actual level-nerrors inE.

We will further define a candidate level-n error S as minimal if S contains exactly 2ⁿ points. Now we will proceed to show by induction that every point in E\En−1 belongs to at least one minimal candidate level-nerrorS. The base case of n = 1 is true provided that Q ≥ a and U ≥ b (so that two points that are (a, a, b)-linked fit inside a box of sizeQ×Q×U).

For the inductive step, let us assume that the hypothesis holds true for (n−1).

Let ~x₁ ≡ (x₁, y₁, t₁) be a spacetime point that is a member of E\En−1; we will show that this point must belong to a minimal candidate level-nerror. The point must be a member ofE\En−2, so by the inductive hypothesis it belongs to some minimal candidate level-(n-1) errorS₁. The minimality implies thatS₁ consists of

5.7. Error decomposition proof 126

two disjoint minimal candidate level-(n-2) errors that are (aQⁿ⁻², aQⁿ⁻², bUⁿ⁻²)- linked. Therefore,S1 must fit into a box of size (a+ 2)Qⁿ⁻²×(a+ 2)Qⁿ⁻²×(b+ 2)Uⁿ⁻². Now, if any point inS₁ were part of an actual level-(n-1) error, then all of S1 would be contained in that actual level-(n-1) error. But~x1∈S1and~x1 ∈/En−1, which means that S₁ must not be contained in any actual level-(n-1) error. Since S1 satisfies condition (iii), it must not satisfy condition (iv). Then there exists some ~x₂ ∈ E \En−1 such that ~x₂ ∈/ S₁ and ~x₂, S₁ are (aQⁿ⁻¹, aQⁿ⁻¹, bUⁿ⁻¹)- linked. Now, it is true that ~x2 must belong to the set of points in E\En−1 that are not in S₁ such that~x₂, ~x₁ are ((a+ 1)Qⁿ⁻¹,(a+ 1)Qⁿ⁻¹,(b+ 1)Uⁿ⁻¹)-linked.

The inductive hypothesis says that there must be a minimal candidate level-(n-1) errorS₂ containing~x₂.

We can also choose S₂ such that S₁ and S₂ are disjoint. There are two cases to be considered. The first case is to suppose that there exists some~x2 such that

~

x₂ is 2(a+ 2)Qⁿ⁻²-separated from ~x₁. Then the size restrictions on S₁ and S₂ force them to be disjoint. The second case is to suppose that the first case does not hold; i.e., there does not exist such an~x₂. Then at worstS≡S₁∪S₂ fits in a box of size 4(a+ 2)Qⁿ⁻²×4(a+ 2)Qⁿ⁻²×4(b+ 2)Uⁿ⁻² which is ((a+ 1)Qⁿ⁻¹− 4(a+ 2)Qⁿ⁻²,(a+ 1)Qⁿ⁻¹−4(a+ 2)Qⁿ⁻²,(b+ 1)Uⁿ⁻¹−4(b+ 2)Uⁿ⁻²)-separated from the rest ofE\En−1. SinceQ≥4(a+ 2) andU ≥4(b+ 2),S satisfies (i) and (iv) in the definition of an actual level-(n-1) error. But S also satisfies (ii) (since it containsS1) and (iii). Then S must be an actual level-(n-1) error. However, we have already shown thatS₁ cannot be in an actual level-(n-1) error, so the second case leads to a contradiction and cannot occur; then S1 and S2 can be chosen to be disjoint, and the set S is a minimal candidate level-nerror containing ~x₁.

This argument actually allows us to place more restrictive bounds on level- n errors. By replacing (n−1) with n in the argument above, we can see that a minimal candidate level-n error S1 fits into a box of size (a+ 2)Qⁿ⁻¹ ×(a+ 2)Qⁿ⁻¹×(b+2)Uⁿ⁻¹. Also, given a point~xinE\E_n−1that is (2(a+2)Qⁿ⁻¹,2(a+

2)Qⁿ⁻¹,2(b+2)Uⁿ⁻¹)-separated from at least one member ofS1, the point~xlies in a second minimal candidate level-n errorS₂ that is disjoint from S₁. If S₁, S₂ are

5.7. Error decomposition proof 127

(4(a+ 2)Qⁿ⁻¹,4(a+ 2)Qⁿ⁻¹,4(b+ 2)Uⁿ⁻¹)-separated and also (aQⁿ, aQⁿ, bUⁿ)- linked, we can see from conditions (iii) and (iv) thatS1 andS2cannot be part of an actual level-nerror. Thus, ifS₁ is part of a actual level-nerror, any point inS\S₁ must be (5(a+2)Qⁿ⁻¹,5(a+2)Qⁿ⁻¹,5(b+2)Uⁿ⁻¹)-linked withS1. This means that the entire error will fit into a box of size 7(a+2)Qⁿ⁻¹×7(a+2)Qⁿ⁻¹×7(b+2)Uⁿ⁻¹).

We can take the intersection of this box with the bounding box of sizeQⁿ×Qⁿ×Uⁿ from condition (i). This reveals that an actual level-n error fits inside a box of size min{Q,7(a+ 2)}Qⁿ⁻¹×min{Q,7(a+ 2)}Qⁿ⁻¹×min{U,7(b+ 2)}Uⁿ⁻¹. We will use this result in the proof demonstrating correction of higher level errors in Section 5.8.2.

Finally, let us define the level-n error rate_n as the probability that a box of sizeQⁿ×Qⁿ×Uⁿ has nonempty intersection with at least one candidate level-n error. A spacetime box of unit size cannot include more than four edges and four vertices of the lattice, so clearly ₀ ≤ 4(p+q), where p and q are the phase flip and measurement error probabilities per time step. An upper bound on n can be found by considering the probability that a box of sizeQⁿ×Qⁿ×Uⁿhas nonempty intersection with at least one candidate level-(n-1) error. This probability must be bounded above by Q²U n−1. Gray asserts that by the Kesten-Vandenberg inequality, since each candidate level-nerror is composed of at least two candidate level-(n-1) errors,

n≤(Q²U n−1)². (5.1)

Then we can see that

_n≤(Q²U)²ⁿ⁺¹⁻²(4(p+q))²ⁿ <(Q⁴U²4(p+q))²ⁿforn >0 . (5.2) That is to say, level-nerrors get progressively more rare (in fact, double exponen- tially) as ngets larger, as long as

(p+q)< Q⁻⁴U⁻²/4. (5.3)