Quantum computation is a hot topic, no doubt. But sometimes I feel that, in order to understand it well, we should go deeper into the physics of classical computation... So, let's go.
According to Landauer, the only step in a computation which requires an expense in energy is deletion. Why that? His argument is simple. Deletion is the only operation in a computer which is irreversible. Therefore it implies an increase in entropy DS. So, we'll have to pay an energy T DS.
So, among the many things that this assertion make me think, this one can be funny. How to store a memory bit, physically? Anyway you decide to do it, it should be a system with two states, 0 and 1. Let's consider a 1D system, characterized by a potential V(x), which in principle is a double well with two equal minima, associated with the values 0 and 1, and separated by a potential barrier.
At temperature T, the probability of finding the system at equilibrium at any position, p(x) will be proportional to the Boltzmann factor exp[-V(x)/kT]. This function p(x) has two equal maxima, also at the 0 and 1 positions.
So, let's assume that the bit is 1. This means that the probability distribution should be highly peaked around 1. Imagine that the potential energy for both values, 0 and 1, are the same. Can this be stable? It can stay for long, if the barrier is large, but will eventually come to the previous function p(x).
Let us assume that the barrier is large enough for our purposes. Now the question is: (a) how to take the probability distribution from one minimum to the other? According to Landauer, we should be able to do this without any expense in energy, and (b) how to delete the information? For this, we should spend a minimum energy which is kT log(2).
2 comments:
Recuerdo haberlo comentado contigo y haber llegado a alguna conclusión, en cierto modo mis recuerdos estan vacios dado que no me acuerdo cuales fueron los comentarios ni las conclusiones a las que llegamos... se que pensamos en el hecho de gastar energia al borrar que era equivalente a... y aqui se acaba... puede que se haya gastado energia en borrarme a mi esa conversación ;-)
Por cierto, creo que borrar no es lo irreversible, dado que se podria volver a escribir borrando el borrado, el problema es que borrar es un estado unico y no consideramos escribir como uno solo... sino que hay tantos estados como posibles escrituras ;-)
Ahora mi duda ¿todos los borrados son iguales?
Ya pongo bien la duda. Escribir sobre blanco es reversible porque puedes volver al blanco, escribir sobre escrito es irreversible dado que no has guardado memoria de lo escrito... ¿hay que borrar para escribir o se sobreescribe directamente?
Bueno, técnicamente el borrado no es un estado aparte. Por convención borrar es poner a cero todos los bits del disco, por lo que no debería ser distinto escribir un 0 porque toque un 0 allí que escribir un 0 para borrar un 1.
Post a Comment