PHYS 705: Classical Mechanics. Canonical Transformation II
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1 1 PHYS 705: Classcal Mechancs Canoncal Transformaton II
2 Example: Harmonc Oscllator f ( x) x m 0 x U( x) x mx x LT U m Defne or L p p mx x x m mx x H px L px p m p x m m H p 1 x m p m 1 m H x p m x m m (n x & p)
3 3 Example: Harmonc Oscllator f ( x) x m 0 x U( x) 1 H p m x m x Applcaton of the Hamlton s Equatons gve, x H p H x p m m m p xm x Combnng the two equatons, we have the standard equaton for SHO: x x 0 Notce that the system s not cyclc wth the orgnal varable x. We wll attempt to fnd a canoncal transformaton from (x, p) to (X, P) such that X s cyclc so that P s constant.
4 4 Example: Harmonc Oscllator Wth the form of H as sums of squares, we wll try to explot Let try sn cos 1 1 H p m x m g P p gpcosx x sn X (*) m Substtutng these nto our Hamltonan, we have X s cyclc n K! 1 g P g P K g Pcos X m sn X m m m Now, we need to fnd the rght g(p) such that the transformaton s canoncal. Try the followng type 1 generatng functon: F F( x, X, t) (wth x and X beng the ndep vars)
5 5 Example: Harmonc Oscllator As n prevous examples, we wll try to wrte p (now the dep var for F) n terms of the (ndep var) (x, X) by dvdng the two equatons n CT: p gp x cos X gp sn X 1 m m cot X For F to be a canoncal transformaton, t has to satsfy the partal dervatve relatons (Type 1 n Table 9.1 n G): F p P x F X The frst partal dervatve equaton (left) gves: F m x p m xcot X F F( x, X, t) cot X x or p m xcot X
6 6 Example: Harmonc Oscllator m x F F( x, X, t) cot X Then, substtutng ths tral soluton of F nto the nd partal dervatve relaton: P F mx 1 mx 1 P X sn X sn X Solvng for x (old) n term of (new), we have: X, P x P sn m X x P sn m X
7 7 Example: Harmonc Oscllator To get the second transformaton for p (old) n terms of (new), we can p m xcot X use our prevous relaton and our newly derved eq: X, P x X, P x P sn m X p m xcot X P p m sn X cot X mpcos X m p m PcosX
8 8 Example: Harmonc Oscllator Snce we explctly calculated ths par of transformatons usng the approprate generatng functon, they are canoncal:,, px, P x X P x P sn m X p m PcosX By comparng wth our orgnal transformaton equaton (Eq. (*)), we have: g P x sn X m p gpcos X gves Then, the transformed Hamltonan s: K m P (*') g P g P m P P m m
9 9 Example: Harmonc Oscllator Wth the transformed Hamltonan, the Hamlton s Equatons gve, K P E K P 0 X P const (X s cyclc) K X P X t depends on IC **Notce how smple the EOM are n the new transformed cyclc varables. In most applcaton, the goal s to fnd a new set of canoncal varables so that there are as many cyclc varables as possble. **P s ust the rescaled total energy E and X s ust a rescaled and shfted tme. P, X ** s an example of an Acton-Angle par of canoncal varables.
10 10 Example: Harmonc Oscllator Usng the nverse transform: x P sn m X p m PcosX The EOM n X, P can be wrtten n terms of the orgnal varable: X t P const x P sn m t p m Pcost
11 11 Example: Harmonc Oscllator Then wth K P E E x sn m t p mecost X t P const p P me x E m Phase Space area s nvarant! E area E X
12 1 Symplectc Approach & Posson Bracet Let consder a tme-ndependent canoncal transformaton: Now, let consder the full tme dervatve of : Q Q q, p P P q, p (We assume the transformaton not explctly dependng on tme ) dq Q Q Q q p (E s sum rule over ) Q dt q p Usng the Hamlton s equatons, we can rewrte the equaton as: Q Q q H p Q p H q (*) (E s sum rule over )
13 13 Symplectc Approach & Posson Bracet Now, f the transformaton s canoncal, we also must have K P (Note: K H F t H snce canoncal trans does not dep on tme explctly) Q q q ( Q, P), p p ( Q, P) Usng the nverse transform, we can consder H (the orgnal Hamltonan) as a functon of Q and P,.e, H P,,, H H q Q P p Q P Then, we can evaluate the RHS of the equaton for n the 1 st lne as: Q Q H H q H p P q P p P (**)
14 14 Symplectc Approach & Posson Bracet Comparng the two expressons for, Eq. (*) and (**), Q Q q H Q H p p q (*) we must have the followng equaltes for all pars: Q p (subscrpts denote the functonal dependence of the quanttes) Q q P H p H q P p AND q P qp, p P QP, qp, QP, Q q (**) Smlar consderatons for P wll gve a dfferent but smlar pars of relatons: P p Q P q AND q Q qp, p Q QP, qp, QP,,
15 15 Symplectc Approach & Posson Bracet Q p q P Q q p P qp, QP, qp, QP, P p q Q P q p Q qp, QP, qp, QP, These four condtons for varous are called the drect condtons for a canoncal transformaton.,
16 16 Posson Bracet For any two functon and dependng on q and p the Posson Bracet s defned as: uq, p vq, p uv,, qp u v u v q p p q (E s sum rule for n dof) PB s analogous to the Commutator n QM: 1 1 uv, uvvu where u and v are two QM operators
17 17 Some Basc Property of Posson Bracets General propertes for Posson bracets wth respect to an arbtrary par of varables (x, y) (not necessarly canoncal varables): uu, 0 uv, vu, au bv, wau, wbv, w uv, wu, wv uv, w u v w v w u w u v,,,,,, 0 (the x, y subscrpts are suppressed here) (ant-symmetrc) (lnearty) (dstrbutve) (Jacob s Identty),, uvware some arbtrary functons of and a & b are some constants. xy,
18 18 Symplectc Approach & Posson Bracet Now, f we substtute the canoncal varables themselves as (u, v) & (x, y), we have: q, q p, p 0 qp, qp, q, p p, q qp, qp, (hw) These are called the Fundamental Posson Bracets. ( q, p) ( Q, P) And, under a canoncal transformaton, we can also show that: Q, Q P, P 0 qp, qp, Q, P P, Q qp, qp, The Fundamental Posson Bracets are nvarant under a CT!
19 19 Fundamental Posson Bracets & CT (E s sum rule on ndex) Q Q Q Q Q, Q q qp, q p p P P P P P, P q qp, q p p Q q Q p Q q P p P P P q P p P q Q p Q Q 0 0 Q P Q P Q, P q qp, q p p Q q Q p Q q Q p Q Q P, Q Q, P qp, qp, Usng the Drect Condtons of the CT The Fundamental Posson Bracets are nvarant under a CT!
20 0 Symplectc Approach & Posson Bracet A general Posson Bracet s nvarant under a CT: (tme-nvarant case)! u, v, QP u Q v P u P uv, qp, q, p Q, P v Q u q u p v q v p q u u p v q v p q Q p Q q P p P q P p P q Q p Q multply out and regroup u v, u v, u v, u v q q q p p q p, p q q q p p q p QP, QP,, QP, QP p u v u v u v u v q p p q q p p q uv,, qp
21 Symplectc Approach & Posson Bracet The nvarant of the Posson bracets under the canoncal varables CT defnes the symplectc structure of the canoncal transformaton. uv, uv, QP, qp, Q, P q, p qp, QP, Thus, n general, f we are calculatng a Posson Bracet wth respect to a par of canoncal varable, we can drop the q,p subscrpt. uv, uv, qp,
22 Symplectc Approach & Posson Bracet Note that ths symplectc structure for the canoncal transformaton can also be expressed elegantly usng the matrx notaton that we have ntroduced earler : Recall that the Hamlton Equatons can be wrtten n a matrx form, H η J η wth q, p ; 1,, n n H H H H, ; η q η p n and J 0 I I 0
23 3 Symplectc Approach & Posson Bracet ζ Now, let the n-column vector correspondng to the new set of canoncal coordnates Q, P And, we can wrte a transformaton from (q, p) (Q, P) as a vector equaton: Then, by chan rule, the tme dervatves of the varables are related by: ζ Mη ζ ζ η where M,, 1,,n (Jacoban matrx) Substtutng the Hamlton Equaton bac nto above, we have: H ζ MJ η η ζ
24 Symplectc Approach & Posson Bracet Consderng the nverse transformaton (Q, P) (q, p) and by chan rule, we can rewrte the rght most factor as: or, n matrx notaton: sum over T H ζ MJM ζ H H T H H M η ζ Puttng ths bac nto our dynamc equaton, we have, ζ η M Prevously, Here, M H H 4
25 5 Symplectc Approach & Posson Bracet Now, f the coordnate transformaton s canoncal, then the Hamltonan equaton must also be satsfed n ths new set of coordnates: H ζ J ζ ζ ζ η Comparng ths equaton wth our prevous equaton: ζ ζ η T ζ MJM ζ We can see that n order for to be a vald canoncal H transformaton, we need to have M satsfyng the followng condton: MJM T J (ths condton s typcally easer to use than the drecton condton for a CT)
26 6 Symplectc Approach & Posson Bracet In terms of these matrx notaton, we can also wrte the Posson bracet as, uv, η u J η v η ζ ζ η And f s a canoncal transformaton, then Fundamental Posson Bracets can smply wrtten as,,, ζζ ηη J η ζ
27 7 Posson Bracet & Dynamcs Consder some physcal quantty u as a functon of the PS varables : - Its full tme dervatve s gven by: u u( q, p, t) du u u u q p dt q p t - Applyng the Hamlton s Equatons:, q H p p H q du u H u H u dt q p p q t uh, qp,
28 8 Posson Bracet & Dynamcs So, we arrved at the followng general equaton of moton for u(t): u du u u, H dt t We have explctly wrtten ths equaton wthout reference to any partcular set of canoncal varables to emphasze the fact that the form of the equaton s nvarant to any canoncal transformatons Obvously, we need to use the approprate Hamltonan n the gven coordnates. Applyng the above equaton wth u = H, we have and: dh dt H t H, H 0
29 9 Posson Bracet & Dynamcs General Comments: u q or p 1. If, we get bac the Hamlton s Equatons: H q q, H p and du. If u s a constant of moton,.e., 0 dt du u u u, H uh, dt t H p p, H q u t u t Specfcally, for u explctly not depends on tme,.e.,, du dt 0 uh, 0 0 (hw)
30 30 Posson Bracet & Dynamcs The PB s analogous to the Commutator n QM: CM uh, uh, (Posson Bracet) (u s some quantum observable here) The Hesenberg equaton of moton for u n QM s gven by: 1 QM (QM Commutator) du (assumer u has no u, H explct tme dt u dependence,.e., 0) t The statement on the vanshng of the Posson bracet for u beng a constant of moton n CM (a conserved quantty) corresponds to the statement that u commute wth H n QM : uh, 0
31 31 Posson Bracet & Dynamcs 3. Posson s Theorem -Let u, v be two constants of moton,.e., uhvh uv,,, 0 Then, s another constant of moton! - Now, let Taylor s expand u(t) around u(0) for t small: (Jacob s Identty) Chec: uv,, H H, uv, u, vh, v, Hu, 4. Assume that u does not explctly depend on tme so that du dt u, v, H v, uh, uh, (*) du t d u ut () u (0) t dt! dt 0 0 0
32 3 Posson Bracet & Dynamcs - Usng Eq (*), we have du dt 0 uh, 0 (evaluated at t = 0) And, du d du uh,, H dt dt dt (evaluated at t = 0) - Substtutng these two terms and smlar ones nto our Taylor s expanson for u(t), we have t ut () u(0) tuh, uh,, H 0 0!
33 33 Posson Bracet & Dynamcs t ut () u(0) tuh, uh,, H 0 0! So, one can formally wrte down the tme evoluton of u(t) as a seres soluton n terms of the Posson bracets evaluated at t = 0! The Hamltonan s the generator of the system s moton n tme! Ths has a drect correspondence to the QM nterpretaton of H: The above Taylor s expanson can be wrtten as an operator eq: Ht ˆ ut () e u(0) u H 0 where Hu ˆ 0, ut e u Ht/ () (0) (QM propagator)
34 34 Posson Bracet & Dynamcs A smple example n applyng to a freely fallng partcle: u z H p m mgz z zh, t zt () z(0) tzh, zh,, H 0 0! Now, we evaluate the dfferent PBs: z H z H p z p p z m zhh zh,, H 1 zh,, z mgg z p p z m zh,, H, H 0
35 35 Posson Bracet & Dynamcs p H m At, we have ntal condtons: 0 zh, 0 We then have, p m mgz z t 0 z0 z, p0 p Substtutng them nto the expanson, zh,, H g zh H H 0 0 t zt () z(0) tzh, zh,, H 0 0! () p m g 0 zt z0 t t 0 0,,, 0
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