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Welcome back to the ah lecture on chemistry
and the introduction to molecular spectroscopy
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so the lecture will continue on the quantum
mechanics that i introduced a little bit in
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the previous lecture the summary of quantum
results that we need to use ah we started
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with the description on the hydrogen atom
and wrote down the schrodinger equation as
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all of this representing the total energy
of the electron acting on the wave functions
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psi e which is a function of the three coordinates
y e z e so this is the h psi is equal to e
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psi e x e y e z e ok
now hydrogen atom due to the specific property
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of the potential energy between the electron
and the nucleus being spherically symmetric
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that is it has same value on a sphere of radius
or on any part on the surface of the sphere
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ah the standard method for solving the hydrogen
atom problem is to use the spherical polar
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coordinate system so one writes the wave function
in terms of the three polar coordinates r
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theta phi and obtains a solutions n
now remember this n when we complete this
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equation e n psi n r theta phi n stands for
a set of three quantum numbers namely a principle
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quantum number n and assume it quantum number
n l l and ah magnetic quantum number m three
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quantum numbers are there therefore the wave
function for the hydrogen atom is given by
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psi n l m r theta phi and the eigen value
for this for the electron of course it is
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only on the principle quantum number therefore
we have degeneracies for the wave function
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psi n l m r theta and phi the standard practice
is to solve the schrodinger equation using
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method of separation of variables and the
energies are obtained in the process as the
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familiar ah expression that neils bohr also
showed earlier using the rydberg constant
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or h is the rydberg constant and in terms
of numbers experimentally as well as ah theoretically
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the number is one not nine six hundred and
seventy seven centimeter inverse please remember
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this is h c nu bar if you recall the first
lecture its ah r h is a wave number and the
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energy is inversely proportional to the square
of the quantum number particle in a box the
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energies proportional to m square harmonic
oscillator the energy is proportional to n
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the hydrogen atom energy proportional to one
by n square so quantum number dependences
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are different for different models ok
the wave functions themselves or ah written
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as a product of a radial and n angular orbital
ok anyway first e n then n is one the energy
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corresponds to minus h c rydberg constant
and thats ah in the electron volt familiar
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electron volt it is minus thirteen point six
electron volts for n equal to one obviously
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for n equal to two its is minus thirteen point
six divided by four and for n equal to three
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it is minus thirteen point six divided by
nine etcetera until n goes to infinity this
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energy level is zero so if you look at these
spectrum of the hydrogen atom in an energy
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scale if you do that and if you write this
as the zero and this as the minus thirteen
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point six then the next energy this is for
n equal to one the next energy is one fourth
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of this so roughly somewhere this is about
half so roughly here n equal to two sorry
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it is one fourth therefore ah its somewhere
here n equal to two and the next one is one
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ninth of this so somewhere here n equal to
three and you can see that the energy levels
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become very very dense as you go long and
the state of zero is for infinitely large
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quantum numbers and when the electron is completely
free from the nucleus
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so this is the wave function this is the energy
level picture and what about the wave function
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picture the wave function picture leads to
the familiar description in terms of the s
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p d f type orbitals in chemistry so far we
have not needed the g orbitals we have not
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discovered elements with sufficiently large
atomic number which warrants g orbitals and
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the wave functions psi n l m r theta phi is
expressed in terms of a radial function which
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depends on the quantum number l and n and
is a function of radial coordinate or and
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an angular function which is a function of
the quantum numbers l and m and it is dependent
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on the angles theta and phi and these are
polar coordinates ok you recall that x e is
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r sin theta cos phi on the spherical system
y e is r sin theta sin phi and z e is r cos
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theta so these are the theta phi or theta
phi dependences of the coordinates therefore
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the wave function is expressed using that
and you see that the radial part you see that
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these are radial part and then there is a
angular part
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the radial part r l n of r is multiplied is
expressed using laguerre polynomials and an
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exponential times an exponential of minus
ah z r by n where n is the principle quantum
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number and the angular part is described by
spherical harmonics
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y l m theta phi therefore psi n l m is the
product of the three product of the two r
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times y ok what are the values for n the mathematics
tells us that the n has the allowed values
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one two three all the way to infinity l has
the allowed value zero one two upto n minus
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one for any given n and the m has the values
zero plus minus one plus minus two until plus
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minus l so there are two l plus one m values
there are n l values and n of course can go
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from one two three to infinity
so the typical wave function picture that
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comes out of this is when n l m and psi if
we write and the energies and the labels we
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have then n is one l is zero m is zero this
is r one zero of r y zero zero of theta phi
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and the energy is minus thirteen point six
e v this is the one s orbital and then n is
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two you have l is equal to zero and l equal
to one for l equal to zero m is equal to zero
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is the only allowed value r two zero of r
y zero zero theta phi this is the l m and
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this is minus thirteen point six by four e
v this is the two s orbital it is actually
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angled independent zero zero is square root
of one by square root of four pi thats a number
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so its independent of theta phi
so all those things which are angled independent
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so theta phi independent they are called the
s orbitals when l is one m can be one zero
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or minus one and these are obviously written
as r two one y one one r two one y one zero
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r two one y one minus one and these are usually
called the two p orbitals both the two s and
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two p orbitals of the same energy because
the energy depends only on the principle quantum
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number and therefore you have four orbitals
for n equal to two and likewise when you do
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n equal to three you got zero one and two
zero is same as what is called the three s
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orbital one is what is called the three p
orbital and then l is two you have got five
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quantum numbers namely two one zero minus
one minus two for the m and there are five
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wave functions all of which both the zero
one and i mean all the three nine functions
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together will have the same energy minus thirteen
point six by nine e v and this five functions
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are called the three d orbitals and the labels
of three d orbitals come from the specific
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features that you are familiar with this harmonics
the probability distributions for the hydrogen
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atom is the last thing that i would like to
write before we move on to the angular momentum
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the probability issues are angular distribution
is due to the y l m theta phi star y l m theta
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phi multiplied by sin theta d theta d phi
this is like the psi star psi d x in the case
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of angles you have sin theta d theta d phi
come in as the angular element and in the
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case of radial probability the rest of it
namely r n l of r star r n l of r r square
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d r remember that in the cartesian space the
coordinate ah x e y e and z e for the electron
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can have values going from minus infinity
to plus infinity for all three of them x e
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ye and z e therefore this is the volume element
that you must have d tau for the ah probability
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distribution and normalizing the wave function
when you write psi n l m r theta phi psi star
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n l m and psi n l m r theta phi the corresponding
thing in the radial the in the spherical polar
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coordinate system not volume element d tau
is given by d r r square and then you have
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d theta sin theta and d phi hm so this is
obviously broken into a radial part a radial
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part and an angular part star and an angular
part and you see that the integrally separable
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its a three poled triple integral and the
triple integral is easily separated into an
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angular distribution and radial distribution
so these are some of the ah basic elements
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that you need to know in the model problems
of ah quantum mechanics which will keep coming
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again and again in the ah spectroscopy and
every other ah area of theoretical chemistry
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now the last thing that i want to do for this
particular course relevant ah and something
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that needs to be recapped is angular momentum
in fact we have already touched upon angular
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momentum and we discussed this spherical harmonics
i mean i did not tell you so but we will see
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that in a minute ok the rotational kinetic
energy of a system moving around a point is
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given by the square of the angular momentum
divided by its moment of inertia and l angular
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momentum has the same dimension as h bar has
the same dimension as h mass length square
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t to the minus one classically the angular
momentum is the position vector r of the particle
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times p the momentum cross p r cross p is
the vector product and ah l is ah directed
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l is the access for the l is perpendicular
to the plain containing r n p is a vector
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cross product
therefore that is a classical definition and
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in quantum mechanics when you replace the
momentum by the corresponding derivative operator
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l becomes minus i h bar y dou by dou z minus
z dou by dou y x unit vector plus z dou by
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dou x minus x dou by dou z y unit vector plus
x dou by dou y minus y dou by dou x z unit
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vector and l is a vector so what you have
is essentially l x x component of the angular
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momentum the y component l y on the angular
momentum times y the z component of the angular
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momentum times z and the square of the angular
momentum l square is l dotted with itself
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and thats given by the quantity l x square
plus l y square plus l z square the components
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of the angular momentum l x r given by minus
i h bar times that l y by minus i h bar times
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that and l z with the minus i h bar times
that ok so these are operator representation
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for the angular momentum in quantum mechanics
and they have this property namely l x does
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not commute with l y and the commutation of
l x and l y it is given by i h bar l z and
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likewise for cyclic ah permutations namely
l y l z is i h bar l x l z l x is i h bar
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l y ok the square of the angular momentum
operator on these ah if you take them and
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any of these components if you take the square
of the angular momentum operator l square
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and its commutation with l x the answer is
zero l square commutes with all the three
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components n l square l y is zero l square
l z is zero and in the spherical polar coordinates
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in spherical polar coordinates the square
of the angular momentum is given by minus
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h bar square one by sin theta dou by dou theta
acting on sin theta dou by dou theta plus
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one by sin square theta dou square by dou
phi square and this is used in the solution
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of the hydrogen atom when you transform the
hydrogen atom kinetic energy you would see
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that this term comes with a minus h bar square
by two m with one term preceding that namely
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one by r square dou by dou r r square dou
by dou r and then you have one by r square
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times everything the minus h bar square is
outside therefore you have one by r square
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sin theta dou by dou theta multiplied by this
derivative and then you have one by r square
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sin square theta multiplied by that therefore
this comes straight from the spherical polar
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coordinate ah representation of the angular
momentum and the operator for the angular
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momentum l square
therefore immediately can be identified to
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have this y l m as the eigen value l square
on y l m theta phi will give you h bar square
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l into plus one y l m theta phi l m theta
phi the operator l z if you again go through
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the algebra of the coordinate representation
and transformation will come out with this
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form minus i h bar dou by dou phi the same
as the phi here and this has l z on y l m
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theta phi the eigen value m h bar y l m theta
phi and ah y l m is simultaneously eigen function
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of both l square and l z because the operators
commute l z and l square commute therefore
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mathematics tells us that its possible to
have the same eigen function for both of them
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but with of course different eigen values
namely m h bar for l z and h bar square l
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into l plus one for the operator l square
and you remember y l m of course has limitations
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in the sense for any l there are two l plus
one y l m s starting from y l l to y l minus
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l therefore there are two l plus one such
functions eigen functions for any angular
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momentum quantum number given by l l is an
integer
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this is something that you are already familiar
with but if you recall the particle in a ring
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which is a one dimensional problem only phi
is the coordinate and the kinetic energy is
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given by minus h bar square by two i d square
by d phi square i is the moment of inertia
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of the particle moving on a ring moment of
inertia and
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the equation that one needs to solve for the
particle in a ring is h psi is equal to e
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psi this is the h therefore the differential
equation that you need to solve is d square
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d phi square plus two i e by h bar square
psi of phi equal to zero and note that the
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dimension of i e by h bar square dimension
if you look at it i is the moment of inertia
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so it is mass time to length square e is the
energy which has the dimension mass in length
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square into time minus two and h bar square
h has the dimension of mass length square
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time to the minus one and its square therefore
i e by h r square is dimensionless because
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you see everything is cancels here m square
cancels with this m square l square l square
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cancels with l four and so does t to the minus
two therefore its a dimensionless quantum
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number dimensionless quantity of course phi
is a angle its dimensionless therefore you
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can see any derivative involving phi should
only be added to another dimensionless quantity
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so you need some dimensions are extremely
important therefore what you see is that the
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solution to this equation turns out to be
d square by d phi square plus a quantum number
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m whose square is included and then you have
a psi m phi is equal to zero and m has the
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values zero plus minus one plus minus two
and so on and you can immediately see that
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the energy which is given by the expression
two i e by h bar square is equal to m square
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gives you energy is equal to h bar square
m square by two i ok this is for one dimension
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for three dimensions or more than one dimensions
you already have that the this one for three
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dimensions you already have this results the
angular momentum square gives you the eigen
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values l into l plus one and you see that
h bar square l into l plus one instead of
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h bar square m square therefore if you have
to do the same problem in three dimension
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from one to three you see that e l is given
by h bar square l into l plus one by two i
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for the specific router for overall square
of the angular momentum and this is what we
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will see as an important quantity later when
we study the diatomic molecular ah spectroscopy
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rotational spectroscopy and also when we study
the rotational spectroscopy of a spherical
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top molecule you will see that the eigen values
for the rotational energies will be a quantum
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number l into l plus one and the units are
h bar square by two i I is the moment of inertia
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for the molecule in that case and therefore
the relation that we have here should be recalled
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when we do the microwave spectroscopy and
also when we do row vibrational spectroscopy
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later ok
the eigen functions for rotational angular
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momentum are obviously given in the case of
a one dimension its easy to write that the
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eigen functions psi m r e to the plus or minus
i m phi that would be the solution for this
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equation namely d square by d phi square of
psi m plus m square psi m is equal to zero
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that would be the solution exponential plus
or minus i m phi and a general solution psi
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for a particle in a one dimensional motion
can be a linear combination of both namely
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a m e to the i m phi plus b minus m e to the
minus e m phi ok just to indicate that there
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are two different constants and there is a
linear combination of both the acceptable
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solutions and this is a degenerate solution
because the energy depends on the square of
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m so it does not matter whether m is plus
or minus the corresponding exponential i m
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phi will be the solution and both of which
are different wave functions
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so for a particle ah on a ring you have degeneracies
for every m other than m equal to zero so
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these are important relations for understanding
the role of angular momentum in rotations
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and also the coupling of rotations with molecular
vibrations and so on and therefore please
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keep this in mind ah i think i have come to
the summary of whatever that minimally quantum
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mechanics we needed before we move on to understanding
ah elementary microwave and infrared spectroscopy
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therefore anything else that we need i would
ah provide that that point of time we will
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ah continue this with another important result
in electronic spectroscopy for measurement
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quantitatively using what is known as the
beer lamberts law and then we move on to the
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first most important topic for spectroscopy
the born oppenheimer approximation this week
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we will complete both of these and then we
move on to the lectures on the molecular ah
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spectroscopic details for rotations and vibrations
until then
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thank you very much