WEBVTT

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PROFESSOR: Today, we have to
discuss harmonic perturbations.

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So we've done Fermi's golden
rule for constant transitions.

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We saw transitions from a
discrete state to a continuum.

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And by integrating
over the continuum,

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we found a nice rule,
Fermi's golden rule,

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that govern the transition
rate for this process.

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So the only thing we
have to do different

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now is consider the case
that the perturbation is not

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just a step that gets
up and stays there,

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but it has a
frequency dependence.

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So that will bring a
couple of novel features.

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But at the end of the
day, as we will see,

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our Fermi's golden rule is
going to look pretty similar

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to the original
Fermi's golden rule.

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A nice application of
Fermi's golden rule

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is the calculation of the
ionization rate for hydrogen,

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in which you take
a hydrogen atom,

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you put it in an electric
field or send a light wave,

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and then suddenly the
electron and the hydrogen atom

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from the ground state ionizes.

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And we can compute already--
we have the technology

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to compute the ionization rate.

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That's a pretty
physical quantity.

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And that will be an example
we'll develop today.

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Those rates have
the funny situation

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that the calculation
can be somewhat

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involved and interesting.

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And the answers, generally,
by the time you simplify them,

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are pretty simple
and pretty nice.

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So it's a good idea.

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You have to have patience
with those calculations

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to simplify it till the end,
and that's pretty instructive.

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So we begin with
harmonic perturbations.

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So we did constant
perturbations already.

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So now harmonic perturbations.

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So our situation is
that in which age H of t

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is equal to a known
Hamiltonian plus delta H of t.

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And this time, delta H
of t is conventionally

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written as 2 H
prime cosine omega t

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for some t between t0
and 0, and 0 otherwise.

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All of us wonder why the 2 here.

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One reason for it-- it's
all convention, of course.

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You have your perturbation,
and what you call E H prime

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or what you call 2 H
prime is your choice.

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But this 2 has the
advantage that when

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you describe the cosine
in terms of exponentials--

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e to the i omega t plus e to the
minus i omega t-- the over 2,

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it cancels this one.

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And that makes Fermi's golden
rule, that will follow also

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and will be valid for
these perturbations,

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take exactly the
same form as it did

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for the case of
constant perturbation.

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So it's fairly convenient to
put that 2, and we'll put it in.

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Some books don't, and then they
have different looking formulas

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for Fermi golden rule
depending to which case you're

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talking about.

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Of course, when we mean that
this is the time dependence,

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we are implying that H
prime is time independent.

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Because the time
dependence is this one.

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That's what we're
interested in considering.

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Of course-- this has
been asked sometimes--

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H prime can depend on all
kinds of other thing-- position

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coordinates, some
other quantities.

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But we're focusing on time
here, so we'll leave it there.

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Moreover, for reasons
of convention,

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just let's always thing
of omega as positive.

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It wouldn't make a difference
if it would be negative here

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with the cosine function, but
let's just set by convention

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that omega is positive.

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Finally, we're going
to do transitions again

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from an initial
to a final state.

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So we will consider
the case when

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we go from an initial
state to a final state.

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And therefore, we will
work in this language

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with this constant
coefficients Cn's, these

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coefficients that multiply
the states in psi tilde.

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Psi tilde is equal to
Cn n of t, sum over n.

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And these Cn's,
at time equals 0,

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will be equal to delta ni,
which means that they are all 0

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except when we're talking
about Ci at 0 is equal to 1

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because we start with
an initial state.

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We had a general formula for
the transition coefficient.

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And Cm of 1 at time equal t--

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or I'll put t0--

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is equal to sum over n,
integral from 0 to t0 e

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to the i omega mn t prime, delta
H mn of t prime over i h bar,

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Cn at time equals 0, dt prime.

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This was our general formula
for transition coefficients.

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The Cm's is the coefficient
or the amplitude for the state

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to be found in the m
eigenstate at time t0

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to first order in
perturbation theory.

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And it depends on
where you started on.

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That's why the sum over n
here with initial states.

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But this sum is going
to collapse because we

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know we start with the state i.

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So when we substitute
Cn equal to this,

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the sum only works
when n is equal to i.

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So we'll put for ni's.

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And, of course, we're going to
also take for the final state

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to be f.

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So the formula now
reads Cf 1 at t0

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is equal to integral from 0
to t0 e to the I omega f--

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m was f-- i t prime.

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And now the delta H. The delta
H is this whole quantity,

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so we have to substitute it.

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So 2 H prime is the only part
that has matrix elements.

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The cosine omega t
is just a function.

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So it's H prime fi cosine
omega t prime, and dt prime.

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There's the i h bar.

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So I think I got
everything right.

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The sum collapsed.

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mn is being replaced
by the right labels.

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mn here, this is the expectation
value between m and n.

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And that becomes
between f and i.

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And it affects this
whole thing, but it just

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ends up affecting the
Hamiltonian H prime here.

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So I think we're OK.

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We have everything there.

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And Hfi, of course, doesn't
have time dependence.

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So we said H prime has
no time dependence.

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So that thing can go
out of the integral.

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So this will go out.

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And the integral is simple
because you have now

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Hfi prime over i h bar.

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And the 2, we leave
it for the cosine.

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So we get two integrals.

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t0 e to the i omega fi
plus omega t prime--

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from the first exponential
in the cosine--

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plus an e to the i omega
fi minus omega t prime--

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from the second
exponential in cosine--

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dt prime.

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Well, that's very nice.

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This is all doable.

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The Hfi doesn't
give us any trouble.

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It's a constant.

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It's out of the integral.

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It's all pretty nice and simple.

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So we can do these
two integrals.

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They're integrals
of exponentials,

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so it's just an exponential
divided by those coefficients.

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So I'll just do it and
evaluated it between t0 and 0.

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So what do we get?

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Minus i Hfi prime over
h bar, e to the i omega

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fi plus omega t0, minus 1,
over omega fi plus omega.

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You can imagine that
e to the i omega

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t integrates to e to the
i omega t over omega.

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So that's why that works.

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And the two limits are t0 and 0.

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Plus e to the i omega fi
minus omega t0, minus 1

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again, over omega
fi minus omega.

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Great.

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Our integral is there.

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It's done.

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And now it's time to
appreciate what it tells us,

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because it tells us something
very important, this formula.

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So you look at this
and you say, well, OK.

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This is the transition
amplitude to state omega--

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I'm sorry-- state
f, final state.

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And it depends on omega fi.

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And omega fi just Ef
minus Ei over h bar.

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So if I know my final
discrete state Ef,

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I can figure out what is
the transition probability.

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Now, these denominators are
intriguing because maybe

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if you adjust the
frequency omega--

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suppose you have an initial
state and a final state here.

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You may adjust the frequency
omega to match them,

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and in that case maybe make
the denominators equal to 0.

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And that's exactly what
kind of happens here.

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So, first of all, if you
look at this expression,

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it's a sum of two terms that are
added together and multiplied

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by a constant.

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As t0 really goes
to 0, completely

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goes to 0, t0, each
factor, actually,

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if you see the Taylor
expansion of the exponential,

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you cancel the 1 and you
then cancel the linear term

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with the denominator.

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This is just i t0.

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And this is also i t0.

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So they're comparable as
time is really going to 0.

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But time going to 0 is
never of interest for us.

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For us, we need to be
the time a little big

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already so that
our calculations,

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as we did in the constant
transitions that had lobes that

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decreases constants over t0--

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we needed the time to
be sufficiently large

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so that the lobes are narrow.

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And that, we could guarantee.

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So t0 going to 0 is
not very interesting.

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We need t0 a little bigger.

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Not too big that the
rate of a process

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overwhelms the probability.

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But we need a little big.

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So, in that case,
the numerators are

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going to be bounded numbers.

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You see, you have an
exponential minus 1.

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So that varies from--

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the magnitude of this thing
varies from 2 to 0, basically.

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In fact, in these
numerators, you

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can see, if the phase is 0
for some particular value--

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if the exponential has a phase
that's proportional to 2 pi,

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this is 0.

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And then sometimes this
exponential is minus 1,

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so it gets to minus 2.

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So it's finite.

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And the same is here.

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So these are bounded numerators.

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On the other hand, you
may have the possibility

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that these things become 0.

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And those are the cases
that are of interest to us,

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the cases when those
terms are going to be 0.