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 that we have been
given giving you for the last six weeks and 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 was given to find out which of these different choices was important
so 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 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 given by a max planck
so this quanta or the 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 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 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 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 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 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
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 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 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 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 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 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 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 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 taking on to the fact that this matter wave
is universal fact the question is raised on how the ratio of that goes between them
and so the ratio of de broglie wavelength for a cricket ball of mass of a given mass
to that of helium atom both of which are traveling at the same speed it just for argument
sake is being looked at and 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 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 so for example 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 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 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 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 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 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 dependent on entire set of three possibilities
so what we are essentially saying is if we consider all possibilities we get an nine
dimensional problem more clearly helium atom has one nuclei as we mentioned and two
electrons each of which would depend on three variables implying that its 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 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 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
for this 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 gets absorbed when as soon
as we say h cross in there by irrespective whatever we have been number 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 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 three times roughly three
times ten to the power ten centimeters per second and 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 rotation constant
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 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 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 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 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 will take that part of
but the solution part 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 i would like to close todays lecture see you
next time
thank you