1
00:00:00,750 --> 00:00:02,478
PROFESSOR: This is the answer.

2
00:00:02,478 --> 00:00:08,430
Tan of 2 k a plus delta.

3
00:00:08,430 --> 00:00:11,640
It's a little messy, no wonder.

4
00:00:11,640 --> 00:00:15,830
Just plotted this functions.

5
00:00:15,830 --> 00:00:26,290
[INAUDIBLE] Sin k plus a
cosh kappa a [INAUDIBLE]

6
00:00:26,290 --> 00:00:37,720
plus a prime over kappa
cosh a prime a sinh kappa a.

7
00:00:37,720 --> 00:00:44,470
You have sin a prime a sinh.

8
00:00:44,470 --> 00:00:48,532
The sinh causes outer
[INAUDIBLE] the same

9
00:00:48,532 --> 00:00:50,740
but this one changes.

10
00:00:50,740 --> 00:00:56,270
Centered here, k prime
over kappa cos k prime

11
00:00:56,270 --> 00:01:02,466
a cosh kappa a.

12
00:01:02,466 --> 00:01:07,140
So it's a well-defined
expression,

13
00:01:07,140 --> 00:01:10,070
if you know the
number of kappa a,

14
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you can calculate everything.

15
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Let me just make sure
you can see that.

16
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If you know-- if you
said, for example, ka,

17
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this is a natural variable.

18
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ka is just-- the k is related
to the energy direction.

19
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So ka has no limits, call it u.

20
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Then you have two parameters
of the square well [INAUDIBLE],,

21
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a depth we don't and
a height we want.

22
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So there's natural to
define just [INAUDIBLE]

23
00:01:53,730 --> 00:01:58,920
would have z 0 squared
which is 2m v 0

24
00:01:58,920 --> 00:02:02,400
a squared over h squared.

25
00:02:02,400 --> 00:02:08,830
You also define z 1 squared
[INAUDIBLE] as 2m v1

26
00:02:08,830 --> 00:02:11,611
a squared over h squared.

27
00:02:16,968 --> 00:02:22,620
And then the energy--

28
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we can do another
thing, we're going

29
00:02:29,046 --> 00:02:34,192
to find the energy divided
say, by v 1 to be little e.

30
00:02:37,168 --> 00:02:41,194
And that's reasonable
because these energies

31
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is compared always with
v 1, and we're solving--

32
00:02:45,630 --> 00:02:48,146
we've solved this
problem in the domain

33
00:02:48,146 --> 00:02:51,730
when the energy
is less than v 1,

34
00:02:51,730 --> 00:02:53,470
and that's why we'll
have kappas here,

35
00:02:53,470 --> 00:02:56,045
and if energy was
bigger than v 1,

36
00:02:56,045 --> 00:03:00,414
you would have trigonometric
functions everywhere.

37
00:03:00,414 --> 00:03:04,670
Now you can switch from
trigonometric to hyperbolic

38
00:03:04,670 --> 00:03:09,590
by analytic continuation,
letting and angle

39
00:03:09,590 --> 00:03:13,740
become imaginary, a
trigonometric function becomes

40
00:03:13,740 --> 00:03:14,674
a hyperbolic function.

41
00:03:18,410 --> 00:03:21,320
Most of us are
rather comfortable

42
00:03:21,320 --> 00:03:27,110
doing this at the beginning,
because of the matter of sine

43
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if you mess up a sine,
it's a big problem.

44
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But at the end of the day, it
actually saves a lot of work.

45
00:03:35,790 --> 00:03:40,840
So a little of that in
the homework you will see.

46
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But at any rate, this is
valued for e in this range.

47
00:03:48,030 --> 00:03:55,010
And therefore this ratio
is h squared k squared.

48
00:03:55,010 --> 00:04:05,722
You can put an a squared
over 2m v 1 a squared.

49
00:04:05,722 --> 00:04:10,266
And you can see
a u squared here.

50
00:04:10,266 --> 00:04:16,010
And-- I'm sorry, not there.

51
00:04:16,010 --> 00:04:19,760
A u squared here, and
the h squared divided

52
00:04:19,760 --> 00:04:21,519
here gives you a z 1.

53
00:04:21,519 --> 00:04:27,910
So this is u squared
over z 1 squared.

54
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A couple of extra things--

55
00:04:29,990 --> 00:04:32,690
I'm just putting it
here because if you ever

56
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have the curiosity of doing
this plots, this may help.

57
00:04:37,940 --> 00:04:42,370
k prime squared is
related to the energy,

58
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therefore u squared
and v 0 z 1 squared.

59
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This is-- c 0 squared,
[INAUDIBLE] 0.

60
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And for the other
one, kappa a squared

61
00:04:56,020 --> 00:05:02,060
is equal to z 1 squared
minus u squared.

62
00:05:02,060 --> 00:05:05,590
So everything becomes a
function of u Wherever

63
00:05:05,590 --> 00:05:09,350
you have a kappa prime a
you have a square root of u.

64
00:05:09,350 --> 00:05:14,900
Now, somebody has to give you
the values of z 0 and z 1.

65
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Those are the data of
the potential-- z 0

66
00:05:17,820 --> 00:05:19,988
and z 1 are numbers
you know then.

67
00:05:19,988 --> 00:05:25,220
Therefore you know kappa a
is a function of u, kappa--

68
00:05:25,220 --> 00:05:32,360
k prime a is a function
of u, and ka, which is u.

69
00:05:32,360 --> 00:05:36,380
Therefore this equation becomes
a function of u [INAUDIBLE]

70
00:05:36,380 --> 00:05:41,330
this arc tangent here,
and so for delta.

71
00:05:44,516 --> 00:05:46,435
It's messy.

72
00:05:46,435 --> 00:05:52,200
There's no-- remember I
mentioned the other day

73
00:05:52,200 --> 00:05:56,070
the fact that you could do
trigonometric identities,

74
00:05:56,070 --> 00:05:58,450
and so for tan delta here.

75
00:05:58,450 --> 00:06:02,275
But then this I don't think
simplifies when you solve that.

76
00:06:02,275 --> 00:06:03,840
It becomes just a bigger mess.

77
00:06:03,840 --> 00:06:07,590
You did solve for tan
delta, it's very messy.

78
00:06:07,590 --> 00:06:10,580
So anyway, let's leave it there.

79
00:06:10,580 --> 00:06:14,160
Are there any questions
in solving this problem?

80
00:06:14,160 --> 00:06:16,680
So in principle, we solved it.

81
00:06:16,680 --> 00:06:18,690
I didn't plot
anything, so you still

82
00:06:18,690 --> 00:06:22,190
don't have any insight
as to what's happening.

83
00:06:22,190 --> 00:06:24,590
But you've learned in
principle how to solve it.

84
00:06:27,410 --> 00:06:28,612
Any questions?

85
00:06:38,940 --> 00:06:41,458
OK, so let's plot this then.

86
00:06:47,802 --> 00:06:51,720
Maybe I should start here.

87
00:06:51,720 --> 00:06:53,010
OK.

88
00:06:53,010 --> 00:06:55,470
I'll start here.

89
00:06:55,470 --> 00:06:58,360
Oh, if I'm going to plot,
I have to choose things.

90
00:06:58,360 --> 00:07:02,500
So we'll choose z 0
squared equals to 1,

91
00:07:02,500 --> 00:07:04,183
or z 0 equals to 1.

92
00:07:04,183 --> 00:07:09,960
And z 1 squared equals to 5.

93
00:07:09,960 --> 00:07:11,440
So we put a big barrier there.

94
00:07:14,660 --> 00:07:22,434
And now let's go delta
as a function of ka or u.

95
00:07:28,540 --> 00:07:38,354
Now, from this equation, we
see that u and the little e

96
00:07:38,354 --> 00:07:43,650
must be at most 1 for our
formulas to be correct.

97
00:07:43,650 --> 00:07:47,690
So u must be up to square--

98
00:07:47,690 --> 00:07:53,890
up to z 1, because you must
keep this ratio less than 1.

99
00:07:53,890 --> 00:07:58,540
So we can plot u up
to square root of 5.

100
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5, here's 2.

101
00:08:06,687 --> 00:08:09,338
And that's it.

102
00:08:09,338 --> 00:08:14,990
And here we're going
to plot delta of u.

103
00:08:14,990 --> 00:08:20,466
And here is minus
pi over 2 minus pi.

104
00:08:20,466 --> 00:08:24,690
And so what does
it do, the phase?

105
00:08:24,690 --> 00:08:29,040
There no way you can guess,
I think, from this formula.

106
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I could not guess
from this formula.

107
00:08:31,560 --> 00:08:36,590
So we could try to imagine--

108
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it's possible to guess,
actually, after you've

109
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solved this week's homework.

110
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You probably will
have a good guess

111
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that the phase shift begins
with minus 2ka, as if

112
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with very little energy
you reflect back here.

113
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So there will be a shift that
is calculable without doing

114
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any work.

115
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So it begins linearly.

116
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And it represents
a time advance.

117
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So it goes linearly
and negative.

118
00:09:08,960 --> 00:09:11,350
That's how it begins.

119
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Because for very little energy,
it's going to bounce back.

120
00:09:17,070 --> 00:09:19,894
And you know that the
delay is proportional

121
00:09:19,894 --> 00:09:21,550
to this derivative.

122
00:09:21,550 --> 00:09:24,270
So must be negative like that.

123
00:09:24,270 --> 00:09:27,040
So then what does it do?

124
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It crosses this point,
which is almost pi,

125
00:09:34,052 --> 00:09:38,830
and here is what something
quite remarkable happens.

126
00:09:38,830 --> 00:09:52,855
That some value you star,
which is about 1.8523, I think.

127
00:09:52,855 --> 00:09:54,316
That's what I calculated.

128
00:09:58,220 --> 00:09:59,696
Let's see [INAUDIBLE].

129
00:10:02,504 --> 00:10:09,820
The face that is almost minus pi
suddenly jumps very, very fast,

130
00:10:09,820 --> 00:10:16,851
crosses pi over 2, and then
about [INAUDIBLE] something

131
00:10:16,851 --> 00:10:18,535
like this.

132
00:10:18,535 --> 00:10:20,750
I don't know what else
it does, because we

133
00:10:20,750 --> 00:10:23,210
haven't calculated it.

134
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But it jumps very fast.

135
00:10:26,300 --> 00:10:29,060
Almost a value of pi.

136
00:10:29,060 --> 00:10:34,070
Now, let me-- this
is quite interesting.

137
00:10:34,070 --> 00:10:37,280
If you think of what we
used to call the scattering

138
00:10:37,280 --> 00:10:43,220
amplitude as squared
was sin squared delta.

139
00:10:43,220 --> 00:10:46,800
The amplitude of the
scattered wave, sin

140
00:10:46,800 --> 00:10:50,630
squared delta, the amplitude
of the scattered wave

141
00:10:50,630 --> 00:10:52,820
is going to be quite
large here, it's

142
00:10:52,820 --> 00:10:58,920
going to be 1, which is the
maximum, as squared is here.

143
00:10:58,920 --> 00:11:02,162
So I want to keep these
two blackboards aligned.

144
00:11:05,540 --> 00:11:07,280
So here it goes like this.

145
00:11:07,280 --> 00:11:15,200
It's going to go up like this,
and do this, and broadly go

146
00:11:15,200 --> 00:11:22,690
down, and then
very sharply go up.

147
00:11:22,690 --> 00:11:24,832
More sharply, at least.

148
00:11:24,832 --> 00:11:29,770
Go up and then
something like that.

149
00:11:29,770 --> 00:11:32,870
So here it is.

150
00:11:32,870 --> 00:11:40,880
This point over here has a
strong scattering amplitude,

151
00:11:40,880 --> 00:11:45,640
but there's nothing too
dramatic happening here.

152
00:11:45,640 --> 00:11:49,810
This sine corresponds to
time advanced, the derivative

153
00:11:49,810 --> 00:11:51,885
is negative.

154
00:11:51,885 --> 00:11:56,730
And time advanced cannot
be too large, as you know.

155
00:11:56,730 --> 00:12:00,780
On the other hand,
here is time delayed,

156
00:12:00,780 --> 00:12:04,190
and apparently they
can't be very large.

157
00:12:04,190 --> 00:12:07,975
So we think this must
be the resonance,

158
00:12:07,975 --> 00:12:11,680
and this is not a resonance,
even though the scattering

159
00:12:11,680 --> 00:12:15,260
amplitude is bigger.

160
00:12:15,260 --> 00:12:24,530
So we continue here,
and plot the time--

161
00:12:24,530 --> 00:12:29,010
interestingly, the
amplitude inside the well.

162
00:12:32,630 --> 00:12:33,910
So here it is.

163
00:12:33,910 --> 00:12:39,753
How does the wave function
become, through this constant,

164
00:12:39,753 --> 00:12:41,820
inside the well?

165
00:12:41,820 --> 00:12:45,675
And indeed, the this confirms
that nothing very special

166
00:12:45,675 --> 00:12:48,650
is happening here.

167
00:12:48,650 --> 00:12:51,945
What happens now is some
sort of behavior like this,

168
00:12:51,945 --> 00:12:58,766
and a big jump here, in which
the amplitude apparently--

169
00:12:58,766 --> 00:13:01,020
I have not quite
confirmed this number,

170
00:13:01,020 --> 00:13:02,730
it's at least a value 3.

171
00:13:06,540 --> 00:13:10,702
[INAUDIBLE] very
large and short.

172
00:13:10,702 --> 00:13:17,630
I should have room for one more
plot [INAUDIBLE] the star plot

173
00:13:17,630 --> 00:13:19,340
of this thing is--

174
00:13:19,340 --> 00:13:21,240
I'll do it compressed here.

175
00:13:24,090 --> 00:13:29,060
The total delay, 1
over a d delta dk.

176
00:13:31,980 --> 00:13:36,500
Well, it begins
negative, and remember

177
00:13:36,500 --> 00:13:41,720
that when we did the
1 over a, d delta dk,

178
00:13:41,720 --> 00:13:42,970
this is a pure number.

179
00:13:42,970 --> 00:13:47,622
It expresses the delay
in terms of the time

180
00:13:47,622 --> 00:13:51,080
that it would take to
travel the inside region.

181
00:13:51,080 --> 00:13:55,540
So how many-- if you would
get a 1, or a minus 1,

182
00:13:55,540 --> 00:13:59,220
it's just a delay of
the size of the time

183
00:13:59,220 --> 00:14:02,210
needed to travel back and forth.

184
00:14:02,210 --> 00:14:05,560
So actually this goes
a little negative

185
00:14:05,560 --> 00:14:08,165
at the beginning-- we
know the derivative is

186
00:14:08,165 --> 00:14:10,410
like that-- and
when you plot this,

187
00:14:10,410 --> 00:14:14,100
you see that it's
very sharp, and it's

188
00:14:14,100 --> 00:14:17,690
a value of about fourteen.

189
00:14:17,690 --> 00:14:22,180
Fourteen times gets delayed from
what you would expect naturally

190
00:14:22,180 --> 00:14:26,799
of the time that it should have
spent travelling back and forth

191
00:14:26,799 --> 00:14:29,733
A gigantic time delay.

192
00:14:29,733 --> 00:14:33,190
A peak in the time delay.

193
00:14:33,190 --> 00:14:36,870
Peak in time delay.

194
00:14:39,590 --> 00:14:46,860
Peak in the amplitude
inside the well.