1
00:00:00,570 --> 00:00:05,020
PROFESSOR: OK, so our
discussion at the beginning

2
00:00:05,020 --> 00:00:08,950
was based on just
taking the state,

3
00:00:08,950 --> 00:00:13,690
those instantaneous
energy island states,

4
00:00:13,690 --> 00:00:18,820
and calculating what phases
would make it satisfy

5
00:00:18,820 --> 00:00:20,500
the Schrodinger equation.

6
00:00:20,500 --> 00:00:22,540
And we found those
are the phases that

7
00:00:22,540 --> 00:00:25,810
came close to satisfying
the Schrodinger equation,

8
00:00:25,810 --> 00:00:27,430
but not quite.

9
00:00:27,430 --> 00:00:37,280
So in order to do this under a
more controlled approximation,

10
00:00:37,280 --> 00:00:43,850
let's do a calculation where
we put all the information in.

11
00:00:43,850 --> 00:00:50,230
So if you have a
state psi of t, we'll

12
00:00:50,230 --> 00:00:56,020
write it as a
superposition of states

13
00:00:56,020 --> 00:00:58,210
of instantaneous eigenstates.

14
00:01:01,890 --> 00:01:04,470
So this is a general solution.

15
00:01:10,190 --> 00:01:15,660
Maybe general [? an ?]
[? sets ?] for a solution.

16
00:01:15,660 --> 00:01:19,960
The wave function-- since
those instantaneous energy

17
00:01:19,960 --> 00:01:23,170
eigenstates are
complete orthonormal,

18
00:01:23,170 --> 00:01:29,650
this form a ON basis,
orthonormal basis at all

19
00:01:29,650 --> 00:01:31,300
times--

20
00:01:31,300 --> 00:01:35,500
at any time, it's an
orthonormal set of states--

21
00:01:35,500 --> 00:01:41,300
we should be able to write our
state as that superposition.

22
00:01:41,300 --> 00:01:43,840
So what we're going
to do is now kind

23
00:01:43,840 --> 00:01:47,770
of re-do the analysis of the
[INAUDIBLE] approximation

24
00:01:47,770 --> 00:01:52,240
more generally so that we see,
in fact, equations that show up

25
00:01:52,240 --> 00:01:54,710
that you can solve in general.

26
00:01:54,710 --> 00:02:01,460
So the Schrodinger equation
is i h bar dvt of psi

27
00:02:01,460 --> 00:02:05,140
is equal to H psi.

28
00:02:05,140 --> 00:02:08,419
So let's look at what
it gives us here.

29
00:02:08,419 --> 00:02:14,560
So we'll have i
h bar sum over n.

30
00:02:14,560 --> 00:02:17,050
And I have to
differentiate this state.

31
00:02:17,050 --> 00:02:19,480
So we get Cn dot--

32
00:02:19,480 --> 00:02:22,150
dot for time derivatives--

33
00:02:22,150 --> 00:02:29,620
psi n plus Cn psi n dot--

34
00:02:29,620 --> 00:02:31,830
this is a time derivative
of this state--

35
00:02:36,360 --> 00:02:38,700
is equal to H of psi--

36
00:02:38,700 --> 00:02:42,915
this is the sum over n Cn of t--

37
00:02:42,915 --> 00:02:51,570
H of psi n, is equal
to E n of t psi n of t.

38
00:02:58,420 --> 00:03:02,020
OK, so that's your equation.

39
00:03:02,020 --> 00:03:06,100
Now let's see in various
components what it gives you.

40
00:03:06,100 --> 00:03:12,790
So to see the various
components, we form an overlap

41
00:03:12,790 --> 00:03:15,440
with a psi k of t.

42
00:03:15,440 --> 00:03:19,360
So we'll bring in a psi k of t.

43
00:03:19,360 --> 00:03:22,630
And what do we get?

44
00:03:22,630 --> 00:03:24,940
Since these states
are orthonormal,

45
00:03:24,940 --> 00:03:28,570
psi k, when it comes here,
this is a function of time.

46
00:03:28,570 --> 00:03:30,080
It doesn't care.

47
00:03:30,080 --> 00:03:32,740
Psi k hits a psi n.

48
00:03:32,740 --> 00:03:34,660
That's a Kronecker delta.

49
00:03:34,660 --> 00:03:35,990
The sum disappears.

50
00:03:35,990 --> 00:03:43,780
And the only term that
is left here is Ck dot.

51
00:03:43,780 --> 00:03:49,380
So we get i h bar Ck
dot from this term.

52
00:03:55,360 --> 00:04:02,430
And let's put the second
term to the right hand side.

53
00:04:02,430 --> 00:04:07,650
So let's just write what we
get from the right hand side

54
00:04:07,650 --> 00:04:09,960
and from this term.

55
00:04:09,960 --> 00:04:12,090
So from the right
hand side, we have

56
00:04:12,090 --> 00:04:15,990
the psi k on that thing
that is on the right.

57
00:04:15,990 --> 00:04:21,899
That, again, hits this state
and produces a Kronecker delta.

58
00:04:21,899 --> 00:04:31,260
So we get Ck Ek of
t from the term that

59
00:04:31,260 --> 00:04:33,370
was on the right hand side.

60
00:04:33,370 --> 00:04:37,110
And here, however, we
don't get rid of the sum

61
00:04:37,110 --> 00:04:42,022
because psi k is not
orthonormal to psi n dot.

62
00:04:42,022 --> 00:04:45,670
Psi n dot is more complicated.

63
00:04:45,670 --> 00:04:47,530
So what do we get here?

64
00:04:47,530 --> 00:04:59,700
Minus i h bar the sum over n psi
k psi n dot inner product Cn.

65
00:05:08,120 --> 00:05:10,970
OK, that's pretty
close to what we want.

66
00:05:10,970 --> 00:05:16,060
But let's write it still in
a slightly different way.

67
00:05:16,060 --> 00:05:21,430
I want to isolate the Ck's.

68
00:05:21,430 --> 00:05:29,620
So from that sum, I will
separate the Ck part.

69
00:05:29,620 --> 00:05:34,480
So we'll have Ek of t.

70
00:05:34,480 --> 00:05:43,670
And there's going to be a term
here, when we have n equal k,

71
00:05:43,670 --> 00:05:58,820
so I'll bring it out there--
minus i h bar psi k psi k dot

72
00:05:58,820 --> 00:06:01,210
Ck.

73
00:06:01,210 --> 00:06:04,300
And the last term
now becomes i h bar

74
00:06:04,300 --> 00:06:16,800
the sum over n different
from k psi k psi n dot Cn.

75
00:06:16,800 --> 00:06:21,570
OK, so this is the form
of the equation that

76
00:06:21,570 --> 00:06:28,060
is nice and gives you a
little understanding of what's

77
00:06:28,060 --> 00:06:30,800
going on.

78
00:06:30,800 --> 00:06:34,550
That's a general
treatment of trying

79
00:06:34,550 --> 00:06:38,870
to make a solution from
instantaneous energy

80
00:06:38,870 --> 00:06:40,430
eigenstates.

81
00:06:40,430 --> 00:06:42,870
Here were your instantaneous
energy eigenstates.

82
00:06:42,870 --> 00:06:45,170
We tried to make a solution.

83
00:06:45,170 --> 00:06:48,980
That is the full equation.

84
00:06:48,980 --> 00:06:50,990
What did we do before?

85
00:06:50,990 --> 00:06:54,290
We used just one of them.

86
00:06:54,290 --> 00:06:58,680
We took one instantaneous
energy eigenstate

87
00:06:58,680 --> 00:07:02,930
and we tried to make a solution
by multiplying by one thing,

88
00:07:02,930 --> 00:07:04,100
and then we tried.

89
00:07:04,100 --> 00:07:06,980
But then it doesn't
work because when

90
00:07:06,980 --> 00:07:12,620
you have just one coefficient,
say k, with some fixed k,

91
00:07:12,620 --> 00:07:14,600
you have this equation.

92
00:07:14,600 --> 00:07:19,340
But then you couple to
all other coefficients

93
00:07:19,340 --> 00:07:23,720
where n is different from k.

94
00:07:23,720 --> 00:07:28,910
So what we did before
was essentially,

95
00:07:28,910 --> 00:07:35,440
by claiming that
this term is small,

96
00:07:35,440 --> 00:07:39,610
just focus on this
thing, and this

97
00:07:39,610 --> 00:07:42,730
is an easily solvable
equation that,

98
00:07:42,730 --> 00:07:48,070
in fact, gives the type
of solution we have there.

99
00:07:48,070 --> 00:07:52,090
When you have C
dot equal to this--

100
00:07:52,090 --> 00:07:56,480
I'll write it in our
previous approximation,

101
00:07:56,480 --> 00:08:05,712
so in the approximation where
the last term is negligible.

102
00:08:10,560 --> 00:08:13,990
And we would see why
it could be negligible.

103
00:08:13,990 --> 00:08:20,770
Then we get just i
h bar Ck dot equals

104
00:08:20,770 --> 00:08:31,180
Ek of t minus i h bar
psi k psi k dot Ck.

105
00:08:31,180 --> 00:08:40,150
And this thing is solved
by writing Ck of t

106
00:08:40,150 --> 00:08:46,930
is equal to e to the 1 over
i h bar integral from 0 to t

107
00:08:46,930 --> 00:08:56,090
of this whole thing Ek of t
prime minus i h bar psi k psi k

108
00:08:56,090 --> 00:09:07,260
dot of t prime dt
prime times Ck of 0.

109
00:09:10,730 --> 00:09:12,790
This is a differential equation.

110
00:09:12,790 --> 00:09:16,990
So i h bar times the
time derivative of this--

111
00:09:16,990 --> 00:09:20,650
if you apply a i h
bar time derivative,

112
00:09:20,650 --> 00:09:25,000
you differentiate with respect
to time, then exponent,

113
00:09:25,000 --> 00:09:29,110
you get 1 over i h bar
that cancels this i h bar.

114
00:09:29,110 --> 00:09:32,350
And the derivative of the
exponent is this factor--

115
00:09:32,350 --> 00:09:38,620
just this standard, first order,
time dependent differential

116
00:09:38,620 --> 00:09:39,710
equation.

117
00:09:39,710 --> 00:09:45,490
So last time we said we
ignored possible couplings

118
00:09:45,490 --> 00:09:49,990
between the different modes
represented by this term,

119
00:09:49,990 --> 00:09:52,850
and we just solved
this equation,

120
00:09:52,850 --> 00:09:56,710
which gave us this, which
is exactly what we've

121
00:09:56,710 --> 00:09:57,970
been writing here.

122
00:10:00,700 --> 00:10:05,240
e to the I theta of k
comes from the first term

123
00:10:05,240 --> 00:10:06,710
on that integral.

124
00:10:06,710 --> 00:10:10,280
And e to the i gamma of k
comes from the second term

125
00:10:10,280 --> 00:10:12,300
on that integral.

126
00:10:12,300 --> 00:10:16,290
These are the same things.

127
00:10:16,290 --> 00:10:21,840
So what is new here is
that there is a coupling.

128
00:10:21,840 --> 00:10:27,480
And you cannot assume that
just Ck evolves in time some

129
00:10:27,480 --> 00:10:30,610
particular k, and
the others don't.

130
00:10:30,610 --> 00:10:33,720
The others will get coupled.

131
00:10:33,720 --> 00:10:44,160
In particular, if you have
that at time equals 0--

132
00:10:44,160 --> 00:10:51,870
t equals 0-- some Ck
of 0 is equal to 1,

133
00:10:51,870 --> 00:10:56,730
but all the other
ones, Ck primes at 0,

134
00:10:56,730 --> 00:11:04,110
are equal to 0 for all k
prime different from k,

135
00:11:04,110 --> 00:11:09,660
so your initial condition is
you are in the state k at time

136
00:11:09,660 --> 00:11:12,570
equals 0.

137
00:11:12,570 --> 00:11:15,930
That's why you have Ck at time
equals [? 0 to ?] [? a 1. ?]

138
00:11:15,930 --> 00:11:20,440
And the others are 0
at different times.

139
00:11:20,440 --> 00:11:24,670
If you look at your
differential equation

140
00:11:24,670 --> 00:11:33,600
and try to see what happens
after a little time, well,

141
00:11:33,600 --> 00:11:38,250
we know Ck dot is
going to change,

142
00:11:38,250 --> 00:11:40,270
is going to have a
non-trivial value.

143
00:11:40,270 --> 00:11:41,770
This is going to happen.

144
00:11:41,770 --> 00:11:45,210
But these things are
not going to remain 0.

145
00:11:45,210 --> 00:11:48,730
Those other states are
going to get populated,

146
00:11:48,730 --> 00:11:55,500
in particular, i h
bar Ck prime dot.

147
00:11:55,500 --> 00:11:57,030
Look at the top equation.

148
00:11:57,030 --> 00:12:04,050
Apply for k equal to k prime
and look at time equals 0.

149
00:12:04,050 --> 00:12:06,960
What do you get
at time equals 0?

150
00:12:06,960 --> 00:12:09,730
Well, you would get
all these factor--

151
00:12:09,730 --> 00:12:15,740
Ek prime times Ck
prime at time equals 0.

152
00:12:15,740 --> 00:12:18,840
Well, that's 0.

153
00:12:18,840 --> 00:12:19,950
That's nothing.

154
00:12:19,950 --> 00:12:23,670
But from the last term,
what would we get?

155
00:12:23,670 --> 00:12:36,440
Minus i h bar sum over n
different from k prime of psi k

156
00:12:36,440 --> 00:12:44,520
prime psi n dot Cn
at time equals 0.

157
00:12:47,542 --> 00:12:53,260
And here, well, the only
one that is different from 0

158
00:12:53,260 --> 00:12:57,160
at time equals 0 is Ck.

159
00:12:57,160 --> 00:13:00,400
And k, when n is
equal to k, is allowed

160
00:13:00,400 --> 00:13:03,040
because we said k prime
is different from k.

161
00:13:03,040 --> 00:13:05,710
So there is one term here.

162
00:13:05,710 --> 00:13:15,670
This is minus i h bar
psi k prime psi k dot.

163
00:13:18,220 --> 00:13:23,560
Only when n equals to k
you get something here

164
00:13:23,560 --> 00:13:25,870
because this is the
only term that exists.

165
00:13:25,870 --> 00:13:27,520
And it's equal to 1.

166
00:13:27,520 --> 00:13:34,840
So immediately, at time equals
0, the other coefficients start

167
00:13:34,840 --> 00:13:35,830
changing.

168
00:13:35,830 --> 00:13:40,855
You start populating the
other instantaneous energy

169
00:13:40,855 --> 00:13:43,060
eigenstates.

170
00:13:43,060 --> 00:13:47,710
So there is real mixing
in this top thing that

171
00:13:47,710 --> 00:13:52,420
says it's not rigorous to claim
that you stay in that energy

172
00:13:52,420 --> 00:13:53,440
eigenstate.

173
00:13:53,440 --> 00:13:57,070
It starts to couple.

174
00:13:57,070 --> 00:14:00,990
And if it starts to
couple, then eventually you

175
00:14:00,990 --> 00:14:02,190
make transitions.

176
00:14:02,190 --> 00:14:07,760
The only hope, of course, is
that that term is really small.

177
00:14:07,760 --> 00:14:13,680
And basically, we can
argue how that term

178
00:14:13,680 --> 00:14:18,730
becomes a little small by
doing a little calculation.

179
00:14:18,730 --> 00:14:22,650
And then we can be rigorous
and take a long time

180
00:14:22,650 --> 00:14:25,770
to get to the conclusion,
or we can just state it.

181
00:14:25,770 --> 00:14:30,600
And that's what we're
going to do after analyzing

182
00:14:30,600 --> 00:14:32,720
that term a little more.