chem 373- lecture 12: harmonic oscillator-ii

8/3/2019 Chem 373- Lecture 12: Harmonic Oscillator-II

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Lecture 12: Harmonic Oscillator-II.

The material in this lecture covers the following in Atkins.

Section 12.4 The energy levels

Section 12.5 The wavefunction

Lecture on-line

Quantum mechanical harmonic oscillator (properties)

(PowerPoint)

Quantum mechanical harmonic oscillator (properties) (PDF format)

Handout for this lectureWriteup on Harmonic Oscillator


2/29

Harmonic oscillator...Quantum mechanically .Hamiltonian

We have for the harmonicoscillator

E kxpot =

= +

1

2

2

where x is the displacementfrom equilibrium.Thus the hamiltonian is given by :H = E + E

H2m

d

dx

1

2kx

kin pot

2 2

22 h

V x kx( ) =

1

2

2

Mass

Displacement

Force constant

Review


3/29

E

1

2h

3

2h

5

2h

72h

x

9

2h

112

h

The energy

evels of a

harmonic

oscillator

are evenly

paced with

eparation, with =

k/m)1/2.

Even in its

owest state,an oscillator

has an

nergy

greaterhan zero.

Harmonic oscillator...Quantum mechanically .Energy levels

v = 0

v = 1

v = 2

v = 3

v = 4

v = 5

v = 6E v= +h( )1

2

Review


4/29

We have the general solution

v v vx Ny

H y( ) exp ( )=

2

2; y = x /

Harmonic oscillator...Quantum mechanically.... Wavefunction

It is readilly shown that

N

v =

=

1

2

1

2

2

1

2

1

2

2

v

v

v

v

v

so

x

v

yH y

!

( )

!

exp ( )


5/29

The graph of the Gaussian function,f(x) = e-x2.


bell shaped Gaussian function


6/29


_________________________

v H

1

1 2y

2 4y - 2

3 8y - 12y

4 16y - 48y +12

5 32y -160y +120y

6 64y - 48y + 72y - 120

_____________________________

v

2

3

4 2

5 3

6 4 2

0

Hermit polynominals

Note

y y

that H for v odd (1,3,5,7,..)

is odd : H = - Hv

v v( ) ( )

Note

y y

that H for v even (0,2,4,6,8...)

is even : H = Hv

v v( ) ( )


7/29


Pr

( )

' !

" '

'

operties

H yH vH

H yH vH

recursionformula

H H e dy v

v v v

v v v

v vy v

vv

of Hermitpolynominals :

+ =

=

=

+

2 2 0

2 2

2

1 1

1

22

Here

H

H

v

v

"

'

=

=

d H

dy

d H

dy

2

v2

v


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v

v

vx

v

yH y( )

!

exp ( )=

1

221

2

2

o xy

H y

y

( ) exp ( )

exp

=

=

1

2

1

2

1

2

2

0

1

2

2

For the groundstate v = 0

of the harmonic oscillatorwe have the wavefunction

The normalized wavefunction andprobability distribution (shown also by

shading) for the lowest energy state ofa harmonic oscillator.


9/29


v

v

vx

v

y H y( )

!

exp ( )=

1

221

2

2

1 1

2

2

21

22

( ) expxy

y=

For v = 1

The normalized wavefunction andprobability distribution (shown also by

shading) for the first excited state of aharmonic oscillator.


10/29

The normalized wavefunctions for the first fivestates of a harmonic oscillator. Note that the numberof nodes is equal to v and that alternate

wavefunctions are symmetrical or antisymmetricalabout y = 0 (zero displacement).


v

v

vx

v

y H y( )

!

exp ( )=

1

221

2

2


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v v vx Ny H y( ) exp ( )=

2

2

Particle canbe foundoutsideclasical region


12/29

The probability distributions for the first five states

of a harmonic oscillator represented by the densityof shading. Note how the regions of highestprobability (the regions of densest shading) movetowards the turning points of the classical motion asvincreases.


v

v

vx

v

yH y( )

!

exp ( )=

1

221

2

2


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Particle confined

in boxParticle confinedin harmonic potential

V kx=1

2

2

V = 0

Potential energyincreases more

suddenly forparticle in a box


14/29

Comparison of energy levels in harmonic oscillator

and particle in a box

Energy

v

v

Spacing

E

levels for

harmonic oscillator

E = (1

2h

h

+

=

=

)

, , ,0 1 2 3

Energy levels

in particle in

box

E =

n2h

h

2

2

2

2

81 2 3

2 18

mLn

E

nmL

=

=

+

, ,

( )

E

12 hhhh

32hhhh

52 hhhh

7

2

hhhh

92 hhhh

112 hhhh

v=0

v=1

v=2

v=3

v=4

v=5

Harmonic oscillatorParticle-in-box

n=1

n=2

n=3

n=4

n=5

h2

8 mL 2

4 h

2

8mL2

9 h

2

8mL2

16 h

2

8mL2

25 h2

8mL2

Zero-point Energy



15/29

Harmonic oscillator...Quantum mechanically.... properties

Expectation

x x dxv v

values

=

-

* ( ) ( )

x x x x dx

Ny

H y xy

H y dx

v v

v v v

=

=

-

-

* ( ) ( )

exp[ ] ( ) exp[ ] ( )22 2

2 2

v v vx Ny

H y( ) exp ( );=

2

2 y = x/

=

N

yH y

x yH y d

xv v v2 2

2 2

2 2

exp[ ] ( )( )exp[ ] ( ) ( )

-

=

N

yH y y

yH y dyv v v

2 2 2 2

2 2 exp[ ] ( ) exp[ ] ( )

-

=

N y H y yH y dy

v v v

2 2 2 exp[ ] ( ) ( )-


16/29


Properties of Hermit

polynominals :

H yH vH

H yH vH

recursionformula

H H e dy

v

v v v

v v v

v v y

v

vv

" '

'

( )

'

!

+ =

=

=

+

2 2 0

2 2

2

1 1

2

1

2

H yH vH

yH H vH

v v v

v v v

+

+

=

= +

1 1

1 1

2 2

1

2

=

N y H y yH y dyv v v2 2 2

exp[ ] ( ) ( )-

=

+

+

N y H y H y dy

N v y H y H y dy

v v v

v v v

2 2 2 1

2 2 21

12

exp[ ] ( ) ( )

exp[ ] ( ) ( )

-

-= + = 0 + 0

< x > = 0

v,v+1 v,v-1


17/29


You

x x x x dx

k v

v v

will show in assigned questions :

2 2

1

2

=

= +

-

* ( ) ( )

( )

h

We note that < x increases with vas the probability to find the particle atthe turning points increases.

Also < x decreases with k

2

2

>

>


18/29


It

V k x k x x x dx

k v k v E

v v

follows

=

= + = + =

1

2

1

2

1

2

1

2

1 1

2

1

2

1

2

2 2=-

* ( ) ( )

( ) ( )h h

We

V T E E T E T E

can find the average kinetic energy from

+ =< > + = =1

2

1

2

We

v

also have T =p

2mthus < T > =

1

2mp

E

2;

or p = 2mE; p = 2m (1

2

x2

x2

x2

x2

)

< >

< > < > +h

You are being asked to shown in assigned problems

< p = 0x >


19/29


In general for a potential

V = a x

It can be shown that

"Virial Theorem"

2

b + 2

bb + 2

b

Tb

V

V E

T E

=

=

=

2


20/29

Harmonic oscillator...Quantum mechanically..

Vibration Spectroscopy

V R V RdV

dR

R

d V

dRR

d V

dRR

e e

e e

( ) ( ) ( )

( ) ( ) ...

= +

+ + +

1

2

1

8

2

22

3

33

Taylor expansion

0

small

0

V Rd V

dRR k R

d V

dRke e( ) ( ) ;( )= = =

1

2

1

2

2

22 2

2

2


21/29

Harmonic oscillator...Quantum mechanically

The force constant is a measure of the

curvature of the potential energy closeto the equilibrium extension of thebond. A strongly confining well (onewith steep sides, a stiff bond)

corresponds to high values of k.

V Rd V

dRR k R

d V

dRk

e e( ) ( ) ;( )= = =

1

2

1

2

2

2

2 22

2


22/29


We note relation between bond energy D ;bond order and force constant k


23/29


24/29


25/29


The three

normal

modes of

H2O. Themode v2 is

predominant

ly bending,

and occursat lower

wavenumber

than the

other two.


26/29


27/29



28/29

What you should learn from this lecture

1. You are not required to remember the hermit polynomials

and their relations. However you should be able to make use of thetwo tables

_________________________

v H

1

1 2y

2 4y - 2

3 8y - 12y

4 16y - 48y +12

5 32y -160y +120y

6 64y - 48y + 72y - 120

_____________________________

v

2

3

4 2

5 3

6 4 2

0 Hermit polynominals

Pr

( )

' !

" '

'

operties

H yH vH

H yH vHrecursionformula

H H e dy v

v v v

v v v

v v y v vv

of Hermit

polynominals :

+ =

=

=

+

2 2 0

2 2

2

1 1

1

22


29/29

What you should learn from this lecture

2. You should remember H is odd for v odd and even for v even.

You should understand the meaning of odd and even functionsv

3. You should understand the problem assigned to thislecture on the vibrating diatomic molecule A - B

chem 373- lecture 12: harmonic oscillator-ii

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