1
00:00:00,499 --> 00:00:01,320
That's a solution.

2
00:00:01,320 --> 00:00:05,790
It's an accomplishment
to have such a solution.

3
00:00:05,790 --> 00:00:08,560
If somebody gives you
a value of the energy,

4
00:00:08,560 --> 00:00:11,340
you can calculate what
is the phase shift,

5
00:00:11,340 --> 00:00:15,860
but we probably want
to do more with it.

6
00:00:15,860 --> 00:00:24,780
So you decide to plot
this on a computer.

7
00:00:24,780 --> 00:00:27,850
Again, there's lots of
variables going on here,

8
00:00:27,850 --> 00:00:32,790
so you would want to figure out
what are the right variables

9
00:00:32,790 --> 00:00:33,750
to plot this.

10
00:00:33,750 --> 00:00:38,190
And the right variables
suggest themselves.

11
00:00:38,190 --> 00:00:45,030
From k squared equal 2 me over
h squared, unit less constant

12
00:00:45,030 --> 00:00:51,510
are things like ka, k
prime a, and that's it.

13
00:00:51,510 --> 00:00:57,310
Well, so ka is a proxy
for the energies.

14
00:00:57,310 --> 00:01:06,995
OK, a squared is really 2me,
a squared over h bar squared.

15
00:01:10,080 --> 00:01:14,250
And so this we
could call anything.

16
00:01:14,250 --> 00:01:17,220
Well, let's call it u.

17
00:01:17,220 --> 00:01:22,280
On the other hand, k
prime squared then--

18
00:01:22,280 --> 00:01:28,680
if you have k prime a squared
that it's also unit free

19
00:01:28,680 --> 00:01:38,880
would be 2me a squared
over h squared plus 2mv0

20
00:01:38,880 --> 00:01:41,980
a squared over h squared.

21
00:01:41,980 --> 00:01:43,990
You probably recognize them.

22
00:01:43,990 --> 00:01:46,850
The first one is just u squared.

23
00:01:46,850 --> 00:01:49,840
I should call this
u squared, sorry.

24
00:01:49,840 --> 00:01:53,755
U squared, and this is
our friend z0 squared.

25
00:01:57,430 --> 00:02:01,840
It's that number that tells you
the main thing you always want

26
00:02:01,840 --> 00:02:05,780
to know about a square well.

27
00:02:05,780 --> 00:02:11,750
That ratio between the energy
v0 to the demand to the energy

28
00:02:11,750 --> 00:02:14,630
that you can build
with h bar m and a.

29
00:02:17,240 --> 00:02:19,970
So here we go.

30
00:02:19,970 --> 00:02:24,110
We have k prime a
given by this quantity,

31
00:02:24,110 --> 00:02:30,010
and therefore let me
manipulate this equation.

32
00:02:30,010 --> 00:02:31,910
Might as well do it.

33
00:02:31,910 --> 00:02:36,410
It probably easier to
consider just tan delta, which

34
00:02:36,410 --> 00:02:38,720
is the inverse of this.

35
00:02:38,720 --> 00:02:41,570
You would have 1
minus the inverse

36
00:02:41,570 --> 00:02:47,710
of this would be k prime a
over ka, put the a's always,

37
00:02:47,710 --> 00:03:05,630
so cot k prime a tan ka over
tan ka plus k prime aka cot k

38
00:03:05,630 --> 00:03:07,500
prime a.

39
00:03:07,500 --> 00:03:14,065
So in terms of
our variables, see

40
00:03:14,065 --> 00:03:17,160
k prime a is the
square root of this,

41
00:03:17,160 --> 00:03:24,140
so k prime a square root of
u squared plus z0 squared,

42
00:03:24,140 --> 00:03:31,580
and k prime a over ka,
you divide now by u.

43
00:03:31,580 --> 00:03:39,070
So it's square root of 1 plus
z0 squared over u squared.

44
00:03:39,070 --> 00:03:40,320
That's this quantity.

45
00:03:40,320 --> 00:03:46,426
So how big, how much space
do I need to write it?

46
00:03:46,426 --> 00:03:47,800
Probably, I should
write it here.

47
00:03:47,800 --> 00:03:53,390
1 minus square root of
1 plus z0 squared over u

48
00:03:53,390 --> 00:04:02,010
squared cot k prime a is the
square root of z0 squared

49
00:04:02,010 --> 00:04:17,649
plus u squared and tan of
k a, which is u over tan u

50
00:04:17,649 --> 00:04:22,440
plus square root
of 1 plus z0 over u

51
00:04:22,440 --> 00:04:33,740
squared cotangent of square root
of z0 squared plus u squared.

52
00:04:33,740 --> 00:04:39,190
OK, it's not terrible.

53
00:04:39,190 --> 00:04:42,560
That's tan delta.

54
00:04:42,560 --> 00:04:46,930
So if somebody gives
you a potential,

55
00:04:46,930 --> 00:04:50,440
you calculate what z0
is for this potential,

56
00:04:50,440 --> 00:04:54,190
you put z0 there, and you
plot as a function of u

57
00:04:54,190 --> 00:04:55,630
with Mathematica.

58
00:04:55,630 --> 00:04:58,450
And plotting as a
function of u is

59
00:04:58,450 --> 00:05:01,540
plotting as a function of ka.

60
00:05:01,540 --> 00:05:04,630
And that's perfectly
nice thing to do.

61
00:05:04,630 --> 00:05:06,810
And it can be done
with this expression.

62
00:05:11,710 --> 00:05:14,500
In this expression,
you can also see

63
00:05:14,500 --> 00:05:21,860
what goes on when u goes to 0.

64
00:05:21,860 --> 00:05:25,090
Not immediately, it takes
a little bit of thinking,

65
00:05:25,090 --> 00:05:27,110
but look at it.

66
00:05:27,110 --> 00:05:29,360
As u goes to 0,
well, these numbers

67
00:05:29,360 --> 00:05:30,920
are 1, that's perfectly OK.

68
00:05:30,920 --> 00:05:37,340
That seems to diverge, goes
like 1 over u, but u going to 0.

69
00:05:37,340 --> 00:05:38,360
This goes to 0.

70
00:05:38,360 --> 00:05:41,280
So the product goes to a number.

71
00:05:41,280 --> 00:05:45,410
So the whole-- the
numerator goes to a number,

72
00:05:45,410 --> 00:05:47,760
some finite number.

73
00:05:47,760 --> 00:05:50,810
On the other hand,
when u goes to 0,

74
00:05:50,810 --> 00:05:53,120
the denominator
will go to infinity,

75
00:05:53,120 --> 00:05:57,770
because while this term
goes to 0 the tan u,

76
00:05:57,770 --> 00:05:59,990
this number is finite.

77
00:05:59,990 --> 00:06:02,480
And here you have a 1/u.

78
00:06:02,480 --> 00:06:05,420
So the denominator
goes to infinity.

79
00:06:05,420 --> 00:06:07,400
And the numerator
remains finite.

80
00:06:07,400 --> 00:06:14,960
So as u goes to 0, tangent
of delta goes to zero.

81
00:06:14,960 --> 00:06:20,810
So you can choose delta
to be 0 for 0 energy.

82
00:06:20,810 --> 00:06:34,580
So as u goes to 0, you get
finite divided by infinity,

83
00:06:34,580 --> 00:06:35,870
and goes to zero.

84
00:06:35,870 --> 00:06:41,010
So tan delta goes to 0.

85
00:06:41,010 --> 00:06:49,680
And we can take delta of ka
equals 0, which is u to be 0.

86
00:06:49,680 --> 00:06:54,040
The phase shift
is 0 for 0 energy.

87
00:06:58,469 --> 00:07:01,160
Let me go here.

88
00:07:01,160 --> 00:07:03,200
So here is an example.

89
00:07:03,200 --> 00:07:09,400
z0 squared equal 3.4.

90
00:07:09,400 --> 00:07:19,630
That actually correspond
to 0.59pi for z0.

91
00:07:19,630 --> 00:07:23,910
z0 equal 0.59pi.

92
00:07:23,910 --> 00:07:26,420
You may wonder why
we do that, but let

93
00:07:26,420 --> 00:07:29,160
me tell you in a second.

94
00:07:29,160 --> 00:07:33,950
So here are a couple
of plots that occur.

95
00:07:33,950 --> 00:07:37,610
So here is u equals ka.

96
00:07:37,610 --> 00:07:41,760
And here's the phase
shift, delta of u.

97
00:07:41,760 --> 00:07:46,260
You have the tangent of
delta, but the phase shift

98
00:07:46,260 --> 00:07:47,550
can be calculated.

99
00:07:47,550 --> 00:07:50,970
And what you find
is that, yes, it

100
00:07:50,970 --> 00:07:54,220
starts at 0, as we mentioned.

101
00:07:54,220 --> 00:07:56,490
And then it starts
going down, but it

102
00:07:56,490 --> 00:08:05,545
stabilizes at minus pi,
which is a neat number.

103
00:08:08,120 --> 00:08:09,670
That's what the
phase shift does.

104
00:08:12,750 --> 00:08:16,130
The so-called scattering
amplitude, well you

105
00:08:16,130 --> 00:08:19,550
could say, when is this
scattering strongest?

106
00:08:19,550 --> 00:08:23,750
When you get an extra wave of
this propagating more strongly?

107
00:08:23,750 --> 00:08:32,840
So you must plot sine squared
delta and sine squared is

108
00:08:32,840 --> 00:08:35,840
highest for minus pi over 2.

109
00:08:35,840 --> 00:08:43,190
So this goes like this, up,
and decays as a function of u.

110
00:08:49,400 --> 00:08:55,190
Third thing, the delay, is 1/a.

111
00:09:03,300 --> 00:09:11,240
The delay is 1/a d delta
dk, as a function of u.

112
00:09:11,240 --> 00:09:15,350
So that, you can imagine,
that takes a bit of time,

113
00:09:15,350 --> 00:09:18,740
because you would have to
find the derivative of delta

114
00:09:18,740 --> 00:09:22,590
with respect to u, and do
all kinds of operations.

115
00:09:22,590 --> 00:09:26,400
Don't worry, you will have
a bit of exercises on this

116
00:09:26,400 --> 00:09:27,500
to do it yourselves.

117
00:09:27,500 --> 00:09:32,240
But here the delay turns
out to be negative.

118
00:09:32,240 --> 00:09:33,920
And this is unit-free.

119
00:09:33,920 --> 00:09:38,210
And here, comes to be
equal minus 4 for equals 0,

120
00:09:38,210 --> 00:09:40,490
and goes down to 0.

121
00:09:44,180 --> 00:09:47,390
So in this case, the
delay is negative.

122
00:09:47,390 --> 00:09:50,780
So the reflected
packet comes earlier

123
00:09:50,780 --> 00:09:54,200
than you would expected,
which is possible,

124
00:09:54,200 --> 00:10:00,420
because the reflected
packet is going slowly here.

125
00:10:00,420 --> 00:10:04,270
Finally, at this point, reaches
more kinetic energy, just--

126
00:10:04,270 --> 00:10:07,360
and then back.

127
00:10:07,360 --> 00:10:11,420
So that's the delay.

128
00:10:15,390 --> 00:10:18,420
And you can plot another thing.

129
00:10:18,420 --> 00:10:23,090
Actually it's kind of
interesting, is the quantity

130
00:10:23,090 --> 00:10:26,980
a, this coefficient here.

131
00:10:26,980 --> 00:10:29,370
That gives you an idea
of how big the wave

132
00:10:29,370 --> 00:10:32,910
function is in the well.

133
00:10:32,910 --> 00:10:36,870
How much does it
stick near the well?

134
00:10:36,870 --> 00:10:41,340
So it peaks to 1.

135
00:10:41,340 --> 00:10:44,460
And it actually goes
like this, and that's

136
00:10:44,460 --> 00:10:45,790
the behavior of this form.

137
00:10:48,760 --> 00:10:50,810
Basically, it does those things.

138
00:10:50,810 --> 00:10:56,510
So, so far so good.

139
00:10:56,510 --> 00:10:58,070
We got some information.

140
00:10:58,070 --> 00:11:02,310
And then you do a
little experiment,

141
00:11:02,310 --> 00:11:07,816
and try, for
example, z0 equals 5.

142
00:11:11,530 --> 00:11:15,520
And you have delta
as a function of u,

143
00:11:15,520 --> 00:11:20,050
and here is minus pi, minus 2pi.

144
00:11:20,050 --> 00:11:23,770
And actually, you find that it
just goes down, and approaches

145
00:11:23,770 --> 00:11:24,970
now minus 2pi.

146
00:11:34,430 --> 00:11:42,200
So actually, if you
increase this z0 a bit,

147
00:11:42,200 --> 00:11:48,410
it still goes to pi, a pi
excursion of the phase.

148
00:11:48,410 --> 00:11:51,860
But suddenly, at
some value, it jumps.

149
00:11:51,860 --> 00:11:55,670
And it now goes to 2pi.

150
00:11:55,670 --> 00:11:58,450
And if you do with a
larger value, at some point

151
00:11:58,450 --> 00:12:02,610
it goes to 3pi and 4pi.

152
00:12:02,610 --> 00:12:04,210
And it goes on like that.

153
00:12:07,600 --> 00:12:15,320
Well if z0 would have been
smaller, like half of this,

154
00:12:15,320 --> 00:12:20,010
the phase would go down
and would go back up,

155
00:12:20,010 --> 00:12:24,020
wouldn't go to minus pi.

156
00:12:24,020 --> 00:12:26,120
It does funny things.

157
00:12:26,120 --> 00:12:31,020
So what's really
happening is that there

158
00:12:31,020 --> 00:12:36,900
is a relation between
how much the phase moves,

159
00:12:36,900 --> 00:12:41,160
and how many bound states
this potential has.

160
00:12:41,160 --> 00:12:44,490
And you say, why in the world?

161
00:12:44,490 --> 00:12:48,780
This calculation had nothing
to do with bound states.

162
00:12:48,780 --> 00:12:52,050
Why would the phase shift
know about the bound states?

163
00:12:52,050 --> 00:12:54,300
Well actually, it does.

164
00:12:54,300 --> 00:12:57,000
And here is the thing.

165
00:12:57,000 --> 00:13:01,290
If you remember, you've
actually solved this problem

166
00:13:01,290 --> 00:13:04,945
in homework, the half
square well, in which you

167
00:13:04,945 --> 00:13:07,560
put an infinite wall here.

168
00:13:07,560 --> 00:13:12,390
And if you had the full square
well, from minus a to a,

169
00:13:12,390 --> 00:13:16,590
this problem has all
the old solutions

170
00:13:16,590 --> 00:13:20,280
of the full square well.

171
00:13:20,280 --> 00:13:22,950
All the old solutions exist.

172
00:13:22,950 --> 00:13:28,140
And if you remember
the plots that you

173
00:13:28,140 --> 00:13:31,620
would do in order
to find solutions,

174
00:13:31,620 --> 00:13:40,050
you have pi/2, pi, 3pi/2, 2pi.

175
00:13:40,050 --> 00:13:44,100
And here is the even solution.

176
00:13:44,100 --> 00:13:46,110
Here is the odd solution.

177
00:13:46,110 --> 00:13:49,740
I'll do it like that.

178
00:13:49,740 --> 00:13:52,620
Here is an even solution.

179
00:13:52,620 --> 00:13:55,512
Here is an odd solution.

180
00:13:55,512 --> 00:13:59,540
And I marked the odd
solutions, because we

181
00:13:59,540 --> 00:14:04,720
care about the odd ones, because
that's what this potential has.

182
00:14:04,720 --> 00:14:12,670
So z0 equals 0.59pi is
a little more than pi/2.

183
00:14:12,670 --> 00:14:16,050
So it corresponds
to one solution.

184
00:14:16,050 --> 00:14:21,360
So there is one bound
state for this z0.

185
00:14:21,360 --> 00:14:26,430
z0 equals 5 is about here.

186
00:14:26,430 --> 00:14:28,840
it's in between 3pi/2 and this.

187
00:14:28,840 --> 00:14:33,450
And there's two nodes,
two intersections.

188
00:14:33,450 --> 00:14:36,450
Therefore, two solutions
in the square well.

189
00:14:36,450 --> 00:14:42,330
And here we have that the phase
has an excursion of, not just

190
00:14:42,330 --> 00:14:44,310
pi for one, but 2pi.

191
00:14:47,240 --> 00:14:53,020
And if you did this
experiment for awhile,

192
00:14:53,020 --> 00:14:54,610
you would convince
yourself there's

193
00:14:54,610 --> 00:15:01,300
a magic relation between how
much the phase shift moves,

194
00:15:01,300 --> 00:15:05,800
and how many bound states
you have in this potential.

195
00:15:05,800 --> 00:15:10,840
This relation is called
Levinson's theorem.

196
00:15:10,840 --> 00:15:14,470
And that's what we're going
to prove in the last half

197
00:15:14,470 --> 00:15:16,740
an hour of this lecture.