However you’re imagining Landauer’s double well procedure raises the energy level of one of the wells conditional on a control signal, you should be able to imagine my procedure being implemented in a similar way.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
Just to give an explicit example, though, you can imagine the control signal being used to determine the position of the particle that encodes the bit.
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. The control signal could come at any time or never and the bit must be stable for very long periods of time.
“Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV. If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
A machine which turns a domino right side up is in fact raising the energy of the “domino tipped over” state. If the domino stayed tipped over, the mechanical arm pushing the domino up would at some point be intersecting with the domino, a state of very high energy indeed. Unless you’re imagining some other way for the domino resetter to function?
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. [...] “Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV.
Yes, you need to be 1eV below extraneous nearby states to be doing anything sensible at all. This is an energy difference and not an energy cost. You’re not dissipating any energy because the particle is not occupying those states. Determining the position of the particle based on a control bit is a reversible operation. If you want to argue that it must cost 1eV, you need to explain either why the operation is not actually reversible, or why it must have a thermodynamically unavoidable energy cost anyways, despite being reversible.
The control signal could come at any time or never and the bit must be stable for very long periods of time.
So the particle is in the far away states by default and the control signal kicks it into the nearby states when the bit has to be erased.
If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
I’m certainly not saying the wheel part is necessary, it’s just a particular example I gave. I do think it’s allowed by thermodynamics to reliably and costlessly alter the particle’s path in ways that are reversible. I’m not trying to recover the energy from the erased bit (I assume you meant “bit” here, as particles aren’t erased). I’m spending the kTlog2 (on average), and then that’s the end of it.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
No, the whole point of controllable erasures is that sometimes the bit isn’t erased. So we need 2 outputs, one for the control wire as you say, and one for the original bit since sometimes the erasure signal is 0 and so the bit isn’t erased. The gate is still irreversible, of course, but it seems like your mental model of the situation here is quite different than mine, so I’m pointing out the difference in the hopes of some clarification.
I do not imagine it this way, as trying to manipulate the barrier just punts the issue. It’s simpler to imagine static wells which the particle/bit moves between, as in dominoes.
This is not a true memory bit storage mechanism unless the particle is in some sort of loop in a low energy potential well about 1eV lower than the background. The control signal could come at any time or never and the bit must be stable for very long periods of time.
“Determining the position of the particle” requires deflecting it into 1 of 2 paths with high reliability, which requires energy on order 1eV. If you can reliably alter the particle’s path—you no longer need the wheel, as one path can just exit the device or thermalize to erase. Any attempt to recover the energy of the erased particle is a dead end which just punts the problem to a new mechanism for erasure.
The gate which is formed from this process has 2 unique bit inputs (memory bit, control wire bit) and only 1 unique bit output (the control wire bit), as the other is erased. It is fundamentally irreversible—it doesn’t matter how many complex additional mechanisms (flywheels) you tack on, they are just distractions.
A machine which turns a domino right side up is in fact raising the energy of the “domino tipped over” state. If the domino stayed tipped over, the mechanical arm pushing the domino up would at some point be intersecting with the domino, a state of very high energy indeed. Unless you’re imagining some other way for the domino resetter to function?
Yes, you need to be 1eV below extraneous nearby states to be doing anything sensible at all. This is an energy difference and not an energy cost. You’re not dissipating any energy because the particle is not occupying those states. Determining the position of the particle based on a control bit is a reversible operation. If you want to argue that it must cost 1eV, you need to explain either why the operation is not actually reversible, or why it must have a thermodynamically unavoidable energy cost anyways, despite being reversible.
So the particle is in the far away states by default and the control signal kicks it into the nearby states when the bit has to be erased.
I’m certainly not saying the wheel part is necessary, it’s just a particular example I gave. I do think it’s allowed by thermodynamics to reliably and costlessly alter the particle’s path in ways that are reversible. I’m not trying to recover the energy from the erased bit (I assume you meant “bit” here, as particles aren’t erased). I’m spending the kTlog2 (on average), and then that’s the end of it.
No, the whole point of controllable erasures is that sometimes the bit isn’t erased. So we need 2 outputs, one for the control wire as you say, and one for the original bit since sometimes the erasure signal is 0 and so the bit isn’t erased. The gate is still irreversible, of course, but it seems like your mental model of the situation here is quite different than mine, so I’m pointing out the difference in the hopes of some clarification.