so in the last lecture we looked at the levels
of implementation that you can do with optical
approaches to quantum computing and there
may be several more of it but what we will
do now is that we will look at those kinds
of implementations with other aspects as we
mentioned in the last class also that there
are other aspects like sprintronics were optics
and other kinds of optical implementations
as well as teleportation we have looked at
it before all those become very interesting
and important those we will deal with it as
we look at those particular kinds of implementations
from todays lecture onwards for this week
for the rest of this week we will be solving
the assignment problems ah that we have been
given giving you for the last six weeks and
ah maybe i will not be able to do it in the
last week but generally as much as possible
we will try to do because while doing these
assignment problems we will also revisit and
refresh some of the problems and some of the
aspects at we were been looking into in this
area of implementation of quantum computing
we will also link you at every time while
we do the problems to the key concepts that
we are looking at while we are doing this
so let us now look at this principle of solving
the problems so the first problem set is the
one which we had given to you in the very
first week and the first question in that
problem is as we mentioned here and the solution
had been given to you but the basic concept
that we were discussing and looking for in
that problem was the principle of the particle
nature of light
so with that let us now look at all the different
aspects that we can do for this particular
problem started with in the first week where
we had these multiple choice questions and
their solutions were given so the first one
was based on the photoelectric effect um which
was questioning the concept of photoelectric
effect that we discussed and so the question
basically ah was given to find out which of
these ah different choices ah was important
so ah it sort of was looking for the particle
nature of light to be understood and in this
case most important parameter or understanding
gain from the discretization of the energy
that was first proposed by max planck in terms
of the max ah in terms of the plancks constant
h um and this is a visual representation of
how the continuous energy picture change to
the discretized energy picture discrete or
quantize energy picture after the principle
in ah given by a max planck
so this quanta or the ah discreteness of the
energy in terms of the plancks constant was
the key to bringing the photoelectric effect
explanation by einstein where he was able
to show that the radiation field is quantized
because no matter how much of a photons you
provide beyond a certain energy level they
will be no electron ejection ah for the photoelectric
effect essentially showing that the light
below a frequency of say in this particular
example of certain hertz or wavelength longer
than six eighty three nanometers would not
eject electrons the fact that this plot was
not dependent on the intensity of the incident
light implied that the interaction was like
a particle which gave all its energy to the
electron and and ejected it with that energy
minus that which took to escape the surface
so this extra energy that was necessary to
make this happen it is often known as the
work function of the um surface so on so forth
but generally speaking this was the basic
concept on the particle nature of light which
was been asked in this first question
the second question goes into the uncertainty
principle which states that if you can measure
position and time simultaneously then you
cannot measure so here um one of the important
things to remember that it actually questions
both position and time and then gives you
choices with respect to that and so the correct
choices essentially the correlation between
these two which we now discuss here so this
was first realized by heisenberg where he
was able to understand in the world of very
small particles one cannot measure any property
of a particle without interacting with it
in some way this interaction introduces an
unavoidable uncertainty into the result and
one can never measure all the properties exactly
now this was the realization of heisenberg
that gave rise to this uncertainty principle
its illustrated here in some detail so if
you shine light on electron and detected reflected
light it just using a microscope then assume
this kind of a way cartoonist picture where
the electron photon collision happens before
and minimum uncertainty in position is given
by the wavelength of light so to determine
the position accurately it is necessary to
use light with short wavelength however ah
a after the electron photon collision by plancks
law we know that there is a photon with short
wavelength has a large energy which would
mean thus it would impart a large kick to
the electron but to determine its momentum
accurately electron must only give a small
kick
so the moment you want to actually determine
the position very accurately you were almost
lost in a terms of its momentum and thats
the basic idea here because of the re coil
of this so this means using light of long
wavelength so the measurement disturbs the
system and ah that is the consequence of this
entire problem so because the particles are
so small the interaction is large enough to
do these issues so the uncertainty principle
essentially means that there is a fundamental
limit to the accuracy of a measurement determined
by the heinsenbergs uncertainty principle
and this is different from the classical uncertainty
where the measurement of uncertainty is due
to limitations of the measurement apparatus
there is nothing to do with the fundamental
ah limit which is there in terms of quantum
mechanics and in this particular case for
example in classical there is really no limit
in principle to how a accurately measurement
can be made it all depends on the instrumental
apparatus that we are talking about
in quantum mechanics we have these two very
important aspects one related to the position
which is ah the position accuracy and the
simultaneous measurement of the linear momentum
its with precision delta p then the product
of the two can never be less than this this
is one part of the uncertainty heisenberg
also recognized the time energy uncertainty
relationship which is related to the energy
of the system as well as the time that is
ah be used and so and there are two different
uncertainties and the problem probe both of
them thats why the answer was for both the
cases when position and velocity are the once
which cannot be measured simultaneously and
similarly time and energy cannot be mentions
simultaneously
when all these four parameters are there thats
the reason why the answer was given in that
particular wave this particular problem is
perhaps easier ah which basically introduces
and looks at the concept of the wave function
which is just a mathematical description of
a quantum entity and in a little bit more
detail manner ah here is the description which
we have already looked at before but in very
precised sense this is the schrodinger's a
proposed equation that accurately describes
the motion of wave that accompanies the electron
this wave represented by the greek letter
psi gave precise predictions for the behavior
of atoms
the wave function is a complex function a
function containing quantities that are complex
numbers and is constructed in such a way that
performing an operation to take the modulus
of the complex function for a given state
results in a real value that represents the
probability density of a result that corresponds
to that state this is how wave function is
transformed into a meaningful physical prediction
so thats the importance of the wave function
that it actually embeds the all the physical
um a important information that are necessary
so the wave function represented by greek
letter psi is a mathematical description of
a quantum entity and thats what the choice
answer would have appealed
similarly ah the next question is on the combined
condition of atomic orbitals which is a molecular
orbital and this is defined in many different
base but this is a particular definition which
is been looked at here and there is a there
is a little right up on if that i have put
up for you now
so that it becomes clear a molecular orbital
is a mathematical function that describes
the wave like behavior of an electron in a
molecule the term orbital was introduced by
mulliken in nineteen thirty two as an abbreviation
for one electron orbital wave function and
thats the reason why the most appropriate
um connection between the molecular orbital
these to the one electron orbital wave function
at an elementary level molecular orbital is
used to describe the region of space in which
the function has a significant amplitude they
are usually constructed by combining atomic
orbitals or hybrid orbitals from each atom
of the molecule or other molecular orbitals
from groups of atoms molecular orbitals can
be calculated quantitatively thats why molecular
orbitals have a lot of ah important positions
and intec interactions to take here and so
i thought a little bit more detail on what
we have done before is necessary and here
the most important part to keep in mind is
that the is a mathematical entity which has
to do with ah one electron orbital wave function
and it sort of like connecting to the molecular
orbital in that sense
there are two questions on de broglie so the
first one is just trying to see that ah you
connect the de broglie to the concept that
he essentially introduced this is a a matter
wave characteristic and so anything related
to the wave like nature of matter is now considered
to be related to de broglie so that was the
point of this particular question and this
is the concept that he introduced through
his thesis these thesis in nineteen twenty
four where he proposed that just like light
both wave like and particle like properties
electrons with momentum p also have wave like
properties like wavelength wave like properties
like ah ah sorry with wavelength lambda such
that lambda is related to the plancks constant
over the momentum
now this relationship is now known to hold
for all types of matter and not just for electrons
although it was initially proposed by him
in his thesis for the electrons but it was
for everything and in the next question ah
taking on to the fact that this matter wave
is universal fact the question is raised on
how the ratio of that ah goes between them
and so the ratio of de broglie wavelength
for a cricket ball of mass of a given mass
to that of ah helium atom both of which are
traveling at the same speed it just for argument
sake is being looked at and ah for to do this
problem its important that you use the de
broglie relation with that relates matter
with momentum and its wavelength and take
a ratio of the two cases and realize that
the wavelength is inversely proportional to
mass and the velocity
now since velocity has been kept same so it
essentially means that you have to only consider
the mass of the system for both of them and
once you do that you just get a ratio of the
two masses in this particular case point four
and ah four coming from the helium is canceled
and you are left with a order of magnitude
value which is simply relating the conversion
hm between the two
so the next question looks at a degeneracy
of a particular state of ah so for example
ah this is a thirteen point six over n square
electron volt is the energy this is the as
you know for the hydrogen atom and it is under
the constraints that the state has all these
other characteristics and you are suppose
to find the solution and this involves realizing
that each quantum level n for an atom ok thats
the principle quantum number there are zero
to n minus one angular momentum states each
of which have a degeneracy of again ah zero
plus minus one all the way to plus minus l
which means that each l state will have two
l plus one degeneracy
so the total degeneracy given that each n
state can have any volume from zero to n minus
one would then mean a summation of this entire
two l plus one over zero to n plus one so
once you do the summation will find that the
answer it's a its series where the answer
comes out to be n squared so thats how this
little maths is suppose to take you there
the next ah couple of questions on this are
essentially looking at the properties of these
molecules to some level but based on very
simplistic arguments and so considering a
harmonic oscillator model for hydrogen molecule
and an hd molecule and assuming that the same
force constants exist for both molecules the
lowest energy state for h two is you have
to find out that to basically you would need
to know the how many oscillator model form
of energy and the harmonic for a harmonic
oscillator energy states are given by this
where b is the quantum number where w can
go from anywhere from zero to whatever number
and then here the way the energy essentially
exist makes sure that there is a zero point
energy and a because at v equal to zero the
energy exits its half h cross omega
h cross by the way as we are a known is h
over to two pi ok now the frequency of vibration
is given by ah the spring constant over that
or the force constant over the a mass or reduce
mass as the system may be and assuming thats
this has been given in the problem that the
same force constant exists for the both the
molecules the lowest energy state for both
will have an inverse relationship with respect
to their masses with respect to their their
respective mass which means that ah you will
have the e zero being proportional to one
over root m and since hydrogen mass is a lower
mass as compare to the hd molecule the lowest
energy for h two will then therefore be higher
than that of hd because a its inversely proportional
so you will be slightly lower in a energy
then a h two so that is the level of identification
that we are suppose to make for these kinds
of the problems
now similarly the other simple question was
based on the ideas to how to looks for wave
functions and realize how many variables are
necessary for describing a wave function so
for a schrodinger wave function have of this
is helium atom am he was asked how many coordinates
or how many variables this depend on so in
order to do that the best thing is to draw
a simplistic picture of say the helium atom
which has one nucleus and two electrons now
the potential part of this will have the electron
electron reflection part where there is a
r one two which is the relative ah distance
between the two electrons and the two columbic
attractions would be between the nuclei and
the nucleus
so the potential energy part will have an
three components one of them is repulsive
and the other two being attractive and therefore
the wave function will depend on each of these
three parameters and each of these parameters
are ah dependent on entire set of three possibilities
so what we are essentially saying is ah if
we consider all possibilities we get an nine
dimensional problem ah more clearly ah helium
atom has one nuclei as we mentioned and two
electrons each of which ah would depend on
three variables implying that its ah a wave
function would therefore depend on all possible
nine variables
so this is how this problem has been looked
at and this is the sort of the way this problems
have been designed and done in this kind of
problem set the there is a one issue in the
last problem of this week one which while
i was going through a solving it i realize
that this question were the lowest rotational
energy gap of the linear carbon dioxide molecule
is being questioned under the assumption that
it's a three d rigid rotor this is a completely
um thought problem because its thought of
as a linear molecule and assume to be a three
d rigid rotor where the carbon is set to be
at the center of the rotating axis and
um and two oxygen atoms exactly at bohr radius
so its taken as fifty three um pecometer on
both sides ah from the central carbon atoms
so the carbon is consider to be the center
and this has been asked with the idea that
whether you can calculate this to be a number
which comes out to be a exactly one of these
unfortunately am perhaps i didn't do it very
with a lot of effort but have when i was doing
it quickly i could not get exact match to
one of these problems and there one of these
answers which i given here and so thats actually
little bit a way a difficulties
so what have decided is that i will open the
problem for you and then we will ah look in
to this problem further into the next lecture
because i dont think will have much time in
this lecture to finish the problem but what
we will do is will open up this problem ah
for this ah now and if we are not able to
get any answer then we will have to see why
the must have been and printing typing error
on my side on while making the question and
let us see
so the question is based on the idea that
we know how the energy obey three d rigid
rotor looks like and that is sort of given
in terms of a this expression where the energy
of the lth state is can be written in this
term very often you are used to the h cross
terms however in this particular case since
we are going to write in terms wave numbers
i have taken out the h cross and therefore
we have gotten these a pi sitting in there
if you write in terms of h cross these pi
square term vanish and its a perhaps sometimes
easier to remember them a in that forms so
we often write them in the h cross terms with
two i so thats roughly what we do because
four pi squared ah gets absorbed when as soon
as we say h cross in there by irrespective
whatever we have been number ah a your energy
is going to look like this and this results
in an energy gap in terms of l to l plus one
let say you will get something of this form
ah and if you write it in terms of in wave
numbers then you are going to divide it divide
the answer by h c thats the idea and in this
particular case when you divide by h c you
have to remember to use your value of hm c
in terms of a centimeters per second
so generally in these cases what is done is
it is taken as ah three times roughly three
times ten to the power ten centimeters per
second and ah this is the velocity of light
which is taken instead of ten to the power
eight three times ten to the power eight meters
per second this is just to make sure that
we can get the wave number unit at the end
so the i is the so we have here the quantum
number l rotational ah rotation constant ah
be i now i is related to the reduced mass
in this case so i is mu r squared r is the
distance um so in our in our particular model
when we have the carbon sitting in the center
and the oxygen sitting linearly on the other
sides the r is basically this distance that
we are looking at which has been given to
us the mu is the reduce mass
now the reduce mass can be taken in many different
ways ah in this particular case the simplest
one is to consider only the carbon and the
oxygen because those are the only two points
that we are looking at the h the rest of the
same however in reality um by just considering
it to be a three body system you can always
utilize the harmonic mean for this system
also in spectroscopy however there is a little
bit more involved manner of doing this thats
why ah the problem were was simply simplified
by providing the assumption that its an a
linear level by this particular case um in
in in the simplest way of looking a three
body kind of a situation like this reduce
mass will just be the three masses in this
case oxygen um carbon and oxygen this could
also be used but to make it even simpler we
just kept between the two to get rough estimate
and we have to always use the unit for conversion
into the kilogram right was it goes and the
r has been given as we mentioned in the question
so the moment of inertia i can then be calculated
although this particular ah way of calculating
the moment of inertia has a lot of assumption
in it and any way by going by this root we
can get some form from where we can get the
lowest rotational energy gap which we have
described here by just plugging in the values
it so turns out that once we plug in these
values in order to get the number in terms
of wave numbers and at least i did not find
it to come close enough to the values that
we are used to the best be came to us i think
not what we have given as choice
so as we had been talking to you about this
problem it does not give any of the results
that we have been talking that we have provided
in this problem thats because ah most of the
so this last problem in that next problem
set had the difficulty that the printed solutions
that were provided as options did not actually
come as any of the answers that we are looking
at here either when you do it in the approximate
sense that we present here or any of the other
sense that we do here and therefore um i dont
think you will find any solution which will
be in connection correct to the option that
are given so in that way what i am going to
do is we are not going to have that as a grade
numbering for you ah will take that part of
but the solution part ah was important to
show you as how to do this problem because
there is more important to realize how to
do a problem rather than to just be able to
get some marks for it
so what i have done here is that i given you
an idea as to how to do this problem in case
something of similar nature comes out it can
be very many very much more complicated as
i have been mentioning here but many times
will be essentially looking at the problem
in a much more simpler way to make sure make
sure that we are getting the main aspects
of the problem so in the next week we will
be looking at the other problems such that
we have been giving you and with this ah i
would like to close todays lecture see you
next time
thank you
