> I assume the energy lost as heat occurs due to the "current" going into the and gate having "nowhere else to go" other than to dissapate as heat?

Not quite, it's a thermodynamic principle that applies to any way you could possibly compute AND. Basically, the laws of physics are reversible, so your computation must be reversible too. There are 4 possible inputs to an AND gate, so to be reversible there must be 4 possible outputs, one for each input. But we only want one output for the rest of our computation, so the other one dump into the environment somehow.

I think we're talking about the same thing, but I explained it absolutely terribly.

If we "dump the other output" back into the power source, such as the battery, does that solve the problem of not implicitly dumping it into the environment? Or is it still destroying information?

I think that you could move those bits to destroy them somewhere else but what I got from pezezin post is that there are many order of improvement to achieve before that loss becomes significant enough to warrant the incredible complexity of shipping wasted bits to the heat sink.

How will you choose which output is the one with the AND result? You'd need some logic to pick which output was the right one, I assume. Then you've got the same problem again.

This is the first time I'm hearing it phrased this way and I wonder why it hadn't occurred to me before. Thanks so much, in any case, you have just increased my understanding quite a bit. :) slapshead

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.