Transcript No Slide Title

Lecture 14: Angular Momentum-II.
The material in this lecture covers the following in Atkins.
Rotational Motion
Section 12.7 Rotation in three dimensions
Lecture on-line
Angular Momentum-II (3- D) (PDF)
Angular Momentum-II (3-D) (PowerPoint)
Handout for this lecture (PDF)
Tutorials on-line
Vector concepts
Basic Vectors
More Vectors (PowerPoint)
More Vectors (PDF)
Basic concepts
Observables are Operators - Postulates of
Quantum Mechanics
Expectation Values - More Postulates
Forming Operators
Hermitian Operators
Dirac Notation Use of Matricies
Basic math background
Differential Equations
Operator Algebra
Eigenvalue Equations
Extensive account of Operators
Extensive account of Operators
Audio-visuals on-line
Rigid Rotor (PowerPoint)
(Good account from the Wilson Group,****)
Rigid Rotor (PDF)(Good account from the
Wilson Group,****)
Slides from the text book
(From the CD included in Atkins ,**)
Classical Angular Momentum
Angular momentum in classical physics
Consider a particle at the position r
Review of classical physics
Position and velocity in 3D
v
r
k
i
j
Where
r = ix + jy + kz
The velocity of this particle is given by
dr
dx
dy
dz
v = dt = i dt + j dt + k dt
Classical Angular Momentum
The linear momentum of the particle with mass m is
given by
dx
p = mv where e.i p x = mvx = m dt
The angular momentum is defined as
Review of classical physics
Angular Momentum in 3D
L = rXp
L
|r| |p| sin

p

r
L =rXp
The angular momentum is perpendicular to the plane
defined by r and p.
Classical Angular Momentum
Review of classical physics
We have in addition
Angular Momentum in 3D
L = rXp = (ix + jy + kz)X (ip x + jp y +kpz)
L = (r ypz -rzpy)i + (rzpx -rxpz)j + (rxpy - rypx)k
or
i
rXp =
rx
px
j
k
ry
rz
py
pz
Classical Angular Momentum
Why are we interested in the angular
momentum
?
Review of classical physics
Angular Momentum in 3D
Consider the change of L with time
F
dL
dr
dp
=
Xp
+
rX
dt
dt
dt
dL
dp
=
mvXv
+
rX
dt
dt
r
dp
= rX dt
dL
d
dr
d2 r
dt = rX dt [m dt ] = rXm dt 2
m
d2r
2
dt
 F (Newtons Law)
Classical Angular Momentum
Review of classical physics
Angular momentum and
central force in 3D
dL
 r F
dt
F
r
For centro-symmetric systems in which
the force works in the same direction as r
we must have
dL
dt = 0 : THE ANGULAR
MOMENTUM IS CONSERVED
Classical Angular Momentum
Review of classical physics
Angular momentum and
r
Examples :
F central force in 3D
movement of electron around nuclei
movement of planets around sun
For such systems L is a constant of motion, e.g. does
not change with time since
dL
dt = 0
In quantum mechanics an operator O representing a
constant of motion will commute with the Hamiltonian
which means that we can find eigenfunctions that are
both eigenfunctions to H and O
Quantum mechanical representation of angular momentum
operator
Rotation..Quantum Mechanics 3D
We have
L = rXp = iLx + jLy + kLz
where
Angular momentum operators
of quantum mechanics
in 3D
Lx = r y pz - rzpy ; Ly = rzpx - rx py ; Lz = rx py - ry px
In going to quantum mechanics we have
x --> x
; y --> y ; z --> z

px --> -i x


; py --> -i y ; pz --> -i z
Thus :




Lx = -i (y z - z y ) ; L y = -i (z x - x z )


L z = -i (x y - y x )
Rotation..Quantum Mechanics 3D
We have
Angular momentum operators
of quantum mechanics
in 3D
L = iLx + jLy + kLz
thus
L.L = L 2 =(iLx + jLy + kLz).(iLx + jLy + kLz)
L 2 = L x 2 + L y 2 + L z2
Rotation..Quantum Mechanics 3D
Can we find common eigenfunctions to
L2 , Lx , Ly , Lz
?
Commutation relations for
angular momentum operators
of quantum mechanics in 3D
Only if all four operators commute
We must now look at the commutation
relations
Th e two operators L1 and L2 will
commute if
[L1,L2 ] f(x,y,z) =(L1L2 - L2L1) f(x,y,z) = 0
Rotation..Quantum Mechanics 3D Commutation relations for
angular momentum operators
of quantum mechanics in 3D
ˆ2  L
ˆ L
ˆ representing
For the quantum mechanical operators L
the square of the length of the angular momentum
as well as the operators representing the three Cartesian
ˆ x ;L
ˆ y; L
ˆz
components of the angular momentum vector L
we have
[Lˆ2 , Lˆ x ]  [Lˆ2 , Lˆ y ]  [Lˆ2 , Lˆ z ]  0
[Lˆ , Lˆ ]  i Lˆ
x
y
[Lˆ y , Lˆ z ]  i
[Lˆ , Lˆ ]  i
z
x
z
Lˆ x
Lˆ
y
Rotation..Quantum Mechanics 3D
Common eigenfunctions for
ˆ and L
ˆ2 .
L
z
How do we find the eigenfunctions ?
The eigenfunctions f must satisfy
L zf = af
and
L 2f = bf
The function f must in other words
satisfy the differential equations
L zf = af
as well as
L 2f
= bf
Rotation..Quantum Mechanics 3D
It is more convenient to solve the equations in
Angular momentum operators
spherical coordinates
(x,y,z)  (r,  

of quantum mechanics in
spherical coordinates in 3D
r

We find after some tedious but straight forward
manipulations
d
Lz = -i  ]
d
d2
d
1
d2
L2 = - 2[
+cot  sin2
]
2

d
d
d
Rotation..Quantum
Mechanics 3D
Common eigenfunctions for
ˆ and L
ˆ2 .
L
z
We must solve :
ˆ (,)  b(,) and L
ˆ 2 (,)  c(,)
L
z
The eigenfunctions to L2 and Lz are given by
(,) = Yl,m ((,)
2l+ 1 (l | m!| |m|
=
Pl (cos)  exp[im]
4 (l | m!|)
Eigenfunctions are orthonormal
2
* (,)Y
Y
(,)
r
sin dd  l,l' m,m'
 lm
l'm'
Rotation..Quantum Mechanics 3D
Common eigenfunctions for
ˆ and L
ˆ2 .
L
z
2l+ 1 (l | m!| |m|
(,) = Yl,m ((,) =
Pl (cos)  exp[im]
4 (l | m!|)
We have that l can take the values : l = 0,1, 2, 3,4..
and the possible eigenvalues for L 2 are 2l(l  1)
ˆ 2  (,)  2 l(l  1) (,)
L
lm
lm
for a given l value m can take the 2l+ 1 values
- l, - l + 1,...,-1,0,1,...,l - 1,l
and the possible eigenvalues for Lz are m
ˆ z lm (,)  m lm (, )
L
Rotation..Quantum Mechanics 3D
ˆz
Common eigenfunctions for L
ˆ 2 . Spherical harmonics
and L
2l+ 1 (l | m!| |m|
(,) = Yl,m ((,) =
Pl (cos)  exp[im]
4 (l | m!|)
l
m
0
0
1
1
0
1
Y lm (,)
1
4
3
cos
4
3
sin exp[i]
4
Rotation..Quantum Mechanics 3D
ˆz
Common eigenfunctions for L
ˆ 2 . Spherical harmonics
and L
2l+ 1 (l | m!| |m|
(,) = Yl,m ((,) =
Pl (cos)  exp[im]
4 (l | m!|)
l
m
Y lm (,)
2
0
5
(3cos2  1)
16
2
1
15
cossin[i]
8
2
15
sin 2 exp[2i]
32
2
ˆz
Common eigenfunctions for L
ˆ 2 . Spherical harmonics
and L
2l+ 1 (l | m!| |m|
(,) = Yl,m ((,) =
Pl (cos)  exp[im]
4 (l | m!|)
Rotation..Quantum
Mechanics 3D
l
3
3
3
3
m
Y lm (,)
0
7
(5cos3  3 cos)
16
1
21
(5cos2  1)sin [i]
64
2
105
sin 2  cos exp[2i]
32
3
35
sin 3  exp[2i]
64
What you should learn from this lecture
1. you should know the definition of angular
momentum as L = rxp.
2. You should be aware of the commutation relations
[Lˆ2 , Lˆ ]  [Lˆ2 , Lˆ ]  [Lˆ2 , Lˆ ]  0
x
y
z
[Lˆ x , Lˆ y ]  i Lˆ z ;[Lˆ y , Lˆ z ]  i Lˆ x;[Lˆ z , Lˆ x ]  i Lˆ y
3. You should realize that the above commutation
has the consequence that we only can find
ˆ 2 and one of the
find common eigenfunctions to L
ˆ z . Thus we can
components , normally taken as L
only know L 2 and L z precisely.
What you should learn from this lecture
4. You are not required to know the exact form of the
eigenfunctions
2l+ 1 (l | m!| |m|
(,) = Yl,m ((,) =
Pl (cos)  exp[im]
4 (l | m!|)
ˆ z and L
ˆ2
to L
5. You should know that l can take the values : l = 0,1, 2, 3,4..
and the possible eigenvalues for L 2 are 2l(l  1)
ˆ 2  (,)  2 l(l  1) (,)
L
lm
lm
6. You should know that
for a given l value m can take the 2l+ 1 values
- l, - l + 1,...,-1,0,1,...,l - 1,l
and the possible eigenvalues for Lz are m
ˆ z lm (,)  m lm (, )
L
We have
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
f
f
Lxf = -i ( yz - zy ) = -i u x
f
f
L yf = -i ( zx - xz ) = -i u y
Next
LxLyf = -i Lxu y
u y
u y
LxLyf = -i [ -i ( y z - z y ) ]
u y
u y
LxLyf = - 2 [ y z - z y ]
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
We have
uy
 f f
z = z (zx - x z )
uy
f
2f
2f
z = x + zzx - x z
Further
uy  f f
= y (zx - x z )
y
uy
y =
2f
2f
z yx - x yz
combining terms
Thus
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
f
2f
2f
2f
2f
LxLyf = - 2[ yx + yz zx - yx z - z2yx +zxyz ]
f
2f
2f
2f
2f
LxLyf = - 2[ yx + yz zx - yx z - z2yx +zxyz ]
It is clear that L xLyf can be evaluated by
interchanging x and y We get:
f
2f
2f
2f
2f
LyLxf = - 2[ xy + xzzy - xyz - z2xy +zyxz ]
using the relations
2f
2f
zy = yz
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2
etc.
We have
f
f


[ LxLy - LyLx] f = - 2[ yx - xy ] = - 2[ yx - xy ]
f


We have: Lz = -i [ xy - yx
Thus: [ LxLy - LyLx] f
]
= i Lz f ; [Lx,Ly] = i Lz
We have shown [L x,Ly] = i Lz
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
By a cyclic permutation
Z
X
C3
z
Y
Y
z
X
X
Y
[ Ly,L z] = i Lx
[ L z,Lx ] = i Ly
We have shown that the three operators L x ,L y ,L z
are non commuting
What about the commutation between L x ,Ly ,Lz and L2
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
Let us examine the commutation relation
between L2 and Lx
We have:
[ L2 ,L x ]  [L2x  L2y  L2z ,L x ]
[ L2 ,L x ]  [L2x ,L x ]  [L2y ,Lx ]  [L2z ,Lx ]
For the first term
[L2x ,L x ]  L2xL x  L xL2x  L3x  L3x  0
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
For the second term
[L2y ,L x ]  L2yLx  LxL2y
 L2yLx  LyL xLy  L yLxL y  L xL2y
 L y [L yL x  L xL y ]  [L yLx  LxL y ]L y
Z
 i LyL z  i LzL y
X
Y
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
For the third term
[L2z,Lx ]  L2zL x  L xL2z
2
2
 L zL x  LzLxL z  LzL xLz  L xLz
 L z[LzL x  L xLz ]  [LzL x  L xLz ]L z
 i L zL y  LyL z
Z
X
Y
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
In total
[ L2 ,L x ]  [L2x  L2y  L2z ,L x ]
0
i L yLz  i LzLy i L zL y  LyL z
Z
X
Y
0
Appendix : Commutator relations for
angular momentum components L x ;L y ;Lz ;L2 .
We have shown
[L2,Lx] = [L x 2+Ly 2+Lz2,Lx ] = O
now by cyclic permutation
Z
X
Y
[Ly 2+Lz2+Lx 2,Ly ] = [L 2,Ly ] = 0
[Lz2+Lx 2+Ly 2,Lz] = [L 2,Lz] = 0
Thus Lx ,Ly ,Lz all commutes with L 2
and we can find common
L2 and Lx
or L 2 and Ly
eigenfunctions for
or L 2 and Lz
the normal convention is to obtain eigenfunctions that are
at the same time eigenfunctions to L z and L2.
How do we find the eigenfunctions ?
Rotation..Quantum Mechanics 3D
We have

-i  S()T() = b S()T()
or

-i S() T()= b S()T()
multiplying with 1/ S( ) from left
T()

=
ib
T()
The general solution is
T() = AExp[
ib
]
A general point in 3-D space is given by
( r,)
Z
rcos
(x,y,z)  (r,  

r

Y
X
rsin 
We have the following relation
x= r sin cos
y= r sin sin 
z= r cos
The same point is represented by (r, +2)
We must thus have
ib
ib
ib
ib
Exp[ ] = Exp[ (  2) = Exp[ ] Exp[ 2]
Thus
Exp[
ib
2b 
2b 
2] = cos
+ isin
=1




This equation is only satisfied if
b
= m
with
m = 0,±1,±2,......
Thus the eigenvalue b is quantized as
b =
m
m = 0,±1,±2,......
The possible eigenfunctions are
T() = AExp[im] , m = 0,±1,±2,......