chem 373- lecture 13: angular momentum-i

8/3/2019 Chem 373- Lecture 13: Angular Momentum-I

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Lecture 13: Angular Momentum-I.

The material in this lecture covers the following in Atkins.

Rotational Motion

Section 12.6 Rotation in two dimensions

Lecture on-line

Angular Momentum in 2D (PowerPoint)

Angular Momentum in 2D (PDF format)

Handout for this lecture


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Tutorials on-line

Vector conceptsBasic Vectors

More Vectors (PowerPoint)

More Vectors (PDF)

Basic conceptsObservables are Operators - Postulates of Quantum

Mechanics

Expectation Values - More Postulates

Forming OperatorsHermitian Operators

Dirac Notation

Use of Matricies

Basic math background

Differential Equations

Operator Algebra

Eigenvalue EquationsExtensive account of Operators


3/29

Audio-visuals on-line

Rigid Rotor (PowerPoint)

(Good account from the Wilson Group,****)

Rigid Rotor (PDF)

(Good account from the Wilson Group,****)


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Motion in 1D

Free translation

Confined translation1D

Harmonic Oscillation1D

Vibrating diatomic molecule


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Motion in 2D

Confined translation2D Rotation in 2D The rigid rotor

Rotating Diatomics Vibrating Diatomics


6/29

Quantum Mechanical Rotation in 2D

let us consider a particle of mass mmoving in the xy plane in a circle

of fixed radius a

k

j

i

x

y

z

a

The position of the particleis given by

r = i x + j y

where

r = r r

vv r

r

= + =x y a2 2

With the velocity given by

dr

dt= i

dx

dt+ j

dy

dt

v = iv + j vx y

vv r

r v r

Position and velocityof circular motion


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k

j

i

x

y

z

r

It is more informative towork in the sphericalcoordinates (r, )

We have

r = i x + j y

r = i rcos + j rsin

The velocity is given by

v = i drcos + j rsin

r r r

r r v

r v

dt

ddt

or

dtddt

since r is constant (r = a)

v = - i rsin d + j rcosr v

Position and velocityof circular motion in

spherical coordinates


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We note thatv | = v v

r v =

+=

a

d

dt

ad

dt

2 2 2 2

2

( ) [sin cos ]

( )

k

j

i

x

y

z

r

v

r r v

r v

r = i rcos + j rsin

v = -i rsin

d

+ j rcos

dt

d

dt

also v r v r v r

ad

dta

d

dt

Velocity

x x y y:

sin cos

sin cos

(v) perpendicular to

position vector (r).

= +

= +

=

2

2 0

s

r

Position and velocityof circular motion in



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k

j

i

x

y

z

r

v

Let us now evaluateAngular momentum:

L = r p = m r vv r r r r

We have

L = r p = ( ix + jy) ( ipxr r r r r r r

+ jpy )

= i i (x i j (x

j i (y

r r r r

r r r r

v v

v r

v r

+

+ + =

=

=

p p

p j j ypkxp kyp

L xp yp k

L m xv yv k

x y

x y

y x

y x

y x

) )

) ( )

( )

( )

Angular momentumof circular motion


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k

j

i

x

y

z

r

v

L

r rL m xv yv k

ory x= ( )

r r v

r v

r = i rcos + j rsin

v = - i rsin dr + j rcos

dt

ddt

r

r

L mrd

dt

mr ddt

k

= +[ cos

sin ]

2 2

2 2

r r

r r r

L m rd

dtk

L m r v k r p k

=

= =

2

| | | | | | | |

Angular momentumof circular motion in



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k

j

i

x

y

z

r

v

L

r rL mr d

dtk= 2

The total energy is E = Ekin +=

E

Epot

pot

0

|p mr ddt= 2

Ep

m

m rd

dt

m

mrd

dt

mrkin= = =

2

2 2 2 22

22 2 2

( ) ( )

EL

mr

EL

I

kinz

kinz

=

=

2

2

2

2

2

or since I = mr is moment of inertia2

Total energy ofcircular motion


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Quantum Mechanical Rotation in 2D...

quantitative derivation

k

j

i

x

y

z

r

x r= cos ; y = rsin

The Hamiltonian is againgiven by

H = E + Ekin pot

or since E = 0pot

Constructing quantummechanical Hamiltonian

of circular motion inCartesian coordinates

Hm x y

= + = +

p

2m

p

2mx2

y2

h2 2

2

2

22


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k

j

i

x

y

z

r

2

2

2

2

2

2 2

2

2

2 22 2

22

1 1

2

1 1

x y r r

Hm r r

+ = + +

= + +

r r

r r ( )

( ) ( ) ( )h

Since does not depend on r,

the first twoderivatives with respect to r can beneglected and we have

H m r=

h2

2

2

22

1

H = - 2m

2h d

dx

d

dy

H E

2

2

2

2+

= ( ) ( )

Next making use ofspherical coordinates

Constructing quantummechanical Hamiltonian

of circular motion inspherical coordinates


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k

j

i

x

y

z

r



or since I = mr2 : HI

=

h2 2

22

Constructing Schrdingerequation of circular motion

in spherical coordinates

Hm r

=

h2

2

2

22

1

The Schrdingerequation is

= = h

h

2 2

2

2

2 22

2

I E orIE

2

22

22

2=

=

m

mIE

l

l

h


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The general solution is m

l

m

l ml lN im

where

IE

( ) exp= [ ]

= 2

hhere m is just

at the moment a real number

l

We must have that

represent the same point inspace

m m

(r, ) ( ,

l l

ce and r

( ) ( )

sin )

= +

+

2

2


quantitative derivationSolutions to Schrdingerequation of circular motion


k

j

i

x

y

z

r

22

2

22

2

=

=

m

mIE

l

l

h


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ml

=

( )

exp ( )

exp exp

+ =

+[ ]

[ ] [ ]

2

1

22

12

2

im

im im

l

l l

Quantum Mechanical

Rotation

in 2D...


= m

m

1

21

21

2

2

exp ( )

( )

i m

m

l

l

ll

[ ]( )

= ( )

Thus for to beequal to

m

m

m

l

l

lml

or

( )

, ; ; ;....;

( ):+

( ) =

=

2

1 1

0 1 2 3

2

k

j

i

x

y

z

r

The general solution is

ml ml lN im( ) exp= [ ]

Solutions to Schrdingerequation of circular motion


Nml=

1

2


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We have that

mm

;

m

ll

l

= =

=

2

20 1 2

2 2IEThus E

Ih

h

; ; , ..

Quantum Mechanical

Rotation

in 2D...


Negative m values corrspondto rotation in one directionpositive m values to rotations

in the other direction. Form values of the same absolutevalue but opposite signs theenergies are the same. The two

states are degenerate

l

l

l

Properties of solutions toSchrdinger equation of circularmotion in spherical coordinates


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Quantum Mechanical

Rotation

in 2D...


The real parts of the wavefunctionsrepresents a particle on a ring.

EI

=

=

h2 2

20 1 2

m;

m

l

l ; ; , ..

Properties of solutions toSchrdinger equation of circularmotion in spherical coordinates

ml l

l l

im

m i m

( ) exp

{cos sin }

= [ ]

[ ] + [ ]

1

21

2

Note that the number of nodes in the realpart of increases with m .lm l ( )


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Quantum Mechanical

Rotation

in 2D...


L xp ypz y x= ( )

Angular momentum of solutions toSchrdinger equation of circularmotion in spherical coordinates

or quantum

xp yp

ix

yy

x

y x

mechanically

L

L

z

z

=

=

s

h s

or in spherical coordinates

Lz =h

i

We

iml l

note that

is an eigenfunction to L

m

z

( ) exp

= [ ]12

with eigenvalues;L m mz l l= = h 0 1 2; ; ;....


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What you should learn from this lecture

1. The angular momentum a particle with poition r

and momentum p is given as L = r p = m r v .

ofv

rv

r r r r 2. For a circular motion in the xy - plane the angular momentumis L (xp yp ) and pointing along the z - axis.z y x=

3

2

2. For a circular motion the total energy is all kinetic and given by

where I = mr is the moment of inertia2EL

Ikin

z=

4. The possible energies for the circular motion of

a particle are withEI

= = h

2 2

20 1 2

mm

ll ; ; , ..

51

2

.

( ) exp

The corresponding wavefunctions

are

the real part has m nodesl

ml lim

where

= [ ]


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What you should learn from this lecture

6

1

2. ( ) expThe wavefunctions

also eigenfunctions to L with eigenvalue mz l

ml limare

= [ ]h

71

2

2

. ( ) exp

.

Thus represents a system

with energy E =m

2Iand angular momentum m

m values represent counter clockwise motion

whereas negative values represent clockwise motion

l2

l

l

ml lim

Positive

= [ ]

hh


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Appendix : Quantum Mechanical Rotation in 2D...ualitative origin of quantization

The angular momentum of a particle of mass m on a circular pathof radius r in the xy-plane is represented by a vector of magnitude

prperpendicular to the plane.

We

EL

Ikin

zhave

=2

and

L prI mr

z == 2

and

L pr

I mr

z == 2


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+ =

[ ]

=

+( )

( , )

cos sin

x t AExpiEt Exp ikx

AExp

i

Et kx kx

h

h

p k= h

Consider a free pariclemoving to the right Its wavefunction is given by

Its momentum is given by

The wavelength isdetermined from the condition

+ += +( , ) ( , )x t x t

AExpiEt kx kx

AExpiEt k x k x

+( ) =

+( ) + +( )( )

h

h

cos sin

cos sin

k = 2 k = 2 /

p h= =h( / ) /2

de Broglie

Appendix : Quantum Mechanical Rotation in 2D...qualitative origin of quantization


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Consider a particle moving at a ring

Its wavefunction ( ) must satisfy

acceptable Not acceptable

( ) ( )= + 2

Same physical situation



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Accordingh

to QM a particle of momentum p

moves as a wave with wave length =p

Standing wave must repete itself

thus wavelength must be a integer

fraction of the circumperence

=2 r

n

de Broglie



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Thusr

m

h

p

2=

=2 r

m

=h

p

We have

Boundary cond.

de Broglie

m = 0 1 2 3, , , ,...

or

phm

r

m = 1,2,3,4,...= 2

0

,


A di Q M h i l R i i 2D


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The angular momentum of a particle confined to aplane can be represented by a vector of length |ml|

units along the z-axis and with an orientation thatindicates the direction of motion of the particle. Thedirection is given by the right-hand screw rule.

phm

r

pr m

=

= =

20

m = 1, 2, 3, 4,...

Lz

,

h

E prI

h mI

Em

I

= =

=

( )( )

,

2 2 2

2

2 2

2

0

hm = 1, 2, 3, 4,...



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Vibrational spectroscopy

E

1

2

h

3

2h

5

2h

7

2h

x

9

2h

11

2h

v = 0

v = 1

v = 2

v = 3

v = 4

v = 5

v = 6 E v= +h( )

1

2

A vibrating diatomicmolecule AB

Has the vibrational energy levels

=k

=

+

m m

m m

A B

A B

A photon of energy h

will be absorbed if

h = h = / 2


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Vibrational spectroscopy

Molecules in general absorbs photonsof specific frequencies (fingerprints)

chem 373- lecture 13: angular momentum-i

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