Good morning. In the last few lectures I have
told you that to describe the behaviour of
microscopic particles it is necessary to think
of them as waves. And the evolution the dynamics
the rules governing this wave is what is known
as quantum mechanics. And we were discussing
the rules of quantum mechanics. So, what I
told you in the last class was the first thing
that I told you in the last class was that
we have to represent the different states
of a system or a particle.
So, states are represented by different wave
function psi so, corresponding to every possible
state of a particle. There is a wave function
psi if I have a different state then I am
going to have a different wave function. And
I also told you that the wave function psi
is governed by the Schrodinger wave equation.
The wave equation governing the evolution
of this wave function is the Schrodinger wave
equation. So, let me again write down the
Schrodinger wave equation just to remained
you i h cross del by del t psi is equal to
minus h cross square by 2 m del del x square
del square x by del x square psi plus v psi.
So, this the wave equation governing the wave
psi the wave function psi in a situation where
the particle is restricted to move only in
1 dimension 1 direction which is the x direction.
And where the particle has a, is in a potential
which is represented by v so, the external
influences on the particle are there in this
potential v. And in the last lecture, we had
considered the situation where the potential
v is 0. That is v have a free particle on
which there is no external influence and for
this particular situation. We had used the
method of separation of variables defined
a solution to this find a solution for the
wave function.
And I also told you that any linear superposition
of these solutions of different such solutions
is also a solution of the Schrodinger equation.
Then I told you that corresponding to every
dynamical observable 
there is a Hermitian operator. So, for every
different possible state of the particle you
have a different wave functions psi. Now,
given a particle there are various possible
dynamical quantities that you can observe
for example the position is one of them. The
momentum is another the energy is a third
possibility so, these are all dynamical observables.
And in quantum mechanics where the state of
e the system is represented by a wave function
every dynamical observable is represented
by a Hermitian operator.
And I also told you what is a Hermitian operator
rather I did not tell you what is a Hermitian,
operator. But the key property of a Hermitian
operator which is the property which is important
for our purposes is the property that the
Eigen values of a Hermitian operator are real.
So, a so, we are not going into the definition
of a Hermitian operator as far as we are concerned
Hermitian operators are operators which have
the property that. The Eigen values of such
operators are real the definitions are little
more complicated and for this the purposes
of this lecture we will not go into that.
So, the second point let me repeat it again
is that corresponding to every dynamical observable.
There is an operator which acts on these wave
functions. And these operators corresponding
to dynamical observables are Hermitian operators.
The crucial property of Hermitian operators
is that there Eigen values are always real
I also told you the. What the operator corresponding
to the position is what the operator corresponding
to the momentum of the particle is. And what
the operator corresponding to the energy or
more precisely the Hamiltonian is and these
operators are all guaranteed to have real
Eigen values. So, this is what we had discussed
in the last lecture. Now, in today’s lecture
we are going to first we going to discuss
mainly focus on what happens when we make
a measurement.
So, the issue that we are going to discuss
is what happens when we make measurements
before going into this let there is a point
which I should make clear?
And it has to do with what I mean what do
we mean by a dynamical observable I have not
written dynamical here what do we mean by
a dynamical observable. So, for a particle
we have various quantities like it is mass
is charge etcetera all of which are attributes
of the particle they do not change. So, for
an electron it has a fixed mass by mass we
always refer to the rest mass of the electron.
It has a charge all of which can be measure
when the electron is at rest and these do
not change for an electron these are fixed
numbers. So, when I talk about observables
I do not refer to such quantities as the mass
or the charge I refer to dynamical quantities
which change as the particle moves around
and the position the momentum and the energy
are examples. So, now, let us get down to
the situation that we wish to discuss and
the question that we are going to discuss.
We going to start our discussion with is as
follows there is a particle which has a state
which is described by a wave function psi.
So, there is a particle which has a state
which is in a state which is described by
a wave function psi. So, what I can say is
that the particle has a wave function psi.
And let us consider a situation where we measure
the momentum of the particle so, we measure
we measure the momentum of the particle. And
the question is what is going to happen when
we measure the momentum of the particle. So,
there are 2 possibilities which we will discuss
there are 2 possible situations which can
arise when I make a measurement of the momentum.
The first possibility is that when psi so,
let we consider the first possibility which
I will refer to as A.
So, the first possibility is if psi is an
Eigen function of the operator corresponding
to momentum. So, the operator corresponding
to momentum is p and if the situation is such
that the wave function of the particle is
an Eigen function of the momentum operator.
So, if this wave function is an Eigen function
of the momentum operator let me p psi is equal
to p 1 psi. If this wave function is an Eigen
function of the operator p then the operator
p acting on psi will give me a number which
I have denoted by p 1 into psi. That is the
property that is what we mean by the statement
if psi is an Eigen function of the momentum
operator. So, if the wave function of the
particle is an Eigen function of the operator
corresponding to the quantity that we are
measuring which in this case is the momentum.
Then the measurement will give a value p 1
where p 1 is the Eigen value so, the measurement.
Is going to give the Eigen value p 1 and in
this measurement process the wave function
is not going to change so, let me write it
here wave function is unchanged.
So, if the question let we rewind you again
the question we are addressing is that we
have a particle which has a wave function
psi. And the question is what happens when
we measure the momentum of the particle. And
we are first discussing a situation where
the wave function psi is an Eigen function
of the momentum operator which is the operator
corresponding to the quantity. That we are
measuring if this psi is an Eigen function
of this operator with an Eigen value p 1.
Then the measurement is going yield a value
p 1 for the momentum the Eigen value is the
value that I am going to get when I measure
the momentum. And the wave function is unchanged
by the act of measurement.
So, if repeat the measurement if repeat and
I measure the momentum again.
I am again going to get the same value p 1,
because the wave is unchanged.
So, this is what happens when I have a particle
in a state which is an Eigen function of the
operator corresponding to the observable that
I am measuring. Now, let we next consider
a situation so, the situation is still the
same so, I have in I have the same I have
a.
Particle with a wave function psi and I measure
the momentum, but in this case B psi the wave
function psi is not an Eigen function 
of P. Before proceeding with this let me just
back track to the situation A which I was
discussing where the wave function is an Eigen
function of the operator p.
So, let we consider an example of this that
is something which I forgot to do. So, let
we consider an example.
The example which I am going to consider is
Psi x t is equal to e to the power minus i
by h cross E 1 t minus p 1 x so, this is the
wave function that we are considering and
the momentum operator. The operator corresponding
to the observable momentum I had told you
is minus i h cross del by del x. So, let us
see what happens when this operator acts on
this wave function so, when the momentum operator
acts on the wave function psi given over here.
What we have to do is we have to multiply
this wave function by minus i h cross and
then differentiate it with respect to x. Now,
when we differentiate this wave function with
respect to x we get the same function multiplied
by minus multiplied by i p by h cross.
So, we get the same function multiplied when
I differentiate we get the same function multiplied
by this is equal to minus i h cross. And the
derivative of this function is i p by h cross
into psi right that you can check easily if
you differentiate. This with respect to x
this is e to the power i by h cross into p
x the minus here and minus here give plus.
So, when i differentiated I will get e to
the power of i p by h cross into psi and this
p 1 into psi because there is a p 1 here.
And this gives me p 1 minus i into i is 1
h cross cancels out this is gives me p 1 into
psi. So, it is clear that this wave function
is an Eigen function of the momentum operator.
And it has an Eigen value p 1 the constant
that appears over here so, for this wave function
if I make a measurement of the momentum.
This wave function is an Eigen function if
I make a measurement of momentum I will get
the value p 1 and if i in this process of
measurement the wave function is unchanged.
So, if I repeat the experiment I will again
get the value p 1 if I repeated again I will
still get the value p 1 so, in this process
the wave function is unchanged.
Now, let us go back to the situation where
the wave function psi is not an Eigen function
of the momentum operator question is what
will happen now. So, let me now give you an
example of this and the example that I will
consider is as follows we have.
A wave function psi x t which is equal to
3 fifth e to the power minus i by h cross
E 1 t minus p 1 x plus 4 fifth e to the power
minus i by h cross E 2 t minus p 2 x. So,
note that this wave function is a super position
of 2 wave functions which looks like this
with different constants the constants being
e 1 and e 2 p 1 and p 2.
So, this itself is an Eigen function of the
momentum operator with Eigen value p 1 this
itself is an Eigen function of the momentum
operator with Eigen value p 2. The question
is now what happens when I make a measurement
of the momentum for this wave function which
is a combination of 2 Eigen function. This
wave function itself is not an Eigen function
of the momentum operator remember the wave
function that. We are dealing with now is
a superposition of 2 different Eigen functions
of the momentum operator. And this is not
an Eigen function of the momentum operator
the questions is what happens when I make
a measurement of the momentum. Now, what happens
is as follows this wave function is a superposition
of 2 Eigen functions each of which this one
has an Eigen value p 1 this 1 has an Eigen
value p 2.
So, what happens when I make a measurement
is that when I make a measurement of the momentum
for the wave function psi I can get one of
2 possible outcomes the p 1 or p 2. So, when
I make a measurement of momentum for a wave
function which is not an Eigen function of
the momentum operator then the value that
I get is one of the Eigen value.
So, what would what I have to do is I have
to express this wave function interns of Eigen
function of the momentum operator which is
what I have done here so, the wave function
psi which is not an Eigen function of the
momentum operator is a super position of 2
different Eigen functions. So, whenever I
make a measurement of the momentum I will
get either this value or this value.
I will get either p 1 or p 2, because this
wave function psi is a superposition of 2
Eigen functions with Eigen values p 1 and
p 2 we could have considered. The situation
where psi is a superposition of 3 Eigen functions
P 1 with value p 1 1 with value p 2 1 with
value p 3 in such a situation if I were to
measure the momentum I would get one of these
3 momentums when I make a measurement of the
momentum one of these 3 values. So, whenever
I make a measurement I will get one of the
Eigen values of the operator so, the act of
measurement.
The act of measurement is going to give me
an Eigen value of the momentum operator 
and the psi. So, the Eigen values the possible
Eigen values here are p 1 and p 2 and the
psi the wave function itself is going to change
if I get p 1 as my momentum. Then the wave
function after the measurement is going to
be psi 1 if I get the momentum p 2 as my result.
Then the wave function after the measurement
is going to be psi 2 the probability that
I get p 1 is the square of this.
Coefficient over here so, in this case the
probability of getting p 1 is 3 by 5 square
which is 9 by 25 so, the probability of getting
the value.
P 1 is 9 by 25 the probability of getting
the value p 2 is the square of this coefficient
which is 16 by 25?
So, let me summarize these here again.
So, there is a probability so, when I do a
measurement of the momentum 
there is a probability 3 by 5 square which
is nine by 25 that I get the value p 1. And
if I get this value p 1 then the wave function
becomes psi 1 after the measurement. There
is a probability 4 by 5 square that I will
get the value p 2 if I get the value p 2.
Then the wave function gets changed it is
becomes psi 2. So, depending on whether I
get p 1 or p 2 after the measurement the wave
function will becomes psi 1 or psi 2 where
psi 1 is this function. And psi 2 is this
function.
So, let me recapitulate the point and trying
to make the point I am trying to make is that
if I have a particle in a wave function which
is not an Eigen function of the operator corresponding
to the dynamical observable I am measuring.
Then the measurement will give me one of the
Eigen values of that operator. So, the momentum
operator for example, can have different Eigen
values the act of measurement is going to
give me one of the Eigen values. And I can
calculate the probability of getting any particular
Eigen value so, what I have to do is I have
to expand this wave function.
In terms of Eigen functions of the momentum
operator which is what I have done for this
particular wave function psi it is possible
in general to always expand any arbitrary
wave function in terms of Eigen functions
of these observable quantities. And now when
I make a measurement I will get one of the
possible Eigen 1 of the Eigen values as my
result. The probability of getting particular
e Eigen value is the coefficient in the expansion
in this case the probability of getting p
1 is the square of this. The probability of
getting p 2 is the square of this note that
I have chosen these coefficients. So, that
the some of these 2 probabilities is 1 in
case the sum is not 1 I can I have to multiply
the wave function with an overall number which
ensures to finally, ensure that. The sum of
all the probabilities is equal to 1 that is
called normalizing the wave function. So,
act of measurement returns a value which is
an Eigen value of the operator corresponding
to the observable. That I am measuring and
the act of measurement also changes the wave
function.
So, if I start with the wave function psi
and I make a measurement of the momentum if
I get a value p 1. Then after the measurement
I can say that the wave function as changed
and it has become psi 1. That is the Eigen
function corresponding to p 1 if instead I
had got the value p 2 I could say the after
the measurement the wave function is in the
as. Now, changed to psi 2 where psi 2 is the
Eigen value Eigen function corresponding to
this Eigen value.
So, the point here very important point is
the act of measurement changes the wave function.
It is only when the wave function is an Eigen
function of the operator that you are measuring
the variable that measuring that. The wave
function does not change in any general situation
if I make a measurement it will change the
wave function. And the wave function will
go over to one of the Eigen functions of the
operator corresponding to the observable that
I am measuring. And I have given you an example
with this particular wave function and where
we make a measurement of the momentum I can
have 2 possible outcomes p 1 p 2. If I get
the outcome p 1 I know that after the measurement
the wave function is this if I get p 2.
Then I know after the measurement the wave
function is this the, normalize the co efficient
outside as to be chosen to normalize the wave
function. So, I have told you what happens
in both the situations 1 where the wave function
is an Eigen function of the operator corresponding
to the observable that we are measuring. The
second where the wave function is not one
of the Eigen functions of the operator corresponding
to the observable. That we are measuring with
this background we can now go back. And interpret
the wave function that we had derived in yesterday’s
lecture, in yesterday’s lecture we had solved
a Schrodinger equation. And the solution that
we obtained for a free particle where there
is no potential we use the method of separation
of variables to obtain a solution and the
solution that we obtained was psi.
A, some normalization constant e to the power
minus I by h cross E 1 t minus p 1 x this
is the solution that we had obtained in the
last class. This is was the solution of the
Schrodinger equation and in this solution
A tilde was an overall constant out side in
addition to this. There was this constant
p 1 and this constant E 1 both of which were
related as E 1 is equal to p 1 square by 2
m. And I could arbitrarily chose either E
1 or p 1 E 1 had to be positive p 1 could
be negative. So, if I fix E 1 and p 1 was
fixed if i fix p 1 then e 1 was fixed one
of them is can be chosen arbitrarily. And
we had not gone into the physical significance
of these constants. Now, let us ask the question
what happens if I measure the momentum of
the particle if it is described by this wave
function. So, there is a particle which is
described by this wave function which we saw
was a solution to the Schrodinger equation
free particle described by this wave function.
Let us ask the question what happens if we
measure the momentum. Now, the operator corresponding
to momentum is minus i h cross del by del
x so, when this operator acts on psi it gives
me. So, what happens if when you differentiate
this exponential with x you pick up minus
I by h cross into p 1 i into h cross I into
p 1 by h cross i into minus i gives you 1.
The 1 by h cross into h cross gives you 1
so, what you have is this is equal to p 1
psi so, what you can say is that if a particle
is described by this wave function. And if
you make a measurement of momentum since this
wave function is an Eigen function of the
momentum operator. It will give you the value
P 1 it has an Eigen value p 1 so, whenever
you measure momentum you will find that you
will get a value p 1.
So, the momentum measurement of momentum will
always give you a value p 1 and this measurement
is not going to disturb the wave, function.
Similarly, if you measure energy 
the operator the Hamiltonian operator whose
Eigen values are the energy? So, strictly
speaking there is a Hamiltonian operator which
is which gives you the energy of the particle
the Hamiltonian is the energy of the particle.
So, the Hamiltonian operator is i h cross
del by del t. And this wave function is an
Eigen function of the Hamiltonian operator
the operator corresponding to energy h psi.
So, when you take the time derivative of this
you will get minus i by h cross E 1 minus
i into I will give you 1 is the 1 by h cross
into h cross will you give once what you get
is that this is equal to E 1 into psi.
So, when if you make a measurement of the
energy of the particle you will also get a
value you will get a value E 1. And this is
not going to disturb the wave function so,
if you repeat the experiment again. And measure
the energy you will get the again the same
wave function multiplied by E 1. So, it is
so, whenever you make a measurement of the
energy it has a well-defined value E 1 if
you make a measurement of the momentum it
has a well-defined value p 1. So, this wave
function corresponds to a particle with energy
E 1 and momentum p 1. Now, the energy of the
momentum are related like this which is what
you expect now what happens when you measure
the particles position.
Now, when the position operator acts on this
it basically multiplies this with x so, it
is clear that this function this wave function
is not an Eigen function of the position operator.
We have discussed what will happen when you
make a measurement of the position. The way
you interpret what you can say about, what
you can predict for a measurement of the position
is that the probability density of finding
the particle at some position is what you
can predict.
So, rho x t is going to give the probability
density of finding the particle at some point.
And this is equal to the mod of psi x t square
or you can write it as psi star x t into psi
x t now, let us look at the wave function.
That we have we have this wave function so,
when I take it is complex conjugate it will
not have this minus sign. And if I multiply
this function with this complex conjugate
this into its complex conjugate is going give
us 1.
So, what am going to get is that the probability
density of finding the particle somewhere
the probability density is a constant and
the constant is the modulus of A square. So,
it is equally probable that we find the particle
anywhere the probability density of finding
the particle at any position the same it does
not depend on x. And it is the same everywhere
so, this wave function has a precise refers
to a particle which has got a precise momentum
p 1 it has precise energy E 1. But, it has
you cannot make any prediction as to the particles
position. It is equally probable that the
particle is there anywhere in the whole of
space. So, this wave function has no position
information it has no information at all about
the particle’s position. So, this is the
interpretation of the wave function which
we had derived in yesterday’s class.
It has very precise information about the
particles momentum you can interpret this
constant as being the particles momentum you
can interpret. This constant as being the
particles energy, but it has no information
at all about the particles position momentum.
And energy are very absolutely well defined,
but the uncertainty in the momentum in the
position is infinite it has no information
about the particles position. So, the first
point is that this wave function is consistence
with a de Broglie with the de Broglie hypothesis
the de Broglie principle. And so, the Schrodinger
equation is a Schrodinger wave equation essentially
incorporates the de Broglie hypothesis. It
incorporate the prediction of de Broglie regarding
the relation between the wave number and the
energy the momentum the angular frequency
and the energy. So, it is consistent with
that we also find that this wave predicted
by the Schrodinger wave equation has very
precise energy and momentum, but it has no
position information.
So, this is quite consistent with the Heisenberg
uncertainty principle which we have which
I had told you about it tells you that. There
is a fundamental restriction to the accuracy
with which you can determine the position
and momentum of the particle of a particle
simultaneously determine the position and
momentum of a particle and the uncertainty.
In the position the uncertainty in the momentum
the product of these will be of the order
of h at least of the order of h may be more
could be more than this, but it cannot be
very less than this. So, the product the on
the uncertainty in the position and momentum
is going to be of the order of h in this particular
case the wave function that we have seen.
There is no uncertainty in the momentum this
uncertainty in the momentum is 0 as a consequence.
The uncertainty in the position has to become
infinite, because the product has to be of
the order of h cannot be less than this. So,
is this is 0 this has to become infinite so,
this uncertainty principle is an, is. So,
you see the wave the wave nature is basically
I mean if you want to have you if you want
the particle to have a wave to have to a to
be describe by a wave. It is absolutely essential
that it is a consequence basically that if
you want to have a wave picture there is going
to be an un uncertainty in the position. And
in the product momentum right let we illustrate
this with one more example consider a situation
where we have a.
Particle which is incident like this in this
direction so, we are supposed to in quantum
mechanics we know that we should thing of
this as a wave. So, we have a wave propagating
like this and the wavefronts now, we know
what the wavefronts look like. The wavefront
are going to be like this so, what we have
is a particle we will call this the x direction
what we have here is a particle incident along
the x axis. And we are considering all 3 we
are considering all 3 directions know.
So, we have a particle incident along the
x axis the particle is straight by a wave
function like this it is now a function of
x y and z all 3, but it depends on only on
x. So, we have a particle incident along the
x axis it has a well-defined energy. It has
a well-defined momentum but, we have a absolutely
no information about the particles position.
It is equally likely the probability of finding
the particle is all the same everywhere so,
this particle is propagating along the x axis.
Now, what we do is we introduce a slit over
here a slit of width delta y so, before the
particle is incident on the slit. The wavefronts
are now parallel to the y this we will call
the y axis this is the x direction the wavefronts
are parallel to the y axis. And probability
of finding the particle anywhere is the same
because it has no why dependence. So, the
mod square of psi is constant actually it
is a constant everywhere now, what happens
when the particle is incident on the slit
when the particle emerges. We know that the
uncertainty in y is of the order of delta
y, because the particle has been blocked here.
And here so, we know that when the particle
emerges the uncertainty in y has been reduced
and it has now of the order of delta y. So,
initially there is no uncertainty in the momentum
we know that p x has a value the x component
of the momentum has a value which is what
we have here.
The Eigen value here is the value.
The y and z component have value 0 so, there
is no uncertainty in the momentum, but there
is an uncertainty in the position it could
be anywhere. Now, when this emerges from the
slit when it is sent through a slit there
is now reduced uncertainty in the y direction
in the y position. And the uncertainty in
the y position is delta y, but as a consequence
of sending this through a slit we know that
when we send a wave through a slit. There
is going to diffraction and the wave that
emerges is going to be spread out over a range
of values. And we know that it is going to
be spread out over a value delta theta where
delta theta is of the order of lambda by the
width slit width which delta y.
This is something we have learnt in diffraction
so; the wave which was initially propagating
along the x axis when it comes out is going
to have a spread in direction. The spread
in direction delta theta is of the order of
lambda by the slit width delta y now if I
multiply this by the wave number of this wave.
So, k into delta theta is a spread in the
wave number in the y direction this is delta
k y. And this is equal to lambda k by delta
y let me also multiply this whole equation
so, this with h cross. And what it tells me
is that the spread in the momentum of the
particle in the y direction in this direction
is this is the spread in the momentum h into
delta k y so, the spread in the momentum.
Delta Py is equal to lambda.
Into the initial momentum k into h cross p
divided by delta y;
Now, the momentum of the particle and the
wavelength the de Broglie wavelength we know
are related as follows the de Broglie wavelength
is h by p. So, what this tells us is that
the uncertainty in the position and the uncertainty
in the y component of the momentum is equal
to h is of the order of the h. right.
So, what have we seen here we see that if
this particle if you represent a particle
as a wave initially this wave has no uncertainty
in the momentums the momentum is precisely
known. But there is absolute total uncertainty
about the position because we have no position
information. Now, what we have done is we
have send the wave through a slit. What the
slit is does is it localizes the particle
the moment you localize the particle you introduce
an uncertainty in the wave vector through,
because of the diffraction. And this uncertainty
in the wave vector gets converted to an uncertainty
in the y component of the momentum.
So, you have reduced the uncertainty in the
y component of the y position of the particle,
but, at the expense of introducing an uncertainty
in the y component of the momentum. And the
product of these two has to be of the order
of h that is what we see so, this is a consequence
of the diffraction. So, what we find is that
this uncertainty the fact that the product
of the uncertainty is cannot be below a certain
number is a direct consequence of the fact
that you have a wave you are representing
the particle by a wave. And if you want the
wave picture to be consistent this kind of
an uncertainty relation has to be valid right.
So, we see that the uncertainty, the uncertainty
relation and the wave nature are at both are
both have to be if you have the wave picture.
Then there should be uncertainty relation
you cannot determine the position and momentum
of the particle both to arbitrary level of
accuracy simultaneously right. So, let us
know go back to the point which were discussing
the point being what happens when we make
a measurement.
So, if I let me know consider a general situation
if I have some A particle in a state where
it has a wave function psi. And I have an
observable o which is represented by an operator
o. So, I am making a measurement on this wave
function I am measuring a quantity corresponding
to which there is an observable. There is
an operator o then if I make this measurement
the outcome of this measurement is going to
be one of the Eigen values of this operator.
And if I can represent this psi as a superposition
of different Eigen functions of this operator
which I can always do it is a theorem that
can always represent any arbitrary wave function
in terms of Eigen functions of this physical
observable. Then if I can represent it as
say c 1 psi 1 plus c 2 psi 2 psi 1 and psi
2 are Eigen functions of this operator psi
1 has an Eigen value o 1. So, psi 1 is an
Eigen function of this operator with Eigen
value o 1. Psi 2 is an Eigen function of this
operator with Eigen value o 2. So, we are
considering a situation where there is a particle
in a state psi which can be represented in
terms of these 2 Eigen functions of the operator
o.
In general there may be 3 4 5 you may require
many more Eigen functions of this operated
to represent this wave function, but you can
always represent it as a superposition of
Eigen functions of this operator the general
situation. So, if in this particular case
I can represent psi as a superposition of
2 Eigen functions 1 with Eigen value o 1 another
with Eigen value o 2 when I make a measurement
of this observable I will get either o 1.
And the probability of getting o 1 is c 1
mod square by c 1 mod square plus c 2 mod
square and the probability of getting o 2
is similarly c 2 mod square divided by c 1
mode square plus c 2 mod square that is the
general situation. Let me know consider a
different question again to do with measurement.
And the question is as follows the situation
is as follows suppose I have a particle in
a state psi.
And I have many replicas of this particle.
So, I have many replicas of the particle I
will draw few of them. And we the question
the situation that we are going to consider
is that I have this particle many replicas
of this particle all in exactly the same state
psi and what I do is I measure the momentum
of the particle. And this state psi is not
an Eigen function of the momentum operator.
So, when I make a measurement of the momentum
let us say that I get a value p 1 here. If
I make a measurement of the momentum I will
get I could get a different value p 2 I will
get the say the first value p 1 here I could
get a different value p 3 here I get p 1 here
I get p 2 p 1 p 2 p 2. So, the situation that
I am considering is as follows I have particle
which is in a certain state psi I have many
replicas of this particle all of them are
in exactly the same state psi.
Now, I go and measure the momentum of the
particle is exactly identical and I have many
replicas of it, but it is not an Eigen the
state psi is not an Eigen function of the
momentum. So, I cannot predict exactly what
outcome I am going to get what I can predict
is this probability of different outcomes
the outcome is going to be an Eigen value
of the momentum operator. So, if I do the
experiment I can predict the what the probability
of getting a particular Eigen value is and
suppose I do it then I will get one of the
Eigen value. So, here I get p 1 a possible
Eigen value of the momentum operator here
I get p 2 which again is an out possible out
Eigen value of the momentum operator here
I get p 1 again here I get p 3 etcetera.
Now, the question is what is the expectation
value or the mean value of the momentum or
the mean value? So, if I do the experiment
if I have many many replicas I can predict
what the mean value I can calculate what the
mean value is going to be if I get p 1 10
times if I get p 2 3 times if I get p 3 4
times. I will multiplied 4 into p 1 plus whatever
10 10 into 4 p 1 plus 4 into p 2 etcetera
and divide by the total number that will tell
me the mean value. The question is how can
I predict what the mean value of the momentum
should be from the wave function? So, the
way to calculate the mean value or the expectation
value the value that expect to get if I do
the experiment only once the way I calculate
that.
This is what we call this the expectation
value or the mean value is through this integral
minus infinity to infinity psi star. The momentum
operator p into psi dx this integral gives
me the makes a prediction for what the mean
value is going to be.
I can now do the experiment have many replicas
of this particle in the state psi. And then
measure the value of p that I get in all of
these experiments. And then calculate the
mean I am going to get p 1 some number of
times p 2 some number of times p 3 some number
of times. So, I am going to get all the possible
Eigen values some number of times and then
I calculate the mean.
I can predict what the mean should be from
the wave function by doing this integral.
So, the integral that I have to do is as follows.
I have the wave function psi which the particle
that is the state in which the particle is
I act with the momentum operator remember
the momentum operator is minus i h cross del
by del psi i multiplied by psi star. So, this
is a function of x I have to integrate this
from minus infinity to infinity. This integral
is going to give me the the mean position
or the expected position of the particle.
Let me stop today’s lecture over here and
take up the question of how to calculate the
uncertainty in the momentum and related issues
in the next lecture.
