So, short ah recap to begin todays ah discussion.
So, we were talking about time dependent perturbation
theory. So, we have a time dependent ah perturbation
which is likely to cause ah transitions of
from one state to another maybe from a ground
state to an excite state or vice versa .
And the the Hamiltonian ah the perturbation
Hamiltonian is written as ah its H prime m
k ah which is a function of t and then we
have a Amk sine omega t that is the perturbation
that we have been talking about. ah K is the
initial state ah where the system was before
the perturbation was switched on and m is
ah say an excited state 
or a different state that than ah compared
to K and we are ah going to talk about computing
this coefficient. So, that is related to the
probability of making a transition for the
system to make a transition from an initial
state K to a final state m, ah we could simply
say that its a final state and if we are ah
resorting to a first order perturbation theory
we will put a one here and the that is ah
going to be the thing ah or the quantity to
compute and this is what we have ah found
yesterday ah or rather the last discussion
that we were having ah that is. So, this is
equal to i A m k ah and 2 h cross and exponential
i t naught omega m k plus omega and ah minus
1 and omega m k plus omega and the exponential
i t naught omega m k minus omega ah minus
1 and ah omega m k minus omega.
And. So, it consists of 2 terms ah the first
term is called as the absorption term, which
we will see immediately the second term is
called as the emission ah term and we will
talk about these ah separately. In fact, ah
we are going to concentrate mostly on the
second term ah which ah as i said that its
related to the emission ah that is ah transition
ah from ah one state to another.
Ah. So, ah let us look at each one of the
terms ah. So, the denominator of the first
term . So, one is first term in RHS that is
inside the bracket ok . So, then ah the poles
of that pole spins were ah the denominator
becomes 0 that is this is the condition and
omega m k as we have said that it is ah E
m minus E k over h cross ah plus omega equal
to 0. So, that immediately tells us that E
m equal to E k minus h cross omega which means
that ah E m is less than E k and. So, E k
is the initial ah state or the unperturbed
level before the perturbation was switched
on. So, ah what it means is that the H prime
m k ah which is of course, related to the
ah A m k and so, on it ah carries away energy
energy and so, its related to the absorption.
So, the field is absorbing energy from the
system. So, that is why its called as a absorption
term and let us look at the second term.
And that term is of course,, with a negative
sign its a ah minus i Am k by 2 h cross and
I have an exponential ah i t naught omega
m k minus omega ah minus 1 divided by omega
m k minus omega and so, on ah we can write
this we can simplify this ah we can write
the ah numerator as a sign ah by taking exponential
i t naught by 2 omega m k minus omega out.
So, that gives ah this one the second term
with one ah exponential minus i t naught omega
m k minus omega by 2.
So, if we do that simplification then we will
ah and that way we can get rid off this i
as well because the sign ah in terms of the
exponential comes with a definition ah which
has a 2 i in the denominator, and once we
do all that which i leave it to you ah its
a simple step to complete. So, this is ah
h cross and we will ah keep a i t naught by
2 omega m k minus omega ah and ah. So, this
is ah exponential ah. So, this becomes really
a sin ah . So, its a sin and then there is
a t naught by 2 omega m k minus omega ah and
divided by omega m k minus omega and we just
come back to this term.
Ah Now look at the denominator the denominator
is of course, the denominator makes sense
when ah or rather it has a pole, when omega
mk equal to omega which means that ah Em now
is a E m minus E k ah its equal to h cross
omega and ah that tells us that the E m is
greater than E k which means that the H prime
m k or the perturbation term ah is it pumps
energy into the system, and this is related
to emission ok
So, ah now. So, basically we will ah more
concentrate on this rather than on the first
term because this term makes a large contribution
close to the resonance that is known omega
m k becomes equal to omega, and in any case
both terms are equally important let me just
ah with a diagram let me explain that. So,
this is ah your omega axis and this is omega
equal to 0, now I have ah plus omega here
ah and a minus omega here. So, these are ah.
So, there is a plus omega and minus omega.
So, there is a a line which are. So, this
corresponds to the ah emission line or the
second term 
and this corresponds to the absorption 
and that is the first term .
Now, of course, these are ah the spectroscopic
lines or these are the lines that will show
up in experiments, and these lines cannot
just be a ah ah you know very sharp lines
which would ah of course, and the width of
the line would depend on the ah basically
the least count of the measuring apparatus
or ah the energy range over which it can measure
ah sort of these or rather these see this
line.
So, let me give a little bit of a weight to
these lines. So, these will be the spectral
lines that will be seen for each one of them
and let us just talk about these ah things
to be ah delta omega and ah delta omega here
as well and this distance or rather the the
2 ah lines are separated by a twice of omega
fi ah which is ah. So, in this particular
case as it ah the way the ah figure is drawn,
its very clear that we need to consider one
line at a time in the limit that your delta
omega is much much smaller than omega fi or
2 omega fi .
So, if this condition is satisfied, then only
ah we need to look at one of the lines at
a time and maybe ah at this moment let us
just consider that we are concentrating on
the second term which is ah the the resonance
term. The other term is called as the which
is in the ah negative minus omega so, that
is called as the anti-resonant term . So,
they are symmetrically placed about 0 and
at this as I said that we are ah just going
to talk about the resonant term at this moment.
And let us tell ah this ah call this quantity
ah let us just write with a different color.
So, let us call this quantity as alpha ok
.
Ah And in which case the. So, the C m 1 t
ah mod square its equal to A m k ah by h cross
square and there is a sin square alpha t naught
by 2 divided by alpha square.
So, that is the form of this and let us in
addition ah take ah beta equal to alpha t
0 by 2. So, then this thing would be written
as Amk by h cross square ah sin square beta
by beta square and since we have taken a t
0 by 2 whole square common, that had to be
multiplied here. So, we will multiply it.
So, t t 0 square by 4 and that will be there
ok.
So, this is our ah form of the transition
amplitude, which is related to the probability
of transition from ah initial state k to a
final state m, and this quantity sin square
beta by beta square is called as the sinc
function and the sinc function or the square
of the sinc function sin ah theta by theta
is called as a sinc function. So, this is
ah if we plot it it looks like that you know
the square will never become negative. So,
ah it will look like 
ah this resembles the diffraction pattern
ah that you may have seen in optics 
ah is nicely symmetric I might not have drawn
it accurately, but it is it has to be you
know ah I mean a little ah sort of all right.
So, this is ah . So, I am plotting a sin.
So, this is my ah C m 
C m 1 t square and this is ah as a function
of ah you know . So, alpha in fact, this is
like sine square beta by beta square apart
from these factors or you can take it to be
proportional to this. So, this is proportional
to this this is actually ah sin square beta
by beta square ah versus alpha that is being
plotted. So, this is at alpha equal to 0 and
this is at 2 pi over t 0, this is at 4 pi
over t 0, this is at 3 pi over t 0 and this
is at 5 pi over t 0 and so on so, forth all
right.
So, this is a plot that we are ah plotting
sin square ah hm beta by beta square versus
alpha and so, ah pi by t 0 corresponds to
ah this entire thing Amk by h cross square
t 0 square by ah pi square ah ah t 0 square
by 4 and. So, at. So, 3 pi by t 0 will ah
corresponds to this entire thing divided by
9 ah and at 5 pi by t 0 it is ah. So, at 5
pi by t 0 it is ah it corresponds to this
entire thing divided by 25 and so, on.
So, ah if you see this the area under the
curve which is of course, related to the probability
of transition, ah it is proportional to t
0 because ah the height ah goes as 1 by t
0 and ah this ah alpha is ah and then of course,
its a ah there is a t 0 square that is there
and there is a 1 over t 0. So, this multiplied
by 1 over t 0 will give a t 0. So, that is
the ah transition probability and the ah.
So, area under the curve we will write it
here. So, area under the curve 
transition probability ok.
So, let me. So, that is the transition probability
which goes as ah of course, as t 0 all right.
So, let us just give a physical picture that
what is happening.
So, you have 2 level system here ah call this
as a phi f which is nothing, but that m ah
by the index m and this is a phi i and so,
this is that term number 1 ah that is the
absorption term or the anti-resonant term
and for the other one its the phi f and a
phi i and so, there is a particle that makes
a transition from phi f to phi i and that
is the second term, which is related to the
emission ok . So, that is the physical picture
for ah these 2 terms and of course, we are
ah more interested in ah talking about this
ah the emission term, that we are talked about
.
So, of course, ah the emission term is more
important 
that is we can neglect the absorption term
when ah as I said earlier that 2 of omega
fi we can put it inside a mod sign that is
it does not matter which one is ah bigger
the as the final state energy is a bigger
of the initial state energies are bigger,
ah this has to be much greater than delta
omega .
Ah Now delta omega that is the line width
ah depends on the time duration for which
the perturbation is switched on. And if we
simply ah go by the order of magnitude using
a or an estimation rather using Heisenberg's
uncertainty principle, then this can be simply
written as 2 pi over ah t 0 ah. So, just to
remind you t 0 is the duration ah for which
the perturbation is switched on ok. So, ah;
that means, that ah the t 0 has to be much
greater than 1 by omega fi ah which of course,
at resonance its equal to 1 over omega which
means that the additional restriction of neglecting
the anti-resonant term is that, the perturbation
term ah must go through several cycles ah
in this interval 0 to t 0 ok. So, that is
the ah the additional restriction that we
have.
Now, if ah the t 0 is of the order of 1 over
omega, we more ah we have a situation where
the perturbation is linearly varying with
time and in which case ah it is not possible
to neglect or rather separate out these 2
terms, such that we can concentrate on 1 at
the expense of another. So, we will take this
to be true that ah that this the perturbation
term ah must go through several cycles. So,
its rapidly oscillating ah during the time
that the perturbation is switched on all right.
Now, this has been told earlier it could happen
that ah during the course of transition the
particle or the atom goes on to a state, ah
the final state is a part of a continuum which
means that it is not a single well defined
state its a ah sort of collection of many
states and its a continuum; which forms a
continuum by a having many ah energy levels
ah having very close by values all right.
In which case we cannot talk about a probability
of this transition and rather we should talk
about a probability density of transition,
and this probability density should be integrated
over in order to get the total probability,
integrated over all possible energy states
ah that line that ah you know in that range.
So, coupling 
with states of the continuous spectrum 
ok and so, if the final state 
falls into a continuum 
ok and so, ah phi f is not well defined. So,
the probability density will be given by ah
phi f and psi t mod square, and this probability
density has to be integrated over and how
does it arise? It arises ah say consider a
collision event, in which a particle with
an initial momentum p i ah which. So, p i
which means that the energy E i equal to p
i square 2 m we are talking about a free particle
and this particle after the collision gets
scattered on to a final momentum pf.
And now the detector by which it is being
captured ah has an has a finite angular aperture,
and the energy selectivity of the sensitivity
may not be perfect, and in which case ah we
talk about the solid angle of the detector
which ah or other detects this particle and
this gives. So,. So, this going to finite
energy sensitivity, the detector measures
ah energies in a solid angle or momentum rather
ah we talked we should talk about momentum
ah in a solid angle d omega and that gives.
So, basically the final state energy ah . So,
which means that ah its between some E f and
E f plus d E f is the detection takes place,
in which case we will have to actually count
the number of states ah which falls into that
continuum of states ah between E f and Ef
plus dEf and so, the ah the probability ah
P of ah say phi f t ah it should be sum over
all these final states and then its a phi
f ah psi t and ah square.
Ah Now this sum over f that is the energy
states that line that range Ef to Ef plus
dEf is nothing, but the density of states
corresponding to that final state energy,
hence this p of phi f and t ah has to be integrated
over and one gets a phi f hm and psi t mod
square and rho epsilon d epsilon, and this
is the final result for that probability for
ah system to make a transition from an initial
state to a final state in presence of a time
dependent perturbation .
So, ah we now equipped to talk about the Fermis
golden rule and let us see what it means in
the present context that is including this
density of states for the final states and
so, on.
So, let us. So, ah let us take again a sinusoidal
perturbation of the form that we had taken.
So, we have taken this H prime to be ah or
rather H prime m k to be equal to ah a mk
ah sin omega t that is what we have been discussing
And now we let us call it as a P fi t just
the same thing as what we have written as
the ah transition probability its nothing,
but Amk a mk is in general hermitian. So,
which means Amk equal to a km star and so,
on. So, this ah divided by h cross square
and a sin square beta by beta square and at
t 0 square by 4 and beta equal to omega f
minus omega i divided by 2 into t . So, its
alpha into ah t and hence this is what we
ah need to calculate and then we if you wish.
So, this is that Pfi t now this P final state
ah phi f r t is equal to 1 by h cross square
ah d epsilon rho epsilon and ah phi f ah H
prime ah phi i square and a sin square beta
by beta square which is just again writing
this Amk, where ah just try to connect ah
that m is equal to the final state and k equal
to the initial state ok. So, this is ah implicit
here and now ah we have to ah perform this
integral.
Now in order to perform this integral it could
be quite complicated because there are ah.
In fact, 3 terms ah which all depend upon
the energy, that is the term which is shear
the term which is the density of states, which
depends upon energy which is usually a smooth
function ah in most of the cases that we are
accepting in some ah pathological cases where
ah this density of states show a divergence
ah close to some bandage or some such thing
ah this is the ah the matrix element square
ah and this is that sin square beta by beta
square of the sinc square function that we
have obtained because of the time dependent
perturbation
Now, its ah not too difficult to understand
that both these quantities are somewhat slowly
varying quantities. The only quantity that
varies very rapidly ah for the resonant term
we are talking about the just the resonant
term is this quantity that is the sin square
beta by beta square. And this sin square beta
by beta square if we just go on to see your
ah the plot that we have drawn earlier is
this. So, if you see this you see the major
contribution of this probability density would
come from this central peak and the ah the
auxiliary peaks are of course,, there,, but
they are you know ah by down by a factor of
9 or down by a factor of 25. So, they are
ah not as important as this and because of
this close to resonance, we can sort of ah
replace it by a delta function ok ah with
the width which is is very small and that
width of course,, we will talk about that
width.
But the fact that the sin square beta by beta
square can be ah replaced by your you know
by a delta function, that one has to be convinced
about first and this is what i am trying to
ah impress upon, that in this particular ah
this term or this integral that we are talking
about ah the one that is most important is
the sin square beta by beta square ok .
So, if we are trying to replace it by a delta
function c what are the conditions corresponding
to that. So, let us take this limit limit
t tending to infinity ah sin square beta by
beta square is nothing,, but a delta function
up to a constant factor, which is E minus
ah Ef i or we can simply write it as a epsilon
minus which is a general energy and this is
2 pi h cross t ah delta of Ef i minus E ok
ah. So, this is ah just like that ah limit
epsilon tending to 0 sin square x over epsilon
divided by x over epsilon is ah can be replaced
by a 4 pi by epsilon delta of x and just you
ah try to connect that x by epsilon is equal
to alpha t by 2 in our cas,e where ah uh x
is equal to alpha and epsilon equal to 2 over
t ok.
So, ah the main ah or the summary of this
discussion is that, that we are going to replace
this sin square beta by beta square by delta
function which is the ah singularly most ah
ah I mean rapidly varying function, the other
functions are somewhat ah slow varying and
can nearly be taken as constant ah in the
vicinity of resonance ok. And this ah delta
function has got a width which is given by
ah 2 pi ah h cross over t and . So, in this
ah width we are going to ah talk about the
density of states in that energy width how
many ah energy states are there, that need
to be incorporated in calculating the transition
probability . So, that is actually ah encoded
into our ah rho epsilon d epsilon .
So, let us just ah introduce that we are in
ah interested in talking about the transition
probability per unit time 
which ah is equal to 1 by t 0 and then we
have a C k t mod square ok ah let us just
write C m because this is what you are ah
C m t square and if we are talking about the
first order perturbation theory we will have
a one on ah in the here and then its integrated
over all final states ok and of course,, these
ah m ah lies in to in the continuum ok .
So, we will introduce as we have said that
we will introduce density of states, we can
write it as a rho m which is ah between ah
epsilon m and epsilon m plus ah dE m ok. So,
ah w will be ah 1 over t 0 ah d E m rho m
c m mod square now the dE m equal to d of
ah epsilon k or rather ah epsilon m minus
epsilon k minus h cross omega this is as we
said that is for the resonant term and this
is equal to ah d of omega k m minus omega
ah h cross and this is e equal to d of omega
k m minus omega and multiplied by a t naught
by 2 just to ah make it look like a beta that
we have introduced and there is a 2 h cross
by t 0. So, this is equal to a 2 h cross by
t 0 tb d beta .
So, ah the transition probability per unit
time becomes equal to 2 pi by t naught and
ah d beta rho m or we can I mean instead of
writing it as rho. So, we can write it as
rho epsilon m and a ah km mod square ah Amk
and A k m are same as I said that they are
mostly Hermitian. So, its sin square beta
by beta square. So, this is ah much slow varying
function 
as compared to this this is a fast varying
function ok . So, this is what it is . So,
these are the transition probability per unit
time and then of course, we can write this
down as.
So, this W is equal to 1 over 2 h cross ah
we take this thing out and take its average
value since ah this is a much slowly varying
function ah as compared to the second one
then we take this and take an average value
of these 2 terms and take them out, and let
us write them as rho epsilon m ah and A km
ah square ah and a minus infinity to plus
infinity ah that is sin square beta by beta
square and there is of course,, a d beta.
Now, this has a value this integral has a
value that value is equal to ah pi ok. So,
this becomes simply equal to pi over 2 h cross
and this ah rho ah now I can I can also split
this 2 terms and so, this is A km mod square
by 2. So, the average ah value of this ah
matrix element can be written as a half of
the ah of these A km square and. So, this
is equal to simply equal to. So, ah if we
write it in words, its equal to the transition
probability per unit time time is ah proportional
to the ah density average density of states
average density of states we write it with
a dos. So, Dos means ah Dos is for density
of states for the final continuum of states
. So, its average density of states a multiplied
by the ah square of the matrix element of
the perturbation term 
between initial and final states ok.
So, ah write it with this this thing and let
me just box it once again this is called as
a Fermi's golden rule all right. So, this
is ah the Fermi's golden rule and its applicable
to a variety of situations especially say
the ionization of the hydrogen atom ah in
the ground state. So, when ah hydrogen atom
or the electron in the hydrogen atom is in
the ground state, and if it is somehow given
ah an energy which is equal to 13.6 electron
volts, ah that is the magnitude of the perturbation
of the time ah independent part of the perturbation
there is a matrix element Amk, if that is
of that order 13.6 electron volt and then
this electron will be ejected or this [hydro/hydrogen]-
hydrogen ion will be ionized, and this electron
will go on to ah ah final continuum of states
and the final ah those density of states of
those final continuum of states will have
to be taken into account in order to calculate
the transition probability by unit time and
it has also to be multiplied by the average
density of states of those final states
So, ah instead of doing that which is ah ionization
of hydrogen atom let us give a short in introduction
to ah the lasers or which is a part of a big
ah or rather discussion called as a interaction
of radiation with matter, we will not do that
ah very rigorously,, but we will just introduce
the a b coefficients and we will tell you
that how ah to calculate these coefficients
Einsteins A,B coefficients ok
So, let us again define a 2 level system ok
this is called level one and this is called
level 2 and this is ah called the A12 and
this is called ah I will write it with a dotted
line which is called as a B 12 and we will
also write with another dotted line which
is called as a B ah 2 1. So, its just the
final state is written ah first and then the
initial state is written. So, ah. So, these
are called as the Einstein's A B coefficients.
So, where ah . So, these are called A fi and
B fi. So, there A f i and Bf i ah. So, the
A coefficient ah it stands for spontaneous
emission 
ok and the b coefficient stands for ah induced
emission also absorption we will tell you
what and in which case the first term that
we are neglected in that original expression
we will have to be brought back.
So, ah the transition probability for 
for induced emission ah for spontaneous emission
say first spontaneous emission 
is equal B 12 u omega ah let me just write
that u omega is the energy ah density or ah.
So, let us have u omega is the energy density
per unit frequency range frequency range,
ah that is between or ah ah or you can say
u omega d omega to be ah the energy density
between omega and omega plus d omega which
is a better ah representation. So, we will
talk about u omega d omega together which
is a energy density in the range ah in the
frequency range ah omega and omega plus d
omega.
So, the transition probability for spontaneous
emission is this I am sorry this is not for
spontaneous emission spontaneous emission
does not depend upon the energy density its
simply equal to A12. Now the transition. So,
this is number o1; number 2 is transition
probability for induced emission 
its equal to B 12 u omega and third is transition
probability 
for induced absorption. Once again to remind
you that this is the first term that we have
ah talked about earlier. So, this is B 21
u omega
So, ah these are the nomenclatures of these
ab coefficients and let us see that at steady
state . 
The number of upward transitions 
is same as as the number of downward transition.
So, that tells that N 2 multiplied by A 12.
So, that is the number of atoms making a transition
ah from ah 2 to 1 ah. So, that is n 2 and
ah n one is the number of atoms making a transition
from 1 to 2. So, its A 1 2 plus B 1 2 u omega
ah this is equal to n 1 B 2 1 u omega. So,
i can write that A 12 plus B 1 to u omega
equal to N 1 plus N 2 B 2 1 u omega. So, as
ah minus E 1 equal to h cross omega ah one
can write down from the Boltzmann distribution
N 1 by N 2 is equal to exponential by e one
ah beta or by k t which is equal to exponential
h cross omega by k t. So, that is N 1 by N
2. So, ah A 1 2 plus ah B 1 2 ah u omega equal
to exponential h cross omega by k t ah B 2
1 u omega. So, u omega is equal to A 1 2 divided
by exponential h cross omega by k t b 21 minus
B 21.
Now, in order to find these coefficients ah
we can do a comparison with a Plancks law
of blackbody radiation. Radiation and where
u omega is given by some constant ah quantities
such as this and the both distribution follows
here and so, comparing ah let us call this
as 1 and 2 . So, comparing 1and 2 B 12 is
equal to B 21 which says that the probability
of of induced emission 
and absorption are the same .
So, ah from one one can get ah u omega equal
to A 1 2 divided B 12 now since they are same
one can simply write it as h cross omega by
k t ah minus 1. So, A 1 2 divided byB 1 2
its equal to h cross omega cube by 4 pi square
c cube. So, that tells that ah A 12 a one
2 equal to h cross omega cube by pi square
c cube ah and into B on12. So, there is a
actually one ah coefficient out of the 3 which
is independent because where the 2 of them
are equal and ah one of them has a relationship
with another. So, ah one independent coefficient
ok.
So, the last thing that one can see here is
the following that if we if we want to prove
that these ah B 1 2 equal to B 2 1 how that
comes about ah in a more rigorous way.
Then let us talk about the absorption term
in presence of the time dependent perturbation.
So, absorption. So, which means that m equal
to A 2 k equal 1 ok and. So, c 2 t in our
original language is equal to minus i A 2
1 by 2 h cross ah exponential i t naught omega
2 1 minus omega minus 1 divided by omega 2
1 minus omega. So, that is the c 2 and the
emission term is m equal to one and k equal
to 2 and. So, that is the c 1 t is equal to
a plus a 1 by 2 h cross and exponential i
t naught omega one 2 plus omega ah minus 1
omega 12 plus 0 ah So, we will prove that
ah c 1 t mod square equal to c 2 t mod square.
So, ah c 1 star t can be written as.
So, c 1 star t can be written as ah minus
i A 1 2 by h 2 h cross exponential minus i
t naught omega 1 2 plus omega minus 1 omega
12 plus omega that is ah also omega one 2
equal to minus omega 2 1. So, that makes the
c star t is equal to minus A 1 2 ah star ah
hm ok. So, I should make it start as well
ah knowing that these are our mission ah we
will we can change it 2 h cross and exponential
minus i t naught omega minus omega 2 1 ah
. So, that is the ah and minus 1 divided by
omega minus omega 22 1.
Ah Since H prime is Hermitian 
ah A m k is equal to a km star which has been
already told. So, ah it is if one can ah see
that now a one t mod square equal to a 2 t
mod square.
And since ah a 1 t mod square is proportional
to B 12 u omega which is the emission term
and a 2 t is B 2 1 u omega which is the absorption
term. So, B 1 2 equal to B 2 1.
Now, ah microscopic derivation of these coefficients
we have not derived this coefficient,, but
those coefficients are very easy to derive
from whatever we have done and each of these,
we know now that there is just only one ah
coefficient that is unknown and that coefficient
can be determined exactly in the same way
that we have calculated this ah these x these
transition probabilities the only thing that
we need to know is that ah the H prime ah
which will come from the interaction of radiation
with matter and we will do it in a tutorial
ah problem this interaction of radiation with
matters
And what is the Hamiltonian for that ah what
you need to know is that a priori without
going into the details now the canonical momentum
now changes from p to p minus e a ah where
a is the vector potential, which gives the
magnetic field ah which be equal to curl a
and hm and of course,, there is a ah gauge
ah choice of gauge that could be ah also discussed
there,, but ah other than that we ah now have
all the necessary tools to calculate the transition
probability which is only a feature of the
time dependent perturbation, as has been stressed
many times for and.
So, as I told that it can also be used to
calculate the ab coefficients, now we just
know need to know one coefficient and the
rest will follow ah or we can also calculate
the the ionization of hydrogen atom ah by
using this ah time dependent perturbation
theory .
