Deleting and entropy

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).