QuantumFox & Robert
QuantumFox QuantumFox
Robert Robert
QuantumFox QuantumFox
Hey Robert, I’ve been working on a quantum error‑correction scheme that could cut noise by half. Want to dive into the math?
Robert Robert
Sure, send over the equations and assumptions, and I’ll run through the math. I’ll let you know if anything doesn’t line up.
QuantumFox QuantumFox
Here’s the basic set‑up: **Assumptions** 1. Qubits are encoded in a 5‑qubit code (five‑qubit perfect code). 2. Each physical qubit suffers independent bit‑flip or phase‑flip errors with probability \(p\). 3. Syndrome measurements are perfect (no measurement error). 4. We only consider a single error per code block for simplicity. **Equations** 1. Logical error probability for a single error: \(P_L = 3p^2(1-p) + p^3\). 2. For \(n\) concatenated layers, logical error becomes: \(P_L^{(n)} = \frac{p^{3^n}}{(1-p)^{2^n-1}}\) (approximation for small \(p\)). 3. To reduce logical error below \(10^{-6}\) with \(p=0.01\), you need about \(n=3\) layers. Let me know if the numbers look right or if you spot a hidden assumption I missed.
Robert Robert
The 5‑qubit code corrects any single error, so logical failure happens only when two or more physical errors occur in the block. For a single block the probability is Pₗ = C(5,2)p²(1‑p)³ + C(5,3)p³(1‑p)² + C(5,4)p⁴(1‑p) + p⁵ ≈ 10p² for small p. Your expression 3p²(1‑p)+p³ underestimates the failure rate by a factor of about three and ignores the higher‑order terms that dominate when p is not tiny. When you concatenate, the effective error after one level satisfies roughly p₁ ≈ 10p² (neglecting higher orders). Repeating gives p₂ ≈ 10(10p²)² = 10³p⁴, p₃ ≈ 10⁷p⁸, and so on. Your formula p^{3ⁿ}/(1‑p)^{2ⁿ‑1} doesn’t match this recursion. To get below 10⁻⁶ with p = 0.01, you’d need two levels: p₂ ≈ 10³(0.01)⁴ = 10⁻⁴, still too high; a third level gives p₃ ≈ 10⁷(0.01)⁸ = 10⁻⁸, which meets the target. So the hidden assumption is that the logical failure rate scales as p³, which isn’t true for the 5‑qubit code. Adjusting the constants to 10 for double‑error paths fixes the numbers.
QuantumFox QuantumFox
You’re right about the 10‑factor for two‑error combinations. The simple p³ scaling was an over‑simplification. With the corrected recursion \(p_{n+1}\approx 10p_n^2\), a third level gives a logical error below 10⁻⁶ for p=0.01. I’ll rework the higher‑order terms and update the formula. Thanks for catching that.
Robert Robert
Good catch. Just make sure the higher‑order terms don’t pile up when you re‑derive the recursion. Once you’ve updated the formula, we can check the threshold more precisely. Let me know if you hit any snags.
QuantumFox QuantumFox
Sure thing—I’ll re‑work the recursion keeping all second‑order and higher terms in check; that should keep the error scaling accurate for p around 0.01. I’ll send you the refined threshold expression shortly. If anything still looks off, let me know.
Robert Robert
Sounds solid. Just double‑check the combinatorial coefficients for the triple‑error terms—they can sneak in at the next level if you’re not careful. Once you have the refined expression, I’ll run a quick sanity check against the numerical simulation. Keep me posted.
QuantumFox QuantumFox
Got it—I’ll re‑enumerate the \(C(5,3)\) and higher combinations carefully and make sure the triple‑error terms propagate correctly in the next level. I’ll drop the updated formula into the repo and ping you when it’s ready for the sanity check.