Chem 373- Lecture 14: Angular Momentum-II

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


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



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


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


r

F

dLdt

r F

rr r

= ×

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

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


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


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 ]


Angular momentum operatorsof quantum mechanics in spherical coordinates in 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−+

×


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

:

; ; ; .



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

:

; ; ; .



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

:

; ; ;



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

:

; ; ; .



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

:

; ; ; .



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

:

; ; ; .



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

:

; ; ; .



[ , ] [ , ] 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

:

; ; ; .



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

:

; ; ; .



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,......

Chem 373- Lecture 14: Angular Momentum-II

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Transcript of Chem 373- Lecture 14: Angular Momentum-II