正则变换
正则变换
中文书P273:8.8
In a gravitational field of acceleration g , a particle of mass m falls down vertically （see the figure ）.
Take the height y above ground as the generalized coordinate of the particle, the Hamiltonian of the particle is
),(y p y H mgy m
p y
+=
22
If the generating function ),(1Q q F of canonical transformation from old variables
),(),(y p y p q = to new variables ),(),(θθp P Q = is taken as
=),(1Q q F )6/(),(31θθθy g mg y F +-=
1) Find the new Hamiltonian ),(*
P Q H
of the particle;
2) Give the canonical equations of Hamilton of new canonical variables ),(P Q ;
3) Find the solution of )(t y in terms of the solutions of ))(),((t P t Q under the initial conditions
h y t ==0, 00
mv p t y
-==.
【Hint 】When the generating function ),(1Q q F from old variables ),
(p q to new variables
),(P Q is given, the equations of canonical transformation are
???
???
?
??-=??=Q Q q F P q Q q F p ),(),(11
【Solutions 】
1) Find the new Hamiltonian ),(*
P Q H
of the particle;
When the generating function =),(1Q q F )6/(),(31θθθy g mg y F +-=
is given, the equations of canonical transformation are
???
???
???-=??=Q Q q F P q Q q F p ),(),(11 y
m
哈密顿正则方程
Where ),(),(y p y p q =, ),(),(θθp P Q =,
i.e.
???
???
?+=??-=-=??=)2/(),(),(211θθθθθθg y mg y F p mg y y F p y i.e.
??
???-=-=2/2
θθθg mg p y mg p y The new Hamiltonian ),(*
P Q H is
)
,(),,()
,(),(θθθθθθp p p p y y y y y p y H p H ==*=
)2/(2)(22
θθθg mg
p mg m mg -+=θp =
2) Give the canonical eqs. of Hamilton of new canonical variables ),(P Q ；
根据哈密顿正则方程
???
??????-=??=
Q H P P H Q *
*
其中，),(),(θθp P Q =, θθθp p H =*
),(.
→??
?==0 1θθ
p
These are just the eqs. of ))(),((t P t Q .
3) Find the solution of )(t y in terms of the solution of ))(),((t P t Q under the initial
conditions h y t ==0
, 00
mv p t y -==.
The solutions of ))(),((t P t Q are
??
?=+=β
αθθ)(
)(t p t t 其中两个待定常数βα,由初始条件决定。由于
??
???-=-=2/2
θθθg mg p y mg p y 故),(),
(y p y p q =的解为
??
???+-=+-=2)(21)()()(αβαt g mg t y t mg t p y 由初始条件h y t ==0, 00
mv p t y -==可定出待定常数βα,，
??
???+==g v h mg g
v 2/200βα 所以),(),
(y p y p q =的最终解为
??
???+-=+-+
=+-=+-=)21()(212)()()/()(20202000gt t v h g v t g g v h t y gt v m g v t mg t p y
中文书P273:8.10
Prove that the following transformations from old variables
),(p q to new variables
),(P Q
????
?==P
kQ p P k Q q sin 2cos /2
are canonical transformations (where k is a constant); If the old Hamiltonian is
)(2
1),(222
q k p p q H += , find the new Hamiltonian ),(*P Q H ，set up the equations
of motion of new canonical variables ),(P Q , give the solution of ),(p q .
【解】如果能够找到上述变换的生成函数，则说明变换是正则变换。因为),(P q 可以自由变化，可作自变量，相应的生成函数为),(2P q F ，下面看),(2P q F ＝？
当生成函数为),(2P q F 时，正则变换方程为
???
?
??
?=??==??=),( 22P q Q P F Q p(q,P)q F p
将具体变换方程?????==P
kQ p P
k Q q sin 2cos /2写成自变量),(P q 的函数，得
??
???=
=P kq P q Q P qk P q p 22cos 2),(tan ),(，
即有
???????=??=??P
kq P F P qk q
F 22
22
cos 2 tan 由于生成函数),(2P q F 的微分为
dP P F dq q F dF ??+??=222dP P
kq Pdq qk 2
2
cos 2tan +=
)tan 2
()(tan 2)2(tan 222P kq d P d kq q Pd k =+=
故变换母函数为P kq P q F tan 2
),(2
2=，能够找到变换的生成函数，说明变换是正则变
换！
新哈密顿量为
),(*P Q H )
,(),,(2),(,(P Q p p P Q q q t P q F p q H ==?
?? ??
??+= ),(),,(222)
(2
1
P Q p p P Q q q q k p ==+=
?????==P
kQ p P
k Q q sin 2cos /2 ， 所以
),(*P Q H kQ =
根据哈密顿正则方程
???
??????-=??=Q H P P H Q *
*
得新正则变量),(P Q 的方程为
???-==k
P Q
0 ),(P Q 解为α=)(t Q ，β+-=kt t P )(，其中两个待定常数βα,由初始条件决定。
由于
????
?==P
kQ p P
k Q q sin 2cos /2
故),
(p q 的解为
????
?-=-=)
sin(2)
cos(/2kt k p kt k q βαβα
【证法二】如果能证明变换后的变量P Q ,满足如下基本泊松括号
??
?
??===1],[0],[0],[P Q P P Q Q 就能证明变换是正则变换。将具体变换方程?????==P
kQ p P
k Q q sin 2cos /2写成显式，有
??
??
?=+
=)/(tan 22122qk p P k p kq Q
注意到泊松括号性质
[]0,=u u ，显然有
??
?==0
],[0
],[P P Q Q
而
]arctan ,221[],[2
2???
? ??+=qk p k p kq P Q
利用泊松括号的线性性
∑∑=j
j j j
j j v u b v b u ],[],[
有
]arctan ,[21]arctan ,[21],[22qk
p
p k qk p q k P Q +=
再利用 [][][]w v u w u v vw u ,,,+=
有
]arctan ,[]arctan
,[],[qk
p
p k p qk p q kq P Q += 注意到
[]q
G G p ??-=,， []p
G G q ??=,
有
q
qk p k p p qk p kq P Q ??
??? ???-????? ?
??=arctan arctan ],[
令qk p x arctan =，则qk
p
x =tan ，注意到
qk x p x x p x x x p x p qk p 2
2cos )(tan cos )(tan )(tan arctan =??=????=??=????? ??? x k q p q x x q x x x q x q qk p 222cos )(tan cos )(tan )(tan arctan ???
? ??-=??=????=??=????? ??? 所以
x k q p k p qk x kq P Q 2
22cos cos ],[???
? ??+= 即
x qk p P Q 2
2cos 1],[???
??????
??? ??+=1cos )tan 1(22=+=x x 即变换后的变量满足基本泊松括号，证明变换?????==P
kQ p P
k Q q sin 2cos /2是正则变换。
中文书P273:8.11
Consider the transformation from old variables ),(p q to new variables ),(P Q , whose
equations of transformation are
?????==p
q P p
q Q ββα
α
sin cos where αand β are two constants.
(1) Using the fundamental Poisson brackets, find the values of αand β for which above
transformation represents a canonical transformation.
(2) Using the values of α and β you found, find the generating function （生成函数）
),(3Q p F of above canonical transformation in terms of equations of canonical transformation
???
???
???-=??-=Q Q p F P p Q p F q ),(),(33 Hint: 错误！未找到引用源。 The Poisson bracket of two functions U(q,p ) and V(q,p ) is
defined as
q
V
p U p V q U V U ????-
????=
],[
② x
dx
x d 2
cos )
(tan =
【解】(1)如果上述变换为正则变换，要求变换后变量),(P Q 满足以下基本泊松括号
0],[=Q Q 0],[=P P
1],[=P Q
即
1
211)(sin )(sin )(cos )(cos ],[1---=+=????-
????==ααααααββαβββββαq p q p q p q p q q
P
p Q p P q Q P Q 要使上式对所有的q 成立，要求
??
?=-=0
121
ααβ
即
??
?==2
2
/1βα
(2) Find the generating function ),(3Q p F
为求变换的母函数),(3Q p F ，将已知的变换关系
?????====p
q p q P p
q p q Q 2sin sin 2cos cos 2
/12
/1ββαα 写成自变量),(Q p 的函数，有
??
?==)
2tan(),()
2(cos /),(22p Q Q p P p Q Q p q
即
???
???
?=??-==??-=)2tan(),()2(cos /),(32
23p Q Q Q p F P p Q p Q p F q
而母函数),(3Q p F 的微分为
dQ p Q dp p Q dQ Q F dp p F dF )2tan()
2(cos 2
2
333--=??+??= 即
[]??
????-=??????--=)2tan(22)2tan()2tan(2222
3p Q d Q d p p d Q dF 故母函数为
)2tan(2
),(2
3p Q Q p F -=