We were discussing this uncertainty principle
and it is derivations. I actually demonstrated
the uncertainty principle as apply to position
and momentum. For example, we show that delta
x into delta p x is greater than or equal
to h cross by 2.
And I also told you that this is actually
a special case of a generalized uncertainty
principle, which says that delta A into delta
B must be greater than or equal to minus 1
by 4 expectation value of the converted of
A with B. So, would we will spend a little
bit of time, I hope I got it right, if not
anyway, I have going to derive it so, we would
get it right. And in the specific case of
x and p x, this actually let to their relationship
that delta x into delta p x must be greater
than or equal to h cross by two.
Now, this is a actually a peculiarity of a
quantum mechanics, same classical mechanics
it is possible for you to have a particle
with a well define possessions as well as
a well defined momentum. If I have this chalk
piece and if I let it grow each instant. I
know where it is and I also can find out what
is the momentum of the particles. So, there
is no uncertainty according to the classical
mechanics, either in possession or momentum.
You can define both at the same time to any
precision that you want, but there is not
what quantum mechanics is says actually if
you look at this relation. This uncertainty
principle what it says that, there is no state
of the system in which the momentum and the
possession are both well defined there is
no such that.
So, if you try to construct a state with a
very little and uncertainty in momentum, then
as we have seen previously in the previous
lecture the momentum becomes very uncertain.
We did derive an expression for delta x and
delta p and if you look at that expression
what we found was, delta x was actually when
we, I am not going to write the wave functions
about delta x. If it I said delta x is equal
to a then I found that delta p x was actually
h cross by 2 a.
So, if you try to reduce the uncertainty in
possession by reducing the value of a, then
uncertainty momentum increases, right. Or
if you try to reduce the uncertainty in momentum
by increasing the value of a, that you can
do then what will happen is the wave function?
That we had it actually becomes very, very
broad in the possession for it uncertain.
So, let us now look at the derivation of this
expression. The derivation is based upon an
inequality with separate us Schwartz inequality.
So, let me first tell you what this is actually?
If you think of any arbitrary function u and
another arbitrary function, which I shall
be note as v, these are two functions. If
you because, we are thinking of quantum mechanics,
we can think of them as acceptable wave functions.
What I will do is? I will take the first function
multiplied by a parameter lambda. So, lambda
into u plus v, I am going to take, lambda
I will assume is a real; it is not a complex
number, lambda is real, but u and v are arbitrary
they may be complex.
So, what I am going to do is, you can take
this function, take it is complex conjugates,
I mean, make thinks simple, I mean, I was
going to say, I will make it one dimensional,
but it is not necessary. So, these are the
functions of any number of variables u and
v. So, lambda u plus v, I am going to take
the function multiplied by it is own complex
conjugate. And then I will multiplied by the
volume element d Tau and integrate over the
entire space. I am just say the parameter
and because you have taking this combination
in the function and multiplying it by it is
own complex conjugate what will happen.
The products this is actually equivalent to
calculating the magnitude of lambda u plus
v and squaring it, that is what this objectives.
And this I know has to be positive and therefore,
if I integrated it over the entire space,
what should I get? I should get a number which
is definitely greater than 0. So, this has
to be greater than 0, but it may so happen
that, have some I was not very careful and
then what might happen is that, I have taken
lambda to be minus 1 suppose and u to be equal
to v.
Suppose I mean by accident I choose that then
what will happen this function, will came
out to be 0, which case I have the possibility
have in this equal to 0 and therefore, I can
very confidently say that this integral has
to greater than or equal to 0 right. And if
that is the way it is I can expand this if
you multiplied it out say lambda well this
star operation you can take it inside. It
is not going to affect lambda because lambda
is assume to be a real. It will affect u as
well as v.
So, I will take this star operation inside
and I am going to get that. lambda is just
a parameter it does not depend on the possession
coordinates, it is something that you can
choose anywhere it may be 1 or 2 or 3 or 3.5
some number. But interestingly whatever value
you choose for lambda, the answer is guarantee
to be greater than or equal to 0 that is important.
So, let us multiply this out you will have
the product of this two, that is going to
be lambda square u star u that is 1 term right.
When you multiply you are going to get that
because lambda into lambda is there and then
you will have the volume element d Tau and
actually the integral we are locally here,
but the integral I am going to put it here
because lambda does not depend upon any possession
coordinates is just a number just a constants.
So, therefore, this will be made first term
and they will have 3 more terms. One of the
next term will be lambda times integral d
u sorry not d u, but d Tau u star v right,
that will come when you multiply these with
that and then you will have another time which
is roughly of the same form lambda times integral
u v star, correct. And the last term will
be how much it is going to be integral v star
v d Tau and this has to be greater than or
equal to 0 right. Actually something very
obvious and what I am going to do is I am
going to say this is actually of the form
lambda square into some capital a where capital
a is this, plus lambda times well, I let me
call this capital b.
Then if you look at this you would realize
that this nothing but B star, the complex
conjugate of B. The way it is this one and
that one are complex conjugates. So, if I
put this equal to B that equal to B star.
So, here I can say lambda times B plus B star
plus, this objects I am just going to denoted
it temporarily by the symbol C and this has
to be greater than or equal to be 0. And it
does not matter, what the value of lambda
is this is greater than or equal to 0.
So, you can give any value either negative
or positive it is guaranteed it should be
greater than or equal to 0 correct. So, you
look at this is a quadratic in lambda. And
you have to remember whatever you study at
regarding quadratics, a quadratic is greater
than is guaranteed that, the quadratic is
greater than or equal to 0. What does this
means? This means that, this object that I
hope you remember discriminates has to be
negative.
You remember that I suppose it is see. If
I have quadratic, quadratic like a x square
plus b x plus c, if it is guaranteed that
is greater than or equal to 0; that means,
actually b square minus 4 a c has to be less
than or equal to 0. This is something that,
I shall assume you know already. And therefore,
if I just applied it here what is the result
that I am going to get? I am going to get
b square while in this case you say this is
small b is to be identified with capital B
plus B star. So, capital B plus B star the
whole square minus 4 times A into C has to
be less than or equal to 0 right.
And that implies that, let me a write it may
be revise, while I will take this to the other
side; that means, for this left hand side
is less than or equal to 4 A C or I will write
it in this portion 4 A C must be greater than
or equal to B plus B star whole square. Or
if I divide by 4, I should get this, right.
Straight forward, but now let me put back
the definitions of and A C and B what is A?
A is actually integral u star u d Tau, that
is A, this is A. C is integral d Tau of v
star v, that is C, is greater than or equal
to 1 by 4, what are the values of B and B
star well you have the definitions here.
So, what will happen is, integral d Tau u
star v plus integral d Tau v star u, there
is a square right. And this inequality is
separate to as the Schwartz inequality. It
is quiet useful in quantum chemistry actually.
Whenever you encounter the later on it, whenever
you encounter overlap integral they will exit
to be you can prove inequalities regarding
overlap integral that overlap integral I have
not introduce. So, starting from here or using
this inequality, I will be proving that inequality.
And the way in this inequality is, we can
actually you can compare the two and you realize
that somehow what I have to have delta A square
here. And how I will have to, if you compare
the two I will have to have delta B square
here. And then I have to see what that happens
to the right hand side, and eventually the
right hand side will work out to be equal
to that, this is what going to happen. So,
how do I mange this things is the question.
Now these will take a few steps. Well let
me start with this operator A. Operator A
of course, we are doing quantum mechanics
who operator A is Hermitian.
So, the moment using A is Hermitian, I know
that, if you allowed d Tau operate upon psi
multiplied it by phi star and then the volume
element d Tau, answer will be equal to this
expression. You allowed A to operate upon
phi multiply by psi star integrate over the
entire space and then take the complex conjugate
right. In fact, if you took this complex conjugation
inside, when you can see what is going to
happen? You will get d Tau, this star and
that star will be make it just psi, then you
will get A star phi star right. This is what
will happen if you took this star of operation
inside. And that is equal to integral d Tau;
see what is happening is that? I have to take
A, take it is star what do you mean by taking
the star, if A contains square roots of minus
1. If it contains i, then you have to replace
that with minus i, A does not contain i, then
you do not have to worry this, just unchanged.
So, if you like you can write this expression
in a slightly different passion, you see I
have this A star it is an operator remember,
operating upon phi star. So, what this say
is take A star allow it operate upon phi star
and then multiply the result by psi and integrate
over the entire space, that is what it says.
But now, what I will do is, you see, I will
put this A star phi star first. And then I
am going to put psi but then of course, this
can be confusing because A should not operate
upon psi. So, therefore, I will put a bracket
here to indicate that this is operating will
be upon phi not on psi.
So, if you look at this expression, if you
compare this two, what is if that you find?
If you had a Hermitian operator and if you
had an expression like this actually what
you can do is? You can allow A to operate
upon phi star instead of allowing A to operate
upon psi, but then of course, not A itself,
but a star has to be operating look at this
is. This operate it can, at the moment it
is operating on psi, but instead of that operate
can be operating on phi star that what is
saying it can be operating upon phi star.
Then it is not the operator itself, but it
is complex conjugate, that has to operate
upon psi star you will get the same answer
right. So, therefore, you can put then the
other way, if I had an A star here operating
upon this phi star. That A star can be transfer
to A, sorry, transfer to psi, but then, when
you transfer it becomes A. So, if you had
an expression like this A star is operating
upon phi star, but then this can be allowed,
I mean it from here it can moved and allow
to operate upon psi.
The answer is going to remain unchanged provided
you take not A star, but A right, this is
useful. So, with this I am going to think
of, let me think of an operator which, I will
denote as delta A, why do I call it delta
A because, I am going to say it is nothing,
but A minus the expectation value is nothing
wrong in that, I mean you are going to give
me an operate any operative I will first calculated
expectation value and I am going to subtract
that expectation value. Expectation value
will be a number, in fact, it will be a real
number because, this is our experimental answer
right. So, this object will be a real number.
So, from A, I will just remove that and then
I have this operative. Now, what I am going
to imagine is that, I want to calculate delta
A operating upon psi 
multiplied by delta A operating upon psi star,
volume element d Tau integrate over the entire
space. Now, you may wonder why do all these
things. The answer is extremely simple, this
is actually nothing but I am going to claim
that this is nothing but delta A square. That
is what I will show, this object and that
is the reason why introduce this operate otherwise
there is no reason for make introduce this
operator.
This operator is such that the uncertainty
in A can be written as delta A operating upon
psi star multiplied by delta A operating upon
psi. How do I show that? Well this is actually
equal to integral of d Tau A minus expectation
value of A into psi, you have to take this
star of that. And you have A minus expectation
value of A right. So, there are actually I
mean if you multiply this, I would you will
realize that, there are four terms, correct.
There are four terms.
Let me look at the terms. So, the first term
will look like what? Integral d Tau well A
operating upon psi star and A operating upon
psi this is the first term. There are 4 terms;
I am just writing the terms are one by one.
So, this is the first term correct. It comes
from this and that. I mean A operating on
psi multiplied with star operations and then
you A operating upon psi that is first term.
And this is actually equal to integral d Tau
A star operating upon on psi star multiplied
by A operating upon psi right.
Straight forwards step, but then I have already
told you that, if I have the A operate A star
operating here, I can actually allowed this
to be transfer there, in general. Because
you look at what I have done here for any
arbitrary psi, if I had A operating upon psi
multiplied by phi star integrated over d Tau.
The answer is equal to this expression and
therefore, I already told you, if I had an
A star here, that A star can be transfer to
psi, but only thing is that in that transfer
process A star becomes A that is all.
So, that is what I am going to do here, I
am going to transfer this A and allow it to
operate upon this part. The only thing as
I have told you is that the transfer process
A star becomes A and therefore, what will
happen? This is nothing but integral of d
Tau psi star A square psi or what is that?
That is nothing but the expectation value
A square that is my first term right. Then
you look at the next term, if there are 4
terms, we will take the next term by which
one should I take, well may be this with that,
this into that, into psi and psi star of course,.
So, that will appear with a negative sign.
The negative sign you are going to minus integral
d Tau what are the terms, that we have expectation
value of A into psi, you have to take it complex
conjugate and then you will have A psi, this
is one of the next term correct. And what
is going to happen is expectation value of
A is just in number right. If you calculate
the expectation value, this is the average
of a large number of measurement, it just
a real number.
So, this number is not affected by the integration.
So, I can take it out. So, what will happen,
I will have this object right, because I have
taken this expectation value outsides. So,
the expectation value is come out then I have
psi A operating upon psi star right. And what
is this? Nothing but expectation value of
A right. So, therefore, this expectation value
of A into expectation value of A, I forgotten
to put the negative signs it should be there.
So, therefore, minus A expectation values
square is what this term fine. And you look
at this next term what is it actually the
we have taken this into that, into that, we
have taken this into that, into that, the
next term will be A psi star, we again this
will going to be with a negative sign.
So, this is the next term right, I hope I
got things right A operating upon psi correct.
I take it from here and this A multiplied
by psi is the term that I take and so, this
is what it is and what is this actually A
I can take out. So, minus expectation value
of A integral A star psi star take the complex
conjugate of this. So, I will get A star psi
star multiplied by psi d Tau fine. And just
as previously I can transfer this A star to
that psi. So, if I transfer that what is that
I am going to get minus A integral psi star
A psi d Tau and what is this. This is nothing
but expectation value of A and therefore,
what is A to get another minus A square. And
what is the last term? Last term is obtained
from this into that right. So, that is actually
quite simple might because, we have the product
of two negatives sign it is actually plus.
So, what will happen is, I will have d Tau
A psi star multiplied by expectation value
of A into psi that is it. That is the last
term, but there you know that the A is a real
number. So, therefore, star does not affect
it, this is the real number. So, this and
that together they may be taken out. So, I
will get A expectation value of A square multiplied
by integral psi star psi, but we are assuming
that psi is a function that is normalized
and therefore, this integral is 1 and therefore,
what happens the final answer that I get is
expectation value of A square.
So, interesting we have four terms, the first
term is actually nothing but the expectation
value of A square. The second two terms are
actually identical right, you get the same
answer for the second two terms and to the
last term is A square. So, if you added the
four terms together, what is the answer that
I had to get? I can write it here. If I added
the four terms what do you get? You get expectation
value of A square minus expectation value
of A square that is all, correct.
Because there were two with the negative sign,
two expectation value of A square with negative
signs and to one with the positive sign and
therefore the net result is this. And therefore,
what is it I mean, whatever I want it to show,
I have achieved in what sense. I claimed that
delta A square if I define it to be like this
it is nothing, but our previous definition.
This is something this is the way; we define
the delta A square in the previous lecture.
So, if that is the way it is, suppose I take
the operator B and I am going to define something
which will be B minus expectation value of
B correct and I will say that I have delta
B which is define to be this right. And if
this delta B operates upon psi 
and delta B operates upon psi ones more you
take it complex conjugate integrate over the
entire space. What will happen? The answer
will be actually equal to delta B square the
uncertainty in B correct. And square of the
uncertainty in B right, if you agree with
that, then I am going to make a slight modification
by introducing an i here.
If I had put an i either, what is going to
happen? I am going to get, I am going to take
it is stars. So, it will come out and it is
going to be minus i, I have to neutralize
that I cannot just have a minus i. So, I will
put another i here. So, that everything combine
together will give me a plus point. So, I
have not done anything. So, therefore, this
also is valid right. I mean this are the simple
mathematical manipulations. The reason for
doing this is not obvious at this moment,
but you will get that answer, which is what
we want any way.
So, what I am going to do now is this is my,
this is correct and now, I have this Schwartz
inequality. This is valid for any u and v,
what I will do? I will say that u is going
to be delta A operating upon psi right. This
is valid for any u and v or let me say, u
is delta A operating upon psi. And what is
v, I will say is i times delta B operating
upon psi, I can always do that, because this
Schwartz inequality is valid for any arbitrary
u and v. So, I am perfectly justified choosing,
whatever you can chose your own u and v, it
has to be valid because this is the general
inequality that is valid. So, if you did that
what will happen they, if you use that then
we have attend whatever we wanted to attend
because, what we want this is equal to? This
should be equal to delta A square that is
what we wanted.
So, if we use yet u to be that, then this
will actually be equal to delta A square and
what will this be? If we made this choice,
this will be equal to delta B square. Uncertainty
in A square and uncertainty in B square and
now, with this choice what will happen to
this part is the question and we will work
that out it is not a difficult calculation,
fairly simple if you work it out powerfully,
I will get that answer. So, let me do that.
So, if you chose u to be this and v to be
equal to that, what will happen to integral
d Tau u star v?
What will that be, it will be actually equal
to integral d Tau, u is how much delta A 
psi star into v. v is i into delta B psi correct.
So, you can take that I outside, that go to
get i then integral d Tau, you are going to
have A minus expectation value of A psi. You
have to take the complex conjugate. So, star
will comes here then you will have to let
it stay here. Because I am taking the complex
conjugate then, delta B will be B minus expectation
value of B into psi. And what will be integral,
what is the other integral v star u right.
So, you are going to get integral d Tau v
star u and that be equal to integral d Tau
v star. v star will give me a negative sign
there will be minus i because u, this is v.
It took the star, it going to give me a minus
i delta B. delta B operating upon what, psi
then you have to take this star then you have
to have delta A operating upon psi, it actually
will be equal to minus i integral d Tau B
minus B expectation value, you have to take
this star of the whole thing, psi star A minus
expectation value of A psi. Now what happen
if I add these two things up? Correct.
I have to add these two things up. So, if
I added these two things up, what is the answer
that, I am going to get is the question, but
before I add the two up I want to do something
more I just want to see this is. This whole
thing is an operator and A is a Hermitian
operator, this we said just a number right.
So, what I can do is, I can actually transfer
and allow them to operate upon this I from
here, I can allow them to transfer to here
and operate and all that what we have to do
is when I transfer I have to take the complex
conjugate of that. So, let me do that as a
intermediate step there. So, here what is
going to happen is, this is going to be equal
to i times integral d Tau, you transfer it
to the allow to operate the upon that. So,
I shall get psi star A minus expectation value
of A there is already a star, but when you
transfer you will put one more star.
So, double star means you get the same number
original number back. So, get A minus expectation
value of A into B minus expectation value
of B into operating upon psi. That is what
happen to this term and if you put at this
term kind of thing with this. What will happen?
You will get minus i integral d Tau right.
You take this you are going to transfer this
to the other side there is already a star
when you transfer you will take one more star
and therefore, what will happen? The star
operations you see back the original the due
to star and back the original, but then what
will happen you will get psi star. This is
going to be transfer and it is going operate
upon here.
So, you are going to get, this you are going
to transfer you see this was original operating
upon psi star, but instead of operating upon
psi star, you will transfer it and you will
allowed it operate upon that. And while to
be actually very, very clear, I should put
a bracket here, because this operator was
operating only upon this part. And this operator
again is operating only here.
So, therefore, when you transfer it. It is
going from here answer getting operated on
that points. So, that is what happens. So,
then if you added these two up, which are
the once I should add up these two things.
You will see something interesting happening,
you will see that if I added these two up
and I am to going get an i integral d Tau
psi star. This is your first term correct,
I have just written the first term and I have
to write the second term, unfortunately there
is no space.
So may be what I will do is, I will write
the whole thing up here. I will write this
term once more. The first term is actually
plus i integral d Tau psi star. This term
let me remove this; this term is going to
be psi star A minus expectation value of A
into B minus expectation value of B operating
upon psi that is the first term. And the second
term is actually with negative sign integral
d Tau psi star B minus expectation value of
B into A minus expectation value of A. So,
these are the two terms. If you just added
the two up I would see that the i is common,
until i there are severally things others
commons, let me take it d Tau psi star and
then you will have this products A minus expectation
value of A into B minus expectation value
of B minus I am combining the two terms.
So, this is what results, right. I have not
done anything great, I mean I have just simplified
things. And then they, if you look at this
expression, what is that you are going to
get? You will get it i times integral d Tau
psi star to multiply things, you are going
to get A into B from here and from there you
are going to get B into A is not in the same
order. You see A into B minus B into A is
what you will get? And then you should remember
that these things are numbers as for us numbers
for concern, you can put them in any order
you like. They are not operators and therefore,
you for example, will get minus A expectation
value of A into B and you will get the similar
kind of term from here.
What is that I said minus expectation value
of A into B correct and from here, you are
going to get minus expectation value of A
into B, but with the positive sign. So, they
will cancel each other similarly all the other
terms you can just expand you will see that
all the other terms except these two terms
will cancel each other and therefore, what
happens you get this as the answer. So, we
have evaluated this part what is the answer?
The answer is that and therefore, if I just
substitute it that to the right hand side
what I am going to get? I am going to get
1 by 4 i times integral d Tau psi star, what
is this actually? It is the A B minus B A
and this is the thing that we have been referring
to us the commutater of A and B. So, therefore,
you get A commutator B, this is just chotta
notation right. And then I have to put psi
here and close the bracket.
So, this is the object that I have written,
but I remember, I have A square here. So,
therefore, we have evaluated the right hand
side. The right hand side evaluates to be
this on the left hand side, this is delta
A square this is delta B square. So, sees
the delta A square into delta B square has
to be greater than or equal to be this object,
that is what it says, but there is i sitting
inside and you have to take square of that.
So, when you take the square of that you can
remove this i and say that there should be
a negative sign, correct. And if I remember
correctly we had actually or I had actually
demonstrated this by taking A to B x and B
to B x that I did yesterday and you should
realize that. This is a very general thing
what does it say, it says that give me any
two operators in quantum mechanics, give me
any two operators and it does not matter what
the state of the system is.
It does not matter it can be any state of
the system this inequality has to be valid
and this is extremely general right. Extremely
general, if you apply it to possession and
momentum you will get what is refer to as
the Heisenberg’s uncertainty principle.
So, this is generalize the uncertain relation,
this was not derived by Heisenberg, or somebody
else. As Heisenberg derive delta x into delta
p x is greater than equal to h cross by 2,
but this generalized uncertain relationship
was derived later and it is valid for any
A and B. That is the perfect time for me to
stop for today; thank you I shall continue
I mean we will use this in the next lecture.
Thank you for listening.
