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DEMONIC Programming Horsman Abramsky Outline Maxwells Demon and - PowerPoint PPT Presentation

DEMONIC Programming Horsman Abramsky Outline Maxwells Demon and Landauers Hypothesis. Thermodynamics in 1 slide. A toy model system. Defining a programming language for the toy system. DEMONIC syntax and semantics. Allowed


  1. DEMONIC Programming Horsman Abramsky

  2. Outline Maxwell’s Demon and Landauer’s Hypothesis. Thermodynamics in 1 slide. A toy model system. Defining a programming language for the toy system. DEMONIC syntax and semantics. Allowed operations expressed in DEMONIC. Formal verification: a computational invariant. The Second Law and Landauer’s Hypothesis proven.

  3. O ¡ ¡for ¡ ¡a ¡ ¡muse ¡ ¡of ¡ ¡fire! Henr/ ¡V; ¡opening ¡words

  4. O ¡ ¡for ¡ ¡a ¡ ¡muse ¡ ¡of ¡ ¡fire! single-particle gas in equilibrium with a heat bath at temperature T!

  5. The thermodynamics of computation “Information is physical” — information processing is necessarily a physical process obeying laws of physics. Thermodynamics: one-way entropy increase, therefore key constraints on what and how information can be processed. The connection between information and entropy/thermodynamics: Landauer’s Hypothesis : erasure of 1 bit requires kTln2 of work. But . . . it has not been proven .

  6. A toy system - thermodynamics 101 Single particle in a box, two pistons, one partition, heat bath T. Variables: pressure p, volume V, entropy then H = − k ( p L ln p L + p R ln p R ) . 1 then the entropy of the box pV H out − H in = k ln V out T = const V in

  7. A toy system - thermodynamics 101 Allowed operations: Insert/remove partition

  8. A toy system - thermodynamics 101 Allowed operations: Insert/remove partition

  9. A toy system - thermodynamics 101 Allowed operations: Insert/remove pistons left and right

  10. A toy system - thermodynamics 101 Allowed operations: Insert/remove pistons left and right

  11. A toy system - thermodynamics 101 Allowed operations: Insert/remove pistons left and right

  12. A toy system - thermodynamics 101 Allowed operations: Insert/remove pistons left and right (NB isothermal compression: requires work kTln2)

  13. Formalising the system What we will do: extract out the logical structure of the state and allowed transitions into a programming language . An allowed operation is a function/basic statement not a primitive. Allowed programs are built out of allowed statements. state variable : s = ( X , A , I , w ) 2 T ⇥ B ⇥ B ⇥ Z . left-hand-side of the box, p . We : prob of being on LHS. Therefore X 2 T : = { 0 , 1 2 , 1 } . respectively. W wkT ln2 A, I : Boolean flags for partition/a piston. w : total work extracted from the system in unit of kTln2.

  14. DEMONIC syntax LProb :: = T : = 0 | 1 / 2 | 1 Part :: = B : = true | false Pist :: = B WUnit :: = Z Field :: = LProb | Part | Pist | WUnit Fieldname :: = X | A | I | w (where W = wkT ln2) s 2 State :: = ( LProb , Part , Pist , WUnit ) BExp :: = B | State . A | State . I S 2 Statement :: = S 1 ; S 2 | S 1 � S 2 | State . Fieldname : = Field | if BExp then S 1 else S 2 | skip

  15. DEMONIC operational semantics ( assign ) h x : = a , s i ) h skip , s [ x 7! a ] i h S 1 , s i ) p h S 0 1 , s 0 i ( comp1 ) h S 1 ; S 2 , s i ) p h S 0 1 ; S 2 , s 0 i ( comp2 ) h skip ; S , s i ) h S , s i ( if1 ) h if B then S 1 else S 2 , s i ) h S 1 , s i if J B K s = true ( if2 ) h if B then S 1 else S 2 , s i ) h S 2 , s i if J B K s = false ( prob1 ) h S 1 � S 2 , s i ) 1 / 2 h S 1 , s i ( prob2 ) h S 1 � S 2 , s i ) 1 / 2 h S 2 , s i

  16. Allowed thermodynamic operations Inserting a partition: PartIn = de f ( s . A : = true ) Inserting a partition changes the partition Removing a partition: PartOut = de f if ( s . A = true ) then (if ( s . I = false ) then ( s . X : = 1 2 ) and ( s . A : = false ) else ( s . A : = false ) ) else skip

  17. Allowed thermodynamic operations Removing a piston (left and right) LPistOut = de f if ( s . I = false ) or ¬ ( s . X = 0 ) then skip else ( if ( s . A = true ) then ( s . I : = false ) else ( s . I : = false ) and ( s . X : = 1 2 ) and ( s . w : = w + 1 ) ) Removing a piston to the left can only be done if there is a piston, and the particle is on the right RPistOut = def if ( s . I = false ) or ¬ ( s . X = 1 ) then skip else ( if ( s . A = true ) then ( s . I : = false ) else ( s . I : = false ) and ( s . X : = 1 2 ) and ( s . w : = w + 1 ) ) Removing a piston to the right is by symmetry with removing to the left.

  18. Allowed thermodynamic operations Inserting a piston is more complicated… Can we insert a piston to the right? No! Would compress to zero volume, requiring infinite work. But for a programming language we have to give the outcome if it were attempted.

  19. Allowed thermodynamic operations Consider this cycle h PartIn ; LPistIn ; PartOut ; LPistOut , ( 1 2 , F , F , w 0 ) i ) h LPistIn ; PartOut ; LPistOut , ( 1 = 2 , T , F , w 0 ) i = ) 1 / 2 h PartOut ; LPistOut , ( 0 , T , T , w 0 ) i = ) h LPistOut , ( 0 , F , T , w 0 ) i ) h skip , ( 1 = 2 , F , F , w 0 + 1 ) i h PartIn ; LPistIn ; PartOut ; LPistOut , ( 1 2 , F , F , w 0 ) i ) h LPistIn ; PartOut ; LPistOut , ( 1 = 2 , T , F , w 0 ) i = ) 1 / 2 h PartOut ; LPistOut , ( 0 , T , F , w 0 � w c ) i ) h LPistOut , ( 1 = 2 , F , F , w 0 � w c ) i ) h skip , ( 1 = 2 , F , F , w 0 � w c ) i Expected work extracted: outcomes, W e = ( w 0 + 1 2 ( 1 � w c )) kT ln2. make the choice that the set of t No perpetual motion implies require W e  W o , i.e. w c � 1.

  20. Allowed thermodynamic operations Therefore… LPistIn = def if ( s . X = 1 ) then ( s . w : = w � 1 ) else ( if ( s . X = 0 ) then ( s . I : = true ) else ( if ( s . A = false ) then ( s . X : = 1 ) and ( s . w : = w � 1 ) and ( s . I : = true ) else [( s . X : = 0 ) and ( s . I : = true )] � [( s . X : = 1 ) and ( s . w : = w � 1 )] )) RPistIn = def if ( s . X = 0 ) then ( s . w : = w � 1 ) else ( if ( s . X = 1 ) then ( s . I : = true ) else ( if ( s . A = false ) then ( s . X : = 0 ) and ( s . w : = w � 1 ) and ( s . I : = true ) else [( s . X : = 1 ) and ( s . I : = true )] � [( s . X : = 0 ) and ( s . w : = w � 1 )] ))

  21. Computational Invariant Statement Probabilistic computational invariants are given over the set of probability distributions over states. This is easy for a physicist: expectation values! An invariant statement is a predicate that is true after a transition if it is true before, and preserved under composition. What is the invariant statement for this single-particle system…?

  22. Computational Invariant Statement h wk ln2 i� 1 2 ( h H ( X ) i + H ( h X i ))  0 Where transitions will satisfy this inequality at every step if x , H ( x ) = � kT ( x ln x + ( 1 � x ) ln ( 1 � x )) , (probabilistic) computational invariant for the basic transitions Every composition of the allowed thermodynamic operations satisfies this invariant afterwards if it satisfies it beforehand. What is that entropic quantity???

  23. The Second Law is a theorem of the system γ Kelvin statement of the second law: 6 9 γ : ( X 0 , A 0 , I 0 , w 0 ) � ! ( X 0 , A 0 , I 0 , h w f i > w 0 ) . Define the zero-point of the work counter as 1 w 0 = 2 k ln2 ( h H ( X 0 ) i + H ( h X 0 i )) then the invariant is satisfied initially. Final invariant gives that h w f i k ln2 � 1 2 ( h H ( X 0 ) i + H ( h X 0 i ))  0, in its Kelvin formulation is then a theorem which straightforwardly implies implies h w f i  w 0 . ements satisfy the for all allowed operations and compositions

  24. Landauer Erasure Two entropies make up the invariant entropy: <H(X)>: average entropy within a branch of the computation. H(<X>): entropy of the probability distribution of the computation (across all its branches). Consider X=1/2, partition=true. <H(X)>=H(<X>)=kln2. Measurement gives in two branches, X=0 and X=1. <H(X)>=0 but H(<X>)=kln2 still. Resetting the result to a known state gives one branch, eg. X=0. <H(X)>=0 and H(<X>)=0.

  25. Landauer Erasure Given the invariant h wk ln2 i� 1 2 ( h H ( X ) i + H ( h X i ))  0 transitions will satisfy this inequality at every step if i Measurement of a bit of information requires at least least 1 2 kT ln2. complicated, in part i Resetting of a measured bit of information requires at least least 1 2 kT ln2. complicated, in part Erasure (measure-then-reset) of an unknown bit of information requires a work cost of at least i i i + = 1 least 1 least 1 2 kT ln2. 2 kT ln2. 2 kT ln2. complicated, in part complicated, in part

  26. Conclusions We have used formal semantics and verification as a process logic for single-particle thermodynamics. Basic transitions and operations are defined, as are their composition, and a new invariant statement found. The Second Law is provably satisfied by any combination of the basic operations. This is not “up for debate”! Landauer Erasure — work cost of measure then reset — is a formal consequence of the logical system.

  27. Further work Lots! Extending to multi-particle states, extend to statistical mechanics, rederive partition function statements, extend definition of Landauer Erasure etc etc etc. What is the new entropy? What’s its connection to the Holevo quantity? What’s the relationship to the Second Law? And finally… Where else in physics can we use these verification tools to prove formal statements about the possible states of a system??

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