Quantum Supremacy Using a Programmable Superconducting Processor

• p. 46 Table VII, column FXEB (%), 14-cycle first row

  2.22 × 10−6

• p. 46 Table VII caption

  fidelity 2.22 × 10−6

Judge's note: Checked Table VII and its caption on the p.46 image. The caption says 288101 paths are needed for F ≈ 0.5% and that 128 paths were computed. Scaling linearly in F, as the caption describes, gives 0.005×128/288101 = 2.22e-6 as a fraction, i.e. 2.22e-4 %. The column header says (%), so this entry is in a different unit from the rest of the column. In the caption, "fidelity 2.22 × 10−6" is consistent as a fraction.

Second check: another model agreed

• p. 8 Sec. IV B, after Eq. (29)

  which is always larger than for the XEB

Judge's note: Page 8 gives linear XEB sd sqrt((1+2F−F^2)/Ns), log XEB sd sqrt((π^2/6−F^2)/Ns) and HOG sd sqrt((log^-2 2 − F^2)/Ns). 1/ln^2 2 ≈ 2.081 > π^2/6 ≈ 1.645, so HOG exceeds log XEB for all F. HOG exceeds linear XEB only while 1+2F < 2.081, i.e. F < 0.54. Elsewhere the paper uses plain 'XEB' for the log version (p.6: 'The standard XEB uses the natural logarithm'; p.8: 'Eq. (23) for XEB, f = log ps'). Under that reading the sentence is true. Since Fig. S9 compares HOG with linear XEB, the wording is ambiguous; I am not treating it as an error.

Second check: another model agreed

• p. 27 Sec. VII B

  We find that our RQCs satisfy this assumption when the circuit depth is larger than 12, see Fig. S35a.

• p. 38 Fig. S35 caption

  Distribution of bitstring probabilities from a 53-qubit 20-cycle circuit.

• p. 27 Sec. VII B, last paragraph

  when the circuit depth is larger than 12, see Fig. S35a

• p. 38 Fig. S35 caption

  Distribution of bitstring probabilities from a 53-qubit 20-cycle circuit

Only the authors can say what is right here.

Judge's note: Fig. S35 on p.38 has histograms of Dp and log(Dp) for one 53-qubit, 20-cycle circuit only. It shows Porter-Thomas-like behaviour at m=20 but cannot support a threshold at depth 12. Fig. S7 (p.8) is the only depth scan (depth 9–14, XEB vs 1/sqrt(D)). The cited figure does not back the specific claim.

Second check: another model agreed

• p. 40 Sec. VIII H

  The value is found to be 4.4% and is used in subsequent analysis.

• p. 40 Sec. VIII H

  The estimate is total σp0 = 1.92×10−4

• p. 40 Sec. VIII H, right column

  The value of σF as well as the ratio σF /F in the range of measured fidelities monotonically decreases. We take max(σF /F) as the estimate of relative systematic uncer- tainty

• p. 40 Sec. VIII H

  The value is found to be 4.4%

Only the authors can say what is right here.

Judge's note: Page 40 states that σF/F decreases monotonically over the measured range, so its maximum is at t=0, where it equals 1.92e-4/5.51e-3 ≈ 3.5%. At t=17.4 h, Eq. (80) gives σF ≈ 1.30e-4 and F ≈ 4.31e-3, a ratio of about 3.0%. The text does not show how 4.4% was obtained (1.92e-4/4.37e-3 would give it). The value carries into the ±0.10 systematic term on p.41. Only the authors can say which inputs were used.

Second check: another model agreed

• p. 41 Fig. S38 panels c and d annotations

  3 years...

• p. 41 Fig. S38 panel d

  150 years...

• p. 57 Table XI

  4 years

Only the authors can say what is right here.

Judge's note: Both annotations appear on the p.41 figure image; Table XI is on p.57. Scaling Table XI's m=20 entry (4^31·2^4 paths, 0.2%) by its own path counts and fidelities gives about 156 years at m=18 and about 3.7 years at m=16. The m=12 (about 2.3 h) and m=14 (about 15.7 days) entries reproduce. Both sources are labelled approximate, and the Ns assumed by Table XI is not stated, so this is minor and for the authors to reconcile.

Second check: another model agreed

• p. 56 Sec. X F, right column

  lated 1.6 × 10^14 core hours for a simulation of the circuit with 0.002 fidelity

• p. 57 Table XI, m=20 row

  10000 years

• p. 41 Sec. VIII I

  requires 5000 years with a classical computer with CPU power equivalent to 1 million cores

• p. 56 Sec. X F, right column

  resulting in a calcu- lated 1.6 × 1014 core hours for a simulation of the circuit with 0.002 fidelity

Only the authors can say what is right here.

Judge's note: Checked p.56 (text, ref [87] "We assume 2 vCPUs per core"), Table XI on p.57 and p.41. The 1.6e14 core-hours reproduces from 4^30×0.002×246 s, and divided over 10^6 cores it gives about 1.8e4 years, not 1e4. Table XI is labelled "approximate". The gap would close if Table XI counts one job per vCPU, but the paper does not say this. The authors should state how the core counts are defined.

Second check: another model agreed

• p. 58 Fig. S51 caption

  XEB on patch and elided supremacy circuits was done at m = 14, 16, 18, and 20.

• p. 41 Fig. S38 panel a / panel f

  m = 12, Ns=0.5M, 100sec per instance

• p. 44 Fig. S41 x-axis

  12

• p. 41 Fig. S38 panels

  m = 12, N_s=0.5M, 100sec per instance

Judge's note: Fig. S38 on p.41 has a panel for m = 12 elided circuits, and panel f plots m = 12–20. Fig. S41 on p.44 plots m = 12–20. The S51 caption on p.58 lists m = 12 among the stars for full supremacy circuits but leaves it out of the patch/elided depths.

Second check: another model agreed

• p. 62 Sec. XI C, right column

  cannot be sampled in time F2(1−o(1)n), unless SETH is false

• p. 62 Sec. XI C, paragraph after Theorem 1

  cannot be sampled in time F2(1−o(1)n), unless SETH is false.

• p. 62 Theorem 1 statement

  there is an AM[LT + 2Lm] protocol

Only the authors can say what is right here.

Judge's note: Page image p.62 checked: the exponent is printed (1−o(1)n), while the line before has (1−o(1))n. That part is a plain typo. On the inference: Theorem 1 turns a time-T sampler into an Arthur–Merlin protocol, not a classical algorithm. Footnote [106] states the assumption as a classical time bound ('takes time no less than 2^{(1−2/k)n} to solve k-SAT'). Getting from the AM protocol to a contradiction with SETH needs the extra assumption that Merlin cannot help. The text only argues this informally ('It is unlikely that Merlin can be of any help'), and that argument is about #P vs AM, not about SETH. The paper does not state or cite an AM/nondeterministic form of the hypothesis. Theorem 1 itself is not affected, so this is an authors' query about the wording of the conclusion, together with the typo.

Second check: another model agreed

• p. 62 Theorem 1 statement, Eqs. (115)–(116)

  for a given quantum circuit U on n qubits and F ≥0.

• p. 63 Eqs. (123)–(126)

  Setting θ = Fλ+(1−F)/2n

Only the authors can say what is right here.

Judge's note: Pages 62–63 checked. Eqs. (115)/(116) and (123)/(125) divide by F. At F = 0, r_{U,0}(x) = 2^{-n} carries no information about U. The boundary case F = 0 is therefore not covered by the proof; the hypothesis should be F > 0. For F > 0 the result is unaffected.

Second check: another model agreed

• p. 63 Lemma 2

  let Hq,t be a family of pairwise-independent linear hash functions mapping {0, 1}^q to {0, 1}^t

• p. 63 Lemma 2

  Let S ⊆{0, 1}^n be arbitrary

• p. 63 Lemma 2

  over the choice of h ∈Hn,t

• p. 63 Lemma 2

  Let S ⊆{0, 1}n be arbitrary with |S| ≥δ−32t. Then with probability larger than 9/10 over the choice of h ∈Hn,t,

Judge's note: Page image p.63 (Lemma 2 statement) checked. n is the qubit count throughout Sec. XI. The application in Lemma 1 uses q = Lm, and the cost 'less than 2Lm' matches 2q. This is a notational slip with an evident fix.

Second check: another model agreed

• p. 6 Fig. S6 caption

  Different colored pots correspond

• p. 6 FIG. S6 caption

  Different colored pots correspond to different values of the swap error

• p. 6

  Different colored pots correspond to different values

Judge's note: The page 6 image shows 'Different col-ored pots correspond' in the Fig. S6 caption. It is a plain misspelling.

Second check: none (judge only)

• p. 7 Sec. IV A, right column

  We normally use mutiple circuits

Judge's note: The crop of page 7 (right column) shows 'mutiple' as printed.

Second check: none (judge only)

• p. 9 Sec. IV D

  we can prepare a bistring

• p. 42 Sec. IX

  is a function of bistrings B

• p. 50 after Eq. (87)

  by generating bistrings distributed uniformly

• p. 54 Sec. X E

  each possible bistring of 0 and 1 values

Judge's note: I saw the misspelling on all four page images: crops of p.9, p.42 and p.50, and the full p.54 (right column, Sec. X E).

Second check: none (judge only)

• p. 12 Sec. VI A 2

  can be see in Figure S10

Judge's note: The crop of page 12 shows 'can be see in Figure S10'.

Second check: none (judge only)

• p. 15 Sec. VI C 1

  A typical control sequence is shown Fig. S14a.

Judge's note: The crop of page 15 (right column) shows the sentence without 'in'.

Second check: none (judge only)

• p. 42 Sec. VIII I, last paragraph

  A false positive is only produced by a additive-type statistical fluctuations

• p. 43 Fig. S40 caption

  seen in c,.

• p. 42 Section VIII I

  produced by a additive-type

Judge's note: The crop of page 42 shows 'by a additive-type statistical fluctuations'. The crop of page 43 shows the caption ending 'fidelity seen in c,.'. Both are cosmetic slips.

Second check: none (judge only)

• p. 49 Sec. X C

  denote by U the set of m states generated by executing the circuit U

• p. 49 Sec. X C

  A the set of M bitstrings generated by JUQCS-A

• p. 49 Sec. X C

  We denote by U the set of m states generated by executing the circuit U.

• p. 49 Sec. X C

  We denote by A the set of M bitstrings generated by JUQCS-A

Judge's note: The p.49 image shows italic lowercase m. The definition of A on the same page says "the set of M bitstrings". Eq. (87) on p.50 normalises by 1/M for X = U, and the text says M = 500000. In the rest of the paper, m is the number of cycles.

Second check: none (judge only)

• p. 50 after Eq. (87)

  by generating bistrings distributed uniformly

Judge's note: Seen on the p.50 page image.

Second check: none (judge only)

• p. 51 Table IX, Peak performance row

  10.4 PFlops/s

Judge's note: Seen in Table IX on the p.51 image: 1.8 PFlop/s, 5 PFlop/s, 10.4 PFlops/s. Cosmetic inconsistency in the unit.

Second check: none (judge only)

• p. 52 Sec. X D 1, after Eq. (100)

  tr(S†_{a,i}S_{b,j}) = δij

Judge's note: The p.52 image shows S†_{a,i}S_{b,j}. The S operators act on qubit b (Eq. 100), and the matching R condition uses a on both factors.

Second check: none (judge only)

• p. 54 Sec. X E, right column

  the treewdith can be used

• p. 54 Sec. X E, first paragraph, right column

  the treewdith can be used to study the complexity of Feynman methods

• p. 54 Section X E, first paragraph

  the treewdith can be used to study

Judge's note: Seen in the crop of the p.54 image; the same paragraph spells "treewidth" correctly elsewhere.

Second check: none (judge only)

• p. 55 FIG. S48 caption

  the resulting contraction ordering it is ignored

Judge's note: Seen in the Fig. S48 caption on the p.55 image.

Second check: none (judge only)

• p. 63 Lemma 1 protocol step 3

  He rejects if any of the three equations is not satisfied.

• p. 63

  He rejects if any of the three equations is not sat-

Judge's note: Page image p.63 checked. Only two equations appear in step 3. The θ inequality is introduced separately afterwards ('Then he checks if θ ≤ ...').

Second check: none (judge only)

• p. 63 after Lemma 1 protocol

  cost to compute is hs(w) is less than 2Lm

Judge's note: The page image confirms the duplicated 'is'.

Second check: none (judge only)

• p. 64 Author list

  Matthew McEwen5,Anthony Megrant1, Xiao Mi1,Kristel Michielsen

Judge's note: Crop of p.64 checked. The page image shows 'McEwen^5,Anthony' and 'Mi^1,Kristel' with no space, while the other names are spaced normally.

Second check: none (judge only)

• p. 64 Ref. [8]

  An introduction to Cartans KAK decomposition for QC programmers

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: the cited preprint's actual title is 'An introduction to Cartan's KAK decomposition for QC programmers'. The page image confirms that the printed text reads 'Cartans', so this is not an extraction loss.

Second check: none (judge only)