Welcome to the ninth lecture on basic quantum
mechanics. Today we will be discussing some
general issues regarding the Eigen values
and Eigen functions, for the one dimensional
Schrodinger equation. In my last lecture we
had solved the problem of an electron or a
proton confined in a potential well of infinite
depth. Essentially, we had solved the Schrodinger
equation for a particular form of the potential.
And as I had mentioned earlier a major part
of quantum mechanics, non relativistic quantum
mechanics is essentially obtaining the solution
of the one dimensional and the three dimensional
Schrodinger equation for different forms of
the potential.
So, as we had discussed in our earlier lectures
that the one dimensional Schrodinger equation
is given by I H cross delta psi by delta t
is equal to H psi psi as a function of x and
time and H psi of x t is equal to H cross
square by 2 mu delta 2 psi by delta x square
plus v of x times psi. This is known as the
one dimensional Schrodinger equation for the
free particle for the particle in a potential
energy function v of x. We assume that the
potential energy function is independent of
time, depends only on the space coordinates.
And then as we had discussed earlier we can
write down the solution using the method of
separation of variables, which is given by
psi of x and t of t. We have done this before
and we found that the variables indeed separated
out and the time dependent part was given
by E to the power of minus I E t by H cross,
where E is a constant. So, if we indeed substitute
this we would obtain the following equation
that minus H cross square by 2 mu d 2 psi
by d x square because now the small psi depends
only on the x coordinate plus v of x.
Psi of x is equal to E of x E of psi. So,
this equation is actually an Eigen value equation
because we can write this equation in the
following form. That H psi is equal to E psi
and where H is the Hamiltonian operator which
is the kinetic energy p square by 2 mu plus
v of x and if I substitute for p is equal
to minus H cross delta by delta x. So, we
will get minus H cross square by 2 mu delta
2 by delta x square plus v of x. This operator
representation for the total energy this is
known as the Hamiltonian this is known as
the Hamiltonian of the system. So, my Schrodinger
equation is usually written as is written
the time independent Schrodinger equation
is written as an Eigen value equation.
So, if I substitute this here. So, as was
obtained a few minutes back minus H cross
square by 2 mu since psi depends only on the
x coordinate. I can replace the partial differential
operator by a total differential operator
d 2 psi by d x square plus v of x psi of x
is equal to E times psi of x. As I had mentioned
earlier this is an Eigen value equation. This
is a even this equation is an Eigen value
equation. We will find that for a given potential
energy distribution only certain discrete
or continuum values of E are allowed those
are the Eigen values of the problem and corresponding
to each Eigen value. There is a wave function
those are known as the Eigen functions on
the system.
So, we have we can write this equation rewrite
this equation just a simple ordering d 2 psi
by d x square plus 2 mu, where mu is the mass
of the particle E minus v of x. In most textbooks
the equation is write at the written either
in this form or in this particular form. Now,
in my last lecture I had assumed v of x to
correspond to a particle in a one dimensional
box. And we had assumed the particle to be
confined between x is equal to 0 and x is
equal to a. Let us suppose and we had found
that the energy Eigen values are given by
E n, which was equal to n square pi square
H cross square by 2 mu a square.
This is equal to n square E 1, where E 1 is
the energy level corresponding to n equal
to 1. So, that is the ground state energy
value 2 pi square H cross square by 2 mu a
square. And the corresponding wave function
was given by psi n of x was equal to under
root of 2 by a sin n by x by a where n takes
the values 1 2 3 etcetera. Today we will discuss
some general properties of the Eigen values
and Eigen functions for a given potential
energy function v of x for example, we will
show that all Eigen values must necessarily
be real. Secondly, we will show that Eigen
functions corresponding to different Eigen
values are necessarily orthogonal. So, let
us first prove that the Eigen values are always
real.
So, we write the Eigen value equation. So,
d 2 psi by d x square. Let us suppose for
the not state in my previous problem. You
remember that we had explicitly determined
the energy Eigen values.So, this is my ground
state, this is the first excited state, this
is second excited state, this is the third
excited state n equal to 1, n equal to 2,
n equal to 3, n equal to 4 etcetera. So, we
assume we do not take a specific form of v
of x for a general v of x. We assume that
that psi n of x is an Eigen function belonging
to the Eigen value 2 mu by H cross square
E n minus v of x psi n of x is equal to 0
I now write the complex conjugate of this
equation.
So, if I take the complex conjugate. So, we
will have d 2 psi n star by d x square mass
and H cross are necessarily real quantities.
So, you will have 2 mu by H cross square and
let us suppose E n can take complex values
E n star, but the potential energy function
is a real function of x. So, the complex conjugate
of that is the same as v star of x is equal
to v of x multiplied by psi n star of x. So,
the second equation number 1 is the Schrodinger
equation corresponding to the not Eigen state
equation number 2 is the complex conjugate
of equation 1. Where we have assumed that
the potential energy function is a real function.
What I do is I multiply the first equation
and usually the convention is you multiply
on the left by psi m psi sorry psi n star.
And the second equation on the right by psi
n and then subtract. So, the left hand side.
So, the left hand side the first term will
be if you see this carefully psi n star d
2 psi n by d x square minus psi n d 2 psi
n star by d x square plus 2 mu by H cross
square. Please see this here also it is psi
n star psi n here also it is psi n star psi
n. So, if we subtract this from this the v
will cancel out. We will have E n minus E
n star multiplied by psi n star psi n equal
to 0 this term. If you write carefully then
it would be a total differential d d x of
please see this psi n star d psi n by d x
minus psi n d psi star by d x.
Straight forward differentiation of this equation
will give the first term will be d psi n star
d x d psi n by d x. And then the second term
will be psi n star d 2 psi n by d x square
and the if you differentiate these two terms,
then it would be minus d psi n by d x d psi
n star by d x. And then this term will come
minus psi n d 2 psi n star by d x square.
So, obviously, these two terms will cancel
out sorry I am sorry this two terms will cancel
out .This term will remain I am sorry. So,
these two terms will be psi n star d 2 psi
n by d x square minus this. So, this quantity
is rigorously equal to this. So, what I do
is I integrate this plus I integrate this
plus I integrate this. So, I will obtain I
will if I if I integrate a total differential
of a quantity multiplied by d x of course.
So, we will get psi n star of x d psi n by
d x minus psi n d psi n star by d x this plus
2 mu by H cross square. I can take this outside
and then integral mod psi n square d x. So,
this is equal to 0. So, from minus infinity
to plus infinity. So, this I take the limits
from minus infinity to plus infinity for any
practical problem. When you have a localized
particle the wave function and it is derivative
must go to 0 at the boundary these are the
boundary conditions of the problem. So, this
quantity at both the limits will be 0. So,
this will be 0.
So, we will obtain this quantity as 0. So,
we will have 2 mu by H cross square which
is just a number, which can be removed E n
minus E n star integral minus infinity plus
infinity mod psi n star d x is equal to 0.
Now, this quantity is positive definite because
the this is the square of a wave function.
In fact, the wave function we always assume
to be normalized. So, that this quantity is
always equal to 1 we can always assume to
be equal to 1.So, therefore, and this is a
constant. So, you must have E n star must
be equal to E n that is all Eigen values of
the Schrodinger equation must necessarily
be real.
All Eigen values all this is an important
result all Eigen values we will come back
to it when we discuss the deducts bra and
ket algebra, but, all Eigen values of H the
Hamiltonian are real and as we will discuss
later that if you make a measurement of energy
then you will obtain one of the Eigen values
of H. So, these are the possible values that
we will measure if we make a precise measurement
of energy.
So, we have proved that the energy Eigen values
are real what we will next prove is that that
if E n, if there are two Eigen values E n
and E m and if these two are not equal then
the corresponding wave functions psi n and
psi m. So, we will have psi m star x psi n
of x d x taken between minus infinity to plus
infinity. This will be equal to 0, when this
condition is satisfied we say that the two
functions are orthogonal. So, this is known
as the orthogonality condition orthogonality
condition.
Now, the proof is very simple we write Schrodinger
equation for the n th state d 2 psi n by d
x square plus 2 mu by H cross square mu is
the mass of the particle E n minus v of x
psi n of x is equal to 0 then I write for
the mth state. So, d 2 psi m by d x square
plus 2 mu by H cross square E m minus v of
x psi m of x is equal to 0. What we do is
we take the complex conjugate of the second
equation. So, I take the complex conjugate
of the second equation. Of course mu and H
cross are real quantities numbers E m we have
proved just now to be real.
So, E m star is equal to E m. So, we do not
have to do anything there and then we take
the complex conjugate here. There is nothing
here sorry. So, because E m is a real number
and then what we do is do the same trick.
As we had done before multiply this by psi
m star on the left the whole equation and
multiply this equation by psi n on the left
right. So, if you subtract if you multiply
the second equation by psi n of x and then
subtract then as we had obtained in my earlier
slide. We will obtain psi m star d 2 psi n
by d x square minus psi n d 2 psi m star by
d x square plus 2 mu by H cross square please
see this psi m star psi n psi m star psi n.
So, v and v are the same. So, these two terms
will cancel out. So, I will be left with E
n minus E m psi m star psi n equal to 0. So,
what we do next is as we had done in the previous
slide I write this as a total differential
d by d x psi m star d psi n by d x minus psi
n d psi m star by d x 
let me leave some space here and the same
quantity 2 mu by H cross square E n minus
E m psi m star psi m I now multiply by d x
and integrate from minus infinity to plus
infinity.
I now integrate from minus infinity to plus
infinity d x this is equal to 0. I had done
this in my last slide that if you expand this
out there will be four terms two of them will
cancel out and the remaining two terms will
be this. So, this will be psi m star d 2 psi
n by d x square minus psi n d 2 psi m star
by d x square and then there will be a terms
like d psi m star by d x d psi n by d and
d psi n by d x these two terms will cancel
out. So, this is the differential of a of
a quantity. So, that the integral is just
this quantity and.
So, therefore, the left hand side after carrying
out the integration will be just psi m star
psi n prime that is the differential of psi
n prime psi n minus psi n psi m star prime.
Take it between minus infinity to plus infinity.
Now, this is 0 because the wave function vanishes
at infinity for to be for them to be square
integrable and normalisable they must vanish
at infinity. So, by square integrable I mean
any function psi n of x is square integral
means mod psi n square d x from minus infinity
to plus infinity must be finite.
And I will choose a multiplicative constant.
So, that this is equal to one. So, a function,
which satisfies this equation or this actually
the equation. That this quantity the intergral
should be finite is known as the square integrable
function and for a square integrable function.
The function itself must vanish at infinity
otherwise it is it will not be a square integrable
function.So, therefore, the wave function
and its derivative must vanish at infinity
and. So, therefore, this term will be 0 this
will be equal to 0.
So, we will have only. So, if this term is
0. So, I can cancel this out I will get this
is equal to zero. So, I will have product
of two terms product of two terms if I write
this E n minus E m multiplied by this, but,
I have initially assumed that E n is not equal
to E m, therefore, this integral must be 0
and this integral minus infinity to plus infinity
psi m star of x psi n of x d x must be 0 when
E n is not equal to E m. This condition as
I had mentioned earlier is known as the orthogonality
condition orthogonality condition.
I am assuming that the wave functions are
normalized therefore, minus infinity to plus
infinity mod psi n square d x is equal to
one. So, this is known as the normalization
condition and I can combine both of them to
write down this equation integral from minus
infinity to plus infinity psi m star of x
psi n of x d x is equal to delta m n. Where
this term delta m n is known as the kronecker
delta function 
and this is equal to 0 if m is not equal to
n and is equal to 1 if m is equal to n.
So, we have derived an extremely important
result. That the wave functions belonging
to different Eigen values wave functions belonging
to different Eigen values are necessarily
orthogonal and this is the orthogonality condition.
There is one more thing that I would like
to mention that we will use this that the
that the wave function that the Eigen functions
of the Hamiltonian form a complete set of
functions; that means.
If psi n of x that is suppose the in a particular
problem we have a set of wave functions, which
are the Eigen functions of the H of the operator
H. Then any arbitrarily function say phi of
x any arbitrarily well behaved. Well behaved
means it has to be single value it should
not be infinite anywhere and it should be
a square integrable function. So, that is
the meaning of the word well behaved. That
it should be single valued function at a particular
value of x phi of x must have an unique value.
It must not have any infinities it must not
go to infinity in at any point and it is a
square integrable function that is phi of
x d x integral from minus infinity to plus
infinity must be less than infinity.This symbol
means this inequality means that this integral
is finite. So, such a function is known as
a well behaved function and I state without
proof that the Eigen functions form a complete
set of function that is arbitrary function
can be. I will give you examples of that c
n psi n of x.
So, this is known as the completeness condition
that any arbitrary function can be represented
as a sum as a linear combination of the Eigen
functions of H. So, how do I determine c n
what we do is I multiply this equation I multiply
this equation by psi m star of x d x and then
integrate. So, what we will have is that the
left hand side becomes psi m star of x phi
of x d x is equal to summation the sum is
over all values of n the c n psi m star of
x psi n d x integral from minus infinity all
limits are from minus infinity to plus infinity.
But this I have just now proved to be equal
to delta m n. So, therefore, this equation
becomes right hand side becomes c n delta
m n sum the summation is over all values of
n. So, only the n equal to m term will survive
because for all other terms this kronecker
delta function is 0. As I had mentioned that
delta m n is equal to 0 for m not equal to
n and is equal to 1 if m is equal to n. So,
therefore, the summation is our n. So, when
n takes the value m that term survives.
So, this will be equal to c m. So, this is
how we will determine the coefficient as I
have indicated in my earlier lectures also.
So, c m is equal to or I can write down c
n, c n is equal to integral. And this limits
are also from minus infinity to plus infinity
psi m star of x phi of x d x sorry n this
should be n.Now, what I do is next step is
I substitute for c n from here to here, but
you must be careful because this is a definite
integral over x. This x should not get confused
with this x, but since this is a definite
integral I can quietly put a prime here.
And then this x and this x can be differentiated.
I hope it is clear because this is a function
of x and c n is a constant and c n is a definite
integral. And in a definite integral it does
not matter ,whether you put write x or y or
z. For example, E to the power of minus x
square d x from minus infinity to plus infinity
is the same as E to the power of minus y square
d y. It does not really matter what you what
is in both of them are equal to square root
of pi.So, this x should not get confused with
this x and. So, therefore, I quietly put a
prime and then I substitute this here. So,
I will substitute this expression for c of
n in this equation.
So, I will obtain please see this carefully
that phi of x phi of x is equal to summed
over n and c n and what is the value of c
n as we had determined earlier minus infinity
to plus infinity psi n star x prime phi of
x prime d x prime multiplied by psi n x.What
I will do is I will first carry out the summation
and then the integration. So, that all these
quantities, which are dependent on n I take
this. So, I get is equal to integral from
minus infinity to plus infinity please see
this carefully phi of x prime phi of x prime
and then some function of x, x prime d x prime.
And as you would have noticed that f of x
prime f of x, x prime is the summation of
psi n star x prime multiplied by psi n of
x summed over n.
If you would recall the earlier lectures 1
of the earlier lectures in, which we had derived
the we in, which we had defined the Dirac
delta function and that was phi of x will
be equal to minus of infinity to plus infinity
phi of x prime delta of x minus x prime d
x prime. So, it picks up the value at x only
no matter what the function phi mean phi is.
So, therefore, this function has to be 0 for
all values of x other than x prime and. So,
therefore, this quantity must be the Dirac
delta function this condition is often known
as the completeness condition. The that is
the wave functions the Eigen functions of
the Hamiltonian operator form a complete set
of orthonormal function.
And we say this through the following equation
that summation psi n star of x prime psi n
of x summed over n is equal to delta of x
minus x prime.So, we have derived two very
important relations this is known as the completeness
condition. And the other is other is that
psi m star of x psi n of x d x is equal to
delta m n. This is known as the orthonormality
condition it combines the orthogonality condition
and the normalization condition. So, this
is known as the orthonormality condition.
So, these limits are also from minus infinity
to plus infinity. So, now, in the example
that we had discussed in my last lecture lecture
8.
We consider a particle in a one dimensional
box 
and the domain that we consider was between
0 and a. And he we had found that the energy
Eigen values are E is equal to E n square
of E 1, where E 1 is equal to pi square H
cross square by 2 mu a square these are the
Eigen values and as you can see all Eigen
values are real.We had also derived that the
wave functions are given by under root of
normalized wave functions sin of n pi x by
a. And if you use this you can immediately
show that psi m star of x psi n of x if m
and n are different then it will be 0.So,
this is equal to delta m n. So, we have E
1 E 2 E 3 E 4 E 5 and so on there are infinite
number of states.
Infinite number of discrete states and any
function, any well behaved function in the
region from 0 to a can be represented by can
be approximated and this is my Fourier series
that c n psi n of x.So, this is something
like the Fourier series that you must have
learnt in your college. So, the wave functions
also form a complete set of functions and.
So, therefore, psi n star x prime multiplied
by psi n of x summed over n equal to 1 2 3
infinity is equal to delta of x minus x prime.
So, this was the particle in a box problem
that we had done yesterday. Let me consider
another problem about 4 5 3 4 lectures back
I had discussed the free particle problem
now in the free particle problem let me do
it on an another sheet v of x is 0 everywhere.
So, it is a free particle and the Hamiltonian
operator, which is equal to minus H crosses
p square by 2 m H cross square by 2 mu d 2
by d x square plus v of x, but, v of x is
0. So, we had solved this equation and we
had found that if I write H psi is equal to
E psi then we will get d 2 psi by d x square
plus 2 mu E by H cross square psi is equal
to 0. So, if I write H psi is equal to E psi
then simple manipulations will give this where
E is now the Eigen value where it is a number.
So, we write this as p square this quantity
as p square and the Eigen functions are psi
of x becomes psi p of x this is equal to E
to the power of I by H cross p times x.Now,
I put because I know this I put a factor 2
pi H cross. So, what are the values of p E
has to be positive E has to be positive because
I have put because if p if E becomes negative
as you can as you can see p square is equal
to 2 mu e. So, if p E becomes negative then
p will become imaginary if p will become imaginary
then times I will be something like either
plus or minus kappa x.
When this happens then the wave function either
goes to infinity at plus infinity or minus
infinity E to the power of plus kappa x will
blow up at x is equal to infinity as x tends
to infinity E to the power of minus kappa
x will blow up as x tends to minus infinity.
So, that is not possible because the if the
wave function becomes infinity it is no more
square integrable you can no more normalize
the wave function.So, therefore, E has to
be positive, but, p can be plus or minus.
So, for a given value of E there are 2 values
of p 1 plus 1 minus we say that there is a
2 fold degeneracy.
So, if you work this out then I can write
down that psi p prime of x will be equal to
1 by 2 pi H cross E to the power of I by H
cross p prime x. Please see this if I take
a star here make the complex conjugate then
this will be minus here and I can write down
psi p prime star x psi p x d x minus infinity
to plus infinity I multiply this. So, I get
1 over 2 pi H cross E to the power of I by
H cross p minus p prime time has x d x from
minus infinity to plus infinity. So, this
is the Dirac delta function delta of p minus
p prime. This is an example, where the Eigen
functions form a continuum. In the previous
example we had a discrete set of energies
we had a discrete set of energies only certain
values of energy were allowed in this case
we have all possible values of energy from
0 to infinity.
So, we say that the energy Eigen values form
a continuum energy Eigen values 
are 0 less than E can take all values between
0 and infinity. So, they form a continuum
and when this happens the orthogonality. Orthonormality
condition is represented by this equation
psi p prime star of x psi p of x d x the kronecker
delta symbol is replaced by the Dirac delta
function.Also these functions of course, these
they are they form a complete set of functions.
So, that any arbitrary function phi of x can
be written as a of p superposition times psi
p of x psi p of x d p and. So, therefore,
these limits are also from minus infinity
to plus infinity. So, what is psi p of x psi
p of x is 1 over root 2 pi H cross integral
a of p E to the power of I by H cross p x
d p.
So, this is the Fourier transform. So, it
is a superposition of the moment of these
wave functions and. So, therefore, a of p
is given by the inverse Fourier transform
which we had discussed in quite a bit of length
this will be phi of x E to the power of minus
I by H cross p x d x all limits are from minus.
So, for any well behaved function phi of x
I can always make this expansion because I
can always find a of p by carrying out the
inverse Fourier transform now I can substitute
for a of p in this equation, but, I have to
be careful once again this is a definite integral
over x and this x should not be confused with
this x. So, I must quietly once again put
a prime here and I when I substitute that
I will obtain if you see this carefully that.
That if I substitute a of p from here there
then you will get phi of x is equal to 1 over
twp pi H cross phi of x prime and then there
will be another function x comma x prime d
x prime where f of x prime comma x prime is
equal to I will take the 1 over 2 pi H cross
here 1 over 2 pi H cross integral E to the
power of I by H cross p x minus x prime d
p.
This we know that this is the delta function.
So, as soon as I substitute it here I get
psi of x. So, this is my completeness condition
is no more a sum it is an integral. So, psi
this condition that I represent this is the
completeness condition integral E to the power
of I by H cross p d p from minus infinity
to plus infinity is equal to delta of x minus
x prime this is the continuum Eigen function
of the of this is the completeness condition
for the Eigen functions of the operator H.
I would like to mention one more thing and
that is let me consider the momentum operator
p op the momentum operator. We had said that
the momentum operator can be represented by
minus I write it as a total differential and
let us try to find out the Eigen functions
of the momentum operator. So, I write this
as op of psi is equal to p psi this is an
Eigen value equation for the momentum operator
where p is an Eigen value which is just a
number. So, we had the Eigen value equation
H psi is equal to E psi this is the Hamiltonian
which is an operator and you find that only
certain values of E are allowed these are
the Eigen values.
So, you have the momentum operator p op and
this is a I want to solve this Eigen value
equation.So, I will have minus I H cross d
psi by d x is equal to p psi where p once
again is a number.So, I multiply by I by H
cross. So, the left had side becomes one.
So, I get 1 over psi d psi by d x is equal
to I by H cross times p.If you integrate this.
So, you get log psi is equal to I by H cross
p times x plus a constant.
So, psi becomes the constant I will choose
as one over under root of 2 pi H cross E to
the power of I by H cross p x. So, these functions
psi of x I write this as psi p of x and I
have put the factor 2 pi H cross E to the
power of I by H cross p x, where p can take
any value from plus infinity to minus infinity
to plus infinity. These are known as the normalized
momentum Eigen functions these are known as
the normalized momentum Eigen functions these
are the Eigen functions normalized Eigen functions
of the momentum operator. Actually these are
simultaneous Eigen functions not only of the
momentum operator, but H psi is equal to E
psi for the free particle.
So, therefore, we say that psi p of x is equal
to 1 over root 2 pi H cross E to the power
of I by H cross p x p going from plus minus
infinity to plus infinity they are simultaneous
Eigen functions, simultaneous Eigen functions
of p op, which is equal to minus I H cross
d by d x and also and also of the Hamiltonian.
For the free particle Hamiltonian for the
free particle is H cross square by 2 mu d
2 by d x square actually the Hamiltonian has
a v of x term. But v of x is 0 this is for
a free particle and these wave functions these
wave functions are therefore, Eigen functions
of p op as well as of this. And these wave
functions are often known as the momentum
Eigen functions momentum Eigen functions.
And they form a orthonormal set that is psi
p prime star x psi p x d x is equal to the
Dirac delta function p minus p prime and they
also all limits are from minus infinity to
plus infinity. And they also form and orthonormal
the completeness psi p of x psi star p of
x p prime psi p of x d x this is equal to
delta of x minus x prime sorry psi p’s x
prime here and this is the orthonoarmality
condition. This is the completeness condition
let me rewrite this again this is not I have
not written this carefully.
So, this is integral psi p star of x prime
time has psi p of x d p is equal to delta
of x minus x prime this is the completeness
condition. So, this completes the analysis
for the free particle problem as well as for
the particle in a box. In our next lecture
we will discuss first the solutions of the
linear harmonic oscillatory problem and then
we will derive those solutions thank you.
