Erasing a bit costs at least kBT ln 2

A qubit meets a reservoir, the open chain of d sites in its thermal state at temperature T, and the pair evolves by a unitary U. Proved in Lean: the heat given to the reservoir is at least k_B T times the entropy the qubit loses, Q ≥ k_B T ΔS. Every point you add is one unitary; none can enter the orange region.

sites d
0.50
0.50

Heat Q against entropy lost ΔS (tap a point)

unitaries Q = k_B T ΔS, Lean forbidden ΔS = S(ρ), all the qubit holds

Populations before (grey) and after (blue)

entropy lost ΔS
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heat Q
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floor k_B T ΔS
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margin Q − k_B T ΔS
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Units with k_B = 1. The reservoir Hamiltonian is the open chain, hopping between neighbouring sites, levels −2 cos(kπ/(d+1)). The unitaries on this page permute the joint energy levels of qubit and reservoir: they are unitary, every state stays diagonal, and all numbers are exact sums. Best erasure puts the largest joint populations on the qubit's first level. At p(0) = 0.5 the qubit holds one bit, ΔS ≤ ln 2; a full erasure needs a cold reservoir, and even then pays at least k_B T ln 2. Lean: MState.landauer, landauer_temperature, landauer_bit, from Sᵥₙ_prod, Sᵥₙ_uConj and Sᵥₙ_le_gibbs (Klein) — source.