So, we were looking at the properties of Hermitian
operators. We saw one property; the property
is that any Eigen value of a Hermitian operator
has to be a real number. Now we want to look
at one more property. This is regarding orthogonality
of the Eigen functions.
So, we imagine that we have this operator
A. It has the Eigen functions which we denote
as psi 1, psi 2, psi 3, etcetera, and the
corresponding Eigen values are a 1, a 2, a
3, etcetera, and what we will prove is that
any one of these Eigen functions belonging
to different Eigen values will be orthogonal.
This is what we will prove? And the way the
proof goes is like this, because A is Hermitian
I know that A operating up on any acceptable
function psi multiplied by phi star integrated
over the whole space must be equal to A operating
up on phi multiplied by psi star volume element
d tau integrated over the entire space star.
So, these two numbers must be complex conjugates
of each other, for any two acceptable state
functions psi and phi. So, now what we will
do is, we will imagine that the psi is one
of these Eigen functions or let us say it
is the nth Eigen function having an Eigen
value which is A n, correct, and phi I will
choose it to be the m th Eigen function, okay,
psi m having an Eigen value which we will
denote as m, and further I will also assume
that a m and a n are different. So, they do
not have the same Eigen value, but they have
different Eigen values. So, a m is not equal
to a n. So, let me evaluate the left hand
side of this equation. The left hand side
is going to be integral d tau psi m star.
I am just evaluating this part A operating
upon psi n; that is the left hand side this
part, and if you look at this expression,
what is going to happen? A is operating up
on psi n; psi n is an Eigen function of this
operator.
So, what is going to happen? The result will
be psi n multiplied by the Eigen value A n.
So, therefore, this will become integral d
tau psi m star a n psi n, but a n is just
a number. So, it is not necessary for me to
keep it here. I can take it from there and
move it to this place. So, therefore, this
will become integral a n multiplied by psi
m star psi n, okay. So, that is the left hand
side. Now you have to evaluate the right hand
side; may be I will start from here. The right
hand side is actually integral d tau; psi
is identified with psi n. This psi instead
of psi I am going to put psi n. So, I will
have psi n inside. Then I will have a. Well,
the psi n should be having a star. Then I
will have a operating upon; the phi is identified
with psi m. So, I will have psi m, and then
of course, the whole thing after evaluation
I have to take the complex conjugate, correct,
and now what is A operating up on psi m? Psi
m is an Eigen function of A.
So, therefore, this is going to be equal to
integral d tau psi n star a m psi m, but this
a m I can move out. So, I will have a m integral
d tau psi n star psi m, the whole thing star,
correct, and this star suppose I perform the
star operation I am going to get a m star,
but a m is an Eigen value of the Hermitian
operator. So, a m is assured to be real. So,
the star operation taking the complex conjugation
does not affect a m. So, what will happen
this actually I could say a m star, but a
m star is the same as a m, because a m is
real, multiplied by the star of this, and
what is going to happen? You are going to
get psi n. This star you take inside psi n
star star that is nothing but psi n, and psi
m you have to take the complex conjugate.
So, we evaluated the left hand side, left
hand side is equal to that. We evaluated the
right hand side; the right hand side is equal
to that. So, therefore, that implies that
these two have to be equal, correct. For a
Hermitian operator this has to be equal to
that and that let us write that equation.
What it says is that a m into integral d tau
psi n psi m star is equal to a n into integral
psi m star psi n. This is what happens, and
so if you took the right hand side to the
left hand side, what are you going to get?
You are going to get a m; you see that this
and that are the same. So, therefore, what
you are going to get is a m minus a n into
integral psi m star psi n d tau must be equal
to 0. This is the result that you are getting,
but then we have already said that we are
interested only in the cases where a m is
not equal to a n. So, therefore, a m minus
a n cannot be 0 and therefore, I am perfectly
justified in dividing throughout by this expression.
If this was 0 then I i cannot divide by that,
but by definition a m minus a n is not 0,
because the two Eigen values are not the same.
So, you divide throughout by this, and what
do you find? You find that the m th Eigen
function must be orthogonal to the nth Eigen
function. We have seen examples of this. For
example, in the case of particle in a one
dimensional box or harmonic oscillator; even
in the case of the hydrogen atom you can show
that this is valid, but you may ask what will
happen if the Eigen functions have the same
Eigen value? Then the proof is not valid,
and they need not be orthogonal, okay. They
need not be orthogonal, but normally because
we want everything to be orthogonal once we
get the Eigen functions we make sure that
they are orthogonal, and that is always possible,
okay. Having seen this I now want to again
go back to the postulate three and discuss
that a little bit.
Now let us imagine that I have an operator
A corresponding to some observable some observable
which I may have to denote by this symbol
A. For example, it could be momentum or it
could be energy; it could be anyone of the
may be angular momentum or some such observable
is there, and I have an operator corresponding
to that. And that operator will have Eigen
functions which we have denoted by the symbols
i 1s, i 2s, i 3s, i 4, etcetera, and the corresponding
Eigen values are a 1, a 2, a 3, a 4, etcetera.
And further I will assume that if I took any
Eigen function may be the m th Eigen function
and multiplied it by the n th Eigen function
and integrated over the entire space, okay.
I am going to now say that okay, this is such
that the answer is equal to 1, when? When
m is equal to n. If they are the same the
answer is 1, and this is equal to 0 if m is
not equal to n. So, I will assume that; I
have already performed this normalization
business.
So, each function is normalized, and each
function I will assume is orthogonal to the
remaining functions. So, this object you see
this is an object which is if m is equal to
n I have 1, and if m is not equal to n I have
0. This usually is written as delta m n. The
definition of delta m n is this. If m is equal
to n it is equal to 1, and if m is not equal
to n it is equal to 0; that is the definition
of this symbol, and it is referred to as the
Kronecker delta; Kronecker is the name of
a person. We have already seen another delta
not delta, but a delta function. This was
introduced by Dirac that we discussed earlier.
Now in addition to this you see every operator
that occurs in quantum mechanics has a very
interesting property. The Eigen functions
always form a complete set. I suspect I have
discussed this a little bit in connection
with particle in a one dimension box.
What it means is that if you give me any arbitrary
function which is an acceptable wave function
then I can expand it in terms of these Eigen
functions. You give me any arbitrary function,
it is possible for me to write it as a linear
combination of psi 1, psi 2, psi 3, psi 4,
etcetera; that means suppose you give me a
function anyone of you gives me a function
phi which is an acceptable function. Then
this phi may be expressed in terms of psi
1, psi 2, psi 3, etcetera, how? You can say
c 1 psi 1 plus c 2 psi 2 plus c 3 psi 3 plus
etcetera. So, if that is possible then you
say that the functions form a complete set,
and actually all the Eigen functions of the
Hamiltonian operator they do form a complete
set. This is something that is rather difficult
to prove mathematically. So, we will leave
this to the mathematicians; we would just
use this information, okay.
So, suppose I have let us say a state function
which I will temporarily normally we denote
it by the symbol capital psi, but temporarily
let me say I denote it by the symbol phi suppose;
this is my function, and suppose I make a
measurement of what? Of the observable which
I have denoted by the symbol A and postulate
three will tell you that the answer has to
be an Eigen value of this operator that is
associated with the observable. So, therefore,
if the operator A has the Eigen values a 1,
a 2, a 3, a 4, etcetera. If you make a single
measurement the answer will be one of these
Eigen values, and I keep on making measurement
again and again and again and again large
number of times. It is not necessary that
I should get the same answer. I may get answer
a 1, then a 2, then may be a 3. So, finally,
if I have made a large number of measurements
then what will I do? I will calculate the
average of all these measurements.
So, this is something that I have already
discussed. I get a 1 n 1 times, a 2 n 2 times,
a 3 n 3 times, etcetera. Then I can calculate
the average; how would I calculate the average?
n 1 a 1 plus n 2 a 2 plus n 3 a 3 plus etcetera
divided by total number of measurements which
is nothing but n 1 plus n 2 plus n 3 plus
etcetera, and quantum mechanics actually gives
you a definite procedure for calculating this
average. What is the procedure it says that
okay, what you have to do is you will take
the state function which temporarily I have
to decide to denote by the symbol phi. This
is my state function allow it to be operated
up on by the operator corresponding to the
observable A. Then multiply the result by
the complex conjugate of phi and integrate
over the entire space, and if your function
is not normalized you will have to divide
this by phi star phi d tau, and this is referred
to as the expectation value of A, and this
average this experimentally measured average
is actually equal to this expectation value.
So, therefore, if you know the state function
it is not necessary to do the measurements
actually. You can use the state function and
calculate the average, and that is the use
of quantum mechanics. You do not have to do
the experiment; if you do not want to do it
you can get the same result by doing this
calculation. Now I will also assume might
at least for this lecture that phi is normalized,
right. So, that means this is equal to 1.
So, therefore, if I wanted to calculate the
expectation value of A; this is the formula
that I have to use. I mean strictly speaking
I should have this other term also which divides
it, but we will assume that the function normally;
normally we always normalize the function.
So, therefore, the integral will be 1.
Now what I am going to do is I am going to
suppose I mean I wanted to calculate the average
of a large number of measurements of the square
of the observable A. Well, if you are doing
the experiment, what you will do is you will
have you would have measured the value of
a 1. Then here in this expression you will
put a; instead of a 1 you will put a 1 square.
There you will put a 2 square, here you will
put a 3 square and so on, right, and so the
same thing. I mean if you wanted to calculate
all that you need to do is instead of putting
A you just have to put A square, and what
will that be? It will be integral d 2 phi
star A phi; oh sorry minor mistake it is not
A but A square. Now just to illustrate the
points what I am going to do is I am going
to say I have a phi. Suppose phi is actually
of the form may be c 1 psi 1 plus c 2 psi
2, okay. To work things out I will take such
a simple example, and then see what will happen
even if I had a general function which may
be written as a combination of c 1 psi 1 plus
c 2 psi 2 plus c 3 psi 3 c 4 psi 4, etcetera.
But to do the calculations may be it is simpler
just to assume that phi is given by c 1 psi
1 plus c 2 psi 2. So, this obviously is not
an Eigen function. You see psi 1 is an Eigen
function of A; psi 2 also is an Eigen function
of A, but this combination is not an Eigen
function of A, right. So, I am imagining that
I have such a situation that my state function
is not an Eigen function of the operator whose
observable I am measuring, correct. So, if
you did this, if I now calculate the value
of expectation value of A; this incidentally
is referred to as the expectation value of
A. This is equal to integral d tau, what is
going to happen? I would have A operating
upon c 1 psi 1 plus c 2 psi 2, and that is
to be multiplied by what object? C 1 psi 1
plus c 2 psi 2 but with a complex conjugate,
fine, and now you can look at this expression.
A is operating up on c 1 psi 1; c 1 is just
a number. It is a constant and c 2 also is
a constant.
So, what will happen? A will simply operate
upon psi 1 or A will operate up on psi 2;
A can be taken inside the bracket and allowed
to operate up on these two things. So, what
is going to happen is I will get integral
d tau c 1 may be the star operation can be
taken inside, c 1 star psi 1 star plus c 2
star psi 2 star multiplied by this A I am
going to take it inside, allow it to operate
up on these things. So, the answer will be
c 1 a 1 psi 1 plus c 2 a 2 psi 2. Now you
see when you have such a product; well, here
I have already used the fact that psi 1 is
an Eigen function of the operator. Now when
you have such a product, obviously, you are
going to get four terms, right, as a result
of multiplying it out.
Let me just write the first term. The first
term is going to be integral d tau; first
term will be obtained by combining these two,
okay. Well, I know that I am going to have
an integral, but then of course, c 1 is a
constant. So, it is not affected by the integration;
a 1 also is unaffected by the integration.
So, let me write those things outside. Imagine
that I have an observable A associated with
I have the operator which we normally denote
by A with a hat on top of it, and I have an
another observable B associated with which
we have the operator B. And for example, A
could be the x coordinate of the particle
and B could be the corresponding momentum,
and the generalized answer to the principle
which we will prove, states that delta A square
into delta B square; these are the uncertainties
in A and B. It states that it has to be greater
than or equal to minus 1 by 4 expectation
value of the commutator of A and B 
the whole square. Now this of course requires
explanation. What do I mean by the commutator
of A and B? Let me first define the commutator.
When I write two operators A and B within
such a square bracket this means that I am
taking off AB minus BA, okay. So, this is
the meaning of writing A comma B within this
kind of square brackets, and just to illustrate
let me imagine that A is actually x, okay,
the operator corresponding to position, and
B is the operator corresponding to momentum
which we normally write as p x. So, therefore,
let me say I want to calculate the commutator
of x with p x.
So, by definition it is actually going to
be, well, these are operators. So, therefore,
what I will have is I will have also have
a psi on which the operators will operate,
psi could be any function. So, the definition
of that is actually x p x operating upon psi
minus p x x operating upon psi, and the question
is what is this? Well, let me calculate this.
How do I calculate that? The answer is I can
say okay, this is equal to x. P x is actually
minus i h cross dou by dou x. So, this has
to operate up on psi, and if it operates up
on psi what happens? You will get the derivative
of psi with respect to x multiplied by x and
of course, multiplied by minus i h cross,
and from there you have to subtract what?
Minus i h cross dou by dou x operating upon
x into psi, and let me try to calculate this,
okay. The first term is actually x into, well,
may be I will expand this and remove the bracket.
So, minus i h cross x into dou psi by dou
x is the first term, and what happens to the
second term? Well, obviously, this minus and
that minus will make it a plus and i h cross
dou by dou x operates upon a product, the
product of two terms and so you have taken
the derivative of two; derivative of a product
of two terms naturally you are going to get
the following. You are going to get plus i
h cross; dou by dou x will operate upon x.
So, the answer will be just psi, and you will
have plus i h cross; dou by dou x will operate
up on psi, and the answer will be x into dou
psi by dou x, and you can see that the first
and the last term.
These two terms cancel nicely, and hence the
result is just i h cross into psi. So, the
commutator of x and p x operating up on any
arbitrary function psi is just the same as
multiplying that psi by i h cross. What does
this mean? This means that the commutator
of these two operators is actually equivalently
multiplication by i h cross, okay; having
calculated the commutator let me use it in
this inequality. So, what will I do? I will
identify a with x, right, a with x and b with
p x.
So, here I am going to have the square of
the uncertainty in x, and this is going to
be the square of the uncertainty in b x, and
that would be greater than or equal to minus
1 by 4 expectation value of the commutator,
but the commutator is just i h cross, right.
So, you will have i h cross expectation value,
but then it is just the expectation value
of a constant and the expectation value of
a constant is just that constant itself. So,
therefore, it is not necessary for me to put
this expectation value there. The answer is
just the constant and then you have to take
the square. So, therefore, you will have the
whole thing square, and if you took the square
what does it mean? I square is obviously minus
1. So, that will cancel this minus sign.
So, you are going to get 1 by 4 h cross square
and this obviously may examine; sorry, there
is a square that I should have put here which
I forgot. So, now, I take the square root
of this, and what happens? I shall get the
relationship delta x into delta p x must be
greater than or equal to h cross divided by
2 which actually is the Heisenberg’s uncertainty
principle. So, using this generalized uncertainty
relationship you can actually arrive at the
Heisenberg’s uncertainty principle which
is the delta x into delta p x must be greater
than or equal to h cross divided by 2. To
make the ideas clear let me take an actual
example.
I am going to say that I have a state function
phi which is equal to e to the power of minus
x square divided by 4a square. I hope the
notation is clear, e to the power of minus
x square by 4 a square divided by 1 by 2 pi
a square to the power of 1 by 4. You may wonder
why this 1 by four answer is extremely simple.
This function is normalized, and this is the
normalization factor; that is all nothing
more to it. If you look at this function,
what will be the appearance of the function?
Answer again is extremely simple. If you made
a plot of this function against x you will
get what is referred to as a Gaussian, okay.
In fact, this wave function is just the wave
function for the harmonic oscillator if you
had chosen a to be equal to some constant,
I do not remember what that constant is, but
if you choose a to be that particular constant
then this is just the Eigen function for the
harmonic oscillator.
Now the width of this function, this is actually
what is referred to as a Gaussian; the width
of this actually is roughly I mean I would
say it is proportional to a, roughly speaking
it is two times a roughly I mean the width
of the function. So, if I had a very small
what will happen is that the function is very
narrow; while if I had a large what will happen
the function is very broad, okay. If it is
very narrow it will be highly peaked, right,
because the function is normalized, right.
So, now, if you had such a function, and suppose
you make a measurement of position such a
wave function implies that the particle will
be found in this region with fairly large
probability.
But if you were somewhere here the probability
of finding it is very small; that is what
such a wave function means and therefore,
I can ask what will be the average value of
its position? If I make a large number of
measurements sometimes I may find the particle
here or sometimes there or sometimes there.
I make large number of measurements. I do
the averaging, and the average value of the
position of the particle will be obtained
by a quantum mechanical calculation as expectation
value of x; what will it be? It is going to
be integral phi star. Well, star has no effect
on this function. So, phi star x phi d tau
or d x because it is a one dimensional problem,
and that means of integration are from minus
infinity to plus infinity. I do not have to
do this calculation, because I suspect that
you would be able to tell me what this is
actually, what the answer will be if you did
the calculation. Because this function is
symmetric about the origin about this point
it is an even function.
So, therefore, what will happen? The probability
that the particle will be found on this side
will be the same as the probability that would
be found on the other side, and therefore,
this when you calculate the average has to
be 0. But then suppose I wanted to calculate
x square, what will be the expression? It
is going to be minus infinity to plus infinity
phi star x square phi d x. I do not have any
choice other than to calculate this, right.
So, how will I calculate this? I will just
substitute for phi 1 by 2 pi a square to the
power of 1 by 2, right; that is coming from
phi star phi. Then I will have x square e
to the power of minus x square divided by
2 a square, and well, I do not have space.
So, maybe I will put a d x here and integrate
from minus infinity to plus infinity. So,
this integral it is necessary to evaluate
it.
Let me go ahead and evaluate it. I shall get
1 by 2 pi a square to the power of half integral
minus infinity to plus infinity dx x square
e to the power of minus x square by 2 a square.
I have just rewritten the integral, because
you see your integrand is an even function.
What I can do is I can I do not have to integrate
from minus infinity to plus infinity. It is
enough if I integrated from 0 to infinity
provided I multiply the result by a factor
of 2, and the next thing that I will do is
I will make a substitution; what is a substitution
that I will do? X square by 2 a square I am
going to say it is equal to y, because I have
e to the power of minus x square by 2 a square;
I want to simplify the appearance of this.
So, x square by 2 a square is equal to y or
maybe what I can do is instead of saying that
I could say x divided by square root of 2
a square is equal to y.
That is my substitution; that is that, x divided
by square root of 2 a square. Oh no no, sorry,
sorry, that is a mistake, x square divided
by 2 a square is equal to y; this is my substitution.
So, that actually means x is equal to square
root of 2 a square into square root of y,
okay, which obviously implies that d x is
equal to square root of 2 a square into square
root of y. Another mistake always I seem to
make two mistakes successively. So, dx will
be equal to that much. So, let me do this
integral 2 divided by 2 pi a square to the
power of 1 by 2 integral 0 to, well, when
x is 0 y has to be 0; when x is infinity y
has to be infinity. So, therefore, the limits
are still from 0 to infinity, dx is equal
to so much. So, you will get square root of
2 a square divided by 2 root y into dy; x
square where is it? X square is equal to x
square by 2 a square is y.
So, therefore, x square may be written as
2 a square y and e to the power of minus y.
So, if I took this square root of 2 a square
out, there is another 2 a square. This two
also I mean please watch carefully; I would
not want to miss any two or anything. So,
this two also I have to take out, and then
what will happen? I will have integral 0 to
infinity dy. There is a y here; there is a
square root of y in the denominator, so therefore,
y to the power of half e to the power of minus
y. So, let me cancel out the two’s, a square
also goes. Suppose this is correct; correct
me if there is a factor of something missing,
right. Now this is an integral that hopefully
will be familiar to you. This is known as
a gamma integral or a gamma function. The
definition of a gamma function I should write
it somewhere, maybe I can write it here.
This is the definition of a gamma function.
It is a function of n, okay, and you can say
it is defined as this integral. Now this is
actually very nice function, because you can
easily show that gamma n is equal to n minus
1 gamma n minus 1. I am only writing the properties
that I myself shall made which can be of course
proved, but this is not a class in mathematics.
So, we will not prove any of these things.
So, you can easily show that gamma n is actually
equal to n minus 1 into gamma n minus 1. In
fact, you can further show that gamma 1 is
something that you can easily evaluate. You
will find it is 1, and gamma half again something
that can be evaluated; the answer is square
root of pi. This is all that we will need.
So, therefore, if you look at this expression,
what is the answer that I am getting? 2 a
square divided by square root of pi.
You can identify that this is a gamma integral,
and in fact it is actually equal to gamma
of 3 by 2; that is all, because if you put
a n is equal to 3 by 2 in this definition
you are going to get that exactly the same
integral, okay, and further gamma of 3 by
2 I can make use of this relationship gamma
of 3 by 2. Let me write it here gamma of 3
by 2 will be if I made use of that this is
actually recursion relation will be equal
to half of gamma half using that relationship,
and gamma half you know it is equal to root
pi. So, therefore, using the information from
mathematics we have been able to get the value
of the integral, and it is just square root
of pi divided by 2 which means that this is
equal to 2 a square divided by root pi into
the gamma of 3 by 2 is nothing but root pi
divided by 2 and therefore, what is the answer
that you get? Nicely this root pi and that
root pi this 2 and that 2 cancel. So, you
get the answer a square.
So, summarizing all these things what has
happened? We found that x average is equal
to 0, x square average you have just now evaluated;
x square average actually equal to this is
what we have evaluated x square average it
started here, it ended only there that is
actually equal to a square, okay. So, this
actually means if I go on measuring the value
of x I will get several different values;
they do not have to be the same. I will go
on getting different values, but the average
of all this will be equal to 0. Then if you
took the square and averaged then obviously,
that will not be equal to 0, and that average
also I have calculated that is equal to a
square. So, therefore, I can define a kind
of uncertainty in the measured values of the
position of the particle. What is that uncertainty?
I mean the way I would define it is I would
say delta x square is equal to x square average
minus x average square.
So, therefore, this is going to be equal to
a square minus 0 square which is just a square,
right and therefore, as far as this particular
wave function is concerned or particular this
state function is concerned I say that there
is some uncertainty in the position of the
particle, and how much is that? It is equal
to a, agreed, and that is not surprising,
because I told you roughly the width is dependant
upon the value of a. If you increase the width
of the function, if you wanted to increase
the width of this function you just have to
increase the value of a. This is something
that I have mentioned at the beginning. You
can make this function narrower and narrower
by decreasing the value of a, and when you
decrease the value of a, what will happen?
You are actually decreasing the uncertainty
in position, because you saw that the uncertainty
in position is actually equal to a according
to this calculation.
So, you can choose your value of uncertainty;
there is no problem, by just adjusting the
width of this a, right, and now I ask the
question suppose I make a measurement of momentum.
Now we were talking about measuring the momentum,
sorry talking about measuring the position;
now suppose I am going to measure the momentum
of the particle. This function is not an Eigen
function of the momentum operator, right.
So, therefore, I am going to get different
values, and therefore, there should be a measured
value of the uncertainty in the momentum of
the particle that experimentally determines
the uncertain in momentum; how much is that?
We will calculate it in the next lecture,
and then finally, we will find that the uncertainty
principle according to the Heisenberg actually
follows, when we calculate that and multiply
the delta x with the delta p; you will find
that the uncertainty principle results.
Thank you for listening.
