1
00:00:01,220 --> 00:00:04,840
PROFESSOR: OK, so
that's our equation.

2
00:00:04,840 --> 00:00:09,400
Three terms-- first term,
spontaneous emission,

3
00:00:09,400 --> 00:00:14,290
second term, stimulated
emission proportional to Nb,

4
00:00:14,290 --> 00:00:18,430
third term increases
Nb's absorption.

5
00:00:18,430 --> 00:00:21,130
Our strategy now is, so
what do you do with this?

6
00:00:21,130 --> 00:00:24,540
This looks like a good
equation, but what are we to do?

7
00:00:24,540 --> 00:00:26,950
We haven't used three, really.

8
00:00:26,950 --> 00:00:31,720
So we can solve for U. And
this time, the equation

9
00:00:31,720 --> 00:00:34,150
won't want to make U equal 0.

10
00:00:34,150 --> 00:00:36,190
It will give us something.

11
00:00:36,190 --> 00:00:40,780
And then we can compare with
a thermal equilibrium relation

12
00:00:40,780 --> 00:00:42,760
to get something.

13
00:00:42,760 --> 00:00:46,000
So that's our goal.

14
00:00:46,000 --> 00:00:49,270
So how do we solve for U?

15
00:00:49,270 --> 00:00:52,780
Let's divide by Nb.

16
00:00:56,510 --> 00:01:02,030
So divide by Nb.

17
00:01:02,030 --> 00:01:08,970
So a 0 here, pass this to
that side, and divide by Nb.

18
00:01:08,970 --> 00:01:13,815
So you get A on the
left-hand side equals--

19
00:01:16,860 --> 00:01:18,740
let's see, should I put the--

20
00:01:18,740 --> 00:01:24,770
well, minus Bba Nb.

21
00:01:24,770 --> 00:01:28,200
Actually, I'll do it like this--

22
00:01:28,200 --> 00:01:29,610
the second term first.

23
00:01:29,610 --> 00:01:41,700
Bab Na over Nb times U--

24
00:01:41,700 --> 00:01:47,430
so I'll put it here,
a U of omega Ba--

25
00:01:47,430 --> 00:01:49,620
minus Bba.

26
00:01:54,080 --> 00:01:58,380
OK, so I moved the first term
to the left-hand side, grouped

27
00:01:58,380 --> 00:02:02,580
the U. Why did I group the U?

28
00:02:02,580 --> 00:02:05,610
Because I said,
let's find what U

29
00:02:05,610 --> 00:02:10,080
is, and try to compare with
other things that we have.

30
00:02:10,080 --> 00:02:20,080
So now we can solve for
the U. U of omega ba

31
00:02:20,080 --> 00:02:24,490
is equal to A divided
by this quantity.

32
00:02:24,490 --> 00:02:30,220
So it's A divided
by a big thing.

33
00:02:30,220 --> 00:02:32,590
Let me factor out the Bab.

34
00:02:36,730 --> 00:02:50,860
So now we have here Na
over Nb minus Bba over Bab,

35
00:02:50,860 --> 00:03:03,670
or A over Bab 1 over
Na over Nb is there,

36
00:03:03,670 --> 00:03:09,550
is e to the beta
h bar omega ba--

37
00:03:12,530 --> 00:03:16,910
It's a minus sign with
respect to number 2--

38
00:03:16,910 --> 00:03:21,185
minus Bba over Bab.

39
00:03:26,050 --> 00:03:28,540
OK, this is great.

40
00:03:28,540 --> 00:03:30,015
We're in very good shape.

41
00:03:34,446 --> 00:03:40,200
In fact, we're here.

42
00:03:40,200 --> 00:03:44,700
Let's compare with
our thermal radiation.

43
00:03:44,700 --> 00:03:46,920
So here is what we got.

44
00:03:46,920 --> 00:03:51,150
And we have to compare
with fact number

45
00:03:51,150 --> 00:03:55,200
three, which is the thermal
radiation formula, which

46
00:03:55,200 --> 00:04:11,870
is h bar over pi squared C
cubed omega cubed 1 over e

47
00:04:11,870 --> 00:04:15,530
to the beta h omega.

48
00:04:15,530 --> 00:04:20,240
And I'm comparing with
the thermal radiation,

49
00:04:20,240 --> 00:04:22,520
U of omega ab.

50
00:04:22,520 --> 00:04:28,550
I canceled the D omegas, because
this is U without the D omegas.

51
00:04:28,550 --> 00:04:33,830
So omega ba minus 1.

52
00:04:33,830 --> 00:04:38,370
All right.

53
00:04:38,370 --> 00:04:41,730
It's perfectly nice and ready.

54
00:04:41,730 --> 00:04:45,120
We need to compare this
formula with this one.

55
00:04:45,120 --> 00:04:48,670
They have to be equal.

56
00:04:48,670 --> 00:04:53,560
Well, this factor, must be
equal to this factor, given

57
00:04:53,560 --> 00:04:57,580
that here you have e to
the beta h bar omega ba.

58
00:04:57,580 --> 00:04:59,350
These are constants.

59
00:04:59,350 --> 00:05:04,450
And the first thing that you
discover is Bba equal Bab.

60
00:05:04,450 --> 00:05:06,880
And so yes, very nice.

61
00:05:06,880 --> 00:05:09,520
This is what we learn
in perturbation theory.

62
00:05:09,520 --> 00:05:14,170
These two processes
have identical rates.

63
00:05:14,170 --> 00:05:19,780
So the first statement
that this equality requires

64
00:05:19,780 --> 00:05:30,100
is that Bab is equal to Bba.

65
00:05:30,100 --> 00:05:34,210
So there's just
one B coefficient,

66
00:05:34,210 --> 00:05:37,810
just like there's
one A coefficient.

67
00:05:37,810 --> 00:05:47,530
And the second thing that
you learn is that A over Bab

68
00:05:47,530 --> 00:05:48,415
is this ratio--

69
00:05:52,320 --> 00:05:58,845
is h bar omega ba cubed
over pi squared C cubed.

70
00:06:04,640 --> 00:06:09,380
So we will be able to
calculate the B coefficients,

71
00:06:09,380 --> 00:06:12,770
because they represent
the familiar properties

72
00:06:12,770 --> 00:06:15,470
of harmonic perturbations
transitions,

73
00:06:15,470 --> 00:06:18,020
and we've done already.

74
00:06:18,020 --> 00:06:21,620
Calculating A is
harder, in principle.

75
00:06:21,620 --> 00:06:25,070
The process of spontaneous
emission is a harder process.

76
00:06:25,070 --> 00:06:28,710
But this relation says that we
don't have to worry about it.

77
00:06:28,710 --> 00:06:34,355
We already, if we know B,
these are constants, we know A.

78
00:06:34,355 --> 00:06:38,780
So many times, even in
problems in the homework,

79
00:06:38,780 --> 00:06:42,980
you will be asked, what is
the spontaneous transition

80
00:06:42,980 --> 00:06:45,170
rate for this decay?

81
00:06:45,170 --> 00:06:46,580
You have an oscillator.

82
00:06:46,580 --> 00:06:48,080
It's in an excited state.

83
00:06:48,080 --> 00:06:51,680
What is the rate of
spontaneous transition?

84
00:06:51,680 --> 00:06:53,720
And then you say, OK,
this is complicated.

85
00:06:53,720 --> 00:06:57,830
But you calculate the
stimulated transition rate,

86
00:06:57,830 --> 00:07:00,230
and then plug in
this coefficient.

87
00:07:00,230 --> 00:07:04,480
So it will not be difficult.

88
00:07:04,480 --> 00:07:12,060
Now, spontaneous emission, as
we discussed there in the top,

89
00:07:12,060 --> 00:07:14,310
doesn't include
the factor having

90
00:07:14,310 --> 00:07:18,750
to do with the density of
photons, the U that tells you

91
00:07:18,750 --> 00:07:20,220
how many photons there are.

92
00:07:20,220 --> 00:07:23,730
Because the photons
play no deep role

93
00:07:23,730 --> 00:07:26,610
in producing the transition.

94
00:07:26,610 --> 00:07:32,190
But at some level,
this transition--

95
00:07:32,190 --> 00:07:35,820
while it's not stimulated by
the photons that are flying

96
00:07:35,820 --> 00:07:37,110
around--

97
00:07:37,110 --> 00:07:44,600
when you calculate it, in a
serious, detailed calculation,

98
00:07:44,600 --> 00:07:47,960
you can think of
them as stimulated

99
00:07:47,960 --> 00:07:52,410
by the vacuum fluctuations
of the electromagnetic field.

100
00:07:52,410 --> 00:07:56,670
So if you were to quantize
the electromagnetic field,

101
00:07:56,670 --> 00:07:59,030
there are vacuum fluctuations.

102
00:07:59,030 --> 00:08:02,750
And those vacuum
fluctuations, you could say,

103
00:08:02,750 --> 00:08:09,420
they are stimulating what we
call "spontaneous emission."

104
00:08:09,420 --> 00:08:12,570
So at the end of the day,
the electromagnetic field

105
00:08:12,570 --> 00:08:14,520
is a quantum field.

106
00:08:14,520 --> 00:08:19,260
And therefore, it has zero
point energies, Casimir effects,

107
00:08:19,260 --> 00:08:21,120
vacuum fluctuations.

108
00:08:21,120 --> 00:08:22,870
You can't get away from it.

109
00:08:22,870 --> 00:08:29,400
So in some sense, everything
is stimulated emission,

110
00:08:29,400 --> 00:08:32,370
stimulated by a lot of
photons or stimulated

111
00:08:32,370 --> 00:08:34,260
by the vacuum fields.

112
00:08:34,260 --> 00:08:38,789
Anyway, we'll still
remain with the name

113
00:08:38,789 --> 00:08:44,610
"spontaneous emission," and
keep those things very distinct.

114
00:08:44,610 --> 00:08:47,610
These rates have
different effects

115
00:08:47,610 --> 00:08:51,520
at different
temperatures, as well.

116
00:08:51,520 --> 00:08:57,900
So we'll consider
that, for example,

117
00:08:57,900 --> 00:09:05,460
in that at low temperatures, the
black body radiation has a very

118
00:09:05,460 --> 00:09:07,360
little number of photons.

119
00:09:07,360 --> 00:09:10,590
So most of the
transitions, if they occur,

120
00:09:10,590 --> 00:09:14,130
are happening due to
spontaneous transitions.

121
00:09:14,130 --> 00:09:17,020
On the other hand, as you
increase the temperature,

122
00:09:17,020 --> 00:09:21,630
the number of photons that
are available per unit volume

123
00:09:21,630 --> 00:09:26,320
increases, and stimulated
emission takes over.

124
00:09:26,320 --> 00:09:32,870
So these are the different
contributions to the rates.

125
00:09:32,870 --> 00:09:35,700
And it gives you a perspective--
at very low temperature,

126
00:09:35,700 --> 00:09:37,590
spontaneous dominates.

127
00:09:37,590 --> 00:09:43,130
At very high temperatures,
stimulated emission dominates.