We now analyze the relation between L and Φec. Recall the following terminology and notations.
A ground state of the memory Hamiltonian will be referred to as avacuum. It will be convenient to perform an overall energy shift such that the vacuum has zero energy. A Pauli operator is an arbitrary tensor product of single-qubit Pauli operatorsX, Y, Zand the identity operatorsI. We will say that a Pauli operator createsmdefectsiffP anticommutes with exactlymstabilizer generators GXc , GZc. Equivalently, applyingP to the vacuum one obtains an eigenvector ofH with energy m.
For example, using the explicit form of the generators, see Figure5.1, one can check that single-qubit X or Z errors create 4 defects, while Y errors create 8 defects. See Figure7.2. We will say that a Pauli error P is corrected by the decoder if Pec(S(P)) = ±P G, where S(P) is the syndrome ofP andGis a product of stabilizer generators. LetN = 2L3be the total number of qubits. Note that N is also the number of stabilizer generators for the cubic code.
Let Γ = (P0, P1, . . . , Pt) be a finite sequence of Pauli operators such that the operatorsPi and Pi+1 differ on at most one qubit for all 0≤i < t. We say that Γ is anerror pathimplementing a Pauli operatorP whenP0 =I andPt =P. Let mi be the number of defects created byPi. The maximum number of defects
m(Γ) = max
0≤i≤t mi
will be called anenergy costof the error path Γ. Given a Pauli operatorP, we define itsenergy barrier∆(P) as the minimum energy cost of all error paths implementingP,
∆(P) = min
Γ m(Γ).
Although the set of error paths is infinite, the minimum always exists because the energy cost is a nonnegative integer. In fact, it suffices to consider paths in which Pi are all distinct. The number of such paths is finite since there are only finitely many Pauli operators for a given system size.
It is worth emphasizing that an operator P may have a very large energy barrier even though P itself creates only a few defects or no defects at all. Consider as an example the 2D Ising model,
H = −12P
huviZuZv, where the sum runs over all pairs of nearest neighbor sites on the square lattice of sizeL×Lwith open boundary conditions. Then the logical-X operator P =N
uXu has an energy barrier ∆(P) = Lsince any sequence of bit-flips implementing P must create a domain wall across the lattice. It is clear that a Pauli operatorP acting nontrivially onnqubits has energy barrier at mostO(n).
A naive intuition suggests that a stabilizer code Hamiltonian is a good candidate for being a self-correcting memory if there exists an error correction algorithm, or a decoder, that corrects all errors with a sufficiently small energy barrier. Errors with a high energy barrier can confuse the decoder and cause it to make wrong decisions, but we expect that such errors are unlikely to be created by the thermal noise. We make this intuition more rigorous.
Let f be the maximum energy barrier of Pauli operators that appear in the expansion of the quantum jump operators Aα,ω or A†α,ωAα,ω of Eq. (8.3). Since Aα,ω act on a constant number of qubits by Proposition8.3.1, we havef =O(1). Belowmis an arbitrary energy cutoff. LetD= ker ˆH be the set of all operators that are commuting with the Hamiltonian H. Every operator in D is block-diagonal in the energy eigenstate basis. Sinceρ(0) is supported on the ground subspace ofH, we haveρ(0)∈ D. Below we only consider states fromDand linear maps preservingD. Define an orthogonal identity decomposition
I= Π−+ Π+
where Π− projects onto the subspace with energy< m+f and Π+projects onto the subspace with energy≥m+f. Introduce auxiliary Lindblad generators
L−(ρ) =X
α
X
ω
h(α, ω)
Bα,ωρBα,ω† −1
2{ρ, B†α,ωBα,ω}
, where Bα,ω= Π−Aα,ω (8.8)
and
L+(ρ) =X
α
X
ω
h(α, ω)
Cα,ωρ Cα,ω† −1
2{ρ, Cα,ω† Cα,ω}
, where Cα,ω = Π+Aα,ω. (8.9)
Simple algebra shows thatL− andL+ preserve Dand theirrestrictions on Dsatisfy
L=L−+L+. (8.10)
It is useful to note that any X ∈ D commutes with Π±. By abuse of notations, we shall apply Eq.(8.10) as though it holds for all operators.
Lemma 8.4.1. Suppose that an error correction algorithm s 7→ Pec(s) corrects any Pauli error P with the energy barrier smaller than m+ 2f. Let Φec be the corresponding decoder defined by
Eq.(8.6). For any timet≥0and for any state ρ0 supported on the ground subspace ofH one has
Φec(eL−t(ρ0)) =ρ0.
Proof. Since all maps are linear, we may assumeρ0=|gi hg|is a pure state. Then,eL−t(ρ0) is in the span of states of form |ψi= Π−En· · ·Π−E2Π−E1|gi, where Ei are Pauli operators that appears in the expansion of Aα,ω or A†α,ωAα,ω. This follows from the Taylor expansion of eL−t. Since Pauli errors map eigenvectors ofH to eigenvectors ofH, we conclude that either|ψi= 0, or |ψi= En· · ·E2E1|gi. Furthermore, the latter case is possible only if the Pauli operatorE≡En· · ·E2E1
has energy barrier smaller thanm+ 2f. Indeed, definition of Π− implies thatEjEj−1· · ·E1creates at mostm+f−1 defects for allj= 1, . . . , n. By assumption, each operatorEj can be implemented by an error path with energy cost at mostf. Taking the composition of all such error paths one obtains an error path forE with energy cost at mostm+ 2f−1 and thus Φec will correctE. Since Φec ◦ eL−tis a TPCP map, we must have Φec(eL−t(ρ0)) =ρ0.
For any decompositionL=L−+L+ one has the following identity:
eLt=eL−t+ Z t
0
ds eL−(t−s)L+eLs, (8.11)
which follows from the identity d
dseL−(t−s)eLs=eL−(t−s)(−L−+L)eLs=eL−(t−s)L+eLs.
Lemma 8.4.2. Assume the supposition of Lemma 8.4.1. Let Qm be the projector onto the (high energy) subspace with at least mdefects. Then
(t)≡ kΦec(ρ(t))−ρ(0)k1≤O(tN)TrQme−βH (8.12)
for any initial state ρ(0) supported on the ground subspace of H. The time evolution of ρ(t) is governed by the Lindblad equation, Eq. (8.1), with the inverse temperature β of the bath.
Proof. Writeρ0 ≡ ρ(0). First, Proposition 8.3.3 saysρ(t) = eLt(ρ0). Applying Lemma 8.4.1 and Eq. (8.11) one arrives at
(t)≤ Z t
0
dskeL−(t−s)L+eLs(ρ0)k1≤t· max
0≤s≤t kL+eLs(ρ0)k1. (8.13) We shall use an identity
L+(X) =L+(QmXQm) (8.14)
valid for any X ∈ D. Indeed, any quantum jump operatorAα,ω changes the energy at most byf, so thatL+(Q⊥mX) =L+(XQ⊥m) = 0 for anyX ∈ D(note that anyX ∈ Dcommutes withQm). We arrive at
(t)≤t· kL+(QmeLs(ρ0)Qm)k1≤t· kL+k1· kQmeLs(ρ0)Qmk1≤O(tN) TrQmeLs(ρ0), (8.15)
where the maximization over s is implicit. We used Proposition8.3.1 and the positivity ofL(ρ0).
in the last inequality. Since the ground-state energy ofH is zero, one has
ρ0=Zβρβρ0, (8.16)
whereZβ is the partition function. It yields
TrQmeLs(ρ0) = Trρ0eL∗s(Qm) =ZβTrρβρ0eL∗s(Qm) =Zβhρ0, eL∗s(Qm)iβ, (8.17) Proposition 8.3.4 implies that the map eL∗s is also self-adjoint with respect to the Liouville inner product. Hence, we have
TrQmeLs(ρ0) =Zβhρ0, eL∗s(Qm)iβ =ZβheL∗s(ρ0), Qmiβ= TreL∗s(ρ0)Qme−βH ≤TrQme−βH, (8.18) where the last inequality is becauseeL∗s is a unital completely positive map andρ0≤I.