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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.

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.