chem 373- lecture 10: particle in a box-ii

Lecture 10: Particle in a Box-II. Lecture on-line Particle in a box-II (PowerPoint) Particle in a box-II (PDF format)

Assigned problems Handout for lecture Writeup on particle in a 3D-box

Translational Motion12.1 A Particle in a Box(d) The properties of the solution

12.2 Motions in two dimensions

Tutorials on-line

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

Audio-visuals on-line Particle in a box (PowerPoint) (good short account of particle in a box by the Wilson group, ****) Particle in a box (PDF) (good short account of particle in a box by the Wilson group, ****) Slides from the text book (From the CD included in Atkins ,**)

X=0

V= ∞

X=l

V= ∞V= 0

V VRegion I Region II Region III

For regions I and III

-2m

x)x) = E x)

-2m

x)x) = E x)

x) = 1

2mx)

h

h

h

2 2

2

2 2

2

2 2

2

d

dxV

d

dx

d

dx

ψ ψ ψ

ψ ψ ψ

ψ ψ

(( (

(( (

((

+

+ ∞

∞

:

-2m

x)x) = E x)

-2m

x)= E x)

For region II

Vh

h

2 2

22 2

2

d

dxd

dx

ψ ψ ψ

ψ ψ

(( (

((

+

Re view particle in1-D box. Schrödinger eq.

The particle in a box...one dimension

Eh n

ml=

2 2

28

Re view particle in1-D box. Energy and wavefunction

Wavefunctionsn

x =2l l

n = 1,2,3,4....nψ π

sin[ ]

The particle in a 1D box...propeties of the solutions

Possible expectation values < A >ˆ

Ai

ddx

= − ∴h Momentum

Expectation value of p

p p dxi l

npl

x ddx

npl

x dx

i l

npl

xnpl

npl

x dx

x

x n x nl

l

<

ˆ

ˆ ˆ sin[ ] sin[ ]

sin[ ]( )cos[ ]

* *>= ∫ = − ×∫

= − ∫

ψ ψ h

h

2

20

0

=2l l

n = 1,2,3,4....nψ π

sin[ ]n

x

< > − = − =∫ ∫ˆ sin cos sinpi

di

dxn n

=h h2

l1l

θ θ θ θ θπ π

0 02 0

Introducing x dx n = nl

; dnl

limits : to θ π θ π θ θ π= = =; 0

= = − = − =h h h

i l i ln

i ln1

22

12

0 212

1 1 00[cos ] [cos cos ] [ ]θ ππ

Expectation value < p for particle in 1-D box

x >

=2l l

= - i2lexp

l + i

2lexp

l

which is a superposition of eigenfunctions to p with

eigenvalues 2l

and -2l

n

x

ψ π π πsin[ ] [ ] [ ]

ˆ

nx i

nx i

nx

nh nh

−

< >Px = 0


x >The particle in a 1D box...propeties of the solutions

Does the particle move ? What would be the outcomeof a meassurement of Px ?


Possible expectation values < A >ˆ

A

md

dx= - Kinetic energy

h2 2

22∴

Does the particle move ? What would be the outcome

of a meassurement of Px2 ?

=2l l

n = 1,2,3,4....nψ π

sin[ ]n

x


x2 >


pl l l

l l l

x2

x2

ˆ

ˆ ( sin[ ]) ( )( cos[ ])

( )( )( )( sin[ ])

ψ π π π

π π π ψ

n

n

d

dx ln

xd

dx

nl

nx

n nl

nx

h n

l

= − = −

= − − =

h h

h

22

22

22 2

2

2 2

12

4


p p p

x2

x2

x2

x2

ˆ

ˆ ˆ ˆ* *< >= ∫ = ∫ =ψ ψ ψ ψn n n ndxh n

ldx

h n

l

2 2

2

2 2

24 4


Wavefunctionsn

x =2l l

n = 1, 2, 3, , 4, 5, 6...

nψ πsin[ ]

We xThus

observe that has n - 1 nodes where (green) has zero; (blue) has one; (cyan) has one

etc..

n nψ ψψ ψ ψ

( ) .= 0

1 2 3

Wel x l x xo o o

also observe that for n odd are symmetrical around midpoint at x = l/2 where x is any value o < l/2We call such functions even

n

n n o

ψψ ψ( / ) ( / )2 2+ = − >

Wel x l x xo o o

also observe that for n even are asymmetrical around midpoint at x = l/2 where x is any value o < l/2We call such functions odd

n

n n o

ψψ ψ( / ) ( / )2 2+ = − − >

The first five normalized wavefunctionsof a particle in a box. Eachwavefunction is a standing wave, andsuccessive functions possess onemore half wave and a correspondinglyshorter wavelength.

Properties of : Nodesand parity

nψ

Why do different solutions and havedifferent number of nodes if n m ?

nψ ψm≠

Because must have both positive an negative

regions for to integrate to zero

n m

-m

ψ ψ

ψ ψn dx∞

∞∫

Two functions are orthogonal if the integral of theirproduct is zero. Here the calculation of the integralis illustrated graphically for two wavefunctions of aparticle in a square well. The integral is equal to thetotal area beneath the graph of the product, and iszero.

The particle in a 1D box...propeties of the solutionsProperties of : Origin of Nodes

nψ

Particle in 2D box

Particle at : r = xe yex yv r r

+

Total Epot Energy of particle E = Ekin +

Em

p p V x y Hx y= + + =12

2 2[ } ( , )

Transition to QM

pi

ddx

pi

ddy

Hm

d

dx

d

dyV x y

x y

ˆ ; ˆ

[ ] ( , )

→ − → −

= − + +

h h

h2 2

2

2

22

Hamiltonian and Schrödinger eq. in 2D

We have for the Schrödinger eq

md

dx

d

dyV x y E

.

h2 2

2

2

22[ ] ( , )

ψ ψ ψ ψ+ + =

Particle in 2D box

We must solve H = EWe assmume :

ψ ψ

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

Thus (x,y) = 0 outside box.ψ

Inside box we must solve

h2 2

2

2

220

md

dx

d

dyV x y E

with V x y

[ ] ( , )

( , )

ψ ψ ψ ψ+ + =

=

Boundary conditions

Separation of variables We shall look for solutions of the form

is a product of two functions dependingon a single variable (separation of variables)

ψψ

( , ) ( ) ( )( , )

x y f x f ywhere x y

= 1 1

For this to be a solution we must haveh2 2

2

2

22md

dx

d

dyV x y E[ ] ( , )

ψ ψ ψ ψ+ + =

or

md f f

dx

d f f

dyV x y f f E f f

h2 21 22

21 22 1 2 1 22

[( ) ( )

] ( , )( ) ( )+ + =

Particle in 2D box

or

mf x f y f x f y Ef x f y

h21 2 1 2 1 22

[ ( ) ( ) ( ) ( )] ( ) ( )" "+ =

or

mf x

f x

f yf y

Eh2

1

1

2

22[

( )

( )

( )( )

]" "

+ =

Particle in 2D box

For a given x this relationmust hold for all x Thus

mf x

f xE

also

mf yf y

E

E E

x

y

x y

. :

E =

− =

− =

+

h

h

21

1

22

2

2

2

"

"

( )

( )

( )( )

]

h21

1

2

22mf x

f x

f yf y

E[( )

( )

( )( )

]" "

+ =

Separation of variables

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

We have

mf x

f xE

f f a

mf x E f x

f xa

na

xh n

ma

x

x

x x

; o x a

= 0

E

x

− = ≤ ≤

=

− =

= =

h

h

21

1

1 12

1 1

1

2 2

2

2

0

22

8

"

"

( )

( )

( ) ( )

( ) ( )

( ) sin[ ];π

Separation of variables Particle in 2D box

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

Separation of variables Particle in 2D box

We have

mf y

f xE

f f b

mf y E f y

f yb

n

bx

h n

mb

y

y

y y

; o y b

= 0

E

y

− = ≤ ≤

=

− =

= =

h

h

22

1

2 22

2 2

2

2 2

2

2

0

22

8

"

"

( )

( )

( ) ( )

( ) ( )

( ) sin[ ];π

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

Particle in 2D box

Thus

Thus (x,y) =4

ab

ψ π πsin[ ]sin[ ]

na

xn

by

Eh n

ma

h n

mb

x y

x y= +2 2

2

2 2

28 8

For a = b

Eh

man nx y= +

2

22 2

8[ ] Dege

nerate−

Degenerate

−

For a = b

(x,y) =2a

ψ π πsin[ ]sin[ ]

[ ]

na

xn

ay

Eh

man n

x y

x y= +2

22 2

8

n n E( )

x xh

ma

2

28

1 1 2 2 1 5

1 2 5

2 2 8

2 3 13

3 2 13

Separation of variables

What you should learn from this lecture 1

1 2 3.

( , , ...).

The solutions to the Schrödinger eq. for a 1Dparticle in a box are not eigenfunctionsto p They represents states for which a measurement

of p would give hn2l

and < p = 0

nx

x x

ψ n =

± >2

1 2 3

4

4

2

22

2

.( , , ...)

.

The solutions to the Schrödinger eq. for a 1Dparticle in a box are eigenfunctions

to p with eigenvalues h

They represents states

fo r which a measurement of p would give h

n

x2

2

x2

2

ψ n

n

ln

l

=

31 2 3

.( , , ...)

*

The solutions to the Schrödinger eq. for a 1Dparticle in a box have nodes tosatisfy the orthonormality condition

nψ

ψ ψ ∂

n

dxm n nm

=

∫ =

What you should learn from this lecture

4.

(x,y) =4

ab

For a 2D box of sides a and b the wavefunctionis zero outside the box and given by

inside

ψ π π( , ) sin[ ]sin[ ] ,nx ny

x yy

x y

na

xn

by n

Eh n

ma

h n

mb

(n positive integers)

The corresponding energy is given by

x

= +2 2

2

2 2

28 8

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

The particle in a 1D box..Appendix on orthonormalization

the two solutions and have the properties that

if m n

We say that and areorthogonal

n m

n m-

n m

ψ ψ

ψ ψ

ψ ψ

*

∞

∞∫ = ≠dx 0

We have

=2l l

u = m,n

2l l

2l l

n m-

u

-

I dx

withu

In

xm

x dx

ce

nm

nm

= ∫

= ∫

= − − +

∞

∞

∞

∞

ψ ψ

ψ π

π π

α β α β α β

*

sin[ ]

sin[ ] sin[ ]

sin sin sin [cos( ) cos( )]12

Thus1l l

1l l

l

l

l

l

-I

n m xdx

n m xdx

or

I

n m x

n m

n m x

n m

I n mn m

nm

nm

o

l

nm

= −∫ − ∫

+

=

−

− −

+

+

= − + − −+ =

∞

∞

−∞

∞cos[

( )] cos[

( )]

sin[( )

( )

sin[( )

( )

{ sin[( ) ]sin[( ) }

π π

π

π

π

π

ππ

0 00

We thus have

if n m

if n m

n m-

n m-

n m-

ψ ψ

ψ ψ

ψ ψ δ

*

*

*

∞

∞

∞

∞

∞

∞

∫ = ≠

∫ = ≠

∫ =

dx

dx

or

dx nm

0

1

The particle in a 1D box...conjugated molecules

AAnn aa cctt uuaa ll ss yy sstt eemm tt hhaa tt aapp pprr oo xx ii mm aatteess tt hhee pp aa rrtt iicc ll ee ss-- ii nn --aa --bb ooxx pp rr oo bb llee mm ss::

A number of molecules have conjugated π-bonds in which the

electrons are delocalized over the whole molecules

H H H H H H

H2C=C– C=C--C=C- C=CH2

(1) They consists in the first place of a chain of atoms heldtogether by a system of σ-bonds

(2) In addition to the electrons that participate in the σ-bonds there

are present in the molecule electrons that are delocalized.These

electrons are refered to as π-electrons.We shal refer to their

number as nπ

Appendix for Lab 5 on Dyes

For the particle in a box we have the energy levels

E

E

E

E

n

n-1

2

1

With the energies

En = h2n2

8ml2

and associated orbitals

ψn(x) =

2

l sin

nπxl



To obtain the many-electron system of lowest energy we begin tofill the levels of lowest energy

E

E

E

E

n

n-1

2

1

For nπ -electrons,where we shal assume that nπ is even, a total

of nπ2 levels will be occupied



E

E

2

1

E nπ2

E nπ2+1

HOMO

LUMO

The total energy of the system is given by

Eground = 2*E1 + 2*E2 + ........ 2*Enπ2



For the first excited state we promote one electron out of theHOMO and into the LUMO

∆E = ELUMO - EHOMO = h2

nπ

2 + 1 2

8ml2 -

h2

nπ

22

8ml2

∆E = h2

8ml2 (nπ+1)

This is the energy required to excite the groundstate to the firstexcited state.We have that the excitation energy ∆E is related tothe frequency at which the excitation occure by

ν = 1

hc ∆E

Thus the absorption energy frequency is

ν = h

8ml2c (nπ+1)



Particle in 2D box

For a = b

Thus (x, y) =2a

fold generate

ψ π πsin[ ] sin[ ]

[ ]

na

xn

ay

Ema

n nma

n n

One

x y

x y

x y

= + =

= =

h h2

22 2

2

282

8

1

The wavefunctions for a particleconfined to a rectangular surface depicted as contoursof equal amplitude.(a) n 1 = 1, n2 = 1, the state oflowest energy

Appendix : Solutions toparticle in 2D - box

Particle in 2D box

Degenerate : States with the sameenergy have different wavefunctions

For a = bThus (x,y) =2a

; n fold generate

ψπ π

sin[ ]sin[ ]

[ ]

na

xn

ay

Eh

man n

h

ma

n n ntwo

Eh

ma

x y

x y

x y x y

= + =

= = = =

=

2

22 2

2

2

2

2

82

8

1 2 2 1

58

The wavefunctions fora particle confined to a rectangular surface depicted as contoursof equal amplitude.(b) n1 = 1, n2 = 2,(c) n1 = 2, n2 = 1,

Particle in 2D box

The wavefunctions for a particle confined to asquare surface. Note that one wavefunction can beconverted into the other by a rotation of the box by90°. The two functions correspond to the sameenergy. Degeneracy and symmetry are closelyrelated.


Particle in 2D box


ψπ π

sin[ ]sin[ ]

[ ]

na

xn

ay

Eh

man n

h

ma

x y

x y= + =2

22 2

2

282

8

n none

Eh

ma

x y= =

=

2 2

88

2

2

fold generate


Particle in 2D box


; n fold generate

ψπ π

sin[ ]sin[ ]

[ ]

na

xn

ay

Eh

man n

h

ma

n n ntwo

Eh

ma

x y

x y

x y x y

= + =

= = = =

=

2

22 2

2

2

2

2

82

8

3 2 2 3

138


chem 373- lecture 10: particle in a box-ii

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