WEBVTT
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PROFESSOR: So what do we do?
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We are going to sum
over final states
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the probability to go
from i to final at time t0
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to first order.
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Since the sum of our final
states is really a continuum,
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this is represented
by the integral
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of the f i t0 1, multiplied
by the number of states
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at every little interval.
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So this will go rho of Ef dEf.
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So this is what we developed
about the number of states.
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So I'm replacing this--
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I have to sum but
I basically decide
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to call this little dN,
the little number of states
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in here, and then I'm going to
integrate this probability, so
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the number of states over
there, and therefore the dN
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is replaced by rho times dEf.
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So then this whole transition
probability will be 4 integral,
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I'm writing now the integral,
VfI squared, sine squared,
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omega f i t0 over 2 Ef minus
Ei squared, rho of Ef, d of Ef.
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And you would say
at this moment,
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OK, this is as far
as you go, so that
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must be Fermi's golden
rule, because we
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don't know rho of Ef, it's
different in different cases,
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so we have to do that integral
and we'll get our answer.
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But the great thing
about this golden rule
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is that you can go far and
you can do the integral.
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Now I don't even know, this
VfI also depends on the energy,
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how am I ever going
to do the integral?
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That seems outrageous.
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Well, let's try to do
it, and part of the idea
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will be that we're going
to be led to the concept
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that we already
emphasized here because
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of this suppression, that
only a narrow band of states
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contribute, and in
that narrow band,
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if the narrow band is narrow
enough, in that region
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VfI high be
approximately constant
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in a narrow enough
region, and rho
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will be approximately constant.
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So we'll take them
out of the integral,
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do the rest of the
integral, and see later
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whether the way we're
doing the integral
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shows that this
idea is justified.
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So I'll just-- you
know, sometimes you
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have to do these things,
of making the next step,
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so I'll do that.
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I'll take these things
out, assuming they're
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constant enough, and then we'll
get 4 VfI squared, rho of E,
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what should I put here?
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E sub i, is that right?
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Because if it's all evaluated
at the initial energy Ei,
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if only a narrow
band will contribute,
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I'll put an h squared here so
that this will become omega fi,
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and now I will integrate over
the sum range of energies
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the function sine squared
omega f t0 over 2,
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over omega fi squared dEf.
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So I just took the thing
out of the integral
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and we're going to hope
for some luck here.
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Whenever you have an integral
like that it probably
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is a good idea to plot
what you're integrating
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and think about it and
see if you're going
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to get whatever you wanted.
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Look, I don't know how far
I'm going to integrate,
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I probably don't
want to integrate
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too far because then these
functions that I took out
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of the integral
are not constants,
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so let's see what
this looks like,
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the integrand,
this function here.
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Well, sine squared
of x over x squared
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goes to 1, you know when--
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this we're plotting as
a function of omega fi.
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Why?
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Time is not really what we're
plotting into this thing,
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we're plotting-- we're
integrating our energy,
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Ef, omega fi is Ef minus Ei,
so omega is the variable you
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should be plotting, and
when omega goes to zero,
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this whole interval goes
like t0 squared over 4,
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and then sine squared of x
over x squared does this thing,
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and the first step here is
2 pi over t0, 2 pi over t0
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and so on.
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And now you smile.
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Why?
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Because it's looking
good, this thing.
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First what's going
to be this area?
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Well, if I look at
this lobe, roughly, I
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would say height t0 squared with
1 over t0, answer proportional
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to t0.
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This whole integral is going
to be proportional to t0.
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The magic of the combination
of the x squared growth,
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t0 squared and
the oscillation is
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making into this
integral being linear
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in t0, which is the probability
the transition [INAUDIBLE]
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is going to grow
linearly is going
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to be a rate, as we expected.
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So this is looking very good.
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Then we can attempt to see that
also most of the contribution
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here happens within this
range to the integral.
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If you look at the integral of
sine squared x over x squared,
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90% of the integral
comes from here.
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By the time you have these
ones you're up to 95%
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of the integral.
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Most of the integral
comes within those lobes.
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And look what I'm going
to say, I'm going to say,
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look, I'm going to try
to wait long enough,
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t0 is going to be long enough so
that this narrow thing is going
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to be narrower and narrower and
therefore most of the integral
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is going to come from
omega fi equal to 0,
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which means the f equals to Ei.
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If I wait long enough with
t0, this is very narrow,
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and even all the other
extra bumps are already
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4 pi over t0 over
here is just going
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to do it without any problem,
it's going to fit in.
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So another way of
thinking of this
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is to say, look, you could
have argued that this is going
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to be linear in t0 if you just
change variables here, absorb
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the t0 into the energy,
change variable,
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and the t0 will go out of
the integral in some way,
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but that is only true if the
limits go from minus infinity
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to plus infinity.
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So I cannot really integrate
from minus infinity to plus
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infinity in the final
energies, but I don't need
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to because most of the integral
comes from this big lobe here,
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and if t0 is sufficiently large,
it is really within no energy
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with respect to the energy, Ei.
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So our next step is
to simply declare
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that a good approximation
to this integral
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is to integrate the whole
thing from minus infinity
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to infinity, so let me say this.
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Suppose here in this range omega
fi is in between 2 pi over t0
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and minus 2 pi over t0.
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What does this tell us that
omega fi is in this region?
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Well, this is Ef minus Ei over
h bar so this actually tells you
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Ef is in between Ei
plus 2 pi h bar over t0,
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and Ei minus 2 pi h bar over t0.
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All right, so this
is the energy range
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and as t0 becomes larger and
larger, the window for Ef
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is smaller and smaller, and
we have energy conserving.
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So let's look at our integral
again, the integral is I,
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that's for the integral,
this whole thing,
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will be equal to
integral dEf sine squared
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of omega fi t0 over 2
over omega fi squared.
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So what do we do?
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We call this a variable, u,
equal omega fi t0 over 2,
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so that du is dEf
t0 over 2 h bar,
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because omega fi is
Ef minus Ei, and Ef is
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your variable of integration.
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So you must substitute
the dEf here
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and the rest of the integron.
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So what do we get from the
dEf and the other part?
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You get at the end h bar
t0 over 2 integral from--
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well, let's leave it, sine
squared u over u squared du.
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So look at this, the
omega fi squared,
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by the time you get here omega
fi goes like 1 over time,
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so when it's down here we'll
give you a time squared,
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but the dE gives you 1 over
time so at the end of the day
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we get the desired linear
dependence on t0 here,
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only if the integral
doesn't have t0 in here,
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and it will not have it if you
extend it from minus infinity
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to infinity.
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And there's no error,
really, in extending it
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from minus infinity to
infinity because you basically
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know that n lobes
are going to fit here
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and are going to be accurate,
because there is little energy
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change if t0 is large enough.
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If t0 is large enough, even a
20 pi h bar and a 20 by h bar
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here, that still will do it.
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So we integrate like
that, we extend it,
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and we get this whole
integral has value pi,
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so we get h bar t0 pi over
2, that's our integral, I.
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So our transition
probability, what is it?
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We have it there,
over there, we'll
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have the sum over final states,
i to f of t0, first order
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is equal to the integral
times this quantity,
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so that quantity
is h t0 pi over 2,
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so it's 4, what do
we have, VfI squared,
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rho of Ei over h squared,
then h bar t0 pi over 2.
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So your final answer
for this thing
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is 2 pi over h bar VfI
squared, rho of Ei t0.
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So let's box, this is
a very nice result,
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it's almost Fermi's
golden rule by now.
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Let's put a time t
here, t0 is a label,
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not to confuse our time
integrals or things like that,
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so we could put
the time, t, here,
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is 2 pi over h bar VfI
squared rho of Ei t.
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From here we have
a transition rate,
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so a transition rate is
probability of transition
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per unit time, so
a transition rate
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would be defined as the
probability of transition
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after a time t, divided by
the time t that has elapsed,
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and happily, this has worked
out so that our transition rate,
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w is 2 pi over h bar
VfI squared rho of Ei,
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and this is Fermi's
golden rule, a formula
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for the transition rate to
the continuum of final states.
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You see, when I see
[INAUDIBLE] it almost
00:17:06.640 --> 00:17:08.579
seemed you still
have to integrate,
00:17:08.579 --> 00:17:12.400
there is a rho of E and
let's integrate [INAUDIBLE]
00:17:12.400 --> 00:17:17.109
but the interval has been
done and it says transmission
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amplitude squared evaluated
at the state initial
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and final with the same
energy and final state,
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and the rho evaluated at the
energy of the initial state.
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You don't have to
do more with that.
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So we have this formula, let's
look at a couple more things.
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Do units work out?
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Yes, this is transition per
unit, this is 1 over time,
00:17:51.540 --> 00:17:57.750
this is energy squared,
this is 1 over energy,
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and this is an h bar, this
will give you 1 over time,
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so this thing goes well.
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How about our assumptions?
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This was calculated using some
time t, we used to call it t0.
00:18:13.170 --> 00:18:15.450
How large does it have to be?
00:18:15.450 --> 00:18:22.680
Well, the larger it is the
more accurate the integral is,
00:18:22.680 --> 00:18:25.230
but you don't want
to take it too large,
00:18:25.230 --> 00:18:29.490
either, because
the larger it is,
00:18:29.490 --> 00:18:32.610
the transition
probability eventually
00:18:32.610 --> 00:18:36.840
goes wrong at first order
of perturbation theory.
00:18:36.840 --> 00:18:43.200
So this argument is valid
if there is a time, t0,
00:18:43.200 --> 00:18:51.150
that is large enough so that
within this error bars, rho
00:18:51.150 --> 00:18:54.490
and the transition matrix
elements are constant so
00:18:54.490 --> 00:18:56.650
that your integral is valid.
00:18:56.650 --> 00:19:02.430
But this t0 being large
enough should be small enough
00:19:02.430 --> 00:19:08.550
that the transition probability
doesn't become anywhere near 1.
00:19:08.550 --> 00:19:15.430
That will happen in general or
if VfI is sufficiently small,
00:19:15.430 --> 00:19:21.660
so when VfI is sufficiently
small, this will always hold,
00:19:21.660 --> 00:19:28.680
and in physical applications
this happens and it's OK.
00:19:28.680 --> 00:19:31.920
So that's our presentation
and derivation
00:19:31.920 --> 00:19:35.250
of Fermi's golden rule,
and we will turn now
00:19:35.250 --> 00:19:38.590
to one application
and we will discuss.