chem 373- lecture 15: angular momentum-iii

8/3/2019 Chem 373- Lecture 15: Angular Momentum-III

1/21

Lecture 15: Angular Momentum-III.

The material in this lecture covers the following in Atkins.

Rotational Motion

Section 12.7 Rotation in three dimensions

Lecture on-lineAngular Momentum-III (3D. Part-2 (PDF)

Angular Momentum-III (3D. Part-2) (Powerpoint)

Handout for this lecture


2/21

Rotation Quantum..

Mechanics 3D

Consider now a particlemoving on the surface of asphere with the radius a

Its Hamiltonian is given by

H = -2m

-2m

the potential energy isuniform over the sphere and canbe put to zero

2 2h h + = 2 2V

cesin

= + +22

2

2

2

2

2

x y z

We must solve

- 2m

2h

=2

E

Schrdinger eq. for particle movingon sphere with radius a


3/21

Rotation Quantum..

Mechanics 3D

Use (r, , )not (x,y,z)

= + 22

2

2

2 2

2

r r r

L

r

h


[sin

sin sin ]

[ cotsin

]

L d

d

d

d

d

d

d

d

d

d

d

d

22

22

2

2 2

2

2

1

1

= +

+ +

= -

2

2

h

h

We must solve

-2m

2h =2 E


4/21

Rotation Quantum..

Mechanics 3D

We must solve -2m

2

h

h = = + 2 2

2

2

2

2 2

2

E

r r r

L

r;

+ =h

h

2 2

2

2

2 22

2

m

a

r r

a

r

L a

rE a[

( , , ) ( , , ) ( , , )] ( , , )

[

( , , )] ( , , )

L a

ma E a

2

22

=

Wavefunction does not depend on r as a variable

( r is a constant r = a). Thus only last term on r.h.s.different from zero


[

( , )

] ( , )

L

ma E

2

22

=

We must have H =1

2 2

2

ma

L


5/21

Rotation Quantum..

Mechanics 3D

We Emust solve H = ( , ) ( , )

We must have H =1

2 2

2

maL

What is ( , ) ?

Thusma

L E

or

L E

1

= 2ma

2

22

2

2

( , ) ( , )

( , ) ( , )

=

Thus the eigenfunctions ( , ) to H must be

the eigenfunctions to L2

.

( , ) = Y ( , ) =2 +1

4P

L Y Y L Y m Y

, l|m|

2 ,m ,m z l,m ,m

ll

ll ll ll

ll ll

ll

ll ll

mm

mim(

( | !|

( | !|)(cos ) exp[ ]

( )

+

= + =h h2 1 ;



6/21

If ( , ) = Y ( , )lm

and what is E ?

Rotation Quantum..

Mechanics 3DSchrdinger eq. for particle movingon sphere with radius a

= 2ma = YThus

ma

= 2 ma E

Finally

E =

2

m

2

2

( , ) ( , ); ( , ) ( , )

( , ) ( ) ( , ) ( , )

( )

( )

L E

L Y Y EY

or

ma

m m m

2

2 2

2

2

2

1 2

1

1

2

ll

ll ll llll ll

ll ll

ll ll

= + =

+

+

h

h

h= ; I = ma

2h2 1

2

ll ll( )+

I


7/21

Rotation Quantum..

Mechanics 3D

H, L and L for a particle

moving on a sphere havecommon eigenfunctions Y

2z

mll

( )L Y ( , ) = Y ( , )2 m 2 mll llll ll h + 1

L Y ( , ) = Y ( , )z m mll ll hm

HY

IYm mll ll

ll ll( , )

( )( , ) =

+h2 1

2



8/21

Rotation Quantum..

Mechanics 3D

H, L and L for a particle moving on a sphere havecommon eigenfunctions Y

2 z

lm


We already knew that L and L had commen

eigenfunctions since [L ,L

z2

z2

] = 0However H =

1

22

2

maL

Thus it is readily shown that [H,L ] = [H,Lz

2 ] = 0

This explains why H,L and L have common eigenfunctionsz2

For Y

m

EI

m

obs obs

obs obs

a state described by measurements of

L L and will give the outcomes

(L (L

= (L2I

each time

z 22

z2

ll

ll ll

ll ll

( , )

,

) ( ); )

( ) ( ) )

= + =

= +

h h

h

2

2

1

12


9/21

Rotation Quantum..

Mechanics 3D

( , ) = Y ( , ) =2l+1

4 P

L Y Y L Y m Y

l,m l|m|

2l,m l,m z l,m l l,m

(

( | !|

( | !|) (cos ) exp[ ]

( )

l m

l m im

l l

+

= + =h h2 1 ;

For

L obs obs

l = 0 we have m = 0;

Y

(L

o,o

z

=

= =

1

4

0 02

( ) ; )

Value of Y is uniform over

sphere

oo

l = 0 m = 0

Properties of solutions toSchrdinger eq. for particle moving

on sphere with radius a


10/21

l = 1 m = 0 1 1, ,

( , ) = Y ( , ) =2l+1

4P

L Y Y L Y m Y

l,m l|m|

2

l,m l,m z l,m l l,m

(( | !|

( | !|)(cos ) exp[ ]

( )

l m

l mim

l l

+

= + =h h2

1 ;

Rotation Quantum..

Mechanics 3DProperties of solutions toSchrdinger eq. for particle moving


ll m Y

2

1 2

- 1 - 2

lm

2

2

2

(L (L

1 0 Y

Y

Y

z2

1,0

1,1

1,-1

) )

cos

sin exp[ ]

sin exp[ ]

obs obs

i

i

=

=

=

3

40

1 38

1

13

81

h h

h h

h h


11/21

L

Lz

Lz

|L| =

LL

Lz

=

L z = -

h

h

= 0

2h

Rotation Quantum..

Mechanics 3D

For L mobs obsl = 1 we have m = -1,0,1; and (Lz( ) )2 2

2= =h h

We have three states all with Lis the length |L | of L is in all

three cases |L |= 2

L is oriented differentlyin the three states withL

2

z

=

=

hr

hr

h h

2 2

0

That

However

, ,

For

will

Lobs z obs

each of the states a

meassurement of L and Lalways give the same

valuee.g l=1,m=1

gives (L

2 z

2) ;( )= =2 2h h




12/21

Rotation Quantum..

Mechanics 3DProperties of solutions toSchrdinger eq. for particle moving


L

Lz

Lz

|L| =

LL

Lz

=

L z = -

h

h

= 0

2h

For L

mobs

obs

l = 1 we have m = -1,0,1;and (Lz

( )

)

2 22==

h

h

What about L or L ?

We have (L L

(L ) - (L ) =

a meassurementof L or L can have as outcome

- ,0,

x y

x y2

2 obs z2 obs

x y

2

2 2 22

+ =

=

)obs

However

h h h

h h


13/21

Forwill

Lobs z obs

each of the states a meassurement of L and Lalways give the same value e.g l = 1,m = 1

gives (L

2

z

2) ;( )= =2 2h h

What

representing

for

each meassurement

about the expectationvalues < L and < L

the avarage valuefrom many meassurementsand what are the possiblevalues for L and L

x y

x y

> >

?

Rotation Quantum..

Mechanics 3D



For L

mobs

obs

l = 1 we have m = -1,0,1;and (Lz

( )

)

2 22=

=h

h

L

L z

L z

|L| =

LL

Lz

=

Lz

= -

h

h

= 0

2h


14/21

Rotation Quantum.. Mechanics 3D

( , ) = Y ( , ) = 2l+14

P

< L L m

l,m l|m|

2z l

( ( | !|( | !|)

(cos ) exp[ ]

( )

l ml m

im

l l

+

>= + < >=h h2 1

m Y (L ) (L )

0 Y

Y

Y

z obs2

obs

2,0

2, 1

2, 2

llm

2

2

2

0 6

1 1 6

2 2 6

=

=

=

15

83 1

158

15

322

2

( cos )

sin cos exp[ ]

sin exp[ ]

h h

m h h

h h

i

i

l = 2 m = 0 1 1 2 2, , , ,


15/21

h

|L| = 6h

L

Lz

Lz

= 0L

L

Lz =2

Lz

= -

hLz

=

h2

hL

z

= -


( , ) = Y ( , ) =2l+1

4P

< L L m

l,ml

|m|

2z l

(( | !|

( | !|)(cos ) exp[ ]

( )

l m

l mim

l l

+

>= + < >=h h2 1

For Ll = 2 we have m = -2, -1, 0,1, 2 < >=2 26h

We have five states all with Lis the length | L | of L is in all

three cases |L |= 6

L is oriented differentlyin th three states with

L

2

z

=

=

hr

hr

h h h h

26

2 0 2

That

However

, , , ,

For

will

Lz

each of the states ameassurement of L and L

always give the same

value e.g l = 2,m = 1

gives < L

2z

2 >= < >=6 2h h;


16/21


(a) A summary of the Fig. 12.31. However, becausethe azimuthal angle of the vector around the z-axisis indeterminate, a better representation is as in (b),where each vector lies at an unspecified azimuthal

angle on its cone.

Properties of solutions to



17/21


The permitted orientations of angular momentumwhen l= 2. We shall see soon that thisrepresentation is too specific because the azimuthal

orientation of the vector (its angle aroundz

) isindeterminate.

( )L Y ( , ) =2 lm 2 h l l + 1

L Y ( , ) =z lm hm

HY

l l

ma Ylm lm( , )

( )

( , ) =

+h2

2

1

same energy

Different orientation


18/21

Electron spin

Stern and Gerlach observed in 1921that a beam of electrons (Ag- atoms)split in an inhomogeneous magneticfield into two beams

They attributed to an internal

spin angular momentumof the electron


19/21

Electron spin

An electron spin (s = 1/2) can take only two

orientations with respect to a specified axis.

An electron (top) is an electron with ms=

+1/2;

a electron (bottom) is an electron with ms= - 1/2.

rS

rSThe length of the spin- angular

momentum is

l s= =1

2m ms= =

1

2

|S = s(s +1) =

1

2(1

2+1)h h

h=

3

2


20/21

What you must know from this lecture

1. For a particle moving on a spherewe have that H, L and L commutes.

Thus they must have common eigenfunctions.

These eigenfunctions are the speherical harmonics

2z

2. The corresponding eigenvalue equations are

L Y ( , ) = Y ( , )

L Y ( , ) = Y ( , )

2 m 2 m

z m m

( )

( , ) ( ) ( , )

ll ll

ll ll

ll ll

ll ll

ll ll

h

h

h

+

= +

1

12

2

m

HYI

Ym m


21/21

What you must know from this lecture

3

1

1

2

2

2

. ( , )

,

) ( ); )

( )( )

a state described by measurements of

L L and will give the outcomes

(L (L

each time

z2

2z

For Y

m

EI

m

obs obs

obs

ll

ll ll

ll ll

= + =

=+

h h

h

4 1. ( , )

(

a given we have 2 eigenfunctions

with the same energy and the same length

L = 1) of the angular momentum.The all have different projections (orientations) L

of the angular momentum onto the z - axis.

2

z

For Y

all

m

mll ll

ll ll

ll+

+ =

h

h

chem 373- lecture 15: angular momentum-iii

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