chem 373- lecture 9: particle in a box-i

8/3/2019 Chem 373- Lecture 9: Particle in a Box-I

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Lecture 9: Particle in a Box-I.

The material in this lecture covers the following in Atkins.

Translational Motion

12.1 A particle in a box

(a) The Schrdinger Equation

(b) The acceptable solution(c) Normalization

(d) The properties of the solutions

Lecture on-line

Particle in a box-I (PowerPoint)Particle in a box-I (PDF format)

Handout for lecture (PDF)

Writeup on solving differential equations (PDF)

Writeup on particle in 1-D box (PDF)


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Tutorials on-line

Basic concepts of importance for the understanding of the postulates

Observables are Operators - Postulates of Quantum Mechanics

Expectation Values - More Postulates

Forming OperatorsHermitian Operators

Dirac Notation

Use of Matricies

Basic math background

Differential EquationsOperator Algebra

Eigenvalue Equations

Extensive account of Operators


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Audio-visuals on-lineParticle 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 ,**)


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The particle in a box...one dimension

For a particle moving in one dimension we have for theHamiltonian :

H =p

2m+ V(x) = -

2m

d

dxV(x)x

2 2 2

2 h +

the particle is described by the wavefunction x) which is asolution to the S.W.E

(

H x) = E x)

-2m

x) x) = E x)

where E is the energy of the particle

( (

( ( ) ( (h2 2

2d

dxV x+


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-2m

x) x) = E x)h2 2

2d

dxV x ( ( ) ( (+

X=0

V= X=l

V= V= 0

V V

We shall consider a case in which V(x) is given by :

Thus inside" the box"between x 0 and x = l we have V = 0=

Outside the "box" V =


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We have:

I : -

2m

x)x) = E x) X < 0

h2 2

2

d

dx

(( (+


X=0

V=

X=l

V= V= 0

V V

I I : -

2m

x)x) = E x) 0 < X < l

h2 2

2

d

dx

V

(

( (+

III: 2m

x)

x) = E x) X > l

h2 2

2

d

dx

(

( (+


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For region I :

-2m

x)x) = E x) X < 0

h2 2

2

d

dx

(( (+


Or :

-2m

x)x)

h2 2

2

d

dxE

([ ] (=

X=0

V= X=l

V= V= 0

V V

Since E


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X=0

V=

X=l

V= V= 0

V V

We have

x) =1

2m

x)

(

(

h

2 2

2

d

dx

Must befinite

Also for region III where X > l

-

hus:( x) = 0 for x < 0

we have x) =1

2m

x)

(

(

h

2 2

2

d

dx

and must require x) = o for x > l(

< > Td

dx

dx= -2m

2h

*

2

2


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X=0

V=

X=l

V= V= 0

V VRegion I Region II Region III

We havex ) = x ) = 0 I III ( (

{ Probability of finding particel in III

"No particle outside box"

Thus

P x) x) = 0

P x) x) = 0

I

III

( ) ( (

( ) ( (

*

*

x

x

I I

III III

=

=

{ Probability of finding particel in I


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X=0

V=

X=l

V= V= 0

V VRegion I Region II Region III

Now for region II :

-2m

x)x) = E x)

-2m

x)= E x)

h

h

2 2

2

2 2

2

d

dxV

d

dx

(( (

((

+

d

dx

mE

d

dx

mE

2

2 2

2

2 2

2

2

((

((

x) x)

or

x)

= - k x) ; k =2 2

= h

h


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We have

cos[kx]cos[kx]

and

sin[kx] sin[kx]

d

dx

k

ddx

k

2

22

2

22

=

=

Thus ( ) cos[ ] sin[ ]x A kx B kx= +

x)= - k x) ; k =2 2

d

dx

mE2

2 2

2

((

h


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X=0 X=l

V V

= 0 = 0

Thus (x) = A cos[kx] + Bsin[kx]

()=0 (l)=0

We must have (x) 0 since the wavefunction must

be continous according to BornII =


Thus x B kx: ( ) sin[ ]=

Thus we get the BOUNDARY condition :II A k B k( ) cos[ ] sin[ ]0 0 0 0= + =

which is only possible if A = 0


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B = 0

We must also have ( ) sin[ ]

:

l B kl

possible solutions

= = 0

X=0 X=l

V V

= 0 = 0

Thus (x) = Bsin[kx]

()=0 (l)=0

or k = 0Both trivial and will make the wave functionzero everywhere

Not acceptable on physical grounds


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X=0 X=l

V V

= 0 = 0

Thus

(x)=

Bsin[kx ]

()=0 (l)=0

Non trivial solution Introducing k = k'l

For x = l :'

'

II l B kl Bk

l

l B k( ) sin[ ] sin[ ] sin[ ]= = =

For x = l' for k' = n n = 1, 2, 3,..... II l B k( ) sin[ ]= = 0


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We x B

n

l xhave the non - trivial solution :

( ) sin[ ]=

n = 1; 2; 3,....

We have further that sin[ -n

lsin[

n

lThus we need only consider solutions n = 1,2,3,...

x x] ]=

Further since

x)

= - k x) ; k =2 2d

dx

mE2

2 2

2

((

h

h; E =

k

2m

2 2

k =n

ln = 1,2,3,4. = Note E 0

; E

n

ml

h n

ml=

h2 2 2

2

2 2

28 8


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The allowed energy levels for a particle

in a box. Note that the energy levelsincrease as n

2, and that their

separation increases as the quantumnumber increases.

Eh n

ml

=2 2

28

Lowest level n = 1, energy not zero why ?


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The first five normalized wavefunctions

of a particle in a box. Eachwavefunction is a standing wave, andsuccessive functions possess onemore half wave and a correspondingly

shorter wavelength.

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

Wavefunctionsn

x=2

l ln = 1 , 2

n

sin[ ]


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Wavefunctions n x= 2l l

n = 1 , 2

n sin[ ]

Pr : *obability

n = 1 , 2Note particle most

likely to be found in

middle for n = 1

n n

Energy:

n=1,2,..Eh n

ml=

2 2

28

(a) The first two wavefunctions, (b) thecorresponding probability distributions, and (c) a

representation of the probability distribution interms of the darkness of shading.

From normalization


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For n = 1n

For n n = large (20)

probability

of findingparticle largest in middle

of box

*n n dx

probability

of finding particle samethroughout box

(classical limit)

*n n dx


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What you should learn from this lecture1. For the particel in a box You should remember the

dependence of the energy on the quantum number nand the box length l. You should also note the lowerlimit (n = 1) for n and the consequence it has for theminimum energy (zero point energy)

Energy :

n=1,2,..Eh n

ml

=2 2

28


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Wavefunctions

nx=

2

l ln=1,2n

sin[ ]

What you should learn from this lecture

2..

.

You should also know roughly the form of thestatefunctions Especially, you should note thatn

n

nodes


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For the solution we can fix B by the normalization requiremet

(x) = B sin[n

lx]

* * * *( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x x dx x x dx x x dx x x dxl

I

l

I II

ll

II IIIl

III

= + + 0

0

o

Thus x dx

Introducing

x dx

xx l n

: I = sin[n

n

ld

n

l

Limits of

0

l

l =

= =

= == =

]

:

:

:

2 1

0 0

Thus

d

I

I B o on

I Bl

n

o

n

I = B

B

2

2

2 2

l

nsin

ln

l

n

2

0

12

14

2

2

0

2

=

= + =

sin

[ ]

Thus B and | B | =2

land B =

2

l

is any real number. Note : B2

l

2

l B

2

2 2

=

= = =

2

2

li

where B B i i l

exp[ ]

exp[ ] exp[ ]*

o

Appendix N: ormalization


23/24

f t Expi

Et( ) =

h

( , ) ( ) ( )x t f t x=


n x t

ni

iEt( , ) sin[ ] exp[ ]=

2

l

l h

n xn

i( ) sin[ ] exp[ ]=2

l l

Phase factor

We shall deal (mostly) with time time independent(stationary) states where the properties are determined

by time independent operators A where :

< >= = +

A x t A x t dx n i i Et A

ni

iEt dx

n n * *( , ) ( , ) sin[ ] exp[ ]

sin[ ] exp[ ]

2l

2

l

l

l

h

h

Appendix : TimeDependence


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< >= = +

A x t A x t dxn

ii

Et A

ni

iEt dx

n n * *( , ) ( , ) sin[ ] exp[ ]

sin[ ] exp[ ]

2

l

2

l

l

l

h

h

sin

exp[ ] exp[ ] sin[ ]]

sin[ ]

ce

A ii

Et ii

Etn

A

ndx

A time - independent2

l

2

l

< >= +

h h l

lThus

2

l

2

l

we need only be concerned with the

real solutions

< >=

=

=

An

An

dx

A dx

n

n n

n

sin[ ]] sin[ ]

sin[ ]

*

l l

l

Appendix : TimeDependence

chem 373- lecture 9: particle in a box-i

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