I don’t buy your ~kT argument. You can make the temperature ratio arbitrarily large, and hence the energy arbitrarily small, as far as I understand your argument.
There’s some intrinsic energy required to erase a bit, kTlog2. This is true no matter how large the temperature ratio is. The point is that we’d normally be paying some extra energy cost in addition to this kTlog2 in order to get the particle over the energy wall, and it’s this that we can make arbitrarily small by changing the temperature ratio.
Overall, I’d say that the “simple straightforward argument” serves mostly as an intuition pump for the idea that a high energy wall is needed only during computation and not erasure. It won’t convince a skeptical reader, the computational study is in the post because it’s actually pretty important.
With your model, I don’t understand why the energy ‘generated’ when swapping isn’t thermalised (lost to heat). When you drop the energy of the destination state and the particle moves from your origin to your destination state, the energy ‘generated’ seems analogous to that from bit erasure; after all, bit erasure is moving a particle between states (50% of the time). If you have a mechanism for harvesting energy, you can always use it.
Basically, there’s two ways the energy of the particle can be lowered:
Lower the energy of a state while the particle is sitting in that state. This energy is recaptured by you.
Due to interaction with the environment, which has lots of thermal noise, the particle jumps from a high-energy state to a low energy state. This energy is dissipated and not recaptured by you.
When we’re lowering the energy of the destination state, we’re recapturing energy through mechanism 1. The process is basically the reverse of a Landauer erasure. We’re accepting a bit of thermal noise from the environment in exchange for being given a little less than kTlog2 of energy. Then in the reverse phase we spend a little more than kTlog2 to push that bit back into the environment.
Anyways, this model can be shown to still respect the Landauer limit. See this. If you perform the process very slowly such that the probability distribution over states matches the Boltzmann distribution then the energy cost integral gives kTlog2.
How would you experimentally realise mechanism 1? It still feels like you need an additional mechanism to capture the energy, and it doesn’t necessarily seems easier to experimentally realise.
With regards to 2, you don’t necessarily need a thermal bath to jump states, right? You can just emit a photon or something. Even in the limit where you can fully harvest energy, thermodynamics is fully preserved. If all the energy is thermalised, you actually cannot necessarily recover Landauer’s principle; my understanding is that because of thermodynamics, even if you don’t thermalise all of that energy immediately and somehow harvest it, you still can’t exceed Landauer’s principle.
How would you experimentally realise mechanism 1? It still feels like you need an additional mechanism to capture the energy, and it doesn’t necessarily seems easier to experimentally realise.
One example (probably not the easiest thing to implement in practice) is the charge-patterned wheel that I mentioned in my thread with Jacob. If we’re lowing the energy of a state while the particle is in that state, then the potential gradient puts a force on the particle which pulls back on the wheel, increasing its energy.
With regards to 2, you don’t necessarily need a thermal bath to jump states, right? You can just emit a photon or something.
If you’re emitting into vacuum (no other photons there at all), then that’s like having access to a thermal bath of temperature 0. Erasure can be done for arbitrarily low cost under such conditions. If the vacuum has temperature higher than 0, then it has some photons in it and occasionally one of them will come along and knock into our particle. So then we pretty much have a thermal bath again.
There’s some intrinsic energy required to erase a bit, kTlog2. This is true no matter how large the temperature ratio is. The point is that we’d normally be paying some extra energy cost in addition to this kTlog2 in order to get the particle over the energy wall, and it’s this that we can make arbitrarily small by changing the temperature ratio.
Overall, I’d say that the “simple straightforward argument” serves mostly as an intuition pump for the idea that a high energy wall is needed only during computation and not erasure. It won’t convince a skeptical reader, the computational study is in the post because it’s actually pretty important.
Basically, there’s two ways the energy of the particle can be lowered:
Lower the energy of a state while the particle is sitting in that state. This energy is recaptured by you.
Due to interaction with the environment, which has lots of thermal noise, the particle jumps from a high-energy state to a low energy state. This energy is dissipated and not recaptured by you.
When we’re lowering the energy of the destination state, we’re recapturing energy through mechanism 1. The process is basically the reverse of a Landauer erasure. We’re accepting a bit of thermal noise from the environment in exchange for being given a little less than kTlog2 of energy. Then in the reverse phase we spend a little more than kTlog2 to push that bit back into the environment.
Anyways, this model can be shown to still respect the Landauer limit. See this. If you perform the process very slowly such that the probability distribution over states matches the Boltzmann distribution then the energy cost integral gives kTlog2.
How would you experimentally realise mechanism 1? It still feels like you need an additional mechanism to capture the energy, and it doesn’t necessarily seems easier to experimentally realise.
With regards to 2, you don’t necessarily need a thermal bath to jump states, right? You can just emit a photon or something. Even in the limit where you can fully harvest energy, thermodynamics is fully preserved. If all the energy is thermalised, you actually cannot necessarily recover Landauer’s principle; my understanding is that because of thermodynamics, even if you don’t thermalise all of that energy immediately and somehow harvest it, you still can’t exceed Landauer’s principle.
One example (probably not the easiest thing to implement in practice) is the charge-patterned wheel that I mentioned in my thread with Jacob. If we’re lowing the energy of a state while the particle is in that state, then the potential gradient puts a force on the particle which pulls back on the wheel, increasing its energy.
If you’re emitting into vacuum (no other photons there at all), then that’s like having access to a thermal bath of temperature 0. Erasure can be done for arbitrarily low cost under such conditions. If the vacuum has temperature higher than 0, then it has some photons in it and occasionally one of them will come along and knock into our particle. So then we pretty much have a thermal bath again.