1 PHYS 705: Classcal Mechancs Canoncal Transformaton II
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 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 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 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 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 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 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 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 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 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
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 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 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 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 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 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 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 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!
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
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,
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
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
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
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)
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
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,
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
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 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 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
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 0 0 0 (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 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 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 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