1
00:00:01,170 --> 00:00:03,350
PROFESSOR: So we go
back to the integral.

2
00:00:03,350 --> 00:00:06,065
We think of k.

3
00:00:09,860 --> 00:00:15,200
We'll write it as k
naught plus k tilde.

4
00:00:15,200 --> 00:00:20,780
And then we have
psi of x0 equal 1

5
00:00:20,780 --> 00:00:26,330
over square root of 2pi
e to the ik naught x--

6
00:00:26,330 --> 00:00:29,180
that part goes out--

7
00:00:29,180 --> 00:00:42,502
integral dk tilde phi
of k naught plus k tilde

8
00:00:42,502 --> 00:00:48,605
e to the ik tilde x dk.

9
00:00:48,605 --> 00:00:49,105
OK.

10
00:00:54,880 --> 00:01:00,145
So we're doing this integral.

11
00:01:00,145 --> 00:01:05,410
And now we're focusing
on the integration

12
00:01:05,410 --> 00:01:10,960
near k naught, where the
contribution is large.

13
00:01:10,960 --> 00:01:16,485
So we write k as k naught
plus a little fluctuation.

14
00:01:16,485 --> 00:01:20,520
dk will be dk tilde.

15
00:01:20,520 --> 00:01:24,520
Wherever you see a k, you must
put k naught plus k tilde.

16
00:01:24,520 --> 00:01:27,950
And that's it.

17
00:01:27,950 --> 00:01:30,350
And why do we have to worry?

18
00:01:30,350 --> 00:01:38,230
Well, we basically have now
this peak over here, k naught.

19
00:01:38,230 --> 00:01:41,020
And we're going
to be integrating

20
00:01:41,020 --> 00:01:45,130
k tilde, which is
the fluctuation, all

21
00:01:45,130 --> 00:01:50,680
over the width of this profile.

22
00:01:50,680 --> 00:01:57,100
So the relevant region of
integration for k tilde

23
00:01:57,100 --> 00:02:07,420
is the range from delta k over
2 to minus delta k over 2.

24
00:02:07,420 --> 00:02:11,265
So maybe I'll make this
picture a little bigger.

25
00:02:18,350 --> 00:02:20,260
Here is k naught.

26
00:02:20,260 --> 00:02:27,310
And here we're going to be going
and integrate in this region.

27
00:02:27,310 --> 00:02:35,080
And since this is delta
k, the relevant region

28
00:02:35,080 --> 00:02:37,870
of integration--

29
00:02:37,870 --> 00:02:50,330
integration-- for k tilde
is from minus delta k over 2

30
00:02:50,330 --> 00:02:52,180
to delta k over 2.

31
00:02:55,260 --> 00:02:58,120
That's where it's
going to range.

32
00:02:58,120 --> 00:03:02,670
So all the integral has to
be localized in the hump.

33
00:03:02,670 --> 00:03:04,810
Otherwise, you don't
get any contribution.

34
00:03:04,810 --> 00:03:08,280
So the relevant
region of integration

35
00:03:08,280 --> 00:03:12,840
for the only variable that
is there is just that one.

36
00:03:16,260 --> 00:03:24,780
Now as you vary this k tilde,
you're going to vary the phase.

37
00:03:24,780 --> 00:03:31,170
And as the phase changes, well,
there's some effect [? on ?]

38
00:03:31,170 --> 00:03:36,890
[? it. ?] But if x is equal
to 0, the phase is stationary,

39
00:03:36,890 --> 00:03:40,910
because k tilde is going to
very, but x is equal to 0.

40
00:03:40,910 --> 00:03:43,850
No phase is stationary.

41
00:03:43,850 --> 00:03:46,730
And therefore, you will
get a substantial answer.

42
00:03:46,730 --> 00:03:48,800
And that's what we know already.

43
00:03:48,800 --> 00:03:53,420
For x is going to
0 or x equal to 0,

44
00:03:53,420 --> 00:03:56,600
we're going to get a
substantial answer.

45
00:03:56,600 --> 00:04:00,240
But now think of the
phase in general.

46
00:04:00,240 --> 00:04:06,350
So for any x that
you choose, the phase

47
00:04:06,350 --> 00:04:08,640
will range over some value.

48
00:04:08,640 --> 00:04:18,040
So for any x different from
0, the face in the integral

49
00:04:18,040 --> 00:04:37,820
will range over minus delta k
over 2x and to delta k over 2x.

50
00:04:37,820 --> 00:04:40,100
You see, x is here.

51
00:04:40,100 --> 00:04:42,710
The phase is k tilde x.

52
00:04:42,710 --> 00:04:48,590
Whatever x is, since k tilde
is going run in this range,

53
00:04:48,590 --> 00:04:53,390
the phase is going to run in
that range multiplied by x.

54
00:04:57,750 --> 00:05:00,070
So as you do the integral--

55
00:05:00,070 --> 00:05:02,250
now think you're
doing this integral.

56
00:05:02,250 --> 00:05:06,520
You have a nice, real,
smooth function here.

57
00:05:06,520 --> 00:05:09,570
And now you have a running
phase that you don't

58
00:05:09,570 --> 00:05:12,390
manage to make it stationary.

59
00:05:12,390 --> 00:05:14,490
Because when x is
different from 0,

60
00:05:14,490 --> 00:05:16,950
this is not going
to be stationary.

61
00:05:16,950 --> 00:05:18,680
It's going to vary.

62
00:05:18,680 --> 00:05:24,820
But it's going to vary from
this value to that value.

63
00:05:24,820 --> 00:05:29,850
So the total, as you
integrate over that peak,

64
00:05:29,850 --> 00:05:36,070
your phase excursion is
going to be delta k times x--

65
00:05:36,070 --> 00:05:48,580
total phase excursion
is delta k times x.

66
00:05:53,490 --> 00:05:58,760
But then that tells
you what can happen.

67
00:05:58,760 --> 00:06:05,010
As long as this total phase
excursion is very small--

68
00:06:05,010 --> 00:06:13,370
so if x is such
that delta k times

69
00:06:13,370 --> 00:06:19,780
x is significantly less than 1--

70
00:06:19,780 --> 00:06:23,530
or, in fact, I could
say less than 1--

71
00:06:32,830 --> 00:06:36,130
there will be a
good contribution

72
00:06:36,130 --> 00:06:45,340
if x is such that--

73
00:06:45,340 --> 00:06:54,090
then you will get
a contribution.

74
00:06:54,090 --> 00:06:57,970
And the reason is because the
phase is not changing much.

75
00:06:57,970 --> 00:07:01,760
You are doing your integral,
and the phase is not killing it.

76
00:07:01,760 --> 00:07:06,350
On the other hand,
if delta k times x--

77
00:07:10,290 --> 00:07:15,550
delta k times x is
much bigger than 1,

78
00:07:15,550 --> 00:07:20,660
then as you range over
the peak, the phase

79
00:07:20,660 --> 00:07:26,590
has done many, many cycles and
is going to kill the integral.

80
00:07:26,590 --> 00:07:37,670
So if k of x is greater than
1, the contribution goes to 0.

81
00:07:40,440 --> 00:07:48,840
So let's then just extract
the final conclusion

82
00:07:48,840 --> 00:07:50,280
from this thing.

83
00:07:50,280 --> 00:08:11,990
So psi of x 0 will be sizable
in an interval x belonging

84
00:08:11,990 --> 00:08:15,380
from minus x0 to x0.

85
00:08:23,340 --> 00:08:29,870
So it's some value
here minus x0 to x0.

86
00:08:33,450 --> 00:08:39,600
If, even for values
as long as x0,

87
00:08:39,600 --> 00:08:44,059
this product is still about 1--

88
00:08:46,830 --> 00:09:07,820
if for delta k times x0, roughly
say of value 1, we have this.

89
00:09:07,820 --> 00:09:20,480
And therefore the uncertainty
in x would be given by 2x0.

90
00:09:24,470 --> 00:09:31,880
So x0 or 2x0, this x0 is
basically the uncertainty in x.

91
00:09:31,880 --> 00:09:36,560
And you would get
that delta k times

92
00:09:36,560 --> 00:09:41,870
delta x is roughly equal to 1--

93
00:09:41,870 --> 00:09:50,630
so delta k delta x
roughly equal to 1.

94
00:09:50,630 --> 00:09:53,420
So I'm dropping factors of 2.

95
00:09:53,420 --> 00:09:57,350
In principle here,
I should push a 2.

96
00:09:57,350 --> 00:10:02,570
But the 2s, or 1s, or
pi's at this moment

97
00:10:02,570 --> 00:10:05,240
are completely unreliable.

98
00:10:05,240 --> 00:10:09,050
But we got to the
end of this argument.

99
00:10:09,050 --> 00:10:15,597
We have a relation of
uncertainties is equal to 1.

100
00:10:15,597 --> 00:10:21,350
And the thing that comes
to mind immediately

101
00:10:21,350 --> 00:10:26,780
is, why didn't Fourier invent
the uncertainty principle?

102
00:10:26,780 --> 00:10:29,540
Where did we use
quantum mechanics here?

103
00:10:36,450 --> 00:10:38,340
The answer is nowhere.

104
00:10:38,340 --> 00:10:40,480
We didn't use quantum mechanics.

105
00:10:40,480 --> 00:10:47,280
We found the relation between
wave packets, known to Fourier,

106
00:10:47,280 --> 00:10:50,700
known to electrical engineers.

107
00:10:50,700 --> 00:10:54,630
The place where quantum
mechanics comes about

108
00:10:54,630 --> 00:11:02,710
is when you realize that these
waves in quantum mechanics, e

109
00:11:02,710 --> 00:11:09,860
to the ikx represent states
with some values of momentum.

110
00:11:09,860 --> 00:11:16,470
So while this is fine and it's
a very important intuition,

111
00:11:16,470 --> 00:11:21,160
the step that you
can follow with is--

112
00:11:21,160 --> 00:11:21,925
it's interesting.

113
00:11:21,925 --> 00:11:29,950
And you say that, well,
since p, the momentum,

114
00:11:29,950 --> 00:11:37,510
is equal to h bar k and
that's quantum mechanical--

115
00:11:37,510 --> 00:11:39,190
it involves h bar.

116
00:11:39,190 --> 00:11:44,290
It's the whole discussion about
these waves of matter particles

117
00:11:44,290 --> 00:11:45,320
carrying momentum.

118
00:11:48,610 --> 00:11:53,920
You can say-- you can
multiply or take a delta here.

119
00:11:53,920 --> 00:11:58,648
And you would say, delta p
is equal to h bar delta k.

120
00:12:02,140 --> 00:12:07,090
So multiplying this
equation by an h bar,

121
00:12:07,090 --> 00:12:14,280
you would find that delta
p, delta x is roughly h bar.

122
00:12:18,700 --> 00:12:20,590
And that's quantum mechanical.

123
00:12:26,100 --> 00:12:31,020
Now we will make the
definitions of delta p and delta

124
00:12:31,020 --> 00:12:37,670
x precise and rigorous
with precise definitions.

125
00:12:41,050 --> 00:12:45,930
Then there is a precise
result, which is very neat,

126
00:12:45,930 --> 00:12:50,590
which is that delta
x times delta p

127
00:12:50,590 --> 00:12:56,110
is always greater than or
equal than h bar over 2.

128
00:12:59,260 --> 00:13:02,570
So this is really exact.

129
00:13:02,570 --> 00:13:07,570
But for that, we need to
define precisely what we

130
00:13:07,570 --> 00:13:14,910
mean by uncertainties, which
we will do soon, but not today.

131
00:13:14,910 --> 00:13:19,640
So I think it's
probably a good idea

132
00:13:19,640 --> 00:13:23,330
to do an example,
a simple example,

133
00:13:23,330 --> 00:13:29,630
to illustrate these relations.

134
00:13:29,630 --> 00:13:35,690
And here is one example.

135
00:13:35,690 --> 00:13:45,290
You have a phi of k of
the form of a step that

136
00:13:45,290 --> 00:13:51,460
goes from delta k over 2
to minus delta k over 2,

137
00:13:51,460 --> 00:13:57,750
and height 1 over
square root of delta k.

138
00:13:57,750 --> 00:14:00,590
That's phi of k.

139
00:14:00,590 --> 00:14:03,860
It's 0 otherwise--

140
00:14:03,860 --> 00:14:06,370
0 here, 0 there.

141
00:14:06,370 --> 00:14:07,760
Here is 0.

142
00:14:07,760 --> 00:14:09,350
Here is a function of k.

143
00:14:13,780 --> 00:14:14,740
What do you think?

144
00:14:14,740 --> 00:14:19,990
Is this psi of x, the
psi x corresponding

145
00:14:19,990 --> 00:14:24,990
to this phi of k-- is it going
to be a real function or not?

146
00:14:35,120 --> 00:14:37,145
Anybody?

147
00:14:37,145 --> 00:14:40,790
AUDIENCE: This equation
[? is ?] [? true, ?] [? but-- ?]

148
00:14:40,790 --> 00:14:42,976
PROFESSOR: Is it true or not?

149
00:14:42,976 --> 00:14:44,410
AUDIENCE: I think it is.

150
00:14:44,410 --> 00:14:46,220
PROFESSOR: OK.

151
00:14:46,220 --> 00:14:47,720
Yes, you're right.

152
00:14:47,720 --> 00:14:49,520
It is true.

153
00:14:49,520 --> 00:14:53,880
This phi of k is real.

154
00:14:53,880 --> 00:14:58,400
And whenever you have
a value at some k,

155
00:14:58,400 --> 00:15:02,930
there is the same
value at minus k.

156
00:15:02,930 --> 00:15:08,060
And therefore the star doesn't
matter, because it's real.

157
00:15:08,060 --> 00:15:10,130
So phi is completely real.

158
00:15:10,130 --> 00:15:13,790
So phi of k is equal
to phi of minus k.

159
00:15:13,790 --> 00:15:18,440
And that should give
you a real psi of x--

160
00:15:18,440 --> 00:15:20,060
correct.

161
00:15:20,060 --> 00:15:28,175
So some psi of x-- have to
do the integral-- psi of x0

162
00:15:28,175 --> 00:15:34,500
is 1 over square root of
2 pi minus delta k over 2

163
00:15:34,500 --> 00:15:37,820
to delta k over 2.

164
00:15:37,820 --> 00:15:45,040
The function, which is
1 over delta k in here--

165
00:15:47,660 --> 00:15:49,210
that's the whole function.

166
00:15:49,210 --> 00:15:52,930
And the integral was supposed
to be from minus infinity

167
00:15:52,930 --> 00:15:53,980
to infinity.

168
00:15:53,980 --> 00:15:58,120
But since the function only
extends from minus delta

169
00:15:58,120 --> 00:16:03,520
k over 2 to plus delta k over
2, you restrict the integral

170
00:16:03,520 --> 00:16:05,600
to those values.

171
00:16:05,600 --> 00:16:12,850
So we've already got the phi
of k and then e to the ikx dx.

172
00:16:17,980 --> 00:16:20,890
Well, the constants go out--

173
00:16:20,890 --> 00:16:24,370
2 pi delta k.

174
00:16:24,370 --> 00:16:31,120
And we have the integral
is an integral over x--

175
00:16:35,472 --> 00:16:37,050
no, I'm sorry.

176
00:16:37,050 --> 00:16:38,260
It's an integral over k.

177
00:16:38,260 --> 00:16:39,820
What I'm writing here--

178
00:16:39,820 --> 00:16:45,230
dk, of course.

179
00:16:45,230 --> 00:16:54,900
And that gives you e
to the ikx over ix,

180
00:16:54,900 --> 00:17:01,250
evaluated between delta k over
2 and minus delta k over 2.

181
00:17:04,010 --> 00:17:12,700
OK, a little simplification
gives the final answer.

182
00:17:12,700 --> 00:17:23,290
It's delta k over 2pi
sine of delta kx over 2

183
00:17:23,290 --> 00:17:27,339
over delta kx over 2.

184
00:17:30,310 --> 00:17:36,520
So it's a sine of x
over x type function.

185
00:17:36,520 --> 00:17:43,150
It's a familiar looking curve.

186
00:17:43,150 --> 00:17:44,215
It goes like this.

187
00:17:44,215 --> 00:17:49,604
It has some value-- it goes
down, up, down, up like that--

188
00:17:55,300 --> 00:17:57,400
symmetric.

189
00:17:57,400 --> 00:18:04,254
And here is psi of x and 0.

190
00:18:04,254 --> 00:18:17,100
Here is 2 pi over delta k, and
minus 2 pi over delta k here.

191
00:18:21,370 --> 00:18:25,400
Sine of x over x
looks like that.

192
00:18:25,400 --> 00:18:33,860
So this function already was
defined with the delta k.

193
00:18:33,860 --> 00:18:36,290
And what is the delta x here?

194
00:18:36,290 --> 00:18:44,910
Well, the delta x is
roughly 2 pi over delta k.

195
00:18:48,930 --> 00:18:52,730
No, it's-- you could say it's
this much or half of that.

196
00:18:52,730 --> 00:18:55,610
I took [? it half ?] of that.

197
00:18:55,610 --> 00:18:58,040
It doesn't matter.

198
00:18:58,040 --> 00:19:00,740
It's approximate
that at any rate now.

199
00:19:00,740 --> 00:19:02,330
So delta x is this.

200
00:19:02,330 --> 00:19:06,720
And therefore the
product delta x, delta k,

201
00:19:06,720 --> 00:19:11,050
delta x is about 2 pi.