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Classical Angular Momentum
Angular momentum in classical physics
Consider a particle at the position r
r
v
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 px = mvx = m dt
The angular momentum is defined as
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
We have in addition
L = rXp = (ix +jy + kz)X (ipx + jpy +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 ?
Consider the change of L with time
F
dL
dr
dp
dt = dt Xp + rX dt
dL
dp
dt = mvXv + rX dt
r
dp
= rX dt
dL
d
dr
d2r
dt = rXdt [mdt ] = mrXdt 2
dL
dt = mrXF
F = force
Classical Angular Momentum
dL
 mr  F
dt
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
F
r
Classical Angular Momentum
Examples :
r
F
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
We have
L = rXp = iLx + jLy + kLz
where
Lx = ry 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 ) ; Ly = -i (z x - xz )


L z = -i (x y - yx )
3D
Rotation..Quantum Mechanics 3D
We have
L = iLx + jLy + kLz
thus
L.L = L 2 =(iLx + jLy + kLz).(iLx + jLy + kLz)
L2 = Lx 2 + Ly 2 + Lz2
Rotation..Quantum Mechanics 3D
Can we find common eigenfunctions to
L2 , Lx , Ly , Lz
?
Only if all four operators commute
We must now look at the commutation
relations
The two operators L x and L y will
commute if
[L xL y - L yL x] f(x,y,z) = 0
Rotation..Quantum Mechanics 3D
We have
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 ]
Rotation..Quantum Mechanics 3D
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
Rotation..Quantum Mechanics 3D
Thus
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
Rotation..Quantum Mechanics 3D
etc.
We have
f
f


[ LxLy - LyLx] f = - 2[ yx - xy ] = - 2[ yx - xy ]
f


We have: L z = -i [ x y - yx
Thus: [ LxLy - LyLx] f
]
= i Lz f ; [Lx,Ly] = i Lz
We have shown [L x,Ly] = i Lz
Rotation..Quantum Mechanics 3D
By a cyclic permutation
Z
X
C3
z
Y
Y
z
Y
X
X
[ Ly,L z] = i Lx
[ Lz,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
Rotation..Quantum Mechanics 3D
Let us examine the commutation relation
between L2 and L x
We have :
[ L2 ,L x ]  [L2x  L2y  L2z ,L x ]
2
2
2
2
[ L ,L x ]  [L x ,L x ]  [L y ,Lx ]  [L z ,Lx ]
For the first term
[L2x ,L x ]  L2xL x  L xL2x  L3x  L3x  0
Rotation..Quantum Mechanics 3D
For the second term
2
2
2
[L y ,L x ]  L yLx  LxL y
 L2yLx  LyL xLy  L yLxL y  L xL2y
 L y [L yL x  L xL y ]  [L yLx  LxL y ]L y
 i LyL z  i LzL y
Z
X
Y
Rotation..Quantum Mechanics 3D
For the third term
2
2
2
[L z,Lx ]  L zL x  L xL z
 L2zL x  LzLxL z  LzL xLz  L xL2z
 L z[LzL x L xLz ]  [LzL x  L xLz ]L z
 i L zL y  LyL z
Z
X
Y
Rotation..Quantum Mechanics 3D
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
Rotation..Quantum Mechanics 3D
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 ] = [L2,Ly ] = 0
[Lz2+Lx 2+Ly 2,Lz] = [L2,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
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
spherical coordinates
(x,y,z)  (r,  

r

We find after some tedious but straight forward
manipulations
d
Lz = -i  ]
d
d2
d
1
d2
L 2 = - 2[
+cot  sin2
]
d 2
d
d 
We must now solve
Rotation..Quantum Mechanics 3D
^
L zY() = b Y()
and
^
L 2Y() = c Y()
First let us solve

^
L zY() = -i  Y() = b Y()
We shall try a solution of the form
Y() = S()T( )
as ^
L z only depends on


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,......