1
00:00:00,590 --> 00:00:03,560
PROFESSOR: So let's do
a connection formula

2
00:00:03,560 --> 00:00:07,970
and use it to solve a problem.

3
00:00:07,970 --> 00:00:10,430
The derivation of such
connection formulas,

4
00:00:10,430 --> 00:00:12,200
we'll face it the next time.

5
00:00:12,200 --> 00:00:15,710
And we'll go further
applications of the methods.

6
00:00:15,710 --> 00:00:18,215
So, yes.

7
00:00:18,215 --> 00:00:23,740
AUDIENCE: So [INAUDIBLE] that
problem like [INAUDIBLE] where

8
00:00:23,740 --> 00:00:25,820
the wave function vanished.

9
00:00:25,820 --> 00:00:29,180
Is that a problem of the
[? our ?] perturbation

10
00:00:29,180 --> 00:00:31,100
or a classical problem?

11
00:00:31,100 --> 00:00:35,430
Because at that point,
then it means the kinetics

12
00:00:35,430 --> 00:00:37,860
and the potential are
equals to each other.

13
00:00:37,860 --> 00:00:42,810
So like the potential cannot
be bigger than the energy,

14
00:00:42,810 --> 00:00:44,295
then that's a classical thing.

15
00:00:44,295 --> 00:00:45,790
That's not a
perturbation problem.

16
00:00:45,790 --> 00:00:47,200
PROFESSOR: Right.

17
00:00:47,200 --> 00:00:50,040
This is not a problem of
the perturbation theory.

18
00:00:50,040 --> 00:00:56,830
It's just our lack of
knowledge, our ignorance of how

19
00:00:56,830 --> 00:00:59,080
the solution looks near there.

20
00:00:59,080 --> 00:01:03,920
So there's nice WKB
formulas for this solution

21
00:01:03,920 --> 00:01:06,010
away from the turning points.

22
00:01:06,010 --> 00:01:07,480
Those are this.

23
00:01:07,480 --> 00:01:10,520
But the solutions near
the turning points

24
00:01:10,520 --> 00:01:14,560
violate the semiclassical
approximation.

25
00:01:14,560 --> 00:01:17,650
Therefore, you have to find
the solution near the turning

26
00:01:17,650 --> 00:01:20,560
points by any method you have.

27
00:01:20,560 --> 00:01:24,220
That is not going to be
the semiclassical method.

28
00:01:24,220 --> 00:01:27,310
And then you will
find the continuation.

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00:01:27,310 --> 00:01:31,330
Now, there's several
ways people do this.

30
00:01:31,330 --> 00:01:33,670
The most down to
earth method, which

31
00:01:33,670 --> 00:01:36,160
is I think the method we're
going to use next time,

32
00:01:36,160 --> 00:01:42,190
is trying to solve the thing
near there, the solution.

33
00:01:42,190 --> 00:01:44,830
People that are
more sophisticated

34
00:01:44,830 --> 00:01:48,400
use complex variables,
methods in which they

35
00:01:48,400 --> 00:01:50,920
think of the solution
in the complex plane,

36
00:01:50,920 --> 00:01:53,170
the x plane becomes complex.

37
00:01:53,170 --> 00:01:55,410
And they're coming
to the turning point

38
00:01:55,410 --> 00:01:58,540
and they go off
the imaginary axis

39
00:01:58,540 --> 00:02:02,200
to avoid it and come
on the other side.

40
00:02:02,200 --> 00:02:05,050
It's very elegant,
very nice, harder

41
00:02:05,050 --> 00:02:10,270
to make very precise, and a
little difficult to explain.

42
00:02:10,270 --> 00:02:12,550
I don't know if
I'll try to do that.

43
00:02:12,550 --> 00:02:14,350
But it's a nice thing.

44
00:02:14,350 --> 00:02:16,450
It's sort of
avoiding the turning

45
00:02:16,450 --> 00:02:18,490
point by going off the axis.

46
00:02:18,490 --> 00:02:20,500
It sounds crazy.

47
00:02:20,500 --> 00:02:25,870
So here is a connection formula.

48
00:02:25,870 --> 00:02:28,570
Here is x equals a.

49
00:02:28,570 --> 00:02:30,580
Here is v of x.

50
00:02:30,580 --> 00:02:36,520
And here is a solution for
x less than a, little a,

51
00:02:36,520 --> 00:02:41,650
I'll write it like
that, p of x cosine x

52
00:02:41,650 --> 00:02:51,280
to a kappa k of x vx prime
minus pi over 4 minus B

53
00:02:51,280 --> 00:03:02,490
over square root of
p of K of x sine x

54
00:03:02,490 --> 00:03:10,170
to a k of x prime vx
prime minus pi over 4.

55
00:03:10,170 --> 00:03:14,546
So look, this is a general
solution in allowed region.

56
00:03:14,546 --> 00:03:16,920
[INAUDIBLE] no, it doesn't
look like what you wrote here.

57
00:03:16,920 --> 00:03:22,020
But sines and cosines are linear
combinations of these things.

58
00:03:22,020 --> 00:03:26,640
And why do I put this
silly minus pi over 4?

59
00:03:26,640 --> 00:03:31,440
Because that's what
my solutions connect

60
00:03:31,440 --> 00:03:33,880
to solutions on the other side.

61
00:03:33,880 --> 00:03:38,340
Which is to say that the
solution on the other side,

62
00:03:38,340 --> 00:03:41,760
people that work this
connection formulas,

63
00:03:41,760 --> 00:03:49,710
discovered that takes
this form, a to x

64
00:03:49,710 --> 00:03:58,470
kappa of x prime vx prime plus
b over the square root of kappa

65
00:03:58,470 --> 00:04:12,005
of x e exponential a to x
kappa of x prime vx prime.

66
00:04:15,270 --> 00:04:16,940
So here it is.

67
00:04:16,940 --> 00:04:19,459
Those numbers, a
and b, are things

68
00:04:19,459 --> 00:04:21,680
you have to keep track now.

69
00:04:21,680 --> 00:04:26,150
They say if your solution
for x greater [INAUDIBLE]

70
00:04:26,150 --> 00:04:28,880
has this decaying exponential
and this growing exponential

71
00:04:28,880 --> 00:04:34,030
with b an a, your
solution far to the left

72
00:04:34,030 --> 00:04:37,140
will look like this.

73
00:04:37,140 --> 00:04:40,540
It's a pretty strange statement.

74
00:04:40,540 --> 00:04:43,240
Sometimes you don't
want b to be 0.

75
00:04:43,240 --> 00:04:45,100
Sometimes you do.

76
00:04:45,100 --> 00:04:50,200
Let's do an example
with this stuff

77
00:04:50,200 --> 00:04:54,510
so that you can
appreciate what goes on.

78
00:05:05,150 --> 00:05:09,410
So that's your first look
at a connection condition.

79
00:05:09,410 --> 00:05:15,020
There's a little bit of
subtleties on how to use them.

80
00:05:15,020 --> 00:05:19,470
We will discuss those
subtleties next time, as well.

81
00:05:19,470 --> 00:05:24,320
But let's use it in a case
where we can make sense of this

82
00:05:24,320 --> 00:05:26,610
without too much trouble.

83
00:05:26,610 --> 00:05:29,720
So here is the example.

84
00:05:29,720 --> 00:05:33,890
And somebody wants you to
solve the following problem.

85
00:05:33,890 --> 00:05:37,340
You have some slowly
varying potential

86
00:05:37,340 --> 00:05:41,690
that grows, grows indefinitely.

87
00:05:41,690 --> 00:05:42,830
v of x.

88
00:05:42,830 --> 00:05:47,330
And you wish to find
the energy eigenstates.

89
00:05:47,330 --> 00:05:52,240
In particular, you wish
to find the energies.

90
00:05:52,240 --> 00:05:56,420
What are the possible
energies of this potential?

91
00:05:56,420 --> 00:06:01,700
This potential with have
a infinite wall here.

92
00:06:01,700 --> 00:06:04,000
So the potential
is infinite here.

93
00:06:04,000 --> 00:06:07,390
And it grows like
that up on this side.

94
00:06:11,010 --> 00:06:16,180
We're going to try to write
the solution for this.

95
00:06:16,180 --> 00:06:24,860
So we'll do this with
the WKB solution.

96
00:06:24,860 --> 00:06:29,710
And let me do it in
an efficient way.

97
00:06:29,710 --> 00:06:32,160
So what do we have here?

98
00:06:32,160 --> 00:06:36,280
This is the analog of
our point x equals a.

99
00:06:36,280 --> 00:06:38,910
Is that right?

100
00:06:38,910 --> 00:06:44,310
This is the point where the
turning solution goes funny.

101
00:06:44,310 --> 00:06:48,390
So we will call it, for the
same reason we did before,

102
00:06:48,390 --> 00:06:52,695
this point a will
make matters clearer.

103
00:06:58,250 --> 00:07:00,740
We truly don't know
what that point

104
00:07:00,740 --> 00:07:04,200
is because we truly don't know
what are the allowed energies.

105
00:07:04,200 --> 00:07:09,740
But so far we can write E as
a variable that we don't know.

106
00:07:09,740 --> 00:07:14,030
And a can be determined if
you know E because you know

107
00:07:14,030 --> 00:07:17,451
the shape of the potential.

108
00:07:17,451 --> 00:07:17,950
OK.

109
00:07:17,950 --> 00:07:23,860
So let's think of the region,
x much greater than a.

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00:07:23,860 --> 00:07:25,420
We are here.

111
00:07:25,420 --> 00:07:27,100
We're in a forbidden region.

112
00:07:27,100 --> 00:07:30,710
And the potential is
still slowly varying.

113
00:07:30,710 --> 00:07:33,850
Let's assume that
slope is small.

114
00:07:33,850 --> 00:07:37,465
So a solution of this
type would make sense.

115
00:07:40,030 --> 00:07:46,710
So we must have a WKB solution
on this right hand side

116
00:07:46,710 --> 00:07:51,450
to the right of x equals a.

117
00:07:51,450 --> 00:07:55,620
And that's the most general
solution we'll have.

118
00:07:55,620 --> 00:07:58,440
On the other hand,
here there's two types

119
00:07:58,440 --> 00:08:02,670
of solutions, a growing
exponential and a decaying

120
00:08:02,670 --> 00:08:03,550
exponential.

121
00:08:03,550 --> 00:08:08,040
And we must have only
the decaying exponential.

122
00:08:08,040 --> 00:08:11,940
Because this
potential never turns.

123
00:08:11,940 --> 00:08:16,800
If this potential would grow
here and then turn down,

124
00:08:16,800 --> 00:08:20,430
you might have what is
called the tunneling problem.

125
00:08:20,430 --> 00:08:22,920
And this has to be rethought.

126
00:08:22,920 --> 00:08:25,840
But this problem is
still a little easier.

127
00:08:25,840 --> 00:08:29,520
We have just this
potential growing forever.

128
00:08:29,520 --> 00:08:34,530
And on the right, we
must have b equal to 0.

129
00:08:34,530 --> 00:08:42,300
For x greater than a
solution, b is equal to 0.

130
00:08:42,300 --> 00:08:49,290
And this solution must
have a different from 0.

131
00:08:49,290 --> 00:08:54,120
But if the solution
has a different from 0,

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00:08:54,120 --> 00:08:57,840
we now know the
solution on this region,

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00:08:57,840 --> 00:09:03,090
x significantly less than a.

134
00:09:03,090 --> 00:09:05,640
In this region we
know the solution is

135
00:09:05,640 --> 00:09:08,170
given by that formula up there.

136
00:09:08,170 --> 00:09:13,590
So our solution for x much
less than a, the solution

137
00:09:13,590 --> 00:09:17,340
must take the form
whatever a is.

138
00:09:17,340 --> 00:09:19,440
1, 2, 3.

139
00:09:19,440 --> 00:09:22,320
I kind of normalized
these things yet.

140
00:09:22,320 --> 00:09:27,780
So the solution, I'll write it,
psi of x, will be of the form 1

141
00:09:27,780 --> 00:09:36,450
over square root of
k of x cosine of x

142
00:09:36,450 --> 00:09:43,900
to a k of x prime vx
prime minus pi over 4.

143
00:09:46,650 --> 00:09:49,080
All right.

144
00:09:49,080 --> 00:09:53,850
That's what WKB predicts because
of the connection condition.

145
00:09:53,850 --> 00:09:58,080
On the forbidden region,
you know which wave exists.

146
00:09:58,080 --> 00:10:02,220
And therefore, far to the
left of the turning point,

147
00:10:02,220 --> 00:10:04,390
you'd also know
the wave function.

148
00:10:04,390 --> 00:10:08,750
It's the term within a.

149
00:10:08,750 --> 00:10:11,550
So have we solved the problem?

150
00:10:11,550 --> 00:10:13,332
Well, I don't think we have.

151
00:10:13,332 --> 00:10:18,020
We still don't know the energy,
so we must have not solved it.

152
00:10:18,020 --> 00:10:22,880
In fact, it doesn't look
like we've solved it at all.

153
00:10:22,880 --> 00:10:28,580
Because at x equals 0,
these wave functions

154
00:10:28,580 --> 00:10:33,870
should vanish because
there's a hard wall.

155
00:10:33,870 --> 00:10:40,590
And I don't see any reason
why it would vanish.

156
00:10:40,590 --> 00:10:49,350
So let's do a little
work here expanding this

157
00:10:49,350 --> 00:10:52,150
and orienting it
a little better.

158
00:10:52,150 --> 00:10:59,820
So I want to write this
integral from 0, x to a.

159
00:10:59,820 --> 00:11:01,140
You're having an integral.

160
00:11:01,140 --> 00:11:05,220
You have 0, x, and a.

161
00:11:05,220 --> 00:11:08,430
Because we are in the
x less than a region,

162
00:11:08,430 --> 00:11:13,300
this integral is the
one that we have here.

163
00:11:13,300 --> 00:11:15,750
I'll write it as
an integral from 0

164
00:11:15,750 --> 00:11:20,220
to a minus an
integral from 0 to x.

165
00:11:20,220 --> 00:11:26,280
So integral from x to
a is equal to integral

166
00:11:26,280 --> 00:11:33,220
from 0 to a minus an
integral from 0 to x.

167
00:11:33,220 --> 00:11:35,155
This will make things
a little clearer.

168
00:11:53,070 --> 00:12:02,270
In fact, we could do things
still a little easier.

169
00:12:02,270 --> 00:12:04,620
So what do we have here?

170
00:12:04,620 --> 00:12:08,690
We have this exponential,
the wave function.

171
00:12:08,690 --> 00:12:11,270
Not an exponential, a
trigonometric function.

172
00:12:11,270 --> 00:12:21,400
1 over square root of k of x
cosine of minus the integral

173
00:12:21,400 --> 00:12:27,105
from 0 to x of k of x dx prime.

174
00:12:31,760 --> 00:12:34,570
This integral gives
rise to two integrals

175
00:12:34,570 --> 00:12:35,555
and I wrote the first.

176
00:12:40,610 --> 00:12:42,560
Then I come with
the other [? sine ?]

177
00:12:42,560 --> 00:12:48,590
plus the integral from 0
to a of k prime of x dx

178
00:12:48,590 --> 00:12:50,960
prime minus pi over 4.

179
00:12:54,290 --> 00:12:58,555
And let's call this thing delta.

180
00:13:07,210 --> 00:13:10,130
So let's explore this wave
function a little more.

181
00:13:10,130 --> 00:13:13,460
So things have
become kind of nice.

182
00:13:13,460 --> 00:13:16,280
There's no x dependence here.

183
00:13:16,280 --> 00:13:19,280
That's very nice about
this part of the formula.

184
00:13:19,280 --> 00:13:22,070
This is an angle.

185
00:13:22,070 --> 00:13:23,210
No x dependence.

186
00:13:23,210 --> 00:13:29,000
And the x dependence is
just here from this term.

187
00:13:29,000 --> 00:13:32,710
And it's a nice x dependence
because it's an integral from 0

188
00:13:32,710 --> 00:13:33,290
to x.

189
00:13:33,290 --> 00:13:35,030
So it's kind of nice.

190
00:13:35,030 --> 00:13:37,750
The upper limit has the x.

191
00:13:37,750 --> 00:13:39,520
It's all kind of elegant.

192
00:13:39,520 --> 00:13:44,830
So trigonometric
function of this cosine

193
00:13:44,830 --> 00:13:55,986
of a difference of things
is equal to its cosine

194
00:13:55,986 --> 00:14:10,710
of the first term, 0 to x k of
x prime dx prime cosine delta

195
00:14:10,710 --> 00:14:21,311
plus sine of the first term 0
to x k of x prime dx prime sine

196
00:14:21,311 --> 00:14:21,810
delta.

197
00:14:30,420 --> 00:14:34,420
You have this quantity
and this delta.

198
00:14:34,420 --> 00:14:38,440
So I use this trigonometric sum.

199
00:14:38,440 --> 00:14:38,940
OK.

200
00:14:38,940 --> 00:14:43,275
But now you can
see something nice.

201
00:14:48,050 --> 00:14:51,000
What did we say about
the wave function?

202
00:14:51,000 --> 00:14:53,305
It had to vanish at the origin.

203
00:14:57,110 --> 00:15:00,870
And let's look at
these two functions.

204
00:15:00,870 --> 00:15:04,080
Which one vanishes
at the origin?

205
00:15:04,080 --> 00:15:07,530
You have cosine of
the integral from 0

206
00:15:07,530 --> 00:15:12,480
to x when x is equal to
0 at the origin is cosine

207
00:15:12,480 --> 00:15:14,490
of 0, which is 1.

208
00:15:14,490 --> 00:15:18,630
On the other hand, when
x is small, x goes to 0.

209
00:15:18,630 --> 00:15:22,720
You get 0 for the integral
and the sine vanishes.

210
00:15:22,720 --> 00:15:25,640
So this is the right term.

211
00:15:28,770 --> 00:15:32,300
So this wave function
would be correct

212
00:15:32,300 --> 00:15:34,310
if this term would be absent.

213
00:15:34,310 --> 00:15:36,335
This is a term that
you must make absent.

214
00:15:39,260 --> 00:15:44,250
Psi of x without that term
would be a valid wave function.

215
00:15:44,250 --> 00:15:46,780
It is the wave function.

216
00:15:46,780 --> 00:15:49,530
Therefore, the
right wave function

217
00:15:49,530 --> 00:15:54,420
if we demand that cosine
delta be equal to 0.

218
00:15:54,420 --> 00:15:59,250
So now we're imposing a
very non-trivial condition.

219
00:15:59,250 --> 00:16:03,030
This wave over there.

220
00:16:03,030 --> 00:16:06,490
This contribution to the
argument, to the angle

221
00:16:06,490 --> 00:16:12,940
here, this delta must be
such that cosine delta is 0.

222
00:16:12,940 --> 00:16:16,510
In which case, this
term would disappear

223
00:16:16,510 --> 00:16:18,700
and we would have a
good wave function.

224
00:16:18,700 --> 00:16:23,350
Psi of x, if this is true,
is 1 over square root

225
00:16:23,350 --> 00:16:29,890
of k of x times sine times
the sine delta, which

226
00:16:29,890 --> 00:16:30,960
is another number.

227
00:16:30,960 --> 00:16:36,945
Sine of 0 to x k of
x prime dx prime.

228
00:16:40,450 --> 00:16:44,560
So we need cosine
delta equal to 0.

229
00:16:44,560 --> 00:16:47,680
You know, this
argument gives you

230
00:16:47,680 --> 00:16:53,440
this wave function in a nice
way here written what it is.

231
00:16:53,440 --> 00:16:58,840
But we would have been able
to find this even faster.

232
00:16:58,840 --> 00:17:03,730
If you just demand that
Psi at x equals 0 is 0,

233
00:17:03,730 --> 00:17:09,310
you must have that this thing,
the integral from 0 to a

234
00:17:09,310 --> 00:17:14,319
of this quantity minus pi
over 4 must have 0 cosine.

235
00:17:14,319 --> 00:17:17,440
And that's the
condition we did fine.

236
00:17:17,440 --> 00:17:19,599
The advantage of
our rearrangement

237
00:17:19,599 --> 00:17:22,480
is that when that happens,
the whole wave function

238
00:17:22,480 --> 00:17:23,950
looks like that.

239
00:17:23,950 --> 00:17:25,329
And that's kind of nice.

240
00:17:25,329 --> 00:17:28,840
That gives you the picture
of the wave function.

241
00:17:28,840 --> 00:17:32,665
A fairly accurate picture of the
wave function in this region.

242
00:17:35,200 --> 00:17:41,050
Not very near the turning
point, but you got that.

243
00:17:41,050 --> 00:17:43,970
So what is this cosine
delta going to 0?

244
00:17:46,630 --> 00:17:57,160
Well then delta must be
2n plus 1 times pi over 2.

245
00:17:57,160 --> 00:18:02,410
So the places where the
cosine is 0 is pi over 2.

246
00:18:02,410 --> 00:18:04,330
That's for n equals 0.

247
00:18:04,330 --> 00:18:06,185
3 pi over 2.

248
00:18:08,800 --> 00:18:10,420
So for n equal 1.

249
00:18:10,420 --> 00:18:11,560
And it just goes.

250
00:18:11,560 --> 00:18:14,720
The vertical axis
and the unit circle.

251
00:18:14,720 --> 00:18:21,730
So this is for n equals 0,
1, 2, 3, goes on and on.

252
00:18:21,730 --> 00:18:27,100
And this is a very
wonderful condition.

253
00:18:27,100 --> 00:18:35,110
This says that the integral
from 0 to a of k of x prime dx

254
00:18:35,110 --> 00:18:38,500
prime minus pi over 4.

255
00:18:38,500 --> 00:18:42,880
So actually, we have here
you can multiply the 2 here.

256
00:18:42,880 --> 00:18:44,010
You get an n.

257
00:18:44,010 --> 00:18:47,510
And then you have pi over 2.

258
00:18:47,510 --> 00:18:50,830
And the pi over 4 that
comes from that term

259
00:18:50,830 --> 00:18:58,770
becomes n plus 3/4 pi.

260
00:18:58,770 --> 00:19:03,990
It's a Bohr-Sommerfeld
quintessential condition.

261
00:19:03,990 --> 00:19:07,290
Look, I box that equation
because that really

262
00:19:07,290 --> 00:19:08,650
gives you the answer.

263
00:19:08,650 --> 00:19:10,170
Now, why?

264
00:19:10,170 --> 00:19:13,590
Because you know what k is.

265
00:19:13,590 --> 00:19:20,040
k p of x is equal to
h bar k is square root

266
00:19:20,040 --> 00:19:23,745
of 2n E minus v of x.

267
00:19:29,050 --> 00:19:32,770
And if you know E, you know a.

268
00:19:32,770 --> 00:19:36,430
So you have now an integral.

269
00:19:36,430 --> 00:19:39,160
And you take, for
example, n equals 0.

270
00:19:39,160 --> 00:19:42,890
So you want the integral
to be equal to 3/4 pi.

271
00:19:42,890 --> 00:19:45,400
But you will have to
start changing the value

272
00:19:45,400 --> 00:19:49,030
of the energy with your computer
if you cannot do the integral

273
00:19:49,030 --> 00:19:52,480
analytically and find, oops,
for this value of the energy

274
00:19:52,480 --> 00:19:57,670
the integral of what I know,
this is a known function,

275
00:19:57,670 --> 00:19:58,870
gives me this value.

276
00:19:58,870 --> 00:20:03,460
So this is a very practical way
of finding the energy levels.

277
00:20:03,460 --> 00:20:07,120
It does give you the
approximate energy levels.

278
00:20:07,120 --> 00:20:11,560
And it's remarkably
accurate in many cases.

279
00:20:11,560 --> 00:20:16,600
It also has a little
intuition this

280
00:20:16,600 --> 00:20:20,820
is for n equals 0 would
be like the ground state.

281
00:20:20,820 --> 00:20:23,650
n equal 1 first take
on excited states.

282
00:20:23,650 --> 00:20:25,570
And indeed, that makes sense.

283
00:20:25,570 --> 00:20:31,780
Because this integral
from 0 to a of k

284
00:20:31,780 --> 00:20:35,680
is the total phase
of the wave function

285
00:20:35,680 --> 00:20:39,160
as you move from 0 to a.

286
00:20:39,160 --> 00:20:43,360
And that wave is a
number of factor n

287
00:20:43,360 --> 00:20:45,670
times pi plus a little bit.

288
00:20:45,670 --> 00:20:49,210
So it has time for m 0s.

289
00:20:49,210 --> 00:20:55,000
The phase as you move from
0 to a in the wave function,

290
00:20:55,000 --> 00:20:59,830
the sine function will
have n 0s if this condition

291
00:20:59,830 --> 00:21:02,330
is satisfied consistent
with that solution.

292
00:21:02,330 --> 00:21:03,550
So that's it.

293
00:21:03,550 --> 00:21:05,170
We'll continue next time.

294
00:21:05,170 --> 00:21:07,840
Find the solution
of WKB exactly.

295
00:21:07,840 --> 00:21:09,770
All right.