Landauer's Erasure Principle

Derivation

Landauer's Erasure Principle

D-418 Home PU-404 Threads energy · chance Depends on second-law-of-thermodynamics, shannon-entropy-from-multiplicity

Statement

Erasing one bit of information stored in a physical system held in contact with a heat bath at absolute temperature T requires the dissipation of at least Q_min = k_BT ln 2 of heat into that bath, and correspondingly the environment's entropy must rise by at least k_B ln 2. The bound is set entirely by T and fundamental constants; it is independent of the storage mechanism, the switching speed, and the circuit technology.

Why it matters

Landauer's principle is the point where information theory and thermodynamics become one subject. It says that logical irreversibility — a many-to-one map on the state of a memory, of which RESET-to-zero is the archetype — carries an unavoidable thermodynamic cost, while logically reversible operations need not. This is the physical content behind the slogan "information is physical."

It also resolves the Maxwell-demon and Szilard-engine paradoxes: the demon can extract work by exploiting information, but to run in a cycle it must eventually erase its memory, and that erasure returns to the bath exactly the entropy the second law demands. The bound is now experimentally confirmed in colloidal particles, single electrons, and nanomagnets, and it fixes the ultimate energy floor of irreversible computation.

Assumptions

A single bit is a physical system with two distinguishable, equally probable macrostates before erasure.If the two states are not equiprobable, the memory carries less than one bit of Shannon entropy and the erasure cost falls below k_BT ln 2; the "ln 2" is precisely the Shannon entropy of a fair bit.

The memory is in contact with a single heat reservoir at fixed temperature T, and erasure is carried out arbitrarily slowly (quasistatically).Drop quasistaticity and additional dissipation from finite-rate operation adds to the bound, so real erasers pay more than k_BT ln 2, never less; the theorem bounds only the reversible ideal.

Erasure is logically irreversible: the map sends both input states (0 and 1) to one fixed output state (say 0), so the final logical state carries no record of the initial one.If the operation is logically reversible (a permutation of memory states, like NOT or a controlled copy into blank memory), no phase-space compression of the accessible states occurs and the bound is zero.

The relevant phase-space volumes are those the system can actually access on the timescale of the operation (ergodicity within each logical state).If the system is non-ergodic — trapped in a sub-region far smaller than the nominal macrostate — the entropies that enter the Clausius argument are not the coarse-grained "one bit," and the accounting must be redone with the true accessible volumes.

Derivation

S_info = k_B ln Ω

Adopt the Boltzmann–Shannon entropy of the memory's accessible microstates. Before erasure the bit is equally likely 0 or 1, so its state occupies Ω_i microstates spread over two equal logical wells. A

S_i = k_B ln(2Ω₀), S_f = k_B ln(Ω₀)

Let Ω₀ be the number of microstates within one logical well. Initially the system may be in either of two equiprobable wells (total 2Ω₀); after RESET it is confined to the single "0" well (Ω₀). This is the phase-space compression that defines erasure. A

ΔS_sys = S_f − S_i = k_B ln(Ω₀) − k_B ln(2Ω₀) = −k_B ln 2

Subtract; the Ω₀ cancels inside the logarithm, so the result is independent of the internal well structure and depends only on the two-to-one collapse. A

ΔS_total = ΔS_sys + ΔS_env ≥ 0

Apply the second law to the isolated composite of memory plus reservoir. Any real process has non-negative total entropy production; equality holds only in the reversible (quasistatic) limit. B

ΔS_env ≥ −ΔS_sys = +k_B ln 2

Rearrange step 4 and substitute the memory's entropy drop from step 3. The environment must absorb at least the entropy the memory shed. B

ΔS_env = Q / T

The reservoir is large and stays at temperature T; heat Q delivered to it raises its entropy by the Clausius relation for a reservoir. B

Q / T ≥ k_B ln 2 ⟹ Q ≥ k_BT ln 2

Combine steps 5 and 6 and multiply through by T > 0 (which preserves the inequality). Rearranged symbolically before any number is inserted. A

Result

Q_min = k_BT ln 2

Reading. Destroying one bit of information — collapsing two equally likely memory states into one — forces at least k_BT ln 2 of heat out of the memory and into the surrounding bath, and raises the bath's entropy by at least k_B ln 2. The cost is per erased bit, scales linearly with temperature, and is completely independent of how the bit is physically encoded. The minimum work that must be supplied to perform the erasure equals this same k_BT ln 2 in the quasistatic limit.

Units check. [k_B] = J K⁻¹, [T] = K, and ln 2 is dimensionless, so k_BT ln 2 has units J K⁻¹ · K = J, an energy — as a heat must be. The companion entropy bound k_B ln 2 carries J K⁻¹, correct for an entropy.

Limiting cases

T → 0: k_BT ln 2 → 0. Erasure becomes thermodynamically free of heat cost as the bath approaches absolute zero — but the third law forbids reaching T = 0, so the cost never truly vanishes.
N independent bits erased: the entropy changes add, giving Q_min = N k_BT ln 2 — the bound is extensive in the information destroyed.
Erase in base d (a d-state memory reset to one state): ln 2 → ln d, so Q_min = k_BT ln d. The "2" is just the multiplicity being compressed.
Reversible logic (NOT, copy into blank, CNOT): ΔS_sys = 0, so Q_min = 0. Only the many-to-one step is charged.
Biased bit, probability p of "1": the cost drops to k_BT · H(p) with H(p) = −p ln p − (1−p) ln(1−p); it is maximal at p = ½.

Breaks when

Non-equilibrium or multiple reservoirs. With two baths at different temperatures, or a memory driven far from equilibrium, the single-T Clausius relation Q = TΔS_env no longer holds and the clean k_BT ln 2 floor must be replaced by a more general entropy-production bound (e.g. a fluctuation-theorem inequality).
Correlated or already-known memory contents. If the erased bit is correlated with a reference system (a side-record), the conditional entropy can be zero or even negative; with quantum side-information the Landauer cost can be driven below k_BT ln 2, and to zero when the record makes the outcome certain. The bound applies to the unconditioned bit.
Finite-time / finite-precision erasure. Real erasure in finite time τ incurs extra dissipation scaling like 1/τ, and imperfect erasure (success probability < 1) has a strictly lower ideal cost. The equality Q = k_BT ln 2 is only the reversible, perfect-erasure limit.
Ballistic / non-ergodic memories. If the memory cannot explore its nominal macrostate on the operation timescale, the coarse-grained "one bit" overstates the accessible entropy and the naive bound misrepresents the true cost.

Failure modes

Charging computation instead of erasure. Believing every logic gate costs k_BT ln 2. Only logically irreversible (many-to-one) steps are charged; reversible gates are free in principle. This is the basis of reversible computing.
Confusing writing with erasing. Thinking that setting a bit to a known value from a known initial value costs energy. Overwriting a bit whose value you already know (RESTORE) is a one-to-one relabeling and costs nothing; the cost comes from discarding unknown information.
Dropping the ln 2. Quoting the bound as k_BT (forgetting the logarithm of the multiplicity), which overstates the floor by ~44%.
Sign/direction error. Claiming the memory's entropy rises. The memory's entropy falls by k_B ln 2 (it becomes more ordered); the environment's entropy rises to compensate.
Believing it forbids a Maxwell demon. The demon can measure and act reversibly; what it cannot do is reset its memory for free. The bound closes the cycle, it does not block the measurement.
Treating it as a technology limit. Assuming better transistors could beat it. It is a thermodynamic bound on any physical implementation at temperature T, independent of hardware.

Discussion

The deepest lesson is that the two-to-one logical map is a physical compression of phase space. Before erasure the memory's representative point could sit in either of two wells; afterward it must sit in one. Liouville's theorem forbids compressing the accessible phase-space volume of an isolated Hamiltonian system, so the "missing" volume — and the k_B ln 2 of entropy attached to it — cannot vanish. It must be exported to the environment as heat. Landauer's principle is precisely the bookkeeping of where that entropy goes.

Notice what is and is not charged. The bound is levied on information destruction, not on information processing, and not even on the measurement that acquires information. Bennett's resolution of the Szilard engine makes this concrete: the demon extracts k_BT ln 2 of work per cycle by measuring which half of a box a molecule is in, but it stores that one bit, and to return to its initial state it must erase the bit — paying back exactly k_BT ln 2. The ledger balances and the second law is safe. This is why erasure, not observation, is the thermodynamically expensive step.

At the frontier the principle sharpens rather than breaks. With a reference memory holding a correlated copy, the relevant quantity is the conditional entropy H(X|R), and the generalized bound reads Q ≥ k_BT · H(X|R). Quantum mechanically this conditional entropy can be negative (entangled side-information), so erasure can in principle extract work while resetting the bit — a striking inversion that has been demonstrated in NMR and superconducting platforms. None of this violates the second law; it refines what "the entropy of a bit" means when correlations are present. Modern stochastic thermodynamics recovers Landauer as the equality case of an integral fluctuation theorem, ⟨e^{−β(W−ΔF)}⟩ = 1, whose Jensen-inequality corollary ⟨W⟩ ≥ ΔF gives the minimum work k_BT ln 2 for the free-energy change of the double-to-single-well confinement.

Common misconceptions. Landauer's bound is not a statement about circuit heating in today's chips (which dissipate ~10⁴–10⁶ times the Landauer floor from resistive and leakage losses). It is not violated by DNA, brains, or quantum computers. And it does not say computation must dissipate — only that forgetting must. A perfectly reversible computer that never erases could, in principle, run at zero energy cost; the price is paid only when accumulated garbage bits are finally cleared.

Worked examples

Q_min = k_BT ln 2 (one bit, room temperature)

Erase a single bit in a memory at T = 300 K. Use k_B = 1.381 × 10⁻²³ J K⁻¹ and ln 2 = 0.6931. Symbolic form first, then substitute. A

Q_min = (1.381 × 10⁻²³)(300)(0.6931) J

Insert the three factors with units J K⁻¹ · K · (dimensionless). A

Q_min = 2.87 × 10⁻²¹ J ≈ 0.018 eV ≈ 0.72 k_BT

Multiply out; convert with 1 eV = 1.602 × 10⁻¹⁹ J. Note 0.018 eV is well below the ~0.026 eV thermal scale, consistent with a sub-k_BT floor. A

Q_min ≈ 2.9 × 10⁻²¹ J per bit at 300 K

Reading. About three zeptojoules — roughly 18 milli-electron-volts — is the irreducible heat of erasing one bit at room temperature. A machine erasing 10¹⁸ bits per second (an exabit/s) could not dissipate less than ~2.9 mW from erasure alone.

P_min = Ṅ k_BT ln 2 (power floor of an erasing processor)

A processor irreversibly erases at a rate Ṅ = 10¹⁶ bits per second while its die sits at T = 350 K (77 °C). Minimum dissipated power is the per-bit floor times the erasure rate. B

P_min = (10¹⁶ s⁻¹)(1.381 × 10⁻²³ J K⁻¹)(350 K)(0.6931)

Substitute; units are s⁻¹ · J K⁻¹ · K = J s⁻¹ = W. B

P_min = 3.35 × 10⁻⁵ W ≈ 34 μW

Multiply the factors. This is the theoretical erasure floor; the same chip's real dissipation is many watts, so Landauer is ~5–6 orders of magnitude below present practice. B

P_min ≈ 34 μW

Reading. Even at 10¹⁶ erasures per second, the fundamental erasure power is only tens of microwatts. The heat modern chips actually produce is dominated by resistive and leakage losses, not by the Landauer bound — which is why the principle is a distant floor, not today's binding constraint.

Problems

(Basic.) Compute the minimum heat to erase one bit at liquid-nitrogen temperature, T = 77 K.
Solution
Q = k_BT ln 2 = (1.381 × 10⁻²³)(77)(0.6931) = 7.37 × 10⁻²² J ≈ 4.6 × 10⁻³ eV. About one quarter of the room-temperature value, as expected from the linear scaling with T (77/300 ≈ 0.257).
(Extensivity.) A memory chip is cleared by erasing 8 GB = 8 × 8 × 10⁹ = 6.4 × 10¹⁰ bits at 300 K. Find the total minimum heat.
Solution
Q = N k_BT ln 2 = (6.4 × 10¹⁰)(2.87 × 10⁻²¹ J) = 1.84 × 10⁻¹⁰ J ≈ 0.18 nJ. Erasing an entire 8 GB module has a Landauer floor under a nanojoule — utterly negligible next to the ~joules such an operation actually costs.
(Base change.) A ternary memory cell has three equally likely states and is reset to one of them. What is the minimum erasure heat at 300 K, and how does it compare to a bit?
Solution
The multiplicity compressed is 3, so Q = k_BT ln 3 = (1.381 × 10⁻²³)(300)(1.0986) = 4.55 × 10⁻²¹ J. Ratio to a bit: ln 3 / ln 2 = 1.585, i.e. a "trit" costs 1.585 times a bit — exactly log₂3, the number of bits of information in one trit.
(Biased bit.) A bit is known to be "1" with probability p = 0.9 before erasure. Find the minimum erasure heat at 300 K and compare to the fair-bit value.
Solution
Use Q = k_BT H(p), H(p) = −0.9 ln 0.9 − 0.1 ln 0.1 = −0.9(−0.1054) − 0.1(−2.3026) = 0.0948 + 0.2303 = 0.3251 nats. Then Q = (1.381 × 10⁻²³)(300)(0.3251) = 1.35 × 10⁻²¹ J. This is 0.3251/0.6931 = 0.469 of the fair-bit cost 2.87 × 10⁻²¹ J: a more predictable bit is cheaper to erase because it carries less information.
(Szilard engine, synthesis.) A single-molecule Szilard engine at T = 300 K measures which half of its container the molecule occupies, then extracts work by an isothermal expansion. (a) How much work can it extract per cycle? (b) Show the cycle is consistent with the second law once memory erasure is included.
Solution
(a) Isothermal quasistatic expansion of one molecule from volume V/2 to V gives W_out = k_BT ln(V/(V/2)) = k_BT ln 2 = 2.87 × 10⁻²¹ J. (b) The engine has stored one bit (which half). To return to its exact initial state and run a true cycle, it must erase that bit, costing at least Q_erase = k_BT ln 2 = 2.87 × 10⁻²¹ J dumped to the bath. Net work per closed cycle: W_net = W_out − W_erase ≤ k_BT ln 2 − k_BT ln 2 = 0. No net work is extracted from a single bath in a cycle, so the Kelvin–Planck statement of the second law holds. The apparent violation was hidden in the neglected erasure step.