Smallest Bitcoin Miner Possible \ stacker news ~science

1696 sats \ 3 comments \ @MaaliMKen 27 Dec 2023 science

Who can help me model, using mathematics and physics, the smallest bitcoin miner theoretically possible as per current knowledge of quantum mechanics? Hint: This article I wrote - https://hackernoon.com/the-smallest-bitcoin-miner-possible I pledge10k sats as a thank you.

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700 sats \ 2 replies \ @elysia 27 Dec 2023

Here is some back of the napkin math. I'll keep working on this to refine it.

Quantum Miner Model

Aspect	Description
Quantum Computer Model	Gate-based quantum computer using superconducting qubits
Qubit Efficiency	Each qubit potentially replaces a classical logic gate in SHA-256
Error Correction	Surface code error correction; assuming 10 physical qubits per logical qubit
Energy Approach	Hamiltonian for energy budget; Landauer's limit for energy per bit change
Quantum Computational Efficiency	Quantum computers can simulate classical computational systems, including Bitcoin mining
Physical Constraints	Qubits require near-absolute zero temperatures, necessitating a cryogenic setup

Calculations

Item	Value
Logical Qubits for SHA-256	256 qubits (theoretical)
Error Correction Overhead	10 physical qubits per logical qubit
Total Physical Qubits	2560 qubits (256 logical x 10)
Space per Qubit	10 mm² per qubit (including control mechanisms)
Quantum Processing Unit Size	25600 mm² (0.0256 m²)
Total Bits for Mining Process	351,522 bits
Energy Consumption per Bit Change	(2.9 \times 10^{-21}) Joules (Landauer's limit)
Total Energy for Mining Process	Calculation Required

Calculation of Total Energy and Final Size Estimation

Description	Calculation	Result
Total Energy Requirement	(351,522 \text{ bits} \times 2.9 \times 10^{-21} \text{ Joules/bit})	(1.019 \times 10^{-15}) Joules
Quantum Processing Unit	0.0256 m²	-
Cryogenics and Electronics	Assumed 10x the quantum processing unit	0.256 m²
Total Size	Quantum Unit + Cryogenics and Electronics	0.2816 m²

The total energy required for mining a Bitcoin block, based on Landauer's limit and the given computational load, is approximately (1.019 \times 10^{-15}) Joules.

0 sats \ 1 reply \ @MaaliMKen OP 25 Feb

Thanks @elysia. Great answer.

0 sats \ 0 replies \ @MaaliMKen OP 25 Feb

Sent you some sats 👌. If you're up for it, we could continue this journey.