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Quantum Mechanics - predicts the properties of small microscopic systems such as atoms
and molecules. This was necessary as Classical Mechanics does not appear to apply to
these systems.
We’ve seen that changing the composition can change the chemical properties. Think of
the change that occurs to cause sickle cell anemia, just one amino acid changes. In the
case of vision, the pigment rhodopsin absorbs visible light and undergoes photochemical
changes that cause electrical signals to be transmitted toy the optic nerves to the brain.
The photoexcitation causes a significant change a cis-trans isomerization in the geometry
of the chromophore at its attachment site occurring rapidly in about 5 ps. Quantum
Mechanics contributes in several ways to the understanding of how this process occurs.
One of the most important considerations is how the light absorption process itself
occurs. Furthermore, there is an energy barrier that lies at 175 kJ/mol or more above the
cis state. This barrier essentially eliminates the thermal background that would be
present do to background from heat. Evidently heat, and visible light are related and
there is energy associated with light as well. In fact, Quantum Mechanics developed out
of the study of “light” and its interaction with matter.
Much of Chemistry is based on this:
...........Observations, Color changes, Qualitative Analysis
...........Spectroscopic Identification
Different types of light probe different properties
Nature of Light
Newton vs Huygens. Wave or particle both can give information
............Diffraction
What is a wave
............Velocity =  .................................... c = 3.00 x 108 m/s = (light)
Maxwell’s Relations ............... Electromagnetic Theory
(Happy Physicists)
Blackbody Radiator (hot poker)
fig. Experiment ...........................................................................Intensity vs wavelength
(temps)
Rayleigh Jeans Law (Derived from Maxwell’s Theory)
described low side of distribution but Ultraviolet Catastrophe
Planck’s Formula
Einstein and the Photoelectric Effect
fig. Experiment
fig. Data
Energy of light could be described as E = h;
h was the same constant that appeared in Planck’s formula for blackbody “packets of
light, quanta”
About this time people interested in Discharge from different gases, their emission was
different from blackbody? How?
Rydberg represented wavelengths of all spectral lines of H atoms by:
........................v/c =..1/ = RH(1/n12 - 1/n22) .......................... RH = 109,737 cm-1
Bohr’s “Old Quantum” Theory Explained this:
1. Fig.
2. Angular Momentum of the electron is “quantized”. Its magnitude can take on values
h/2,
.... 2h/2, 3h/2etc.
3. Electrons lie at ceratin discrete orbits (violates Maxwell’s Theory)
2
2
2
.......................... r = (h4
/
/ = h/2
0/(m0e ) * n ..= ..a0 n ............................. h
4. Energy is also quantized:
.................................. Potential .......... V = -e2/(40r)1/n2 = - mee4/((40)2(h/2)2) 1/n2
.................................. Kinetic H = -1/2 V
TOTAL = -1/2 - me e4/((4o)2(h/2)2) = -mee4/(8o2h2) 1/n2
From Ephoton = E (Energy difference between the states involved in the emission h and
c=)
E = hc/
1/E/hc 1/ = (En1(final) - En2(initial) /hc =
Bohr calculated the constant and it was Rydberg’s constant
Bohr's theory worked for hydrogen spectra but not many others. Also Classically, an
electron in a circular orbit should
Davidson and Germer Slit experiment for electrons.
...... only light had particle duality, but so did ?
De Broglie’s Hypothesis:
................................. = .................................................. p = momentum
Example:
Calculate the wavelength of a hydrogen atom traveling at 1000 m/s.
Calculate the wavelength of a fast ball traveling at 95 mph.
Quantum Theory - it is not possible to know simultaneously the position and momentum
of a small particle such as an electron?????? (Heisenberg Uncertainty Principle).
........x...* ...... p ...... ≥ ...... h/4..........ALSO .........E ...... * ........ t.....≥ ........ h/4
Classical Physics and classical mechanics say that given the initial condition, we can
predict the condition of an object at any later time.
However, electrons have wave character. Need a new mechanics of motion of electrons
and other small particles.
On Nov. 25, 1925, Erwin Schrodinger gave a presentation on deBroglie’s thesis work in
Zurich. In the discussion that followed, Peter Debye commented that he thought that this
approach to wave-particle duality to be somewhat childish. “After all”, said Debye, “ to
deal properly with waves one had to have a wave equation .... “
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 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 
2
where ~
v is known as the Laplacian operator and is given by 2/x2 + 2/y2 + 2/z2 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 atom, and the results fit with
experimental observations and the results of the Bohr atom.
Note that the differential eqn. above is and eigenvalue eqn. The functions satisfying such
an eqn. are known as eigenfunctions, and the quantities of energy are called eigenvalues.
Remember however, that we can’t know exactly how the electrons move in an atom.
Bohr’s model of the hydrogen atom is fuzzier, no longer discrete orbits. Don’t know
exactly where we’ll find electron. _____________ uncertainty and ______________
hypothesis both point to this. This suggests that there is only a probability that the
electron exists in a particular place.
Born interpretation of particle-wave duality is:
Since this duality exists for all objects, there is a certain probability associated with the
motion of any object.
Revolutionary - goes against Determinism
Study Operators since in Quantum Mechanical Eqn.
An operator A is any mathematical entity which operates on any function and turns it into
a different function.
Examples
A(x) = x
so that ..... A (x) = x (x)
or
A = d2/dx2
so that A = d2(x)/dx2
Consider an operator that is a function of both x and /x
A= /x x
Let this operate on  and see what you get!
To each operator such as A(x,d/dx) there exists a set of numbers an, and a set of functions
Un(x), such that A Un(x) = anUn(x).
an is called an eigenvalue and Un is the corresponding eigenfunction.
Eigenfunctions of an operator are those special functions that are not altered by the
operation of the operator except by multiplication by the eigenvalue.
Consider A = -id/dx
If Un(x) is periodic in the range 0 to L.
What happens when two operators operate successively?
Define the commutator of two operators.
[A,B] = AB - BA (difference between operating first with B and then with A, AND first
with A and then with B.
Consider A = x and B = d/dx
Find commutator by letting it operate on a function.
Thus the main properties of operators are:
1) with each operator is associated a set of numbers, the eigenvalues through the
eigenvalue eqn.
2) in general, the commutator of two operator is not zero, although it can be.
These are important properties which will be used to derive the postulates of Quantum
Mechanics.
Classical mechanics breaks down when applied to
systems. Why?
“One interpretation”
In the physics of classical systems it is assumed that the operations of observation don’t
appreciably disturb their motion.
Perhaps the simplest observation is made by turning on a light to look at an object. This
involves striking the object with photons. The photons should be of wavelength
sufficiently small or high frequency that the position can be determined accurately. If the
object is large, then a blow with a photon
Dirac “ there is a limit to the fineness of our powers of observation and the smallness of
the accompanying disturbance - a limit which is inherent in the nature of things, and can
never be surpassed by improved techniques.
Since the operations of observation affect the physical system then they should appear
explicitly in the theory. These operations have two main properties:
1. To each observation (observing energy, observing momentum, observing position,
etc.) there belongs a set of numbers - the possible results of the observation. These
numbers may be over a continuous range, or set of discrete values.
2. If we have two types of observation A and B (A might be position, B might be
observing momentum). If observation B is followed by observation A, then this is
denoted AB. The BA is the same type of observation carried out in the opposite order.
Since each observation may disturb the results of the other, the two procedures may yield
different results if AB-BA is not zero. The magnitude of the value of this expression
must be related to the magnitude of the disturbances.
If AB - BA = 0, then the operators corresponding to A and B are said to
they have
.
and
Correspondence Principle
The properties of the physical observations correspond to the mathematical properties of
operators developed before. To each belongs a set of numbers, and the effect of either
pair may depend on the order in which they are applied.
So assume observations A are represent by operators A, there being one operator for each
observable property, ----- the energy, the position, the momentum etc.
The function on which the operators operate represent the sate of the system, and are
known as state functions or wavefunctions. If the state function is an eigenfunction, then
it is referred to as an eigenstate.
This leads to THREE INTERPRETIVE ASSUMPTIONS!!
I. The possible results of an observation A are the corresponding eigenvalues an.
II. An observation A on a system in an eigenstate Un certainly, leads to the result an.
III. The average value of repeated observations on a set of systems, each one in an
arbitrary state. ..... y is:
..... <A> =
Thus if a system is in a general state (say in one dimension x) y(x), an observation
corresponding to A must (by postulate 1) have as its result one of the eigenvalues of A.
Repeated observations of A on a set of systems, each in a state (x) will produce a
statistical distribution of different eigenvalues.
 must therefor be a linear combination of the eigenstates of the system Un(x)
............................. (x) = c1 U1(x) + c2 U2(x) + c3 U3(x) + .....
These together allow us to interpret and redefine the Schrodinger equation derived by
Schrodinger.
Let H be known as the Hamiltonian. It is the energy operator.
H=T+V
T = p2/2m ........ and V is the potential energy.
H Un = En Un ......................Schrodingers time independent Equation. It is an eigenvalue
equn. as it has the form:
(operator) X (function) = (numerical factor) X (function)
Values of En derived here represent the ONLY possible values that measurement of the
energy will yield.
Wavefunction must be _________________________
How do we find it?
Applications:
Our interest lies mostly in solving Schrodinger’s Eqn. for the energies associated with a
molecule.
Think about the energy a molecule possesses:
1. Energy of Movement of the whole molecule:
ENERGY OF TRANSLATION
2. Energy of Movement associated with molecular tumbling: ENERGY OF ROTATION
3. Energy of Movement associated with the atoms in the molecule moving back and forth
relative to one another: ENERGY OF VIBRATION
4. Energy associated with the electronic arrangements or electronic states of the
molecule. ELECTRONIC ENERGY
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 Interpretation, 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 (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 from (r) e-iEt//h
Postulate 7
Heisenberg Uncertainty Principle
Proposed by W. Heisenberg in 1927 suggested the use of operators in the 1st
Place.
Expression x p ≥ h/2
/
suggests that some observables cannot be
simultaneously precisely specified.
What are the conditions under which the two observables may be specified
simultaneously with as much precision as desired?
Postulate: If two observables [A,B] are to have simultaneous precisely
defined values, then their corresponding operators must commute. [A,B] =
0.
MODELS FOR TRANSLATIONAL ENERGY
Application 1: Free Particle
Case where the potential energy is 0.
What does Schrodinger’s eqn become?
This is a differential equation, second order (highest derivative) ordinary (one indep
variable x)
Rearranges to:
d2/dx2 + ........................ = 0
In general a differential eqn. of the form:
d2U/dz2 + p dU/dz + q u = 0 ...................................... p and q are both constants
Many solutions to this differential eqn. How do we pick the right solution?
Ask what is physically acceptable.
Born Interpretation
Analogy with wave theory of light A2 = Intensity
The square of the wavefunction (if real) or the product of the complex conjugate of the
wave function time the wave function (if imaginary) is proportional to the prabability of
finding the particle at each point in space.
If the amplitude of the wave function of a particle is  at some point x,
then the probablility of finding the particle between x and x + dx is proportional to:
For our present case: A ekx =  is a solution of this if k satisfies the condition
............ k2 + pk + q = 0
or .........k2 + 0k + .......... = 0
So  = A ek1 x + B ek2 x............ where
k1 = ........................................and k2 = ..................................................(1)
or  = Aeikx + Be-ikx
since by Eulers relation:
(x) = A cosK + B i sin K + B cos K - BisinK
....... = (A+B)cosK + (A-B) i sin K
an alternative solution is  = C cosK + D sinK
Now from (1) E =
since E is just the kinetic energy,
Thus we get both the wavefunction and allowed energy values by solving the
Schrodinger Eqn.
There need to be some restrictions on the wavefunction, the eigenfunction, which is a
solution tot he Schrodinger Eqn. (in this case for the free particle)
OVER the DEFINED INTERVAL:
a) If  is a solution to the eqn. so is N.
But according to the Born Postulate, the wavefunction squared or the wavefunction times
its complex conjugate tells us where the particle is, so
..................................................= 1
Normalization of the wavefunction.
b) the wavefunction must be continuous,.
c) the wavefunction must have a continous slope.
d) the wavefunction, and its 2nd derivative should be finite everywhere in defined region.
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.
Looked at freely translating molecule before. Consider a particle “confined” by potential
energy.
Application 2: 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!!
THUS WE Can now define the energy and the energy wavefunction for the particle in the
box:
E=
and
=
Correspondence Principle States that classical mechanics emerges from quantum
mechanics as high mechanics as high quantum numbers are reached.
Let us examine the energies (energy eigenvalues) and the energy eigenfunctions and
eigenstates.
Also, the 2 plot ....................... (spends equal time at all points in the box at large “n”)
Plots:
Note that the lowest energy level is not for n=0, but rather for n=1, and the energy is
not zero! This is known as the ZERO POINT ENERGY.
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
Application 4: Particle in Box applied UV-Vis (electronic) absorption of Dye
Molecules
Fig. Bottom of page 459 for Carotene
 electrons in conjugated “tether” can move freely, how many pi electrons are there?
Carotene has 22  electrons.
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!
It is difficult to measure directly the energy of the bound electrons in a molecule such as
-Carotene, but an energy difference can be measured from the absorption spectrum of
the molecule.
What is the lowest energy transition of the electron to the higher state?
Application 5: Tunneling
If the potential energy is not infinite outside of the box, then the wavefunction can have a
nonzero value outside of the box, in the walls, and if the wall is thin enough, then the
wavefunction may remain nonzero to the other side of the wall. This is known as
TUNNELING!
Figure:
The probability of tunneling through the barrier is given by this relationship!
T=
The following figures give the tunneling probability as a function E/V value.
Note that the tunneling probability depends on:
the ____________ of the particle, the ________________ of the potential barrier, and the
_______________ of the potential barrier.
The mass is important as the wavelength is proportional to the momentum, MV, and
because the kinetic energy is equal to ½ mv2 then
.................................= h/(2mE)1/2
Electron tunneling occurs in many oxidation-reduction reactions catalyzed by
______________. Rapid electron transfer between donor-acceptor sites separated by 1020 angstroms in a protein has been established.
Proton tunneling can occur as well. An example is oxidation of ethanol by acetaldehyde:
Classically, the reactants need to have an energy equal or larger than the activation
energy to react, but the proton can tunnel through the activation barrier instead of going
over it. This increases the speed of the reaction.
Can look at speeds of hydrogen, deuterium, and tritium and compare them. See figure
9.12.
Application 6: 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 toa wall with a ball attache to the other side.
By Hooke’s Law, the force is linearly related to the displacement x from equilibrium.
.............................................. F(x) = -kx
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.
One 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(-ax2) where v is the quantum number for this system. v = 0,1,2,3,...
And Hv is the Hermite polynomial, and Nv is the normalization factor.
Examples:
o =
1 =
THIS RESULT SAYS THAT the wavefunctions, which are solutions to the energy
eigenvalue eqn., the Schrodinger eqn. for the Harmonic Oscillator, ARE
ORTHOGONAL
What is orthogonal?
The Harmonic oscillator wavefunctions when normalized represent an
ORTHONORMAL set.
What do the wavefunctions look like? What about the probabilities of the square of the
wavefunctions?
What about the energies? What is there value. Are they the same or do they differ from
the particle in the box energies?
Both quantum mechanics and classical mechanics give the same fundamental vibration
frequency o of the oscillator, however.
.................................... o = 1/2 (k/) or  = (k/)1/2
Where,  is the reduced mass of the oscillator and k is the force constant for the bond.
Is there a zero point energy here?
Separation between adjacent levels, are related to the light absorbed to cause a transition
between vibrational levels in a molecule
Excitation from the lowest level (ground state) to the next highest level involves
absorption of light. Therefore, light will be absorbed if its frequency,, is equal to a
vibrational frequency of a harmonic oscillator. A harmonic oscillator is a good
approximation for the vibration of diatomic molecules and the vibration of bonds in
polyatomic molecules. Vibrational frequencies are characteristic of a particular type of
bonds such as C-H, C-C, C=C, C-N, but they can also be used to determine force
constants k for the particular vibration.
In harmonic oscillator excitation from the first excited state to the next highest level
involves?
What about emission?
Examples: