lecture 14: angular momentum-ii. the material in this lecture covers the following in atkins
DESCRIPTION
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) - PowerPoint PPT PresentationTRANSCRIPT
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 ,**)
Angular momentum in classical physicsAngular momentum in classical physics
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 sinΦ
r
p
The angular momentum is perpendicular to the plane .defined 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 = r x 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
md2r
dt2=F (Newtons Law)
Classical Angular Momentum
r
F
dr L
dt=
r r ×
r F
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
Quantum mechanical representation of angular momentumQuantum mechanical representation of angular momentumoperatoroperatorWe 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 --> -ihδδx ; p y --> -ih
δδy ; p z --> -ih
δδz
:Thus
L x = -ih( yδδz - z
δδy ) ; L y = -ih( z
δδx - x
δδz )
L z = -ih( xδδy - y
δδx )
Rotation..Quantum Mechanics 3D
Angular 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 commutation
relations
The 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 operatorsof quantum mechanics in 3D
[ˆ L 2,ˆ L x]=[ˆ L 2,ˆ L y]=[ˆ L 2,ˆ L z]=0
[ˆ L x,ˆ L y]=ihˆ L z[ˆ L y,ˆ L z]=ihˆ L x
[ˆ L z,ˆ L x]=ihˆ L y
For the quantum mechanical operators ˆ L 2=ˆ L ⋅ ˆ L representing the square of the length of the angular momentum
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 ˆ L x ; ˆ L y; ˆ L zwe have
How do we find the eigenfunctions ?
The eigenfunctions f must satisfy
Lzf = af and L 2f = bf
The function f must in other wordssatisfy the differential equations
Lzf = af
as well as
L2f = bf
Rotation..Quantum Mechanics 3DCommon eigenfunctions for ˆ L z and ̂ L 2.
It is more convenient to solve the equations in
spherical coordinates
Θ
φ
r
(x,y,z)→
(r, Θ,φ )
We find after some tedious but straight forward
manipulations
Lz = -ih [ddφ
]
L2 = -h2[ d2
dΘ2 +cotΘ
d dΘ
+ 1
sin2Θ d2
dφ2 ]
Rotation..Quantum Mechanics 3D
Angular momentum operatorsof quantum mechanics in spherical coordinates in 3D
Rotation..Quantum Mechanics 3D
The eigenfunctions to L2 and Lz are given by
ψ(φ,θ)=Yl,m((φ,θ)
=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
We must solve :
ˆ L zψ(θ,ϕ)=bψ(θ,ϕ) and ̂ L 2ψ(θ,ϕ)=cψ(θ,ϕ)
Common eigenfunctions for ˆ L z and ̂ L 2.
Eigenfunctions are orthonormal
Ylm*(ϕ,θ)Yl'm'(ϕ,θ) ∫ r2sinθdθdϕ =∂l,l'∂m,m'
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 mhˆ L zψ lm(φ,θ)=hmψ lm(φ,θ)
ψ(φ,θ)=Yl,m((φ,θ)=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
We have that l can take the values:l=0,1,2,3,4..
and the possible eigenvalues for L2 are h2l(l+1)ˆ L 2ψlm(φ,θ)=h2l(l+1)ψ lm(φ,θ)
Rotation..Quantum Mechanics 3DCommon eigenfunctions for ˆ L z and ̂ L 2.
l m Y lm(ϕ,θ)
0 0 14π
1 0 34π
cosθ
1 ±1 m 34π
sinθexp[±iϕ]
Rotation..Quantum Mechanics 3DCommon eigenfunctions for ˆ L z
and ̂ L 2. Spherical harmonics
ψ(φ,θ)=Yl,m((φ,θ)=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
Rotation..Quantum Mechanics 3DCommon eigenfunctions for ˆ L z
and ̂ L 2. Spherical harmonics
l m Y lm(ϕ,θ)
2 0 5
16π (3cos2θ−1)
2 ±1 m158π
cosθsinθ[±iϕ]
2 ±2 1532π
sin2θexp[±2iϕ]
ψ(φ,θ)=Yl,m((φ,θ)=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
Rotation..Quantum Mechanics 3D
Common eigenfunctions for ˆ L z
and ̂ L 2. Spherical harmonics
l m Y lm(ϕ,θ)
3 0 7
16π (5cos3θ−3cosθ)
3 ±1 m2164π
(5cos2θ−1)sinθ[±iϕ]
3 ±2 10532π
sin2θcosθexp[±2iϕ]
3 ±3 m 3564π
sin3θexp[±2iϕ]
ψ(φ,θ)=Yl,m((φ,θ)=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
What you should learn from this lecture
1. you should know the definition of angular momentum as
r L =
r r x
r p .
2. You should be aware of the commutation relations
[ˆ L 2,ˆ L x]=[ˆ L 2,ˆ L y]=[ˆ L 2,ˆ L z]=0
[ˆ L x,ˆ L y]=ihˆ L z;[ˆ L y,ˆ L z]=ihˆ L x;[ˆ L z,ˆ L x]=ihˆ L y
3. You should realize that the above commutationhas the consequence that we only can find
find common eigenfunctions to ˆ L 2 and one of thecomponents , normally taken as ˆ L z. Thus we can
only know L2 and Lz precisely.
What you should learn from this lecture
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 mhˆ L zψ lm(φ,θ)=hmψ lm(φ,θ)
4. You are not required to know the exact form of the eigenfunctions
ψ(φ,θ)=Yl,m((φ,θ)=2l+14π
(l−|m!|(l+|m!|)
Pl|m|(cosθ)×exp[imφ]
to ˆ L z and ̂ L 2
5. You should know that l can take the values:l=0,1,2,3,4..
and the possible eigenvalues for L2 are h2l(l+1)ˆ L 2ψlm(φ,θ)=h2l(l+1)ψ lm(φ,θ)
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
We have
Lxf = -ih( yδfδz - z
δfδy ) = -ih ux
Ly = -f ih( zδfδx - x
δfδz ) = -ih uy
Next
LxLy = -f ih Lxuy
LxLy = -f ih [ -ih( yδuyδz - z
δuyδy ) ]
LxLy = -f h2 [ yδuyδz - z
δuyδy ]
We have
δuyδz =
δδz (z
δfδx - x
δfδz )
δuyδz =
δfδx + z
δ2fδzδx - x
δ2fδz2
Further
δ u
y
δ y
= δδy (z
δfδx - x
δfδz )
δuyδy = z
δ2fδyδx - x
δ2fδyδz
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
combining terms
Thus
LxLyf = -h2[ yδfδx + yz
δ2fδzδx - yx
δ2fδz2 - z2
δ2fδyδx +zx
δ2fδyδz ]
LxLy = -f h2[ yδfδx + yz
δ2fδzδx - yx
δ2fδz2 - z2
δ2fδyδx +zx
δ2fδyδz ]
It is clear that LxLy f can be evaluated by :interchanging x and y We get
LyLx = -f h2[ xδfδy + xz
δ2fδzδy - xy
δ2fδz2 - z2
δ2fδxδy +zy
δ2fδxδz ]
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
using the relations
δ2fδzδy =
δ2fδyδz .etc
We have
[ LxL y - LyLx] = -f h2[ yδfδx - x
δfδy ] = -h2[ y
δδx - x
δδy ]
f
We have: Lz = -i h[ xδδy - y
δδx ]
Thus: [ LxLy - LyLx] f = ihLz f ; [Lx,Ly] = ihLz
We have shown [Lx,Ly] = ihLz
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
By a cyclic permutation
[ Ly,L z] = ihLx
[ Lz,Lx] = ihLy
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: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
Let us examine the commutation relation
between L2 and Lx
We have:
[ L2,Lx]=[Lx2 +Ly
2 +Lz2,Lx]
[ L2,Lx]=[Lx2,Lx]+[Ly
2,Lx]+[Lz2,Lx]
[Lx2,Lx]=Lx
2Lx −LxLx2 =Lx
3 −Lx3 =0
For the first term
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
For the second term[Ly
2,Lx]=Ly2Lx −LxLy
2
=Ly2Lx −LyLxLy +LyLxLy −LxLy
2
=Ly[LyLx −LxLy]+[LyLx −LxLy]Ly
=−ihLyLz −ihLzLy
Y
Z X
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
For the third term[Lz
2,Lx]=Lz2Lx −LxLz
2
=Lz2Lx −LzLxLz +LzLxLz −LxLz
2
=Lz[LzLx −LxLz]−[LzLx −LxLz]Lz
=ihLzLy +hLyLz
Y
Z X
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
[ L2,Lx]=[Lx2 +Ly
2 +Lz2,Lx]
In total
−ihLyLz −ihLzLy=0 +ihLzLy +hLyLz =0
Y
Z X
Appendix: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
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: Commutator relations for
angular momentum components Lx;Ly;Lz;L2.
Rotation..Quantum Mechanics 3DWe have
-ihδδφ S(Θ)T(φ) = b S( Θ)T(φ)
or
- ihS(Θ)δ
δφ T(φ)= b S( Θ)T(φ)
multiplying with 1/ S( Θ) from left
δT(φ)
δφ =
ib
hT(φ)
The general solution is
T(φ) = AExp[
ib
hφ]
A general point in 3-D space is given by ( r,Θ, φ)
Θ
φ
r
( , , )x y z → ( ,r Θ,φ )
X
Y
ZrcosΘ
rsinΘ
We have the following relation = x rsinΘ cosφ = y rsinΘ sinφ = z rcosΘ
( ,The same point is represented by rΘ,φ+2π)
We must thus have
Exp[
ibh
φ] = Exp[
ibh
(φ+2π) = Exp[
ibh
φ] Exp[
ibh
2π]
A general point in 3-D space is given by ( r,Θ, φ)
Θ
φ
r
( , , )x y z → ( ,r Θ,φ )
X
Y
ZrcosΘ
rsinΘ
We have the following relation = x rsinΘ cosφ = y rsinΘ sinφ = z rcosΘ
( ,The same point is represented by rΘ,φ+2π)
We must thus have
Exp[
ibh
φ] = Exp[
ibh
(φ+2π) = Exp[
ibh
φ] Exp[
ibh
2π]
Thus
Exp[
ibh
2π] = cos
2πbh
⎛ ⎝
⎞ ⎠+ isin
2πbh
⎛ ⎝
⎞ ⎠=1
This equation is only satisfied if
bh= = 0,±1,±2,......m with m
Thus the eigenvalue b is quantized as
= b h m = 0,±1,±2,......m
The possible eigenfunctions are
(T φ) = [AExp imφ] , = 0,±1,±2,......m