Now so let’s go back and look at few of
the problems in the sheets given earlier.
We start with set p which was on the harmonic
oscillator. I hope some of you have got copies
and after writing down the Schrodinger equation
plus the solution and so on, there is a set
of problems that are given. So let’s go
through these systematically. The first one
refers to a particle in a box, the orthonormality
condition, the completeness relation, etc.
You are supposed to verify these. So they
are just straight verifications. And then
i have to also ask for the momentum space
wave function.
So the problem was the particle in a box,
0 to L, the energy levels En squared pi squared
h cross squared over 2m L squared and the
normalized wave functions where n is 1, 2,
etc and the wave function is zero outside
the box. you are also asked to find the momentum
space wave function corresponding to this
and that has also been written out here as
phi n tilde (p) is equal to integral, - inifinty
to infinity, dx p x phi n (x) and this quantity
here the normalized value p x is 1 over square
root 2 pi h cross e to the ipx over h cross.
as you can see, if this thing had been just
e to the ipx or something like that or a sin
or cosine from - infinity to infinity, then
this would be just a super position of a discrete
number of momentum states but, because the
wave function gets cut off outside the box,
it’s localized to within this box. So on
the x axis, the particle doesn’t ever go
outside the box.
So this means, the uncertainty in the position
can at no stage be bigger than L itself. Definitely
it is inside the box which means the uncertainty
in the momentum cannot be smaller than the
order of h cross over L. so that immediately
says that the momentum space wave function
is actually spread out unlike the position
space wave function. And you can compute this
once you put this in and then take mod squared
of this to find the momentum density. In the
case of the ground state, if you plot phin
tilde (p) mod squared, you would get something
like this
It is a peak here and another peak here at
these values and this will be h cross pi by
L and - h cross pi over L. n corresponds to
1 here. So it’s phi1. I leave this to you
to actually work it out. So it is an extended
wave function in any case. It’s a kind of
a statement if we have function which has
compact support, then the Fourier transform
has support from - infinity to infinity. The
more you try to localize in position, the
more it spreads out in the momentum. So it
is not a momentum eigenstate for sure. In
fact, there is a result given for this. it
is cos squared pL over 2h cross and then there
is a rational function. it is easy to calculate
the position and momentum uncertainty. You
have to simply calculate some integrals here
and then show that delta x delta p for any
eigenstate n is certainly much bigger than
h cross over 2 including the ground state.
The next problem was a particle moving freely
in 1 dimension but it was given in the position
basis by Gaussian wave packet. So the idea
is to try to localize the particle near some
point x 0. The question is what happens to
this as a function of time, and compute the
momentum space eigen function etc. so you
are supposed to calculate delta x delta p,
find the wave function at any instant of time
greater than zero, find what the corresponding
momentum space wave function is, the uncertainties,
etc.
In particular, for this wave packet, which
is an interesting wave packet psi (x, 0) = 1
over pi sigma squared to the power 1/ 4, these
are normalization factors, e to the power
– i k¬0 x. e to the – (x - x 0) whole
square over 2 sigma squared. so mod psi (x,
0) squared is a Gaussian peaked about the
point x 0 and for this Gaussian wave packet,
we are supposed to compute the uncertainty
in position, momentum, etc and then show that
at any instant of time, x average is equal
to x 0 + pt over m. so again for this Gaussian
wave packet, the mean value of x moves like
a classical particle. It is a free particle.
So classically the momentum is constant and
x would be x 0 + pt over m but now Ehrenfest
theorem kicks in and you have expectation
values moving according to this rule. You
can do this directly by integrating or writing
down the Hamiltonian which is just p squared
over 2m using Heisenberg’s equation of motion.
You are also asked to find the spread of the
wave packet as a function of time.
This wave packet will spread and the physical
reason it spreads is because different components
of different wavelengths don’t travel at
the same velocity. The wave velocity and the
group velocity are not the same in this case
because the energy is dependent quadratically
on the wave number. It is a free particle.
So E is p squared over 2m and p is h cross
k. E is h cross omega. So omega is proportional
to k squared which means d omega by dk is
not omega over k. that immediately leads to
dispersion. The rest were some identity in
the harmonic oscillator for the linear harmonic
oscillator what is the expectation value in
any of the energy eigenstates? What is the
expectation value of the kinetic energy and
that of the potential energy?
They happen to be equal in the case of the
harmonic oscillator. These 2 are equal because
that problem is extremely symmetric under
the interchange of x and p. in some sense,
its quadratic in x and quadratic in p and
the Hamiltonians are very symmetrical and
then there were some Baker Campbell Hausdorff
type of identities. The only significant thing
you must note in this problem set is our definition
of coherent state.
So an oscillator coherent state was defined
in this following way where alpha is any complex
number summation n = 0 to infinity, alpha
to the power n over root n factorial n where
these were the oscillator eigenstates. We
know that a on alpha is alpha on alpha, alpha
is any complex number but you can also rewrite
this in another form.
We can also write this as e to the power - half
mod alpha squared. We know that n itself can
be found by taking a dagger and acting on
the ground state. Therefore this becomes summation
n = 0 to infinity, alpha a dagger to the power
n over n factorial acting on the ground state.
because a dagger to the n on 0, divided by
square root of n factorial is in fact the
excited state n. so this is equal to e to
the - half mod alpha squared, e to the power
alpha a dagger acting on the ground state.
Now we know the commutation relation between
a and a dagger, so we know the one between
e to the alpha a dagger and e to the alpha
star a as well. And if you use that commutation
relation or any of these relations which have
been given to you earlier, this can also be
written as e to the alpha a dagger - alpha
star a acting on 0.
So, there is a certain operator called the
displacement operator; D (alpha), it should
be D (alpha, alpha star) because it is linearly
independent of each, but just for convenience
of notation I will call it D of alpha. This
operator which is the exponential of this
combination here acting on the ground state
gives you the coherent state. Now the reason
this is called the displacement operator is
the following. First of all, this operator
is a unitary operator. That is easy to see
that D D dagger = D dagger D = the identity
operator. So it’s unitary transformation
on the ground state that gives you the coherent
state. And physically what it corresponds
to is the following. The wave function in
the position bases corresponding to this alpha
would be found in the usual way.
It is x on alpha and that would be x on n
and what is x on n? By definition it is phi
n (x), it is the normalized eigen function
in the position basis. And what is phi n (x)
equal to for the harmonic oscillator? There
is a Gaussian, so its = e to the - x squared
over 2 in units of h cross over m omega or
whatever. so really you should have m omega
over 2h cross and so on sitting here multiplied
by the Hermite polynomials Hn (x) again in
those units and then there some normalization
constant here, 1 over 2 n factorial square
root and so on. so for you to find this, you
have to put in here and compute it which doesn’t
look like a very trivial exercise because
you have to remember that if i put this in,
you have to find alpha to the n and then Hn
but now you use the expression for the generating
function of Hermite polynomials and then this
sum collapses and you can’t compute this
sum and the answer is again a Gaussian but
centered not at the origin like this one is
but centered at the real part of alpha.
What about the imaginary part of alpha? What
would that correspond to? Would that play
a role because I am claiming this is a Gaussian
of the form e to the – (x - alpha 1) whole
squared over 2 and other factor and there
are phase vectors and so on but alpha 1 is
the real part of alpha. What you think is
a role played by the imaginary part of alpha?
it would be the center of the Gaussian wave
packet in the momentum space wave function
because you know that the momentum space wave
function is a Fourier transform of the position
space wave function and the Fourier transform
of a Gaussian is a Gaussian once again. So
alpha 2 would appear if you did p with alpha.
You would get p - alpha 2 and so on. What
does that imply physically?
It says a Gaussian wave packet has been essentially
shifted in the sense that the harmonic oscillator,
classically not quantum mechanically would
correspond to an elliptic orbit in x and p.
this is an oscillator oscillating about the
point (0,0) in x and p. the equilibrium point
at (0,0) once you apply this displacement
operator e to the alpha a dagger - alpha star
a, you get a new state which is also again
an oscillator but would correspond to something
like this. It is simply displaced from this
origin to a new equilibrium point. This is
alpha 1 that is alpha 2 about there. That
is why i use the symbol D and that is why
it is called the displacement operator. It
displaces this oscillator. What is the uncertainty
product delta x delta p for the ground state
of the harmonic oscillator?
It is exactly h cross over 2. It’s a Gaussian
wave packet. so it is the minimum uncertainty
state but so is every coherent state for arbitrary
alpha. it is again Gaussians. so once again
the uncertainty product is a minimum value.
in fact, if you plot delta x verses delta
p, delta x is in units of square root of m
omega over h cross and delta p is in units
of 1 over square root of m omega h cross.
Then, both these are dimensionless quantities.
Delta p delta x must be greater than or equal
to ½. It’s equal to ½ for this hyperbola.
And symmetrically at this point 1 over square
root of 2e. For each of the two. This is where
the ground state of the harmonic oscillator
is and this is where all coherent states are
for all of them. You have this minimum uncertainty
with this symmetry between x and p. the excited
states in the harmonic oscillator would be
(delta xn) (delta pn) = (n + ½) h cross.
so the ground state n = 0 is 1/2 h cross and
the excited states, I have set units such
that h cross disappears on both sides. Those
would correspond to the states here here.
They are square root of 3 over 2 squared,
five over 2 and so on.
Let’s take e to the alpha a dagger - alpha
star a and apply it on 0. i get alpha which
is a coherent state. What happens if i apply
this on some excited state r and not on 0?
It’s the same operator acting on an excited
state of the oscillator. This is called the
generalized coherent state. It has interesting
properties and they are not minimum uncertainty
states including the ground state. This is
denoted by |r, alpha> and it is called a generalized
coherent state. They again have a lot of applications
in quantum optics which we will not get into
here.
What is interesting to note is that these
operators D(alpha) D(alpha prime) have interesting
multiplication properties among themselves
and that has been given here as an exercise.
i have given D (alpha) D(beta) = D(alpha +
beta) e to the I, imaginary product of (alpha
beta star). Then it is 
fairly straight forward to verify this relation
here. This is again another way of writing
the Weyl commutation relation between x and
p. recall that earlier when we said x commutator
p is ih cross unit operator, what is e to
the i ax e to the i bp? , where a and b ordinary
numbers and that again had a form similar
to this. a and a dagger if you like are linear
combinations of x and p and are non Hermitian.
so this is essentially the same Heisenberg
algebra being rewritten in different ways.
The important point about the coherent states
is that they are not orthogonal to each other
and this is crucial because ||
mod squared is e to the - mod |alpha – beta|
squared. so certainly if it is normalized;
if alpha is equal to beta you get 1. Otherwise
it is not zero since they are not orthogonal
states. They are an over complete set. it
is not a complete set and are not orthogonal
to each other. The next problem in problem
set 3 was on finding the time dependent propagator
for the simple harmonics oscillator. So let’s
go back to the actual time dependent Schrodinger
equation and try to solve this equation completely
by using a Green’s function for it and that
is what i call the propagator. So let’s
write it down and you are just asked to verify
this because it requires a little bit of mathematics
to work it out.
So the equation was i h cross delta psi /delta
t (x, t) = - h cross squared over 2m d 2 psi
over dx 2 + 1/2 m omega squared x squared
psi. And now the point is, if you specify
adequate boundary conditions, and in this
case we want the wave function to be normalized.
so that means psi ( x, t) must go to zero
as x goes to +/ - infinity together with an
initial condition. then it is possible to
write the solution at any time as integral
dx prime and then a kernel which is the function
of x, x prime and t; psi ( x prime, 0).
This quantity is called the propagator because
it helps you go from the wave function at
time zero to the wave function at any later
instant of time. That’s called the Feynman
propagator 
and i given an expression for the propagator
here. It’s a fairly complicated expression
and the idea is to verify whether it’s true
or not. By the way, what is the propagator
for the free particle? Suppose you had no
oscillator at all, what does the propagator
look like? As a free particle propagator,
there is no potential. Certainly you can still
write the wave function at time t in this
form. It is an initial value problem. You
are interested in psi of x comma t for all
t greater than equal to zero.
You are given psi (x, t) at t=0 and you are
trying to find what it is at later instance
of time. So you are interested only in the
half line in t namely t greater than 0. It’s
a first order differential equation in t.
so what is the automatic thing to do? You
would use Laplace transforms. And for the
x variable that is defined from - infinity
to infinity, we would use Fourier transforms.
So you do a Fourier transform in x, Laplace
transform in t and the job is done and you
compute the solution. What do you think is
going to happen? that is the free particle
propagator and we could do this very painfully
but you can see very easily that if you took
a Laplace transform in t, if s is the conjugate
variable to it, this is going to be s times
a transform - the initial value and if you
do a Fourier transform in x and k is the conjugate
variable, differentiation with respect to
x corresponds to multiplication by k.
So it is clear that the transform is going
to have a k squared multiplying it and it
is going to have an s multiplying this and
there is an in homogeneous term. So eventually
this transform is going to be 1 over some
(s + k) squared or (s – k) squared acting
on the initial value and the inverse transform
of that is an exponential. So it will have
an e to the - k squared which is a Gaussian
in k and if you take the inverse transform
of a Gaussian, you would get an e to the - x
squared. So I would expect in this problem
that i am going to get something like e to
the - x squared over t acting on the initial
state. That would be my transform. Are you
familiar with the diffusion equation? We know
the solution to that.
I will write this down by inspection because
remember, the diffusion equation says delta
rho / delta t for any concentration rho is
D times d 2 rho/ dx 2. That is the diffusion
equation and what is the solution to this?
So if you say rho (x, 0) = delta (x - x prime),
some special point x prime where every all
the entire molecular species is concentrate
at some point x prime and then you let it
diffuse as a function of time, the answer
is some kind of Gaussian. Then rho ( x, t)
= 1 over square root of 4 pi Dt e to the – (x
- x prime) squared over 4 Dt. that is the
standard solution to the diffusion equation.
So it says if on the x axis, you start with
an initial concentration at the point x prime
as a function of t, after some time we are
going to get this and then it is going to
get widened out and so on. So it is a Gaussian
concentrated about that point. Well this equation
looks pretty much like that because this - h
cross squared could be written as ih cross
squared. So it is essentially the same equation
on both sides. so let us write this as - h
cross squared and bring the i h cross here.
So this goes away and there is 1 over i here
which gives you ih cross. It’s the same
solution. So this means that if the initial
particle was concentrated at the time x prime,
that Gaussian is a solution but if it’s
given by distribution by itself; psi(x prime,
0), all you have to do is to put this corresponding
solution in here. So let’s put that in.
e to the - x - x prime whole squared over
4 times D. Now D in our problem is ih cross
over 2m. It is imaginary and it is not real
diffusion of course. It is just the Schrödinger
equation. The resulting equation is this.
Had this been some initial point t 0, then
this would be just (t - t 0) for all t greater
than t 0. And then the normalization, square
root of 4 pi, D is ih cross over 2m, root
m by 2pi h cross. That’s the exact solution
to the time dependent Schrodinger equation
for an arbitrary initial distribution. You
specify any psi (x prime, 0) and that is the
solution at any time t. this quantity this
thing here together with this factor is called
the propagator. In this case the free particle
propagator. So what I have done in writing
this solution here is exponentiating the second
derivative operator. Because it says pretend
that the right hand side is just some number,
and then it says the following.
delta rho over delta equal to some lambda
times rho, lambda is an operator which acts
on the x variable etc but the solution here
is rho at time t equal to e to the lambda
t rho at time 0. So going back to the original
x variable, it says rho x comma t = e to the
power Dt d 2over dx 2 rho (x, 0). Dt has dimensions
of length squared. So this is dimensionless.
So the question is what is this equal to?
If this was a d over dx, what would this be?
If this were a first derivative, then you
can use Taylor's theorem and this would be
just rho (x) displaced by D. So this quantity
would just be root Dt d over dx, this would
be just x + square root of Dt but when you
have a second derivative, then the answer
is some integral operator because it must
depend on all x because all derivatives are
acting on it, arbitrarily high orders.
Therefore it must depend on rho not just at
one x but at all sorts of places and it becomes
an integral operator whose kernel is given
by this. so that is the reason you get the
Gaussian multiplying the psi and then in integrating
over x prime, because you are really exponentiating
the second derivative operator. And the answer
is that it is an integral operator and the
kernel is a Gaussian. Now for the harmonic
oscillator, you are exponentiating not just
d 2over dx 2 but you are exponentiating the
operator d 2over d x 2+ x squared and those
two parts don’t commute with each other.
So this is Green function and it is a very
non trivial object. That’s not always possible
to write it down for all problems.
The path integral way of doing quantum mechanics
starts at this point. It tells you how to
find these propagators by what is called time
slicing but we are not going to that in this
course. However, this expression for the harmonic
oscillators Green function is sufficiently
important that you need to know. So that was
a reason for giving it. The idea is that you
give me the Schrodinger equation and you specify
the initial wave function, then the future
wave function is obtained by some propagator
acting on the initial wave function. This
takes you from any point x prime to any other
point x.
The next problem is had this harmonic oscillator
Hamiltonian in suitable units + lambda (a
+ a) dagger in suitable units, what does this
Hamiltonian correspond to? Lambda is a real
constant. You are also asked to find the eigenvalues
of this Hamiltonian. One way to do it this
is we are asking what does this correspond
to physically, so we should go back to the
physical variables x and p. if you go back
to the original variables, the Hamiltonian
was equal to 1/2 p squared + 1/2 x squared.
We have set h cross as 1, omega as, 1 m as
1 and so on + some constant lambda (a + a
dagger). But what is (a + a dagger)? It is
root 2 x. what does that correspond to? But
physically what has been done to the oscillator?
You put a constant electric field or just
a spring which goes up and down and you add
gravity to it, so it has moved the center
of oscillation. So the potential is no longer
like this but this potential has gone over
to something like this.
All that’s happened is you shifted the oscillator.
So would you expect the energy levels to be
changed? You wouldn’t expect the spacing
to be changed but certainly there has in added
constant. So you should complete squares in
this case. You end up with some x prime - 
some constant. So all you have done is to
shift this some other position 
and got a potential like this. So you can
change the energy levels by adding an overall
constant. So this problem is trivially solvable.
It is just a linear perturbation on a quadratic
Hamiltonian and it is trivially solvable in
this case. What value should lambda take in
order that the ground state energy be exactly
0? Well, you can fix this and you know what
this quantity is. You know how much you are
going to add. That depends on lambda and that
must be exactly equal to half h cross omega
in this problem half. So once you fix that,
then the ground state energy is exactly at
0. so that is a trivial problem.
The next one was the problem of a free particle
in a constant field of force. So the potential
was - fx and I worked this problem out explicitly
in class. The solutions in the position basis
are airy functions but the solutions can be
written in terms of the momentum basis. The
equation is easy to solve because the Fourier
transform has things like e to the power of
a p cubed term and the p term. And the normalization
is energy normalization. So that was for problem
set 3. And then in set 4, there were problems
on the harmonic oscillator eigen functions
themselves.
One of the points i wanted to point out was
the following. The differential equation satisfied
by the position space wave functions in the
harmonic oscillator and that satisfied by
the momentum space wave functions are exactly
the same in suitable units. Therefore the
solutions are exactly the same in functional
form.
The solutions in x if you put in the appropriate
constants and so on, then the solutions in
x are of the form phi n (x) which is some
normalization constant, e to the - x squared
over 2 and then Hn(x). in the momentum basis,
phin (p) tilde for the same eigen state would
go like some An e to the - p squared over
2m Hn (p). So the functional forms are exactly
the same. On the other hand, I know that the
momentum space wave function is a Fourier
transform of the position space wave function.
So I know also 
that phi n tilde (p) = Fourier transform of
phi n tilde, and this is equal to integral
dx e to the ipx over root 2pi setting h cross
= 1 times phi n (x). so i could regard this
as an integral operator with kernel e to the
ipx acting on a function in L 2 to produce
another function in L 2 because this 2 is
square integrable. This is normalized to 1,
mod phi n (p) whole squared dp is 1 and so
is mod phi n (x). We know that the Fourier
transform of an L 2 function is also L 2.
In this case, we got something better. It
says phi n(p) has the same functional form
as phi n (x). So this is like an eigen value
equation and apart from an overall multiplicative
constant whose modulus is 1, this and that
are exactly the same. This is some other function;
modulus of A n is modulus A n prime.
So apart from that, these 2 are exactly the
same function. So it says that these functions
are something special. They could also be
regarded as eigenfunctions of the Fourier
transform operator because you take the function
and you have to do a Fourier transform on
it. You get another function in the same space
and is again in L 2 but that function has
the same form as the original function. Therefore
this is an eigen function of the Fourier transform
operator. And the eigenvalue must have unit
modulus. So in general, the eigenvalue is
with some complex number of unit modulus.
now what about the 4ier transform of Gaussian?
It gives you exactly the Gaussian and in fact,
if you work with the ground state H 0 is 1,
you would discover that it is equal to itself
in functional form. So the eigenvalue is 1
in this case. So the ground state in fact
has eigen value + 1. So the ground state wave
function of a harmonic oscillator is also
an eigen function of the Fourier transform
operator and has eigen value + 1. What about
the excited states? Well, the first excited
state you will discover in functional form
that phi 1 tilde(p) is i times phi 1 of x
which implies that it is an eigen function
of the Fourier transform operator with eigen
value + i. the next excited state would have
- 1 as eigen value the third excited state
has eigen value - I and the forth 1 comes
back to + 1 and this is because the forth
power of the 4ier transform operator acting
on functions in L 2 is the identity operator.
So F to the power 4 f = f itself. You can
verify this. In fact you can verify that f
squared acting on f (f squared f) of x = f
(- x). In other words, the square of the Fourier
transform operator is a parity operator. so
you take a function of x, you do the Fourier
transform of the Fourier transform; not the
inverse, once again Fourier transform of the
Fourier transform and you get a f (- x). That
is easy to verify. Therefore you do it 4 times
you get back the original function. this immediately
suggest that the Fourier transform operator
itself has eigen values equal to the 4th roots
of unity; 1, i, - 1 and –i. and the harmonic
oscillator energy eigenfunctions are also
Fourier transform operator eigenfunctions
such that the ground state has eigenvalue
+ 1. The first excited state i, second excited
state – i, third excite - 1 third excited
state - i and 4th one back to 1 and that keeps
going. So this is an extremely interesting
property of these functions.
So they are not arbitrary functions as you
can see. It is the symmetry between x and
p that gives you some profound inside into
what’s happening with regard to the Fourier
transform operator itself. Well, the obvious
question to ask is if these functions of the
eigenfunctions of the Fourier transform operator,
then the Fourier transform operator is like
the square root of the parity operator. The
parity operator is the like the square root
of the unit operator, then what about the
square root of the Fourier transform operator?
Would the harmonic oscillator eigenfunctions
also be eigenfunctions of this operator with
eigenvalues at the 8th roots of unity on the
unit circle and so on? Then you can ask for
the square root of that operator the 16th
roots of unity and so on. I leave you to verify
whether this is true or not.
It is an interesting exercise in mathematical
physics and you have to verify this is true
or not and it has implications in other areas
of mathematics but we won’t go into that.
But up to the level of the Fourier transform
operator, I have given here what the expected
answers are, so please check that out. Then,
last questions I have to do with reflex and
transmission. These are things I have already
done in class. I have just written them down
systematically here. So, that you have the
exact formulas with the correct notation and
so on. And finally the last question is an
extension of the uncertainty principle.
I already mention that delta a delta b, if
you had 2 arbitrary operators then delta A
squared delta B whole squared is greater than
equal to 1/4th the expectation value of the
commutator of A with B squared. this was the
generalized uncertainty principle I mentioned,
where A commutator B is some - i C or whatever,
where C is the Hermitian operator but you
can also ask what is the exact relation and
it turns out to be greater than equal to this
commutator squared + a similar expectation
value squared of the anti-commutator. And
since that is again a non negative quantity,
you play it safe by saying this is certainly
bigger than this. So there is an extra term
in here. It is actually bigger than this +
another term here which could to be 0 under
circumstances. And that is the actual best
result that you can get. That is the generalization
of the uncertainty principle. It is an extension
of the uncertainty principle. I believe it
is called the Schrodinger Robertson uncertainty
principle. i can’t swear to that right now
because i don’t remember for sure but that
extra term is sometimes used also and plays
a role. So that should take care of these
problem sets. Please go through this completely
and make sure you solve all of them and if
there is any problem, just let me know. So
let me stop here. Thank you!
