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Chemistry 736Lecture 2

Chemistry 736

The Schrödinger Equationg qThe Particle in a BoxUncertainty principley p p

NC State Uni e sitNC State University

Derivation of the Schrödinger Equationg qThe Schrödinger equation is a wave equation.

Just as you might imagine the solution of such an equationJust as you might imagine the solution of such an equationin free space is a wave. Mathematically we can express awave as a sine or cosine function. These functions are

ll f ll d hoscillating functions. We will derive the wave equation infree space starting with one of its solutions: sin(x).

Before we begin it is important to realize that bound statesmay provide different solutions of the wave equation thanthose we find for free space Bound states includethose we find for free space. Bound states include rotational and vibrational states as well as atomic wavefunctions. These are important cases that will be treated

h f d t l d t di f th i i fonce we have fundamental understanding of the origin ofthe wave equation or Schrödinger equation.

The derivativeThe derivative of a function is the instantaneous rate of change.

The derivative of a function is the slopeThe derivative of a function is the slope.

We can demonstrate the derivative graphically. W id th f ti f( ) i ( ) h b lWe consider the function f(x) = sin(x) shown below.

The derivative of sin(x)( )The derivative of a function is the instantaneous rate of change.


At sin(0) the slope is 1 as shown by the blue line.



At sin(/4) the slope is 1/2 as shown by the blue line.



At sin(/2) the slope is 0 as shown by the blue line.



At sin(3/4) the slope is -1/2 as shown by the blue line.



At sin(3/4) the slope is -1/2 as shown by the blue line. Th l f ll li h f l d bl kThe slopes of all lines thus far are plotted as black squares.



At sin() the slope is -1 as shown by the blue line.Th l f ll li h f l d bl kThe slopes of all lines thus far are plotted as black squares.



At sin(5) the slope is -1/2 as shown by the blue line.Th l f ll li h f l d bl kThe slopes of all lines thus far are plotted as black squares.



We see from of the black squares (slopes) that the d i i f i ( ) i ( )derivative of sin(x) is cos(x).

The derivative of sin(x)( )

dsin(x) = cos(x)

ddx

The derivative of cos(x)( )

dcos(x) = -sin(x)

ddx

The second derivative of sin(x)( )

ddsin(x) = -sin(x)

ddx

ddx

The second derivative of sin(x)( )

d2sin(x) = -sin(x)d2

dx2

Sin(x) is an eigenfunction( ) g

If we define as an operator G then we have:d2

dx2-

sin(x) = sin(x)d2-

dx2

sin(x) sin(x)dx2

which can be written as:

G sin(x) = sin(x)

This is a simple example of an operator equation thatis closely related to the Schrödinger equation.is closely related to the Schrödinger equation.

Sin(kx) is also an eigenfunction( ) g

We can make the problem more general by includinga constant k This constant is called a wavevectora constant k. This constant is called a wavevector.It determines the period of the sin function. Now we musttake the derivative of the sin function and also the

sin(kx) k cos(kx)d

function kx inside the parentheses (chain rule).

sin(kx) = -k cos(kx)dx

-

d2sin(kx) = k2sin(kx)d2

dx2-

Here we call the value k2 the eigenvalue.

Sin(kx) is an eigenfunction ( ) gof the Schrödinger equation

The example we are using here can easily be expressedas the Schrodinger equation for wave in space. We onlyhave to add a constanthave to add a constant.

sin(kx) = sin(kx)d2-

h2- h2k2-

In this equation h is Planck’s constant divided by 2 and

sin(kx) sin(kx)dx22m 2m

-In this equation h is Planck s constant divided by 2 andm is the mass of the particle that is traveling through space.The eigenfunction is still sin(kx), but the eigenvalue in this equation is actually the energythis equation is actually the energy.

The Schrödinger equationg qBased on these considerations we can write a compactform for the Schrödinger equationform for the Schrödinger equation.

H = E

d2

d 2-

h2

2m

-H = Energy operator, Hamiltonian

dx22m

h2k2-E = h2k2

2mEnergy eigenvalue, Energy

= sin(kx) Wavefunction

The momentumThe momentum is related to the kinetic energy. ClassicallyThe kinetic energy is:

E = mv21

E = mv2

The momentum is:2

p = mv

So the classical relationship is:pp2

E = 2m2m

If we compare this to the quantum mechanical energy:

h hkE h2k2-we see that: p = hkE =

2m

The general solution to the gSchrödinger equation in free space

The preceding considerations are true in free space.Since a cosine function has the same form as a sinefunction, but is shifted in phase, the general solutionfunction, but is shifted in phase, the general solutionis a linear combination of cosine and sine functions.

Asin(kx) + Bcos(kx) Wavefunction

The coefficients A and B are arbitrary in free space.

= Asin(kx) + Bcos(kx) Wavefunction

However, if the wave equation is solved in the presenceof a potential then there will be boundary conditions.of a potential then there will be boundary conditions.

The Schrödinger equationThe Schrödinger equation for a free particle

The solutions are:

eikx

e-ikx

The particle in a box problemp pImagine that a particle is confined to a region of space.The only motion possible is translation. The particle hasy p ponly kinetic energy. While this problem seems artificial atfirst glance it works very well to describe translational motion in quantum mechanicsmotion in quantum mechanics.

0 LAllowed Region

The solution to the Schrödinger gequation with boundary conditions

Suppose a particle is confined to a space of length LSuppose a particle is confined to a space of length L.On either side there is a potential that is infinitely large.The particle has zero probability of being found at the boundary or outside the boundary.

0 LAllowed Region

The boundary conditions determine the values for the constants A and B

sin will vanish at 0 since x = 0 and sin 0 = 0.s a s at 0 s ce 0 a d s 0 0sin will vanish at a if kL = n.Therefore, k = n/L.e e o e, /

= sin nxLNot LNot

Normalized !

The particle in a box hasThe particle in a box has boundary conditions

(0)= 0 (L)= 0

L

The solution to the Schrödinger equation with boundary conditionsequation with boundary conditions

The boundary condition is that the wave function willb t 0 d t Lbe zero at x = 0 and at x = L.

(0) = Asin(k0) + Bcos(k0) = 0From this condition we see that B must be zero.This condition does not specify A or k.The second condition is:The second condition is:

F thi diti th t kL Th diti(L) = Asin(kL) = 0 or kL = arcsin(0)

From this condition we see that kL = n. The conditionsso far do not say anything about A. Thus, the solutionfor the bound state is: (x) = Asin(nx/L)Note that n is a quantum number!

n(x) = Asin(nx/L)

The solutions to the particle in a boxp

h2k 2 E =

h k2m

2 n 2

=h2 n

L2

22m

=h2n2

=8mL2

The appearance of the wave functions

Note that the wave functions havenodes (i e the locations where theynodes (i.e. the locations where theycross zero). The number of nodes isn-1 where n is the quantum numberf th f ti Thfor the wave function. The appearanceof nodes is a general feature of solutions of the wave equation in bound states. By boundq ystates we mean states that are in a potential such as theparticle trapped in a box with infinite potential walls. Wewill see nodes in the vibrational and rotational wave functionswill see nodes in the vibrational and rotational wave functionsand in the solutions to the hydrogen atom (and all atoms). Note that the wave functions are orthogonal to one another.Thi th t th i t t d d t f t f thThis means that the integrated product of any two of these functions is zero.

The probability interpretationp y pThe wave function is related to the probability for findinga particle in a given region of space The relationship isa particle in a given region of space. The relationship isgiven by:

P = 2dV

If we integrate the square of the wave function over a given volume we find the probability that the particle isg p y pin that volume. In order for this to be true the integralover all space must be one.

1 2dV

If this equation holds then we say that the wave functioni li d

1 = 2dVall space

is normalized.

The normalized bound state wave function

For the wave function we have been consideringFor the wave function we have been considering, all space is from 0 to L. So the normalization constant Acan be determined from the integral:

1 = 2dx0

L

= A2 sin nxL

2dx

0

L

= A2 sin nxL

2dx

0

L

The solution to the integral is available on the downloadable MAPLE worksheet. The solution is just L/2.Thus we have:Thus, we have:

A th ll d li ti t t h

1 = A2 L2 , A2= 2

L , A = 2L

As you can see the so-called normalization constant hasbeen determined.

The probability of finding thep y gparticle in a given region of space

Using the normalized wave functionUsing the normalized wave function

x = 2L sin nx

L

one can calculate the probability of finding the particle in any region of space. Since the wave function is

L L

y g pnormalized, the probability P is a number between 0 and 1.For example: What is the probability that the particle isbetween 0 2L and 0 4L This is found by integrating overbetween 0.2L and 0.4L. This is found by integrating overthis region using the normalized wave function (see MAPLE worksheet).

2d0.4L 2 nx 2

d0.4L

2P = x 2dx0.2L

= 2L sin nx

L dx0.2L

0.25

The appearance of the probability 2

The uncertainty principleWhen we measure the properties of very small particles, wecannot help but affect them. The very act of measuring ca ses a change in the pa ticle’s p ope ties The efo e thecauses a change in the particle’s properties. Therefore, the description of the the measurement is a probability rather than a fixed value. We have seen the Born interpretation of the square of the wavefunction as a probability density.The consequence of this is that certain variables are linkedBy the uncertainty that is inherent in the measurement.By the uncertainty that is inherent in the measurement.Position and momentum are two such conjugate variables.Note that the units of position is the reciprocal of the momentum (if we factor out Planck’s constant)momentum (if we factor out Planck’s constant).x has units of meter, k has units of meter-1

Momentum is p = hk.t has units of time, has units of time -1Energy is E = h.

Where is the particle in the box?pSince we are using a probability function we do not really know exactly where the particle is. We know that the highestknow exactly where the particle is. We know that the highestprobability occurs for the position L/2. We can guess that this is the average position in the box. However, themore precisely we specify the location of the particle themore precisely we specify the location of the particle theless information we have about how fast the particle ismoving. This is a statement of the famous Uncertainty Principle.

xp > h/2

Let’s look at the Uncertainty Principle using the particle-in-a-box example. If we know that the particle is in the lowestlevel then Uncertainty in its position is approximately equallevel then Uncertainty in its position is approximately equalto the width of the probability distribution.

The location of a particle in free space is not defined

Consider a superposition of a wavewith moment hk and h(1.1k)

The sum has a characteristic envelopef t ( )/2frequency at (2 - 1)/2

EnvelopeEnvelope

The sum has a characteristic beatf t ( )/2frequency at (2 + 1)/2

Beats

As we add more frequencies wecan speak of a bandwidth k

k = 0.1

3 added cosines

As the bandwidth increases thepo ition in p e be ome mo e definedposition in x-space becomes more defined

k = 0.2

5 added cosines

The superposition of waves in spacel d t th d i ti f l tileads to the description of a location

k = 0.7

15 added cosines

Relevance of the example

Although the function used in the example is periodic it isrelevant. Since in a given region of space (i.e. where a g g p (measurement can be made) the probability of observing the particle in a given region of space is dependent uponthe number of contributing waves If more wavesthe number of contributing waves. If more waves contribute then the momentum of the particle is lesscertain. Thus, the we can know that moment precisely if

t t ll t i f th iti A b i twe are totally uncertain of the position. As we begin to specify the position more precisely we find that the momentum is less well known. Since p = hk, we can p ,also express this condition as:

xk > 1/2

Fourier transform related pairsPosition and momentum are related by a Fourier transform.

x pTime and energy are related by a Fourier transform.

t Et EThere is an uncertainty relationship for both of these related pairs. Thus, for time and energy we have

t E > h/2

as well. These pairs can be related by a probability function that gives the width of the distribution in eachspace Gaussian functions are particularly useful sincespace. Gaussian functions are particularly useful sincethe Fourier transform of a Gaussian is also a Gaussian.

Gaussian FunctionsA Gaussian function has the form exp{ -(x – x0)2 }.The Gaussian indicated is centered about the point x0.Th F i t f f G i i iThe Fourier transform of a Gaussian in x-space is a Gaussian in k-space. Since p = hk we also call this momentum space. The figure shows the inverse p grelationship.

x k

Free electron model for electronic spectrafor electronic spectra

Before the advent of computers, models such as particle-in-a-box were used for linear polyenes. The idea of such a model is that the electrons from the p-orbitals in a moleculefrom the p orbitals in a molecule are particles and the molecule is the “box”. For example, we can think of ethylene as a shortcan think of ethylene as a short box with two electrons as shown in the figure. Although there are an infinite number of states, only two of them are really important in ethene since there are only two electrons. Thein ethene since there are only two electrons. The HOMO and LUMO are shown on the left and the representation of the two lowest electronic states is shown.

Polyenes

The first four members of the class of polyenes are shown.We can treat the p-system of these molecules using the modelthese molecules using the modelwith 2, 4, 6 and 8 electrons,respectively. As a general rulethe model can be applied to anythe model can be applied to anynumber of p-orbitals, with the assumption each p-orbital willcontribute one electron to the total. Then we populate thelevels calculated using the particle-in-a-box and determinethe transition between the HOMO and LUMO. This isillustrated for butadiene on the next slide.illustrated for butadiene on the next slide.

ButadieneButadiene has 4 -electrons. There is one electroncontributed the each carbon atom (one for each p orbital).

Application to aromatic l l bmolecules: benzene

NODES3

2

1

0

We can construct molecular orbitals of benzene using the

six electrons in orbitalssix electrons in orbitalsH

C

C CHH

C

C

CHH

C

H Electrons arespin-paired

Benzene Structure Electronic Energy Levels

The Free Electron Model Applied to BenzeneBenzene

The system approximates circular electron pathelectron path.

5-5C

C C

H

HH

44

-5C

C

C

C

C

HH

3

4

-3

-4

h2 2 E

C

H

1

2

3

1-2

3– h2meR

22 = E

h2m2 1 i

m=1m=3

m = 01-1E = h m

2meR2 , = 1

2eim

The Perimeter ModelThe aromatic ring has 18 electrons.The system approximates circular electron pathelectron path.

N N 5-5

NN44

-5m=1m=9

3

4

-3

-4

h2 2 E

1

2

3

1-2

3– h2meR

22 = E

h2m2 1 i m = 0

1-1E = h m2meR

2 , = 12

eim

Application to benzene1. Determine number of p electrons (here it is 6).2. Place the electrons in the energy level diagram for the

particle on a circle If the number of electrons followsparticle on a circle. If the number of electrons follows the 4n+2 rule then the levels will be filled exactly.

3. Identify the allowed and forbidden transitions.4. Calculate the energies using the particle on a circle model.

Huckel MO Theory1. Determine the molecular point group.2. Determine the normalized symmetry adapted linear

combinations (SALCs) using the projection operator methodcombinations (SALCs) using the projection operator method.If there is only one set of SALCs then these are also the MOs.

3. Use the SALCs to determine the elements of the Huckelsecular determinant

det |Hij – SijEo| = 04. Simplify the mathematical formalism by letting Hii = and

Hij = . Divide each element by and let x = - E)/5 For each eigenvalue solve for the eigenvectors (coefficients)

j j

5. For each eigenvalue solve for the eigenvectors (coefficients).These are the contributions of each SALC to the MO.

|Hij – SijEo|Cijj = 1

N= 0

j = 1

Cij2

j = 1

N= 1

Butadiene2 3

1 4 2 and 3 are symmetry related1 and 4 are symmetry relatedB th t fBoth transform as

E C2v (xz) (yz) 2 0 -2 0

ButadieneE C2v (xz) (yz)

2 0 -2 0A 1 1 1 1

E C2v (xz) (yz) 2 0 -2 0A 1 1 1 1A1 1 1 1 1

Total 2 + 0 + -2 + 0 = 0A2 1 1 -1 -1Total 2 + 0 + 2 + 0 = 4

E C2v (xz) (yz) 2 0 -2 0

E C2v (xz) (yz) 2 0 -2 0

B1 1 -1 1 -1Total 2 + 0 - 2 + 0 = 0

B2 1 -1 -1 1Total 2 + 0 + 2 + 0 = 4

The order is 4 so the character sums are divided by 4. The irreducible representations are:The irreducible representations are:

= a2 + b2

ButadieneSALC14(b2) = p1 + p4 + p1 + p4

SALC14(b 2) = 12

p1 + p4SALC14(b2)

SALC14(a2) = p1 – p4 – p4 + p1SALC14(a2) p1 p4 p4 p1

SALC14(a2) = 1 p1 – p4SALC14(a2)SALC14(a2) 2p1 p4SALC14(a2)

ButadieneSALC23(b2) = p2 + p3 + p3 + p2

SALC23(b 2) = 12

p2 + p3SALC23(b2)

SALC23(a2) = p2 – p3 – p3 + p2SALC23(a2) p2 p3 p3 p2

SALC (a ) = 1 p pSALC ( )SALC23(a2) = 12

p2 – p3SALC23(a2)

Construct symmetrized Huckel DeterminantThe unsymmetrized Huckel determinant is a 4x4 matrix.The solution is a fourth order polynomial. By factoring according to symmetry the determinant can be simplified intoaccording to symmetry the determinant can be simplified into2 2x2 matrices.

H = 1 p + p H p + p dH11 = 12 p1 + p4 H p1 + p4 d

= 12 p1Hp1d + 1

2 p4Hp4d + 12 p1Hp4d + 1

2 p4Hp1d

H = 1 p + p H p + p d

2 2 2 2=

H22 = 12 p2 + p3 H p2 + p3 d

= 12 p2Hp2d + 1

2 p3Hp3d + 12 p2Hp3d + 1

2 p3Hp2d2 2 2 2= +

Construct symmetrized Huckel Determinant

H12 = 12 p1 + p4 H p2 + p3 d

1 H d + 1 H d + 1 H d + 1 H d= 12 p1Hp2d + 1

2 p4Hp2d + 12 p1Hp3d + 1

2 p4Hp3d

=

H21 = H12

H33 = 12 p1 – p4 H p1 – p4 d

= 1 p Hp d 1 p Hp d 1 p Hp d + 1 p Hp d= 2 p1Hp1d – 2 p4Hp1d – 2 p4Hp1d + 2 p4Hp4d

=


H34 = 12 p1 – p4 H p2 – p3 d

1 1 1 1= 12 p1Hp2d – 1

2 p4Hp2d – 12 p1Hp3d + 1

2 p4Hp3d

=

H43 = H34

H44 = 12 p2 – p3 H p2 – p3 d

= 1 p Hp d + 1 p Hp d – 1 p Hp d – 1 p Hp d= 2 p2Hp2d + 2 p3Hp3d – 2 p2Hp3d – 2 p3Hp2d

= –

H13 = H31 = H24 = H42 = H14 = H41 = H23 = H32 = 0


H11 – E H12 H13 H14

H21 H22 – E H23 H2421 22 23 24

H31 H32 H33 – E H34

H41 H42 H43 H44 – E

– E 0 0 + E 0 0 + – E 0 00 0 – E 0 0 – – E

– E

0 0 E

Divide by and let x = E

Solve each symmetry block separately

x 1 0 01 x + 1 0 01 x + 1 0 00 0 x 10 0 1 x – 1

= 0det

x 1

b2 block

x 11 x + 1

= x x + 1 – 1 = x2 + x – 1 , x = – 1 ± 1 + 42 = – 5 ± 1

2

x 11 1

= x x – 1 – 1 = x2 – x – 1 , x = 1 ± 1 + 42 = 5 ± 1

2

a2 block

1 x – 1, 2 2

Determine the orbital energies

5 + 12 = 1.62

5 – 1 = 0 622 0.62

– 5 + 12 = – 0.62

5 – 1 1 62– 5 12 = – 1.62

Determine the eigenvectorsO l th SALC f th i d ibl t tiOnly the SALCs from the same irreducible representationCan contribute to the same MO.For the b2 block and x = -1.62 we have: 2

– 1.62 11 – 1.62 + 1

c11

c21,

– 1.62 11 – 0.62

c11

c21= 0

21 21

– 1.62c11 +c21 = 0 c11 – 0.62c21 = 0c21 = 1.62c11 c11 = 0.62c21

and c112 + c21

2 = 1 so c112 + 1.622c11

2 = 3.62c112 = 1

c112 = 1/3.62 = 0.276 and finally c11 = 0.525

thi i li th t 1 62(0 525) 0 85this implies that c21 = 1.62(0.525) = 0.85

The MO is 0.53 0.707 p1 + p4 +0.85 0.707 p2 + p3

= 0.37 p1 + p4 + 0.60 p2 + p3

= 0.37p1 +0.60p2 + 0.60p3 + 0.37p4

Determine the eigenvectorsO l th SALC f th i d ibl t tiOnly the SALCs from the same irreducible representationCan contribute to the same MO.For the a2 block and x = 0.62 we have: 2

0.62 11 0 62 + 1

c11

c,

0.62 11 1 62

c11

c= 0

1 0.62 + 1 c21 1 1.62 c21

0.62c11 +c21 = 0 c11 + 1.62c21 = 0c21 = –0.62c11 c11 = –1.62c2121 11 11 21

and c112 + c21

2 = 1 so c112 + 0.622c11

2 = 2.38c112 = 1

c112 = 1/2.38 = 0.722 and finally c11 = 0.85

thi i li th t 0 62(0 85) 0 53this implies that c21 = –0.62(0.85) = – 0.53

The MO is 0.85 0.707 p1 + p4 – 0.53 0.707 p2 + p3

= 0.60 p1 + p4 – 0.37 p2 + p3

= 0.60p1 –0.37p2 – 0.37p3 + 0.60p4

SUMMARY

E =

2a2

0.37-0.37

0.60 -0.60

2a2E = + 1.62

E =

a2

0.37 0.37

2

E = + 0.62

2b2

-0.60 -0.60

0.37 -0.37

2b2

E =

1a2

0.60 -0.601a2

E = - 0.62

E =

1b

0.37 0.37

0.60 0.60

1bE = - 1.62

1b20.37 0.371b2

Bond orderThe bond order can be calculated from the MO coefficients.The p-bond order is the product of the coefficients of two adjacent atoms times the number of electrons in the respectiveadjacent atoms times the number of electrons in the respectiveorbitals.

Bond order = n c coccup

Where nk is the number of electrons in the MO and cik and cjk

Bond order = nkcikc jkk = 1

Are the coefficients.

For butadiene the bond order for p1-p2 is:For butadiene the bond order for p1 p2 is:

The bond order for p2 p3 is:

Bond order = 2(0.37)(0.60) + 2(0.60)(0.37) = 0.888

The bond order for p2-p3 is:Bond order = 2(0.60)(0.60) + 2(0.37)( –0.37) = 0.446

Bond orderThe bond order of the butadiene double bond is 88.8% aslarge as that of ethylene.

However the total bonding energy is 2(0 888)+0 446 = 2 22However, the total -bonding energy is 2(0.888)+0.446 = 2.22,which is greater than two ethylene molecules. This is an example of stabilization by delocalization.

The total energy is the energy of each MO times its occupancy.occup

For butadiene this is:

Total energy = nkEkk = 1

occup

For butadiene this is:(units of )E = 2( – 1.62) + 2( – 0.62) = – 4.48

BenzeneMost of the spectroscopy and reactivity of benzene is attributable to its system. To a good approximation the and orbitals can be separated and orbitals can be separated.

We consider how the 6 p-orbitalstransform under D6h symmetry.

In this case we do not need toconstruct SALCs since there isonly one type of orbital.

D6h character table

Projection operator approachThe operation required to carry the reference p1 orbital intoany of the others.

C6b2g


C3b2g


C2b2g


C32b2g


C65b2g

Molecular orbitals constructed using the projection operator approach

Porphine orbitals

e eeg eg

a2 a1a2u a1u

Nodes in Porphine orbitals

The four orbital model is used toThe four orbital model is used to represent the highest occupied and

lowest unoccupied MOs of porphyrinslowest unoccupied MOs of porphyrins

Th t hi h t i d eg The two highest occupied

orbitals (a1u,a2u) are nearly equal in energy The eequal in energy. The egorbitals are equal in energy.Transitions occur from:

M1

a1u a2u

a s t o s occu oa1u eg and a2u eg.

1u2u

The transitions from ground state orbitalsThe transitions from ground state orbitalsa1u and a2u to excited state * orbitals eg

can mix by configuration interactioncan mix by configuration interaction

T o elect onic t ansitions eg Two electronic transitions

are observed. One is verystrong (B or Soret) and thestrong (B or Soret) and the other is weak (Q).The transition moments are:

M1 M2

a1u a2u

MB = M1 + M2MQ = M1 - M2 0

1u2u

Absorption spectra for MbCO and deoxy MbSoret Band Q Band

The spectrum of the heme has two bands. The B band or SoretBand is allowed and therefore intense. The Q band is forbidden.It is observed because of vibronic coupling with the Soret band.

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