# Poisson brackets and canonical transformations

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1 rof O B Wrght Mechancs Notes osson brackets and canoncal transformatons osson Brackets Consder an arbtrary functon f f ( qp t) df f f f q p q p t But q p p where ( qp ) pq q df f f f p q q p t In order to smplfy the notaton defne the osson bracket 1 1 g g 1 g g g g Usng ths notaton wth g1=f g= we fnd q p p q df f f t Constants of the moton df If f s a constant of the moton (we also say an ntegral of the moton ) then 0 An example s n a problem where the angular momentum of the system under consderaton s conserved A way of vsualzng the meanng of df s to move wth the partcle and check the 1

2 rof O B Wrght Autumn 007 value of f from tme to tme If t s constant then In contrast the quantty f t df 0 s the change n f when we are fxed n one place wth a fxed momentum (You mght thnk ths means that nothng should change but sometmes the externally appled potentals or constrants vary wth tme) From above the conon for f f t df 0 can also be wrtten as If n adon f s not an explct functon of tme that s f both df 0 and f 0 t then f 0 We can use ths equaton to fnd the conon that f should be a f constant of the moton n the case when 0 t Example: free partcle n 1D p m dp Is p a constant of the moton? That s s 0? Let us fnd p p p p q q p =0 dp 0 and p=constant Example: partcle n a potental n 1D p V ( q) (where q=x) m

3 rof O B Wrght Autumn 007 p dp p dv dv p q dq dq 0 In ths case p s not constant but s constant snce general (see last week p 3) and n ths case 0 t d t n Formulas for osson brackets: f g g f f C 0 (where C s a constant) f f g f g f g 1 1 f f g f f g f f g f g f g g f t t t d df dg f g g f Also f q k f p and f p k k f q k (These can be proved usng kl kl q q k l 1for k l and 0 for k l ) kl qk and 0 where p l Specal cases of these relatons are q q 0 p p 0 q p k k There s also an mportant relaton called Jacob s dentty: (where 1for k and 0 for k ) k k f g h g h f h f g 0 (Ths can be proved by wrtng all the terms) 3

4 rof O B Wrght Autumn 007 There s also a theorem called osson s theorem: If f and g are constants of the moton then f g s also a constant of the moton That s df dg d f g Useful for fndng new constants of the moton roof: d df dg f g g f d f g 0 g f 0 0 df dg from above But f 0 and 0 then Canoncal transformatons We are free to choose the generalzed coordnates q 1 q n Whatever we choose d Lagrange s equatons wll apply: ( q1 qn q1 qn t) and q q Consder choosng a new set of coordnates Q Q ( q t) where 1 n ( Q Q Q Q t) and So 1 n 1 n d Q Q If these two Lagrangans lead to the same equatons of moton then for example or df( q t) where F s an arbtrary functon (as proved n Lecture 11 p 4) (The Lagrangan moton Can you see why?) where s a constant also gves the same equatons of 4

5 rof O B Wrght Autumn 007 We can do somethng smlar wth amlton s equatons ( q q p p t) Now consder a new set of n generalzed coordnates 1 n 1 n Q Q ( qp t) ( qp t) If the new Q obey Q Q for the new amltonan ( Q t) then we call the change of coordnates from q p to Q a canoncal transformaton The mechancal moton of the system descrbed by s the same as that descrbed by It s also possble to show see end of ths lecture that for a canoncal transformaton Q k k (where 1for k and 0 for k ) k Ths equaton can be used to test for a canoncal transformaton k ow to do a canoncal transformaton Use the defnton of the amltonan pq The defnton of s Q where ( Q t) For the two amltonans to descrbe the same problem try the general form df( Qq t) ( Q Q t) ( q q t) (1) 5

6 rof O B Wrght Autumn 007 where F s some functon of q Q and t (If there s a constant other than 1 multplyng we do not call t a canoncal transformaton) We can see that the new Lagrangan descrbes the same problem because on ntegratng t t F( q( t1) Q( t1) t1) F( q( t) Q ( t) t) t1 t1 So applyng amlton s prncple t d 0 d s the same as applyng t1 d d t t1 0 so the two Lagrangans gve the same moton df We can wrte Eq (1) as pq Q or df p dq dq ( ) F F F Therefore p q Q t ere F F( qq t) s called the generatng functon for the canoncal transformaton If F F( qq ) only wth no explct dependence on t then we see that That s ( q p) ( Q ) ossble types of problem: 1) We are gven and q q( Q t) p p( Q t) Fnd and F( q Q t ) ) We are gven and F( q Q t ) Fnd and q q( Q t) p p( Q t) 6

7 rof O B Wrght Autumn 007 Example of a canoncal transformaton Consder the smple case of a system wth one q and one p In general for ( q p) we know that q p p q Consder the transformaton to the new coordnates and Q where q=- p=q Check that ths s a canoncal transformaton and fnd the generatng functon F Method 1 to check f t s a canoncal transformaton If t s a canoncal transformaton then Q () Q The relatons between q p and Q do not depend on t so we know that F/ t 0 Therefore ( Q ) ( q p) But and p Q q q Q Q p So Eqs () are satsfed and so the transformaton s canoncal Method to check f t s a canoncal transformaton A transformaton s canoncal f Q 1 Q Q that s f 1 Evaluatng ths drectly we get q p p q Q Q 0 (1 1) 1 so the transformaton s canoncal q p p q 7

8 rof O B Wrght Autumn 007 Fnd the generatng functon F Use F p from p 6 where F F( q Q) But p=q so q F q Q Therefore F Qdq Qq f ( Q) F Also But =-q so Q F q so F qdq Qq g( q) Q For both these solutons for F to be consstent F Qq arbtrary constant because t does not change the moton) ( We do not need to add an Also we know that because F/ t 0 Example For the same canoncal transformaton q=- p=q let as n a sngle partcle n 1D n a potental V Q Then V ( ) m p V ( q) m Utlty of canoncal transformatons Sometmes a partcular choce of p q mght mean that nether p or q are constants But by convertng to Q coordnates we mght be able to make or Q constant Also canoncal transformatons are very useful n quantum mechancs roof that Q 1 mples that the transformaton s canoncal Let us use the notaton of Jacobans for general functons F and G: F F F G F G q p q p q p p q G G q p F G F G 8

9 rof O B Wrght Autumn 007 where means the determnant of the matrx Ths matrx s called the Jacoban A shorthand for the Jacoban s F F q p ( F G) G G ( pq ) q p There s a chan rule for Jacobans: multplcaton ( F G) ( F G) ( u v) where ths s matrx ( x y) ( u v) ( x y) In partcular f we set F G= x y n ths equaton we fnd that ( x y) ( u v) I where ( u v) ( x y) I From the general rule that AB s the product of the determnants) we know therefore that A B (the determnant of a product of two matrces ( x y) 1 ( uv ) ( uv ) ( x y) (3) Our job s to evaluate Q ( Q ) p q ( pq ) Q Q p q owever from Eq (3) we know that Q q p Q ere p q p Q q Q 1 1 ( pq ) ( Q ) q p Q 9

10 rof O B Wrght Autumn 007 Let us frst fnd q p Q and then use ths to fnd Q To go further we shall assume the exstence of the generatng functon F defned n p 6 q p Q p p Q q p q p q q Q Q Q F F Also F F( q Q t) F( q( Q ) Q t) And p q Q Let us frst calculate p p and Q so that we can fnd q p Q p F( q( Q ) Q t) F q q q where q q( Q ) p F( q( Q ) Q t) F F q Q Q q q Q q Q Ths allows us to fnd q p Q F Q Q q Q qq q Q q q p q p q F q F F q qq F F But because q qq Q q p 1 Therefore But Q Q 1 Therefore q p Q Q 1 as requred Note that the quantty F qq always needs to be non-zero for a canoncal transformaton 10

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