I am still to write the problem set which
I will do today and send it you by email about
Harmonics oscillator and all the eigenfunctions.
I’ll start with a couple of quick points
which I had left out last time. The wave functions
themselves phi n (x) for the oscillator are
of the form a normalization constant multiplied
by e to the - x squared over 2 and this was
measured in units of m omega or h cross. So
2 h cross and then m omega x squared. So Gaussian
function of this kind and then Hn (x) root
of m omega by 2 h cross which is a dimensionless
quantity and these Hn’s are the Hermite
polynomials.
Hn (x) equal to polynomial of order n H 0
(x) is 1 and H 1(x) is 2 x and then the even
ones are all even functions of every power
of x and the odd Hn’s have only odd powers
of x and these functions here from a complete
set of mutually orthogonal functions and it’s
a family of orthogonal polynomials. So you
have a relation of the form integral - infinity
to infinity, dx in the weight factor is x
e to the - x squared. so that’s a converse
multiplied by Hn (x) Hn (x) equal to something
times delta nm. i don’t recall the exact
factor here but this is the orthonormality
relation between Hn’s. there is also a generating
function for this Hn. orthogonal polynomials,
linear vector spaces, Legendre polynomial,
Laguerre polynomials, etc. well, these are
all like unit vectors in function space if
you like and the weight factor here is this
quantity Gaussian for the Hermite polynomials.
You could have any range or weight factor
here and similar orthonormality relations,
and the Legendre functions polynomials run
from - 1 to 1. You have integral Pl (x) P
m (x) is delta lm and so on apart from some
constant.
Now this thing here is given by a Rodrigues
formula as are all these orthogonal polynomials.
so Hn (x) = e to the - x squared - d over
dx to the power n e to the x squared. So you
differentiate e to the x squared n times - 1
to n times that and then multiplied by e to
the - x squared to get rid of the overall
e to the x squared factor and you get a polynomial
and they are Hermite polynomials. then there
is a generating function and that relation
says e to the power 2 t x - t squared where
you could expand in powers of t because this
is an entire function and you have only non
negative powers of t. if you expand this exponential
out and you collect powers of t together but
remember there is a t here as well as a t
squared there, so the coefficients would be
some functions of x. and these coefficients
are precisely the Hermite polynomials 0 to
infinity Hn (x) over n factorial t to the
power n. so this is a generating function
this function here generates a power series
in t whose coefficients are just the Hermite
polynomials divided by n factorial. It’s
a very useful relation. And then there are
these usual recursion relations and so on
and so forth. The Hermite polynomials are
the regular solutions of the following second
order differential equation.
d 2 Hn over dx 2 - 2 x d Hn over dx + 2 n
Hn =0. so they satisfy this second order differential
equation. There are 2 solutions and they are
linearly independent ones. One of them would
have singularities and other one is polynomial
like in the Legendre case. So I presume you
are a little familiar with the theory of second
order differential equations. So you can solve
these equations by the power series method
or by the Frobenius method and then the regular
solution that you get is the Hn. of course,
there is also a singular solution but that’s
not the one that occurs physically in the
harmonic oscillator problem.
So this is a very useful piece of information
to know. You can use all the properties of
Hn in order to write down what the properties
of the Eigen functions are. recall also that
the energy Eigen values corresponding to these
Eigen functions were these things and n = 0,
1, 2, etc. so this finishes the harmonic oscillator
everything you need to know is now known because
you can take an arbitrary state of the system
of the oscillator and you can expand it uniquely
in terms of these Eigen functions of the Hamiltonian.
Now recall that the Hamilton for this problem
was p squared over 2 m + 1 half m omega squared
x squared where x and p are operators; the
position momentum operators. apart from these
coefficients its quadratic in x and its quadratic
in p. so there is a complete symmetry between
the position space wave functions and the
momentum space wave functions in this case
because what appears here is completely symmetric
in x and p.
So if you were to write this out in the position
basis namely; acting on wave functions phin(x)
in the position basis, then this is - h cross
squared over 2 m d 2 over d x 2 + ½ m omega
squared x squared in the position basis. In
the momentum basis, if you are acting on wave
functions in momentum space, then p remains
as it is. so this is p squared over 2 m, just
multiplication, but x is replaced by + i h
cross d over dp and when you square it, it
becomes – 1/2 m omega squared h cross squared
d 2 over dp 2. so you see apart from these
constants which can be scaled away, you have
an operator which is essentially x squared
- d 2 over dx 2 in the position basis and
the same operator in the momentum basis is
p squared - d 2 over dp 2. so it shouldn’t
be surprising that the solutions you get for
the momentum space wave functions would also
be Hermite polynomial times e to the - p squared
over 2 in proper units and it will satisfy
exactly the same equations in the case of
the x basis.
So the a momentum space wave functions would
be phi n tilde(p) for the same energy eigenvalues,
these would be some suitable normalization
constants multiplied by some Bn e to the power
-, not m omega x squared over 2 x but p squared
over 2 m omega h cross because that’s the
dimensionless momentum, times Hn (p) over
square root 2 m omega h cross. So even without
calculation, you are absolutely guaranteed
that this identifies what the wave functions
would be and they look exactly the same in
both bases.
the ground state wave function in each case
is very special. The ground state wave function
corresponds to n=0 and apart from some constants,
it’s e to the - x squared over 2 H0 but
H 0 is 1, so this goes away and you just get
a Gaussian. So in the ground state of this
harmonic oscillator, the wave function is
a Gaussian both in x as well as in p. and
that’s the reflection of a well known fact
that the Fourier transform of a Gaussian is
also a Gaussian function because we know these
two bases are related by Fourier transform.
so the ground state wave function phi 0 (x)
as a function of x is some kind of Gaussian
and similarly, phi 0 tilde(p) as a function
of p is also a Gaussian. so in the ground
state, whether energy is half h cross omega,
the wave function is extended and runs all
the way from - infinity to infinity. Recall
that the potential itself was just this, this
is V (x) and the wave function itself is this.
so even though classically, at energy 1/2
h cross Omega the particle can only run between
2 finite points quantum mechanically it can
be found anywhere on the x axis from - infinity
to infinity with ever decreasing probability
and dies on extremely fast as you go out.
But it’s a nonzero probability that the
particle in the ground state is actually in
the region outside the classically allowed
region. Now we generated all the higher order
terms by the operator method. We generated
the solutions in x and all higher order solutions.
We would like to find out what the uncertainties
in the position and the momentum are. Let’s
see if you can calculate this.
So what was need to calculate is delta x in
the n’th state, so let me call it n, this
by definition is x - x average in the n’th
state always. So we would like to calculate
the variance and then take the square root
of this quantity. so this would be delta x
and similarly delta p. and you shouldn’t
be surprise that delta x and delta p would
be exactly the same in both cases because
each of the n’th wave function in the position
basis is a Fourier transform of the wave function
in the momentum basis. and they are exactly
the same in functional form. So I expect a
great symmetry between delta x and delta p
in this case. So, in fact calculate just delta
x and then i write down delta p by just changing
units. x is measured in units of h cross over
m omega square root and p is measured in units
of square root of m omega h cross. so that’s
all that you need but there is an easier way
to do this and the way to do this is to ask
this quantity here. We would like to calculate
in the n’th basis. so let me just call this
the ket vector corresponding to the n’th
energy level En and let me first find out
what is this quantity. a was essentially x
+ ip over square root of 2 x in appropriate
units and a dagger was x – ip. So what’s
a with n on both sides?
If you recall a was = x square root m omega
by 2 h cross + ip over root 2 m omega h cross
and a dagger will be = x root m omega by 2
h cross – ip over 2 m omega h cross, where
x and p were the position and momentum operators
and if you recall the Hamiltonian was h cross
omega a dagger a + ½. And we know that in
the n’th eigen state, a dagger a acting
on it is just n on n. this was the number
operator acting on the ket n and we also found
out what a acting on n does. a acting on n
gives you square root of n times n – 1.
It lowers and a dagger acting on n takes you
up. Notice this factor is not n – 1, its
n. so when n is 0, a acting on the vacuum
on a ground state just annihilates a vacuum.
It gives you 0. And then, there are no levels
below that. So given that, what’s the diagonal
element? Well these levels n are all orthonormal.
The moment a acts on n, it lowers you to n
- 1 and then n – 1, overlap with n is 0
because they are perpendicular to each other
in the Hilbert space. So this straight away
gives you 0 immediately and what is this?
Well of course its Hermitian conjugate would
also be zero that’s the number complex number
so its complex conjugate is also zero immediately.
So if you represent a and a dagger as matrices
which you can, you need a infinite dimensional
matrix in each case because n goes all the
way to infinity. It starts at zero and the
label goes all the way to infinity. So you
can represent the position and momentum operators
as infinite dimensional matrices. So what
looks like a d over dx operator acting on
function space, the same operator in the energy
basis looks like an infinite dimension matrix.
So these are abstract operators and they wear
many clothes at different times. So you shouldn’t
be surprised that what looks like a derivative
here; d over dx, really ends up looking like
you know another basis which is just multiplication
by p and then yet another basis. It’s really
an infinite dimensional matrix. Now this tells
you that neither a nor a dagger is a diagonal
matrix because all diagonal elements are 0.
So a + a dagger and a - a dagger which would
be essentially x and p also don’t have diagonal
elements at all. They only have off diagonal
elements and now we can compute these uncertainties.
So for the moment, let me set m= 1, h cross
=1 and omega = 1. I can always do this. I
choose a scale of mass such that m = 1, I
choose a scale of time such that omega = 1
and then I choose a scale of length such that
h cross = 1. I will put them back later on
by dimensionless arguments.
So a is x + ip over root 2 and a dagger = x
- ip over root 2. M, omega and h cross are
equal to 1. We will restore those units later
on. So what’s x equal to? if i add these
two, x is a + a over root 2 and p = a - a
dagger over i root 2. They have to be Hermitian
x and p. so a and a dagger are not Hermitian.
a and a dagger are not Harmitian operators
but x and p are physical operators. They are
Hermitian. They must have real eigenvalues.
So I am in business.
What’s n x n? That is zero straight away
and n p n is zero. That’s sort of obvious
because you have this oscillator classically
going back and forth. Therefore its mean position
is at the centre of oscillation. it should
be zero. And it’s not going anywhere. It’s
simply bouncing back and forth in the potential.
So the mean momentum is also zero but the
squares are not zero. we need to compute that
and you can either do this painfully by doing
it on the basis on messing around by differentiating
Hermite polynomials or you can read neatly
by finding out what’s n x squared n?
This is equal to n x squared, all you we have
to do is square that, a + a dagger squared
over 2 acting on n and that’s equal to ½
n and what are these terms inside? So we will
get 
an a squared + a a dagger + a dagger a + a
dagger squared. That’s very important to
remember. since a a and a dagger don’t commute,
you have to keep both terms + a dagger squared
on n but a squared when it hits n is going
to lower to n – 1 and to n – 2. And that’s
orthogonal to this. The first and the last
terms don’t contribute. We want only the
diagonal elements. This is equal to half and
now the matter is exceedingly simple. a dagger
a acting on n is just n. it comes out but
what’s a a dagger? We use the commutation
relation. So a a dagger equal to 1 + a dagger
a. so this gives you n + n + 1. That’s 2
n + 1 over ½. That’s equal to n + ½. So
you right away have these answers for what
the mean square value of the position is in
the n’th normalized eigen state. If you
restore all the other dimensional factors,
then you get n + 1/2 times 
h cross over m omega. It’s got to have dimensions
of length squared.
Similarly, n p squared n but that’s equal
to the same because the i squared and the
minus cancel out with each other. So you see
how symmetric it is and what’s the ground
state answer? In the ground state, so we have
this very interesting answer which says the
following. If I plot delta x here versus delta
p, these are standard deviations. So they
can’t be negative. I am going to do this
for each of the states delta p delta x.
Delta x delta p in the n’th state = n +
1/2 h cross. Now delta in the ground state
is exactly h cross over 2 but the uncertainty
principle between x and p says this is the
least value could ever have. So the ground
state of the harmonic oscillator is a minimum
uncertainty state. Actually if we plot delta
x delta p = h cross over 2. That’s a rectangular
hyperbola in this fashion. So delta p over
m omega h cross and delta x over root h cross
by m omega. Let me measure it in these units.
Then the ground states, its right here at
the value (1/ root 2, 1/ root 2) at the mid.
That’s the nearest point on this rectangular
hyperbola from origin and is symmetric. And
what do the excited states do? They are just
n + ½. so this becomes 3 ½, 5 ½, etc. they
also sit on these hyperbolas. This is the
first excited state. We are guaranteed by
the uncertainty principle that no matter what
state of the harmonic oscillator you look
at, you cannot come down below the curve or
outside this curve.
Every point on this corresponds to delta x
delta p exactly equal to h cross over 2 and
you can’t go below that. However, it’s
possible that there are states of this oscillator
which you don’t know to construct so easily
at the moment but you could be here. You are
below the 1 over root 2, the ground state
uncertainty in the momentum but you are much
bigger than that in the position. So this
is possible and in quantum optics, this kind
of state has been realized and they are called
squeeze states because what you are doing
is squeezing the uncertainty in one of the
variables down to practically zero but you
are in the expense of increasing the uncertainty
in the other variable. So it’s possible
to have those states. We will see that there
is actually an infinite number of states of
the harmonic oscillator all of which are sitting
here, not just the ground state. We don’t
have infinite number of states all of which
would sit on that and you will see how to
generate these states. In a nut shell, after
all, once I have a harmonic oscillator, I
can move its center of oscillation to any
point. i can shift the potential and nothing
would happen. Those states also sit right
in that point and we will see where they come
in and they are called coherent states. We
will talk about that next but you see the
operator way of doing this is much faster.
So you can compute anything extremely fast
once you know how to use this number operator
basis. This number operator states are also
called 
Fock states after Vladimir Fock who first
derived these in Russia long ago. he solved
a large number of problems and the number
operator states he introduced are called Fock
states in his honor and this space itself
is called Fock space. I will explain what
Fock space means a little later. One of the
most important things you have to note about
the harmonic oscillator is that the energy
levels are equally spaced. Therefore when
you quantize, the fields and look at actual
fields, including the electromagnetic field,
it turns out that the quanta, every time you
add one more quantum of the field, you are
adding the rest energy. Therefore you are
adding the constant amount of energy if the
quantum is at rest.
Therefore the harmonic oscillator plays a
fundamental role in quantum field theory and
the space in which those states exist are
called Fock space. We will comeback to this
but now I would like to ask the following
question. We found the uncertainty in the
position and the momentum for the ground state,
what about eigenstates of a and a dagger themselves?
They are not Hermitian but after all, they
are related to x and p which are physical
operators. So the question is can I find Eigen
states of these operators. So let’s try
to do this.
The first thing you must understand is that
these are not Hermitian operators. So I don’t
expect the eigenvalues to be real. It may
be complex. it doesn’t matter after all
if l over x + ip where x and p are physical
and have real eigen values, the number x +
ip is complex in general. So i can live with
that but the question is can I find eigen
functions or not? Now a dagger = x – ip.
Let’s use the units in which m, omega and
h crosses are equal to 1. we put them back
later on. And in the position basis, this
is the same as saying 1 over root 2 (x - d
over dx). And I said h cross equal to 1. So
if this quantity has an eigen state and let’s
call it some chi (x) in the position basis,
then the idea is a dagger on chi(x) should
be equal to lambda on chi(x). if such a chi(x)
exist, lambda could be complex in general.
so it says that you must have x - d over dx
on chi(x) = lambda on chi(x). That’s not
a hard differential equation to solve. It’s
a first order differential equation that can
be trivially solved. We now find the solution
and now see if you can normalize this or not.
So I leave that to you as an exercise to find
out but again we can do this problem also
in the operator basis. So let’s see if that
works in the operator notation.
So I have a dagger on chi = lambda on chi.
that’s the statement here that i have made
except i am working in the energy basis. i
can always expand this chi as summation n
= 0 to infinity, some coefficients cn times
n. you know you can expand it in this form
uniquely. Every state in the square integral
normalizable states can be expanded uniquely
in the form which forms the number operator
basis. So what does this do? Because this
implies chi = c 0 0 + c 1 1 + c 2 2 + … etc.
Now what is a dagger on this? It’s a square
root of n + 1 times n + 1. So square root
n here is 0. so this becomes c 0 square root
of 1 and 1 + c 1 square root of 2 on 2 + …. This
is all the way to infinity but this must be
equal to lambda times chi. so it must be equal
to lambda c 0 1 0 + lambda c 1 n 1 + lambda
c 2 on 2 +… etc. therefore each coefficient
must be equal on both sides. There is no other
choice because you require 2 vectors.
Therefore they must be equal component by
component. There is no zero here. So the conclusion
is c 0 must be 0. So that gets killed and
so on down the line. What is that telling
you? There are no eigen functions. But I can
solve this equation. i can find the solution
which is not trivial. You are telling me that
chi(x) must be zero there. Chi must be the
null vector but that’s not true. This can
be solved. They will not be normalizable.
They won’t be in the space of square integrable
states. So that’s the crucial point. Remember
that this expansion here has given you all
the states in the space of normalizable states.
And there are no normalizable eigen states
of a dagger. There are eigen states in this
space but they are not normalizable. so they
will not be the total probability and will
not be conserved. Now we do the same thing
with a and let’s ask the same thing. So
a is x + ip.
Can we have a on some eigen state, alpha is
alpha on alpha. Alpha is some complex number.
i play the same game as before and now i have
x upon i have a 1 over root 2 x and this is
an i and this is - i from there. So this is
x + d over dx acting on this wave function.
Let’s call it alpha(x) for want of a better
name, is equal to alpha times alpha(x). This
alpha(x) is nothing but in the position basis,
the representative of this state alpha. you
can solve this equation and you can ask if
they are normalizable eigen functions in this
case because it is a + sign here. So it will
immediately turn out that you get things which
are normalizable because this is e to the
- x squared sitting up there immediately.
So we put alpha a on alpha to get alpha on
alpha. I put alpha equal to these coefficients
equal to c 0 etc and I put a on this alpha
and now we are in business because this is
+ alpha here and that gives 0. It annihilates
it and pushes it down but then acts on one
and brings you down to zero and it acts on
2 and brings you down to 1. So you can match
the series since this series is not bounded
from upper side, you can equate these 2 series
because this spectrum is bounded from below.
a dagger doesn’t have an eigen state. Had
this spectrum and bounded above but not from
below then you would have had eigen states
of a a dagger and not of a. You must have
something unbounded on one side or the other,
then the magic works. so this quantity here
gives you c 1 acting on 1 that is root 1 times
0+ c 2 root 2 on 1 + c 3 root 3 on 2 + etc.
So we start of by saying c 1 = alpha over
root 1 c 0, c 2 will be, you equate this term
with this quantity, that’s equal to alpha
over root 2 on c 1 which is equal to alpha
square over root 1 acting on c 0 and so on.
And the n’th coefficient is alpha to be
n over square root of n factorial on c 0.
So this fixes all the coefficients and in
fact, it tells you that this is perfectly
reasonable.
This is equal to a summation from n = 0 to
infinity, there is a coefficient c0 that comes
out and then there is an alpha to the n acting
on n over square root of n factorial. So we
have a state. What are the allowed values
of alpha? It can be any complex number. So
it has a double continuous infinity of eigenvalues.
All numbers in the complex plane are eigenvalues
of the annihilation operator a. and they have
corresponding normalizable eigen states. Let’s
normalize them.
So alpha alpha = 1. i am going to impose that.
so that gives you mod c 0 squared and then
i have an alpha to the n on n and the other
one would be alpha star to the m with root
m factorial. but when you have an n and an
m, you have a delta nm. so this whole thing
becomes summation n=0 to infinity mod alpha
to the power 2 n, that’s alpha alpha star
to the power n over n factorial. But what’s
this series? It is e to the power mod alpha
squared. Therefore that tells you that c 0
= e to the – 1/2 mod alpha squared. i take
the square root of it and I have set the phase
of c 0 to be 0. The overall phase of wave
function doesn’t matter.
Therefore, we have a very important statement
which says a has normalizable eigenstates.
This is a little surprising, non Hermitian
operator has normalizable eigenstates and
these eigenstates are labeled by a complex
number, alpha. And this alpha, the eigenstate
is e to the – 1/2 mod alpha squared summation
n=0 to infinity, alpha to the n over square
root of n factorial on n. what does alpha=
0 correspond to? If we put 0 here, you get
the ground state because a on 0 is 0 actually
and you do get that because I put alpha= 0,
only the n=0 term contributes.
All other n equal to zero alpha to be n is
1 so everything else goes away and you get
the vacuum and this becomes unity. so these
states which are eigen states of a, one state
coincides with an eigen state of a dagger
a and not surprisingly that’s the state
such that a on that state gives you 0. So
alpha = 0 is the vacuum. It is a number operator
state 0 but all other complex alphas are different
states all together. You also know what is
this equal to in terms of a ground state?
I keep raising this. So I act a dagger on
the ground state which gives you a state 1
with the 1 over square root of 1 here and
then i act once again on it, a dagger squared,
and that’s again going to give you a 2 here
but it’s going to give you a square root
of 1 times square root of 2 here and so on.
So it is clear that you can also write this
n as a dagger to the power n over square root
of n factorial acting on the ground state.
You can also write it in that form.
So this implies that this eigen state alpha
can be written as e to the – 1/2 mod alpha
squared summation n = 0 to infinity alpha
a dagger to the power n over n factorial acting
on this ground state; acting on the vacuum.
But now you can do the summation formally.
It’s again an exponential series. So this
is equal to e to the – 1/2 mod alpha squared
e to the power alpha a dagger acting on the
ground state. It’s a very compact form of
writing the eigen state of a. these eigen
states are called coherent states. They play
a fundamental role in quantum optics. When
you apply this to the electromagnetic field
and you quantize it, the coherent states play
a fundamental role. In fact, if you take ideal
single mode laser light with a given polarization,
this is in fact the state of the radiation
field is a coherent state. This has mathematically
identical properties. You don’t have a position
and a momentum there for the photon or anything
like that. This is quantizing the electromagnetic
field itself and then you end up with a state
which has exactly the properties of a coherent
state of this kind.
It’s not an accident that the exponential
of this operator acts here of this operator
a. because, notice that i said a and a dagger
form a Lie algebra. a a dagger in the unit
operator that commentator of a with the unit
operator 0. a dagger with that is 0 and a
with a dagger is 1, the unit operator itself.
They form a Lie algebra. This is called the
Heisenberg algebra and from this Lie algebra,
you can form a group by exponentiating just
as we did in the case of rotations. So you
exponentiate the generators with parameters
and you end up with the group elements. so
that’s exactly what these quantities are
and when it acts on this 0, you get each of
the coherent states. in fact, you can even
get rid of this. You can show it’s essentially
e to the power alpha star a with the - sign
+ alpha a dagger acting on 0. So with a little
reduction i will give this is an exercise.
You can write this as alpha a dagger - alpha
star a acting on 0. It is this operator, please
remember that if a and a dagger don’t commute,
then e to the a dagger + a is not equal to
the product of the individuals terms. So this
can be reduced to that form and then you end
up with this operator acting on 0 on the ground
state which gives you the coherent states.
Physically what does coherent state mean?
What does the wave function look like and
what are normalization conditions and so on?
Well we normalize these states to 1. Remember
that’s how we found this factor here but
you could ask would you expect them to be
orthonormal. The question is the following.
In the space of the harmonic oscillator, all
normalizable states could be expanded uniquely
in terms of number operator states. They form
an orthonormal basis. So there is no question
that we can expand.
Everything in terms of this number operator
which incidentally is a dagger a + ½, that
is just 1/2 times unit operator multiplied
by h cross omega. Essentially in terms of
the eigen states of a dagger a. a dagger a
acting on n was n times the same state and
n runs from 0, 1, 2, 3, etc. so all states
in this space could be expanded in this form.
So every normalizable state psi could be written
in the form cn n, n = 0 to infinity. Since
this is a unique expansion, specification
of this state is the same as specifying that
infinite set of numbers c 0, c 1, c 2 etc.
these are the components. So if you give me
an infinite number of coefficients, i give
you the state uniquely. In the case of alpha,
alpha is a doubly continuous variable, both
the real and imaginary parts.
So you would expect that alpha also forms
the basis and they are normalized. the question
is, if you have a different alpha, is the
inner product of this alpha with this beta
equal to zero or not, if the alpha is not
beta what could you expect? Then I would say
this is orthonormalized. This is what we would
like to find out. Which would you expect?
What would your intuitive expectation be?
You seem to have many more doubly infinite
number of labels, both the real and imaginary
parts of alpha. So if you can uniquely expand
the state in terms of this doubly infinite
labels, how come it suddenly get compressed
to just the nonnegative integers? So it doesn’t
look right. So let’s compute this overlap.
Let’s do beta with alpha = summation over
n summation over m and this state alpha was
= alpha to the n over root n factorial, there
was an n on right side and there is a beta
star to the m over root m factorial. you shouldn’t
forget beta with alpha equal to e to the – 1/2
mod beta squared + mod alpha squared those
who are the 2 normalization factors for the
state and i took the blob data so you got
a put a beta star here over m over this and
then there was an m n but this is delta nm,
that’s orthonormal. So therefore the summation
collapse is to a single sum and you get moderate
+ square + mod beta squared and then you get
alpha beta star to the power n over n factorial.
This is 
what you get. if alpha= beta, you get a 1
of course because you want alpha with alpha
to be 1, you we normalize it.
But this number has a modulus less than unity
but it’s not zero. This is the complex number
here but you can take its phase and its real
part and so on and take out modulus times
e to the power whatever it is and you can
compute what the modulus of this quantity
is. So calculate mod beta alpha. Mob alpha
beta squared will turn out to be proportional
to e to the - alpha - beta whole squared times
some number. So in the complex plane of these
eigen values, mod alpha - beta is the distance
between those 2 points. Now the 2 points coincide
then of course you get unity as normalization
but in the difference you get some number
less than 1, e to the - some positive number.
So it is not zero in general. Its exponential
doesn’t vanish. So this is not an orthonormal
set of states. What to do you expect regarding
completeness? Once again it’s not a complete
set of states but it’s over complete. It
is actually doing much more than that. And
this is another thing. I am gone a leave it
as an exercise.
You can show that integral and now you have
to sum overall possible alphas and alpha is
a complex number so you have to integrate
over the real part of alpha and the imaginary
part from - infinity to infinity. so that’s
over the entire complex planar, let me call
it d 2 alpha to show it’s a 2 dimensional
integral. Alpha alpha over pi = the identity
operator. Again I’m leaving that you as
an exercise to show. It’s slightly harder
in this case but this should be fairly straight
forward and I will indicate how to do that
in the problem sector. These states are said
to be over complete. So the set alpha is an
over complete set. But now comes a key question.
What does the wave function look like for
alpha? Again I will go back to the definition.
I have alpha = summation n = 0 to infinity,
alpha to the n over root n factorial on n
and i would like to look at it in the position
basis. This is the wave function in the position
basis corresponding to this coherent state
alpha. So this is what i call alpha(x) labeled
by this x. the wave function must be some
square integrable function and that would
be equal to this but what’s x with the n?
This is our famous phi n (x). This is phi
n (x) but that = apart from some normalization
constant, summation n = 0 to infinity, alpha
to the n over root n factorial and then this
normalization constant An multiplied by e
to the - x squared over 2 in those units that
we chose, Hn(x). Now you are stuck with this.
You have to do this summation but you are
saved by the fact that this An has a 1 over
square root n factorial and that will kill
this and then the rest would be an exponential
which you can actually sum.
So I am going to leave this to you as an exercise
to do. And what could you expect once you
sum it? You get an exponential which involves
this, remember the expression we have for
the generating function. That’s where it
comes in use and it will turn out this is
a displaced Gaussian. So it’s not e to the
x squared but it’s e to the x - alpha1 squared
where alpha1 is a real part of alpha multiplied
by some phase factor. Similarly the momentum
space eigenfunction would be e to the p - alpha
2 but alpha 2 is the imaginary part of alpha.
so all that’s happening by applying this
operator e to the alpha a dagger - alpha star
a on 0 is that the center of oscillation which
was 0 for the ground state and for all other
states as well and the momentum eigen value
was also 0 that gets shifted.
So classically what it means is that in the
x-p phase plane, the original oscillator corresponded
to an oscillation like this. Once you apply
this operator, the phase trajectory is somewhere
else at this point shifted by alpha1 on this
side and alpha 2 on that side. It’s just
a displaced oscillator. So you would expect
nothing new has happened by doing this. So
in fact you can start with now a state alpha,
treat that like the vacuum and apply a dagger
on it or the equivalent of the dagger and
create another set of excited states everywhere.
So all these statements are deep implications
in quantum optics and I will explain a few
of them as we go along. let me stop here today
and next time we will discover some more properties
of this and what I should like to do is to
give you some insight into what does this
quantity do. It’s a unitary operator and
it’s called the displacement operator for
a few reasons. We will study a few properties
of these. This will also help us understand
the algebra a little better. So we will stop
here. Thank you!
