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:
- lower the barrier — the hill sinks and the two valleys merge into one broad
basin; the particle, wherever it started, now wanders the whole basin;
- tilt — raise the right side, and the particle drifts obediently into the left
half;
- restore the barrier — the hill rises again, trapping the particle in the left
valley. Level the tilt, and the landscape looks exactly as it began.
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.
- a stored bit is a particle held in one of two stable states separated by an
energy barrier \Delta U (the double-well picture);
- thermal noise flips it at a rate suppressed by
e^{-\Delta U/kT} — barriers of tens of kT
make memories that last years;
- holding a bit costs nothing; it is reshaping the landscape
(switching, erasing) that can dissipate energy;
- erasure = a blind protocol ending in one standard valley regardless of the
start — a two-states-into-one compression, and therefore logically irreversible.
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 10^9–10^{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.