lecture 14: angular momentum-ii. the material in this lecture covers the following in atkins

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


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



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


md2r

dt2=F (Newtons Law)


r

F

dr L

dt=

r r ×

r F

Review of classical physics Angular momentum andcentral force in 3D


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


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


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


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 ]


Angular momentum operatorsof quantum mechanics in spherical coordinates in 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!|)


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


(l−|m!|(l+|m!|)


Rotation..Quantum Mechanics 3DCommon eigenfunctions for ˆ L z


l m Y lm(ϕ,θ)

2 0 5

16π (3cos2θ−1)

2 ±1 m158π

cosθsinθ[±iϕ]

2 ±2 1532π

sin2θexp[±2iϕ]


(l−|m!|(l+|m!|)



Common eigenfunctions for ˆ L z


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]


(l−|m!|(l+|m!|)


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


(l−|m!|(l+|m!|)


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



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 ]



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



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



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



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



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



[ L2,Lx]=[Lx2 +Ly

2 +Lz2,Lx]

In total

−ihLyLz −ihLzLy=0 +ihLzLy +hLyLz =0

Y

Z X



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



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

lecture 14: angular momentum-ii. the material in this lecture covers the following in atkins

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