It's a thermodynamic argument. Suppose you have a system of n bits each in an arbitrary state, so it has 2^n microstates. You set one bit to 0, regardless of wh

It's a thermodynamic argument. Suppose you have a system of n bits each in an arbitrary state, so it has 2^n microstates. You set one bit to 0, regardless of what it was originally. It now has 2^(n-1) microstates. Finding entropy S for a given number of microstates N, S = k_b ln N, and so dS = k_b * (ln 2^(n-1) - ln 2^n) = k_b * ln 2 * (n - 1 - n) = -k_b ln 2, i.e. entropy has decreased by a constant amount. But by the second law of thermodynamics, entropy cannot decrease in a closed system, and the way it's dissipated is as heat. How much heat has been released? dE = T*dS, so dE >= k_b * ln 2 * T. Note the dependence on temperature: you can reduce how much heat is released by having it operate at a lower temperature. Even at room temperature, however, this is a billion times less than existing heat dissipation, so there's lots of room for improvement before we start hitting the Landauer limit. This can be worked around by introducing ancilla bits to maintain the number of states in the system, but the instant you destroy the ancilla bits (e.g. by feeding them back to the power source), you dissipate energy. The exact mechanics of this would depend on the implementation of the device you're talking about, but you'd inevitably encounter it and be unable to overcome it.