Welcome back to the lectures in chemistry,
on the topic of Atomic Structure and Chemical
Bonding. My name is Mangala Sunder and I am
a professor of chemistry in the department
of chemistry in Indian institute of technology
Madras and you can see the email addresses
here for you to contact ok . So, in this lecture,
which we will continue from the previous matrix
introduction and also properties of matrixes.
This lecture is, on one of the important aspects
in quantum mechanics namely; natrix eigenvalue,
eigenvector problem. The term eigen means
characteristic. It is a property of the matrix
Eigenvalue eigenvector problem is fundamental
in quantum mechanics in the sense that, the
hamiltonian operator acting on the wave function
giving you the wave function back with a constant,
which is the energy the total energy. This
is an eigenvalue equation for energy. Ok.
Therefore, most of the solutions of quantum
mechanics of real problems, at some point
of time or other will definitely involve solving
them using computers and numerical methods
and for that process the matrix eigenvalue
problem is one of the best known numerical
techniques. For to understand the basic principlesfor
this course and then use them to analyze the
quantum mechanical properties and the measurements
etcetera; the concept of eigenvalues and eigenvectors
is important ok.
Now, before we do this, let me justrecall
with an elementaryequation a system of linear
equations .
Let me write down two simple equations 4x
plus 5y is equal to 0 and we have homogeneous
equations. And then let me write another equation
2x plus 3y is equal to 0. Let us assume that
x and y are such that, that these two equations
have to be solved. You will see that the first
equation gives you 4x is equal to minus 5y
and therefore, x is equal to minus 5 by 4y
.
Now, if I substitute that x here, then what
I get minus 5 by 2 x5 by 2y because, x is
minus 5 by 4 and 2 x is minus 5 by 2 y plus
3 y is equal to 0, which is essentially half
y is equal to 0 or y is equal to 0. If y is
0 quite obviously, x is 0 . So, this is a
trivial. Result it is not a solution. The
reason that this system of equation does not
have a non 0 or a non trivial solution is
that, these two are both independent of each
other they are linearly independent . If you
have two equations in two variables and both
are linearly independent; the only solution
that you have is that variables of 0. Therefore,
it is important for these equations to be
linearly dependent.
And I am sure you have done this in yourprevious
algebra, that the determinant of the coefficients,
which is written from this equation as 4,
5, 2, y, 2, 3. We write this in the matrix
form, this is what the equation is. The determinant
of the coefficient matrix is the determinant
4, 5, 2, 3 is nonzero. It is 12 minus 10,
which is 2 is equal to 2. hm Therefore, for
a homogeneous system ofcoupled equations the
determinant if it is not 0 such a system does
not have a solution a non trivial solution
ok and the trivial solutions are not relevant.
So, let me do the same thing now. If I look
at another the same equation slightly different;
let us assume that it is 2x plus 3y is equal
to 0 and let us assume that 6x plus 9y is
equal to 0 ok. Now it is obvious by looking
at it straight away that, the second equation
is three times the first equation 2x plus
3y is equal to 0. Therefore, these equations
are not linearly independent they are linearly
dependent. And you know if, you write the
determinant you write a determinant then it
is 2 3,6,9, that is equal to 0 because, you
know one row is nothing but the multiple of
another row and determinants are such determinants
of 0 values.
The moment you have this this is actually
one equation, would not lead to x plus 3y
is equal to 0. Therefore, you have a solution
x is equal to minus 3 by 2y. All it tells
you is that x is in terms of y, but it does
not tell you what or x and what x and y are.
Therefore, one variable is undetermined or
indeterminate, you have to give a some some
number and then you have to use x; if y is
5, x will be minus 15 by 2 , but there is
a solution. The solution is that unless y
is 0, x is not 0. So, this is a linear system
of equations ok.
Now homogeneous equations; if you have a a
system of inhomogeneous or non homogeneous
equations, you do not have a problem you have
to actually have the inverse of the matrix
to be taken non homogeneous system .
So, let us do that 2x plus 3y is equal to
5 and 4x plusit cannot be 2x plus 3y and 4x
plus 6y. Let us write 5y is equal to 4 ok.
Now, these have a solution in the sense that
this is a matrix equation 2, 3, 4, 5 x, y,
is equal to 5, 4 and therefore, if you call
this as A and you call this as phi is equal
to some constant, then you only have to multiply
this with an A inverse A phi is equal to C.
Therefore, it is important for you to have
an A inverse here there is also A inverse
ok.
So, A inverse is defined only if the determinant
is nonzero. Therefore, you see this system
of equations cannot have the determinant to
be 0, if it is non homogeneous. So, these
things are elementary algebraic equations
and things you must remember. And now when
we go to theeigenvalue eigenvector problem,
we are generalizing this to a much larger
and a specific type ofsystems.
and the specific type of system is that; if
you have a matrix A and if you have a column
something like x y hm A being A 2 by 2 matrix.
So, let us write to this in terms ofah specific
forms ok. a 11, a 12, a 21 a 22 .
The eigenvalue eigenvector problem is a specific
set of equations which is, a constant lambda
times x and y . So, what you have is if, you
write to this as A and if you write this x
as a column vector, x y as a column vector
given by the operator by the vector x and
you write this as lambda x then, this is called
an eigenvalue equation. And the eigenvalue
equation is such that, x is known as the eigenvector
and lambda is known as the eigenvalue ok.
So, what you have is you are trying to find
out your solution x and y such that, A acting
on that solution gives you a constant time
that, solution itself that is a very special
case and those special cases are most important
in quantum mechanics, including the hamiltonian
equation.
So, let us do in this case itself. Let us
try and find out what the eigenvalues are
ok. Now please remember if you expand this
a 11 x plus a 12 y is equal to lambda x. First
row you remember two matrices are equal only
if, the elements on both sides are corresponding
elements are equal. So, the second one is
a 21 x plus a 22 y is equal to lambda y. And
therefore, you have a 11 minus lambda, x plus
a 12 y is equal to 0, a21 plus a22 minus lambda
this is x and this is y is equal to 0. Now
we are back to the homogeneous equation and
therefore, you see the reason for the previousanalysis
is that, we used that right away that we have
to have the determinant of the coefficient
matrix to be 0 .
Which is a 11 minus lambda, a 12 a 22 a 21
and a 22 minus lambda, this determinant should
be 0, for this to have a non trivial solution.
x and y, that is both x and y are not 0, that
is what is non trivial. Now this is very easy
the determinant is easy to expand.
So, you have a 11 minus lambda times a 22
minus lambda minus a 12 a 21 is equal to 0,
which is now, a quadratic equation in lambda.
So, this is lambda square minus lambda times
a 11 plus a 22. Thenplus you have a 11 a 22
ok, minus a 12 a 21 that is equal to 0. This
is thestandard form of a x square plus b x
plus c. Therefore, the solution you know lambda
can be written as minus b where the coefficient
is this one. Therefore, it is a 11 plus a
22 by 2 minus b by 2 plus or minus b square
minus 4 a c if you remember c is a is 1 and
c is this.
You remember a x square plus b x plus c is
equal to 0. x is minus b by 2 a plus or minus
square root of b square minus 4ac by 2a n
ok. Keep that in mind, then you have the solution.
Therefore, this is plus or minus square root
of b square is a 11 plus a 22 whole square
minus 4 a 11 a 22 plus 4 a 12 a 21. These
are the two solutions to the plus or minus.
Immediately you see that if, for example,
we started with the matrix a 11 with a 12
being 0 and a 21 being 0, if we started with
a diagonal matrix you knew that the diagonal
elements or the eigenvalues.
Because if these two are 0, you can see that
this goes to 0. This goes to 0 and therefore,
this one is a 11 minus a 22 whole square and
the square root of this is going to give you
a 11 plus a 22 1 by 2 plus or minus a 11 minus
a 22 1 by 2 and the two solutions are a 11
and a 22. If these elements were 0 you have
you are back to the original solution, but
if these elements are not 0 which is what
we are interested in therefore, let me undo
this highlight ok. 4 a 12 a 21, then the actual
solution is what I have written down .
So, there are two solutions and therefore,
there are two eigenvalues namely, lambda 1
is a 11 plus a 22 by 2 plus square root of
a 11 minus a 22 whole square, plus 4 a 12
a 21, 1 by 2 ok. And the other solution is
lambda 2 is a 11 plus a 22 by 2 minus one
half square root of the same thing, a 11 minus
a 22 whole square plus 4 a 12 a 21 ok. So,
these are the two eigenvalues .
So, if we go back to the original matrix equation,
we had a 11, a 12, a 21, a 22, x y is equal
to 0. Now, sorry is equal to now lambda 1
x y and there is also another solution a 11,
a 12, a 21, a 22. Now we may want to call
it as x prime, y prime, lambda 2 x prime y
prime since, lambda 1 and lambda 2 are not
the same ok.
Yet we will worry about what is known as a
non degenerate eigenvalue system that is,
the eigenvalues are all different they are
notequal to each other or few of them equal
to each other. All those cases we will not
worry about it. Let us assume that the eigenvalues
are different .
Therefore we have two solutions. x y is 1
eigenvector, associated with eigenvalue lambda
1. x prime y prime is another eigenvector,
associated with another eigenvalue lambda
2 lambda 2 . So, a 2 by 2 system has two eigen
values. In general this is true that an n
by m system has n possible eigenvalues. Some
of which may be equal to each other the others
and it may also have a completely non degenerate
solution and so on. But this is what, is the
eigenvalue eigenvector problem.
Now having done this, what are the eigenvectors?
How do we get x and y? How do we get them,
x prime and y prime? Ok. .
So, quite obviously, it says that a 11 minus
lambda 1 x plus a 12 y is equal to 0. That
is the first eigenvalue equation. And then
the second row equation will be a 21 x plus
a 22 minus lambda 1 y is equal to 0. It is
not necessary for us to verify that the determinant
is 0 because, that is how we got the eigenvalues
in the first place ok lambda 1 because, the
determinant is what we had it a 11 minus lambda
1 a 12 a 21 a 22 minus lambda 1 lambda 1.
This determinant is anyway 0 because, lambda
1 is a solution of this equation therefore,
we do not need to worry about it. Therefore,
if you have to write it, you can write x is
equal to minus a 12 by a 11 minus lambda one
times y. The second solution is not independent
of the first one. You do not need to solve
the second equation. This is all you need
to solve.
Therefore we know x in terms of y and now
you have to choose a y. Eigenvectors are always
or most of the time in linear algebra and
also in quantum mechanics or defined in such
a way that if, you have an eigenvector x we
use what are called normalized eigenvectors.
The normalization if, x and y are real, the
normalization essentially means x square plus
y square is chosen to be 1.
Therefore, no matter what I choose x or y
to be, I have to set to the square of x and
the square of y the sum to be one and therefore,
that constant goes away, when we when you
do that choice. We will see some numerical
examples in the next session and you see this
requirement that the eigenfunctions be normalized,
removes the uncertainty or the indeterminacy
of these variables; one of the variables x
or y. It makes it very clear that x and y
are proportional and they also satisfy the
second condition that x square plus y square
is equal to 1. Therefore, x and y have very
specific values therefore, the eigenvector
is precisely defined.
And it is defined as two different eigenvectors
for two different eigenvalues. This is what
I wanted to do this is a very simple system
2 by 2. And let us stop here and in the next
part of it we will see some examples of doing
this for the very elementary matrices, that
I havealready introduced to you, the pauli
spinangular momentummatrices; the sigma x
sigma y and sigma z. And let us look at the
eigenvalue properties as well as the eigenvectors
and learn a little bit more and also generalize
this to other higher dimensions ok until then
.
Thank you we will stop here for a break.
