In the last lecture we had obtained a physical
interpretation of the wave function. The Schrodinger
equation has written in the terms of wave
function psi and we had said we had derived
an equation of continuity from which we could
interpret mod psi square d tau as the probability
of finding the particle between in the volume
element d tau.
Actually we obtain that this is this should
be proportional to the probability but then
if we normalize this in the entire space,
so then we can interpret mod side square d
tau as the probability of finding the particle
in the volume element d tau. So associated
with the particle if I want to find out the
expectation value of any quantity x then I
just have to multiply by the probability density
function and integrate over the dais space
actually the integration is something like
this that the all the integration is over
the entire space x mod side square d tau divided
by integral mod psi square d tau. Because
mod psi square d tau is proportional to the
probability but, if this integral is 1 which
will assume from now on words so will assume
from now on words that the function psi is
normalized and if the wave function is normalized
then mode psi square this integral over the
entire space is 1 and therefore the denominator
is one. So I will obtain x psi star psi and
I will write this for reasons which will become
clear in a moment as x is equal to integral
psi star x psi d tau. Why do we write it this
particular way it will become clear in 1 minute.
Now we have the time dependent Schrodinger
equation with 3 dimensional Schrodinger equation
i h cross delta psi by delta t is equal to
minus h cross's square by 2 m del square
psi plus V psi, what I do is that I multiplication
multiply this equation by psi star and then
integrate. So I obtain so I obtain on the
left hand please see this carefully. This
integration is over the entire space so the
integral is really at 3 dimensional integral
psi star multiplied by i h cross delta by
delta t into psi d tau integration is over
the entire space and then we will have psi
star multiplied by minus h cross square by
2 m del square psi d tau plus psi star V times
psi. On the last term this term is just the
potential energy function multiplied by the
probability density function integrated over
the entire space, so this quantity is the
expectation value of the potential energy.
Classically we know that the total energy
is the kinetic energy plus the potential energy
so we should expect that the expectation value
of the total energy must be equal to the expectation
value of kinetic energy plus the expectation
value of the potential energy. So therefore,
this quantity should be the expectation value
of total energy and this quantity should be
the expectation value of the kinetic energy.
And now we see that the quantity within the
bracket within this within the bracket is
the operator corresponding to the energy e.
Similarly p square is p x square plus p y
square plus p z square so this is minus h
cross square by 2 m delta 2 by delta x square
plus delta 2 by delta y square plus delta
2 by delta z square, so this quantity within
the brackets is the operator corresponding
to p square by 2 m. So therefore, we get the
recipe that if I want to find the expectation
value for example, p p x then what I have
to do is I will write down psi star p x psi
d tau where p x is the operator representation
of p x. So therefore from above we obtain
the recipe that if I want to find out the
expectation value of x y and z this equal
to this all these integrals are 3 dimensional
integral psi star psi x d tau they are normalized,
similarly for y, similarly for z for p x the
expectation value of p x is equal to psi star
p x psi d tau which is equal to psi star minus
i h cross delta psi by delta x d tau
If I want the expectation value of p x square
then p x square as we know is p x times p
x so minus I by h cross delta by delta x into
minus i h cross that is minus h cross square
delta 2 by delta x square, so the p x square
the expectation value is psi star minus h
cross square delta 2 psi by delta x square.
And the general recipe if I want to measure
a dynamical variable o then the expectation
value of this I represent that by the corresponding
operator and put between psi star and psi
o psi d tau this is the general recipe for
determining the expectation value. So therefore,
we must remember these 2 3 equations the expectation
value is of course very simple the expectation
value of p x is psi star on the left and psi
on the left right the p x represented by this
differential operator representation and this
is for p x square and if you want write the
kinetic energy then I have to just divide
by 2 m. In general if I want to make a measurement
of a dynamical variable represented by o and
if I want to find out the expectation value
then all that I have to do is to integrate
psi star o psi d tau. I am assuming that the
wave function is normalized with this we will
try to give an exact formulation of the uncertainty
principle between x and p x that I in in statistical
theories if there is a variation of a parameter
then the spread is measured in terms of mean
square deviation and that is delta x is defined
as x square the average of x square minus
x average square under the root so this is
the spread the uncertainty in x
Now for example, let me give you an example
of an experiment I measuring the acceleration
due to gravity I use a simple pendulum and
in three random measurement I change the length
of the pendulum and in three random measurements
I get nine seventy nine meter per second square
nine eighty and nine eighty one. The mean
of that is nine eighty centimeter per second
square now I use a kater's pendulum to make
a measurement of acceleration due to gravity
but, there I get 3 values like 979.99 80.0
and 980.1 if I take a mean of that again I
will get nine eighty. But there is a difference
between these two here the uncertainty is
much more than this here the uncertainty of
the order of 1 and here the uncertainty is
the order of 0.1 how do we express this uncertainty
this we express in terms of carrying out let
us suppose the spread will be you will write
down is will be the average value of g square
minus the square of the average value and
then take this square root. In this case it
will come out to be point 8 or 0.9 and in
this case it will come out to be 0.08 or 0.09
or 0.1 above something like that so this spread
the variation if I make a large number of
measurement the variation is usually expressed
by this quantity delta x I take the average
of the square minus square of the average.
Similarly if I make the measurement of the
momentum then I will write it as the square
root of the average of the square of the momentum,
I am considering only the x component I only
considering the x component so p x square
minus p x whole square I will show that for
any wave function the product delta x delta
p x is greater than or equal to h cross by
2 so this is an exact statement for the uncertainty
relation.
Now in order to derive that we will first
prove an inequality and this is known as Schrodinger
inequality and inequality I have f star f
d tau multiplied by g star g d tau where all
functions f and g are well behaved square
integrable functions
this is always greater than or equal to one
by four brackets integral f star g plus f
g star d tau whole square. Now this quantity
we denote by a this quantity we denote by
c and we write down that b is equal to integral
f star g d tau therefore, b star is equal
to the complex conjugate of this is f g star
d tau. All this integrals are over the entire
space and they are all three dimensional integrals.
Now in order to prove this inequality where
f g are all arbitrary function, arbitrary
means single valued and integrable functions
and in order to prove that I consider a parameter
lambda which is real lambda I assume to be
real and I have I consider the integral lambda
f plus g mod square d tau. Now since lambda
is real so this quantity is positive because
it is the square of the modulus and integrated
over the entire space so this must be always
greater than or equal 0 positive always positive
because this is the integrant is a positive
indefinite quantity. So let me write down
the integrant so that is equal to lambda f
plus g multiplied by lambda f star plus g
star, because lambda is itself a real quantity
so I multiply this out so I get lambda square
f star f I can write one before that lambda
times f star g plus f g star 
plus g star g. So I multiply integrate this
d tau and this is always positive definite,
so I integrate it here I integrate it here
so I integrate it here all integrals over
entire space so this is always positive definite
so this is a as I have renounced here and
this is b plus b star and this is c, so I
obtain lambda square a plus lambda b plus
b star plus c will be always greater than
zero.
This means if you plot this as a function
of lambda, it will have no zeros and therefore
for example, as you will know that you see
if I consider an equation like this a x square
plus b x plus c the roots of this equation
are x is equal to minus b plus minus under
root of b square minus four a c divided by
two a but, if this quantity is always positive
then there are no real roots and therefore,
this square must be equal to must be less
than four a c. So that this quantity is negative
and therefore, it will have no real roots,
so this quantity b plus b b star whole square
must be less than or equal to less than equal
to four a c so you get a c must be greater
than one by four b plus b star whole square
and that is my inequality that f star a was
equal to f star that is f star f d tau c was
equal to g star g d tau this must be greater
than one by four b plus b star whole square
that is integral f star g plus f g star d
tau whole square. So therefore, we have been
able to to to derive what is known as Schrodinger
inequality
Now we assume that we assume that f is equal
to p psi and p psi is equal to minus i h cross
delta psi by delta x and g is equal to i times
x psi, then you can see we will discuss the
easiest case first g star g d tau g star s
d tau g star will be i times minus i plus
1, so this will be g star will be x into psi
star so this will be psi star x square psi
d tau. So this is my the expectation value
of x square we will we are considering the
special case although the more general case
can be considered very easily that the expectation
value of x is 0 I can always choose the origin
in such a way that expectation value of x
is zero. we further assume the expectation
value of p x is zero. So g star g d tau I
x psi multiplied by minus i x psi star so
i times minus i becomes one so psi star x
square psi d tau so this is the expectation
value of of x square. Similarly, f star f
d tau will be equal to if you see this f star
will be i h cross delta psi star by delta
x multiplied by f that is minus ih cross delta
psi by delta x multiplied by d tau that is
d x d y d z. So I times minus i is one so
h cross square if I take outside h cross square
and I just consider the x part of the integration
the y and z part is also there so I get delta
psi star by delta x into delta psi by delta
x d x actually into d y d z but, I omitting
that in this moment. So this I integrate by
parts and I obtain h cross square this I take
as the integrating factor psi star delta psi
by delta x from minus infinity to plus infinity
minus integral psi star delta two psi by delta
x square d x of course multiplied by d y d
z and d y d z integrations are also there.
Now for a localized wave packet for a particle
which is localized in a small region of space
this wave function has to vanish at infinity
so this term vanishes at the upper and lower
limit and I obtain only this so I finally,
obtain I finally, obtain integral f star f
d tau is equal to psi star minus h cross square
I take this inside delta 2 psi by delta x
square d x similarly, d y and d z. So this
is the expectation value of p x square because
you remember that p x is minus i h cross delta
by delta x and p x square will be minus i
h cross d by delta x so p x square psi is
minus i h cross delta by delta x into psi,
so this will minus minus plus h cross minus
minus plus and then this I square is minus
so this is equal to minus h cross square delta
two psi by delta x square. So this quantity
becomes the expectation value of p x square
but, I will just write p square 
and the third term on right side will be if
you remember the therefore, so f star f d
tau is p square g star g d tau we are just
now obtained g star g d tau was equal to x
square will be the term on the right hand
side.
You see in the inequality we had we had 1
over this is f star f this is g star g is
greater than equal to one over four f star
g plus f g star so let me evaluate this so
f star g let me do it in a fresh page that
one by four integral f star g plus f g star
again d tau is d x d y d z, so f as we may
recall was equal to minus i h cross so f as
you may recall was minus i h cross delta psi
by delta x and g was equal to i x psi so you
will have so let me just consider the integrant
the integrant is equal to one over f star
g plus f g star and this is equal to one over
four f star that is i h cross delta psi star
by delta x into g that is i x psi plus f that
is f is minus I h cross delta psi by delta
x you have to do this little carefully and
g star is minus i so minus minus becomes plus
so i x psi star, so this is i times i is minus
one and this is i times i is minus one so
this becomes minus h cross by four delta by
delta x psi star x i plus delta psi by delta
x x psi star.
Now if you write down this equation this expression
delta by delta x psi star x i then this will
be please see delta psi star by delta x into
x i plus psi star into please see this carefully
delta by delta x of these two terms the product
of these two terms. And if I differentiate
this if I differentiate this first I will
get psi because the differential coefficient
of x is one plus x delta psi by delta x. So
you obtain so you obtain if you these two
terms are appearing here if you see this carefully
this is delta psi star delta x multiplied
by x psi plus psi star psi plus psi star x
delta psi by delta x, so if you see this this
term is this term let me encircle them with
red so this term is the same as this term
and this term is the same as this term, so
this plus this which is the quantity inside
the square brackets here is this minus this.
Therefore, a integrand will become the integrand
will become minus h cross by for delta by
delta x i has to just patiently work this
out to straight forward but, it requires a
little patience minus psi star psi. If I integrate
this so I have to integrand if I had to integrate
this with respect to x and this with respect
to d x and then d y d z then the integration
over x will be psi star x psi from minus infinity
to plus infinity, so the wave function is
going to vanish at both plus infinity and
minus infinity therefore this term will be
zero and if the wave function is normalized
so psi star psi d tau is one is one if the
wave function is normalized this your minus
sign sitting outside so you will get h cross
by four.
So you will get the relation that a that a
if I substitute that in a let me let me put
it that f star we had obtained that f star
f d tau multiplied multiplied by g star g
d tau is greater than or equal to one fourth
of this integral f star g plus f g star d
tau whole square. So you will obtain so you
obtain that that this was equal to if I remember
it rightly that 
one was g star g was equal to I have found
that out g star g was equal to x's square
and this was equal to p square. So as you
know that delta p is equal to under root of
p square minus p average whole square i'm
assuming p average to be zero therefore, this
is equal to I am taking the special case of
p average being zero, so this is square root
of p square and delta x is equal to under
root of x's square minus x average square
which is under root of x's square I am assuming
I can this I can always choose the origin
of my coordinates such that the x average
is zero. So this from if I substitute this
here I will get the uncertainty relation that
delta p delta x is always greater than or
equal to h cross by two 
this is the uncertainty relation that we have
proved from the first principles and let me
let me give you a few examples. So from the
Schwarz's inequality which is written here
we have and then by assuming that f is given
by g is given by i x I, g was equal to i x
i and similarly, we had assumed an expression
for f we have been able to prove that only
thing that we have assumed is that the expectation
value of p is zero and expectation value of
x is zero but, expectation of value of x zero
is does not mean anything because you can
always choose the origin to be such that the
expectation value of x is zero but, expectation
value of p is zero one can also generalize
this for the case when the expectation value
of p is not equal to zero.
Now let me go back having proved the uncertainty
principle let me go back to a simple wave
packet so the simple wave packet is a one
dimensional packet that we had considered
that psi of x is equal to I consider this
as a super position of plane waves under root
of two pi h cross minus infinity a one dimensional
wave packet a of p E to the power of i by
h cross p x d p. Now the expectation value
of p as we had discussed earlier is equal
to psi star p psi where p is the operator
representing p the variable p and as we all
know now that what we have been discussing
that this is equal to minus i h cross minus
i h cross psi star delta psi by delta x d
x so what I do is I substitute for psi star
here I substitute for psi here it is little
complicated but, if you work this out it is
worth it. So I psi star will be complex conjugate
of this so let us do do this carefully i h
cross and then two pi h cross will come here
two pi h cross. And now there are three integrals
one for psi star another for psi and the third
one is on d x, so please see this first integral
is over d x going from minus infinity to plus
infinity then we write down the expression
for psi star psi star will be two pi h cross
I have already taken outside so this will
be a star p a star p E to the power of minus
i by h cross p x d p the limits are from minus
infinity to plus infinity and then delta psi
by delta x so let me do it with a red pen
so two pi h cross, I have already taken outside
so I will have minus infinity to plus infinity,
if I differentiate with respect to x then
I will get a p inside so I get p times a of
p e to the power of i by h cross p x d p.
Now here I have to be very careful till now
it is all right if I first integrate this
and then integrate this and then carry out
this integration but, I am going to jumble
up the integration then this p has to be distinguished
from this p and therefore, I put a prime here
now I can write all the three integrals together
then this p prime will not get confused with
this p.
So let me write it down that that this p p
will be equal to minus i by h cross by two
pi h cross, I carry out this d p and then
d p prime collect all the terms which involve
only p and p prime and these are a star p
prime p a of p and then there is an integral
over x, let me write this down what is the
integral over x.
So the integral over x is minus infinity to
plus infinity d x e to the power of i by h
cross p x minus p prime x d x. So if I take
the factor to pi h cross this will be minus
infinity to plus infinity I am sorry I have
written d x twice so e by e to the power of
i by h cross p minus p prime into x into d
x and this as you know is delta of p minus
p prime. So therefore, in this integral in
this integral I have taken this factor here
and this factor this integral and I obtain
I will obtain minus I am sorry in I i that
is what I made a mistake that when I differentiated
delta psi by delta x I will not get just p
I will get I by h cross p so this will be
i by h cross sorry. So so i by h cross, so
that I i will multiply by i will become minus
one and this will become plus h cross h cross
I hope you understand what I am trying to
say that we had here delta psi by delta x
and when I differentiate this with respect
to x then I will get an i by h cross times
p into E to the power of I by h cross p x.
So I had forgotten the factor i by h cross
so i times i is minus one minus minus is plus
and the h cross will cancel out with this
and then so this two factors will go away
and this will become plus and this will become
two pi h cross.
Now this two pi h cross I have taken here
so I will obtain for expectation value of
p this remarkable result that minus infinity
to plus infinity d p and if I write down p
the a p star a star p prime then delta of
p minus p prime d p prime so if I carry out
this integration over p prime then this is
just a star p so I get this as minus infinity
to plus infinity d p p mod a p square.
So so we have for any wave function psi for
any wave function psi if I write as one over
two pi h cross integral a p e to the power
of i by h cross p x d p from minus infinity
to plus infinity then then the expectation
value of p will be given by psi star p psi
that is minus i h cross delta psi by delta
x into d x and if you we just carried out
the entire manipulation entire algebra and
we found that this is equal to p into a of
p mod square d p therefore. Similarly if I
calculate p square I will find that this is
equal to minus infinity I leave this is an
exercise you have to differentiate this twice
it is very easy it is very straightforward
and you will have a p square d p.
Thus we can interpret a p square d p as the
probability of finding the momentum between
p and p plus d p, I must add probability of
finding the x component of the momentum between
p x p and p plus d p therefore, this is also
physically obvious because what is an integral
integral is the superposition of this wave
functions and therefore, in this case e this
represents the momentum spread of the wave
function the the the quantity a p square represents
the momentum spread of the wave function.
So if a particle is localized within a distance
of the order of sigma if this is the wave
function then the corresponding momentum spread
function a of p mod square this is say mod
psi square the mod p a p square will be localized
also within a distance within momentums spreads
function will be of the order of h cross pi
sigma.
So therefore, let me we note down the two
terms back we wrote down for the free particle
the most general solution was given by this
of the time independent of the time dependent
Schrodinger equation and here a p represents
we have now physically interpreted a p represents
the momentum distribution function.
And this is how from the given form of psi
of x I can determine the corresponding a of
p by taking the inverse Fourier transform
and interpret psi of x mod square d x as probability
of finding the particle between x and x plus
d x and a p square d p will be the probability
of finding the particle or finding the momentum
between p and p plus d p.
I conclude today's talk by writing the wave
function which is psi x at t equal to 0 let
us suppose we wrote down that we we consider
this two terms back pi sigma zero square raised
to the power of four e to the power of minus
x's square by two sigma zero square e to
the power of i by h cross p naught x. If you
calculate this first of all you can easily
show that minus infinity to plus infinity
mod psi square d x is one so the function
is normalized I have done this quite a few
times I do not want to do. Secondly the expectation
value of x will be the will be x times mod
psi square d x so mod psi square will be e
to the I can write down under root of pi sigma
zero square and then x into x E to the power
of minus x square by sigma naught square the
mod square of this function. This is this
will become one and this will become e to
the power of minus x square by sigma this
limits are from minus infinity to plus infinity
and of course, this is an odd function of
x and therefore, the integral is zero.
So the expectation value of x is 0 the expectation
value of x square is x square here x square
here and here you can easily integrate this
in terms of gamma function and then you will
get one over under root of pi sigma zero square
two integral zero to infinity and then if
you put x square is equal to say y and then
carry out the integration you will obtain
this to be equal to sigma zero square pi two.
So we had we had x average is zero and x square
average is half sigma zero square therefore,
uncertainty in x will be under root of x square
minus x average square so this is zero, so
this will be sigma zero by root two this is
the uncertainty that if you make a measurement
of the x coordinate of the particle then it
will be I showed you a wave packet. So the
wave packet is such that that it is localized
within a distance of sigma zero.
Now for the wave function that I had written
down for the wave function I can write it
as I can write it as equal to one over two
pi h cross integral a p E to the power of
i by h cross p x d p and then take the inverse
Fourier transform of of the function.
And you will finally, get a p is equal to
sigma I leave that an exercise sigma zero
square raised to the divided by pi h cross's
square raised to the power of one by four
raised to the power of one by four e to the
power of minus p minus p 0 whole square sigma
naught square by h cross's square. So this
is if you plot the momentum distribution function
then you will find that p mod a p square d
p. This will be the average value of p this
will come out to be p naught and average value
of p square will be p square of this thing
and if you carry out this integration one
can show that this will come out to be p 0
square plus h cross's square by two sigma
naught square, so delta p will be this minus
square of this so delta p will be under root
of h cross square by two sigma naught square.
So this will be h cross by a root to sigma
naught, so if I take the product of the two
so delta p delta x it will be of the order
of h cross of two, so the uncertainty principle
the uncertainty principle is contained in
the solution of the wave function. So I end
by showing the wave packet that I had shown
last time that this is the let us suppose
I increase the so the particle is described
by this wave function which evolve with time
as shown in this diagram as shown in this
in the evolution of the wave packet the particle
is localized somewhere here and as it propagates
it broadens with time that I have not explicitly
shown. But you can you can carry out the calculations
and show that that how the wave function will
evolve with time and at each step the product
of delta x and delta p is always greater than
each cross by two. So we have considered the
the the definition the the of the uncertainty
principle we have proved the uncertainty principle
given a physical interpretation of the wave
function through the equation of continuity
and interpreted the wave function corresponding
to a Gaussian wave packet. Thank you.
