1
00:00:01,030 --> 00:00:02,800
PROFESSOR: Delta
function potential.

2
00:00:02,800 --> 00:00:06,120
So it's still a one-dimensional
potential-- potential

3
00:00:06,120 --> 00:00:08,960
is a function of x.

4
00:00:08,960 --> 00:00:12,490
We'll write it this
way-- minus alpha delta

5
00:00:12,490 --> 00:00:17,190
of x, where alpha is positive.

6
00:00:17,190 --> 00:00:22,030
So this is a delta function
in a negative direction.

7
00:00:22,030 --> 00:00:23,910
So if you want to
draw the potential--

8
00:00:23,910 --> 00:00:27,350
there's no way to draw really
nicely a delta function.

9
00:00:27,350 --> 00:00:32,970
So you just do a thick
arrow with it pointing down.

10
00:00:32,970 --> 00:00:36,750
It's a representation of
a potential that, somehow,

11
00:00:36,750 --> 00:00:42,070
is rather infinite
at x equals zero--

12
00:00:42,070 --> 00:00:45,010
but infinite and negative.

13
00:00:45,010 --> 00:00:49,990
It can be thought of as the
limit of a square well that

14
00:00:49,990 --> 00:00:53,350
is becoming deeper and deeper.

15
00:00:53,350 --> 00:00:55,900
And, in fact, that
could be a way

16
00:00:55,900 --> 00:00:59,200
to analytically
calculate the energy

17
00:00:59,200 --> 00:01:03,010
levels-- by taking carefully
the limit of a potential.

18
00:01:03,010 --> 00:01:07,180
It is becoming thinner and
thinner, but deeper and deeper,

19
00:01:07,180 --> 00:01:11,290
which is the way you define or
regulate the delta function.

20
00:01:11,290 --> 00:01:13,360
You can imagine
the delta function

21
00:01:13,360 --> 00:01:16,450
as a sequence of
functions, in which it's

22
00:01:16,450 --> 00:01:18,040
becoming more and more narrow--

23
00:01:18,040 --> 00:01:19,850
but deeper at the same time.

24
00:01:19,850 --> 00:01:24,350
So that the area under the
curve is still the same.

25
00:01:24,350 --> 00:01:27,970
So, at any rate, the delta
function potential is

26
00:01:27,970 --> 00:01:31,690
a potential that should be
understood as 0 everywhere ,

27
00:01:31,690 --> 00:01:36,250
else except at the delta
function where it becomes

28
00:01:36,250 --> 00:01:38,000
infinite.

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And there are all kinds
of questions we can ask.

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00:01:41,700 --> 00:01:42,500
OK.

31
00:01:42,500 --> 00:01:44,330
Are there bound states?

32
00:01:44,330 --> 00:01:46,430
What are bound
states in this case?

33
00:01:46,430 --> 00:01:53,270
They are energy eigenstates
with energy less than zero.

34
00:01:53,270 --> 00:02:02,160
So bound states, which
means e less than zero.

35
00:02:02,160 --> 00:02:04,216
Do they exist?

36
00:02:04,216 --> 00:02:08,190
Does this potential
have bound states?

37
00:02:08,190 --> 00:02:11,130
And, if it does, how
many bound states?

38
00:02:11,130 --> 00:02:12,480
1, 2, 3?

39
00:02:12,480 --> 00:02:16,110
Does It depend on the intensity
of the delta function?

40
00:02:16,110 --> 00:02:20,590
When you get more bound states,
the deeper the potential is.

41
00:02:20,590 --> 00:02:23,580
Well, we'll try to figure out.

42
00:02:23,580 --> 00:02:26,550
In fact, there's a lot
that can be figured out

43
00:02:26,550 --> 00:02:29,280
without calculating, too much.

44
00:02:29,280 --> 00:02:36,240
And it's a good habit to try
to do those things before you--

45
00:02:36,240 --> 00:02:39,010
not to be so impatient
that you begin,

46
00:02:39,010 --> 00:02:41,100
and within a second
start writing

47
00:02:41,100 --> 00:02:43,830
the differential equation
trying to solve it.

48
00:02:43,830 --> 00:02:49,860
Get a little intuition about
how any state could look like,

49
00:02:49,860 --> 00:02:54,210
and how could the answer
for the energy eigenstates--

50
00:02:54,210 --> 00:02:57,060
the energies--
what could they be?

51
00:02:57,060 --> 00:03:00,160
Could you just reason
your way and conclude

52
00:03:00,160 --> 00:03:01,620
there's no bound states?

53
00:03:01,620 --> 00:03:02,820
Or one bound state?

54
00:03:02,820 --> 00:03:03,960
Or two?

55
00:03:03,960 --> 00:03:05,840
All these things
are pretty useful.

56
00:03:05,840 --> 00:03:10,355
So one way, as you can
imagine, is to think of units.

57
00:03:15,800 --> 00:03:20,710
And what are the
constants in this problem?

58
00:03:20,710 --> 00:03:23,340
In this problem we'll
have three constants.

59
00:03:23,340 --> 00:03:29,730
Alpha, the mass of the
particle, and h-bar.

60
00:03:29,730 --> 00:03:32,340
So with alpha, the
mass and the particle,

61
00:03:32,340 --> 00:03:36,450
and h-bar you can ask, how
do I construct the quantity

62
00:03:36,450 --> 00:03:38,910
with units of energy?

63
00:03:38,910 --> 00:03:42,100
If there there's only one
way to construct the quantity

64
00:03:42,100 --> 00:03:46,510
with units of energy, then
the energy of a bound state

65
00:03:46,510 --> 00:03:48,880
will be proportional
to that quantity--

66
00:03:48,880 --> 00:03:52,840
because that's the only quantity
that can carry the units.

67
00:03:52,840 --> 00:03:55,150
And here, indeed,
there's only one way

68
00:03:55,150 --> 00:03:58,000
to construct that quantity
with units of energy--

69
00:03:58,000 --> 00:04:00,730
from these three.

70
00:04:00,730 --> 00:04:02,020
That's to be expected.

71
00:04:02,020 --> 00:04:08,290
With three constants that
are not linearly dependent--

72
00:04:08,290 --> 00:04:11,830
whatever that is
supposed to mean--

73
00:04:11,830 --> 00:04:17,019
you can build anything that has
units of length, mass, or time.

74
00:04:17,019 --> 00:04:18,700
And from that you
can build something

75
00:04:18,700 --> 00:04:20,000
that has units of energy.

76
00:04:20,000 --> 00:04:25,030
So you can now decide, well,
what are the units of alpha?

77
00:04:25,030 --> 00:04:32,350
The units of alpha
have give you energy,

78
00:04:32,350 --> 00:04:36,070
but the delta function has
units of one over length.

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00:04:36,070 --> 00:04:38,522
This has one over length.

80
00:04:38,522 --> 00:04:42,940
, Remember if you integrate over
x the delta function gives you

81
00:04:42,940 --> 00:04:43,440
1.

82
00:04:43,440 --> 00:04:45,960
So this has units
of 1 over length.

83
00:04:45,960 --> 00:04:48,150
And, therefore,
alpha has to have

84
00:04:48,150 --> 00:04:52,140
units of energy times length.

85
00:04:54,690 --> 00:04:59,120
So this is not quite enough
to solve the problem,

86
00:04:59,120 --> 00:05:03,010
because I want to write e--

87
00:05:03,010 --> 00:05:06,810
think of finding how do
you get units of energy

88
00:05:06,810 --> 00:05:08,040
from these quantities?

89
00:05:08,040 --> 00:05:11,610
But l-- we still don't
have a length scale either.

90
00:05:11,610 --> 00:05:13,900
So we have to do a
little more work.

91
00:05:13,900 --> 00:05:21,550
So from here we say that units
of energy is alpha over l.

92
00:05:21,550 --> 00:05:23,340
There should be a
way to say that this

93
00:05:23,340 --> 00:05:25,800
is an equality between units.

94
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I could put units or
leave it just like that.

95
00:05:34,540 --> 00:05:37,660
So in terms of units, it's this.

96
00:05:37,660 --> 00:05:40,120
But in terms of units, energy--

97
00:05:40,120 --> 00:05:42,490
you should always remember--

98
00:05:42,490 --> 00:05:44,340
is p squared over m.

99
00:05:44,340 --> 00:05:48,800
And p is h-bar over a length.

100
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So that's p squared
and that's m.

101
00:05:53,200 --> 00:05:56,170
So that's also units of energy

102
00:05:56,170 --> 00:06:05,100
From these two you can get
what has units of length.

103
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Length.

104
00:06:06,860 --> 00:06:09,830
You pass the l to
this side-- the l

105
00:06:09,830 --> 00:06:11,871
squared to this left-hand side.

106
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Divide.

107
00:06:12,370 --> 00:06:20,950
So you get l is h
squared over m alpha.

108
00:06:20,950 --> 00:06:25,970
And if I substitute
back into this l here,

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00:06:25,970 --> 00:06:33,660
e would be alpha over l, which
is h squared, alpha squared, m.

110
00:06:33,660 --> 00:06:38,500
So that's the quantity
that has units of energy.

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00:06:38,500 --> 00:06:44,730
M alpha squared over h
squared has units of energy.

112
00:06:44,730 --> 00:06:46,830
If this has units of energy--

113
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the bound state energy.

114
00:06:50,080 --> 00:06:53,460
Now, if you have a
bounce state here,

115
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it has to decay in order
to be normalizable.

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In order to be normalizable
it has to decay,

117
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so it has to be in the
forbidden region throughout x.

118
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So the energy as we
said is negative,

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energy of a bound
state-- if it exists.

120
00:07:14,410 --> 00:07:16,910
And this bound
state energy would

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have to be negative some number
m alpha squared over h squared.

122
00:07:25,970 --> 00:07:28,850
And that's very
useful information.

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The whole problem
has been reduced

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to calculating a number.

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It better be and the answer
cannot be any other way.

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There's no other way to
get the units of energy.

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So if a bound state
exists it has to be that.

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And that number could be
pi, it could be 1/3, 1/4,

129
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it could be anything.

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There's a naturalness
to that problem

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in that you don't expect
that number to be a trillion.

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00:08:01,610 --> 00:08:05,900
Nor do you expect that number
to be 10 to the minus 6.

133
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Because there's no way-- where
would those numbers appear?

134
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So this number should be
a number of order one,

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00:08:12,110 --> 00:08:16,370
and we're going to wait
and see what it is.

136
00:08:16,370 --> 00:08:21,660
So that's one thing we know
already about this problem.

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The other thing we can do is
to think of the regulated delta

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function.

139
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So we think of this as a
potential that has this form.

140
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So here is v of
x, and here is x.

141
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And for this potential--

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if you have a bound state--

143
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how would the wave
function look?

144
00:08:52,260 --> 00:08:54,480
Well, it would have to--

145
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suppose you have
a ground state--

146
00:08:57,820 --> 00:09:01,378
it's an even potential.

147
00:09:01,378 --> 00:09:03,700
The delta function is even, too.

148
00:09:03,700 --> 00:09:04,590
It's in the middle.

149
00:09:04,590 --> 00:09:05,610
It's symmetric.

150
00:09:05,610 --> 00:09:09,610
There's nothing asymmetric
about the delta function.

151
00:09:09,610 --> 00:09:12,890
So if it's an even
potential the ground states

152
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should be even, because
the ground state

153
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is supposed to have no nodes.

154
00:09:18,930 --> 00:09:23,100
And it's supposed to be even
if the potential is even.

155
00:09:23,100 --> 00:09:24,540
So how will it look?

156
00:09:24,540 --> 00:09:28,440
Well, it shouldn't be
decaying in this region.

157
00:09:28,440 --> 00:09:32,610
So, presumably, it decays here.

158
00:09:32,610 --> 00:09:36,530
It decays there-- symmetrically.

159
00:09:36,530 --> 00:09:42,580
And in the middle it curves
in the other direction.

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It is in an allowed region--

161
00:09:44,520 --> 00:09:47,730
and you remember that's
kind of allowed this way.

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So that's probably
the way it looks.

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00:09:51,290 --> 00:09:56,240
Now, if that bound
state exists, somehow,

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as I narrow this and go down--

165
00:10:00,490 --> 00:10:06,170
as it becomes even more narrow,
very narrow now, but very deep.

166
00:10:06,170 --> 00:10:09,510
This region becomes smaller.

167
00:10:09,510 --> 00:10:13,020
And I would pretty much
expect the wave function

168
00:10:13,020 --> 00:10:15,810
to have a discontinuity.

169
00:10:15,810 --> 00:10:18,390
You basically don't
have enough power

170
00:10:18,390 --> 00:10:21,630
to see the curving
that is happening here.

171
00:10:21,630 --> 00:10:24,940
Especially because the
curving is going down.

172
00:10:24,940 --> 00:10:27,120
The distance is going down.

173
00:10:27,120 --> 00:10:33,490
So if this bound state
exists, as you approach

174
00:10:33,490 --> 00:10:36,220
the limit in which this
becomes a delta function

175
00:10:36,220 --> 00:10:42,240
the energy moves a little, but
stays finite at some number.

176
00:10:42,240 --> 00:10:46,550
And the curvature that is
created by the delta function

177
00:10:46,550 --> 00:10:50,150
is not visible, and the thing
looks just discontinuous

178
00:10:50,150 --> 00:10:53,350
in its derivative.

179
00:10:53,350 --> 00:10:55,890
So this is an intuitive
way to understand

180
00:10:55,890 --> 00:10:57,480
that the wave
function we're looking

181
00:10:57,480 --> 00:11:02,330
for is going to be
discontinuous on its derivative.

182
00:11:02,330 --> 00:11:05,050
Let's write the
differential equation,

183
00:11:05,050 --> 00:11:07,550
even though we're still
not going to solve it.

184
00:11:07,550 --> 00:11:09,430
So what is the
differential equation?

185
00:11:09,430 --> 00:11:16,270
Minus h squared over
m, psi double prime,

186
00:11:16,270 --> 00:11:20,420
is equal to E psi.

187
00:11:20,420 --> 00:11:24,170
And, therefore-- and I
write this, and you say,

188
00:11:24,170 --> 00:11:25,730
oh, what are you writing?

189
00:11:25,730 --> 00:11:28,220
I'm writing the
differential equation

190
00:11:28,220 --> 00:11:31,500
when x is different from 0.

191
00:11:35,610 --> 00:11:38,380
No potential when x
is different from 0.

192
00:11:38,380 --> 00:11:41,710
So this applies for
positive x and negative x.

193
00:11:41,710 --> 00:11:44,140
It doesn't apply at x equals 0.

194
00:11:44,140 --> 00:11:46,280
We'll have to deal
with that later.

195
00:11:46,280 --> 00:11:50,050
So then, no potential
for x different from 0.

196
00:11:50,050 --> 00:11:51,960
And this differential
equation becomes

197
00:11:51,960 --> 00:11:59,980
psi double prime equals minus
2m e over h squared psi.

198
00:11:59,980 --> 00:12:03,040
And this is equal
to kappa squared

199
00:12:03,040 --> 00:12:09,880
psi, where kappa squared is
minus 2me over h squared.

200
00:12:09,880 --> 00:12:13,350
And it's positive.

201
00:12:13,350 --> 00:12:16,000
Let's make that positive.

202
00:12:16,000 --> 00:12:19,180
It's positive because
the energy is negative

203
00:12:19,180 --> 00:12:22,450
and we're looking
for bound states.

204
00:12:22,450 --> 00:12:26,690
So we're looking for
bound states only.

205
00:12:26,690 --> 00:12:28,560
Kappa squared is positive.

206
00:12:28,560 --> 00:12:32,470
And this differential
equation is just this.

207
00:12:32,470 --> 00:12:35,690
I'll copy it again here.

208
00:12:35,690 --> 00:12:38,990
Kappa squared psi.

209
00:12:38,990 --> 00:12:42,920
And the solutions of this
equation are-- solutions--

210
00:12:46,130 --> 00:12:52,640
are e to the minus kappa
x and e to the kappa x.

211
00:12:52,640 --> 00:12:59,840
Or, if you wish, cosh
kappa x and sinh kappa x--

212
00:12:59,840 --> 00:13:00,830
whichever you prefer.

213
00:13:13,920 --> 00:13:16,560
This is something we
now have to use in order

214
00:13:16,560 --> 00:13:17,550
to produce a solution.

215
00:13:20,200 --> 00:13:24,900
But now, let's see
if I can figure out

216
00:13:24,900 --> 00:13:26,940
how many bound states there are.

217
00:13:33,240 --> 00:13:37,840
If there is one bound state,
it's going to be even.

218
00:13:37,840 --> 00:13:39,180
It's the ground state.

219
00:13:39,180 --> 00:13:40,975
It has no nodes.

220
00:13:40,975 --> 00:13:45,800
It has to be even, because
the potential is even.

221
00:13:45,800 --> 00:13:50,990
If I have the first excited
state after the ground state,

222
00:13:50,990 --> 00:13:53,330
it will have to be odd.

223
00:13:53,330 --> 00:13:59,320
It would have to vanish at x
equals 0, because it's odd.

224
00:13:59,320 --> 00:14:01,010
There is it's node--

225
00:14:01,010 --> 00:14:02,390
it has to have one node.

226
00:14:04,910 --> 00:14:10,800
For an odd bound state--

227
00:14:14,200 --> 00:14:17,310
or first excited state--

228
00:14:21,250 --> 00:14:26,590
you'd have to have psi
equals 0 at x equals 0.

229
00:14:32,670 --> 00:14:37,920
And the way to do that would
be to have a sinh, because this

230
00:14:37,920 --> 00:14:39,530
doesn't vanish at zero.

231
00:14:39,530 --> 00:14:40,870
This doesn't vanish at zero.

232
00:14:40,870 --> 00:14:43,050
And cosh doesn't vanish at zero.

233
00:14:43,050 --> 00:14:51,210
So you would need psi of
x equals sinh of kappa x.

234
00:14:56,730 --> 00:14:59,230
But that's not good.

235
00:14:59,230 --> 00:15:06,420
sinh of kappa x is
like this and blows up.

236
00:15:06,420 --> 00:15:08,030
Blows down.

237
00:15:08,030 --> 00:15:10,590
It has to go like this.

238
00:15:10,590 --> 00:15:13,230
It is in a forbidden
region, so it has

239
00:15:13,230 --> 00:15:17,230
to be convex towards the axis.

240
00:15:17,230 --> 00:15:18,300
And convex here.

241
00:15:18,300 --> 00:15:19,930
But it blows up.

242
00:15:19,930 --> 00:15:23,370
So there's no such solution.

243
00:15:23,370 --> 00:15:24,240
No such solution.

244
00:15:26,750 --> 00:15:29,630
You cannot have an
odd bound state.

245
00:15:29,630 --> 00:15:33,870
So since the bound states
alternate-- even, odd, even,

246
00:15:33,870 --> 00:15:36,270
odd, even, odd--

247
00:15:36,270 --> 00:15:37,560
you're stuck.

248
00:15:37,560 --> 00:15:41,720
You only will have a ground
state-- if we're lucky--

249
00:15:41,720 --> 00:15:47,970
but no excited state that is
bound, while a finite square

250
00:15:47,970 --> 00:15:48,470
will.

251
00:15:48,470 --> 00:15:51,590
You remember this
quantity z0 that

252
00:15:51,590 --> 00:15:56,420
tells you how many bound
states you can have.

253
00:15:56,420 --> 00:16:00,080
Probably you're anticipating
that in the case

254
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of the delta function
potential, you can only

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have one bound state, if any.

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The first excited
state would not exist.

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So, enough preliminaries.

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Let's just solve that now.