As I told you this is the polar plot in the
xz plane, and if you wanted to get the three-3
dimensional appearance you have to take this
figure and rotate it about the z axis. , okay.
And further I also told you that in the direction
with theta equal to cos inverse of 1 by root
3, you can actually calculate it; it will
turn out to be 54.73 degrees. You will find
that the value of the function is 0zero. So,
if you marked that direction in this figure,
what you are going to get is z line, okay,
along this line such that this angle is equal
to 54.73. The value of the function is 0,
and similarly along this direction, but why
along those two directions only, you think
of a direction downward like that over a direction
there on the other side, right, such that
this angle, this angle is how much? It is
actually 180 minus 54.7.
Again you will find that cosine of theta will
be equal to 0., okay. So, when you get the
three 3 dimensional appearance by rotating
about the z axis, you think of what will happen
to these lines. These lines actually will
describe a cone something like that, such
that the internal angle; this angle of the
cone is 54.73 degrees, and similarly these
lines will also describe another cone. And
you should realize now that for any point
lying along the surfaces of these cones, the
angular part of the wave function will be
identically equal to 0.
And if the angular part is identically equal
to 0, what will happen? The wave function
itself will be equal to 0, and therefore,
if I think of the total wave function, the
total wave function has two nodal surfaces;
they are surfaces the two nodes, where are
they? One is actually along the surface of
a cone, right, such that the interior angle
this angle within the one is equal to 54.70
degrees, and the vertex of the cone actually
is pointing down. And similarly there is another
cone whose vertex is pointing up along the
surface of that cone also the wave function
will be identically equal to 0, and hence
we have another nodal surface, okay. So, with
this information let me show you a polar plot
of this orbital; this is the appearance which
is obtained by rotating this figure about
the z axis, okay.
So, these two things actually they are describing
something that is a shape like a donut; though
not exactly but you can see the appearance
from the figure. If you want I can rotate
it a little bit. See this is how it would
appear, and if you looked from above roughly
that is how it looks like, from any angle
you can view it. So, this is 3 d is z square
orbital; the orbital is referred to as 3 d
z square, and you can also make a density
plot if you like. So, how will you make the
density plot?
I can do it here. Well the first thing that
I will now do is label the axes; this is let
me say the y axis if you like, and this is
the z axis. Now in these directions actually
the wave function is positive, but then along
to a special direction, these are the special
directions, along them the wave function is
0, and similarly along these directions also
the wave function is zero. So, why do I mark
the directions along which the wave function
is 0, because you see if the function is positive
as soon as I cross this line the function
will change sign. So, on this side what will
happen? The wave function is going to be negative.
This is positive, and then when again I cross
this line what will happen? The function will
again change sign. So, therefore, in this
region the function will be positive, and
then again in this region the function will
be negative. So, that is how the appearance
of the function would look like; if you made
a density plot, this is positive, and that
is negative; this is positive, and that is
negative. So, using the software Mathematical,
I do have a plot of this function a density
plot. Let me just draw it.
Okay, here it is made in the xz plane, but
it does not matter whether it is in the xz
plane or whether it is in the yz plane, the
appearance will be the same. You will see
that in this region the function is positive;
in this region the function is negative, sorry
again positive. In these blue regions anyway
they are represented in these blue regions
you can see actually the value is negative.
That is clear when I point to one of the lines;
you see this is the line along which the value
of the function is negative, and it has a
value of minus 0.3. So, this is the appearance
of the 3 d z square atomic orbital, okay.
Then we have to worry about other atomic orbitals.
I mean I will not be actually writing down
the functions, but perhaps, I can just show
you the polar plots of these other functions,
and that should be enough.
Any orbital that you want to see it is easy
to make a polar plot. This is actually the
orbital that is referred to as a dxy, how
will you get them? Either the procedure is
just the same as that I have outlined.
You have to put other values of m, psi 32
plus or minus 1, but if you put plus or minus
1 the result is going to be complex. So, therefore,
you will take either the cosine function or
the sin function and then proceed, okay, and
the results are actually five orbitals in
total, because you have m equal to 0 plus
or minus 1 plus or minus 2, you have five
orbitals in total. And those five orbitals
actually the conversion is to label them as
3 d xy, 3 d yz, 3 d xz, then 3 d z square
I have described, and there is another orbital
which is denoted as 3 d x square minus y square,
and you can make polar plots of all these
orbitals. So, 3 d xy; xy is the one that we
can see here as you will see in the picture.
See, this is your z direction; you have the
x and y axes, and you can see that these has
4 lobes, right, which are clear in the picture,
and these 4 lobes are pointing in between
the axis. The important thing is to realize
that the orbital is in the xy plane; it is
in the xy plane. It is lying in the xy plane,
and there are 4 lobes, maybe I can draw this
on the board.
Imagine this is the xy plane. So, you have
one lobe here, another one there, another
one here, and the fourth one there, and it
so happens that this and that are plus; in
these directions the wave function is positive,
while in these directions the wave function
is negative, okay. So, that is how it is.
So, 3 d xz what will happen? The appearance
is the same except that it is in the xz plane;
the orbital is lying in the xz plane. It is
very similar to xy except that the everything
is now in the xz plane and of course, this
has a two both of these orbitals has two nodal
planes; where are the nodal planes? Just to
illustrate in the case of this orbital which
I have denoted xy, in this case remember the
z axis is standing perpendicular to the plane
of the board in this fashion. So, if you think
of such a plane which is what plane? It is
the xz plane, or you think of such a plane
which is nothing but the yz plane, then these
are actually nodal planes. Similarly, if you
think of this orbital you would realize that
the xy plane, right, the xy plane is a node,
and the yz plane also is a node for this orbital.
So, then we have seen xy and xz.
We should also see yz; almost the same except
that this is now in the yz plane, okay.
Then we have only one more orbital left, what
is that? It is actually the orbital that is
called x square minus y square dx square minus
y square, this one; how does it appear? You
can see that this is the appearance. The important
thing is to notice that these are not lying
in between the axes, but if you look at it
you would see that these are actually lying
along the x; these are lying along the x and
the y axis, right. Then now draw the dx square
minus y square orbital showing you clearly
the positive and the negative lopes of this
orbital.
So, these are my axes x and y, and the orbital
that I am drawing is d x square minus y square;
it will have a positive lobe like this and
another one in exactly opposite direction,
and it also has two negative lobes. One is
like that, and the other one is like this,
and let me put the signs of these orbitals,
signs of these lobes; there is a negative
sign here, negative sign there. These are
negative regions for the wave function, and
you have a positive lobe here, another positive
lobe there. And the important thing is to
notice that if you performed the operation
of inversion the orbital would remain unchanged,
and people normally say that this is an orbital
that is symmetric under the inversion operation.
So, that is what, that is what regarding these
orbitals. So, how many of them? In total you
have five of them corresponding to m equal
to 0 plus or minus 1 plus or minus 2. Then
when you think of n equal to 4 again I am
not going to describe n equal to 4 in all
its details.
When n is equal to 4 you would have the 4
s atomic orbital; I am not going to describe
it. You will have 4 p whereas atomic orbital
there is only one 4 p, there are 3 of them;
they will be 4 p x for p y and 4 p z, okay.
Then we will have l equal to 2 you will have
4 d; that will be five of them. Their appearance
everything is similar to the to the 3 d orbitals
that we have just now described, why? Because
the angular part of the wave function is just
the same, but then when you have 4 f this
actually will correspond to l equal to 3.
When l is equal to 3 you have the 4 f orbitals;
there are how many of them? There are actually
seven of them, why? Because m can take on
the values 0 plus or minus 1 plus or minus
2 and plus or minus 3. So, there are seven
atomic orbitals; all of them have precisely
the same energy.
In fact, in the case of the hydrogen atom
all these orbitals are precisely the same
energy, and if you counted them how many are
there? 7 f orbitals, 5 d orbitals, 3 p orbitals
and 1 s orbital; so, so the total number is
sixteen which is actually equal to n square
where n is the principle quantum number, n
in this case is 4. So, if you have n particular
value then there are n square orbitals having
precisely the same energy. Then what about
the 4 f orbitals; what about their shapes?
Well, the shapes are actually given in many
a text books; I can use Mathematica and show
you the shapes of these seven f orbitals.
I will show you the shape of one particular
set of the seven f orbitals; I mean if you
look into a book the text book like the book
by Huheey on inorganic chemistry you will
find that he gives you two different sets
of f orbitals.
So, people have I mean two sets of f orbitals,
one set of seven and another set of seven.
These are actually not independent of each
other, because one can be expressed in terms
of the other. So, what I am going to show
you is referred to as the cubic sets. So,
the cubic set actually consists of these seven
orbitals which I will list first, and then
show you how their shapes are? They are actually
written as f z cube, f x cube, f y cube. Then
the next orbitals are f x into y square minus
z square f y into z square minus f square
and f z into x square minus y square, three
of them and then finally, f xyz. So, these
are the seven orbitals that I will show to
you, but before I show it to you that actually
you may ask, why do you use these symbols?
For example, in the case of d xy why do I
write the orbital as d xy, right? That is
a natural question that one would have. The
answer is actually very simple. What you should
do is you should take x x x is given by this
expression, and y is given by that expression
you multiply the two.
So, what will be x y, x into y? It is actually
going to be equal to r square sin theta or
sin square theta actually cos phi sin phi.
That is the result, and you just look at the
angular dependence of this product x y. The
angular dependence is sin square theta cos
phi sin phi and that precisely is the y function.
Remember any orbital would have a y part,
and this is just the y part of this orbital.
So, when you write d xy actually you are implying
that you take the expression for x, you take
the expression for y, multiply the two, and
remove the radial dependence; you will get
the spherical harmonic corresponding to that
function.
So, when you see a symbol like d f xyz it
is actually very simple if you wanted to find
out the angular dependence of this function;
all that you need to do is multiply these
things, right, and then throw out the r dependence,
what is left is the angle dependence of the
function. Now this is actually useful for
this one, it is useful for this one, but with
this it is a little bit more complex. See
what I have told you does not give you that;
it is not applicable for d z square also,
it is not applicable for d z square, but it
is applicable for other orbitals, for all
the other orbitals this is what I have told
you is applicable. So, let me now look at
the f orbitals.
First I will plot f z cube; I have to execute
the command. So, you will see that this is
f z cube. It is similar to your d z square,
but there are also differences. What are the
differences? You can say well, this lobe will
be positive if you like this is going to be;
what about this one? If this is positive this
I would assume is negative and so somewhere
in between is a cone on the surface of which
the wave function will be 0. Then if this
is negative the next one will be positive;
again in between there will be a surface on
which the function will be 0, and that in
this case is very simple. It is nothing but
the x y plane, right, and then if this is
positive, what will happen? This other thing
will be, well, let me repeat plus this is
plus, this is minus, that is plus, and the
last one will be negative, correct, and so
how many nodal surfaces?
Well, the answer is one here, another one
in between these two, and third one between
those two; so, so therefore, three nodal surfaces.
So, this is the appearance of the f z cube.
Notice that it is oriented along the z direction,
right. So, if I put x cube there all that
will happen is that the appearance is going
to be similar except for the fact that it
is oriented along the x direction, and that
is what you are going to see. This is oriented
along the x direction.
Then you can have f y cube. This would be
oriented along the y direction.
Then let us look at f xyz, interesting shape.
Well, it is actually an interesting question
to ask how many nodal planes are there. It
has three nodal planes; where are they? You
can see if you think of the x z plane, x z
plane it is a nodal plane. If you think of
the y z plane again you would realize that
it is a nodal plane, and similarly the x y
plane also is a nodal plane. So, that is how
this orbital is; I mean if you are not convinced
actually I can rotate it in any direction
I like say for example, here I am looking
down along the z axis, right; you are just
looking down along the z axis. So, there are
actually eight lobes for this orbital, and
four lobes are sitting on top of the other
four lobes, and therefore, you can see only
four lobes essentially, and you will see that
along this direction, right, remember red
color actually denotes the x axis; blue color
is the z axis.
, okay.
So, this is actually the blue color along
with this x direction is actually the x z
plane, and you can see that the function vanishes.
Similarly this is the y axis the green color.
So, that is the y axis, and again the z and
the y axes together they form a nodal plane.
So, similarly I can rotate it in any direction,
and you would realize that there are three
nodal planes. So, let us now look at another
orbital which may be what we want to call
it let us say f x y square minus z square.
Oh Mmaybe what I will do is instead of this
I will first plot another one.
See if we look at this figure this is your
f xyz which we have looked at and understood,
and this is the other orbital which I am plotting
now; what is it? You have f z, the orbital
is f z x square minus y square; that is the
orbital that is being plotted, and if you
looked at the two orbitals what you should
realize is that the appearances are roughly
the same. In fact, if you inspect it closely
you would realize that this second orbital
can be obtained from the first if you like
how? By rotating this orbital about the z
axis you just take this orbital rotate it
about the z axis by 45 degrees it might need
some time to realize that, but all that you
need to do is take this orbital, rotate it
by 45 degrees about the z axis, you are going
to get this orbit, okay. That is how this
was.
Therefore, the shapes everything is the same
except that one is obtained from the other
by rotating by 45 degrees about the z axis.
Similarly if you took this orbital and rotated
it about the y axis by 45 degrees, you will
get this orbital. If you rotate about the
y axis by 45 degrees you will get this orbital;
while if you rotated it about the x axis by
45 degrees you are going to get this orbit,
okay. I mean if you want to see I can obviously
show them.
Again I will show just one of them f x; that
means I am rotating by the x axis. So, that
is how these orbitals are. So, there are seven
of them. So, you have seen once I have told
you what is referred to as the cubic set of
f orbitals according to the to the book by
Huheey; the book is inorganic chemistry by
Huheey, okay. Let me summarize what I have
told you regarding the hydrogen atom. We have
completely solved the Schrodinger equation
for the hydrogen atom, and we found that there
were three quantum numbers arising in the
process of solution., okay.
There were three quantum arising in the process
of solution. They are actually the quantum
numbers that are referred to as the n, l and
m; n is referred to as the principle quantum
number. It can take values like n is equal
to 1, 2, 3, 4, etcetera, right. The other
quantum number is l; it is referred to as
the Azimuthal quantum number. It can take
the values 0, 1, 2, 3, etcetera up to n minus
1. This condition actually arose automatically
when we try to solve the equation, and the
third quantum number is m referred to as the
magnetic quantum number. It can take the values
0 plus or minus 1, plus or minus 2, etcetera
and to the maximum value actually will be
plus or minus l, the last possibilities either
plus l or minus l.
So, these actually arose out of our solution,
but the energy depended only up on the principle
quantum number. It does not depend up on the
values of l and m; it simply depends only
up on the value of n. And this is actually
the case with the hydrogen atom, okay, and
then if I represented the allowed energy levels
n is equal to 1 has the lowest possible energy
that is the ground state. This will give me
the 1 s atomic orbital, n is equal to 2 there
are four allowed orbitals corresponding to
l being 0 and l being 1; this is n is equal
to 2, four orbitals. The next one is n is
equal to 3; you can have l equal to 0 1or
2, and correspondingly how many orbitals do
you have? In total you have nine orbitals.
So, I should represent that by drawing nine
lines very close to each other that is difficult.
So, I will not do it, but there are nine orbitals;
what are they? They are actually the 3 s 3
p and 3 d orbitals. In the case n is equal
to 2 you will have 2 s and 2 p orbitals, and
when n is equal to 4 you have 4 s 4 p 4 d
and 4 f, sixteen orbitals, right; they are
all degenerates. So, typically I mean if I
do not represent the degeneracy the way I
would represent this let me not worry about
the degeneracy because that is difficult to
represent. This is n is equal to 1, this is
n is equal to 1; that is n is equal to 2,
then n equal to 3, may be I am not drawing
it correctly, because the energy separation
has to be realistic.
So, let me draw one more figure. This is n
is equal to 1, then I have n equal to 2. This
is one fold degenerates which means it is
non-degenerate. This is four fold degenerate;
then I have n equal to 3 nine fold degenerate,
n equal to 4 sixteen fold degenerate, then
n is equal to 5, 6, 7, 8. As you go up the
separation between the energy levels gets
closer and closer and closer. In the limit
n tends to infinity the separation becomes
0. So, therefore, somewhere here is energy
0, right; here the energy is 0. We found only
these solutions. Remember in our analysis
we said that the total energy of the system
is negative, but then you may ask what will
happen if the total energy is positive, okay?
So, the answer is actually extremely simple.
If we have actually taken a situation where
we have the proton and if the electron is
at infinite distance there is no interaction
between the two and the energy of the system
is 0, okay. Now how can I have an energy which
is greater than 0? The answer is that the
electron has to move with some energy. Then
it will have nonzero energy, right. It has
to move from infinity suppose I just push
it then it will move. It will have some kinetic
energy and the kinetic energy of the electron
suppose the electron is at infinity I just
give it some kinetic energy, it will come
and it will be sit.
So, to say or it will collide, or it will
get scattered by the proton and after the
scattering what will happen? It will go away,
and the amount of kinetic energy that I can
give is decided by me; I can give it any kinetic
energy and therefore, I can solve the Schrodinger
equation with the electron having some kinetic
energy starting with; it starts at infinity
with some kinetic energy comes near the proton,
and it goes away, and as I have told you the
kinetic energy that of the electron initially
is determined by me. So, I can choose it to
have any energy and therefore, above this
0 actually you can give the electron any energy,
and the energy levels are not quantized. It
is possible to have any energy starting from
0 to infinity. These are what are referred
to as the scattering states in which the electron
is simply scattered from the proton.
It comes hits the proton; it spends some time
in the vicinity of the proton and goes away.
So, this is how the allowed energy levels
of the hydrogen atom are, okay, and again
I should also remind you that we have just
spoken of the bound states and the scattering
states. But in addition to this you have the
hydrogen and the proton. Suppose they form
the bound state which is what is referred
to as the hydrogen atom then the hydrogen
atom can execute translational motion as a
whole, right. That is something that we have
not worried too much, but that also is there
just to remind you. But then of course, you
see you have already studied something about
the nature of the hydrogen atom, and you would
know the three quantum numbers are not enough
to describe the state of the hydrogen atom.
I mean in the case of hydrogen atom three
quantum numbers are not enough, you actually
need four quantum numbers, and this fourth
quantum number arises from the idea of spin.
Something that I will describe later, but
before I do that I want to do some general
things, because we now have enough background
to do a few general things. Now you see we
have solved the Schrodinger equation for which
system. So, particle in a box we have solved
the Schrodinger equation for the harmonic
oscillator. We now have solved the Schrodinger
equation for the hydrogen atom. In all the
cases if you remember it was possible to normalize
the wave function; wherever we had the solution
we can normalize it; that was not difficult.
Then further in the case of the particle in
a one dimensional box I did demonstrate you
that any two wave functions, right. They are
orthogonal to one another, and I demonstrated
that for the harmonic oscillator also; for
the hydrogen atom I have not demonstrated
it, but it is possible to show that you give
me a 1 s atomic orbital, this 1 s atomic orbital
will always be orthogonal to the 2 s, it will
be orthogonal to the 2 p, right. So, therefore,
if you give me any two wave functions which
are solutions of the Schrodinger equation.
Then I take the first one which I may denote
as may be psi 1; I will take the second one
I may denote it as psi 2. For example, this
may be the 1 s atomic orbital, this may be
the 2 s, or it may be the 2 p; it does not
matter. Take another atomic orbital; multiply
the two together and by the volume element
d tau integrate over the entire space. The
answer is going to be 0, and we say that these
two functions are orthogonal. We have already
seen this. We have also in our postulates
we have said that any operator that occurs
in quantum mechanics has to be a Hermitian
operator. We made that postulates; I mean
in the postulates this was actually a part
of the postulate.
So, what is a Hermitian operator? The Hermitian
operator is an operator let me say A for which
what I will do is I will allow this to operate
up on A an acceptable wave function like psi
which I denote as psi; multiply it by the
complex conjugative of another acceptable
wave function which I denote as phi, and then
integrate over the entire space. Because I
have done integration about the entire space
the answer is going to be a number; the number
may be complex, okay. Now suppose instead
of doing this I take A, allow it to operate
on phi instead of operating on psi and then
multiply it by the complex conjugate of psi
and integrate over the entire space.
, okay.
If it so happens I mean or if it is so that
this number and that number are complex conjugates
of one another for any acceptable psi and
phi, then I define A as a Hermitian operator.
This was our definition of a Hermitian operator,
okay, and what are the properties of Hermitian
operators that are of interest to us? The
most important thing is that their Eigen values
are always real. Now I am going to prove that
if I have a Hermitian operator then any given
Eigen value of that Hermitian operator has
to be a real number; that is what I am going
to proof now, right. I mean I could have done
the proof earlier, but I thought it is best
to do it at this point, because you do have
some familiarity with orthogonality, normalization
and related things.
So, it will not appear as abstract as it would
have appeared earlier if I had done it. So,
let me say that I have an Eigen function;
well, maybe I mean this operator A would have
several I mean in principle an infinite number
of Eigen functions. So, I need some notation.
Let me say that psi 1, psi 2, etcetera; this
is a least it goes to infinity. These are
Eigen functions of the operator A. What it
means is that if A operate up on psi 1 the
answer is going to be psi 1 itself multiplied
by the corresponding Eigen value which I will
denote by the symbol small a 1, and if it
operate up on psi 2 the answer is going to
be a 2 times psi 2, or I can generalize and
say okay, if a operated up on psi i the i
th Eigen function, the answer is going to
be a subscript i multiplied by psi I, right.
This is how the Eigen value equation would
look like and what is it that I want to prove?
I want to prove that all a i’s are real
numbers. They cannot be complex; that is what
I want to prove. So, how will I prove it?
The answer is actually extremely simple, okay.
Well, what I am going to do is you see in
this equation that I have this is guaranteed
if the operator A is Hermitian this is guaranteed.
So, in this equation what I am going to put
is this psi I am going to put it as psi subscript
I; this psi I am going to put it as psi subscript
I, because what do I want to have? I want
to have A operating up on psi i, right, and
this phi also I am going to identify with
psi i. suppose I did that, what will happen?
This psi is going to be identified with psi
i then I will have A operating on it, and
then here I will have psi i star, right, and
then I would have the volume element d tau,
and I have to integrate over the entire space.
And the fact that A is the Hermitian operator
implies that I can do the same thing here
on the right hand side. So, instead of phi
what would I have? I would have A operating
up on psi I, because phi is identified with
psi i and this psi is identified with psi
i itself.
So, I would have psi I star, and then the
volume element d tau integrate over the entire
space, and then I will have the star, right.
So, all that I am doing is this is satisfied
by a Hermitian operator for arbitrarily acceptable
psi and phi. If I just say that these are
both this psi and phi is both the same Eigen
function of that operator that Eigen function
being psi I, right. So, if that is the way
it is what will happen to the left hand side?
A operating up on psi I is going to give you
the corresponding Eigen value which is small
a i which is just a number multiplied by psi
i. Then I have to multiply by psi I star d
tau integral, right; that is the left hand
side, and what will happen to the right hand
side? There is this star. This A operating
up on psi i is going to do exactly the same
thing.
I am going to get a I, then psi i, then psi
i star, volume element d tau integrated over
the entire space, but now you look at this?
This is just a number. So, therefore, you
see here I have integration with respect to
possession coordinates and so on. But a number
does not depend up on possession coordinate.
So, I can just move it out, and therefore,
what will happen? I will have a i coming outside,
then integral d tau psi i star psi I, right,
and if your wave function is if your Eigen
function is normalized then this will actually
be equal to unity. At the moment I will not
say it is normalized, because I do not need
it, okay, and what will happen to the right
hand side? See this a i is a number. So, I
can just move it out of the integral sign;
I will have it here, but then I have to take
the complex conjugate.
So, what will happen? Then I will have a i
star, and what do I have left? I would have
left integral d tau psi i star psi I the whole
thing complex conjugated, right, and this
complex conjugation I can take it inside if
I like. So, if we took it inside, what will
happen? I will have psi I star star, double
star; double star is equivalent to doing no
starring. So, therefore, this star and that
star I can remove, but when I take the star
inside it is going to get here. So, that is
what happens, and then when you look at this,
then would realize that this integral and
that integral they are the same, right. If
they are the same I can just remove them,
and so what is the result? I get the result
that a i is equal to a i star. What does that
mean? Any given Eigen value is equal to its
own complex conjugate, and this can be valid
only if a i is a real number, right. If a
i is a complex number this will not be satisfied.
So, therefore, any given Eigen value a i has
to be real. So, that is a very general result,
and this is the reason why in postulate 
number two we said that the operator has to
be Hermitian, because the Eigen value is guaranteed
to be a real, right. In postulate number three
we talked about measurements, and we said
that if I make a measurement the answer has
to be an Eigen value of the corresponding
operator, right, and we know that when we
make measurements, the answer is that we get
our real numbers, right, and therefore, it
is necessary that the Eigen values should
be real, and for a Hermitian operator the
Eigen values are guaranteed to be real. So,
this is a general result; I will prove one
more general result later.
Thank you for listening.
