Landauer Principle

Landauer’s principle connects information erasure to thermodynamic cost. In its simplest form, erasing one maximally uncertain classical bit in contact with a heat bath at temperature $T$ requires dumping at least

$$k_B T\ln2$$

of heat into the bath, or equivalently supplying at least that much work in an ideal isothermal reset with no net change in the memory’s internal energy.

The principle is not a statement that “information is heat.” It is a statement about physical processes that map many possible memory states to one standard state. Erasure lowers the entropy of the memory; the missing entropy must be exported to other degrees of freedom if the full dynamics is physical.

Sign Convention

This volume uses $\dot Q$ for heat current into the system. If the memory is the system, then heat dumped to the bath is

$$Q_{\mathrm{bath}} = - Q_{\mathrm{mem}}.$$

For a memory coupled to a single ideal bath at temperature $T$, the entropy-production balance is

$$\Sigma = \Delta S_{\mathrm{mem}} - \frac{Q_{\mathrm{mem}}}{T} \ge 0.$$

Equivalently,

$$Q_{\mathrm{bath}} \ge - T\Delta S_{\mathrm{mem}}.$$

During erasure, $\Delta S_{\mathrm{mem}}$ is negative. Thus the lower bound on heat dumped to the bath is positive.

For a one-bit memory initially maximally mixed and finally in a pure standard state,

$$\Delta S_{\mathrm{mem}} = - k_B\ln2,$$

so

$$Q_{\mathrm{bath}} \ge k_B T\ln2.$$

This is the most familiar Landauer bound.

Work Cost

If the memory Hamiltonian begins and ends in the same form and its internal energy returns to its original value, the first law gives

$$\Delta E_{\mathrm{mem}} = Q_{\mathrm{mem}}+W = 0.$$

Therefore

$$W = - Q_{\mathrm{mem}} = Q_{\mathrm{bath}}.$$

In an ideal reversible erasure process,

$$W_{\min} = k_B T\ln2.$$

More generally, for an isothermal process at temperature $T$, the average work obeys the free-energy inequality

$$W \ge \Delta F = \Delta E-T\Delta S.$$

For a degenerate bit with no energy splitting, $\Delta E=0$. Erasing a maximally mixed bit to a pure state gives

$$\Delta F = k_B T\ln2.$$

The bound can be approached only by a quasistatic, carefully controlled process. Fast, uncontrolled, or noisy erasure dissipates more.

Biased Classical Bit

If a classical bit is $0$ with probability $p$ and $1$ with probability $1-p$, its Shannon entropy in nats is

$$H(p) = - p\ln p - (1-p)\ln(1-p).$$

Resetting it to a definite standard value lowers the memory entropy by

$$\Delta S_{\mathrm{mem}} = - k_B H(p).$$

The heat dumped to the bath must satisfy

$$Q_{\mathrm{bath}} \ge k_B T H(p).$$

The cost is maximal for an unbiased bit, where $H(p)=\ln2$. It vanishes in the ideal limit when the bit value is already certain and the reset operation is matched to that known state.

If entropy is measured in bits,

$$H_{\mathrm{bits}}(p) = - p\log_2 p - (1-p)\log_2(1-p),$$

then the same bound is

$$Q_{\mathrm{bath}} \ge k_B T\ln2\, H_{\mathrm{bits}}(p).$$

Quantum Memory

For a qubit memory with density operator $\rho$, the relevant entropy is the von Neumann entropy

$$S(\rho) = - k_B\operatorname{Tr}(\rho\ln\rho).$$

If the qubit is reset to a pure standard state and the Hamiltonian is effectively degenerate at the beginning and end, then

$$Q_{\mathrm{bath}} \ge T S(\rho).$$

If $\rho$ has eigenvalues $\lambda$ and $1-\lambda$, then

$$S(\rho) = - k_B \left[ \lambda\ln\lambda + (1-\lambda)\ln(1-\lambda) \right].$$

The erasure cost depends on the entropy of the physical state, not on the labels used to describe it. A known pure qubit state can in principle be mapped to a standard pure state reversibly by a unitary control. An unknown ensemble with a mixed density operator has entropy that must be exported if it is reset to one pure state.

Correlations matter. If the memory is correlated with another system, erasing the memory can destroy mutual information. Then the thermodynamic accounting must include the reference system, controller, and record. The Mutual Information page gives the correlation measure used in such statements.

Physical Erasure Cycle

A standard classical picture is a particle in a symmetric double-well potential. The left well represents $0$ and the right well represents $1$.

An ideal erasure cycle can be described schematically:

1. Lower the barrier between wells so the particle can equilibrate.

2. Slowly compress the accessible phase-space volume to the standard side while in contact with a bath.

3. Raise the barrier again to restore a stable memory.

In the quasistatic limit, the work cost approaches $k_B T\ln2$ for an initially unbiased bit. If the compression is fast or uncontrolled, additional entropy is produced and the heat dumped to the bath is larger.

The key point is not the double-well model itself. The key point is many-to-one logical reset implemented by an underlying physical process that must preserve total entropy once the bath and controls are included.

Measurement, Feedback, and Demons

Landauer’s principle often appears in discussions of Maxwell-demon cycles. A measurement can create correlations between a system and a memory. Conditional feedback can then extract work from the system. A complete thermodynamic cycle must also return the memory and controller to their initial states.

If a demon records one bit and later erases that record at temperature $T$, the erasure cost is at least $k_BT\ln2$ when the record is unbiased. This does not mean every measurement immediately dissipates $k_BT\ln2$. The cost is attached to resetting a physical memory, and it depends on the record entropy, correlations, and the chosen thermodynamic cycle.

For measurement terminology, see Selective and Nonselective Measurements and Quantum Instruments.

What the Principle Does Not Say

Landauer’s principle is sometimes overstated. It does not say:

• every logical operation dissipates heat;
• every measurement costs $k_B T\ln2$;
• heat is made of abstract information;
• the bound is automatically reached in real devices;
• erasure cost can be computed without specifying the memory, bath, and control protocol;
• quantum coherence alone violates the principle.

Logically reversible operations can be implemented with arbitrarily small dissipation in ideal limits, though practical devices have other sources of loss. Logically irreversible reset has a thermodynamic cost because it reduces the entropy of the physical memory.

Reversible Computation owns the logical construction and its workspace ledger; this page retains the thermodynamic cost of erasure, reset, and complete physical cycles.

Relation to Entropy Production

Landauer’s principle is a direct application of nonnegative entropy production. With one bath,

$$\Sigma = \Delta S_{\mathrm{mem}} + \Delta S_{\mathrm{bath}} \ge 0.$$

For an ideal bath,

$$\Delta S_{\mathrm{bath}} = \frac{Q_{\mathrm{bath}}}{T}.$$

Therefore

$$Q_{\mathrm{bath}} \ge - T\Delta S_{\mathrm{mem}}.$$

This derivation makes the assumptions visible. If the bath is finite, nonthermal, strongly coupled, or initially correlated with the memory, the simple bath entropy formula may fail and the full entropy balance must be modeled explicitly. See Entropy Production.

Common Mistakes

1. Saying “erasing information destroys energy.” Erasure exports entropy and generally requires work; it need not destroy energy.

2. Forgetting the sign convention. In this volume, heat dumped to the bath is $Q_{\mathrm{bath}}=-Q_{\mathrm{mem}}$.

3. Assigning a fixed $k_BT\ln2$ cost to every measurement. The bound concerns erasure of an unbiased one-bit record.

4. Ignoring correlations with a reference system or controller.

5. Treating the bound as a typical engineering dissipation rather than a reversible lower limit.

6. Using the entropy of an observer’s ignorance without identifying the physical memory state.

Exercises

Biased Bit

A classical bit is $1$ with probability $p$ and $0$ with probability $1-p$. What is the minimum heat dumped to a bath at temperature $T$ when the bit is reset to $0$?

Solution

The initial entropy is

$$S_i = - k_B \left[ p\ln p + (1-p)\ln(1-p) \right].$$

The final state is definite, so $S_f=0$. Therefore

$$\Delta S_{\mathrm{mem}} = S_f-S_i = k_B \left[ p\ln p + (1-p)\ln(1-p) \right].$$

Landauer’s bound gives

$$Q_{\mathrm{bath}} \ge - T\Delta S_{\mathrm{mem}} = k_B T \left[ - p\ln p - (1-p)\ln(1-p) \right].$$

For $p=1/2$, this becomes $k_BT\ln2$.

Qubit Erasure

A qubit memory has eigenvalues $\lambda$ and $1-\lambda$ and a degenerate Hamiltonian. Find the minimum heat dumped to the bath when it is reset to a pure standard state.

Solution

The von Neumann entropy is

$$S(\rho) = - k_B \left[ \lambda\ln\lambda + (1-\lambda)\ln(1-\lambda) \right].$$

The final pure state has entropy zero, so

$$\Delta S_{\mathrm{mem}} = -S(\rho).$$

Thus

$$Q_{\mathrm{bath}} \ge - T\Delta S_{\mathrm{mem}} = T S(\rho).$$

Explicitly,

$$Q_{\mathrm{bath}} \ge k_B T \left[ - \lambda\ln\lambda - (1-\lambda)\ln(1-\lambda) \right].$$

Sign Convention Check

A memory is erased so that $\Delta S_{\mathrm{mem}}=-k_B\ln2$. Use $\Sigma=\Delta S_{\mathrm{mem}}-Q_{\mathrm{mem}}/T\ge0$ to find the bound on $Q_{\mathrm{mem}}$ and $Q_{\mathrm{bath}}$.

Solution

Insert $\Delta S_{\mathrm{mem}}=-k_B\ln2$:

$$\Sigma = - k_B\ln2 - \frac{Q_{\mathrm{mem}}}{T} \ge 0.$$

Therefore

$$- \frac{Q_{\mathrm{mem}}}{T} \ge k_B\ln2,$$

or

$$Q_{\mathrm{mem}} \le - k_BT\ln2.$$

The memory releases at least $k_BT\ln2$ of heat. Since $Q_{\mathrm{bath}}=-Q_{\mathrm{mem}}$,

$$Q_{\mathrm{bath}} \ge k_BT\ln2.$$

References

• R. Landauer, “Irreversibility and heat generation in the computing process,” IBM Journal of Research and Development 5, 183, 1961.
• C. H. Bennett, “The thermodynamics of computation: a review,” International Journal of Theoretical Physics 21, 905, 1982.
• C. H. Bennett, “Demons, engines and the second law,” Scientific American 257, 108, 1987.
• T. Sagawa and M. Ueda, “Minimal energy cost for thermodynamic information processing: measurement and information erasure,” Physical Review Letters 102, 250602, 2009.
• D. Reeb and M. M. Wolf, “An improved Landauer principle with finite-size corrections,” New Journal of Physics 16, 103011, 2014.
• J. M. R. Parrondo, J. M. Horowitz, and T. Sagawa, “Thermodynamics of information,” Nature Physics 11, 131, 2015.