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Advanced Physical Chemistry
NOTES 1
Review of Developments leading to Quantum Mechanics
Nature of Light
Black Body Radiation
Experimental Results led to:
Stefan-Boltzman Law
M = T4
56.7
2
K4 ) independent of the material
Wien's Displacement Law
mzx T = constant
Could Electromagnetic Theory explain these?
Rayleigh Jeans
Energy in a region divided by the volume of the region is the Energy density, E(
AND dERd

dEis the contribution from radiation in the wavelength range  to  + d

R is the spectral density of states and can be thought of as the distribution of energy
emitted a particular wavelength, (the relative amount of the total energy or fraction of the
total energy emitted at a particular wavelength  ).
FROM Electromagnetic Theory, Rayleigh and Jeans derived a theoretical value of the
spectral density of states. The formula is:
RkT/4
where k is
Note what happens as  approaches zero.
This did not match the experimentally derived values and it became known as the
Ultraviolet Catastrophe
Planck however derived the spectral density distribution of a black-body radiator which
was essentially based on the idea that an oscillation of the electromagnetic field of
frequency , could only be excited in steps of energy of magnitude h, due to the fact that
the radiator was an oscillator with oscillations of energy hv, h being a new constant of
nature, Planck's constant.
With this he derived
Rhc/5 ) (e- h c /
λkT
/ (1-e- h c /
λkT
))
h = 6.626x10- 3 4 Js
This fit the entire distribution well.
Heat Capacities
The idea that there were oscillations in a material that gave rise to radiation also led to a
correct value for the temperature dependence of Cv
Dulong and Petit said that the atoms of simple bodies all have the same heat capacity of
25 J/(K mol), no temp. dependence. Found to be incorrect.
Cv,m = 3RfD(T) where fD(T) = 3(T/D)3 0 ∫/ T x4 ex 1 / (ex -1)2 dx
D is the Debye Temperature and is related to the maximum frequency of oscillation that
can be supported by the solid.
This complemented Planck's ideas, and lent credence to the idea that matter was
quantized.
Photoelectric Effect
Einstein studied the emission of electrons when exposed to electromagnetic radiation.
Found that the emission was instantaneous when the radiation was applied even at low
radiation intensities, but there was no emission, whatever the intensity of the radiation
when the frequency of the radiation was below some threshold value. The threshold
value was different for different metals.
The experiment was performed as below
And a plot of the Kinetic Energy as a function of the frequency of light gave the
following plot
Thus there is a linear relationship between the kinetic energy of the emitted electron and
the frequency of the radiation, and a cutoff frequency, below which the electrons are not
emitted.
An energy balance says:
Energy of light = Kinetic Energy of e- + work function of metal (ionization energy)
or
The slope of the line in the plot was equal to
So photons have particle character. If so they should possess a momentum and should
be able to transfer their momentum.
Compton Scattering
The relativistic expression for particle's energy
E2 = m2 c4 + p2 c2
c is the speed of light, and based on photoelectric effect E=h and m = 0 so
(h2 = p2 c2 and c = 
so p = h/
Partial transfer of the momentum of a photon during a collision, should appear as a cange
in the wavelength of the photon.
d = f - i
when the radiation was scattered through an angle  and assuming photon as momentum
h/:
d = 2c sin2 (½)
where c = h/(mec) is called the Compton wavelength of the electron (c = 2.426 pm).
Hydrogen Spectra and Bohr's Theory
Scientists were interested in explaining why H emitted discrete spectral lines, and not a
broad spectrum of visible light.
Balmer experimentally determined that visible lines of H followed
' = RH (1/22 - 1/n2 )
n = 3, 4, 5, ...
RH is Rydberg constant for hydrogen know as RH = 1.097 x 105 cm- 1
Suggested that the energies of atoms were confined to discrete values
Bohr showed that this could be explained by assuming that the angular momentum of an
electron around a central nucleus was only allowed to possess certain values
corresponding to circular orbits at certain discrete radii away from the nucleus
Based on this deduced
En = -e4 /(8h2 o2 ) (1/n2 ) n = 1, 2, 3, ……
is the reduced mass of the atom and o is the vacuum permittivity
Thus this marriage of the model of mechanics and radiation was able to quantitatively
account for the know spectrum of hydrogen including the Lymann, Balmer, and Paschen
lines.
deBroglie's Relation
Since light which was thought to be strictly a wave, can act as both a particle and wave,
deBroglie suggested that fundamental quantities that were thought to be just particles,
could also possess wave properties. The wavelength of these "particles" should be given
by:
 = h/p
(just as for photons)
Davisson and Germer used a crystal of nickel to show that electrons shot at the crystal
yielded a diffraction pattern consistent with the electrons having a wavelength as
proposed by deBroglie.
If all particles have wave-like character, then there should be observational consequences.
Just as a wave of definite wavelength cannot be localized at a point, one should expect
that one cannot expect a definite linear momentum to be defined at a single point.
Is this important for all particles, large and small?
Compare wavelength to the size of the particle.
Heisenberg Uncertainty Principle
p x ≥ h/4
Complementarity of pairs of observables
The mutual exclusion of the specification of one property by the specification of another
A days few later, around Christmas, Schrodinger left Zurich for vacation in the Swiss
Alps. There he developed an equation to take into account this wave particle duality.
Impossible to ‘rigorously derive’ the equation (the Quantum Mechanical Eqn.) that
described microscopic particles which show both wave and particle duality, but from his
notes we can determine that it was derived similar to the following:
Starting point:
Well known Equation of Classical Wave Motion (interrelates the space and time
dependence of waves)
Separate into to further eqns.: one dealing only with spatial variations of the waves, the
other with the time. For waves oscillating in three dimensions, the spatial wave eqn takes
the form.
................................. ~
v2= -k2 
where ~
v2 is known as the Laplacian operator and is given by d2/dx2 + d2/dy2 + d2/dz2 in
Cartesian coordinates, and k, is the waver vector, and is equal to 2, where  is the
wavelength. There is a whole range of functions (called wavefunctions) that satisfy the
eqn., ranging from simple sine and cosine functions to more complicated functions.
But according to de Broglie,  = h/p and if p = mv then
..................................... k = 2/ = 2p/h = 2mv/h
and hence: ...........................~
v2= (42m2v2/h2)*  ............................................ (A)
Also the total energy of a particle E is the sum of its kinetic and potential energies:
................................. E = ½ mv2 + V
rearranging: ......... mv2 = 2(E-V)
and inserting this into Expression (A) gives:
....................................... ~
v2= or
....................................... -h/ 2/(2m) ~
v2 + V = E .................................where h/ = h/2
What is E?
Known as three dimensional Schrodinger Equation and is one of the fundamental
relationships in quantum mechanics, first described by Heisenberg and Schrodinger in
1926.
Schrodinger was able to apply his eqn to the hydrogen atoms and obtained the results
consistent with Bohr's results.
An essential characteristic of Schrodinger's Eqn was that an observable like the energy of
the system can be represented as an operator and the possible results of that observation
are the eigenvalues of that operator in an eigenvalue equation:
In general if for the operator A, an eigenvalue eqn of the form:
AUn = an Un
can be formed for the operator, then:
Un are said to be the eigenfunctions or eigenstates of the operator A and
an are the allowed observable values of the operator operating on each of the eigenstates
In the case of Schrodinger's eqn this is usually written as
H = E

or
H n = En n
where n are the energy eigenstates for the Hamiltonian which is the Energy operator and
n are the energy eigenvalues or the possible energy values that can be observed.
Example:
Determine if the wavefunction 2 = (2/L)1 / 2 sin (2x/L) is an eigenfunction of the
operator d2 /dx2 and if it is, what is the eigenvalue corresponding to this eigenstate?
An important concept to note is that in general any function can be expanded in terms of
a linear combination of all of the eigenfunctions (a complete set) of a particular operator.
These are called the basis functions.
So if Un are the eigenfunctions corresponding to the operator then a general function,
 can be written as a linear combination of the Un as:
 =  cn Un
cn are the coefficients and the sum is over the complete set of the eigenfunctions of the A
operator. The sum is over the defined space.
The advantage of this type of expansion is that it allows one to determine the effect of an
operator on a function that is not simply an eigenfunction of that operator.
The operators for the physical observables such as the position, the momentum, and the
energy must reflect the ideas laid out by the Heisenberg uncertainty principle. There are
two ways that this can be checked.
1. Commutation
Based on the Heisenberg uncertainty principle, it would seem to matter whether the
position is measured first, and then the momentum, or if the momentum is measure
before the position. Thus if x is the position operator and px is the momentum operator in
the x direction, in one dimension if the Heisenberg Uncertainty principle is valid:
(x px - px x)g  0 g
where g is the function describing the system. Thus since
(x px - px x)  0 , then x and px are said to "not-commute"
if on two operators say A and B are two operators that commute, then -BA = 0
The commutator of x and px is defined as [x, px] = (x px - px x) or more generally, the
commutator of two operators Aand B is [A,B] = AB-BA
Thus by the Heisenberg uncertainty principle, the commutator of the operators of two
observables that cannot be specified simultaneously, should not commute, and if they do
commute then they can be specified simultaneously.
Example:
if the operator for the position is multiplication by x or x = x. and the operator for the
momentum, px, is px is ħ/i x. ħ = h/(2) determine whether or not these operators
commute.
There are different representations of quantum mechanics which define the form of the
operators corresponding to the observables. Two common ones are the
Position Representation
Momentum Representation
The energy operator, the Hamiltonian (H) is constructed from the sum of the operators
corresponding to the kinetic energy (T) and to the potential energy.(V). Thus
H=T+V
In one dimension, in the x representation:
T = px2 / (2m) = 1/(2m) (ħ/i)2 (d/dx)2 or -ħ2 /(2m) d2 /dx2
And in 3 dimensions
T = -ħ2 /(2m) (δ2 /δx2 + δ2 /δy2 + δ2 /δz2 ) = -ħ2 /(2m) V2 where V2 is the Laplacian and is
the sum of the three second partial derivatives, at least in the Cartesian Coordinate system
The potential energy typically depends only on the position coordinates and so it is
basically just multiplication by the potential function.
So
H = T + V = -ħ2 /(2m) V2 + V(x)
So to construct operators in the position rpresentation, one should
1. Write the classical expression of the observable in terms of the position coordinates
and the linear momentum.
2. Replace x by multiplication by x and replace px by ħ/i x and similarly for higher
dimensions or other coordinates.
Using the operator to get the value of the observable
The relation between the outcome of an experiment and a calculation done using the
operator is most generally given by:
I = ∫g* O g d / ∫ g* g d(gives the average or expectation value of the observable
corresponding to the operator O)
where g* is the complex conjugate of the function g, and g is the function describes the
state of the system, which may or may not be an eigenfunction of the Operator, O, d is
the volume element corresponding to the defined space. If one dimensional it is dx or if
three dimensional then it is dx dy dz in Cartesian coordinates. Remember in spherical
polar coordinates it is r2 sin dr d d. Also the denominator is related to the overlap
integral, S, but in this case the functions can be completely different functions. If they
are the same and if g is a normalized function, then the denominator is equal to one. It
may be useful to think of the overlap integral.
Example:
The first excited state of a particle in a box is 2 = N sin(2x/L). In one dimension the
defined space is along the x axis from x=0 to x=L. Determine the expression for the
normalization constant.
As you might expect, there are many instances where integrals similar to the ones shown
above for the expectation value of an observable will be needed. These can be simplified
to using the following notation which is related to the matrix formulation of Quantum
Mechanics.
Called Dirac Bracket Notation
Integrals are written as follows:
<m|O|n> = ∫ m* O n doften abbreviated as Omn
<m| is called the bra
|n> is called the ket
or
<m|n> = ∫ mn d
note that <m|n> = <n|m>*
For a normalized function like n, in terms of the bracket notation
<n|n> = 1
For an orthonormal set of function like m,n, o, …..
Then
<m|n> =
zero if m  n and
one if m = n
That is n, m, o, … are all orthogonal (they have no overlap) to one another and they are
individually normalized.
<m|O|n> is often called a matrix element of the operator O. A diagonal matrix element in
the bracket form is <n|O|n>.
Will encounter sums over products of the Dirac brackets that have the form
s <r||s> <s| |c> = s rs sc = ()rc = <r||c>
The sum is equal to te single matrix element (bracket) of the product of operators .,
note also that
s |s><s| = 1
Hermitian Operators
Definition of a Hermitian Operator is on that satisfies
OR
<m|O|n> = { <n|O|m> }*
∫gm* O gn d = {∫ gn* O gm d }*
An alternative way of stating this is:
<m|O|n> = O|m>* |n>
OR
∫gm* O gn d = ∫ (O gm)* gn d
There are some special properties that are associated with Hermitian Operators
Property 1
The eigenvalues of Hermitian operators are real.
Property 2
Eigenfunctions corresponding to the different eigenvalues of a Hermitian operator are
orthogonal.
THE USE OF QUANTUM MECHANICS IS BASED ON A GROUP OF
HYPOTHESES OR POSTULATES WHICH ARE STATED HERE.
These postulates delineate ways to get physical insight about microscopic systems given
they possess both wave and particle properties. For instance how do we know what the
momentum or average momentum of the system is? Or how do we know the energy or
average energy of the system or its position, again given that these systems act like both
particles and waves.
Postulate 1 (The Wavefunction)
State of the System described fully by (r1,r2, ..., t)
ri s are spatial coordinates, t is time.
Many Applications just a(r), b(r) etc. a & b are the quantum #s and a
b are the eigenstates. State of the system is specified by the quantum #’s
that define it.
According to the Born Interpertation, All information that can be obtained
about the state of a mechanical system and the properties of the system that
are open to experimental determination is contained in a wavefunction.
Born Interpretation
Probability that a particle will be found in the volume element d at the
point r is proportional to *  d



can be likened to a probability density
Also∫ * d
must be less than infinity!
The probability of finding a particle in the whole of space is finite

∫*d =
for a normalized wavefunction.
Stipulations on the wavefunction to be in accord with physical reality.
1. For these functions to be in accord with physical reality, , should be
continuous.
a) 1st derivative should be continuous and finite
b) 2nd derivative should be finite, as well.
2. The function should be single valued
3. The function should have an integrable square. (The function is
everywhere finite;  goes to zero at +/- infinity.
Restriction of integrable squares is simply the requirement that the
probability of finding the system in all space must be finite.
Postulate 2 (Operators)
Operators are associated with physical observables.
To every mechanical variable, there is a Hermitian mathematical operator in
one-to-one correspondence
1) Eigenvalues of hermitian operators are real!
2) Two eigenfunctions of a hermitian operator with different
eigenvalues are ....orthogonal to each other
3) If two hermitian operators commute, then they have a set of
common ....eigenfunctions.
ORTHOGONALITY: ∫ f* g dq = ∫ g*f dq = 0 (integral over defined space)
HERMITIAN OPERATORS
x tells about position, px tells about _________________ , Hx tells about
_______________________________
The operators are chosen to satisfy
pq = -ih/ ∂/∂q and
q=q
H=
Postulate 3
Link between wavefunctions and operators - calculations and experimental
observations.
Physical Properties of observables are inferred from the operator. How?
a) If a mechanical variable A is measured without experimental error, the
only possible measured values of a variable A are eigenvalues of the
operator A. If several measurements are made then the average value is
essentially an average over the prepared eigenvalues.
More formally, for a system described by a wavefunction, the mean value
of some observable,, in a series of measurements is equal to the
expectation, (average or mean), value of the corresponding operator <>.
<> = ( ∫* d( ∫ * d )
d = ?
The ∫is over ?
If the wavefunction is normalized this expression is different. What is
normalized?
If  is normalized, then∫ * d = ?
Why should it be normalized?
How do I make a normalized wavefunction?
Postulate 4
If  is a normalized eigenfunction of the operator,  say a , such that 


 a = wa a
(this is a
eqn.)
then determination of the expectation value of the observable corresponding
to , <>, yields one result!
<> =
Postulate 5
If the system is in some state , which is not an eigenstate of the operator,
 the wavefunction may be expressed as a linear combination of the
eigenfunctions of the n . The Probability that a particular eigenvalue is
measured is equal /cn /2, (cn is the coefficient of the eigenfunction  in the
expression for the linear expansion of the wavefunction in terms of the
eigenfuntions) What is /cn /2 ?
n cn n
eignvalues)
 n = wn nwn are the
Look at <> in terms of the expansion?
Thus the probability that the result is a particular eigenvalue wn is related to
cn cn* or /cn /2 called the norm squared or the modulus squared.
Postulate 6
Heisenberg Uncertainty Principle
Proposed by W. Heisenberg in 1927 suggested the use of operators in the 1st
Place.
It was in the pursuit of the idea that since all "particles" have wave-like
character, then there should be observational consequences that Heisenber
developed his uncertainty principle
Expression x p ≥ h/2
/
suggests that some observables cannot be
simultaneously precisely specified.
Certain pairs of observables are said to be complementary with one another,
when exact specification of one property excludes exact specification of the
other.
A very general form of the uncertainty principle was developed by H. P.
Robertson. For two complementary variables A and B:
AB ≥ ½ |<[A,B]>|
where A is the root mean square deviation of A
A = {<A2 > - <A>2 }1 / 2
Where <A2 > = <|A2 |and <[A,B]> = <|[]|
Example:
Apply the generalized rule to the position and momentum operators. Calculate the value
of xpx for the 1st excited state of a particle of mass and in a box of length L. Here 2=
What are the conditions under which the two observables may be specified
simultaneously with as much precision as desired?
subpostulate: If two observables [A,B] are to have simultaneous precisely
defined values, then their corresponding operators must commute. [A,B] = 0.
There is an "energy-time" uncertainty relation which is of the form Et ≥ ћ. This is
actually a relationship between the uncertainty in the energy of a system that has a finite
lifetime , and is better written as:
E = ћ/
Example:
In addition to finding which operators are complementary, the commutator of two
operators also plays a role in determining the time-evolution of the expectation values for
observables. For operators, like O, that do not have an intrinsic dependence on the time
such that dO/dt = 0
d<>/dt = i/ћ <[,O]>
Note that if the operator for the observable commutes with the Hamiltonian, then the
expectation value of the operator
And the expectation value of that observable does not change with time and is a constant
of the motion. It is said to be conserved.
Example:
Apply this to the rate of change of the expectation of the linear momentum, and get
Ehrenfest's Theorem.
Ehrenfest Theorem essentially says that Classical Mechanics deals with average
(expectation) values, and quantum mechanics deals with the details. So for the
momentum
1. d/dt <px> = <F> where F = -dV/dx where V is the potential energy
2. d/dt <x> = <px>/m
which shows that the rate of change of the mean position can be identified with the mean
velocity along the x-axis.
Postulate 7 (Time dependence of the wavefunction)
Postulate - The wavefunction (r,t) evolves in time according to the
equation:
i h/ ∂(r,t)/∂t = H (r,t)
Time dependent Schrodinger Equation
H is the Hamiltonian operator corresponding to the total energy.
Often the potential energy does not depend on time, then the time dependent
wavefunction can be constructed by (r,t) = (r) e-iEt//h
e-iEt//h can be written as cos (Et/ħ) - i sin (Et/ħ)
Thus the time-dependent factor oscillates periodically from 1 to -i to -1 to i and back to 1
with frequency E/ħ.
Note also that (r,t)* (r,t) = ((r)* ei E t / ħ (r) ei E t / ħ = (r)* (r)
The (r) are called stationary states since they are independent of time
Note also that the value of the value of (r,t)* (r,t) is independent of time.
Remember that much of what we learned in P-chem class had to do with finding
and solving Schrodinger's Eqn. for the various model systems such as:
Translational Energy (Free Particle, Particle in a Box)
Rotational Energy (Particle on a Ring, Particle on a Sphere, Rigid Rotor)
Vibrational Energy (Harmonic Oscillator, Morse Oscillator)
Electronic Energy (Particle in a Box, Hydrogenic and Multielectron Atoms, Molecules)
Most often we solved the time-independent Schrodinger Eqn,
We can notice some general features of these solutions to Schrodinger's Eqn.
Let's look at the general eqn. for a particle of mass m in the x dimension:
 = -ћ2 /(2m) d2 /dx2 + V
So Schrodinger's Eqn  = E is
(-ћ2 /(2m) d2 /dx2 + V)  = E
-ћ2 /(2m) d2 /dx2 + V = E
and rearranging
-ћ2 /(2m) d2 /dx2 = EV
ћ2 /(2m) d2 /dx2 = VE
d2 /dx2 = 2m/ħ2 (VE)
If we know the values of V-E and  at a particular point, then we can look at the
variation of the curvature in the wavefunction
Curvature and Amplitude of 
Sign of Curvature and  and V-E
The curvature and the relative size of the total energy and the potential energy
Remind ourselves of Shrodinger's Eqn. and Various Models for Various Energy
Types
Free Particle:
What is the form of H
k = (2mE/ћ2 )1 / 2
The general solution to this is given by:
Relationship between the energy of a free particle and the linear momentum.
E = p2 /(2m) and rearranging k = (2mE/ћ2 )1 / 2
E = k2 ħ2 /(2m)
Then k = p/ћ but by de Broglies relationship 1/ = p/h = p/(2ħ) or 2 = (p/ħ)
Thus /k
Consider now the effect of the momentum operator on both of the components of the
wavefunction
Wavepackets
Scattering and Penetration from Barriers
Infinitely thick Barrier
Two Zones
Barrier of Finite Width
Three Zones
Must have continuity at the boundaries of the zones
Wavefunctions and slope of the wavefunctions much match at the boundaries.
Define the Reflection probability, R, and the Transmission probability, T:
For E<V
T = 1 / {1 + [((eκ L -e- κ L )2 ) / [16(E/V)(1-(E/V))]]}
κ = {2mV (1- (E/V))}1 / 2 / ћ
For Barriers that are high and wide, κL >> 1,
T = 16 (E/V) (1- (E/V)) e- 2 κ L
Plot of T vs E/V for several values of [L (2mV)1 / 2 ] / ħ for E<V
Tunneling
ForE>V
T = 1 / {1 + [(sin2 (k' L)) / [4 (E/V) ((E/V)-1)]]}
k' = [2mV ((E/V)-1)]1 / 2 / ћ
Plot of T vs E/V for several values of [L (2mV)1 / 2 ] / ħ for E>V
Note that at even when E>V, there is still a chance that the particle will be reflected?
Know as anti-tunnelling or non-classical reflection.
These rectangular potential barriers are not realistic, there are only a few realistic
potentials which can be solved and lead to analytical equations for the reflection and
transmission probabilities. One is the Eckart potential barrier.
V(x) = 4Voeβ x / (1+eβ x )2
of inverse length.
here Vo and  are constants. Vo has units of energy,  units
Figure of Eckart potential barrier
T = [cosh{4(2mE)1 / 2 /ħβ} - 1] / [cosh[4(2mE)1 / 2 /ћβ] + cosh{2|8mVo-(ħβ/2)2 |1 / 2 /ћβ}
Figure showing how T varies with E/Vo
Note that the particle may still be reflected even when E>Vo, but there are no oscillations
in the transmission probability.
Particle in a Box
These restrictions on the wavefunction make it such that acceptable solutions to the
Schrodinger eqn. do not exist for arbitrary values of the ENERGY.
Particles may possess only certain energies. ENERGIES ARE QUANTIZED.
(Bohr’s theory for the hydrogen atom)
Free Particle Solution to the SCHRODINGER EQN. REPRESENTS ONE CLASS
(NO POTENTIAL ENERGYCONTRIBUTION TO THE HAMILTONIAN. A
SECOND BROADER CLASS ALLOWS FOR A POTENTIAL ENERGY
CONTRIBUTION.
CONSIDER THE SOLUTION TO THREE POTENTIAL ENERGY SITUATIONS
WHICH ARE IDEAL CASES REPRESENTING THE POTENTIALS WHICH
CONTRIBUTE TO THE TRANSLATIONAL MOTION, VIBRATIONAL MOTION,
AND ROTATIONAL MOTION.
Looked at freely translating molecule before. Consider a particle “confined” by potential
energy.
1 dimensional Particle in a box!
Fig.
Two huge potential walls at each end of a line
Between walls potential energy V = ; At walls V = infinity (infinite repulsive force)
3 regions
Region 1: ...... x<0
Region II ..... 0<x<a
Region III .... a<x
Solve time-independent Schrodinger eqn in each of the regions. In regions I and III
H = E .......... becomes?
Now if  > 0 and V > E. In the wall V = infinity
then d2 / dx2 must go to infinity in the walls
Or if  < 0 and V>E, in the wall V = infinity
then
However one of the tenants was that the derivatives of  must be bounded, so the only
solution is that  = 0 in the walls.
Reasonable since the particle is effectively confined to region II.
Eqn above applies to Region II and it simplifies to the free particle since V=0 in that
region. However, have boundary conditions:
Remember for free particle we found:
(x) = C .....................................
Use boundary conditions to solve for coefficients C and D.
For ((0) = 0, coefficient C must be ............................
For (a) = 0, D sin(ka)=0 only if k=0 or multiple of /a(since sin0, sin, sin(2), sin(3),
etc = ?)
Thus k = na ........................... where n = 1,2,3, .......
Can “n” be zero? Why? Why not?
If n = 0, k = 0 and ....................., always!!
Correspondence Principle States that classical mechanics emerges from quantum
mechanics as high mechanics as high quantum numbers are reached.
2 plot ....................... (spends equal time at all points in the box at large “n”)
Application 3: 2 dimensional particle in a box
fig:
L1 and L2 are the length of the box in the x and y directions respectively.
In region 2 where free particle exists, Schrodingers eqn. is:
Note that this is a 2nd order differential eqn., but that there are two variables, so it is not
ordinary.
Use Separation of Variables Trick
Divide the complete Eqn. into two or more ordinary differential eqns. such tath there is
one eqn. for each variable.
Start by writing the wavefunction as a product of two function, one depending only on x
and a one depending only on y.
..........................................(x,y) = X(x) Y(y)
Now:
product rule:
Again
So Schrodinger’s Eqn. becomes:
Dividing by X(x) Y(y)
E is just a number so the left hand side of the eqn must be constant. The sum is a
constant. So each part must be equal to a constant.
So expect: ............... E = Ex + Ey
Separate into two parts:
Both are essentially the same as the eqn. for a 1 dimensional system.
Same boundary conditions?
Should have the same solutions:
Total wavefunction is a product of the wavefunctions in the x and y directions. The
energy
is the sum of the energies in the x and y directions.
If rectangular box, then L1 and L2 are different, if square box then they are the same.
Does this mean anything special?
Degeneracy: (Two or more eigenstates that have the same energy)
Consider 12 versus 21 and E12 vs E21
Particle in Box applied UV-Vis (electronic) absorption of Dye Molecules
Fig.
 electrons in conjugated “tether” can move freely, how many pi electrons are there?
Length of box is approximated by 1.4 Angstroms per bond.
Only two electrons are allowed in each electronic level of the “electron into the
conjugated box!
What is the lowest energy transition of the electron to the higher state.
Vibrational Motion:
One dimensional harmonic oscillator:
A particle undergoes harmonic motion if there is a strong restoring force where the
restoring force depends on the displacement from equilibrium. One thins of a spring
attached to a wall with a ball attached to the other side.
By Hooke’s Law, the force is linearly related to the displacement x from equilibrium.
.............................................. F(x) = -kx
k is the
The potential is derivable from the force as dV(x)/dx = -F(x)
..................... so
Schrodinger’s eqn. for this particle in one dimension is:?
By making a few substitutions this differential equation can be rearranged to give
something more recognizable in terms of a solution
Let y = .................. dy = ........................................................dy2 =
 = ............................................  = ............................... and =
With these substitutions in the Schrodinger eqn. on gets:
This differential eqn. is much easier to solve: Expect to get both the wavefunction (x)
and the Energy.
Use power series expansion method.
Trick - get asymptotic solution and multiply by a power series,.
For any value of the energy, a value of y or x can be found such that for this value and all
larger values of y,  is negligibly small relative to y2,
Thus the asymptotic form of the wavefunction becomes:
The solution to the asymptotic form is:
But exp(+/-y2 /2) is not an acceptable function. WHY?
Now have a solution that works for large values, but want one that works throughout the
defined space?
This is done based on the asymptotic solution by introducing, as a factor, a power series
in y, and the coefficients of the power series are determined by a substitution in the wave
equation.
Rearranging terms in decreasing order of differentiation.
dividing through by exp(-y2/2)
Two ways to solve this
1. Represent f as a power series.
Substituting these expressions into eqn above gives:
In order for this series to vanish for all values of y, (for f(y) to be a solution), the
coefficients of the individual powers of y must vanish separately.
Thus
power of y ...................Equation
Or in general, for the coefficient of yn
Note that if a0 of a1 is 0 then we only have odd or even terms. This is called a recursion
formula since it allows the coefficients a2,a3, and a4 ...... to be calculated successively if a0
and a1 are assigned.
For arbitrary values of  (which is related to the energy), this series of eqn above has an
infinite # of terms. Is it a satisfactory wavefunction? Why?
Almost like exp(-y2) = 1 + y2 + y2/2! + .............................. which is an infinite series
Can we get this to work as a wavefunction?
remember ...............  = exp(-y2/2) (f(y))
What if f(y) broke off after a certain amount of terms?
That is choose values of E and therefore  which break off after a finite # of terms. Then
the negative exponent would keep the wavefunction finite.
The value that causes break off after the vth term is:
Want to truncate after the even or odd nth term for En. Thus let coefficient of the (n+2)th
term be
zero, an+2 = (-2n-1)/[(n+2)(n+1)]an = 0
So ...........  =
Thus to truncate after the vth term:
With a1 or a0 =0 since as v is even of odd and the value of can only cause the even or
odd series to break off but not both.

Thus Ev = (v+1/2) h .................. v =
This is the Energy Expression for the energy associated with the vth level of a harmonic
oscillator.
Quantum # is v.
The eigenfunctions are the functions given by the power series expansion with the
coefficients in the power series expansion determined from the RECURSION
relationship.
Thus  =
where an+2 =
As said previously, another way to solve this is to recognize that the differential eqn. is
similar to one that has been solved previously using HERMITE POLYNOMIALS.
MORE PRECISELY, the wavefunctions which are solutions to the harmonic oscillator
eqn. are:
v = NvHvexp(-y2/2) or v = NvHvexp(-2 x2/2) where v is the quantum number for this
system. v = 0,1,2,3,...  = (mk/ħ2 )1 / 4 and Nv = (/[2v v!1 / 2 ])1 / 2 (normalization factor)
And Hv is the Hermite polynomial (see table), Also y is a dimensionless variable y=x
The Hermite Polynomials satisfy the equation: H”v - 2yHv' + 2vHv = 0
“ means
And ‘ means
The Harmonic Oscillator Wavefunctions are orthogonal to one another
The Normalization can be expressed as - ∫+  [Hv(y) exp(-y2 /2)]2
Example:
Give the expressions for 0 and 1, the ground state and 1st excited state harmonic
oscillator wavefunctions.
As we said before, the harmonic oscillator energy eigenfunctions have the following
properties:
The integral
-
∫+  Hv’ Hvexp(-y2)dy = [(0 if v not equal v’) or (1/2 2v v! .... if v = v’)
THIS RESULT SAYS THAT the wavefunctions, which are solutions to the energy
eigenvalue eqn., the Schrodinger eqn. for the Harmonic Oscillator, ARE
ORTHONORMAL.
with Nv = (1/22vv!)1 / 2
What do the wavefunctions look like?
.
What about the energies? How do they differ from the particle in the box energies?
Separation between adjacent levels?
Excitation from the lowest level (ground state) to the next highest level involves
absorption of light. In harmonic oscillator excitation from the first excited state to the
next highest level involves?
What about emission?
What are the properties of the oscillator?
<x> = 0 Why?...... Mathematically? ..
.... Physically? ............
Equally likely to be on either side of
x=0
Property of Hermite polynomials and even and odd functions.
<x2> =
........  =
since (k/)½ =
Mean potential energy of the oscillator
<V> = <1/2kx2>
so <V> =
The total energy is:
The Kinetic Energy is:
This result is a special case of the Virial Theorem
“If the potential energy of a particle has the form V = axb, then its mean potential and
kinetic energies are related by:
2<Ek> = b<V> for a harmonic oscillator, b = ?
Thus the total mean energy is
Zero Point Energy here?
Classical Limit:
Highest probability of being found at its maximum displacement, the turning points of
its classical trajectory, why
Note that at high v, the harmonic oscillator wavefunctions have their dominant maxima
close to the turning points.
Non-harmonic Oscillator - Morse Potential
Turns out that Schrodinger's eqn. can be analytically solved for vibrational motion using
a Morse potential instead of a Harmonic potential. In a harmonic potential well, even at
large displacements from equilibrium, the potential is still parabolic and energies are still
equally spaced. However, in real molecules an anharmonic potential well with narrowing
in the energy spacing between energy levels is more appropriate, especially at higher
energies
Figs.
A Morse potential matches the true potential energy over a wider range, but is still an
approximate function.
V(x) = hcDe{1-exp(-ax)}2 a = k/(2hcDe)1 / 2
De is the depth of the minimum of
the potential curve and x = R - Re, the displacement from the equilibrium displacement.
Use of this potential in Schrodinger's eqn. yields:
Ev = (v+½)ћ - (v+½)2 ħxe
where xe = a2 ћ / (2)
 = k/)1 / 2
xe is known as thee anharmonicity constant. Note that the second term subtracts and
more important at larger v.
The number of bound energy levels in now finite v = 0,1,2, … vmax
where vmax< hcDe/(ħ/2) - ½
What is the zero point energy?
Still just an approximation to the true anharmonic potential, and suggests that the actual
vibrational energies of a real molecule could be expressed by a series:
Ev = (v+½)ћ - (v+½)2 ħxe + (v+½)3 ћye + ….
The spectroscopic constants (, xe, ye, etc.) may be treated as empirical parameters
and can be obtained by fitting the eqn. to experimentally observed spectral transitions.
Gross Vibrational Selection rules
selection rules for the vibrational transition v' <- v are based on the electric dipole
transition moment <,v' |  | ,v> where  is the dipole moment of the molecule when it is
in the electronic state and not taking into account rotational transitions.
 = o + (d/dx)o x + ½ (d2 /dx2 )o x2 + …
where o is the dipole moment when the displacement is zero. Transition matrix
elements can therefore be seen to be:
The term with o is zero because the vibrational wavefunctions are orthogonal.
Thus the GROSS Selection Rule is that the transition matrix is non-zero only if the
molecular dipole moment varies with the vibrational displacement otherwise d/dx,
d2 /dx2 , etc. are equal to zero.
So
In order for a molecule to have a vibrational absorption, it must have a dipole moment
that varies with extension.
What about the specific selection rules. For small displacements, ignore the terms with
the higher order derivatives.
so look at: v',v = (d/dx)o <v'|x|v>
When is this matix element non-zero?
Depends on the properties of the Hermite Polynomials, the recursion relations.
so x|v> is proportional to xHv(x), using the recursion relation we find
produces two terms one term with |v+1> and the other with |v-1>. thus the only non-zero
contribution is when v' is v + /- 1. Thus under a harmonic approximation
v = +/- 1
(known as the fundamental transition)
Creation and Annihilation Operators
These give the selection rules as well.
In reality, anharmonicities need to be accounted for and this yields several transitions
with slightly different wavenumbers
vv+1<-v =  - 2(+1)xe + …
For large displacements, the terms in x2 and higher may contribute allowing Transitions
with v = + /- 2.known as the first overtones
Also if the oscillator is significantly anharmonic, then the overtones can occur as well.