lecture 10: particle in a box-ii. lecture on-line

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

QuickTime™ and aGIF decompressor

are needed to see this picture.

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

-h2

2md2ψ(x)

dx2+Vψ(x)=Eψ(x)

-h2

2md2ψ(x)

dx2+∞ψ(x)=Eψ(x)

ψ(x)= 1∞

h2

2md2ψ(x)

dx2

For region II :

-h2

2md2ψ(x)

dx2+Vψ(x)=Eψ(x)

-h2

2md2ψ(x)

dx2=Eψ(x)

Review particle in1-D box. Schrödinger eq.

The particle in a box...one dimension

E =h2n2

8ml2

Review particle in1-D box. Energy and wavefunction

Wavefunctions ψn =2lsin[

nπl

x]

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

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

Possible expectation values < ˆ A >

ˆ A =−hi

ddx

∴ Momentum

Expectation value of ˆ p x

<ˆ p x >= ψn*

∫ ˆ p xψn*dx=−

hi2l

sin[npl

x]×

0

l∫

ddx

sin[npl

x]dx

=−hi2l

sin[npl

x](npl

)cos[npl

x]dx

0

l∫

ψn =2lsin[

nπl

x]

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

<ˆ p x > =−

hi2l

sinθ0

nπ∫ cosθdθ=−

hi1l

sin2θ0

nπ∫ dθ0 =

Introducing θ= nπl

x ; dθ=nπl

dx; limits: θ=0 to θ=nπ

=

hi

12l

[cos2θ]0nπ=

hi

12l

[cos0−cos2nπ]=hi

12l

[1−1]=0

Expectation value <px > for particle in 1-D box

ψn =2lsin[

nπl

x] = -i2lexp[i

nπl

x] +i2lexp[−i

nπl

x]

which is a superposition of eigenfunctions to ˆ p x with

eigenvalues nh2l

and -nh2l

<ˆ P x > = 0

Expectation value <px > for particle in 1-D box


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








Possible expectation values < ˆ A >

ˆ A =-

h2

2md2

dx2 ∴ Kinetic energy

Does the particle move ? What would be the outcome

of a meassurement of Px2 ?

ψn =2lsin[

nπl

x]

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

Expectation value <px2 >

for particle in 1-D box

Expectation value of ˆ p x2

ˆ p x2ψn =−h2 d2

dx2(

2lsin[

nπl

x])=−h2 d

dx(nπl

)(2lcos[

nπl

x])

=−h2(nπl

)(−1)(nπl

)(2lsin[

nπl

x])=h2n2

4l2ψn

Expectation value of ˆ p x2

<ˆ p x2 >= ψn

*∫ ˆ p x

2ψndx=h2n2

4l2ψn

*∫ ˆ p x

2ψndx=h2n2

4l2


Wavefunctions ψn =2l

[sinnπl

x]

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

We observe that ψn has n- 1 nodes whereψn(x) =0.Thus ψ1 ( ) ;green has zero ψ2 ( ) ;blue has one ψ3 ( ) cyan has one

..etcWe also observe that ψn for n odd are symmetrical around midpoint at x= /2l ψn(l / 2 + xo) =ψn(l / 2 −xo ) where xo is any value o< xo > /2l

We call such functions evenWe also observe that ψn for n even are asymmetrical around midpoint at x= /2l ψn(l / 2 + xo) =−ψn(l / 2 −xo) where xo is any value o< xo > /2l

We call such functions odd

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 ψn : Nodesand parity

Why do different solutions ψn andψm have different number of nodes if n≠ ?m

Because ψn ψm must have both positive an negative

regions for ψn-∞

∞∫ ψmdx to integrate to zero

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 ψn : Origin of Nodes

Particle in 2D box



Particle at : v r =x

r e x +y

r e y

Total Energy of particle E=Ekin+Epot

E =1

2m[ px

2+py2}+V(x,y)=H

Transition to QM

ˆ p x→ −hi

ddx

;ˆ p y→ −hi

ddy

H =−h2

2m[d2

dx2+

d2

dy2]+V(x,y)

Hamiltonian and Schrödinger eq. in 2D

We have for the Schrödinger eq.

h2

2m[d2ψ

dx2+

d2ψ

dy2]+V(x,y)ψ=Eψ

Particle in 2D box

We must solve Hψ =Eψ We assmume:

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

Thus ψ(x,y)=0 outside box.

Inside box we must solve

h2

2m[d2ψ

dx2+

d2ψ

dy2]+V(x,y)ψ=Eψ

with V(x,y)=0

Boundary conditions



Separation of variables We shall look for solutions of the form ψ(x,y)=f1(x)f1(y)where ψ(x,y) is a product of two functions dependingon a single variable (separation of variables)

For this to be a solution we must haveh2

2m[d2ψ

dx2 +d2ψ

dy2]+V(x,y)ψ =Eψ

orh2

2m[d2(f1f2)

dx2 +d2(f1f2)

dy2 ]+V(x,y)(f1f2)=E(f1f2)

Particle in 2D box

orh2

2m[f1

"(x)f2(y)+f1(x)f2"(y)]=Ef1(x)f2(y)

or

h2

2m[f1"(x)

f1(x)+

f2"(y)

f2(y)]=E

Particle in 2D box

For a given x this relationmust hold for all x. Thus:

−h2

2mf1"(x)

f1(x)=Ex

also

−h2

2mf2"(y)

f2(y)]=Ey

E=Ex+Ey

h2

2m[f1"(x)

f1(x)+

f2"(y)

f2(y)]=E

Separation of variables

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

We have

−h2

2mf1"(x)

f1(x)=Ex ; o≤ x ≤a

f1(0)= f1(a) =0

−h2

2mf1"(x)=Exf1(x)

f1(x)=2a

sin[nxπa

x]; Ex =h2nx

2

8ma2

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

−h2

2mf2"(y)

f1(x)=Ey ; o≤ y ≤b

f2(0)= f2(b) =0

−h2

2mf2"(y)=Eyf2(y)

f2(y)=2b

sin[nyπ

bx]; Ey =

h2ny2

8mb2

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞



Particle in 2D box

Thus

Thus ψ(x,y)=4ab

sin[nxπa

x]sin[nyπ

by]

E =h2nx

2

8ma2+

h2ny2

8mb2

For a=b

E =h2

8ma2[nx

2 +ny2]

Dege−nerate

Dege−nerate

For a=b

ψ(x,y)=2a

sin[nxπa

x]sin[nyπ

ay]

E =h2

8ma2[nx

2 +ny2]

nx nx E( h2

8ma2)

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. The solutions to the Schrödinger eq. for a 1Dparticle in a box ψn(n=1,2,3...) are not eigenfunctionsto px. They represents states for which a measurement

of px would give ±hn2l

and <px > =02. The solutions to the Schrödinger eq. for a 1Dparticle in a box ψn(n=1,2,3...) are eigenfunctions

to px2 with eigenvalues

h2n2

4l2. They represents states

fo r which a measurement of px2 would give

h2n2

4l2

3. The solutions to the Schrödinger eq. for a 1Dparticle in a box ψn(n=1,2,3...) have nodes tosatisfy the orthonormality condition

ψm* ψn∫ dx=∂nm

What you should learn from this lecture

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

ψ(nx,ny)(x,y)=4ab

sin[nxπa

x]sin[nyπ

by] (nx,ny positive integers)

insideThe corresponding energy is given by

E =h2nx

2

8ma2+

h2ny2

8mb2

V = 0

0 ax

y

V = ∞

V = ∞

V = ∞

V = ∞

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

the two solutions ψn andψm have the properties that

ψ*n ψm

-∞

∞∫ dx =0 if m≠n

We say thatψn andψm areorthogonal

We have

Inm = ψ*n ψm

-∞

∞∫ dx

with ψu =2l

[sinuπl] u= ,m n

Inm =2l

[sinnπl-∞

∞∫ x]

2l

[sinmπl

x]dx

since sinα sinβ =12[ (cos α −β) − (cos α +β)]

Thus

Inm =1l

[cos(n−m)πx

l-∞

∞∫ ]dx−

1l −∞

∞∫ [cos

(n+m)πxl

]dx

or

Inm =[sin(n−m)πx

l(n−m)π

l

−[sin(n+m)πx

l(n+m)π

l

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥o

l

Inm ={0 −0 + [(sin n−m)π] −[(sin n+m)π} =0

We thus have

ψ*n ψm

-∞

∞∫ dx =0 if n≠m

ψ*n ψm

-∞

∞∫ dx =1 if n≠m

or

ψ*n ψm

-∞

∞∫ dx =δnm

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

An An acac tutu aa l l syssys ttem em thth aat t aapppp rrooxxiimama ttes es thth e e pparar titi cc lleses -in--in- aa--bobo x x pprroblobl emsems ::

A number of molecules have conjugated π-bond s i n whic h the

electron s aredelocalize d o ver th e who le molecules

H H H H H H

H2C=C– C= --C C= - C C=CH2

(1) Th ey consist s i n th e firs t p lace of a chai n ofato ms held

togeth er b y a syst em of σ-bonds

(2) In addition to th e electron s tha t participat e i n th e σ-bond s there

are prese nt in th e molecu le electron s tha t ar e delocalized.These

electron s arerefere d to as π-electron .s W e sh al ref er to their

numb er asnπ

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 si n

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,wher e w e sha l assum e that nπ i s ev , en a total

of nπ2 leve ls w ill b e occupied



E

E

2

1

En

π

2

En

π

2

+1

HOMO

LUMO

T he tota l ener gy of thesyste m is give n by

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



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

ΔE = ELUMO - EHOMO = h2⎝

⎜⎜⎛

⎠⎟⎟⎞nπ

2 + 1 2

8ml2 -

h2⎝⎜⎜⎛

⎠⎟⎟⎞nπ

22

8ml2

ΔE = h2

8ml2 (nπ+1)

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

ν = 1hc ΔE

Thus the absorption energy frequency is

ν = h

8ml2c (nπ+1)



Particle in 2D box

For a = b

Thus ψ( ,x )y =2a

[sinnxπa

x] [sinnyπ

ay]

E =h2

8ma2[nx

2 +ny2 ] =2

h2

8ma2

nx =ny =1One fold generate

The wavefunctions for a particleconfined to a rectangular surface depicted as contoursof equal amplitude.(a) n1 = 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 ψ(x,y)=2a

sin[nxπa

x]sin[nyπ

ay]

E =h2

8ma2[nx

2 +ny2]=2

h2

8ma2

nx =1 ny =2 ; nx =2 ny =1two fold generate

E =5h2

8ma2

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°. T he tw o function s corres pond t o the sameener . gy Degenera cy a nd symmetr y ar e closelyrela .ted


Particle in 2D box


sin[nxπa

x]sin[nyπ

ay]

E =h2

8ma2[nx

2 +ny2]=2

h2

8ma2

nx =2 ny =2 one fold generate

E =8h2

8ma2


Particle in 2D box


sin[nxπa

x]sin[nyπ

ay]

E =h2

8ma2[nx

2 +ny2]=2

h2

8ma2

nx =3 ny =2 ; nx =2 ny =3two fold generate

E =13h2

8ma2


lecture 10: particle in a box-ii. lecture on-line

Documents

eq isui

eq fcl

eq f8p2

repulsion of n electrons

molecule electrons

vi3the electrons

dy2 eq fd2

dx2 eq fd2