So, we shall continue the particle in a 2D
box, but for the moment let us consider a
little bit on this famous principle called
The Uncertainty Principle which was first
put forward by Werner Heisenberg .
Now, there is a very beautiful lecture on
Heisenberg's uncertainty principle by Professor
V. Balakrishnan and it is there in the NPTEL
website under basic courses or in physics
, this is on quantum mechanics. The very 1st
lecture is on Heisenberg's uncertainty principle.
I would like everyone, I would like to recommend
that to every one of you to go through that
lecture, but this is very preliminary. It
is not anything like what was there, but you
would appreciate that lecture far more when
you listen to Professor Balakrishnan's account
of how Heisenberg's uncertainty principle
is to be understood.
We will do a much simpler exercise since you
are beginning. This is meant for the introductory
very 1st year students .
Now, uncertainty delta x in any measurement
, measurement quantity x is given by this
simplestatement that it is the difference
between the average of the square of that
variable minus the square of the average square
of the average of that variable and this whole
thing is under a square root, ok. This is
the angular brackets tell you that the average
value .
What is inside is the one for which the average
is taken and therefore, the average is taken
for the square of that value x. Here the average
is taken for the value itself and then, it
is squared the difference between the two.
The square root of this is called the uncertainty
. Average of the square minus square of the
average that is the this I do not know how
to say it in English. It is the square root
, ok or you can write within bracket square
root likewise the uncertainty. This is for
the position variable and this is for the
momentum variable. I have introduced this
in a separateaccount. I might tell you how
this formula comes about and so on , but let
us just introduce these things as defined
in textbooks the delta p is again the average
of the square of the momentum minus the momentum
square, ok .
Delta x delta p the product of the two is
greater than or equal to h bar by 2 , ok . This
is the Heisenberg's statement about the uncertainty
between x and p. What it means is that if
for some preparation of the states, we are
able to minimize this by making sure that
this average and this squared average are
very close to each other. Therefore, we are
able to measure the position very very very
accurately. If we do that what uncertainty
principle tells you that is in the denominator
. Therefore, the uncertainty in delta p is
very large . It is not possible for us to
control the uncertainties to both of them
to absolute minimum except not to violate
this particular relation, ok.
Therefore, this is one of the statements thatyou
might see in textbooks very often regarding
theuncertainty in the position measurement
and uncertainty in the momentum measurement.
What it also means is that position and momentum
cannot be simultaneously used as variables
for describing the state of a particle as
as independent quantities for describing the
state of the particle. The state of the particle
can either be very precisely stated using
the position or very precisely stated using
its momentum, but not both and therefore,
this brings down the whole structure of classical
mechanics where one would imagine in the solution
of the Newton's equation, the precise statement
for the position and velocity of a particle
at one instant of time and be able to solve.
Therefore, if you can specify the velocity,
obviously you can also specify the momentum
of the particle. Therefore, position and momentum
can be simultaneously used as descriptors
for defining the state of a classical particle,
but they cannot be used as descriptors for
the state of a quantum particle and the relation
between the two is given by this famous Heisenberg's
uncertainty principle and Professor Balakrishnan's
lecture tells you how to generalize the Heisenberg's
uncertainty statesin using other classical
formulations and eventually what is known
as the commutator, ok .
Now, let us use the wavefunction psi of x
for the one-dimensional box root 2 by l sin
pi x by l . We will take the n equal to one
case quantum number and if we try to calculate
the average value x for the particle in this
state whose wavefunction and the probability
of the particle at various points is symmetrically
, the same on either side of l by 2 , ok.
It should be immediately clear that the average
value for the particle position given that
these are the probabilities for the particles
position being here or here or here or here.
By looking at this being a symmetrical graph,
you can immediately say x should be l by 2
, but that is also the expectation value or
the average value . This is called the average
value in quantum mechanics for any variable
A in the state , psi is given by psi star
A on psi d tau which is the volume element
or the area element or the length element
similar to whether it is a one-dimensional
box or two or three dimensional divided by
the integral psi star psi d tau , ok. This
is a postulate . I do not want totell you
how this can be arrived at using arguments.
You will find such things in physics books,
but for the particular course that you have
started taking, this is the postulatory introduction
for the expectation value of any variable
A whose corresponding representation as an
operator is given by this A hat and A hat
is between the wavefunction psi and the complex
conjugate psi star if psi is a complex function,
otherwise both of them are psi.
This prescription must be kept in mind. This
is introduced as a postulatory form and let
me calculate the x , but the particle it is
very easy now .
Therefore, the average value x is given by
the integral root 2 by l root 2 by l because
it is psi star psi and you have sin pi x by
l x sin pi x by l dx between 0 and l for the
particle in the quantum state with the quantum
number 1 which is what we call as psi 1 , ok
and x of course, does not change anything.
I mean it simply multiplies to this, therefore
this integral is 2 by l 0 to l sin square
pi x by l multiplied by x dx . Calculate this
integral and show that the answer is l by
2 ok . That is for you to do the exercise
.
What about the momentum ? You have to be careful
in ensuring that the momentum operator which
is a derivative operator is placed as written
here, namely 2 by l that comes from the two
constants psi star psi , then you have sin
pi x by l between 0 to l and the momentum
operator is minus i h bar d by dx acting on
sin pi x by l dx . See that the operator is
sandwiched between the wavefunction and the
complex conjugate of the wavefunction . Here
the wavefunction is real, therefore you do
not see the difference between the two.
What is this ? It is very easy to see that
this will give you the, derivative will give
you cos and sin cos will give you a sin 2
pi x by l and that in this interval is actually
0 . What about the average value p square
? The average value p square is given by 2
by l . Again sin square sin pi x by l and
now you remember it is minus h bar square
d square by dx square for the operator p square
sin pi x by l dx and it is between 0 and l
. did not write the denominator because we
have chosen the wavefunction by ensuring that
the wavefunction isthe integral of the square
of the wavefunction is actually one in the
entire region. Therefore, I did not write
the denominator that is 1.
This of course you know is nothing, but 2
m e . The total energy this is p squared on
the wavefunction. You remember p square by
2 m on the wavefunction gave you e. Therefore,
this is 2 m e, therefore you see that p square
is immediately given by the energy that we
know . You can write that , ok. What about
d x square? If I have to do x square, all
I need to do the same thing . Write x square
on psi 1 and I have the integral that needs
to be evaluated is integral 2 by l sin squared
pi x by l times x square dx .
Therefore, you know the value of x square,
you know the value of x , you know the value
of p and 0, you know the value of p square
has nothing, but 2 m E . This is the only
integral that I have not calculated. Once
you have done that, you can calculate .
Delta x delta p has nothing, but the square
root of p square minus p. Of course, you know
that is 0 times x square minus x whole square
, ok and you should be able to verify that
this answer is greater than or equal to h
bar by 2, ok. So, this is the statement of
Heisenberg's Uncertainty Principle for the
particle in a one-dimensional box.
Now, exactly the same statement can be I mean
it can be extended to particle in a two-dimensional
box except that now you have x and y as two
independent coordinates, px and py as two
independent coordinates. Therefore, you have
a corresponding uncertainty relation in two
dimensions with one exception, namely x and
y are independent coordinates. Therefore,
x and p y can be simultaneously measured or
can be described as a property to the system
y, and p x can be simultaneously specified
for the particle x, and y can be specified
px and py can be specified, but not x and
px and y and py. That is the only thing you
have to remember. The independence of the
the degrees of freedom ensures that the operators
corresponding to those degrees of freedom
commute with each other and if I have not
spoken to you much about commutation that
will be in the next lecture, but in this part
I would simply want you to calculate the Heisenberg's
uncertainty principle as given this is one
simple way of doing it. You can find similar
treatments for the uncertainty when you go
tostudy the other systems like Harmonic Oscillator,
Hydrogen Atom and so on.
What is key to remember is the definition
for the delta x i gave you and the definition
for the delta p i gave you. Those are fundamental
I have not told you where they come from.
Maybe in a separate lecture or in the class
when we discuss these things through elaborations,
I will tell you what the origin of the delta
x and delta p, but these are definitions which
you have to start with working and then, feel
more comfortable. Go back and look at the
whole process of the derivation .
We will continue this exercise to complete
what is known as the introductory, but postulatory
basis of quantum mechanics for this course
in the next part of this lecture which is
the 3rd part for the particle in a two-dimensional
box. With that we will complete the two simple
models particle in 1D and 2D box. We will
meet again for the last portion of the particle
in 2D box lecture the next time.
Thank you
