We define the entropy for a system as:

1) where k = 1.38 e-23 (boltzman’s constant)

We also have these equations from Thermodynamics

2) dU = dW + dQ

3) where dQ = TdS

kT has units of energy such as Joules, eV

Bernd:

if you erase a bit in a box (set it to zero) that was previously in a maximally indeterminate state (ie. Had a 50% probability of being a one or 50% chance of being a zero) have I gained info or lost info?

Rich:

The entropy is now zero, , whereas it was maximal before, ie. , so the entropy has decreased.

Ebeth:

Yes, we have decreased the volume of the phase space

Yael:

If we decrease the logical entropy of a system, as in the above example, then you must have heat flow out of the system!

Rich:

Could you use this heat to do work- perhaps to power the computer that is decreasing in logical entropy?

Ben and Ed:

THE SUPER IMPORTANT MORAL OF THE STORY:

THIS IS UNUSABLE HEAT because: imagine one bit of the heat that comes out. If you know the information stored in this bit, then you could use it to restore the bit in your original system. So then the original bit was not really erased!

But we said we are going to do doing the irreversible erasure. We defined what we are doing as irreversible erasure. SO if we keep the bit of heat somehow separate, able to be read someday, or if we use it to create energy by using only the high bits to power a heat engine (we had to read them to know they are high bits), then we also could have used it to set our original bit back to its original state, so its information wasn’t lost. It’s information wasn’t erased.

So the idea is that to perform an irreversible erasure within some box, you have to throw some of the information out of the box and agree that you can never to use it again within the box.

so let's reserve the word erasure for irreversible destruction of info.

Bennett says reversible computing only costs as much entropy as the difference in phase space volume of the initial state and the final state. A good example would be a billiard ball computer which has perfectly elastic collisions (and thereby uses no energy) except for the change in phase volume between the input register and the output register. How is there an actual use of energy by this input output discrepancy? Some of the energy you used to set 10 billiard balls in motion at 10 mph in the beginning is unnecessary at the end when you must only measure the state of 5 billiard balls going 10mph. So a bunch of energy gets wasted. 5 billiard balls go careening off into space at 10mph.

To give another example of where the heat actually goes, use gas in a cans. Imagine truth table with four cans to put gas in->

00 -> 00

01 -> 10

10 -> 10

11 -> 11

The two cans leak into each other so when you dump stuff into one of them it leaks together with the other and you lose information. It would take energy to make one can have zero gas in it and the other have gas in it again.

ben: so if you have an image of a jungle scene that may include a jaguar in it, and you have to decide jaguar or no jaguar ie. Run away or don’t run away. This is clearly an important kind of computation to do – pattern recognition. The size of your input register is seemingly 1024x768 pixels and your ending register is just 1 bit. Actually your initial phase space has two states. You might run away or you might stay put. Then you analyze the picture and you make a decision and thereby collapse the phase space to one possibility or the other. so you are ordering the register, and entropy has decreased in your computer. You pay for this with the heat you lose.

Yael:

So we've talked about logical entropy and thermodynamic entropy, and we've linked them to energy via eq’s 2 and 3.

Now the 2^nd Law of Thermodynamics states that entropy must increase,

dS >= 0

So in Neil’s paper he discusses three types of entropy/energy (remember energy and entropy are interchangeable using dU = dW + dQ.) that exist in a modern day computer at different scales, but which will eventually merge into a single kind of entropy/energy in the final optimally efficient computer of the future:

1) electrostatic charging energy:

If a system is moved away from equilibrium ( if you reduce a system’s entropy) it will need to dissipate some energy in order to return to maximum entropy.

for example:

mosfet push in voltage on gate to charge it up to a 1

then as you read out answer you discharge capacitor and as it does this some energy will be used up by the flow of heat.

To understand charging a capacitor faster than equilibrium thermodynamics (adiabatic charging) in more detail, you need to understand "non-equilibrium linear thermodynamics."

If 10e7 transisters dump their energy to ground on every clock cycle at 100MHz, then the energy being dissipated is:

The electrostatic energy per bit is still seven orders of magnitude greater than room temperature thermal energy of

kT = .02 eV.

2) Thermodynamic entropy:

So now there is another kind of entropy in the system which is much smaller scale - the fermi energy in the transister – due to the distribution of electrons = p(x)

So the entropy will be = p(x) log p(x)

is on the order of 1 eV or 10e-20J/K

where T = 300K

T = 10e5K

3) logical entropy (destroying a bit):

entropy of an erasure is k log 2

∆Q = 10^-21 J/bit = meV

So when will the three energies be comparable = with 1 electron transisters?

let's do the math for 1 e trans.

C = Q/V

U = .5 C V^2 = .5 Q V = 1.5 eV

so logical entropy becomes the same expression kT log 2 for a 1 electron trans as the fermi energy entropy story because now we will have 1 electron in a box in either a ground state or 1 excited state which is 2 available states.

so for 1 electron in a box with 2 avail states

logical bit entropy = thermodynamic entropy = kT log 2

so yael says there are now room temperature 1 electron transisters = how?!!!!

because John Meyers says that such a system will be "played like a piano" by the heat bath.

Well the how, is it uses two states a "ground state" which is below the energy of temperature kT, and some high, high energy state which is above the energy of kT so that it is robust to perturbation by ambient heat.

Yael's question:

How do we know if computing a particular function requires energy and how much?

Ben:

try defining it with "1 electron transisters" in other words assume a maximally efficient computer and see what the necessary costs are.

Note:

Kolgomorov complexity = algorithmic complexity

reference Chaitin of Aukland University in New Zealand - algorithmic info theory.

Note on computers complexity and randomness:

so a computer can't make random numbers because a computer can only have a finite algorithmic complexity, but the algorithmic complexity of a truly random number is infinite.

John Meyers lectures on why ordinary computers work:

short answer:

1) regenerative amplifier

take a not gate

put in a value in a small region in the bottom gaurantee that you get something out in a SMALLER region near the top

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I keep funneling things by using dynamics of the gate

2) other reason

is accurate clocking

in regular computation you have a flip flop:

use dynamics to fix errors in timing skew

by horses coming into gate and you don't open gate until they all get into the gate

But in quantum computing we can’t do these kinds of dynamics because they require nonunitary operations which destroy our superposition. We need a breakthrough in how we implement quantum computing if we want to solve this problem.

Note: John suggests a paper by pippinger and fisher: equivalence between quantum computational circuits and quantum turing machines