Welcome back to the lectures on chemistry
as part of the National Programme on Technology
Enhanced Learning funded by the Ministry of
Human Resource Development. This is the series
of lectures given to students in their college
years in engineering and in basic sciences.
The Chemistry module we will continue today
is the same as the atoms and molecules. In
the last lecture if you recall, I introduced
the idea that Quantum Chemistry is fundamentally
important in understanding why molecules form,
why molecules undergo transformations to form
various newer species and how do we understand
the structure, properties and the chemical
transformations of molecules, how do we do
this from first principles. If you ask these
questions then Quantum Chemistry provides
the answers. Therefore, the first model on
the atoms and molecules is meant to expose
the use of Quantum Chemistry techniques. I
was categorical in the last lecture that we
are not trying to understand Quantum Chemistry
or Quantum Mechanics but rather we are trying
to follow through a prescription or a procedure
by which we use the rules of the mechanics
and hopefully over a period of time that we
will understand why such rules came into place.
Before I begin my lecture let me welcome you
with this acknowledgement to the National
Programme on Technology Enhanced Learning,
my name is Mangala Sunder, Department of Chemistry
Indian institute of technology Madras in Chennai.
We will start studying model problems in Quantum
Chemistry and we will start with the simple
free particles in one and two dimensions.
But before we do that, let me just recollect
to you some of the most important names in
Quantum Mechanics. In today's lecture I will
give you names of some of the important founders
of the mechanics and in subsequent lectures
I will also introduce you to the illustrious
Chemists who have contributed to all our knowledge
as of today.
The people who made our understanding of the
atoms and molecules possible, a few names
most important ones Max Planck, we introduced
Max Planck in the last lecture through his
constant and his phenomenon of the black body
radiation. Albert Einstein, undoubtedly the
most famous Physicist ever to have lived in
the last century whose contribution to Physics
in general and to Quantum Mechanics in particular
are unparalleled.
Niels Bohr, a Danish Physicist whose name
is familiar to all of you from your high school
textbooks on the planetary model of the hydrogen
atom the orbits of the electrons in the hydrogen
atoms. We recall Niels Bohr who was the first
to apply the quantum hypothesis to understand
the spectra of hydrogen, the line spectra
of hydrogen in terms of the lyman series,
the balmer series, the paschen series and
so on. We recall that there was a famous constant
called the Rydberg constant which Neils Bohr
explained through fundamental methods. Louis
de Broglie, a French Physicist brought in
the idea that matter can be treated as both
particulate as well as wavelike. So this wave
particle duality eventually led Erwin Schrödinger
the Austrian Physicist to ask questions such
as what are the governing dynamical equations
for substances which behaves both wave like
and particle like. Louis de Brogile was the
founder of this matter the wave ideas.
Max Born is seen on the left hand side. Max
Born's contributions in Physics in general
are very well known but in this particular
reference to the Chemistry it is his interpretation
of the quantum mechanical wave function. This
absolute square of the wave function is being
identified with the probability density of
the system. We will see those things a little
bit more in detail. Werner Heisenberg, very
well known for his uncertainty principle,
we must have heard the joke that the Heisenberg
may have been here to indicate the uncertainty
associated with position and velocity, position
and momentum, it is a corner stone of Quantum
Mechanics.
Erwin Schrödinger, the most important name
for all the chemist because all the chemists
quantum chemist, computational chemist, the
ones who do theory the ones who tries to understand
Chemistry from first principles have to solve
the Schrödinger equation , have to understand
how to solve Schrödinger equations both analytically
and through the use of computers.
Paul Dirac, the man on the right hand side
contributed to the development of Quantum
Mechanics through His relativist theory of
electrons. His contributions, in fact he made
such a famous statement that all of the mechanics
as is necessary for the applications of the
Chemistry are well known and it is only a
question of computational difficulty that
we will have to surmount. Paul Dirac's name
is very famous among computational chemistry
for the statements he made and for the developments
in computational chemistry which has happened
in the last 50 or 60 years. People understand
that Dirac gave the basic formulation in terms
of unifying Quantum Mechanics through his
ever famous book called the Principles of
Quantum Mechanics.
Wolfgang Pauli is the other person that again
Chemists know. Pauli's exclusion principle
in your high sdhool when you build up the
electronic structure of an atom by adding
electrons to various orbitals you are told
that no more than two electrons for any orbitals,
no two electrons can have all the same four
quantum numbers. That is one of the many important
things that Wolfgang Pauli did and which is
fundamental to the atomic structure in Chemistry.
Let us go back to the lecture that we have
today, namely the model solutions of the quantum
problems that we introduced, the Schrödinger
equation that needs to be solved for simple
models and from the models we try and understand
the results in terms of simple physical pictures.
And we see that the pictures that come out
of the solutions are quite different from
what we are familiar from our realization
of what happens around us, what happens in
the atoms, in sub atomic, in microscopic domain,
microscopic dimensions seen so strange that
we have to see this only through the Mathematics
and the corresponding solutions of the mathematical
equation as proposed by Erwin Schrödinger.
So, let we summarize the contents of today's
lecture before we go into the details. We
will solve the free particle in the box in
one dimensional model problem and also in
a two dimensional problem. And given some
time, I would like to give you pictorial representations
of the wave functions and squares of the wave
functions. Let me recollect from the previous
lecture the Schrödinger equation that we
wrote down for a particle in the one dimensional
box. You recall that there was a kinetic energy
term −ħ2/2m (d2/dx2), the operator corresponding
to the kinetic energy plus the potential energy,
V(x) of the particle or of the system that
we are interested in the potential energy
experienced by the particle. All of this was
called the Hamiltonian, H which represents
in classical mechanics this total energy of
the system. Quantum mechanical kinetic energy
plus the quantum mechanical potential energy
acting on the wave function Hψ(x) = E ψ(x).
This is the equation governing the wave function
ψ(x). If we know the particle's mass and
if we know the potential energy that the particle
is facing then we are able to formulate the
Hamiltonian of the system but then we have
to solve the differential equation Hψ = Eψ
where E is a constant. This is a model problem.
Now, why is particle in a 1D box for chemist?
Think about a simple system like conjugated
polyenes, alternating single and double bond
structures. Polymeric if you want to the simplest
structure for this is butadiene then of course
you can put the hydrogen to ensure that the
chemical bonding of carbon is taken care of.
The π-electrons which are in the double bonded
structure, the π electrons are easily modeled
to a first approximation by an equation similar
to that of a particle in a 1D box equation.
If you want to know something about what these
molecules do when they absorb light or when
you excite them where the electrons go and
things like that then to a certain approximation
we can understand all of them by solving the
Schrödinger equation for a one dimensional
particle in a box. Therefore, the relevance
is in trying to understand a corresponding
chemical system.
The particle in a 1D model is of course −ħ2/2m
(d2 ψ(x)/dx2)+V(x) ψ(x) = E ψ(x). But with
a 
specific requirement that this potential vanishes
in a small region 0< x <a is the region or
the box or the length of the box. If the particle
moves in one dimension x the length of box
is a, you see that the potential is 0 in that
region and the potential is infinite otherwise
so that the particle does not escape the box.
Simple pictorial representation of the box
is something like what you see here. The particle
moves inside this region, what I had is 'a'
there in the equation is replaced now by L
is length of the box, 0 to L, x is the variable
the particle is moving in a one dimensional
motion. We do not know where it is, we all
know that it is inside the box it cannot escape
the box in that the potentials at the ends
of the box are infinity, infinite, repulsive
so that the particles stays inside the box.
This is the model we want to solve. Now let
us go back to that equation. Since V is 0
inside the box we can simply rewrite this
equation as −ħ2/2m (d2ψ/dx2) =E ψ or
we can bring the constant to other side and
then bring the whole thing back to write this
equation as d2ψ/dx2 +(2mE/ ħ2) ψ(x) = 0.
The energy of a moving particle is always
positive and therefore this particular constant
is 2mE/ ħ2 where ħ is h/2π where h is a
Planck's constant. So this equation is, to
represent it in simple term is d2ψ/dx2 +
k2ψ = 0. The particle in the one dimensional
box gives you with this kind of an approximation
gives you the simplest Schrödinger equation
to solve. Even from the simplest Schrödinger
equation some of the consequences that you
draw from the solution are rattling, they
are difficult to recognize as anything familiar.
Therefore, even a simple model like this has
in it the newness, the discreteness of the
energy coming out of it as we see in a few
minutes. The solution of this equation is
a simple mathematical equation. It is ψ(x)
=A cos(kx)+Bsin(kx), using the real functions,
where k is of course I have told you k2 is
2mE/ ħ2. If you do not know how to get this
solution, my suggestion is that you substitute
this solution in the above equation and see
that this ψ(x) satisfies this equation.
dψ/dx ,if you recall when ψ is A cos(kx)+Bsin(kx).
You know what is dψ/dx. It is -k A sin
kx + kB cos kx. Therefore, the derivative
d2 ψ/dx2, the second derivative is - k2A
cos kx -k2B sin kx. Which is nothing other
than - k2 ψ(x). Therefore, you see that
the equation is the same as what we started
with d2ψ/dx2 + k2ψ = 0. Therefore the solution
makes sense, the method of solving how we
get all these things is not necessarily part
of the Chemistry course but you have to learn
Mathematics at the same time to solve simple
differential equations, let me leave that.
Our point is to make sense out of the solutions
as said earlier.
Now, given that the wave function is A cos
kx + B sin kx. And given the model if you
recall the model V is ∞ at the ends of the
box, x = 0 and x = L. And if we are solving
the Schrödinger equation for the particle
inside the box and we say that ψ(x) is the
solution. Then the presence, this is infinite
potential means the wave function goes to
0 at x = 0, the wave function vanishes, goes
to 0 at x = L that is a model. Therefore the
two conditions that we have ψ(x) =0 at x
= 0 and ψ(x) = 0 at x= L. These are the two
conditions that we have to remember. Since
these two conditions refer to the motion of
the particle with respect to a boundary. You
can also call these 
as boundary conditions because the particle
is bounded by these two limits.
Therefore, let us apply these the boundary
conditions to the solution that we have ψ(x)
=A cos(kx)+B sin(kx) is such that it vanishes
at both the boundaries of the box. Let us
take x = 0, ψ(0) should be 0 and that means
A cos(0) + B sin(0) and sin(0) is 0 therefore
this term goes away. cos(0) is 1. Therefore
this is A implies A = 0. One of the two constants
is removed by one of the boundary conditions.
The next condition is ψ(L) = 0 and that implies
now since A is 0 you recall ψ(x) is B sin
kx only. Therefore what is ψ(L)?. ψ(L) is
B sin(kL) and this has to be 0 as required
by the boundary condition. ψ(L) is 0, the
wave function vanishes at the other boundary.
Again it is a very simple choice if you want
to make, now B can be 0, but the point is
if B is 0 and A = 0 then ψ is a trivial solution
is a useless solution. After all this differential
equation is always satisfied by the trivial
requirement that ψ is 0. We are not looking
for trivial solutions. We are looking for
non trivial solutions. Therefore with the
condition that B is not necessarily 0 then
ψ(L) goes to 0 only if sin(kL) goes to 0.
And that is easy to understand because you
what a sin function does. A sin function when
you plot sin of x as you go from x = 0 or
to all the way up you see sin(0) is 0, sin(π/2)
is 1, sin(π) is 0, sin(2π) is 0, sin(3π)
is 0 and so on. Therefore you see that the
condition kL sin(kL) = 0 implies kL = nπ,
where n = 1, 2, 3, 4 etc. Again n = 0 is not
acceptable for the same reason that B = 0
is not acceptable. So kL = nπ is the requirement
that you get by solving the Schrödinger equation
with the boundary conditions that the wave
function vanishes at one end of the boundary
as well as at the other end of the boundary.
So let us write the solution carefully. Therefore
k= nπ/L, k2 = n2π2/L2. Now these are purely
mathematical equations. Now, the 
equation was ψ(x) = B sin (nπ /L)x.
Interesting thing is k2 which we wrote earlier
for 2m E/ħ2 is now n2π2/L2 where n = 1,
2, 3 etc. Values of n other than integers
are not allowed and what is this I told you
the total energy, E = ħ2n2π2/L2 2m. Now
you know that ħ2 is h2/4π2, where ħ is
h/2π. Therefore, the final expression for
the total energy, E = n2 h2/8mL2. So the energy
of the particle is the possible solution for
the equation Ĥψ =Eψ.
Now for this particle in a box model with
the boundary conditions that ψ(x) is a sin
function vanishing at the ends of the boundary.
But the possible energies that the particle
can have now are discrete values, not possible
for any value of n but only n = 1, n = 2,
n = 3 and so on. So the energy solution for
the particle is h2/8mL2 times n2, an integer.
Therefore it is 1, 4, 9, 16, 25 times this
constant h2/8mL2. These are the possible energies
for the particle. Therefore the discreteness
of the energy of the particle appears immediately
as a requirement of the boundary conditions.
The quantization of the energy of the particle,
the quantization of the moving electron in
a conjugated diene, the electron cannot have
any arbitrary energies but if it is bounded
by the requirement that it stays with the
molecule that it does not fly off from the
molecule. That boundedness requires the electrons
to have only specific energies. We are talking
about Chemistry that we are not talking about
the particle in a box only. That is some model
which is applied to the chemical problem should
give you results which are in tandem with
the chemical observations.
Therefore the simple model which gives you
discretization or quantization of the energy
arises, this discretization arises because
of the requirement of the boundary conditions.
Therefore boundary conditions impose quantization.
The second thing, we have solved for ψ(x)
as some constant, we still do not know what
that constant is, we have B sin nπx/L and
energy is given by that. But before I do that
let us just check when we write the energy
as h2/8mL2 times n2. h is Planck's constant,
typically of the order of 10 raised to -- 34
joules second, h is joules sec and you recall
joule is kg (meter)2(second)−2. Therefore
h is kg (meter)2(second)−1. h2 has a dimension
that [kg(meter)2(second)−1]2 , m mass of
the particle in kilograms.
If you talk about atoms if you talk about
electrons the mass is approximately of the
order of 10 raised to -- 30 kilograms. If
you talk about atoms the masses are approximately
in the range of 10 raised to - 27 kilograms
again extremely small. But you see that these
numbers are no longer so small or so big compared
to the height of the Planck's constant. m
has the dimension of kilogram, L length with
the atomic dimensions, all of you know is
of the order of the Angstroms 10 raised to
-- 8cms, i.e. 10 raised to -- 10 meters. this
unit h square by m L square leave the eight
out what is the dimension kilogram square
meter to the power four second to the minus
two by kilogram mass L square is meter square
so get all this things out you a get kilogram
you get a meter square kilogram meter square
per second square . So, if L has the dimension
of meter and h is kg (meter)2 (second)−1
and mass is kilogram. By substituting all
these dimensions, we get the unit as kg (meter)2
(second)−2. Therefore that has the same
dimension as the energy.
This is the unit of energy for the particle
in a box, h2/8mL2 is the unit times n2. The
quantum number, n the discrete integer with
1, 2, 3 as the possible values which we call
as quantum numbers because it represents the
quantization of energy gives you the energy
of the particle inside the box. Then what
about the wave functions? Here, ψn(x) = B
sin (nψx/L). Let us do this for n = 1, ψ1(x)=
B sin 1πx/L meaning that whatever is the
value of n that is the same value. So, ψ1(x)
= B sin πx/L.
It is easy to plot this equation, and it is
a sin graph between the limits 0 and L. It
is a half sin wave vanishing at both ends
and by this we get the value of B. What is
the meaning of ψ? Nobody knows that ψ has
an interpretation. When we solve the Schrödinger
equation we will have a lot of difficulty
in trying to understand the solutions of the
Schrödinger equation but with a clear conscious
mind that ψ does not have an interpretation
and that the solution we have obtained does
not have a meaning. Using ψ but rather the
ψ2, i.e. ψ × ψ and if the wave function
is complex using the absolute ψ2, Max Born
provided the interpretation that ψ2 evaluated
in a small region of space is same as the
probability of finding the system in that
space. What is the meaning ψ?
If the wave that we have right now is a sin
function but if it is a complex for the some
other problem the absolute square of ψψ*
calculated in a small region of space dτ
represents the probability of finding the
system in a small space dτ. This is the interpretation
provided by the Max Born. And if you think
it is hard you must also understand that Erwin
Schrödinger who proposed this equation for
solving the atomic systems dynamics as well
as the stationary energy levels. Schrödinger
himself did not get the role of ψ correctly.
His proposal of what ψ means was in fact
turned down. And several years later it was
Max Born who gave the correct interpretation
which is now accepted universally as the interpretation
for the wave function. The ψ does not have
a meaning but the absolute value of square
has a meaning related to the probability.
Now, we talk about this small space here,
if you recall the particle moves in the box
between x = 0 to L. Therefore for this coordinate
any small region is dx, the length dx is intermediate.
Therefore you calculate the wave function
ψ(x) at that point and then interpret the
absolute square of the wave function. |ψ(x)|2dx
represents the probability of finding the
particle in the small space between x and
x + dx. The meaning of ψ is none, the
square of the absolute value of ψ (since
in our case it is a real function) is ψ2(x)dx
, represents the probability of finding the
particle 
in the region bounded by x and x + dx. It
is ψ2(x).
Therefore we have a wave function ψ which
is a sin function. So we can calculate ψ2
which is nothing but the (sin)2 function and
we plot (sin)2 as a function of x and then
you see that in each small region or a strip
narrow space that the value of ψ2 gives you
the probability of finding the particle in
that region. If ψ(x) is B sin (nπx/L) where
B is a constant then ψ2 is obviously B2 sin2(n
π x/L). So, if you plot that, this is nothing
again another oscillatory function, it is
a sin2 therefore it is between 0 and L which
is probably something like that. Now, x-axis
is x = 0 and x = L. What is on y-axis? The
maximum value on y-axis is B2 now. The probability
of locating the particle in a small region
between x and x+ dx, and for x is equal to
some value here. This small region, ψ2dx
representing the area of the ψ2 graph in
this small space.
Since x is a continuous variable you can represent
the probability in any small region of space
by choosing that value of variable. Therefore
what happens is if we add all these probabilities
that particle here is anywhere in these anyone
of the small regions. Let us add all these
probabilities what should we get? That is
important because you recall that the potential
was taken to be infinite at the ends of the
box to ensure that the particle stays in the
box.
Therefore, if you calculate all these probabilities
in every region if you calculate the total
area of this graph, it is like finding the
particle anywhere between x = 0 and x = L
which is absolute value absolutely certain
value can be one. Therefore, when you add
all these small areas that is equivalent to
∫ψ2(x) dx between the limits x = 0 and
x = L and that should give you an absolute
value of 1, certainty. There is no leakage
of this particle outside through the boundary,
the boundary is too steep. Therefore this
rigid model first of all gives us a picture
of how things are different in the quantum
world as suppose to the classical mechanical
world.
Unfortunately for a chemist all the atoms
and molecules that we talk about are microscopic
in dimensions and they do follow the quantum
principles. You have to excite the atoms and
molecules to a very large value before you
think about applying the classical physical
laws. Spectroscopy for example you have to
study this quantum principle in order to understand
basic standards of spectroscopy. Now, ψ2(x)
dx = 1. This tells you immediately that between
x = 0 to x = L. B2 sin2(n π x/L) should be
equal to 1. This is an elementary integral.
All of you should know how to solve that.
So I leave that an exercise but the answer
is B2 ×L/2 = 1. The value of the integral
is L/2. Therefore what is B? It is equal to
√ 2/L.
So, simple particle in a one dimensional box
model now what you have is these two ideas
namely that the energy is quantized the energy
of the particle if you measure it should be
one of the values that we wrote down h2 n2/8mL2.
The wave function is has the meaning that
the absolute square of the wave function in
a small region gives you the probability of
finding the particle in that region. Let me
summarize that, therefore the solutions of
the particle in a one dimensional box are
solutions for the Hψ = Eψ, ψ(x) = √ 2/L
sin(nπx/L) and E is h2 n2/8mL2 where n = 1,
2, 3 etc. Therefore when you try to solve
this Schrödinger equation Hψ = Eψ, you
did not get one solution. You got an infinitely
many solutions. So n can be any value from
1, 2, 3,..... ∞. Therefore the corresponding
wave functions are also there are infinite
numbers of such wave functions. Therefore
the advantage in solving the Schrödinger
equation is that you will obtain all the solutions
you need to know.
Therefore ∫|ψ(x)|2dx= 1in between the limits
x = 0 to 
x = L is nothing is often known in quantum
mechanics as a normalization condition. The
normalization condition merely refers to the
fact that the total probability of finding
the system with that wave function everywhere
in that system should be unity. That is the
normalization condition adding all the probabilities
and that immediately gives you a value for
the B in terms of the dimensions of the problem
in the 1/√ L, the length.
You can see here the wave functions plotted
for various values. n = 1, already drew this
as a half wave, sin wave. When n = 2 you see
that wave function is a full wave function.
Let us plot the wave functions, sin π x/L
is a half sin wave, sin 2πx/L for n = 2 when
x goes from 0 to L then sin 2πx/L goes from
sin 0 to sin 2π therefore it is a full sin
wave function. Therefore it goes 0 at half
of the point. And the next one is sin 3πx/L
which is a one and half sin wave. So what
you have is a one and half sin wave.
Now that is for the sin wave. What about the
sin2 wave which represents the probability?
Here there is no negative part. The sin square
is looking like that, This is sin square pi
x/L. The sin2 (πx/L) is looking like sin
πx/L. Therefore sin2(2πx/L) now has a node
which is at the middle, no negative part with
two equal halves and sin2(3πx/L) has three
such equal halves, one thirds and sin2(4πx/L)
has four one fourths and so on which is what
you see in this slide.
The above slide shows the sin2 wave, here
the first one is sin2 (πx/L) and second is
sin2(2πx/L) and then sin2(3πx/L) , sin2(4πx/L)
and so on. sin pi x sin 2 pi x/L sin square
pi x sin square 2 pi x/L 3 sin square 3 4
sin square 4 and so on. So this goes on and
on as the n value increases to very large
values. You will see that the graph has practically
lots of oscillations but very tightly placed
that when you try to measure the probability
in any small region for particles with such
large values of n you will get a uniform probability
independent of the region which is the same
as the classical ideas.
That is for large values of n we do not need
to solve the quantum mechanical equations
the classical mechanical equations and the
concepts of classical mechanics makes sense.
This you must have also recognized something
as the Bohr's correspondence principle where
quantum mechanics and classical mechanics
meet. Anyway these are not important for us
right now. We are looking at it from the point
of view of the solutions of the chemical systems.
The summary for the 1D box now is that the
wave function is √2/L sin nπx/L and the
energies are (h2/8mL2) n2. Particle energies
are discrete and particle position inside
the box is given by a probability description.
Let us digress for a minute the probability
idea, let me play this movie for you and explain
what this movie does. Let us assume that we
are able to follow the motion of the particle
by looking at a narrow window inside the box
at any point of time. Assuming that all other
things to be identical if the particle moving
with a constant velocity inside the box as
we would expect if there is no potential energy,
Newton's first law tells you particles in
motion will continue to be in motion, particles
in rest will continue to be in rest, there
is no potential there is no force therefore
the particle will continue to move with its
velocity constant.
Therefore if we try to locate the particle
by looking at any small region, ∆x1 which
is given by this band the probability that
we will find the particle in ∆x1 is given
by this ratio ∆x1 the window divided by
the total length of the window that is the
box length itself. So, P1 = ∆x1/L. This
is the classical picture that we have in terms
of locating the particle in terms of the finding
where the particle is probably to be present.
There is one more movie, and what is important
is that this P1 = ∆x1/L is independent of
where the ∆x1 is and that is what the strip
which is moving around will tell you whether
the ∆x1 is centered at this point or it
is centered here or it is centered here it
does not matter as long as the ∆x1 is same
the with the probability is the same value.
If we increase that band obviously the probability
is now much larger and if we can look at the
whole box we will anyway locate the particle.
But what is important in the classical idea
is, that is independent of the particles velocity,
independent of the location of the particle
inside the box the probabilities are the same
so long as you are looking at the narrow region
of the same length in various elements, this
is the classical idea.
Now, what you get out of the quantum mechanical
picture? What you have here is on this side
the right hand side of your picture where
you see that this graph represents the particles
the ψ2. Therefore this is the probability
density graph this is not the same everywhere
meaning the likelihood of locating the particle
in this small region is different from this
region if the particles energy is E1 namely
h2/8mL2, n is 1. If n is 2 the particles energies
are now different and the probability density
curve is also different. It is not only not
the same in all regions it is also different
for different energies of the particle. So
you see that the quantum mechanical picture
if the electron to be closely to be located
from the hydrogen nucleus the likelihood of
finding the electron in a region close to
the nucleus the likelihood of finding the
same electron in a region farther away from
the nucleus they are not going to be one and
the same if we have to follow this wave function
picture.
Second, even if the electron has the same
energy in one orbit for example the analog
of boron, in different regions of the electrons
domain with that same energy the electron
is likely to be located with different probabilities.
Therefore you see the quantum idea the Schrödinger
equation introduces concepts which are foreign
to us. Quantum mechanics itself is a sort
of a strange idea and the derivations from
that give us really strange results which
we cannot comprehend.
The only way to see that these results are
no longer strange is by knowing more and more
about similar systems by solving more difficult
problems and interpreting our results along
these lines and then finally coming to the
agreement that even if I do not understand
Quantum Mechanics I know how it works. Given
a problem I know how to solve that problem
and I would be able to obtain the solutions
for electron densities in a molecule. Eventually
that is what we want to do as Chemists is
two map out the electron densities of the
electrons in various atoms and molecules and
show where are the electron depleting regions
where are electron enriched regions where
is the bonding is the bonding directly related
to electron density present in that region
or a depletion of electron density means no
bonding, anti-bonding.
But how do we interpret them using the quantum
mechanical ideas? Since we want to get to
that point as quickly as possible we have
to show the basic Mathematics. This is particle
in the box model it gives you two important
concepts namely discretization of the energy
as a function as a result of boundary condition
and that the probability densities being different
for different regions.
If we have to extend this idea of the motion
of the particle in a one dimension to the
motion of the particle in a two dimensions
namely in a plane not along a line but in
a plane if you have to do that then corresponding
expressions for the kinetic energy of the
particle if you recall that --ħ2/2m d2/dx2
represented the momentum (px)2. And in two
dimensions what you have is the momentum of
electron or the particle moving in a plane
is not (px)2 the momentum p2 is (px)2 +(py)2.
You recall the momentum p is a vector with
px as the component along the x direction
and py as the component along the y direction
and the absolute square, p square is nothing
but the p dotted p vector.
So now you have two components of momenta
in two mutually orthogonal directions. Therefore
this term which is unique for a particle in
one dimension has to be modified to be written
such that you have the two dimensional kinetic
energy --ħ2/2m (∂2/∂x2 + ∂2/∂y2).
This is the analog now of the classical mechanical
p square for a two dimensional system. It
is a partial derivative now meaning that whenever
you evaluate the function under this derivative
you keep the other variable constant. This
is the kinetic energy operator for a particle
in a two dimensional box. And obviously the
particles potential energy is now a function
of both the coordinates x and y because the
position of a point in a plane is given by
two coordinates. Therefore the potential that
the particle experiences at that point is
given by its value for both these coordinates.
Therefore V is now a function of x and y.
This V(x ,y) is the potential energy.
Therefore the Schrödinger equation for this
particle is now becomes for the solution Hψ
= Eψ. To written explicitly the equation
is -ħ2/2m (∂2/∂x2 + ∂2/∂y2)ψ(x
,y). The wave function is the function of
both the variables x and y plus V( x, y) times
ψ(x ,y) equals to Hψ = E ψ( x, y). This
is the equation that we have to solve, a motion
in two dimensions. This is for a particle
in a two dimensional box.
Some of you might be worried about the fact
I am drawing a two dimensional box and I call
this as a one dimensional model and therefore
if I have to draw a two dimensional box model
it has to be a cubic what does this mean?
The one dimension here represents the variable
with respect to which we are solving the Schrödinger
equation. We are solving an equation in one
variable that is what we call as the dimension.
This is only a boundary to indicate that the
particle cannot escape the box. Therefore,
in a two dimensional motion likewise what
you need to do is not to write a line but
a box somewhat like that. And perhaps can
just tell you that this is something like
a box. So now you talking about the motion
of the particle in two dimensions referring
to the fact that one of them is the x variable
the other is the y variable the position of
the particle in this plane or any plane is
what is meant by the model problem particle
in a two dimensional box.
So let me summarize today's lecture. We introduced
to the model problem of particle in a one
dimensional box and its relevance to Chemistry
in terms of following the electron energies
in a specific example being that of a conjugated
dyeing system. There are many other similar
instances in Chemistry where this model has
to be solved. But our purpose is having associated
the solution with a corresponding chemical
problem we went through the motion of the
solution of the system and what is the meaning
associated with the wave functions or its
absolute square and the fact that the energy
is a discrete quantity how do we arrive at
that saying that the boundary conditions are
important for the discretization of energy.
These are the things we have to remember as
the consequences of solving the model problem
and the results being different from our classical
perception of the dynamical motion of the
particle in the sense of classical methods.
With these differences in the next lecture
I would continue to illustrate the particle
in a two dimensional box and there will be
one surprise namely the concept of degeneracy
that will arise when we solve the particle
in two dimensional box in addition to everything
that we had done now namely the quantization
of energies, the probability description etc.
But in addition to that we will have what
are called the degeneracy that will arise
and therefore that is important to recognize
as the next important step and till then thank
you very much.
