Lecture 4 2010

26
The Quantum Theory of Atoms and Molecules Particles in boxes and applications Dr Grant Ritchie

Transcript of Lecture 4 2010

Page 1: Lecture 4 2010

The Quantum Theory of Atoms and Molecules

Particles in boxes and applications

Dr Grant Ritchie

Page 2: Lecture 4 2010

The free particle

ψψ Edxd

m=− 2

22

2h

ikxikx BeAe −+=ψ

What is the appropriate form of the time independent Schrödinger equation?

V

= 0, so that TISE is:

Solutions are of the form: Eigenfunctions (A,B

are constants)

mkE2

22h= Eigenvalues

Probability density example: Suppose that B

= 0, then ψ (x)

= A eikx

. Where is the particle? ψ *ψ = (A

eikx)* (A

eikx) = (A* e– ikx)(A

eikx

) = |A| 2

. This is independent of x, in other words, we cannot predict where we will find the particle.

N.B. k = 2π/λ and as k is known exactly then so is p. cf. Δx Δ px ≥ ħ

/ 2.

There is no restriction on k

or allowed energies ⇒ energy is continuous!

Page 3: Lecture 4 2010

Particle in a 1-d box (or infinite square well!)

kxDkxCBeAe

Edxd

mikxikx sincos

2 2

22

+=+=

=−

−ψ

ψψh

What are the wavefunctions

(eigenfunctions) and energies (eigenstates/eigenvalues) of a particle confined in a very deep potential well?

0 L

V(x)+ ∞

V(x)

→∞

+ ∞

V(x)

= 0

V(x)

→∞

Inside the box: V

= 0, so that SE is:

Outside the box: V

= ∞, so that outside the box ψ

= 0, and because ψ

is continuous, ψ

must also be 0 at the edges of the box, i.e.ψ(0) = 0 and ψ(L) = 0.

For ψ(0) = 0 then C

= 0, and so ψ (x) = D sin(kx).

For ψ(L) = 0 then either D

= 0 (SILLY!) or ψ(L) = D sin(kL) = 0.

sin(nπ) = 0 where n

= 1, 2, 3,…

and so k

is now restricted as follows: L

nk π=

Page 4: Lecture 4 2010

Quantisation of energy

,....3,2,182 2

2222

=== nmL

hnm

kE h

. d d 00 L

DxkxDxLL 21sin* 22 =⇒== ∫∫ ψψ

Boundary conditions lead to wavefunctions

(eigenfunctions) of the form ψ(x) = D

sin

kx

with quantised

values of k (= nπ/L)

and results in quantised

values for the energy E:

To completely specify the eigenfunctions

we need to

normalise

them:

Eigenfunctions:

Eigenvalues:

,....2,1sin2)( =⎟⎠⎞

⎜⎝⎛= n

Lxn

Lx πψ

Page 5: Lecture 4 2010

Particle in box properties

Ψ(x)

0x

L

1.

Increasing n

indicates increasing K.E. (increasing curvature of the wavefunction).

2. Lowest energy: E1 = h2/8mL2 > 0 ; ZERO POINT ENERGY. If E

→ 0, then p, δp

→ 0 and requires δx

→∞ (Heisenberg)

3.

Difference between adjacent energy levels:

ΔE

= En+1

En

= (2n

+ 1) h2/8mL2

= (2n

+ 1) E1

4. The more the particle is confined, i.e. L

gets smaller, the greater ΔE

and the K.E. –

hard to compress matter!

5. If L

→∞ , ΔE

→ 0, continuous: translational energy is not quantised. Back to classical physics!

Consistent with De Broglie: The longest wavelength is λ

= 2 L

and the higher modes have wavelengths given by λ

= 2L/n

(n

= 1,2,...). Therefore the De Broglie relationship yields momentum p

= ±

(nh/2L) = ±

h/λ.

Page 6: Lecture 4 2010

Orthogonality

Importantly, note that the eigenfunctions

for the particle in a box are orthogonal to one another.

Indeed they are said to be orthonormal

in that each function is also normalised.

Often we are interested in integrating products of wavefunctions. If the integral turns out to be zero then the functions, ψn

and ψm

are said to be orthogonal:

mnif

mnifxa

xma

xnL

xxxL

mn

≠=

==⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛= ∫∫

d d 0-

0

1sinsin2)()(* ππψψ

Examples: (i) Show pictorially that the hydrogen atom 2s and 2pz

orbitals

are orthogonal;

(ii) Shown that the two lowest energy wavefunctions

for a particle in a box are orthogonal.

Page 7: Lecture 4 2010

Where is the particle most likely to be?

We can calculate the most probable position of the particle from

knowledge of ψ*ψ. For example, for n

= 1

i.e. particle is most likely to be found at the centre of the box.

This results is clearly at odds with classical expectations where each position in the box is equally likely.

However as n

increases the wavefunction

begins to have so many nodes that in the limit as n

→∞ each position is equally probable

⇒ Correspondence Principle.

0.0 0.2 0.4 0.6 0.8 1.0

1

0

1

x/a0.0 0.2 0.4 0.6 0.8 1.0

-1

0

1

ψ

x/a

n

= 1ψ*ψ

Page 8: Lecture 4 2010

Further comparisons between classical and quantum results

3)(

21)(

2

0

22

00

LxxPxx

LxxL

xxxPx

L

LL

==

===

∫∫

d

d d

⎟⎠⎞

⎜⎝⎛=

Lx

Lx πψ sin2)(

Classically

we expect that the probability density is uniform i.e. all positions in box are equally likely. Thus for a box of length L, the

probability density P(x)

= ψ*ψ = 1/L and so the

average values of

x and x2

are:

Compare with the quantum case where n

= 1 i.e.

then

2

22

0

222

0

2

23sin2

2sin2

πLLx

Lπxx

Lx

LxLπxx

Lx

L

L

−=⎟⎠⎞

⎜⎝⎛=

=⎟⎠⎞

⎜⎝⎛=

d

d

As n

→∞, find that <x2> → classical result.

Page 9: Lecture 4 2010

Momentum of the particle

ψψ .cos)sin(ˆ conskxDi

kkxDdxd

ipx ≠==

hh

0)(sin

))(cos(sin

)(sin

)(sin)(sin

*

ˆ*ˆ

0

22

0

2

0

22

0

2

====

∫∫

L

L

ik

L

L

dxd

ix

x

xkxD

xkxkxD

xkxD

xkxkxD

x

xpp

d

d

d

d

d

d hh

ψψ

ψψ

Given the wavefunction

ψ

= D sin(kx),

what is the particle’s momentum?

Use the momentum operator….

ψ

is not an eigenfunction

px

(see later)

but we can evaluate the

expectation value (see last lecture).

<px

>

= 0 –

this does not mean that the kinetic energy is 0!

mp

m

pE xx

2

ˆ 22

≠=

Page 10: Lecture 4 2010

Eigenfunctions

of the momentum operator

ψψψψ

pdxd

i

ppx

=

=h

ˆConsider the following eigenvalue

equation:

(where p is a constant)

General solution: ( ) negative.or positive becan NB. kAeAe ikxxip

== hψ

Thus Aeikx

are eigenfunctions

of the momentum operator with eigenvalues

p

= ±kћ.

The particle in a box wavefunction

ψ

= D

sin kx

can be expressed as a linear combination of momentum eigenfunctions, i.e. ψ

= D

sin kx

= D′ (eikx

+ e-ikx).

A single measurement of the particle’s momentum must give a definite result of ±kћ. However, since sin kx

contains equal amounts of e±ikx, the average value of the momentum <px > = 0.

Page 11: Lecture 4 2010

An application of the particle in a box problem -

the UV absorption spectrum of cyanine dyes

(Taken from www.jce.divched.org

)

3 2 3 2

3 2 3 2

( ) ( ) ( )

( ) ( ) ( )

k

k

CH N CH CH CH N CH

CH N CH CH CH N CH

+

+

− − = − − =

= − − = − −

&&

b

&&

Dye I: k

= 1

Dye II: k

= 2

Dye III: k

= 3

Page 12: Lecture 4 2010

Electronic structure of cyanine dyes

Have both σ

and π

electrons ⇒ σ electrons have largest probabilities in the plane of the molecules while π

electrons are most like to be found above and below plane of molecule.

Decouple π

electrons from σ

framework ⇒ treat π

electrons as being delocalised over the length of the molecule between the N atoms.

UV and visible light can then be absorbed and energy used to cause transition of π

electrons from one energy level to another.

NB: Each carbon contributes one electron to the π

system

while the two N atoms contribute 3 electrons.

Page 13: Lecture 4 2010

Is the particle in a box model justified for this problem?

Length of the box, L

= bβ +2δ, where b

is the number of bonds.

Page 14: Lecture 4 2010

Resonance condition

Absorption of a photon occurs when the energy of the photon (= hν) matches the difference

in the energy between the two states involved in the transition (ΔE):

( )2

2 22

8photon f i f ihE h E E E n nmL

ν= = Δ = − = −

where ni

and nf

are quantum numbers for the initial and final states respectively.

Which values of ni and nf have to be used? Depends on the number of π

electrons and the Pauli Exclusion Principle which allows a maximum of 2 electrons per orbital. Electron pairs must have opposite

spins.

Thus, for Dye I, k

=1 and we have a total of (3+3) = 6

π

electrons that will pair in levels n

= 1, 2 and 3.Therefore the highest occupied molecular orbital

(HOMO) has n

= 3, while the lowest unoccupied molecular orbital

(LUMO) has n

= 4.

Hence lowest energy transition involves promotion of a π

electron from

n

= 3 → n

= 4.

Example: If the length of the box L

is 8.5Å, what is the peak absorption wavelength for dye I?

Page 15: Lecture 4 2010

Are all transitions possible? –

Selection rules

EU .μ−=

Must always obey Pauli exclusion principle.

Transitions are electric dipole transitions

the oscillating electric field component of the radiation interacts with electrical charges, i.e. the positive nuclei and negative electrons that comprise an atom or molecule, and cause the transitions observed in uv-visible absorption and emission spectroscopies.

The interaction energy, U, between a system of charged particles and an electric field, E, is given by:

The dipole moment is defined for a collection of charges by: ∑=i

iirqμ

where

ri

is the position vector of charged particle i. ( See electrostatics lectures in Michaelmas

term).

Page 16: Lecture 4 2010

The transition dipole moment

∫= τψμψμ diffi ˆ*

In order to obtain the strength of interaction that causes a transition between two states, the transition dipole moment

is used rather than the dipole moment.

For a transition between and initial state, ψ i , to a final state ψ f , the transition dipole moment

integral is.

Just like the probability density is given by ψ*ψ, so the probability for a transition

(as measured by the absorption coefficient) is proportional to μfi

*

μfi.

.

If μfi

= 0

then the interaction energy is zero and no transition occurs –

the transition is said to be electric dipole forbidden. Conversely, if μfi is large, then the transition probability and absorption coefficient are large.

The intensity of the transition is thereforeproportional to.

2* ˆ∫ τψμψ djk

Page 17: Lecture 4 2010

Transition dipole moment integral for particle in a box

∫−=L

iffi xxxxe0

* )()( dψψμ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ +−⎟⎟

⎞⎜⎜⎝

⎛ −−=

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Lifif

Lif

fi

xL

xnnL

xnnx

Le

xL

xnxL

xnLe

0

0

)(cos

)(cos

sinsin2

d

d

ππ

ππμ

Need to consider the transition dipole moment integral for one electron. The dipole moment operator for an electron in one dimension is –ex

and so

Now evaluate μfi

for various wavefunctions

to see which are allowed (μfi

0) and which are forbidden (μfi

= 0).

Example: Is n

= 1 → n = 2 an allowed transition?

⎭⎬⎫

⎩⎨⎧

−ΔΔ

+−−−ΔΔ

⎟⎠⎞

⎜⎝⎛−= )sin(1)sin(1)1)(cos(1)1)(cos(1

22

2

πππππ

μ tottot

tottot

fi nn

nn

nn

nn

LLe

(If you have time!)

consider the generalised transition ni

→ nf

:

If we then define Δn

= nf

ni

and ntot

= nf

+ ni

then the above integral becomes:

Looks bad!!!!!

However, if Δn

is even then ntot

is even and overall μfi

= 0 –

Forbidden!

Page 18: Lecture 4 2010

Δn

is odd is allowed

If Δn

is odd then ntot

is also odd and overall μfi

0 and is given by

⎟⎟⎠

⎞⎜⎜⎝

−=⎟⎟

⎞⎜⎜⎝

Δ−−= 2222222 )(

8112

if

fi

totfi nn

nneLnn

eLππ

μ

The general selection rule is Δn

is odd.

However, we only really see a single peak in the absorption spectrum for each dye because other allowed transitions have very much smaller transition moments.

*Also note that longer molecules have larger absorption coefficients because μfi

increases with the length of the molecule (see uv

spectra earlier in notes).

Example: For dye 1, compare the values of the transition dipole moment integrals for the two transitions n

= 3 → n = 4 and n

= 3 → n

= 6.

Page 19: Lecture 4 2010

Using symmetry to evaluate integrals

)()( xx nn −±= ψψ

parity oddeven ; )( sin2

parityeven odd ; )( cos2)(

−−−=⎟⎠⎞

⎜⎝⎛=

−−+=⎟⎠⎞

⎜⎝⎛=

nxL

xnL

nxL

xnL

x

n

nn

ψπ

ψπψ

otherwise )(2/2/for 0)(

∞=≤≤−=

xVLxLxV

An alternative to evaluating integrals is to use symmetry……Firstly, consider the parity of the particle in a box wavefunctions

by shifting the positions of the potential barriers from 0 and L

to –L/2 to L/2.

02L

−2L

n = 1 n = 2

2L

2L

− 0

* For symmetric potentials V(x)

= V(-x) the ψ

has a definite parity, i.e.

2/' Lxxx −=→

Page 20: Lecture 4 2010

Symmetry II

For transition n

=1 → n

= 2 the transition dipole moment integral is:

( )

dd ∫∫−−

−=⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

2/

2/

2/

2/21 2

cossinL

L

L

L

xxfLex

Lxx

Lx

Le ππμ

Clearly ∫

f (x)dx

0 and the transition is allowed. In contrast, for the transition n

= 1 → n

= 3, ∫

f(x) dx

= 0 and transition is forbidden.

× ×0

2L

−2L

n = 1

2L

2L

− 0

x n = 2

2L

2L

− 0

→2L

2L

− 0

/ 2

/ 2

( ) 0L

L

f x−

≠∫

×0

2L

−2L

n = 1

×2L

2L

− 0

x

2L

2L

− 0

n = 3

2L

2L

− 0

/2

/2

( ) 0L

L

f x−

=∫

Conclusion: if integrand is odd / antisymmetric

/ ungerade

then ∫

f(x) dx

= 0 and transition is forbidden.

Page 21: Lecture 4 2010

Particles in “round”

boxes (or on a ring!)

⎟⎟⎠

⎞⎜⎜⎝

∂∂

+∂∂

+∂∂

−= 2

2

22

22 112

ˆφrrrrm

H h

⎟⎟⎠

⎞⎜⎜⎝

∂∂

+∂∂

−= 2

2

2

22

yxmH h

2

22

2

2

2

2

21

φφ ∂∂

−=∂∂

−=Irm

H hh

x

y

z

r

mp

Jz

φ

Hamiltonian for a particle of mass m

constrained to move in a circular path of radius r

in the xy-plane (V

= 0 everywhere) is:

In polar co-ordinates x

= r

cosφ , y

= r

sinφ and the Hamiltonian becomes:

But radius of ring is fixed and so derivatives in r

are 0 and Hamiltonian simplifies to:

where I

= mr2

is the moment of inertia of the mass m

on the ring of radius r.

SE is of the (familiar) form:

2

222

2

2)(

2

h

h

IEmAe

mIEdxd

lim

m

l

ll

±==

−=−=

withφφψ

ψψψ N.B. ml

has nothing to do with mass m, it is the angular momentum quantum number.

Page 22: Lecture 4 2010

Particle on a ring solutions

,....3,2,1,02

22

±±±== ll mI

mE h

dd d 000 πφφφψψ

ππφφ

π

21***

222

=⇒== ∫∫∫ − AAAeeAA ll imim

,....3,2,1,021)( ±±±== l

imm m e

πl

l

φφψ

Eigenvalues:

Again, boundary conditions lead to wavefunctions

(eigenfunctions) with a restricted range of values for ml

(eigenfunctions) and results in quantised values for the energy E:

|mL

| = 1

|mL

| = 2

mL

= 0

Normalisation of wavefunctions:

Eigenfunctions :

Cyclic boundary conditions:

ψ has to be single-valued!

1)1(..

)2()(

22

2)2(

=−=

==

+=+

ll

llll

ll

mim

imimimim

mm

eei

eAeAeAeπ

πφπφφ

πφψφψ

Therefore ml

can only take the values 0, ±1, ±2, ±3,…

Page 23: Lecture 4 2010

Properties of the solutions

1. Energy gaps decrease as moment of inertia increases. ΔE

→ 0 as I →∞ and so recover classical mechanics.

2. Zero point energy is 0 (when ml

= 0). Does this contravene Heisenberg?

Probability density: ψ* ψ

= (1/2π) i.e. independent of position on ring.

Position on ring is completely uncertain and so Heisenberg allows us to know precisely the angular momentum ⇒ zero point energy can be zero!

3 Two wavefunctions

with different quantum numbers can have the same energy. For example wavefunctions

with ml

= 1 and −1 have the same energy, ћ2/2I. This is known as degeneracy.

Example: Let us treat the π-electrons in benzene as particles of mass m

moving around the circumference of a flat disc of radius r. How well does this simple model reproduce the actual behaviour of the π-electrons in benzene?

Page 24: Lecture 4 2010

Particle in a 3-d box

( ) ⎟⎟⎠

⎞⎜⎜⎝

∂∂

+∂∂

+∂∂

−=++= 2

2

2

2

2

22222

2ˆˆˆ

21ˆ

zyxmppp

mT zyx

h

ψψψ EzyxVm

=+∇− ),,(ˆ2

22h

In 3D the particle momentum is a vector with 3 components, px

, py

and pz

. The kinetic energy operator is therefore

• Therefore the SE equation becomes

And is often abbreviated to

),,(),,(),,(ˆ),,(2 2

2

2

2

2

22

zyxEzyxzyxVzyxzyxm

ψψψ =+⎟⎟⎠

⎞⎜⎜⎝

∂∂

+∂∂

+∂∂

−h

Kinetic energy

Potential energy

Total energy

∇2

is the Laplacian

operator, also known as “del squared”.

Page 25: Lecture 4 2010

Separation of variables

)()()(),,( zZyYxXzyx =ψ

EXYZdzdZXY

dydYXZ

dxdXYZ

m=⎟⎟

⎞⎜⎜⎝

⎛++− 2

2

2

2

2

22

2h

EdzdZ

ZdydY

YdxdX

Xm=⎟⎟

⎞⎜⎜⎝

⎛++− 2

2

2

2

2

22 1112h

Once again let’s assume that the potential energy inside the the box is 0.

Now we assume that the wavefunction

is separable:

and so SE becomes:

Now divide by XYZ:

First term only depends on x, and must be constant because the RHS does not contain x. (Similarly for the other terms). Now we have three 1d equations to solve….

ZEdzdZ

mE

dzdZ

Zm

YEdydY

mE

dydY

Ym

XEdxdX

mE

dxdX

Xm

zz

yy

xx

=−⇒=−

=−⇒=−

=−⇒=−

2

22

2

22

2

22

2

22

2

22

2

22

21

2

21

2

21

2

hh

hh

hh

where EEEE zyx =++

The interpretation of Ex

is the component of kinetic energy from motion in x

direction etc..

Page 26: Lecture 4 2010

Eigenfunctions, -values and degeneracy in a 3-d box

,....2,1,,sinsinsin),,( =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= zyx

z

z

y

y

x

x nnnL

znL

ynL

xnNzyx πππψ

Each of the 3 separated equations is a 1d equation for a particle in a box, whose solutions are known. Inserting appropriate boundary conditions (e.g. ψ

= 0 when x

= 0 and Lx

) yields:

,....3,2,1,,8

),,( 2

2

2

2

2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛++= zyx

z

z

y

y

x

xzyx nnn

Ln

Ln

Ln

mhnnnE

Eigenfunctions

Eigenvalues

( ) ,....3,2,1,,8

),,( 2222

2

=++= zyxzyxzyx nnnnnnmLhnnnE

For the case of a cubic box, Lx

= Ly

= Lz

= L

and so the eigenvalues

are:

nx ny nz

1 1 12 1 11 2 11 1 2

1 level

3 levels

Degeneracy –

different

wavefunctions

which the same

energy.

Reflects symmetry of the box.