I attached a picture, where the energy dissipation (entropy increase) on information erasure is explained. Is the explanation correct?

  1. "RESTORE TO ONE" - is it correct to identify the operation as "information erasure"? It looks like a negation, not like "information erasure".
  2. I would say that the erasurement would be performed, when the "ball" (the black dot near "0" in the picture) would be returned to the meta-stable point between 0 and 1.
  3. Maybe the intention of the author was to show, that "negation" has "erasurement" embedded in it? Then, he states that "computing depends on information erasure" - this statement is too strong? - we have reversible computation that do not erase information, correct?
  4. The picture and point 2. (above) seem to be a very good metaphore of Landauer's principle. Does it match Landauer's principle closely, or maybe even perfectly - or is it just a picture to help only conceptually understand the principle?

(PS. A better title is welcomed..)

Dissipation on negation?


1 Answer 1


You're confused because the example didn't cover all possible cases, and you're extrapolating improperly.

Negation is the operation

0 → 1,
1 → 0.

This is reversible, and can theoretically be performed using an arbitrarily small amount of energy.

Restore to one is the operation

0 → 1,
1 → 1.

This is not reversible, and unavoidably dissipates $k T \ln 2\;$ energy.

  • $\begingroup$ Oh right, RestoreToOne is a general operation, not action performed in pictured case. Next, I think that the model in the picture does not allow to show why negation preserves energy while RestoreToOne dissipates it? (dissipation is the effect of lost degree of freedom, which happens outside of the "model") $\endgroup$
    – Mooncer
    Mar 23, 2012 at 1:24
  • $\begingroup$ @Boordet: the picture certainly doesn't explain why negation can preserve energy (it won't always, you have to implement it carefully) while RestoreToOne necessarily dissipates it. For that, you should read a good survey article ... maybe somebody can suggest one. $\endgroup$ Mar 23, 2012 at 1:29

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