Fault-Tolerant Quantum Computation with Constant Error

• p. 5 Section 2.0.1, top of page

  collapse to |β⟩ = 1/p0 Σ_{i,i|q=0} c_i|i⟩

• p. 5 Section 2.0.1, basic measurement

  the state of the system will collapse to |β⟩= 1/p0 Σ i,i|q=0 ci|i⟩

Judge's note: Read the page 5 image: the factor is a plain fraction 1/p0 with no square root. Reworked: the projected vector has squared norm p0, so dividing by p0 leaves squared norm 1/p0, which is not a unit vector unless p0 = 1. Example: (|0⟩+|1⟩)/√2 gives p0 = 1/2 and the printed formula yields √2|0⟩, squared norm 2. The paper defines a pure state as a unit vector. The slip is confined to this recalled definition; later arguments use density matrices.

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• p. 7 Lemma 1, induction step

  so the density matrix is kl + d/2-deviated.

• p. 13 Appendix C, induction assertion 1

  the density matrix is (r, d) deviated from φʳ(g ◦ ρ₀).

• p. 14 Appendix C, proof of claim 1 for r+1

  applying an r−correction stage followed by an r−computation stage, on a matrix which is (r, kl)-deviated

• p. 13 Appendix C, claims 1 and 2

  the density matrix is (r, d) deviated from φr(g ◦ρ0)

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Judge's note: I viewed pp.6, 7, 13 and 14. Spread is defined on p.6 as "no qubit or gate effects more than l outputs", so a deviated qubit in a control block propagates into the target block. I reworked the case k=l=1, d=2 (which satisfies d≥2kl) with the bitwise CNOT of Theorem 3. At the end of the previous period the control block has one deviated coordinate (1) and the target block has one (2). Bitwise CNOT carries coordinate 1 into the target, and one fault in the rectangle hits target coordinate 3. The target block is then 3-deviated, which exceeds kl+d/2=2 and the code's d=2. The printed step is therefore false, and Lemma 1 (and the matching claim 1 of Appendix C at r=1) is left without a valid proof as stated. Second note: claim 2 maps (r,d)→(r,kl) and claim 1 maps (r,kl)→(r,d). Correction followed by computation therefore ends (r,d)-deviated, not (r,kl). The invariant needs the order computation then correction. The same sentence appears again in the proof of claim 2. This second point is repairable and would by itself be S3.

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• p. 9 Section 5.1, code construction

  Given two linear codes C₁ and C₂

• p. 10 Theorem 3 proof

  To apply the procedures of all the gates, except the Toffoli gate, we simply apply the gate bitwise on all the qubits

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Judge's note: I viewed pp.9-10. The construction on p.9 allows any {0}⊂C2⊂C1. I checked the reviewer's example. The C2 generators 1010101010, 0110011001, 0001111000 and 0000000111 are independent (dim 4), and C1 adds 1110000000 (dim 5). x=1001100000 has even overlap with every C2 generator, so x∈C2⊥. But x∉C1: the coordinates 8, 9 and 10 force the coefficients of g4, g1 and g2 to be zero, and {1,4,5} is not a combination of {4,5,6,7} and {1,2,3}. Also dim C2⊥=6 > dim C1=5. So bitwise H leaves the codespace. A self-duality-type condition (C1=C2⊥) is needed but not stated.

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• p. 10 Section 5.1, after the definition of |Ca>

  both have minimum weight d, then the quantum code can correct for t = ⌊(d−1)/2⌋

• p. 10 Section 5.2

  To correct d errors, set m = 4d + 1

• p. 10 Section 5.2, polynomial-code parameters and V₁

  To correct d errors, set m = 4d + 1 and set p > m + 1.

• p. 10 Section 5.2, definition of V₁

  degf(x) ≤ d

Only the authors can say what is right here.

Judge's note: I viewed p.10. C2 is the evaluation code of polynomials of degree ≤ d vanishing at 0 (dimension d), so C2⊥ is a GRS code of dimension m−d=3d+1 and distance m−(3d+1)+1=d+1. That corrects only ⌊d/2⌋ phase errors. I checked the reviewer's d=1 case directly: f=a+tx gives 2f(1)−f(2)=a, so L=Z1^2 Z2^-1 multiplies |Sa> by ω^a, and Z1^2χ0 = Z2χ1 for the orthogonal words χ0 and χ1=Lχ0. Separately, the first note is right that §5.1 reuses d for minimum weight ("both have minimum weight d, then the quantum code can correct for t=⌊(d−1)/2⌋") on the same page where §5.2 uses d for the number of errors corrected. That is a minor notational clash.

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• p. 17 Appendix E (4), displayed gate

  |a >7−→ Σ_{b∈Fp} w_l^{ab}|a >

• p. 10 Section 5.2, gate list

  |a >7−→ 1/√p Σ_{b∈F} w^{rab}|b >

• p. 17 Appendix E, item (4), displayed gate

  |a >7−→ Σ b∈Fp w^{ab}_l |a >

Judge's note: Page 17 image shows the ket inside the sum as |a>. As printed the map is a scalar multiple of |a>, and that scalar Σ_b w_l^{ab} is 0 for a ≠ 0, so it is not a Fourier transform. The Section 5.2 gate on page 10 is |a> ↦ (1/√p) Σ_b w^{rab}|b>, and the next display on page 17 has kets |b1,..bm>, so |b> is evidently intended.

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• p. 14 Appendix C, proof of claim 1

  can effect only l(kl + l) number of r−blocks in each (r + 1)−block, at the end of the (r + 1)−stage- so if l(kl + l) < d

• p. 13 Lemma 2 (Appendix C)

  2kl ≤ d

• p. 14 Appendix C, proof of claim 1 for r+1

  so if l(kl + l) < d, we have that the final density matrix

Only the authors can say what is right here.

Judge's note: Read on the p14 and p13 images. The induction step for claim 1 is conditional on l(kl+l) < d, which is not a hypothesis of Lemma 2. Counterexample within the lemma's assumptions: l=1, k=1, d=2 gives 2kl = 2 ≤ 2 but l(kl+l) = 2, which is not < 2; l=2, k=1, d=4 gives 2kl = 4 ≤ 4 but l(kl+l) = 8. Theorem 2 uses k = d/2l, i.e. 2kl = d, where the condition becomes l(d/2 + l) < d; this fails for every l ≥ 2 and for l = 1, d = 2. The preceding sentences count k bad rectangles and kl bad blocks, which would suggest l(k+kl) rather than l(kl+l); that form also fails at l=1, k=1, d=2. So the lemma as stated is not covered by its printed proof. Whether the lemma is true under 2kl ≤ d with a tighter count cannot be settled from the files.

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• p. 17 Appendix E (3), displayed proof

  Π_{i=1}^m w^{c_l f(α_l)}|f(α1), ..., f(αm) >

• p. 17 Appendix E (3), displayed proof, second line

  w^a|f(α1), ..., f(αm) >

• p. 17 Appendix E (3), statement

  |Sa >7−→ w^{ca}|Sa >

• p. 17 Appendix E, item (3)

  is done by on the l′th qupit the gate |a >7−→wcla|a >

Only the authors can say what is right here.

Judge's note: Read on the page 17 image. Reworked: with the interpolation coefficients c_l defined in item (2) by f(0) = Σ c_l f(α_l), the product is Π_l w^{c_l f(α_l)} = w^{Σ c_l f(α_l)} = w^{f(0)} = w^a. This matches the last printed line, but equals the stated target w^{ca} only when c = 1. For general c the per-qupit gate would need exponent c·c_l·a. So the procedure formula is wrong as printed for c ≠ 1, though it is evidently repairable. The product is indexed by i while the factor uses l, and the normalisation disappears after the arrow.

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• p. 17 Appendix E, procedure (4), claimed encoded Fourier transform

  We claim that this indeed gives the desired operation.

• p. 18 Appendix E, Case 2

  We claim that the sum vanishes for the b’s in this case.

• p. 18 Appendix E, (4), CASE 2

  We claim that the sum vanishes for the b′s in this case

• p. 18 Appendix E, (4), CASE 2

  Picking f(x) = a + cx2d−r+1

Only the authors can say what is right here.

Judge's note: I viewed pp.17-18. Write f=a+g with g∈V2. The g-dependent part of the exponent is u·(g(α_l)) with u_l=e_l α_l^{2d} b_l. The sum over V2 is nonzero exactly when u⊥C2. Because e_l and α_l are nonzero (as the paper asserts), b↦u is a bijection, so the sum survives for |C2⊥|=p^{m−d}=p^{3d+1} vectors b. Only p^{d+1} of these have deg b ≤ d. For d≥1 there are surviving terms with deg b > d, and the output leaves the code (C1 has dimension d+1). For r=2d+1 the printed choice gives f=a+c, so f(0)≠a, and for larger r the exponent is negative; the proof only covers d<r≤2d. Theorem 4's Fourier procedure is left without a valid proof.

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• p. 8 Proof of Theorem 1

  is smaller than the probability for not more than d/2l faults in two rectangles of M1

• p. 3 Section 1

  The effective noise rate of M1 is thus the probability for more than k errors in a procedure

• p. 8 Proof of Theorem 1

  probability for not more than d/2l faults in two rectangles of M1

• p. 8 Proof of Theorem 1

  smaller than the probability for not more than d/2l faults in two rectangles of M1

Only the authors can say what is right here.

Judge's note: Read on the p8 page image. The first sentence of the proof says the simulation succeeds when the fault path is d/2l-sparse, so failure needs more than d/2l faults in a rectangle. The next sentence ("The probability for a rectangle to have more than k faults ... choose k+1 points") and the theorem's binomial (a choose d/2l+1) both match "more than". The introduction on p3 also defines the effective noise rate as "the probability for more than k errors in a procedure". As printed, the bounding event is the success event, whose probability is close to 1. The stated theorem is unaffected, hence S3.

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• p. 13 Proof of Lemma 3, top of page

  since σ1 reduced to A equals ρ1

• p. 6 Definition 1

  reduced on all qubits except those in A they are equal: ρ1|B−A = ρ2|B−A

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Judge's note: Reworked from Definition 1 on p6: d-deviated means there is a d-sparse A with ρ1|B−A = ρ2|B−A, so σ1 and ρ1 agree on the complement of A. A has at most d < m/2 qubits per block and so cannot determine a majority; B−A has more than m/2 per block and can. The proof never redefines A, so the step as printed is false under the paper's only definition, and it is correct with A replaced by B−A. Lemma 3 itself stands.

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• p. 10 Theorem 3, universal gates and bitwise implementation

  |a >↦ (i)ᵃ|a >

• p. 16 Appendix D, procedure (4)

  is achieved on the encoded state by applying bitwise the gates on each of the qubits in the block.

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Judge's note: I viewed pp.10 and 16. With C2=span(1111000,1100110,1010101), the C2 weights are 0 and 4, so the phase is i^0=i^4=1. The coset 1111111+C2 has weights 3 and 7, so the phase is i^3=i^7=−i. The bitwise gate therefore implements the inverse logical phase, which differs from the claimed gate by a logical Z. The procedure is repairable (apply the inverse gate bitwise), so the existence result stands. For general linear codes, bitwise phase need not be a logical gate at all.

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• p. 10 Section 5.2, evaluation-point condition

  has a non-zero coefficient of x²ᵈ.

• p. 17 Appendix E, Fourier-transform construction

  there are fixed non zero e₁, ..., eₘ

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Judge's note: I viewed pp.10 and 17. e_i is the x^2 coefficient of the Lagrange basis polynomial, which is proportional to the elementary symmetric sum e2 of the other four points. For i=3, {1,2,4,5} gives 49≡0 mod 7. For i=5, {1,2,3,4} gives 35≡0 mod 7. For i=1, {2,3,4,5} gives 71≡1. The x^2 coefficient of G is −e3(1..5)=−225≡6 mod 7, which is nonzero, and p=7>m+1=6. All stated conditions hold, yet e3=e5=0. Then w_l=1 for those l and the coordinate 'Fourier transform' is not unitary. A stronger condition on the αi is needed.

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• p. 13 Lemma 4, statement

  Then, A ∩B is (r, l1 + l2) sparse.

• p. 13 Lemma 4, proof

  so the union is (r + 1, l1 + l2) sparse

• p. 13 Proof of Lemma 4

  the union of A and B is (r, l1 +l2) sparse

• p. 13 Proof of Lemma 4

  the union of A and B is (r, l1 +l2) sparse by induction

Only the authors can say what is right here.

Judge's note: The p13 image clearly shows ∩ in the statement. Reworking the use: if ρ1 and ρ2 agree outside A, and ρ2 and ρ3 agree outside B, then ρ1 and ρ3 agree outside A∪B, so the triangle inequality for the deviation needs sparsity of the union. The proof's induction does establish the union statement. The printed intersection statement is trivially true (a subset of a sparse set is sparse) but is neither what is proved nor what is needed. The intended result stands, hence S3.

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• p. 14 Appendix C, proof of claim 1 for r+1

  these r−rectangles and r−blocks can effect only l(kl +l) number of r−blocks

Only the authors can say what is right here.

Judge's note: Page 14 image reads l(kl+l). The preceding sentence gives k bad r-rectangles and kl bad r-blocks. With spread l the count is l·k + l·kl = l(k+kl), or kl + kl = 2kl if the already-deviated blocks do not spread; the latter matches the lemma's 2kl ≤ d on page 13. Neither equals the printed kl² + l² unless k = l. Check with k=2, l=1: printed gives 3, l(k+kl) gives 4, 2kl gives 4. Also, for k=l=1, d=2 the lemma's hypothesis 2kl ≤ d holds but the printed l(kl+l) < d fails. The lemma itself is probably unaffected; the authors should confirm the intended expression.

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• p. 15 Appendix D, initialization syndrome extraction

  apply controlled not from the block we are initializing to each of the “cat states” bitwise

• p. 15 Appendix D, copying the cat registers

  to imitate a measurement on them.

• p. 16 Appendix D, initialization conclusion

  The corrections will then take this state to an equal distribution over all the basic states in Cperp.

Only the authors can say what is right here.

Judge's note: I viewed pp.15-16. After the CNOTs, the register for row g holds (|x_S>+|x_S⊕1^S>)/√2, and copying it reveals x_S up to complement. That amounts to measuring weight-2 Z_iZ_j operators, which are not stabilizers of |S0>. Even a correct |S0> input is disturbed. The procedure has no Hadamard rotation of the cat state (as in Shor's method), and p.16 itself describes the result as "an equal distribution over all the basic states in Cperp". Correction to the reviewer: x and its complement give identical cat registers, so the circuit does not distinguish all 128 inputs as claimed. The core defect, that more than the syndrome is recorded, still holds. The step is repairable with the standard rotated cat, so the stated result stands.

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• p. 17 Appendix E, procedure (2), interpolation coefficients

  the cₗ are the interpolation coefficients

• p. 17 Appendix E, procedure (3)

  The proof that this achieves the desired operation is

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Judge's note: I viewed p.17. Procedure (2) defines c_l by f(0)=Σ c_l f(α_l). Hence Π w^{c_l f(α_l)} = w^{f(0)} = w^a, exactly as the last displayed line says, rather than w^{ca}. As printed, the procedure implements only c=1. Multiplying each coordinate exponent by c fixes it.

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• p. 18 Appendix E, CASE 2

  The power of w in the sum is the coefficient of x^d in this polynomial.

• p. 18 Appendix E, CASE 2, later

  Looking at the coefficient of x^{2d} of h(x) we have aB0 − cBr g2d

• p. 17 Appendix E (4)

  f_{2d} = Σ e_i f(α_i)

• p. 18 Appendix E, (4), CASE 2

  The power of w in the sum is the coefficient of xd in this polynomial

Judge's note: Page 18 image shows exponent d. Page 17 defines the e_i by f_{2d} = Σ e_i f(α_i), which extracts the x^{2d} coefficient. CASE 1 says 'coeff of x^{2d} in h(x)', and CASE 2 two sentences later says 'Looking at the coefficient of x^{2d} of h(x)'. The x^d is a slip; the argument itself uses x^{2d}.

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