Mechanics
Physics 151
Lecture 22
Canonical Transformations
(Chapter 9)
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Mechanics Canonical Transformation 3, Lecture Notes - Physics, Study notes of Mechanics
Mechanics Physics Canonical Transformations Infinitesimal CT Poisson Bracket Hamilton’s equations Momentum Conservation Angular Momentum Integrating ICT Rotation CT Free Fall Harmonic Oscillator Liouville’s Theorem
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Mechanics
Physics 151
Lecture 22
Canonical Transformations
(Chapter 9)
What We Did Last Time ^
Direct Conditions^
Necessary and sufficientfor Canonical Transf. ^
Infinitesimal CT
^
Poisson Bracket^
Canonical invariant Fundamental PB ICT expressed by Infinitesimal time transf. generated by Hamiltonian^ Æ
Hamilton’s equations
,
,
j
i j^
i^ Q P
q p
p
Q q^
P
⎛^
⎞^
∂⎛
⎞
∂^
=
⎜^
⎟^
⎜^
⎟
⎜^
⎟ ∂^
∂⎝
⎠
⎝^
⎠^
,
,
j
i j^
i^ Q P
q p
q
Q p^
P
⎛^
⎞^
∂⎛
⎞
∂^
= −
⎜^
⎟^
⎜^
⎟
⎜^
⎟ ∂^
∂⎝
⎠
⎝^
⎠
,
,
j
i j^
i^ Q P
q p
p
P q^
Q
⎛^
⎞^
∂⎛
⎞
∂^
= −
⎜^
⎟^
⎜^
⎟
⎜^
⎟ ∂^
∂⎝
⎠
⎝^
⎠^
,
,
j
i j^
i^ Q P
q p
q
P p^
Q
⎛^
⎞^
∂⎛
⎞
∂^
=
⎜^
⎟^
⎜^
⎟
⎜^
⎟ ∂^
∂⎝
⎠
⎝^
⎠
[^
] ,
i^
i^
i^
i
u^
v^
u^
v
u v
q^
p^
p^
q
∂^
∂^
∂^
≡^
∂^
∂^
∂^
[^
,^
]^
[^
,^
]^
i^
j^
i^
j
q^
q^
p^
p
=^
=^
[^
,^
]^
[^
,^
]
i^
j^
i^
j^
ij
q^
p^
p^
q
δ
=^
−^
[ ,
]^
u
u^
u G
t t
δ^
=^
Dynamic View of CT ^
A system evolves with time^
At any moment,
q
and
p satisfy Hamilton’s equations
^
The time-evolution must be a Canonical Transformation!^
Static View = Coordinate system is changing Dynamic View = Physical system is moving
0
0
(^
),^
(^
q t
p t
q t
p t
q
p
This movement
is a CT
Infinitesimal Time CT ^
Infinitesimal CT^
We know that the generator = Hamiltonian Integrating it with time should give us the “finite” CT thatturns the initial conditions
q
( t^0
),^
p (
t ) into the configuration^0
q (
t ),
p
( t ) of the system at arbitrary time ^
That’s a new definition of “solving” the problem
q t
p t
(^
),^
(^
q t
dt
p t
dt
+^
[ ,
]^
u
du
dt u H
dtt ∂
=^
∂^
[ ,
]
q^
q H ^ =
[^
,^
]
p^
p H =
Hamiltonian is the generator of the system’s
motion with time
Hamiltonian is the generator of the system’s
motion with time
Conservation ^
Consider an ICT generated by
G
^
Suppose
G
is conserved and
has no explicit
t -dependence
^
How is
H
(without
t -dependence) changed by the ICT?
^
A transformation that does not affect
H
Æ
Symmetry of the system
Æ
Generator of the transformation is conserved
[ ,
]^
u
u^
u G
t t
δ^
=^
[^
,^
]^
G H
[^
,^
]^
H
H
H G
t t
δ^
=^
+^
If an ICT does not affectHamiltonian, its generator
is conserved
Momentum Conservation ^
Simplest example:^
What is the ICT generated by momentum
p
? i
^
That’s a shift in
q
by i
ε^ Æ
spatial translation
^
If Hamiltonian is unchanged by such shift, then Æ
Momentum
p
is conserved i
^
This is not restricted to linear momentum
[^
,^
]
j^
j^
i^
ij
q^
q^
p
δ^
=^
=^
[^
,^
]^
j^
j^
i
p^
p^
p
δ^
=^
[^
,^
]^
i H p
Hamiltonian isunchanged by a shift of acoordinate
q
The generatorof the ICT isthe conjugatemomentum
p
p^
is conserved
[^
,^
]^
H
p
Angular Momentum ^
The generator
is obviously
^
i.e. the
z -component of the total momentum
^
Generator for rotation about an axis given by a unit vector
n
should be
^
We now know generators of 3 important ICTs^
Hamiltonian generates displacement in time Linear momentum generates displacement in space Angular momentum generates rotation in space
i^
iy^
i^
ix
G
x p
y p
=^
−^
(^
z^
i^
i^ z
L^
=^
×
r^
p
G
L n
Integrating ICT ^
I said we can “integrate” ICT to get finite CT^
How do we integrate
^
First, let’s rewrite it as^
We want the solution
u
(^ α
) as a function
of
with the initial condition
u
u 0
^
Taylor expand
u
(^ α
) from
α^
[ , = 0
]
u^
u G
δ^
[ ,
]
du
d^
u G α =^
[ ,
]
du
u G
d^ α
2
2
3
3
0
2
3
0
0
0
(^
)^
du
d u
d u
u^
u^
d^
d^
d
α
α
α
α^
α
α
α
=^
+^
+^
+^
This is
[ u
, G
]^0
What can I do with these?
Rotation CT ^
Let’s integrate the ICT for rotation around
z
^
Let me forget the particle index
i
^
Parameter
α^
is
θ^
in this case
^
Let’s see how
x changes with
θ
^
Evaluate the Poisson Brackets ^
Where does this lead us?
y^
x
G
xp
yp
=^
2
3
0
0
0
0
[ ,
]^
[[ ,
],^
]^
[[[ ,
],^
],^
]
x^
x^
x G
x G
G
x G
G
G
θ
θ
θ
θ
=^
+^
+^
+^
[ ,
]
x G
y = −
[[ ,
],^
]
x G
G
x =^
−^
[[[ ,
],^
],^
]
x G
G
G
y
[[[[ ,
],^
],^
],^
]
x G
G
G
G
x =^
Repeats after this
Rotation CT
^
Similarly
2
3
0
0
0
0
2
3
4
0
0
0
0
0
2
4
3
5
0
0
0
0
[ ,
]^
[[ ,
],^
]^
[[[ ,
],^
],^
]
cos
sin
x^
x^
x G
x G
G
x G
G
G
x^
y^
x^
y^
x
x^
y
x^
y
θ
θ
θ
θ
θ^
θ^
θ
θ
θ^
θ
θ^
θ
θ
θ
θ
=^
+^
+^
+^
=^
−^
−^
+^
+^
⎛^
⎞^
⎛^
=^
−^
+^
−^
−^
−^
+^
⎜^
⎟^
⎜^
⎝^
⎠^
⎝^
=^
2
3
0
0
0
0
0
0
[ ,
]^
[[ ,
],^
]^
[[[ ,
],^
],^
]
cos
sin
y^
y^
y G
y G
G
y G
G
G
y^
x
=^
+^
+^
+^
=^
Infinitesimal Rotation ^
ICT for rotation is generated by^
We’ve studied infinitesimal rotation in Lecture 8 Infinitesimal rotation of
d
θ^
about
n
moves a vector
r as
^
Compare the two expressions
^
Equation
holds for any
r
that rotates
together with the system^
Several useful rules can be derived from this
G
L n
d^
d^ θ
=^
×
r^
n^
r
[ ,
]
d^
d^
d
=^
⋅^
=^
×
r^
r L n
n^
r^
[ ,
]
⋅^
=^
×
r L n
n^
r
[ ,
]
⋅^
=^
×
r L n
n^
r
Scalar Products ^
Consider a scalar product
of two vectors
^
Try to rotate it ^
Obvious: scalar product doesn’t change by rotation ^
Also obvious: length of any vector is conserved
⋅ a b
[^
,^
]^
[ ,
]^
[ ,
]
(^
)^
(^
(^
)^
(^
⋅^
⋅^
=^
⋅^
⋅^
+^
⋅^
=^
⋅^
×^
+^
⋅^
×
=^
⋅^
×^
+^
⋅^
×
a b L n
a^
b L n
b^
a L n
a^
n^
b^
b^
n^
a
a^
n^
b^
a^
b^
[ , n
]
⋅^
=^
×
r L n
n^
r
Angular Momentum ^
Imagine two conserved quantities
A
and
B
^
How does [
A ,
B ] change with time?
^
Poisson bracket of two conserved quantities is conserved
^
Now consider^
If 2 components of
L
are conserved, the 3
rd^
component must
Æ
Total vector
L
is conserved
[^
,^
]^
[^
,^
]^
A H
B H
=^
[[^
,^
],^
]^
[[^
,^
],^
]^
[[^
,^
],^
]^
A B
H
B H
A^
H A
B
=^
−^
−^
Jacobi’s identity
[^
,^
]
i^
j^
ijk^
k
L^
L^
L
Angular Momentum ^
Remember the Fundamental Poisson Brackets?^
Now we know Poisson brackets between
L
,^ x Ly
,^ L
are non-zero z
^
On the other hand,
, so |
L | may be a canonical
momentum ^
QM: You may measure |
L | and, e.g.,
L
simultaneously, but z
not
L
and x
L
, etc. y
[^
,^
]^
[^
,^
]^
i^
j^
i^
j
q^
q^
p^
p
=^
=^
[^
,^
]^
[^
,^
]
i^
j^
i^
j^
ij
q^
p^
p^
q
δ
=^
−^
PB of two canonical momenta is 0[^
,^
]
i^
j^
ijk^
k
L^
L^
L
[^
,^
]^
i L^
L^
Only 1 of the 3 components of the angular momentum
can be a canonical momentum