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.

Entropy Lost by the Memory

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

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

Assume the memory and bath are initially uncorrelated, the bath starts thermal at temperature $T$, and no accessible side information is consumed. In the ideal reservoir limit, 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

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

In the usual isothermal implementation, approaching the bound requires reversible control. A practical reset also has a nonzero error probability. Exact reset of an initially full-rank memory to a pure state is impossible by a joint unitary with a finite-dimensional, finite-temperature, full-rank bath: the joint state remains full rank, so its reduced memory state cannot become rank one. The ideal pure reset is therefore a limit, not an exactly attainable finite-bath operation.

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.

Accessible side information can reduce the cost. For a classical reference $R$, a reversible average reset conditioned on $R$ costs $k_BT H(M|R)$ for a degenerate memory, rather than $k_BT H(M)$. For example, a perfectly correlated reference bit allows a controlled reversible reset while the reference retains the information. Restoring the reference completes the cycle and restores the corresponding cost.

Quantum side information can have negative conditional entropy. In the asymptotic reversible setting, entanglement can then permit work extraction during reset by consuming correlations; this is a different resource setting from the uncorrelated bound above. See del Rio et al. and Mutual Information.

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 can preserve joint von Neumann entropy while transferring the memory information into the bath and correlations.

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.

Reversible Logic and Physical Reset

The bound concerns entropy reduction in a physical memory. It does not assign a fixed heat cost to every gate or measurement. A known pure state can be rotated reversibly, while an unknown bit must transfer its uncertainty elsewhere during reset. Reversible Computation explains how logical operations can preserve information in additional registers; eventually resetting those registers is again a thermodynamic task.

Finite Baths and Entropy Production

The bound has an exact microscopic form. Let the initial state be $\rho_M\otimes\tau_B$, with $\tau_B=e^{-\beta H_B}/Z_B$, fixed $H_B$, and joint unitary evolution. Writing $Q_{\mathrm{bath}}=\Delta E_B$, the global correlation identity gives

$$\beta Q_{\mathrm{bath}} =-\frac{\Delta S_M}{k_B} +I(M:B)_{\mathrm{final}} +D(\rho_B^{\mathrm{final}}\Vert\tau_B).$$

Both corrections are nonnegative, so the usual Landauer inequality survives a finite bath and strong coupling under these assumptions. Finite bath disturbance generally increases the required heat. Strong coupling does require interaction energy in the separate work balance; it does not invalidate this entropy identity. Initial correlations or a nonthermal bath change the starting assumptions. Reeb and Wolf derive finite-size improvements.

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 $k_BT\ln2$.

Reset with a Small Error

An unbiased degenerate bit is reset to $0$ with probability $1-\delta$. Find the entropy-based heat bound, and evaluate it at $\delta=0.01$.

Solution

The final entropy is $k_BH(\delta)$, so

$$Q_{\mathrm{bath}}\ge k_BT[\ln2-H(\delta)].$$

At $\delta=0.01$, $H(\delta)\simeq0.05600$ nats, giving a bound of about $0.63715\,k_BT$, below the perfect-reset limit $0.69315\,k_BT$. Residual uncertainty lowers the entropy reduction.

Signs of Memory and Bath Heat

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

• C. H. Bennett, “Demons, engines and the second law,” Scientific American 257(5), 108–116 (1987).
• C. H. Bennett, “The thermodynamics of computation: a review,” International Journal of Theoretical Physics 21, 905–940 (1982).
• L. del Rio, J. Åberg, R. Renner, O. Dahlsten, and V. Vedral, “The thermodynamic meaning of negative entropy,” Nature 474, 61–63 (2011). arXiv:1009.1630.
• R. Landauer, “Irreversibility and heat generation in the computing process,” IBM Journal of Research and Development 5, 183–191 (1961).
• J. M. R. Parrondo, J. M. Horowitz, and T. Sagawa, “Thermodynamics of information,” Nature Physics 11, 131–139 (2015).
• D. Reeb and M. M. Wolf, “An improved Landauer principle with finite-size corrections,” New Journal of Physics 16, 103011 (2014).
• T. Sagawa and M. Ueda, “Minimal energy cost for thermodynamic information processing: measurement and information erasure,” Physical Review Letters 102, 250602 (2009).