Bits in Double Wells

We have been talking about bits as abstract 0s and 1s. But a bit in a real machine is a physical thing: a dab of charge, a patch of magnetisation, a voltage held on a wire. Before we can put a price on erasing a bit, we need an honest physical picture of what storing one involves. The picture physicists reach for — the one Rolf Landauer used — is beautifully simple: a bit is a ball resting in one of two valleys.

The double-well potential

Imagine a landscape with two dips separated by a hill, described by an energy curve U(x) — a double-well potential, such as

U(x) = x^4 - b\,x^2,

where the parameter b controls the height of the central hill (the barrier). A particle settles into one valley or the other: left valley = 0, right valley = 1. That's a memory. Every real storage technology is a dressed-up version of this — in a flash cell the "particle" is a packet of electrons and the barrier is an insulating oxide layer; on a hard disk the two valleys are two directions of magnetisation.

Play with the landscape below. Drag the barrier down and watch the two valleys merge into one — with no hill between the states, the memory ceases to be a memory. Tilt the landscape and one valley becomes preferred.

What keeps a bit a bit

The particle is not sitting still. At temperature T it is jiggling, constantly kicked by thermal noise with characteristic energy kT (about 25\ \mathrm{meV}, i.e. 4 \times 10^{-21} joules, at room temperature). Following the Boltzmann distribution, the chance of a kick big enough to carry it over a barrier of height \Delta U is suppressed by the famous exponential factor

P_{\text{hop}} \;\sim\; e^{-\Delta U / kT}.

This factor is ferociously effective. A barrier of just 1\,kT is hopped constantly — useless as a memory. At 40\,kT (about 1 electron-volt), the suppression is e^{-40} \approx 10^{-17}, and a particle rattling against the barrier a billion times a second still waits years to escape. Flash memory hides its electrons behind an oxide barrier of roughly 3\ \mathrm{eV} \approx 120\,kT — which is why your photos survive a decade in a drawer with no power at all.

So a good bit needs a barrier of many kT: stability is bought with energy scale. Note what the barrier does not cost: while the particle just sits in its valley, no energy is being spent. Storing information is free. It is changing the landscape that can cost — and that is where erasure comes in.

Erasure: squeezing two valleys into one

To erase a bit means: whatever valley the particle is in now, end with it in the left valley (a known standard state, "reset to 0"). You are not allowed to look first — erasure must work blind, by manipulating the landscape alone. The classic protocol has three moves, and you can drive it with the slider below:

The landscape ends exactly where it started — but something irreversible has happened. Before, the particle could have been in either valley: two possible states. After, it is in the left valley: one state. The protocol squeezed two possibilities into one — precisely the many-to-one merge that the previous lesson warned about, now happening in a physical energy landscape. Run the film backwards and it makes no sense: the reversed protocol would have to know which valley to return the particle to, and that information no longer exists anywhere in the machine. Its escape route — and the heat that escorts it out — is the subject of the coming lessons.

Because of that exponential. The average time before a thermal kick defeats a barrier is roughly the rattle time (the particle bumps the barrier around 10^910^{12} times per second) multiplied by e^{+\Delta U/kT}. At \Delta U = 20\,kT that's a retention of minutes — hopeless. At 45\,kT it's centuries. The exponential is so steep that the practical difference between "evaporates before lunch" and "outlives your grandchildren" is barely a factor of two in barrier height. Engineers speak of the "40 kT rule of thumb" for a decade of retention — and it also explains why heat kills data: raise T and every stored bit's barrier shrinks in units of kT, which is why an old SSD left in a hot car forgets sooner, and why archives are kept cold.

A tempting thought: dissipation usually comes from rushing (friction, sloshing charge), so surely lowering the barrier infinitely slowly makes the erase protocol cost nothing? No — and the distinction is the whole point. Moving the particle around gently can indeed be made as frictionless as you like; that waste is engineering, and patience removes it. But the protocol compresses two possible states into one, and that logical compression carries an irreducible price — kT \ln 2 of heat, as we will soon derive — no matter how slowly you go. Slowness buys off the friction; it cannot buy back the lost information. An operation that doesn't compress states (like gently carrying the particle from a known valley to another known valley) really can approach zero cost.