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 ,**)
AAnngguu llaa rr mmoommeennttuumm iinn cc ll aassss iiccaa ll pphhyyss iiccss
Consider a particle at the position r
i
k
j
rv
Where
r = ix + jy + kz
The velocity of this particle is given by
v = drdt = i
dxdt + j
dydt + k
dzdt
Classical Angular Momentum
Review of classical physics Position and velocity in 3D
Classical Angular MomentumThe linear momentum of the particle with mass m isgiven by
p = mv where e.i px = mvx = mdxdt
The angular momentum is defined as
L = rXp
L
L = r X p
Φ
|r| |p| sin Φ
r
p
The angular momentum is perpendicular to the planedefined by r and p.
Review of classical physics Angular Momentum in 3D
Classical Angular Momentum
We have in addition
L = rXp = (ix +jy + kz)X (ipx + jpy +kpz)
L = (rypz -rzpy)i + (rzpx -rxpz)j + (rxpy - rypx)k
or
i j k
rXp = rx ry rz
px py pz
Review of classical physics Angular Momentum in 3D
Classical Angular Momentum
Why are we interested in the angular momentum?
Consider the change of L with time
dLdt =
drdt Xp + rX
dpdt
dLdt = mvXv + rX
dpdt = rX
dpdt
dLdt = rX
ddt [m
drdt ] = rXm
d2rdt2
r
F
Review of classical physics Angular Momentum in 3D
md r
dtF
2
2 = (Newtons Law)
For centro-symmetric systems in whichthe force works in the same direction as rwe must have
dLdt = 0 : THE ANGULAR
MOMENTUM IS CONSERVED
Classical Angular Momentum
r
F
dLdt
r F
rr r
= ×
Review of classical physics Angular momentum andcentral force in 3D
Classical Angular Momentum
Examples :
movement of electron around nuclei movement of planets around sun
For such systems L is a constant of motion, e.g. doesnot change with time since
dLdt = 0
In quantum mechanics an operator O representing aconstant of motion will commute with the Hamiltonianwhich means that we can find eigenfunctions that areboth eigenfunctions to H and O
rF
Review of classical physics Angular momentum andcentral force in 3D
QQuuaannttuumm mmeecchhaanniiccaa ll rreepprreesseennttaatt iioonn ooff aanngguullaarr mmoommeennttuummoo pp ee rr aa tt oo rrWe have
L = rXp = iLx + jLy + kLz
where
Lx = rypz - rzpy ; Ly = rzpx - rxpy ; Lz = rxpy - rypx
In going to quantum mechanics we have
x --> x ; y --> y ; z --> z
px --> -ihhhhδ
δx ; py --> -ihhhhδ
δy ; pz --> -ihhhhδδz
Thus :
Lx = -ihhhh(y δδz - z
δδy ) ; Ly = -ihhhh(z
δδx - x
δδz )
L z = -ihhhh(x δ
δy - yδ
δx )
Rotation Quantum.. Mechanics 3DAngular momentum operatorsof quantum mechanics in 3D
We have
L = iLx + jLy + kLz
thus
L.L = L2 =(iLx + jLy + kLz).(iLx + jLy + kLz)
L2 = Lx2 + Ly2 + Lz2
Rotation Quantum.. Mechanics 3D
Angular momentum operatorsof quantum mechanics in 3D
Can we find common eigenfunctions to
L2 , Lx , Ly , Lz ?
Only if all four operators commute
We must now look at the commutationrelations
The two operators L1 and L2 willcommute if
[L1,L2 ] f(x,y,z) =(L1L2 - L2L1) f(x,y,z) = 0
Rotation Quantum.. Mechanics 3D
Commutation relations for angular momentum operatorsof quantum mechanics in 3D
[L L [L L [L L
[L L L
[L L L
[L L L
2x
2y
2z
x y z
y z
z x y
ˆ , ˆ ] ˆ , ˆ ] ˆ , ˆ ]ˆ , ˆ ] ˆ
ˆ , ˆ ] ˆ
ˆ , ˆ ] ˆ
= = =
=
=
=
0
i
i x
i
h
h
h
For the quantum mechanical operators L L L representing the square of the length of the angular momentum
2ˆ ˆ ˆ= ⋅
Commutation relations for angular momentum operatorsof quantum mechanics in 3D
Rotation Quantum.. Mechanics 3D
as well as the operators representing the three Cartesian components of the angular momentum vector L ;L ;Lwe have
x y zˆ ˆ ˆ
How do we find the eigenfunctions ?
The eigenfunctions f must satisfy
Lzf = af and L2f = bf
The function f must in other wordssatisfy the differential equations
Lzf = af
as well as
L2f = bf
Rotation Quantum.. Mechanics 3DCommon eigenfunctions for
L and Lz2ˆ ˆ .
It is more convenient to solve the equations inspherical coordinates
Θ
φr
(x,y,z) → (r,Θ,φ )
We find after some tedious but straight forwardmanipulations
Lz = -ihhhh [ddφ ]
L2 = -hhhh2[ d2
dΘ2 +cotΘd dΘ +
1 sin2Θ
d2
dφ2 ]
Rotation Quantum.. Mechanics 3D
Angular momentum operatorsof quantum mechanics in spherical coordinates in 3D
Rotation Quantum.. Mechanics 3D
The
l ml m
im
eigenfunctions to are given by L and L
( , ) = Y ( , )
=2l+14
P
2z
l,m
l|m|
ψ φ θ φ θ
πθ φ
(
( | !|( | !|)
(cos ) exp[ ]−+
×
We
b c
must solve :
L and Lz2ˆ ( , ) ( , ) ˆ ( , ) ( , )ψ θ ϕ ψ θ ϕ ψ θ ϕ ψ θ ϕ= =
Common eigenfunctions for
L and Lz2ˆ ˆ .
Eigenfunctions
Y Y r d dlm l m l l m m
are orthonormal
*' ' , ' , '( , ) ( , ) sinϕ θ ϕ θ θ θ ϕ ∂ ∂∫ =2
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 L are mz h
hˆ ( , ) ( , )Lzψ φ θ ψ φ θlm lmm=
ψ φ θ φ θπ
θ φ( , ) = Y ( , ) =2l+14
Pl,m l|m|(
( | !|( | !|)
(cos ) exp[ ]l ml m
im−+
×
We have that l can take the values : l = 0,1,2,3,4..
and the possible eigenvalues for L are
L
2
2 h
h
2
21
1
l l
l llm lm
( )ˆ ( , ) ( ) ( , )
+= +ψ φ θ ψ φ θ
Rotation Quantum.. Mechanics 3DCommon eigenfunctions for
L and Lz2ˆ ˆ .
l
i
m Y
0 1
4
0 3
4 cos
3
4 sin
lm( , )
exp[ ]
ϕ θ
π
πθ
πθ ϕ
0
1
1 1± ±m
Rotation Quantum.. Mechanics 3DCommon eigenfunctions for L
and L harmonicsz
2ˆ
ˆ . Spherical
ψ φ θ φ θπ
θ φ( , ) = Y ( , ) =2l+14
Pl,m l|m|(
( | !|( | !|)
(cos ) exp[ ]l ml m
im−+
×
Rotation Quantum.. Mechanics 3DCommon eigenfunctions for L
and L harmonicsz
2ˆ
ˆ . Spherical
l
i
i
m Y
0 5
16 (3cos
1 158
cos sin
15
32 sin
lm
2
2
( , )
)
[ ]
exp[ ]
ϕ θ
πθ
πθ θ ϕ
πθ ϕ
2 1
2
2 2 2
−
± ±
± ±
m
ψ φ θ φ θπ
θ φ( , ) = Y ( , ) =2l+14
Pl,m l|m|(
( | !|( | !|)
(cos ) exp[ ]l ml m
im−+
×
Rotation Quantum.. Mechanics 3D
Common eigenfunctions for L
and L harmonicsz
2ˆ
ˆ . Spherical
l
i
i
i
m Y
0 7
16 (5cos
1 21
64 (5cos
10532
sin
35
64 sin
lm
3
2
2
3
( , )
cos )
)sin [ ]
cos exp[ ]
exp[ ]
ϕ θ
πθ θ
πθ θ ϕ
πθ θ ϕ
πθ ϕ
3 3
3 1
3 2 2
3 3 2
−
± − ±
± ±
± ±
m
m
ψ φ θ φ θπ
θ φ( , ) = Y ( , ) =2l+14
Pl,m l|m|(
( | !|( | !|)
(cos ) exp[ ]l ml m
im−+
×
What you should learn from this lecture
1. you should know the definition of angular momentum as L = rxp.
r r r
2
0
.ˆ , ˆ ] ˆ , ˆ ] ˆ , ˆ ]ˆ , ˆ ] ˆ ; ˆ , ˆ ] ˆ ; ˆ , ˆ ] ˆ
You should be aware of the commutation relations
[L L [L L [L L
[L L L [L L L [L L L
2x
2y
2z
x y z y z z x y
= = =
= = =i i x ih h h
3.
ˆˆ
You should realize that the above commutationhas the consequence that we only can find
find common eigenfunctions to L and one of thecomponents , normally taken as L . Thus we can
only know L and L precisely.
2
z2
z
What you should learn from this lecture
6.
ˆ ( , ) ( , )
You should know that
Lz
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 L are mz h
hψ φ θ ψ φ θlm lmm=
4.
(( | !|( | !|)
(cos ) exp[ ]
ˆ ˆ
You are not required to know the exact form of the eigenfunctions
L and Lz2
ψ φ θ φ θπ
θ φ( , ) = Y ( , ) =2l+14
Pl,m l|m|l m
l mim
to
−+
×
5. You should know that l can take the values : l = 0,1,2,3,4..
and the possible eigenvalues for L are
L
2
2 h
h
2
21
1
l l
l llm lm
( )ˆ ( , ) ( ) ( , )
+= +ψ φ θ ψ φ θ
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
We have
Lxf = -ihhhh( yδfδz - z
δfδy ) = -ihhhh ux
Lyf = -ihhhh( zδfδx - x
δfδz ) = -ihhhh uy
Next
LxLyf = -ihhhh Lxuy
LxLyf = -ihhhh [ -ihhhh( yδuyδz - z
δuyδy ) ]
LxLyf = -hhhh2 [ yδuyδz - z
δuyδy ]
We have
δuyδz =
δδz (z
δfδx - x
δfδz )
δuyδz =
δfδx + z
δ2fδzδx - x
δ2fδz2
Further
δδu
yy =
δδy (z
δfδx - x
δfδz )
δuyδy = z
δ2fδyδx - x
δ2fδyδz
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
combining terms
Thus
LxLyf = -hhhh2[ yδfδx + yz
δ2fδzδx - yx
δ2fδz2 - z2
δ2fδyδx +zx
δ2fδyδz ]
LxLyf = -hhhh2[ yδfδx + yz
δ2fδzδx - yx
δ2fδz2 - z2
δ2fδyδx +zx
δ2fδyδz ]
It is clear that LxLyf can be evaluated byinterchanging x and y We get:
LyLxf = -hhhh2[ xδfδy + xz
δ2fδzδy - xy
δ2fδz2 - z2
δ2fδxδy +zy
δ2fδxδz ]
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
using the relations
δ2f
δzδy = δ2f
δyδz etc.
We have
[ LxLy - LyLx] f = -hhhh2[ yδfδx - x
δfδy ] = -hhhh2[ y
δδx - x
δδy ]
f
We have: Lz = -i hhhh[ xδδy - y
δδx ]
Thus: [ LxLy - LyLx] f = ihhhhLz f ; [Lx,Ly] = ihhhhLz
We have shown [Lx,Ly] = ihhhhLz
Appendix
L L Ly z
:
; ; ;
Commutator relations for
angular momentum components Lx2
By a cyclic permutation
[ Ly,Lz] = ihhhhLx
[ Lz,Lx] = ihhhhLy
We have shown that the three operators L x,L y,L zare non commuting
What about the commutation between Lx,Ly,Lz and L2
Y
Z X
X
Y
z
X
Yz
C3
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
Let us examine the commutation relationbetween L and L2
x
We have :
[ , ] [ , ] L L L L L Lx x y z x2 2 2 2= + +
[ , ] [ , ] [ , ] [ , ] L L L L L L L Lx x x y x z x2 2 2 2= + +
[ , ]L L L L L L L Lx x x x x x x x2 2 2 3 3 0= − = − =
For the first term
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
For the second term[ , ]L L L L L Ly x y x x y
2 2 2= −
= − + −L L L L L L L L L Ly x y x y y x y x y2 2
= − + −L L L L L L L L L Ly y x x y y x x y y[ ] [ ]
= − −i L L i L Ly z z yh h
Y
Z X
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
For the third term[ , ]L L L L L Lz x z x x z
2 2 2= −
= − + −L L L L L L L L L Lz x z x z z x z x z2 2
= − − −L L L L L L L L L Lz z x x z z x x z z[ ] [ ]
= +i L L L Lz y y zh h
Y
Z X
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
[ , ] [ , ] L L L L L Lx x y z x2 2 2 2= + +
In total
− −i L L i L Ly z z yh h= 0 + +i L L L Lz y y zh h = 0
Y
Z X
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
We have shown[L2,Lx] = [Lx2+Ly2+Lz2,Lx] = O
now by cyclic permutation
[Ly2+Lz2+Lx2,Ly] = [L2,Ly] = 0
[Lz2+Lx2+Ly2,Lz] = [L2,Lz] = 0
Thus Lx,Ly,Lz all commutes with L 2
and we can find common eigenfunctions for
L2 and Lx or L 2 and Ly or L 2 and Lz
the normal convention is to obtain eigenfunctions that areat the same time eigenfunctions to Lz and L2.
How do we find the eigenfunctions ?
Y
Z X
Appendix
L L Ly z
:
; ; ; .
Commutator relations for
angular momentum components Lx2
Rotation Quantum.. Mechanics 3DWe have
-ihhhhδδφ S(Θ)T(φ) = b S(Θ)T(φ)
or
-ihhhhS(Θ)δ
δφ T(φ)= b S(Θ)T(φ)
multiplying with 1/ S(Θ) from left
δT(φ)
δφ =
ibh
T(φ)
The general solution is
T(φ) = AExp[
ibh
φ]
A general point in 3-D space is given by ( r,Θ, φ)
Θ
φr
(x,y,z)→ (r,Θ,φ )
X
Y
ZrcosΘ
rsinΘWe have the following relation x= r sinΘ cosφ y= r sinΘ sinφ z= r cosΘ
The same point is represented by (r,Θ,φ+2π)
We must thus have
Exp[
ibh
φ] = Exp[
ibh
( )φ π+ 2 = Exp[
ibh
φ] Exp[
ibh
2π]
A general point in 3-D space is given by ( r,Θ, φ)
Θ
φr
(x,y,z)→ (r,Θ,φ )
X
Y
ZrcosΘ
rsinΘWe have the following relation x= r sinΘ cosφ y= r sinΘ sinφ z= r cosΘ
The same point is represented by (r,Θ,φ+2π)
We must thus have
Exp[
ibh
φ] = Exp[
ibh
( )φ π+ 2 = Exp[
ibh
φ] Exp[
ibh
2π]
Thus
Exp[
ibh
2π] = cos
2πbh
+ isin
2πbh
=1
This equation is only satisfied if
bh
= m with m = 0,±1,±2,......
Thus the eigenvalue b is quantized as
b = hhhhm m = 0,±1,±2,......
The possible eigenfunctions are
T(φ) = AExp[imφ] , m = 0,±1,±2,......