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The following Q&A about reversible computing is available here. It has listed a number of practical scenarios where a reversible circuit can still be dissipating heat. Let's assume that none of these is present in the following discussion.

Okay, then can reversible computing really make the energy dissipation of a computation be an arbitrarily small non-zero amount?

Only insofar as the computer can be arbitrarily well isolated from unwanted interactions, errors, and energy leakage (which are all really different ways of describing the same thing). Even a superconducting reversible computer with the best possible thermal insulation still faces the possibility of being struck by a high-energy cosmic ray - or an uncharted asteroid (an asteroid can be thought of as an extremely high-energy cosmic ray). Even if you bury the computer at the center of a cold planet, a black hole could always swing through the solar system and swallow it up. Objects like rouge black holes passing through can be viewed as merely a very extreme variety of thermal noise, one which illustrates the impossibility of ever getting rid of unwanted interactions completely. Even with total control over the state of the universe, there will always a small residual rate of quantum tunneling out of any non-equilibrium state space. However, it might be possible that as the universe grows and cools, we may be able to compute with ever-smaller lower-bounds on the rate of energy dissipation per operation, and thus perform infinitely many computations with only finite energy, a scenario explored by [Dyson '79], and objected to by [Krauss & Starkman '99]. [Dyson '01] thinks the argument may hinge on whether a true analog basis for computing is possible. However, even if Dyson is correct in the long run, if one chooses a particular era (the next billion years, say), it is not clear whether we could ever control enough of the environment to achieve any desired nonzero rate of dissipation, however small. But, despite all these caveats, it may yet be possible to set up reversible computations that dissipate such amazingly tiny amounts of energy that the dissipation is not a barrier to anything that we might wish to do with them - I call such computations ballistic. We are a long way from achieving ballistic computation, but we do not yet know of any fundamental reasons that forbid it from ever being technically possible.

When we have a pure quantum state $|0\rangle$, we can calculate the Shannon entropy to be zero. That is there is no information. But when a quantum state is an equal superposition of $|0\rangle$ and $|1\rangle$, it's Shannon entropy is $- ln \frac{1}{2} = 0.69$.

Let me draw a quantum circuit as follows.

enter image description here

In this circuit, the output of a Hadamard gate is fed as it's input. Let's assume in the first clock cycle the input is $|0\rangle$. Now at the output the Shannon entropy increases.

My question is: can this scenario be considered as erasure of information as referred to in Landauer's principle?

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    $\begingroup$ You would absolutely love this paper: arxiv.org/abs/0904.3273 $\endgroup$ – Goodies Oct 30 '14 at 7:24
  • $\begingroup$ @Goodies, after a quick review I failed to find a section in that paper which is discussing about the heat dissipation of a reversible circuit. The author didn't discuss about the inapplication of Landauer's principle for any implementation of reversible circuits. Am I missing anything? $\endgroup$ – Omar Shehab Nov 1 '14 at 21:13
  • $\begingroup$ I can't even ask the author as he is no more alive. :( $\endgroup$ – Omar Shehab Nov 1 '14 at 21:18
  • $\begingroup$ I don't actually see a question here. As you say, your circuit is reversible, so there's no lower bound on heat dissipated. The Q&A is about a completely different topic - namely implementation - claiming, as you say, that there will always be heat dissipation due to the fact that it's not possible to completely shut out influence from the rest of the universe. $\endgroup$ – Martin Nov 16 '14 at 11:23
  • $\begingroup$ @Martin, I have rephrased and changed my question. $\endgroup$ – Omar Shehab Nov 24 '14 at 20:17
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Can this scenario be considered as erasure of information as referred to in Landauer's principle?

No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place.

Where's the problem? It's here: The state

$$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$

has also zero Shannon entropy. Why? Because it's a pure state. The corresponding density matrix

$$\frac{1}{2} (|0\rangle+|1\rangle)(\langle 0|+\langle 1|)$$

is rank one (eigenvector is the state with eigenvalue 1) Your error in thinking is that a superposition is dependent on your choice of basis, it has nothing to do with entropy.

The Hadamard gate is a unitary matrix, as such, it implements a unitary evolution and must therefore send pure states to pure states. In whatever basis you consider your pure state to be in, the result will also be a pure state (otherwise, the process would not be reversible).

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