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00:00:00,394 --> 00:00:01,560
PROFESSOR: So what do we do?

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00:00:01,560 --> 00:00:05,070
We are going to sum
over final states

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00:00:05,070 --> 00:00:10,180
the probability to go
from i to final at time t0

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00:00:10,180 --> 00:00:11,165
to first order.

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00:00:14,100 --> 00:00:20,830
Since the sum of our final
states is really a continuum,

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00:00:20,830 --> 00:00:26,280
this is represented
by the integral

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00:00:26,280 --> 00:00:37,930
of the f i t0 1, multiplied
by the number of states

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00:00:37,930 --> 00:00:40,010
at every little interval.

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00:00:40,010 --> 00:00:47,590
So this will go rho of Ef dEf.

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00:00:47,590 --> 00:00:51,795
So this is what we developed
about the number of states.

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00:00:54,920 --> 00:00:56,830
So I'm replacing this--

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00:00:56,830 --> 00:01:00,760
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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00:01:05,260 --> 00:01:09,890
in here, and then I'm going to
integrate this probability, so

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00:01:09,890 --> 00:01:14,770
the number of states over
there, and therefore the dN

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00:01:14,770 --> 00:01:19,450
is replaced by rho times dEf.

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00:01:19,450 --> 00:01:26,350
So then this whole transition
probability will be 4 integral,

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00:01:26,350 --> 00:01:31,890
I'm writing now the integral,
VfI squared, sine squared,

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00:01:31,890 --> 00:01:46,421
omega f i t0 over 2 Ef minus
Ei squared, rho of Ef, d of Ef.

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00:01:50,029 --> 00:01:53,400
And you would say
at this moment,

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00:01:53,400 --> 00:01:55,800
OK, this is as far
as you go, so that

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00:01:55,800 --> 00:01:58,500
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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00:02:07,620 --> 00:02:10,800
But the great thing
about this golden rule

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00:02:10,800 --> 00:02:14,630
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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00:02:26,270 --> 00:02:29,580
That seems outrageous.

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00:02:29,580 --> 00:02:34,130
Well, let's try to do
it, and part of the idea

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00:02:34,130 --> 00:02:38,930
will be that we're going
to be led to the concept

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00:02:38,930 --> 00:02:41,180
that we already
emphasized here because

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00:02:41,180 --> 00:02:45,410
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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00:02:57,770 --> 00:03:00,530
in a narrow enough
region, and rho

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00:03:00,530 --> 00:03:03,020
will be approximately constant.

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00:03:03,020 --> 00:03:06,150
So we'll take them
out of the integral,

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00:03:06,150 --> 00:03:09,750
do the rest of the
integral, and see later

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00:03:09,750 --> 00:03:12,000
whether the way we're
doing the integral

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00:03:12,000 --> 00:03:15,970
shows that this
idea is justified.

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00:03:15,970 --> 00:03:18,330
So I'll just-- you
know, sometimes you

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00:03:18,330 --> 00:03:24,130
have to do these things,
of making the next step,

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00:03:24,130 --> 00:03:26,110
so I'll do that.

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00:03:26,110 --> 00:03:29,940
I'll take these things
out, assuming they're

47
00:03:29,940 --> 00:03:40,900
constant enough, and then we'll
get 4 VfI squared, rho of E,

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00:03:40,900 --> 00:03:42,490
what should I put here?

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00:03:42,490 --> 00:03:44,530
E sub i, is that right?

50
00:03:44,530 --> 00:03:49,520
Because if it's all evaluated
at the initial energy Ei,

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00:03:49,520 --> 00:03:52,240
if only a narrow
band will contribute,

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00:03:52,240 --> 00:04:02,140
I'll put an h squared here so
that this will become omega fi,

53
00:04:02,140 --> 00:04:08,120
and now I will integrate over
the sum range of energies

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00:04:08,120 --> 00:04:15,650
the function sine squared
omega f t0 over 2,

55
00:04:15,650 --> 00:04:21,140
over omega fi squared dEf.

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00:04:24,680 --> 00:04:27,780
So I just took the thing
out of the integral

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00:04:27,780 --> 00:04:31,340
and we're going to hope
for some luck here.

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00:04:37,190 --> 00:04:40,640
Whenever you have an integral
like that it probably

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00:04:40,640 --> 00:04:44,270
is a good idea to plot
what you're integrating

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00:04:44,270 --> 00:04:46,650
and think about it and
see if you're going

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00:04:46,650 --> 00:04:49,265
to get whatever you wanted.

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00:04:53,100 --> 00:04:59,930
Look, I don't know how far
I'm going to integrate,

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00:04:59,930 --> 00:05:01,460
I probably don't
want to integrate

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00:05:01,460 --> 00:05:04,310
too far because then these
functions that I took out

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00:05:04,310 --> 00:05:07,920
of the integral
are not constants,

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00:05:07,920 --> 00:05:11,390
so let's see what
this looks like,

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00:05:11,390 --> 00:05:15,050
the integrand,
this function here.

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00:05:27,590 --> 00:05:31,190
Well, sine squared
of x over x squared

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00:05:31,190 --> 00:05:34,160
goes to 1, you know when--

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00:05:34,160 --> 00:05:39,650
this we're plotting as
a function of omega fi.

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00:05:39,650 --> 00:05:40,590
Why?

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00:05:40,590 --> 00:05:44,520
Time is not really what we're
plotting into this thing,

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00:05:44,520 --> 00:05:47,050
we're plotting-- we're
integrating our energy,

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00:05:47,050 --> 00:05:53,850
Ef, omega fi is Ef minus Ei,
so omega is the variable you

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00:05:53,850 --> 00:05:58,830
should be plotting, and
when omega goes to zero,

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00:05:58,830 --> 00:06:05,850
this whole interval goes
like t0 squared over 4,

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00:06:05,850 --> 00:06:15,860
and then sine squared of x
over x squared does this thing,

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00:06:15,860 --> 00:06:25,670
and the first step here is
2 pi over t0, 2 pi over t0

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00:06:25,670 --> 00:06:26,440
and so on.

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00:06:35,470 --> 00:06:39,490
And now you smile.

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00:06:39,490 --> 00:06:40,810
Why?

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00:06:40,810 --> 00:06:43,960
Because it's looking
good, this thing.

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00:06:43,960 --> 00:06:48,180
First what's going
to be this area?

84
00:06:48,180 --> 00:06:51,820
Well, if I look at
this lobe, roughly, I

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00:06:51,820 --> 00:06:59,380
would say height t0 squared with
1 over t0, answer proportional

86
00:06:59,380 --> 00:07:00,820
to t0.

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00:07:00,820 --> 00:07:04,480
This whole integral is going
to be proportional to t0.

88
00:07:04,480 --> 00:07:08,500
The magic of the combination
of the x squared growth,

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00:07:08,500 --> 00:07:11,710
t0 squared and
the oscillation is

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00:07:11,710 --> 00:07:15,910
making into this
integral being linear

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00:07:15,910 --> 00:07:19,160
in t0, which is the probability
the transition [INAUDIBLE]

92
00:07:19,160 --> 00:07:21,010
is going to grow
linearly is going

93
00:07:21,010 --> 00:07:23,510
to be a rate, as we expected.

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So this is looking very good.

95
00:07:28,980 --> 00:07:37,140
Then we can attempt to see that
also most of the contribution

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00:07:37,140 --> 00:07:43,000
here happens within this
range to the integral.

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00:07:43,000 --> 00:07:55,260
If you look at the integral of
sine squared x over x squared,

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00:07:55,260 --> 00:07:59,320
90% of the integral
comes from here.

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00:07:59,320 --> 00:08:02,730
By the time you have these
ones you're up to 95%

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00:08:02,730 --> 00:08:04,530
of the integral.

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00:08:04,530 --> 00:08:07,095
Most of the integral
comes within those lobes.

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00:08:10,970 --> 00:08:13,380
And look what I'm going
to say, I'm going to say,

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00:08:13,380 --> 00:08:18,330
look, I'm going to try
to wait long enough,

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00:08:18,330 --> 00:08:30,230
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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00:08:35,900 --> 00:08:39,500
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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00:08:55,030 --> 00:08:59,730
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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00:09:04,010 --> 00:09:10,690
is to say, look, you could
have argued that this is going

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00:09:10,690 --> 00:09:16,640
to be linear in t0 if you just
change variables here, absorb

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00:09:16,640 --> 00:09:20,210
the t0 into the energy,
change variable,

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00:09:20,210 --> 00:09:27,200
and the t0 will go out of
the integral in some way,

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00:09:27,200 --> 00:09:32,570
but that is only true if the
limits go from minus infinity

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00:09:32,570 --> 00:09:34,320
to plus infinity.

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00:09:34,320 --> 00:09:38,360
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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00:09:55,190 --> 00:09:59,482
with respect to the energy, Ei.

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So our next step is
to simply declare

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00:10:03,000 --> 00:10:06,570
that a good approximation
to this integral

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00:10:06,570 --> 00:10:10,110
is to integrate the whole
thing from minus infinity

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00:10:10,110 --> 00:10:15,720
to infinity, so let me say this.

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00:10:15,720 --> 00:10:24,540
Suppose here in this range omega
fi is in between 2 pi over t0

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00:10:24,540 --> 00:10:28,320
and minus 2 pi over t0.

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00:10:28,320 --> 00:10:33,270
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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00:10:42,170 --> 00:10:51,220
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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00:11:05,860 --> 00:11:13,990
All right, so this
is the energy range

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00:11:13,990 --> 00:11:20,180
and as t0 becomes larger and
larger, the window for Ef

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00:11:20,180 --> 00:11:24,740
is smaller and smaller, and
we have energy conserving.

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00:11:24,740 --> 00:11:33,210
So let's look at our integral
again, the integral is I,

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00:11:33,210 --> 00:11:38,810
that's for the integral,
this whole thing,

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00:11:38,810 --> 00:11:44,570
will be equal to
integral dEf sine squared

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00:11:44,570 --> 00:11:52,370
of omega fi t0 over 2
over omega fi squared.

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00:11:52,370 --> 00:11:53,660
So what do we do?

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00:11:53,660 --> 00:12:02,660
We call this a variable, u,
equal omega fi t0 over 2,

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00:12:02,660 --> 00:12:13,310
so that du is dEf
t0 over 2 h bar,

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00:12:13,310 --> 00:12:18,380
because omega fi is
Ef minus Ei, and Ef is

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00:12:18,380 --> 00:12:20,610
your variable of integration.

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00:12:20,610 --> 00:12:25,880
So you must substitute
the dEf here

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00:12:25,880 --> 00:12:30,300
and the rest of the integron.

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00:12:30,300 --> 00:12:36,530
So what do we get from the
dEf and the other part?

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00:12:36,530 --> 00:12:42,320
You get at the end h bar
t0 over 2 integral from--

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00:12:42,320 --> 00:12:49,080
well, let's leave it, sine
squared u over u squared du.

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00:12:52,020 --> 00:12:58,970
So look at this, the
omega fi squared,

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00:12:58,970 --> 00:13:05,550
by the time you get here omega
fi goes like 1 over time,

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00:13:05,550 --> 00:13:09,190
so when it's down here we'll
give you a time squared,

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00:13:09,190 --> 00:13:13,940
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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00:13:19,610 --> 00:13:25,220
only if the integral
doesn't have t0 in here,

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00:13:25,220 --> 00:13:29,510
and it will not have it if you
extend it from minus infinity

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00:13:29,510 --> 00:13:32,040
to infinity.

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00:13:32,040 --> 00:13:34,430
And there's no error,
really, in extending it

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00:13:34,430 --> 00:13:38,240
from minus infinity to
infinity because you basically

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00:13:38,240 --> 00:13:42,350
know that n lobes
are going to fit here

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00:13:42,350 --> 00:13:46,010
and are going to be accurate,
because there is little energy

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00:13:46,010 --> 00:13:50,120
change if t0 is large enough.

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00:13:50,120 --> 00:13:55,640
If t0 is large enough, even a
20 pi h bar and a 20 by h bar

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00:13:55,640 --> 00:13:59,770
here, that still will do it.

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00:13:59,770 --> 00:14:02,770
So we integrate like
that, we extend it,

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00:14:02,770 --> 00:14:08,800
and we get this whole
integral has value pi,

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00:14:08,800 --> 00:14:16,210
so we get h bar t0 pi over
2, that's our integral, I.

169
00:14:16,210 --> 00:14:27,280
So our transition
probability, what is it?

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00:14:27,280 --> 00:14:35,950
We have it there,
over there, we'll

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00:14:35,950 --> 00:14:47,740
have the sum over final states,
i to f of t0, first order

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00:14:47,740 --> 00:14:53,990
is equal to the integral
times this quantity,

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00:14:53,990 --> 00:15:00,580
so that quantity
is h t0 pi over 2,

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00:15:00,580 --> 00:15:06,160
so it's 4, what do
we have, VfI squared,

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00:15:06,160 --> 00:15:17,680
rho of Ei over h squared,
then h bar t0 pi over 2.

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00:15:17,680 --> 00:15:20,950
So your final answer
for this thing

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00:15:20,950 --> 00:15:32,290
is 2 pi over h bar VfI
squared, rho of Ei t0.

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00:15:32,290 --> 00:15:35,290
So let's box, this is
a very nice result,

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00:15:35,290 --> 00:15:40,720
it's almost Fermi's
golden rule by now.

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00:15:40,720 --> 00:15:44,470
Let's put a time t
here, t0 is a label,

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00:15:44,470 --> 00:15:48,350
not to confuse our time
integrals or things like that,

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00:15:48,350 --> 00:15:50,740
so we could put
the time, t, here,

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00:15:50,740 --> 00:15:58,722
is 2 pi over h bar VfI
squared rho of Ei t.

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00:16:17,590 --> 00:16:20,470
From here we have
a transition rate,

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00:16:20,470 --> 00:16:24,430
so a transition rate is
probability of transition

186
00:16:24,430 --> 00:16:26,920
per unit time, so
a transition rate

187
00:16:26,920 --> 00:16:31,000
would be defined as the
probability of transition

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00:16:31,000 --> 00:16:35,500
after a time t, divided by
the time t that has elapsed,

189
00:16:35,500 --> 00:16:41,080
and happily, this has worked
out so that our transition rate,

190
00:16:41,080 --> 00:16:51,910
w is 2 pi over h bar
VfI squared rho of Ei,

191
00:16:51,910 --> 00:16:56,740
and this is Fermi's
golden rule, a formula

192
00:16:56,740 --> 00:17:01,840
for the transition rate to
the continuum of final states.

193
00:17:01,840 --> 00:17:06,640
You see, when I see
[INAUDIBLE] it almost

194
00:17:06,640 --> 00:17:08,579
seemed you still
have to integrate,

195
00:17:08,579 --> 00:17:12,400
there is a rho of E and
let's integrate [INAUDIBLE]

196
00:17:12,400 --> 00:17:17,109
but the interval has been
done and it says transmission

197
00:17:17,109 --> 00:17:24,190
amplitude squared evaluated
at the state initial

198
00:17:24,190 --> 00:17:28,089
and final with the same
energy and final state,

199
00:17:28,089 --> 00:17:33,470
and the rho evaluated at the
energy of the initial state.

200
00:17:33,470 --> 00:17:38,250
You don't have to
do more with that.

201
00:17:38,250 --> 00:17:42,870
So we have this formula, let's
look at a couple more things.

202
00:17:42,870 --> 00:17:45,390
Do units work out?

203
00:17:45,390 --> 00:17:51,540
Yes, this is transition per
unit, this is 1 over time,

204
00:17:51,540 --> 00:17:57,750
this is energy squared,
this is 1 over energy,

205
00:17:57,750 --> 00:18:02,460
and this is an h bar, this
will give you 1 over time,

206
00:18:02,460 --> 00:18:06,240
so this thing goes well.

207
00:18:06,240 --> 00:18:08,610
How about our assumptions?

208
00:18:08,610 --> 00:18:13,170
This was calculated using some
time t, we used to call it t0.

209
00:18:13,170 --> 00:18:15,450
How large does it have to be?

210
00:18:15,450 --> 00:18:22,680
Well, the larger it is the
more accurate the integral is,

211
00:18:22,680 --> 00:18:25,230
but you don't want
to take it too large,

212
00:18:25,230 --> 00:18:29,490
either, because
the larger it is,

213
00:18:29,490 --> 00:18:32,610
the transition
probability eventually

214
00:18:32,610 --> 00:18:36,840
goes wrong at first order
of perturbation theory.

215
00:18:36,840 --> 00:18:43,200
So this argument is valid
if there is a time, t0,

216
00:18:43,200 --> 00:18:51,150
that is large enough so that
within this error bars, rho

217
00:18:51,150 --> 00:18:54,490
and the transition matrix
elements are constant so

218
00:18:54,490 --> 00:18:56,650
that your integral is valid.

219
00:18:56,650 --> 00:19:02,430
But this t0 being large
enough should be small enough

220
00:19:02,430 --> 00:19:08,550
that the transition probability
doesn't become anywhere near 1.

221
00:19:08,550 --> 00:19:15,430
That will happen in general or
if VfI is sufficiently small,

222
00:19:15,430 --> 00:19:21,660
so when VfI is sufficiently
small, this will always hold,

223
00:19:21,660 --> 00:19:28,680
and in physical applications
this happens and it's OK.

224
00:19:28,680 --> 00:19:31,920
So that's our presentation
and derivation

225
00:19:31,920 --> 00:19:35,250
of Fermi's golden rule,
and we will turn now

226
00:19:35,250 --> 00:19:38,590
to one application
and we will discuss.