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One conserved bit, two equal blocks

Write the parity operator $P=\prod_i Z_i$. A computational basis state $\lvert x\rangle$ is its eigenstate with value $(-1)^{\operatorname{popcount}(x)}$ — plus one if it has an even number of ones, minus one if odd. Because the XY term swaps $\lvert 01\rangle\leftrightarrow\lvert 10\rangle$ it leaves that count's parity fixed, so $[H_{XY},P]=0$. The Hilbert space therefore breaks into an even and an odd sector of equal size $2^{n-1}$, and the dynamical Lie algebra generated by the Trotter terms breaks with it.

$$\mathrm{DLA}(H_{XY}) \;=\; \mathfrak{su}(2^{n-1})\;\oplus\;\mathfrak{su}(2^{n-1}),\qquad \dim\mathrm{DLA}=2^{2n-1}-2$$

The two blocks are the two parity sectors; nothing the Hamiltonian does connects them. That is the exact statement the panel constructs — the sector each basis state belongs to, the equal counts, and the algebra's dimension — for any qubit number you pick.

Live — parity sectors of $n$ qubits, and symmetric-noise leakage
even sector (P=+1) odd sector (P=−1) leaked out
even states: odd states: dim DLA: leakage (either sector):

The left grid is every computational basis state, coloured by the sector its excitation-count parity assigns it; the two colours come out in exactly equal number, $2^{n-1}$ each. The right panel starts a state in the even sector and, separately, in the odd sector, then applies an independent bit-flip of probability $p$ to each qubit: the leaked fraction is $\tfrac12\bigl(1-(1-2p)^n\bigr)$, and — because the channel treats every qubit alike — it is identical for the two sectors. A symmetric model gives no even–odd preference.

On real hardware the two sectors do not leak equally. The Phase 1 campaign measured the odd sector of the XY dynamical Lie algebra to be the more decoherence-resistant of the two — an asymmetry a symmetric channel like the one above cannot produce. That departure from the symmetric baseline is the finding; its measured size, statistics and raw counts live on the Phase 1 results page, not in this panel.
Deeper: leakage as the observable, and why the sectors are equal
The library defines parity leakage as twice the out-of-sector weight of a state — the probability that has escaped the sector the ideal, parity-conserving Hamiltonian should have kept it in. On a device that leakage is a direct fingerprint of decoherence. The sectors are exactly equal because the number of $n$-bit strings with an even population equals the number with an odd one, both $2^{n-1}$; the leakage formula is the chance that an even number of independent flips occurred, $\tfrac12(1+(1-2p)^n)$ to stay. Both facts are exact and were checked against direct enumeration; this panel computes them, not a hardware result.

Encoding the logical bit into a protected sector

A conserved parity detects errors for free, but to protect a logical qubit you go one step further: encode it so every codeword lives entirely inside one parity sector. A repetition code of odd distance $d$ does exactly that — each logical bit becomes $d$ physical copies, and because $d$ is odd the block's $Z$‑parity equals the logical bit. The whole logical word therefore sits in a definite physical‑parity sector, and any single physical bit‑flip flips that parity, so the check flags it with certainty. The panel encodes one representative block, lets you inject a flip, and reads out the surface‑code resources the $N$‑oscillator register would cost.

$$P_{\text{block}}=(-1)^{\,d\,b}\;\overset{d\ \text{odd}}{=}\;(-1)^{b},\qquad q_{\text{surf}}=N\,(2d^{2}-1),\qquad p_L=A\left(\tfrac{p}{p_{\text{th}}}\right)^{(d+1)/2}$$

Here $b\in\{0,1\}$ is the logical bit, $q_{\text{surf}}$ the flat rotated‑surface‑code qubit count for $N$ logical oscillators at distance $d$, and $p_L$ the sub‑threshold logical error rate with threshold $p_{\text{th}}=0.01$ and prefactor $A=0.1$. The step‑fidelity figure is a deliberately conservative planning bound, $\exp(-N d\,p_L)$: one logical‑error opportunity per QEC round for every oscillator.

Live — one encoded block, and the surface-code resource roadmap
flat surface-code qubits
repetition scaffold qubits
QEC rounds / Trotter step
wall-clock µs / step
logical error rate / round
step-fidelity bound

Rows are integer or closed-form — recomputed here, verified bit-exact against the library's resource model.

block parity: encodes logical: parity check:
Resource estimate only. The library sets parity_survival_claim_allowed = false: it does not claim the XY DLA-parity signal survives a surface-code encoding, and it makes no hardware submission. Whether the physical parity structure is preserved under the logical code is an open theory question the roadmap keeps blocked until the representation-theory and simulation prerequisites close. What this panel shows is the cost and the detection guarantee for a single physical flip — not that the encoding protects the DLA-parity phenomenon.
Deeper: why odd distance, and what the fidelity bound does and does not say
A block of $d$ identical copies of a logical bit $b$ has $Z$-parity $(-1)^{db}$; only when $d$ is odd does this equal $(-1)^{b}$, so the physical parity tracks the logical bit and a lone bit-flip is always odd-weight and therefore visible to the parity check. The step-fidelity number $\exp(-N d\,p_L)$ charges every one of the $N$ oscillators one independent logical-error opportunity in each of the $d$ syndrome rounds — a planning upper bound on the failure exposure, not a simulated fidelity. The flat surface-code column, $N(2d^{2}-1)$, is the honest qubit budget; the repetition scaffold column, $N(2d-1)$, is shown only as the cheaper detection-code comparison, not as a fault-tolerant substitute.

Why the split is worth exploiting

A conserved parity is a free error check: any weight that crosses between sectors could only have come from noise, so the leakage is a built-in decoherence probe that needs no extra measurement. And if one sector really is hardier than the other, encoding the information there is a hardware-level advantage for nothing. The platform's DLA-parity work is the apparatus for finding and quantifying exactly that.

QuantityValueMeaning
Parity operator$P=\prod_i Z_i$excitation count mod 2
Sector size$2^{n-1}$ eacheven and odd, equal
DLA$\mathfrak{su}(2^{n-1})\oplus\mathfrak{su}(2^{n-1})$two independent blocks
DLA dimension$2^{2n-1}-2$generators of the dynamics
Parity leakageout-of-sector weightdecoherence fingerprint

Evidence boundary: this panel computes exact, checkable facts — the parity sectors, their equal sizes, the DLA dimension $2^{2n-1}-2$, and the symmetric-channel leakage — live in your browser, verified against direct enumeration. It deliberately does not compute the measured even–odd asymmetry: that is a hardware observation, with its data on the results page. The symmetric model here is the null baseline the observation departs from.