# Is the Landauer limit reversible

As I understand it, the Landauer limit, $$kTln(2)$$, is the minimum amount of energy to erase a bit. Is it also the minimum amount to create a bit? I'm asking statistical, like Avogadro's number of bits, i.e., $$RTln(2)$$.

My thinking: $$RTln(2)$$ is the work required, probably both ways (create and erase), but I don't know whether that is the same as the energy. By "same" I mean classical like heat, like something that can be used in $$E=mc^{2}$$.

• How are bits created? Can they be created? Oct 5, 2016 at 2:14
• @endolith I think the OP will probably agree that he/ she means "initialized" rather than created. It's a well formed question and the OP's intuition, that it does require work to initialize, is correct. Sep 19, 2017 at 13:21

So, to initialize (rather than creating) a bit does indeed require the input of a minimum quantity of work of $k\,T\,\log 2$ joules, in accordance with Landauer's principle, so that the storage system and environment's state changes comply with the second law of thermodynamics.
To create a bit in some defined state like 0 takes energy $kT\ln(2)+W$, where $W$ is the work required to create whatever the bit is made of (a flip-flop, a particle with spin, etc.). The first factor is just the Landauer cost of setting it in a definite state. As you say for a mole of bits you will have to pay at least $RT\ln(2)$ Joules.
$W$ will be problem dependent, and could in principle be zero: we can just name existing particles with spin as our bits without doing anything to them ("That electron over there will be bit 1, the other one bit 2, and the one over on Mercury bit 3..."). That kind of arbitrary naming still has the problem that we need to remember which electron is which bit, paying a Landauer cost for the naming, but often the indexing of bits can be done implicitly like when they are all spins along a crystal lattice we just found and decided to make our memory.
Since the work $W$ of making a bit typically involves irreversibly changing multiple degrees of freedom while the Landauer limit is about just one degree of freedom, most of the time $W \gg kT\ln(2)$.