In the last lecture, I had spoken in some
detail about the 3rd postulate of quantum
mechanics which pertain to measurement, basically
expectation values and Eigen values of operators
which represented observables. We spoke about
measurement outcomes, repeated measurements
and how exactly expectation values are obtained
and related to measurement outcomes in an
experiment.
Now, in this context a very important question
arises. How well can you make simultaneous
measurements of two observables? You can always
make simultaneous measurements of observables.
Question is; can you measure both of them
simultaneously, to arbitrarily high precision?
And this is really the content of the uncertainty
principle, primarily due to Heisenberg, and
it is the uncertainty principle, that I will
discuss in some detail today.
So, the topic today is the uncertainty principle.
The history of the uncertainty principle is
rather interesting. Dirac and Jordan had proposed,
the transformation theory. The transformation
theory by Dirac and Jordan, was really based
on the theory of linear vector spaces. And
the theory helps one understand both wave
mechanics due to Schrodinger, primarily and
matrix mechanics primarily due to Heisenberg,
within the mathematical structure of liner
vector spaces. This was due to Dirac and Jordan.
Now Heisenberg, in an attempt to understand
transformation theory, had a glimpse of the
uncertainty relation in the following sense.
He realized that it was not possible to simultaneously
measure, to arbitrarily high level of precision
certain observables. For instance, the position
of an object along with the corresponding
linear momentum cannot be simultaneously measured
to an arbitrarily high level of precision.
So, he constructed a Gedanken experiment,
a thought experiment to prove this point.
Now, the thought experiment was all about
a gamma ray, electron microscope, where the
gamma rays interacted with the electrons.
Except that the experiment as conceived by
Heisenberg, assumed that the interaction was
like collisions between mechanical objects,
which is not quite the truth.
However, although the experiment itself as
conceived in its original form by Heisenberg
is not precise. The outcome of the experiment
is certainly true, and it is really to the
attributed to Heisenberg, the fact that simultaneous
observation in quantum mechanics is very different
from simultaneous observation of two variables
in classical physics. So, it was in 1927 that
Heisenberg suggested this, that there was
this relation between position and linear
momentum and so on. But, in 1927 later Kennard,
actually produced a very precise mathematical
expression for this uncertainty relation.
The Kennard relations talks of from the fact,
that x and P x position and the corresponding
linear momentum, do not commute with each
other.
I had spoken about this towards the end of
my last lecture and just using the fact that
this is the commutation relation, between
X and P x. Kennard showed, that the variance
in x if the variance in x is delta x square,
and the variance in P x, is delta P x square,
where the variance of course, is defined as
the expectation value, of X minus the average
value of X the whole squared and so on. So,
this can be easily checked to be expectation
value of X squared that is the first term,
minus expectation value of X the whole squared.
So, this is the variance similarly, I can
give a definition for P sub x.
What Kennard showed was that delta X delta
P x, was greater than or equal to h cross
by 2. This is a very interesting statement.
Because, however good the experimental measurements
may be however good the instruments may be,
there is always a minimum uncertainty delta
X delta P x equal to h cross by 2, in some
state, which is a minimum uncertainty state
and for other states it is greater than this
minimum value.
In 1929 Robertson, derived this uncertainty
relation in a more general setting. Where
two operators a and b were considered, Hermitian
operators. Because, they represent observables
and Robertson showed that delta A delta B,
was greater than or equal to half. The modulus
of the expectation value of the commutator
of A with B It is pretty clear,. that if A
and B commuted with each other then this quantity
became 0. And you have delta A delta B, greater
than or equal to 0. But, in general this is
true.
In 1930, Schrodinger generalized it to give
his uncertainty principle, which can be stated
in this fashion. Certainly the first term
is present. And then there is an addition,
which involves the anticommutator, of a minus
expectation value of A, with B minus expectation
value of B. The anticommutator is defined
between two operators A and B, as A B plus
B A. Sometimes one uses the notation, A B
plus for the anticommutator and just A B for
the commutator, the square bracket A B for
the commutator and A B with the plus out there
for the anticommutator. So, you see there
are two terms that contributed to delta A
square delta B square. These are the variances
in A and B in this particular state.
So, it is clearly a state dependent statement.
And in a given state the product of these
variances is greater than or equal to this
contribution one from the commutator and the
other from this anticommutator. So, since
these are all positive quantities, it is pretty
clear that, delta A square delta B square
is definitely greater than the first term
and that is why you get the relation which
Robertson gave. So, this is due to Schrodinger.
This is a Schrodinger’s uncertainty principle.
In order to derive this uncertainty principle
and study the various ramifications of the
principle, one really starts with the Schwarz
inequality. So, i will first derive the Schwarz
inequality and then proceed to explain the
uncertainty principle in this context.
So, Schwarz inequality is merely this statement.
Suppose, you have two states psi and phi,
norm of psi times norm of phi is greater than
or equal to the modulus of the inner product
of psi with phi. It is clear that if phi with
the null vector, or if psi with the null vector,
then the equality holds, because this side
is zero and so is this side. But, in general
the nontrivial statement would pertain therefore,
two vectors which are not null vectors. Let
me recall that the norm is the positive square
root of the inner product of psi with psi.
It is a length of the vector. So, generalization
of the modulus or the length of a three vector,
a vector in usual three dimension space.
So, let me first consider, this object ket
psi plus a phi, where a is some constant a
complex constant Now clearly the bra is this.
These are states in the linear vector space.
I wish to find the inner product of psi plus
a star phi with psi plus a phi. Now, this
object is clearly the inner product of a state
with itself.
And therefore, is greater than or equal to
0 and since these are not null vector psi
and phi, it is greater than 0. In any case
this object, in general is greater than or
equal to zero or I can expand this and I therefore
have the following, if i did a term by term
expansion. This quantity is greater than or
equal to 0, a is an arbitrary constant. Therefore,
I choose a specific a. Let me choose a to
be this object, minus phi psi by the inner
product phi phi. So, all I have to do is substitute
for a and a star, it is clear that a star
is minus of psi phi, divided by phi phi.
So once I substitute, I have psi plus a star
phi or anyway that is greater than 0. 0 is
less than or equal to, psi psi substitute
for a. This object is in general a complex
number and this is its complex conjugate so
that is like z z star. The 3rd term is the
same as the 2nd term because these are numbers
and therefore, they become commuted across.
Then I have another term modulus of a squared
which is a a star. That multiplies phi phi.
So, I can score that out and now i multiply
throughout, by this inner product. That is
from the first term, I am just clubbing the
2nd and the 3rd terms 
and the 3rd term is simply this, the same
as the 2nd term, half of the 2nd term. And
therefore, I have 0 is less than or equal
to this is, norm psi square, multiplied by
norm phi square. Between these two I just
have minus psi phi, with phi psi, and that
is simply the modulus of psi with phi the
whole square. I could have well written this
as phi here and psi there, I should write
this clearly that is a psi ok. Therefore,
taking it to the other side, I have got Schwarz
inequality.
Therefore, norm psi norm phi, is greater than
or equal to modulus of the inner product of
psi phi, and that is Schwarz inequality. With
this is something we use in order to derive,
the uncertainty principle, which is what I
let him to do now.
For this purpose, I consider two operators
A and B. Hermitian operators, representing
observables. I look at this state A minus
expectation value of A, acting on psi. If
I want to find a variance, that is the expectation
value of a minus expectation value of A, the
whole square. So, I can well write this as
psi and this is Hermitian so this is what
it is. I could called this object psi 1, just
for notation and therefore, this is psi one
psi 1 inner product which is norm of psi 1,
the whole square. Similarly, I can construct
a state psi 2. So, delta B square is obtained
in the same manner, as I did delta A square
and that is psi 2 square. The idea is to find
out what is delta A square delta B square.
In other words, norm psi 1 square norm psi
2 square, so this is like psi 1 and that is
like psi 2. And therefore, I use this Schwarz
inequality right way.
And because of this, I know that norm psi
1 square norm psi 2 square, it is greater
than or equal to the modulus, of the inner
products of psi 1 with psi 2 the whole square.
This is a complex number. It has a real part
and an imaginary part. So, the real part of
this complex number 
is this. And the imaginary part, so this complex
number can be written in terms of its real
part and its imaginary part. We wanted modulus
squared, the modulus squared is clearly going
to be the real part squared plus a imaginary
part squared. So, let me consider that.
So, if i want the modulus square that is as
good as saying, has two parts. So, if we consider
first the real part. What i expand it, it
is simply half the inner products psi 1 psi
2 plus psi 2 psi 1, plus B minus expectation
value of B, A minus expectation value of A.
You could put hats on all these objects to
show that they are operators. But, in this
notation it should be pretty clear, what the
operators are. But, this is simply half of
anticommutator of A minus expectation value
of A, with B minus expectation value of B.
We are interested in the square of this and
therefore this is quarter. Modulus of the
expectation value of the anticommutator of
A minus expectation value of A, with B minus
expectation value of B, squared. Which is
the structure that we see here, this term
from the Schrodinger uncertainty relationship
has already appeared, we have accounted for
the anticommutator. So, next thing is to look
at the commutator which is the imaginary part.
In a similar manner, we can show this, commutator
of A with B, the expectation value of that
mod square.
So, this really comes from the imaginary part
of an appropriate inner product and this comes
from the real part of the inner product. So
putting the two together I get dealt A square
delta B square is greater than or equal to
this term plus that term. And that is the
proof of the Schrodinger uncertainty principle.
Now it is good to discuss certain cases in
this setting. First of all, let us go back
to the two level atom model. And since the
uncertainty relation involves commutators,
it would be good to look at objects which
do not commute.
So, I first example would be the two level
atom, and just to remind you S plus takes
us from ket g to ket e, and S minus takes
us from the excited state down to the ground
state, and S plus is S x plus i S y, and s
minus is S x minus i S y. I want to look at
an Eigen state of S z. So we know that these
two are the bases states which are Eigen states
of S z. And just to refresh your memory, these
are the corresponding Eigen value equations.
We also have a commutator algebra, in terms
of S plus S minus and S z, they look like
this.
So, I am interested in finding out delta S
x squared delta S y squared. The product of
the variances in S x and S y and the state
that I consider, is an Eigen state of S z,
say ket e. So, I use the uncertainty relationship.
Now, the first term is the commutator, so
look at the left hand side of the uncertainty
relationship, first of all the commutator
of A with B.
In this case it is the commutator of S x with
S y, which is i h cross S z. I want to find
the expectation value of the commutator of
S x with S y, in the state e and take the
mod square of that expectation value. That
is the same as modulus I pulled out the i
h cross, S z e gives me, I can forget the
i because I am taking the modulus. So, that
gives me an h cross the S z gives me another
h cross by 2 and this is to be squared. So,
I basically have h cross to the power 4, out
there and there is an h cross by 2 here and
therefore, I have h cross to the 4 by 4. So
this is what I have.
So this is the contribution from the first
term. So delta S x square delta S y square,
is greater than or equal to quarter of this,
plus h cross. Now, this comes from the anticommutator.
This is a contribution from the anticommutator,
of S x with S y. So, let us find out what
that is?
I wish to find out the expectation value,
of S x minus expectation value S x, in this
same state e. The anticommutator of that with
S y minus expectation value of S y in that
state, and then of course the mod square and
so on. First of all in the state e, the mean
value of S x is zero. The simplest way of
seeing this is to write S x as S plus plus
S minus. And since S plus acts on e to destroy
it, there is no contribution from that term
and S minus acts on e to give me g, ket g
and ket e and ket g are orthogonal to each
other. Therefore, there is no contribution.
Similarly, this average value is also 0. So
these terms drop out and all I have, is to
find out the modulus of the anticommutator,
its average value and square it. I need to
estimate what is S x, S y plus S y, S x. Now,
apart from factors, numbers which I can pull
out. This quantity is S plus plus S minus,
S plus minus S minus, plus S plus minus S
minus, with S plus plus S minus. So, the first
term is an S plus squared which doubles up.
So, I have a 2 S plus square but when S plus
acts on e it destroys it and therefore, there
is no contribution from there.
Similarly, there is a minus S minus squared
and that comes twice over but again there
is no contribution from that, because s minus
acting on e twice. First time gets it down
to g and the next time destroys it. Then there
are cross terms, there is an S minus S plus,
plus S plus S minus and that cancels out.
Then there is an S plus, S minus with a negative
sign and that cancels out with the term here
and therefore, this contribution is 0. Does
it turns out that the anticommutator does
not contribute at all and any contribution
that comes, to the right hand side of the
uncertainty relation is from the commutator.
And so I have delta S x, delta S y it is greater
than or equal to h cross squared, by 4. So,
here is an example, where I have objects that
do not commute with each other and therefore,
I have a minimum uncertainty value of h cross
square by 4. This happens in the state e,
which is an Eigen state of S z. You can construct
states where it is greater, where this uncertainty
product is greater than h cross square by
4. It would be a good exercise to see, what
is the value of this product delta S x delta
S y? In the other base state ket g, as also
in super positions of ket e and ket g. Now,
let me look at another example. My next example
is to find out again in this two level model.
The uncertainty product delta S x, delta S
z, in the Eigen state of S z, that is ket
e. So, what is it that I have here? This is
an Eigen state of S z. Therefore, delta S
z is equal to 0. And therefore, delta S x,
delta S z, is equal to 0. But, that does not
been that delta S x is 0, in this state ket
e. This is in fact finite and we can check
this out. So, what is that variance of S x
in this state?
So, delta S x squared, in the state ket e,
is expectation value of S x squared, minus
expectation S x the whole square. This is
0 as we have seen earlier but this object
is not 0. There is a spread in the measurement
of S x. We can compute that right away. Therefore,
S x square this quarter, S plus, plus S minus
times S plus, plus S minus. Now, it is clear
that when these operators are sandwiched,
in the following manner, between the states
ket e and bra e. This does not contribute
and this does not contribute. So the contributions
are 0 from here. But, this task because S
plus, S minus out here S minus acting on e,
gives me g and S plus acting on g, takes it
back to e and there is an e on this side.
Similarly, here I have an S plus acting on
e which will annihilate and therefore, not
make a contribution. But, I do have this term,
which states. So, essentially this is quarter
expectation e S plus S minus e and S minus
e takes it to g, but S pus takes it back to
e. So, apart from a number which is non zero,
there is a non zero value for expectation
S x square. And therefore, delta S x the whole
square, is not equal to 0, although, the mean
value of S x in the state ket e is equal to
0. The following points emerge. We have an
uncertainty product of this kind. That if
you make a measurement of observables A and
B. Look at their average value their variances
and so on. Then the product of the variances
satisfies this relation, there is an inequality
here, could become equal for some states.
We have seen the following cases. If delta
A delta B product, is greater than zero, then
clearly the contribution has come because
this or these two are non zero. It is clear
therefore, that delta A is not equal to 0,
and delta B is not equal to 0. So, this was
the first example we had. Example: A was S
x and B was S y and the state considered was
ket e, which was an Eigen state of S z. Suppose
delta A delta B is equal to 0, which is what
we had in the second example, A was S x and
B was S z. So, delta S z was equal to 0. The
state considered was ket e, it does not mean
it does not imply that delta A is 0 and delta
B is zero, it does not imply this at all.
It simply means that one of them 0 in general
and that is what we have seen in this example.
Now, it is possible that delta A, delta B
greater than 0, equal to a finite quantity
and A, B commutator is not equal to 0. And
that is how the contribution came. It is possible
to make delta B 0, which means you measure
B with infinite precision. In that case the
spread in A is going to be very large, such
that the product of 0 and infinity gives me
a finite quantity. So, it is possible to be
in an Eigen state of B, such that there is
infinite precision in the measurement of the
B. But, then you have really very little idea
about the actual value of A, when you measure
it in that state. So, I look at another example
right now and that has to do with the position
momentum, uncertainty relationship. Relation
which is what Heisenberg gave us and Robertson
formulated precisely.
In that case, looking at the general uncertainty
relationship I have variance, delta X square
delta P x square, is greater than or equal
to quarter of the modulus of the expectation
value in the state considered of the commutator
of x with P x, whole square plus quarter,
of the anticommutator expectation value, the
anticommutator being X minus expectation value
of X, with P x minus expectation value of
P x, in that state. Now, if you get this,
X p x is i h cross identity and therefore,
delta X square delta P x square, is greater
than or equal to quarter h cross squared,
telling us definitely greater than this because
this could also contribute perhaps.
So delta X, delta P x is greater than or equal
to h cross by two. The minimum uncertainty
state is 1, corresponding to which delta X
delta P x, is equal to h cross by 2. And we
will see that the ground state of the oscillator
for instance, is in minimum uncertainty state.
In optics zero photon state is a minimum uncertainty
state and so on, minimum uncertainty in these
variables.
In order to actually make this estimation
and given example a concrete example. I have
to consider a specific system and it will
be good right now to move on to a slightly
more complicated system, the system which
does not have two discrete levels, as its
Eigen spectrum for the Hamiltonian, but an
infinite number of discrete levels for the
Eigen spectrum. In other words, it would be
good to take an example; which pertains to
an infinite dimensional linear vector space.
A discrete infinity of levels, the simplest
thing that one can consider is a multilevel
system, where the Eigen vectors are equally
spaced and indeed this is a situation with
the harmonic oscillator. So, it is a good
thing now to consider the next level of complexity,
which is an infinite dimensional linear vector
space but with a discrete equally spaced energy
Eigen spectrum.
So, if you consider this simple harmonic oscillator
and quantize it, before we actually proceed
to do the problems, solve the problem explicitly,
we take cognizance of the following. A classical
Hamiltonian assuming that the oscillator is
in along the X axis. This P x square by 2
m, oscillator of mass m, plus half m omega
square X square. I am going to drop the suffix
x, simply because there is no other momentum
in consideration. In terms of operators, that
would be p square by 2 m, in quantum physics
this is h classical, h quantum is an operator,
plus half m omega square, X square. We will
see that there are an infinite number of energy
levels for this Hamiltonian. When you quantize
a system, the ground state of this system
happens to be a minimum uncertainty state
in X and P x, and if you wish to give a position
representation to the ground state. It will
turn out to be a Gaussian, a Gaussian in X,
as also a Gaussian in P. Simply, because I
will establish shortly in the course of the
following lectures.
That the position space and the momentum space
are Fourier transforms of each other. And
therefore Fourier transform of a Gaussian
being a Gaussian, when you go to the momentum
space the structure of the wave function the
form of the function as a function of momentum
does not change. So, the Gaussian turns out
to be a minimum uncertainty state. Now, X
and P happen to be infinite dimensional matrices.
I could construct operators out of these which
would take me from one energy Eigen state
of the oscillator to the next energy Eigen
state of the oscillator. And these operators
are going to be analogues of S plus and S
minus in the spin system. With of course,
the important difference, that there is an
infinite set of energy levels in the case
of the oscillator, as we will set out to prove
shortly.
Suppose we forget all the constants here and
I consider a generic system. Where, I have
set for convenience I set m equal to 1, omega
equal to 1 h cross equal to 1 and so on. If
I did that well there are only these three
in this context, then my Hamiltonian the quantum
Hamiltonian, is simply X square plus P square
by 2. I put back the m and the omega and the
h cross later, with the commutation relation
that X P is equal to i times the identity
operator. I could find combinations, I could
find a combination a which is X plus i p and
a dagger which is X, minus i p, and obtain
the commutator of a with a dagger. I could
represent this algebra in terms of the commutator
between a and a dagger. Now, first of all
let us find out what is a dagger a? It is
clear that a and a dagger are individually
not Hermitian operators.
So, a dagger a is x plus i p, x minus i p,
times x plus i p, that is x square minus p
x, plus minus i p x, plus i x p, plus p square,
which is a same as x square plus p square,
plus i commutator of x with p which is x square
plus p square and x p is i h cross. I have
set h cross equal to 1. So, that is x square
plus p square minus one. Therefore, the quantum
Hamiltonian, can be written as a dagger a,
plus 1 by 2.
If you wish, it is easy to absorb, a root
two in a and a dagger. So, that a dagger a
already has a half here. So, in standard notation
this is what we do, multiply the whole thing
by a half, so that the quantum Hamiltonian
can be written as a dagger a plus half. But,
it is a Hamiltonian and if you put in the
h cross the m and the omega properly. You
actually land up with the oscillator Hamiltonian,
to be a dagger a plus half, h cross omega.
That has got the dimensions of energy in any
case. One could study the quantum oscillator
using the Hamiltonian a dagger a plus half
h cross omega. I would formally define it
properly in terms of X and p putting in the
h cross the m and the omega, in the suitable
places.
But, the algebra x p is equal to i h cross,
is equivalent to the statement. That a a dagger
commutator, given that x p is equal to i h
cross. In this case I am writing i because
h cross is equal to 1. The commutator of a
with a dagger, is the commutator of 
X plus i p, with x minus i p. I have the commutator
of X with minus i p and that gives me an i
h cross, out there. Then I have the commutator
of p with x, which is minus i h cross. But,
I have said h cross equal to one therefore,
i am going to get rid of that. That is giving
me 1 here.
Instead of writing x p is equal to i. I could
have well written the commutator of a with
a dagger is equal to 1. So, these are equivalent
ways of writing the algebra of the commutators.
Even as we wrote in the spin system, the commutation
relation S x, S y commutator is i h cross
S z cyclic or equivalently in terms of S plus
S minus commutator. I could work with X p
is equal to i or with a a dagger is equal
to 1.
My next lecture I will use this Hamiltonian,
a dagger a plus half h cross omega for the
quantum oscillator Hamiltonian, and show what
exactly is the minimum uncertainty state;
in the case of the oscillator the equality,
the Heisenberg relation equality, delta X
delta p is equal to h cross by 2, would be
established for the ground state of the oscillator.
