The connection matrix is symmetric and has entries that are 0 or 1, and it represents a graph (you can put in entries that are fractions to represent a case where states are partially connected (i.e. connected but the maximum flow rate between them is lower than 1/τ)). Non-zero diagonal entries represent a state being connected to itself, which is kind of pointless since the flow is going to be 0 in any case.
And looking at your steady state condition it looks like it converges to a transition probability of 1 if the energy levels are the same?
For E1=E2, the steady state condition is:
p1p2=e−βE1e−βE2=1
The normalized distribution which satisfies this requirement is p1=p2=12.
Regardless, the code comment above seems to indicate you are not allowing the particle to stay in its starting state, which seems probably fatal.
It’s a continuous-time differential equation, so the outward flow of probability is continuous in the model. In a small timestep Δt, most particles are not going to change which state they’re in.
Any such potential energy wells are only created by yet other particles which themselves must also be in equally stable low energy configurations for the well to be stable, and so the argument applies recursively—manipulating the wells is just equivalent to moving particles out of wells.
I ask again: If any manipulation of wells recursively creates another bit that we have to erase, then what’s the difference between my well manipulations and Landauer’s? Or for that matter, your domino resetting device? Why doesn’t you model imply that all 3 get mired in an eternal cycle of “create a bit to erase a bit”?
Our initial particle has an entropy associated with it, since it has two possible states, and these allow it to represent a bit. The other particles producing the energy wells do not need to have any entropy, they can be sitting in a single state at the bottom of their own well with high energy walls all around. Because those parts of the device require no information, they can operate ballistically. To take a concrete model, we can build a big wheel with charges glued to various points along its circumference. The wheel spins and charges move past the set of states where we’re keeping our particle, and depending on which charge is close, the energy of each states is raised and lowered. If we’ve designed a good bearing for the wheel, then the slowing of the wheel will be entirely due to its interaction with the particle.
It’s a continuous-time differential equation, so the outward flow of probability is continuous in the model. In a small timestep Δt, most particles are not going to change which state they’re in.
This still seem fatal, as it implies that any arbitrarily tiny transition energy will converge to the bit exiting the bit well, which obviously is not physically correct (and contradicts the claim that the bit well was stable against noise). Once you include the required diagonals I think your mechanism no longer works.
I ask again: If any manipulation of wells recursively creates another bit that we have to erase, then what’s the difference between my well manipulations and Landauer’s? Or for that matter, your domino resetting device? Why doesn’t you model imply that all 3 get mired in an eternal cycle of “create a bit to erase a bit”?
The difference is the domino resetting device (or any practical actual current CMOS computer, or bio neural network etc) erases/thermalizes bits, so it doesn’t have to continue keeping track of them.
None of the published reversible computers designs I’ve seen claim to erase reliably for much less than 1eV btw. That’s not the focus at all, instead they save energy by eliminating unnecessary erasures.
If we’ve designed a good bearing for the wheel, then the slowing of the wheel will be entirely due to its interaction with the particle.
Therein is the problem as this slowing of the wheel will need be about ΔE≈40KBT per bit to reliably eject bits.
To actually reliably erase all the bits, the wheel needs to eject them from their low energy wells, which thus requires ΔE≈40KBT per bit, because each bit had to be in a well with energy that much lower than the neighborhood. So to the extent your wheel works it’s just equivalent to the domino erasure mechanism (which could be having all the dominoes linked to axle rods and levers to a motor etc).
You could try to use the wheel to implement a temporary well, but that doesn’t work because it is not sufficient to simply lower the well as that doesn’t give the particle enough energy to move out of its location reliably.
You can move bits around ballistically for free (within some limits on not being able to aim them very precisely), but again moving bits around is not erasing them.
This still seem fatal, as it implies that any arbitrarily tiny transition energy will converge to the bit exiting the bit well, rather than staying put against noise until the erasure signal, which contradicts the model.
I’m not 100% sure what you’re trying to say here, so forgive me if this is a wrong interpretation. The point is that a given energy wall height gives you a certain length of time in which to do computations. The flow out of a state can be suppressed exponentially by energy wall height, so in practice this can be a really long time. Also, flow out of a state is compensated by flow back into that state from other states so that overall things converge to the Boltzmann distribution. This isn’t relevant to how long it takes a bit to thermalize, though.
Also, “erasure signals” aren’t a thing in my model, see below.
The difference is the domino resetting device (or any practical actual current CMOS computer, or bio neural network etc) erases/thermalizes bits, so it doesn’t have to continue keeping track of them.
My procedure also erases/thermalizes bits so that they no longer need to be tracked, that’s the entire point of it. I think I’m having a hard time understanding what you’re getting at here.
This interaction now just moves a bit around, and we now need a full model of the wheel to see where that bit went. Energy/information is conserved and can’t be destroyed. Any time you lose track of it in your model you have erased/thermalized it, so I can just keep asking where the ~1eV bit went.
The particle jumps around between the states because it has some coupling to the environment, and the environment is full of thermal noise / fluctuations. Because physics is reversible, we could in theory chase the bit into the environment: In principle you can always keep track of where the bit went (at least if you’re Laplace’s demon). But let’s say that once we’ve put the bit into the environment, that’s good enough.
But we still used some energy to do that. We drained some energy from our battery to accomplish this task (maybe it’s a conventional battery or capacitor, or maybe it’s something more exotic, like a spinning flywheel).
The next thing you might worry about is that I cheated at the bit-erasing task by allowing some information to escape to the battery instead of the environment. Obviously there’s some information going into the environment, since coupling to the environment is the entire reason the particle is jumping between states. But maybe the battery is picking up most of the information. That would be bad! But it seems like it could happen, since the amount of energy drawn by the device is random, and then that introduces randomness into the level of energy reserves left in the battery.
First, a meditation: Consider the double slit experiment. Momentum is conserved, so if the photon is refracted through the lower slit and heads back up to hit the detector, some downward momentum must have been transferred to the card with the slits. Likewise, if the photon goes through the upper slit, the card must have gained some upward momentum. Quantum mechanics is very sensitive to leaked information. If even a single bit of information (which slit the photon went though) were leaked to the environment, then the interference pattern would disappear. Why doesn’t the “which way” information encoded in the momentum of the card have this effect?
Anyways, think about erasing a lot of bits. By the central limit theorem, the ratio of variance in the amount of energy left in the battery relative to the number of bits erased goes to zero.
Or say we already have a lot of thermal uncertainty about the amount of energy stored in the battery. Say it’s a Gaussian distribution. This has entropy:
12(1+log2π+logσ2)
If performing 1 erasure costs a small quantity of energy with similarly small variance ϵ2, then the change in entropy of the battery is going to look like:
12(log(σ2+ϵ2)−logσ2)≈ϵ22σ2
The larger the uncertainty in battery energy, the less entropy a little additional information is going to carry.
In any case, note that Landauer’s double-well erase procedure also has an energy cost that varies randomly. If the bit happens to be a 1, then it draws ~1eV, if the bit happens to be a 0 then it’s free. Neither Landauer nor I are secretly depending on being able to shuffle information into the battery, though.
Your device also no longer makes sense to me—how does it erase a bit (or not) on demand from an incoming wire signal? As the wheel now controls the bit well and the erasure, and the wheel isn’t controlled by the input signal, the device no longer works (at least not without yet more complexity).
My model (obviously other models are possible) of a reversible computer looks like a reversible circuit with some small number of irreversible gates. Those gates produce bits that need to be erased on every clock tick (the clock can be synced with the spinning flywheel). In order to do long serial computations, the rest of the output bits are looped back around to the input of the circuit so that more computation can be done on them in the next clock tick. So the flywheel bit erasing device just needs to be able to erase the bits coming out of the circuit on each clock tick, and there’s always the same number of them. No need for an extra separate “erasure on demand” signal / feature.
You can move bits around ballistically for free (within some limits on not being able to aim them very precisely), but again moving bits around is not erasing them.
Yes, I also agree that moving bits around is not erasing them.
Also, “erasure signals” aren’t a thing in my model, see below.
You need an erasure signal to erase a bit.
Let’s step back for a second and more formally define “erasure” and “bit”.
I see several meanings of “bit”:
A True Bit—A bit in computer RAM/registers etc—read/write controlled by arbitrary signal bits on signal lines.
A signal line bit propagating on a wire
A temporary ‘bit’ which is not necessary for 1 or 2 above, and thus isn’t really an irreducible bit
I believe to the extent your spinning wheel thing works, it is only ‘erasing’ type 3 non-bits, which is irrelevant. You need to show erasure of at least type-2, and really type-1 bits.
A reversible computer can already (in simplified theory at least) avoid all type 2 and 3 erasures completely, so it’s irrelevant how much energy they use. You need to show reliable type 1 bit erasure, or you should change your post to add disclaimers that your claimed ‘erasure’ doesn’t actually work for bits in RAM, and can’t actually reduce energy use for a reversible computer.
The entire wheel apparatus in your scheme is just a complex way of reversible propagation of bits on signal lines. We could just simplify that down to a single particle propagating on the signal line.
Yes, I also agree that moving bits around is not erasing them.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM, though I personally don’t think anyone’s going to ever build a reversible computer that works that way. Anyways, what I’m saying is my procedure would suffice to erase a Cannellian Type-1 bit.
The point of the wheel example is that you claimed that: “manipulating the wells is just equivalent to moving particles out of wells”. The wheel device provides a counterexample, since it changes the energy levels of the various states just by spinning. (It’s also an example of a situation where an “erasure signal” is not necessary to erase a bit, since it erases a bit on every clock tick.)
But I’m in general totally fine with having an erasure signal. When I say it’s not in my model, I mean that the basic erasure procedure I describe is independent of that kind of external detail.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Where is it moving to in your opinion? I provided you an argument that the answer isn’t the wheel. What’s left other than the environment?
Just also replying to a chunk of your earlier comment:
To actually reliably erase all the bits, the wheel needs to eject them from their low energy wells, which thus requires ΔE≈40KBT per bit, because each bit had to be in a well with energy that much lower than the neighborhood. So to the extent your wheel works it’s just equivalent to the domino erasure mechanism (which could be having all the dominoes linked to axle rods and levers to a motor etc).
You could try to use the wheel to implement a temporary well, but that doesn’t work because it is not sufficient to simply lower the well as that doesn’t give the particle enough energy to move out of its location reliably.
After lowering the wall between the two states, I do then actually raise up one of the states high enough to kick the particle out of it with high reliability. The reason this doesn’t cost the full 40kT is that as the energy of the state gets higher, the probability of the particle occupying that state is already decreasing. I only pay any energy cost if the particle happens to be occupying the state I’m currently lifting. So you do an integral to get the expected energy cost of kicking this particle all the way out, and it comes out to way less than 40kT. In fact, it comes out to exactly kTlog2 if we’re moving so slowly that the particle is always distributed over states according to the Boltzmann distribution. But if you want to finish quickly, it comes out to a little bit more.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM,
You haven’t described any mechanism to do this yet.
Your flywheel thing (to the extent I now understand it), just seems to be equivalent to representing a signal bit moving down a wire, which could just be simplified to a single minimal particle. You haven’t described anything that is capable of true bit erasure.
True bit erasure would require having a control signal which then controls the wheel, or something like that.
You haven’t described any mechanism to do this yet.
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. What obstacle do you run into when you try to do this?
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. You have some states close to the wheel, and some far away. Far away states aren’t affected by the spinning of the wheel and if the particle is in the far states rather than the near states, then the wheel isn’t slowed down and the bit isn’t erased.
Choosing whether the particle goes into the near states or the far states conditional on a control bit is a reversible process, of course, so that completes the implementation.
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.
The connection matrix is symmetric and has entries that are 0 or 1, and it represents a graph (you can put in entries that are fractions to represent a case where states are partially connected (i.e. connected but the maximum flow rate between them is lower than 1/τ)). Non-zero diagonal entries represent a state being connected to itself, which is kind of pointless since the flow is going to be 0 in any case.
For E1=E2, the steady state condition is:
p1p2=e−βE1e−βE2=1
The normalized distribution which satisfies this requirement is p1=p2=12.
It’s a continuous-time differential equation, so the outward flow of probability is continuous in the model. In a small timestep Δt, most particles are not going to change which state they’re in.
I ask again: If any manipulation of wells recursively creates another bit that we have to erase, then what’s the difference between my well manipulations and Landauer’s? Or for that matter, your domino resetting device? Why doesn’t you model imply that all 3 get mired in an eternal cycle of “create a bit to erase a bit”?
Our initial particle has an entropy associated with it, since it has two possible states, and these allow it to represent a bit. The other particles producing the energy wells do not need to have any entropy, they can be sitting in a single state at the bottom of their own well with high energy walls all around. Because those parts of the device require no information, they can operate ballistically. To take a concrete model, we can build a big wheel with charges glued to various points along its circumference. The wheel spins and charges move past the set of states where we’re keeping our particle, and depending on which charge is close, the energy of each states is raised and lowered. If we’ve designed a good bearing for the wheel, then the slowing of the wheel will be entirely due to its interaction with the particle.
This still seem fatal, as it implies that any arbitrarily tiny transition energy will converge to the bit exiting the bit well, which obviously is not physically correct (and contradicts the claim that the bit well was stable against noise). Once you include the required diagonals I think your mechanism no longer works.
The difference is the domino resetting device (or any practical actual current CMOS computer, or bio neural network etc) erases/thermalizes bits, so it doesn’t have to continue keeping track of them.
None of the published reversible computers designs I’ve seen claim to erase reliably for much less than 1eV btw. That’s not the focus at all, instead they save energy by eliminating unnecessary erasures.
Therein is the problem as this slowing of the wheel will need be about ΔE≈40KBT per bit to reliably eject bits.
To actually reliably erase all the bits, the wheel needs to eject them from their low energy wells, which thus requires ΔE≈40KBT per bit, because each bit had to be in a well with energy that much lower than the neighborhood. So to the extent your wheel works it’s just equivalent to the domino erasure mechanism (which could be having all the dominoes linked to axle rods and levers to a motor etc).
You could try to use the wheel to implement a temporary well, but that doesn’t work because it is not sufficient to simply lower the well as that doesn’t give the particle enough energy to move out of its location reliably.
You can move bits around ballistically for free (within some limits on not being able to aim them very precisely), but again moving bits around is not erasing them.
I’m not 100% sure what you’re trying to say here, so forgive me if this is a wrong interpretation. The point is that a given energy wall height gives you a certain length of time in which to do computations. The flow out of a state can be suppressed exponentially by energy wall height, so in practice this can be a really long time. Also, flow out of a state is compensated by flow back into that state from other states so that overall things converge to the Boltzmann distribution. This isn’t relevant to how long it takes a bit to thermalize, though.
Also, “erasure signals” aren’t a thing in my model, see below.
My procedure also erases/thermalizes bits so that they no longer need to be tracked, that’s the entire point of it. I think I’m having a hard time understanding what you’re getting at here.
The particle jumps around between the states because it has some coupling to the environment, and the environment is full of thermal noise / fluctuations. Because physics is reversible, we could in theory chase the bit into the environment: In principle you can always keep track of where the bit went (at least if you’re Laplace’s demon). But let’s say that once we’ve put the bit into the environment, that’s good enough.
But we still used some energy to do that. We drained some energy from our battery to accomplish this task (maybe it’s a conventional battery or capacitor, or maybe it’s something more exotic, like a spinning flywheel).
The next thing you might worry about is that I cheated at the bit-erasing task by allowing some information to escape to the battery instead of the environment. Obviously there’s some information going into the environment, since coupling to the environment is the entire reason the particle is jumping between states. But maybe the battery is picking up most of the information. That would be bad! But it seems like it could happen, since the amount of energy drawn by the device is random, and then that introduces randomness into the level of energy reserves left in the battery.
First, a meditation: Consider the double slit experiment. Momentum is conserved, so if the photon is refracted through the lower slit and heads back up to hit the detector, some downward momentum must have been transferred to the card with the slits. Likewise, if the photon goes through the upper slit, the card must have gained some upward momentum. Quantum mechanics is very sensitive to leaked information. If even a single bit of information (which slit the photon went though) were leaked to the environment, then the interference pattern would disappear. Why doesn’t the “which way” information encoded in the momentum of the card have this effect?
Anyways, think about erasing a lot of bits. By the central limit theorem, the ratio of variance in the amount of energy left in the battery relative to the number of bits erased goes to zero.
Or say we already have a lot of thermal uncertainty about the amount of energy stored in the battery. Say it’s a Gaussian distribution. This has entropy:
12(1+log2π+logσ2)
If performing 1 erasure costs a small quantity of energy with similarly small variance ϵ2, then the change in entropy of the battery is going to look like:
12(log(σ2+ϵ2)−logσ2)≈ϵ22σ2
The larger the uncertainty in battery energy, the less entropy a little additional information is going to carry.
In any case, note that Landauer’s double-well erase procedure also has an energy cost that varies randomly. If the bit happens to be a 1, then it draws ~1eV, if the bit happens to be a 0 then it’s free. Neither Landauer nor I are secretly depending on being able to shuffle information into the battery, though.
My model (obviously other models are possible) of a reversible computer looks like a reversible circuit with some small number of irreversible gates. Those gates produce bits that need to be erased on every clock tick (the clock can be synced with the spinning flywheel). In order to do long serial computations, the rest of the output bits are looped back around to the input of the circuit so that more computation can be done on them in the next clock tick. So the flywheel bit erasing device just needs to be able to erase the bits coming out of the circuit on each clock tick, and there’s always the same number of them. No need for an extra separate “erasure on demand” signal / feature.
Yes, I also agree that moving bits around is not erasing them.
You need an erasure signal to erase a bit.
Let’s step back for a second and more formally define “erasure” and “bit”.
I see several meanings of “bit”:
A True Bit—A bit in computer RAM/registers etc—read/write controlled by arbitrary signal bits on signal lines.
A signal line bit propagating on a wire
A temporary ‘bit’ which is not necessary for 1 or 2 above, and thus isn’t really an irreducible bit
I believe to the extent your spinning wheel thing works, it is only ‘erasing’ type 3 non-bits, which is irrelevant. You need to show erasure of at least type-2, and really type-1 bits.
A reversible computer can already (in simplified theory at least) avoid all type 2 and 3 erasures completely, so it’s irrelevant how much energy they use. You need to show reliable type 1 bit erasure, or you should change your post to add disclaimers that your claimed ‘erasure’ doesn’t actually work for bits in RAM, and can’t actually reduce energy use for a reversible computer.
The entire wheel apparatus in your scheme is just a complex way of reversible propagation of bits on signal lines. We could just simplify that down to a single particle propagating on the signal line.
Right, and your flywheel thing so far is just moving an (irreducible) bit.
Generally, I don’t see a problem with using my procedure to erase a register in a computer. Seems like it would work fine for that, just like Landauer’s double well procedure would. It would also work fine to erase data directly in RAM, though I personally don’t think anyone’s going to ever build a reversible computer that works that way. Anyways, what I’m saying is my procedure would suffice to erase a Cannellian Type-1 bit.
The point of the wheel example is that you claimed that: “manipulating the wells is just equivalent to moving particles out of wells”. The wheel device provides a counterexample, since it changes the energy levels of the various states just by spinning. (It’s also an example of a situation where an “erasure signal” is not necessary to erase a bit, since it erases a bit on every clock tick.)
But I’m in general totally fine with having an erasure signal. When I say it’s not in my model, I mean that the basic erasure procedure I describe is independent of that kind of external detail.
Where is it moving to in your opinion? I provided you an argument that the answer isn’t the wheel. What’s left other than the environment?
Just also replying to a chunk of your earlier comment:
After lowering the wall between the two states, I do then actually raise up one of the states high enough to kick the particle out of it with high reliability. The reason this doesn’t cost the full 40kT is that as the energy of the state gets higher, the probability of the particle occupying that state is already decreasing. I only pay any energy cost if the particle happens to be occupying the state I’m currently lifting. So you do an integral to get the expected energy cost of kicking this particle all the way out, and it comes out to way less than 40kT. In fact, it comes out to exactly kTlog2 if we’re moving so slowly that the particle is always distributed over states according to the Boltzmann distribution. But if you want to finish quickly, it comes out to a little bit more.
You haven’t described any mechanism to do this yet.
Your flywheel thing (to the extent I now understand it), just seems to be equivalent to representing a signal bit moving down a wire, which could just be simplified to a single minimal particle. You haven’t described anything that is capable of true bit erasure.
True bit erasure would require having a control signal which then controls the wheel, or something like that.
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. What obstacle do you run into when you try to do this?
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. You have some states close to the wheel, and some far away. Far away states aren’t affected by the spinning of the wheel and if the particle is in the far states rather than the near states, then the wheel isn’t slowed down and the bit isn’t erased.
Choosing whether the particle goes into the near states or the far states conditional on a control bit is a reversible process, of course, so that completes the implementation.
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.