We started looking at more than one single
mode, that is multi-mode states and we started
looking at two more states and took some examples.
Do you have any questions?
Sir, we expressed general ket function psi
phi as a product of 2 n 1 and n 2, does it
hold a general case summation on n 1 and n
2 c n 1 and 2 and n 1 and n 2?
Actually, for example, when we quantize electromagnetic
radiation, I find a discrete set of infinite
number of oscillators, each oscillator can
have energy of n plus half h cross omega,
each oscillator is characterized by its annulation
creation operators, each oscillator is decided
by its ket state etcetera etcetera. So, the
complete such state is given by, actually
if I have n one photons in mode 1, n 2 photons
in mode 2, n 3 photons in mode 3, n l photons
in mode l infinite. This is 1 set 1 ket for
example, where there is 1 state in which there
are n 1 photons in mode 1, n 2 photons in
mode 2 etcetera and this I written as n 1
n 2 n l. So, this product means, it is essentially
this is a simplified form of writing the state.
This is fine. What I am asking is that if
we consider two mode states? A general 2 mode
state will be summation over n 1 n 2 c n 1
n 2 and this. So, I have for example, have
c n1 n 2 n 1 going from 0 to infinity and
n 2 going from 2 to infinity.
Now, and then what conditions can be separated
into a form of a product of over summation
of n 1 which are now the states n 1 and this
thing? It depends on this expansion coefficient
here. If you had for example, I would imagine
that if this could be written as a product
of a constant which is n 1 and n 2, then I
can split this into 2, product of two sums,
means I can write this as, sigma n 1 c n 1
n 1 sigma n 2.
Is there any physical significance of being
able to split into these two parts? No, this
is a subset of this. This is more general
than this. This is more general. This is one
set of states, in which, the overall state
is represented as product of state one and
state 2 mode 2 and mode 2. If you do any measurements
of any observable on mode one, you do not
touch mode 2. Though measurements on mode
1 will not influence any of our observations
of mode 2, if you detect for example, whether
there is suppose this is a horizontal and
vertical polarization states of two modes,
same frequency same propagation direction,
if I measure, if I pass this entire state
through a polarizer which is horizontally
oriented corresponding to frequency omega
1. I split the omega 1 and omega 2 for example,
then I will not influence any measurement
of the second state, that does not happen
in a state in which you cannot identify the
product state.
What will be the identity of that? No, because
what happens is any measurement done on one
of the states influences the result of observation
on the second state and those are the entangled
states where this kind of correlation exists
between measurements performed on mode 1 and
mode 2 and this is the most general, this
is a specific type of multi-mode state or
2 mode state.
So, before we look at beam splitter problem,
I want to generalize this into a multi-mode
state. So, for example, suppose I take in
classical picture, a monochromatic wave, a
monochromatic wave occupies time from t is
equal to minus infinity plus infinity, because
it is exponential I omega t or cos omega t
or sin omega t forever for all times. The
moment you make a finite duration wave, which
means I say it is exponential I omega t or
cos omega t from t is equal to 0 the capital
t. It no more remains monochromatic, because
you can take a fourier transform and show
that this contains many frequencies.
So, it is also I can also build a wave which
has a finite duration by adding waves of multiple
frequencies meeting between various frequencies.
So, that becomes a non-monochromatic wave
then. I cannot have, I cannot say, I have
a pulse, which is monochromatic, because the
moment you make a pulse of a finite duration,
it is no more monochromatic, it has a finite
spectral width. Now, the single mode state
that you have looked at, is a single frequency
state, because single mode means single frequency
single polarization state and one propagation
direction k vector. This is this is state
which exists for all times. If I take a single
photon in that state, that single photon can
be anywhere from t is equal to minus infinity
or plus infinity, if I try to detect that
photon, I can, it may be anywhere, I do not
know.
Now, I can form what are called as wave packets
by superposing modes of different frequencies.
So, if I take for example, if I take two frequencies,
I will have meeting between these 2 frequencies
and I will have a distribution which is not
uniform right across time, but it will be.
So, if the limit what can I do is, I can write
the state, a single photon state, what this
implies is this is c 1110203 plus c 2011203
plus c 3010213.
How many photons are there in this state?
So, if I operate by the total number operator,
you let me use some other index m a m. So,
n psi, this is equal to sigma m sigma l c
l a m dagger a m one l. I should be careful
I used different indices for the state and
the number operator here. The total number
operator is consists of a sum of all the number
operators corresponding to each mode. What
is the value of this? If m is not equal to
l, then the corresponding value of the number
of photons in this. This is a short form.
This is like this. So, if m is equal to 2
and this is first 1 it is 0 because 0 there
is 02. So, if m is not equal to l, this gives
me 0 and if m is equal to l, this gives me
11 l. So, what this gives me is, so, this
is simply delta m l 1 l. So, what happens
to this state sigma l? If m is not equal to
l, a m dagger a m is the number operator for
mth mode, and if this particular state corresponds
to a value of l which is different from m
and there are no photons in that mth mode
and in that state and that it 0. Only m is
equal to l terms will survive and if m is
equal to l, then a m dagger a a l dagger a
l 1 l is 1 l.
And. So, I get back c l. So, this means that
this state is a single photon state, total
number of photons is one and you can show
that this is not an Eigen state of a m dagger
a m. That means, let me try to find out whether
this is an Eigen state of the number operator
of one of the states one of the modes. So,
a m dagger, a m dagger a m of psi is equal
to sigma l c l am dagger a m 1 l, which is
how much and this is not equal to psi.
This is not an Eigen state. What it means
is, if you were to measure the number of photons
in the mth mode, you can not define. Every
time there will be a different value, you
can calculate expectation value. What will
be the expectation value? So, mod c m square
defines the probability of getting finding
m photons in this state, in the m, that photon
in the mth mode. That photon, there is only
1 photon, but that is in a superposition state.
So, these modes could be different frequencies.
These modes could correspond in this summation.
These modes in the summation could correspond
to this same propagation direction same polarization,
but different frequencies.
So, by a choice of c l amplitudes, I can build
a state, where just like in classical picture
where I can have different frequencies adding
up to give me a wave packet, I can build a
single photon wave packet by choosing appropriate
values of this coefficient c l. c l depends,
if l represents various frequencies, because
this l now corresponds to various frequencies.
That is various modes corresponds to various
frequencies, all propagating in the same direction
having the same polarization state then by
an appropriate choice of c l s mod c l square,
I can add this up and it will become localized.
It will become little more confined in time
than in an infinite time space,
Just like in classical, if I have a single
frequency wave, that wave extends from minus
infinity to plus infinity, but if I add multiple
frequencies appropriately, in fact, a continuous
range of frequencies from omega 0 to omega
0 plus delta omega, I will get a wave packet,
whose duration will be one by delta omega.
If the spectral width is delta omega, I will
have a duration of a pulse, which is one by
delta omega, of the order of one by delta
omega. So, depending on the amplitude c l
that I choose here, the combination I choose
here, I can actually build a a single photon
wave packet. So, it is this kind of wave packet
that emerges when an atom inside is excited
and emits spontaneously. Because when an atom
is an excited state and it emits by jumping
down, it emits one photon. And that one photon
is a wave packet like this it is coming out,
its frequency is not well defined, because
the energy levels for a finite width natural
broadening etcetera etcetera we discussed.
So, what is going to happen is, this photon
which comes out is a single photon, but in
a form like this, it is not in a single mode,
it is in a superposition of modes. And as
you will be able to measure, you can find
the probability of detecting it after a certain
time. Once you've excited the atom, you can
put a detector here and find out what is the
probability of detecting after one nanosecond,
what is the probability after detecting the
photon after one point one nanosecond etcetera
etcetera and you can form and you will get
a certain distribution with respect to time
which is exactly the spontaneous decay, that
you can measure in spontaneous emission process.
So, these are multi modes single photon states,
and when we say that, we have a photon incident
on a beam splitter or on a device single photon
incident, it is usually, this kind of a wave
packet that comes in. There's only 1 photon,
but it is in a state in which it is a superposition
of many modes, many frequencies. If you choose
a larger spectrum of frequencies, you will
have a narrower wave packet. If you choose
a smaller spectrum of frequencies, you will
have a larger wave packet, which means, you
have less uncertainty in where to find the
photons within the wave packet, if you have
a narrower spectrum. And finally, when into
the monochromatic, it is everywhere.
So, this is what is meant by a. In fact, generalization
of a 2 mode state in to multi-mode state and
I can still have 1 photon in a multi-mode
state, which means, it is occupying, it is
in a superposition state of many frequencies
if I take a photon propagating in one direction
with a one polarization state, it would be
essentially, it is this concept and, so this
state does not have a single frequency, it
has many frequencies. As you can see here
the probability of detecting a photon in the
mth mode, that means, your particular frequency
is mod c m square and the appropriate choice
of mod c m square I can build a wave packet
yes.
When in a monochromatic medium that thing
is that intensity in the time domain. Yes
yes. That means I will not take make exactly
at width. Yes yes, which means if I were to
suppose I take a a quasi monochromatic single
photon state, quasi monochromatic means it
is almost monochromatic. So, suppose here
it corresponds to a a delta p of say one millisecond
duration. So, the corresponding frequency
width is one by one millisecond which is one
kilohertz or something like it approximately.
So, one kilohertz bandwidth will give you
one millisecond time. So, I prepare n large
number of identical one millisecond duration
single photon states and I do experiments
on this.
So, what I would say, what I will say is,
within this one millisecond you will not know
when you will detect a photon. So, I let this
gamma-logistics detector, I may first detect
it right in the beginning, second detector
may detect it after some time, third one third
state, please remember one photon is detected
only once, I am talking about multiple states,
multiple one an ensemble of systems. So, this
first one single photon state comes I measure,
I find the duration within the one millisecond
duration somewhere I find it, the next one
I find another time.
Sir, that means, within this one millisecond
duration any detector can detect? Will detect,
if this is the size we define in the duration.
Usually there is a wave packet which has a
tail like this. So, there is also a probability
of detecting much later also, but the probability
is very very small. So, this mod c m square
distribution will determine which detector
will be taken this time with what property
suppose I could choose, suppose in the limit
of the quantization module. You see, what
is the frequency spacing between the various
modes? it depends on the value of l that I
have chosen, when I quantize the electromagnetic
field.
If I were to increase l, the frequency spacing
become smaller and smaller and the frequency
has become almost continuous. In that case,
I can actually extend the sigma into an integral.
Now, if I take mod c m square as a function
of frequency to be like a Gaussian function,
that means, there are various frequencies
which are present with amplitude which depend
on the Gaussian function , the corresponding
wave packet in the time domain will also be
a Gaussian function, because the Fourier transform
of a Gaussian is a Gaussian. So, I will have
a Gaussian single photon wave packet generated
by a mod c m square, which is again a Gaussian
in frequency space. So, in that Gaussian for
example, I would have, I could have this is
some value of m. So, these are values of m
which correspond to different values of frequencies,
each value of m corresponds to one particular
frequency.
Do we have something like a discrete step?
If l is finite, as l tends to infinite values,
as I increase my quantization volume, these
frequencies get closer and closer. But then
l over here refers to the different frequencies,
I mean every l refers to the frequency? Yeah,
in this summation, I am taking a combination
of modes, for which each l corresponds to
a different frequency. If we increase l let
it, Not this l, I am taking about capital
l, the volume ,the quantization volume, if
I increase the frequency spacing becomes smaller
and smaller and finally, I will have all the
frequencies this is continuous almost becomes.
So, there will be a one particular value of
m in this value, which will have its peak
value, which corresponds to one particular
frequency and there will be different values
of m where the frequencies which corresponding
frequencies where the amplitude will fall
down. All it implies is the probability of
detecting a photon with this frequency largest
within this wave packet the probability of
detecting a photon with this frequency is
very very small. This will lead in the time
domain to a wave packet which exactly looks
like a Gaussian, whose width will be inversely
dependent on this width in the frequency space.
The larger this width is, the smaller is the
time width. The smaller this width is, the
larger is the wave packet width in the time
domain.
So, in a spontaneous emission, the exponential
decay we know, and the spectrum is what the
Fourier transform of an exponentially decaying
function, Lorentzian. So, this spectrum which
you will get lorentzian spectrum, which also
look like a Gaussian, it is much broader.
It is a lorentzian spectrum and the probability
of detecting a photon at this instant will
be maximum and as the time delays, the probability
will fall down to almost 0. And you will define
a natural, a life time of a state etcetera
etcetera. The life time of the state defines
a spectrum width of the emission. So, there
are all related to this. So, this is a very
interesting state, because this is still a
single photon state, but that photon is in
a superposition of many frequencies. You can
only define the probability of detecting a
particular frequency for that photon.
Sir, with different probability, I will get
different energy photons? Yes, Unknowingly
uncertain here, because the frequencies are
uncertain. If I have to precisely define the
energy, it has to be a precise frequency.
So, there is uncertainty in the frequency,
there's uncertainty in the energy, there is
uncertainty in the position, where when you
will detect it when it comes to you. In fact,
I will come to this detection a little later
through another operator, but before that.
So, I thought I will introduce this and this
is a very interesting concept where I can
generalize from a single mode state to a multi-mode
state. In fact, I can further generalize to
not having the same propagation direction
or same polarization in to this sum could
represent different propagation directions
and different frequencies etcetera etcetera.
It is very general.
Now, what I want to do today is because now
I want to introduce this beam splitter. This
is a very important component in many experiments
which deal with quantum aspects of light.
I can build a interferometer with a beam splitter
I can do lot of things and beam splitter is
also used as a model to incorporate losses
or scattering etcetera in quantum optics discussions.
So, what is a beam splitter? I have a device
in which light is incident from here partially
reflected and partially transmitted.
Now, to simplify the mathematics I am going
to assume a beam splitter which is completely
symmetric. That means, the reflected, the
amplitude reflection of this wave is the same
as the amplitude reflection of this wave on
the other side. So, this wave comes reflects
this wave when it comes when I am drawing
double arrows here, its partly deflected and
partly transmitted. So, the single arrow gives
you a single arrow reflected light a single
arrow transmitted light, a double arrow incident
light gives me a double arrow reflected light
and a double arrow transmitted light. And
by symmetry, I am assuming that if this reflection
coefficient is small r this reflection coefficient
is also small r. If this transmission coefficient
is small t, this transmission coefficient
is also small t. I can actually generalize
to asymmetric situations, but we will restrict
this.
So, let me call the amplitude reflection coefficient
as small r and the amplitude transmission
coefficient as small t. So, let me call this
the port one, port two, port three and port
four. There are two input ports and there
are two output ports. This is a very general
device actually. A lot of devices have 2 inputs
and 2 outputs. Please note even if you send
like only in one port, the other port is still
there. You may not be considering the second
port, but there is a second port and I will
show you that in classical, you can forget
about the second port, in quantum, you cannot
forget the second port because vacuum is incident
from second port all the time. Even if you
do not send a light beam from here, there
are modes all modes are occupied all the time
with vacuum at least. So, you cannot forget
the fact that there is vacuum in this input.
Incidentally and that matters a lot in quantum
mechanics, because if you neglect that you
have problem in computational relations.
Now, so first let me do a classical analysis
of this beam splitter and gets some general
relationships and then we will quantize that
through a particular procedure and I will
tell you the procedure of getting the quantized
relations between the annulation operators
of these fields 3 and 4 with respect to the
fields 1 and 2. So, if I have incident waves,
let me call this E1 E2 are incident E3 and
E4 are the output ports. So, what is E3? E3
will be r times E1 plus E times E2 and E4
will be t times E1 plus r times E2. If the
r and t s were not the same I have to write
t prime and r prime for the second 1. And
please note r and t are in general complex.
it is an amplitude reflection coefficients
and amplitude transmission coefficients not
energy reflection coefficients. So, these
are complex
Now, I need to satisfy energy conservation.
So, the sum of the energies in 3 and 4 must
be equal to the sum of the energies in 1 and
2. The energy, certain energy is incident
from here, certain energy is incident from
here,that must be either in 3 or 4, I am neglecting
any absorption or any other scattering or
anything else. So, let me try to calculate.
So, the energy is proportional intensity which
is proportional to the mod modulo square of
the electric field. So, let me calculate what
is mod E3 square plus, mod E4 square.
So, mod E3 square plus mod E4 square is equal
to modulus of r E1 plus t E2 whole square
plus modulus of t E1 plus r 2 whole square.
This is equal to modulus r square modulus
E1 square and please note, E1 and E2 could
also be complex. They are phases plus mod
t square mod E2 square plus r t star E1 E2
star plus r star t E1 star E2 plus mod t square
mod E1 square plus mod r square mod E2 square
plus t r star E1 E2 star plus e star r E1
star E2. So, let me collect the E1 square
term. So, this is mod E1 square into mod r
square plus mod t square, this and this term
plus this and this gives me mod E2 square
into mod r square plus mod t square plus E1
E2 star into r t star E1 E2 star E1 E2 star
plus t r star plus E2 E1 star into r star
t plus t star r. So, this is same as this
is r t star plus t r star r t star plus t
r star. So, this is actually mod E1 square
mod r square plus mod t square plus mod E2
square mod r square plus mod t square plus
r t star plus t r star into E1 E2 star plus
E2 E1star.
Now, what is the condition mod E3 square plus
mod E4 square must be equal to mod E1 square
plus mod E2 square irrespective of the values
of E1 and E2. I can choose E1 is equal to
0, I can choose E2 is equal to 0, I can choose
E1 is equal to E2, any combination I can choose
and still all the time I must have mod E3
square plus mod E4 square as equal to mod
E1 square plus mod E2 square. So, what are
the conditions I get, this must be equal to
1 and this must be 0. Other wise I cannot
satisfy this equation for all values of any
combination you need to. So, I get just simply
by energy conservation. I get Mod r square
plus mod t square is equal to 1 and r t star
plus r star t is equal to 0. So, let me write
r is equal to mod r exponential I phi r and
t is equal to mod t exponential I phi t. So,
this is mod r into mod t into exponential
I phi r minus phi t plus exponential minus
I phi r minus phi t 
r t star plus r star t.
Mod r and mod t are not 0 need not be 0. So,
this tells me cosine of, though this means
phi r minus phi t must be equal to, this is
symmetric situation. If you do not have a
symmetric situation, you get a different combination
of phases of r r prime t t prime etcetera,
you will have another equation there. But
here I get and this is in general, I have
not assumed anything. This is actually a general
4 port device. 2 input port, 2 output ports
the phases of the reflected in the transmitted
components must be different by phi by 2,
if it is a symmetric device.
So, for example, if my beam splitter was a
3 d 50 percent beam splitter, which means
it reflects 50 percent of the light and transmits
50 percent of the light, so, what I can do
is I choose for example, phi t is equal to
0 and phi r is equal to pi by 2 so as to satisfy
this condition. So, t is real and r is imaginary,
phi r is pi by 2. So, if I have a 50 percent
beam splitter, I have r is equal to what is
the magnitude of r, 1 by root 2 and there
is an exponential I phi r so, this is I by
root 2 and t is equal to, this is for a 50
percent beam splitter.
The phase difference is independent of whether
you are choosing 50 percent beam splitter
or 80 percent beam splitter, it is always
pi by 2 in this symmetric situation. So, if
I go back and write the earlier equations,
I find E3 is equal to I by root 2 E2 plus
1 by root 2 E2 and E4 is equal to 1 by root
2 E1 plus I by root 2 E2.
Now, when we do phi t equal to 0 pi I equal
to pi by 2 for all you're given, if you have
the matching generated to it? Yeah, but symmetric
what I am saying is, I have assumed r is equal
to r prime is equal to t t prime, if you do
not have symmetry for example, if I have glass
air interface and interface is not symmetric
situation. So, the reflection coefficient
from here and transmission coefficient complex,
I am talking about complex reflection and
transmission coefficient could be different.
So, I have to work it out and it is very simply.
It is just the same analysis except that I
need to ensure that when I write this equation,
I do not write I write r and t and t prime
and r prime here, that is all, and I do the
same analysis and you can find out the relationship
between the phases of the various coefficients.
So, in fact, this is a in a matrix form E3
E4 is equal to I by root 2 1 by root 2 1 by
root 2 and I by root 2. So, these are classical
relations, you know quantum mechanics in this
these are completely classical equations.
Now, use the following procedure to understand
the property of the interferometer in the
quantum terms. So, these electric fields now
become operators and I will for example, I
have a relationship between the annulation
operators of the mode coming out at 3 and
the modes 1 and 2, similarly the annulation
operators for 4 1 and 2. So, by quantization
I would say, that the annulation operator
a 3 corresponding to the field coming out
in the port 3 is actually I by root 2 a 1
plus 1 by root 2 a 2 and a 4 is 1 by root
2 a 1 plus I by root 2 a 2. A quantum mechanical
operation of the beam splitter is written
in terms of the annulation operators of the
fields coming out in 3 and 4 ports, their
relationship with the annulation operators
on the fields coming in 1 and 2.
Sir, why is it the same? I am just using,
the sort of proposing here, that I will quantize
this by this operation. And I will show you
that a 3 and a 4 satisfy the commutation relations
that are required for a 3 and a 4. a 3 must
satisfy a 3 a 3 dagger must be 1 and a 4 a
4 dagger must be 1. That is also satisfied
from this equation. So, actually what I am
trying to do is, I am just replacing, which
I expected to be all right, replacing the
classical electric field operator by quantum
electric field operators, classical electrical
fields by quantum electric field operators.
Electric fields are replaced by operators.
Some negation condition might be satisfied
by some other combination for a 3 and a 4.
No, once I, this is the quantum equivalent
of this beam splitter, which has this classical
relationships. Of with quantum expression,
which expression this expression I have got
classical. This is classical I replace the
electric fields by the corresponding electric
field operators.
I can see that but. I do in, but why isn't
it exactly that. Actually I must in a in a
completely correct picture I must represent
the beam splitter by a unitary operator and
work out its effect on the input wave function
or the input ket state that is coming in,
that is much more complex, a complex analysis
has to be done, but this gives me the same
result. I am not showing you, I am not proving
that this is right, but I am actually expanding
these classical equations into corresponding
quantum equations and this quantum equations
will predict the results of all measurements
that I do with is beam splitter for example.
This is where I have actually gone from a
classical picture of beam splitter to quantum
picture of beam splitter.
Sir, but if we in that expression for e 3
and e 4, that we write classically, if we
replace the quantum operators, the operators
were e 3 and e 4? No, assume that this is
like postulate, I have actually postulated
here in this in this here, but normally, I
would have to understand the quantum mechanical
operation beam splitter by taking it as a
operator operating on input ket state etcetera
etcetera. That's much more complex, but this
is a procedure which I employ and I find everything
is consistent. For example, I will tell you,
suppose I had not taken the second port into
account and I had said that let me look at
the following problem, I have a wave incident
on beam splitter, E1 is incident, generates
E2 and E3, E3 and E4, I would have got E 3
is equal to r E1. So, I would have replaced
this by. Now, if I calculate the commutation
relation a 3 a 3 dagger is equal to r a 1
r star a 1 dagger, which is equal to mod r
square a 1 a 1 dagger, it is a mod r square.,
that is not 1, it is not a correct relationship.
So, this why I said I cannot neglect this
port.
But it is, see, I think just writing the unity
replace the x for the electric field operator
poised at x we still have a dagger into b
r dagger I will have different k for all the
4 inputs. So, it is not intended for drag
correctly, you know, replace the electric
field by the operator, exact operator that
we derived? Yes . Then it is not introduced
still to write in the previous expression.
No, what I am trying to say is, I am giving
you a procedure, let me assume, that I give
you a procedure, to get a quantum equations
of the beam splitter from the classical equations.
This is done in many other areas where I get
the classical equations and I replace the
electric the variables by corresponding quantum
mechanical operators and I get a relationship
and of course, the final thing is whether
this is correct or not will be predicted by
whether will this equation predict correct
experimental observations. There must be surly
more complicated procedures of getting these
equations from by analyzing the beam splitter
as a quantum mechanical operator etcetera
etcetera. But I think it is much more complicated
and finally, it seems the result is this is
the right result.
So, let me assume that this is the corresponding
operator equations representing, connecting
the operators of the transmitted and reflected
fields to the operator in the input fields
1 and 2 and the input 1 and 2. In fact, I
will do a similar thing when I come to parametric
down conversion. Remember in parametric down
conversion we had, did a obtained an equation
relating the electric field at the output
with the electric fields at the input of the
signal and idler. I can go into the quantum
mechanical picture by replacing those electric
fields by annulation of creational operator.
And I will show you that is the equation I
get there is the equation which is feasible.
Although this is not a very rigorous procedure,
I am not following a rigorous procedure of
deriving these equations, but because of these
classical equations, I am sort of applying
a procedure I should say, which is, which
seems to be correct, that I replace these
electrical operators by the corresponding
annulation of electric fields by the corresponding
annulation operators and get a relationship
between the annulation operators in ports
3 and 4 with respect to ports 1 and 2.
Now, you can check for example, here a 3 a
3 dagger is I by root 1 a 1 plus 1 by root
2 a 2 minus I by root 2 a 1 dagger plus 1
by root 2 a 2 dagger. So, this is equal to
half a 1 a 1 dagger plus I by 2 a 1 a 2 dagger
minus I by 2 a 2 a 1 dagger plus half a 2
a 2 dagger. Now, please note the inputs 1
and 2 are completely independent of each other.
So, the annulation operator a 1 and a 2 commute.
This is 0, this is 0, this is equal to 1,
this is equal to 1 and I get a 3 a 3 dagger
is equal to 1.
Similarly, you can show a 4 a 4 dagger is
1. If I neglected the second port at the input,
as I showed you, I am not able to satisfy
the commutation relation at the output annulation
operator and that is not a correct transformation.
So, let me look at one example of this operator
equations and that is an example where I assume
a single photon is incident on a beam splitter
from port 1. So, this is a 1 photon state
and here it is vacuum. So, I want to find
out what comes out of this beam splitter.
The input psi 11, which is actually, and this
is actually a 1 dagger, because 11 is a 1
dagger 01. Now, as far as the states are concerned,
I know that if there is vacuum incident in
1 and vacuum incident in 2, I will get vacuum
coming out of 3 and vacuum coming out of 4.
So, this beam splitter transforms this state
to an output which is both vacuum and the
beam splitter transforms a 1 in terms of a
3 and a 4. So, I have to invert this equation
and calculate a 1 in terms of a 3 and a 4.
So, if I multiply this, I can show that a
1 is equal to, I get a 1, I multiply this
by I invert this equation. So, let me give
you the inverted equation here. Minus I by
root 2 a 3 plus 1 by root 2 a 4. Estimated
we find out a 2 from here in terms of a 3
and a 4. So, a 1 depends on a 3 and a 4 and
so, if this is the input state 0102 goes to
0304 a 1 dagger gets replaced by 3 and 4 operators
and the output state becomes minus I by root
2 a 3. So, it is a dagger. So, plus I by root
2 a 3 dagger plus one by root 2 a 4 dagger
operating on 0304. So, I replaced a 1 dagger
by the corresponding operators in terms of
a 3 and a 4 and I have replaced the 0102 state
by 0304.
So, what is this? This is I by root 2 a 3
dagger on this. 1304 plus 1 by root 2 a 4
dagger on this gives me a 3 dagger operates
on 03 it gives me 13, A 4 dagger operates
on 04 to give me 04. What is the state? This
is a superposition state which we had discussed
earlier what it implies is the output state
is a superposition of the photon being present
in port 3. So, if I have a single photon coming
from here, you generate a superposition state
of 3 and 4, where the photon could be either
here or here, actually in both, which probability
is half, because of this 1 by root 2 amplitudes
here. The probability of detecting 11304 is,
which is 1 by 2 and similarly the probability
of detecting 0314 is also half.
So, the single photon now goes in to a superposition
state of being in this part of the beam splitter
and in this part of the beam splitter. It
will continue to remain in the superposition
state unless it is disturbed and please note
that I can go on for a billion miles and still
be in a superposition state of both arms and
if you were to detect, put a detector on one
of the arms, you will suddenly detect it or
not detect it.
Now, look at a wave packet, see this is a
single photon state, but single photons can
be in a wave packet. So, as if problem is,
it is as if the photon is in a superposition
state in a both arms, but suddenly collapses.
This is a problem in quantum mechanics. There
is a collapse of the wave function. A measurement
collapses in the wave function and that collapse
has no space variable in this picture. So,
it is not that the beam splitter is either
reflecting or transmitting the photon. It
is reflecting and transmitting. It is in both
arms. In fact, what I will do in next class
is to build a mach zehnder interferometer
from here and what you will find is that there
is an inter phase effect.
The single photon enter this beam splitter
the mach zehnder interferometer and interferes
with itself, the probability amplitudes actually
are interfering. It will not be the part or
the particle is going here, part of the particle
going here. The probability amplitudes of
the both the arms are interfering to produce
a probability of being detected at the output
in either of the arms. I think we will we
will stop here. Any questions?
Thank you.
