PHYS 705: Classical Mechanics. Hamilton-Jacobi Equation

Transcription

1 1 PHYS 705: Classcal Mechancs Hamlton-Jacob Equaton

2 Hamlton-Jacob Equaton There s also a very elegant relaton between the Hamltonan Formulaton of Mechancs and Quantum Mechancs. To do that, we need to derve the Hamlton-Jacob equaton. The Hamlton-Jacob equaton also represents a very general method n solvng mechancal problems. To start Let say we are able to fnd a canoncal transformaton takng our n phase space varables q, p drectly to n constants of moton,.e.,, Q P

3 3 Hamlton-Jacob Equaton One suffcent condton to ensure that our new varables are constant n tme s that the transformed Hamltonan K shall be dentcally zero. If that s the case, the equatons of moton wll be, Q P K P 0 Recall that we have the new and old Hamltonan, K and H, relatng through the generatng functon K Q 0 F by: K H F t

4 4 Hamlton-Jacob Equaton Wth K dentcally equals to zero, we can wrte: F Hq, p, t 0 (*) t For convenence n later calculatons, we shall take F to be of Type so that ts ndependent varables are the old coordnates and the new momenta. In the same notatons as prevously, we can wrte Now, n order to wrte H n terms of ths chosen set of varables, we wll replace all the n H usng the transformaton equaton: p P F q, Q, P, t F ( q, P, t) QP p F qpt,, q q qpt,,

5 5 Hamlton-Jacob Equaton F F Then, one can formally rewrte Eq. (*) as: note that snce t t F F F Hq q t 1,, n;,, ; 0 q1 qn t Ths PDE s known as the Hamlton-Jacob Equaton. F qpt,, F( qpt,, ) QP p F q,, t q Notes: P -Snce are constants, the HJ equaton consttutes a partal dfferental equaton of (n+1) ndependent varables: - We used the fact that the new P s and Q s are constants but we have not specfed n how to determne them yet. - It s customary to denote the soluton by S and called t the Hamlton s Prncpal Functon. F q,,, 1 qn t

6 Recall that for a Type generatng functon, we have the followng partal dervatves descrbng the canoncal transformaton: Hamlton-Jacob Equaton Suppose we are able to fnd a soluton to ths 1 st order partal dfferental equaton n (n+1) varables where we have explctly wrte out the constant new momenta: 1 1,,,,, ; n n F S S q q t P,, (1) S q t p T q,, ( ) S q t Q T 6 S p q S Q P

7 7 Hamlton-Jacob Equaton After explctly takng the partal dervatve on the RHS of Eq. (T1) p and evaluatng them at the ntal tme, we wll have n equatons that one can nvert to solve for the n unknown constants n terms of the ntal condtons,.e., S q,, t q q p t Smlarly, by explctly evaluatng the partal dervatves on the RHS of Eq. t (T) at tme, we obtan the other n constants of moton 0 Q q p, t 0, 0 0 0, 0, 0 S q,, t t 0 qq, tt 0 0

8 8 Hamlton-Jacob Equaton, Wth all n constants of moton solved, we can now use Eq. q (T) agan to solve for n terms of the at a later tme t., q q,, t,,and q S q,, t Then, wth known, we can use Eq. (T1) agan to evaluate, n terms of at a later tme t. p p,, t p S q,, t,, t q p The two boxed equatons consttute the desred complete solutons of the Hamlton equatons of moton.

9 9 Hamlton Prncpal Functon and Acton Now, let consder the total tme dervatve of the Hamlton Prncpal functon S, ds q,, t S S q dt q t Usng Eq. (T1) and the Hamlton-Jacob equaton tself, we can replace the two partal dervatves on the rght by p and H, S ds p pq H L q dt S H t So, the Hamlton Prncpal Functon s dffered at most from the acton by a constant,.e., S Ldtconstant 0

10 10 Example: Solvng Harmonc Oscllator wth HJ f ( q) kq m 0 x U( q) kq Recall, we have: 1 m H p m q E The Hamlton-Jacob equaton gves: km S S Hq, ; t 0 q t 1 S S m q m q t 0 Recall that under ths scheme, we have: K 0 Q and P

11 11 Example: Solvng Harmonc Oscllator wth HJ f ( q) kq m 0 x U( q) kq 1 m H p m q E km When H does not explctly depend on tme, we can generally try ths tral soluton for S:,,, S q t W q t (q and t are separable) S t Wth ths tral soluton, we have and the HJ equaton becomes: 1 S m q t S m q 0 1 m W q m q recall : LHS so H H

12 Example: Solvng Harmonc Oscllator wth HJ m W m q m q The prevous equaton can be mmedately ntegrated: 1 W m q m q W m m q q W m m q dq and S m m q dq t 1

13 13 Example: Solvng Harmonc Oscllator wth HJ m We don t need to ntegrate yet snce we only need S S, q to use n the transformaton eqs: #1 eq: S q,, t S Q m m q dqt 1 dq m dq m t t m m q 1m q Lettng, we can rewrte the ntegral as: t x 1 m q dx 1 x 1 1 m t arcsn x arcsn q

14 14 Example: Solvng Harmonc Oscllator wth HJ m 1 1 m t arcsn x arcsn q Invertng the arcsn, we have wth ' m sn t ' q t The nd transformaton equaton gves, p S q,, t,, t q # eq: S q q p m m q dq t m m q

15 15 Example: Solvng Harmonc Oscllator wth HJ m Substtutng qt, we have, p t m m 1 m 1sn ' p t m t cos ' p t m t cos ' p t m t wth sn ' t '

16 16 Example: Solvng Harmonc Oscllator wth HJ m m sn t ' q t cos ' p t m t Completng the calculatons, we stll need to lnk the two constants of, moton to ntal t 0, q q, p p 0 0 Recall that the constant of moton s H=E of the system so that we can calculate ths by squarng q and p at tme : 1 m p m q 0 0 t 0 0 t And, we can calculate at tme 0 by dvdng the eqs: q p 0 1 tan ' m

17 17 Another example n usng the Hamlton-Jacob Method (t-dependent H) (G 10.8) Suppose we are gven a Hamltonan to a system as, p H matq where A s a constant m Our task s to solve for the equaton of moton by means of the Hamlton s prncple functon S wth the ntal condtons t0, q 0, p mv The Hamlton-Jacob equaton for S s: S S Hq; ; t 0 q t

18 18 Example n usng the Hamlton-Jacob Method Recall that the Hamlton s prncple functon S s a Type generatng functon wth ndependent varables S S q, P, t wth the condton that the canoncally transformed varables are constants,.e., Q and P For a Type generatng functon, we have the followng partal dervatves for S: qpt,, p S q,, t q (1) T Q S q,, t ( T)

19 19 Example n usng the Hamlton-Jacob Method Wrtng the Hamlton-Jacob equaton out explctly n H, we have 1 S m q matq S t 0 (*) S Hq; ; t q Here, we assume a soluton to be of the form:, ;, S q t f t q g t Substtutng ths nto the above Eq (*), we have, 1 f, t matq f ', tq g' t 0 m ( ' s tme dervatve here)

20 0 Example n usng the Hamlton-Jacob Method 1 f m t matq f ', tq g' t Concentratng on the q dependent terms, they have to be ndependently add up to zero So, we have,, 0 f ', t mat mat f t f, And, requrng the remanng two terms addng up to zero also, we have ' 1 m g t f t Substtutng f (t) from above, we have 4 m A t mat 1 g' t f mat f f m m 4 0

21 1 Example n usng the Hamlton-Jacob Method Integratng wrt tme on both sdes, we then have, 5 ma t Af0 3 f0 g t t tg 40 6 m Puttng both f (t) and g(t) back nto the Hamlton s prncple functon, we have, Snce the Hamlton-Jacob Equaton only nvolves partal dervatves of S, can be taken to be zero wthout affect the dynamcs and for smplcty, we wll take the ntegraton constant to be smply,.e., 0 f mat ma t Af Sq, ; t f q t tg 40 6 m f 5 mat ma t A 3 Sq, ; t q t t 40 6 m 0 f 0 g 0

22 Example n usng the Hamlton-Jacob Method We can then solve for usng Eq. T, Solvng for q, we then have,,, 3 S q t A Q q t t 6 m A 3 qt t t 6 m From the ntal condtons, we have Ths mples that q0 0 And, p 0,, t0 5 mat ma t A 3 Sq, ; t qq t t 40 6 m t0, q 0, p mv S q t mat mv q t0 0

23 3 Example n usng the Hamlton-Jacob Method Puttng these two constants back nto our equaton for q(t), we fnally arrve at an explct equaton of moton for the system: A 3 qt tvt o 6, Usng Eq. T agan, we have, pt,, S q t mat q We just found that mv 0, mat pt mv 0

24 4 Connecton to the Schrödnger equaton n QM (Ths connecton was frst derved by Davd Bohm.) We frst start by wrtng a quantum wavefuncton n phaseampltude form, A r, r, t A t e r, t Sr, t where and are real functons of the poston r and t. Let say ths quantum partcle s movng under the nfluence of a conservatve potental U and ts tme evoluton s then gven by the Schrodnger equaton: t m S r, t U r,t

25 5 Connecton to the Schrödnger equaton n QM Now, we substtute the polar form of the wavefunton nto the Schrodnger equaton term by term: S A : e t t t 1 S A t S : e A AS S S e S A A S e A AS S 1 S e SA A S e A AS A S A S S e A A S A S r, r, t A t e S r, t color code: Re Im color code: Re Im

26 6 Connecton to the Schrödnger equaton n QM Substtutng these terms nto the Schrodnger equaton, dvdng out the common factor S e and separatng them nto real/magnary parts, IM (black): t A 1 AS A S t m Smplfyng and : A Regroupng: A A 1 A A t m A S A t m A S A S

27 7 Connecton to the Schrödnger equaton n QM Now, for the real part of the equaton, we have, RE (blue): t U S A A A S UA t m Smplfyng: Rearrangng: S A 1 S U t m A m 1 S A S U m t m A

28 8 Connecton to the Schrödnger equaton n QM 0 Takng the lmt, we have, 1 S S U 0 m t Ths s the Hamlton Jacob equaton f we dentfy the quantum phase S of wth the Hamlton Prncpal Functon or the classcal Acton. To see that explctly, recall the Hamlton Jacob equaton s, S S S Hq1,, qn;,, ; t 0 q1 qn t

29 9 Connecton to the Schrödnger equaton n QM Now, for a partcle wth mass m movng under the nfluence of a conservatve potental U, ts Hamltonan H s gven by ts total energy: p H m U Usng the nverse canoncal transformaton, Eq. (T1), we can wrte H as, 1 S S H U 1 S U mqq m Substtutng ths nto the Hamlton-Jacob Equaton, we have, S p q 1 S S U 0 m t (Ths s the same equaton for the phase of the wavefuncton from Schrödnger Eq)

30 30 Connecton to the Schrödnger equaton n QM Notes: 1. The neglected term proportonal to s called the Bohm s quantum potental, A Q m A Note that ths potental s nonlocal due to ts spatal dffusve term and t s the source of nonlocalty n QM.. The real part of the Schrödnger equaton can be nterpreted as the * contnuty equaton for the conserved probablty densty A wth the velocty feld gven by : t v vs m p/ m A S A t m

31 31 Hamlton s Characterstc Functon Let consder the case when the Hamltonan s constant n tme,.e.,, 1 H q p Now, let also consder a canoncal transformaton under whch the new momenta are all constants of the moton, AND H s the new canoncal momentum, P 1 (the transformed Q are not restrcted a pror. ) H q, p 1 Then, we seek to determne the tme-ndependent generatng functon Wq, P (Type-) producng the desred CT.

32 3 Hamlton s Characterstc Functon Smlar to the development of the Hamlton s Prncpal Functon, snce WqP, p Q s Type-, the correspondng equatons of transformaton are W q, q W q, WqP, (1) T ( T) (Note: the ndces nsde W(q, P) are beng suppressed.) Now, snce s tme-ndependent, and we have W q, t 0 W Hq, q W t K 1

33 33 Hamlton s Characterstc Functon WqP, s called the Hamlton s Characterstc Functon and W Hq, 1 0 q s the partal dfferental equaton (Hamlton-Jacob Equaton) for W. Here, we have n ndependent constants (wth ) n determnng ths partal dff. eq. 1 H And, through Eqs T1 and T, W generates the desred canoncal transformaton n whch all transformed momenta are constants!

34 34 Hamlton s Characterstc Functon In the transformed varables,, and the EOM s gven by the Hamlton s Equatons, P Q K Q 0 Q, P K 1 or K 1, 1 0, 1 By ntegratng, ths mmedately gves, Q1 t1 Q, 1 P (as requred) where are some ntegraton constants determned by ICs. Note that s bascally tme and ts conjugate momenta s the Hamltonan. Q1 P1 1 K

35 35 Hamlton s Characterstc Functon By solvng the H-J Eq, we can obtan Then, to get a soluton for, we use the transformaton equatons (T1 & T), p W q, q q, p Q W q, W q, t, 1 1, 1 p can be drectly evaluated by takng the dervatves of W q can be solved by takng the dervatves of W on the LHR and nvertng the equaton condtons q 0, p 0 The set of n constants are fxed through the n ntal,

36 36 Hamlton s Characterstc Functon When the Hamltonan does not dependent on tme explctly, one can use ether - The Hamlton Prncpal Functon Sq, P, t or - The Hamlton Characterstc Functon W q, P to solve a partcular mechancs problem usng the H-J equaton and they are related by:,,, 1 S q P t W q P t

37 37 Acton-Angle Varables n 1dof - Often tme, for a system whch oscllates n tme, we mght not be nterested n the detals about the EOM but we just want nformaton about the frequences of ts oscllatons. - The H-J procedure n terms of the Hamlton Characterstc Functon can be a powerful method n dong that. - To get a sense on the power of the technque, we wll examne the smple case when we have only one degree of freedom. - We contnue to assume a conservatve system wth beng a constant H 1

38 38 Acton-Angle Varables n 1dof Let say we know that the dynamc of a system s perodc so that qt T qt Recall from our dscusson for a pendulum on dynamcal systems, we have two possble perodc states: lbraton rotaton p p

39 39 Acton-Angle Varables n 1dof - Now, we ntroduce a new varable J pdq called the Acton Varable, where the path ntegral s taken over one full cycle of the perodc moton. - Now, snce the Hamltonan s a constant, we have, 1 H q p - Then, by nvertng the above equaton to solve for p, we have p p q, 1

40 40 Acton-Angle Varables n 1dof - Then, we ntegrate out the q dependence n J pq, dq One can then wrte that J s a functon of alone or vce versa, - Now, nstead of requrng our new momenta to be, we requres 1 1 H H J P - Then, our Hamlton Characterstc Functon can be wrtten as 1 P 1 J (another constant nstead of 1 ) W W q, J

41 41 Acton-Angle Varables n 1dof - From the transformaton equatons, the generalzed coordnate Q correspondng to P s gven by Eq. T - Enforcng our new momenta P to be J and callng ts conjugate coordnate, we have w W Q P W w J - Here, n ths context, s called the Angle Varable. w

42 4 Acton-Angle Varables n 1dof - Note, snce the generatng functon W(q, P) s tme ndependent, the Hamltonan n the transformed coordnate K = H so that 1 H HJ KJ - Correspondngly, usng the Hamlton Equatons, the EOM for s, w K J J vj v J - Now, snce K s a constant, wll also be a constant functon of J. w

43 43 Acton-Angle Varables n 1dof The above dff. eq. can be mmedately ntegrated to get w v J where s an ntegraton constant dependng on IC. wvt - Thus, ths Angle Varable w s a lnear functon of tme!

44 44 Acton-Angle Varables n 1dof Now, let ntegrate w over one perod T of the perodc moton. w wdw dq q Usng the transformaton equaton,, we then have, T T W w J W w qj T dq

45 45 Acton-Angle Varables n 1dof Snce J s a constant wth no q dependence, we can move the dervatve wrt J outsde of the q ntegral, W d W w dq dq Jq dj q T T W p q Then, usng the other transformaton equaton,, we can wrte, d dj w pdq 1 dj dj T J pdq (recall )

46 46 Acton-Angle Varables n 1dof w 1 Ths result means that w changes by unty as q goes through a complete wvt perod T. Then, snce we have, we can wrte wwt () w(0) vt1 1 v J T - gves the frequency of the perodc oscllaton assocated wth q! - and t can be drectly evaluated thru v J K J J wthout fndng the complete EOM (*)

47 47 Acton-Angle Varables n 1dof: Example Let consder a lnear harmonc oscllator gven by the Hamltonan, H Solvng for p, we have, p m m q Moton s perodc n q s the const total E km s the natural freq p mm q Then, we can substtute ths nto the ntegral for the Acton Varable, J m m q dq T

48 48 Acton-Angle Varables n 1dof: Example Usng the followng coordnate change, q sn m The ntegrand can be smplfed, mm q dq mm sn cos d m m m 1 s c m n osd m cos cosd cos d m

49 49 Acton-Angle Varables n 1dof: Example Puttng ths back nto our ntegral for the Acton Varable, we have, Ths gves, J 0 J cos d Invertng to solve for n terms of J, we have, J HK

50 50 Acton-Angle Varables n 1dof: Example Fnally, applyng our result for usng Eq. (*), we have v Thus, K J v J J v 1 k m Here, we have derved the frequency of the lnear harmonc oscllator by only calculatng the Acton Varable wthout explctly solvng for the EOM!