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Trying to understand the physical limits to computation, I notice that among these we have two types of limits that constrain the minimum allowable energy for a computation.

  • Limits that constrain the product of energy and time taken, which include Bremermann's limit and Margolus-Levitin. These two laws pretty much state the same thing as far as I can tell, with the only difference being a constant which doesn't concern me since only order of magnitude matters for my own purposes.
  • Limits that constrain only energy required for computation, notably Landauer's principle.

So let's just write these. I'll utilize the Bremermann's limit since it seems to be more commonly referenced.

$$ E \ge \frac{ 2 \pi \hbar }{ \Delta t } $$

$$ E \ge k T \ln{(2)} $$

If you believe what these two equations are telling you literally, then obviously there is a certain time range where the energy-time limit will be more restrictive and a range where the energy-alone limit will be more restrictive. With trivial algebra, we set the RHS equal to each other and find the time for computation where the crossover happens. I used the temperature of space here. For room temperature, it is much smaller/faster.

$$ \Delta t = \frac{ 2 \pi \hbar }{ k T \ln{(2)} } \approx 2 \times 10^{-13} s $$

If we're considering serial computations, this would leave us in the $\text{THz}$ range. It's not anywhere near Planck time or anything like that, but it also doesn't seem like much of a practical limitation.

Is the energy-time limitation only discussed for academic purposes, with the understanding that it will be swamped by the more restrictive limit? Or is there some deeper reason why it should matter?

Alan Rominger
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  • I don't think that it is the same $E$ in the two formulae. In the first formula, it is the total energy/mass of the quantum system/computer, while, in the second formula, it is the energy given to the environment. – Trimok Jan 10 '14 at 18:11
  • @Trimok Well I thought the entire point of the Margolus-Levitin limit was to apply the quantum uncertainty principle to the performance of a computation. I guess maybe the Landauer's principle may be referring to something else, but I can't quite understand what, other than the energy "lost" due to doing the computation. – Alan Rominger Jan 10 '14 at 19:14
  • Yes, OK, but the total energy/mass $E$ of the computer/system is extremely higher than the energy $\Delta E$ given to the environment (due to the erasing of one bit and the necessary transfert of entropy to the environment). – Trimok Jan 10 '14 at 19:30
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    The answer to the question is obviously no, since a reversible computer need not draw any energy (and Landauer's is inapplicable) while still has to obey Bremermann's limit. – user23660 Jan 10 '14 at 20:27

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