1
00:00:00,237 --> 00:00:00,820
PROFESSOR: OK.

2
00:00:00,820 --> 00:00:04,000
So let's try to solve this.

3
00:00:04,000 --> 00:00:10,810
So my classical approximation is
about solving these equations.

4
00:00:14,190 --> 00:00:15,600
So let's see what we get.

5
00:00:18,290 --> 00:00:21,455
Well, the first equation
is kind of simple.

6
00:00:26,830 --> 00:00:30,190
I think everybody has the
temptation there to just take

7
00:00:30,190 --> 00:00:34,170
the square root, and
that's what we should do,

8
00:00:34,170 --> 00:00:38,970
s0 prime is equal to
plus minus p of x.

9
00:00:44,600 --> 00:00:54,200
And therefore s0 of x is equal
to plus/minus the integral up

10
00:00:54,200 --> 00:00:58,775
to x of p of x prime dx prime.

11
00:01:02,470 --> 00:01:05,630
You see p of x is
pretty much known.

12
00:01:05,630 --> 00:01:07,990
If you know the energy
of your particle,

13
00:01:07,990 --> 00:01:14,680
then it's completely known,
and it depends on e minus v.

14
00:01:14,680 --> 00:01:18,310
So this is a solution
in terms of p of x.

15
00:01:18,310 --> 00:01:22,420
We should think of solving
the differential equation

16
00:01:22,420 --> 00:01:25,390
in terms of p of x.

17
00:01:25,390 --> 00:01:29,450
Now, as a first order
differential equation,

18
00:01:29,450 --> 00:01:31,340
there's a constant
of integration.

19
00:01:31,340 --> 00:01:35,240
And we'll pick it up to
be a number here, x0.

20
00:01:35,240 --> 00:01:38,150
So we start integrating
from some place.

21
00:01:38,150 --> 00:01:40,040
If you integrate
from another place,

22
00:01:40,040 --> 00:01:42,710
you're shifting the
constant of integration.

23
00:01:42,710 --> 00:01:45,230
The main thing is
that the x derivative

24
00:01:45,230 --> 00:01:48,650
here acts as the upper
limit and gives you

25
00:01:48,650 --> 00:01:52,770
the p of x of that equation.

26
00:01:52,770 --> 00:01:54,260
So this is our solution.

27
00:01:54,260 --> 00:01:58,580
Even the plus minus
should not disturb us.

28
00:01:58,580 --> 00:02:01,280
If you have the p
squared, you don't

29
00:02:01,280 --> 00:02:04,440
know if the particle is moving
to the left or to the right.

30
00:02:04,440 --> 00:02:08,720
So that ambiguity is
perfectly reasonable.

31
00:02:08,720 --> 00:02:12,470
Particle can be moving to
the left or to the right.

32
00:02:12,470 --> 00:02:15,630
Now, we look at the
second equation.

33
00:02:15,630 --> 00:02:20,828
So s1 prime is
equal to i over 2.

34
00:02:20,828 --> 00:02:25,910
s0 double prime, if you
have s0 double prime,

35
00:02:25,910 --> 00:02:29,360
you have here s0 prime, so
you take another derivative.

36
00:02:29,360 --> 00:02:37,520
So that's a plus minus p prime
of x divided by s0 prime,

37
00:02:37,520 --> 00:02:40,940
which is plus minus p of x.

38
00:02:40,940 --> 00:02:42,300
That's kind of nice.

39
00:02:42,300 --> 00:02:45,690
The sine is going to cancel.

40
00:02:45,690 --> 00:02:55,500
So we have here i over 2 p
prime of x over p of x or i

41
00:02:55,500 --> 00:03:05,800
over 2 logarithm
of p of x prime.

42
00:03:05,800 --> 00:03:12,360
The derivative of the logarithm
is 1 over the function,

43
00:03:12,360 --> 00:03:17,200
and then by chain rule,
you get the p prime there.

44
00:03:17,200 --> 00:03:21,550
So if s1, the prime
derivative is the derivative

45
00:03:21,550 --> 00:03:25,000
of this thing, so
s1 is going to be

46
00:03:25,000 --> 00:03:30,550
i over 2 log of p of
x plus a constant.

47
00:03:39,450 --> 00:03:41,670
So let's reconstruct
our solution.

48
00:03:45,440 --> 00:03:47,970
That's not hard.

49
00:03:51,180 --> 00:03:53,610
We wrote the answers up there.

50
00:03:53,610 --> 00:04:01,230
So the wave function is e
to the i over h bar times s,

51
00:04:01,230 --> 00:04:04,210
and s is what we had there.

52
00:04:04,210 --> 00:04:11,860
So our wave function is
e to the i over h bar s,

53
00:04:11,860 --> 00:04:18,315
and s was s0 plus
i plus h bar is 1.

54
00:04:23,320 --> 00:04:24,350
There's more.

55
00:04:24,350 --> 00:04:24,980
Is it right?

56
00:04:24,980 --> 00:04:26,840
But we're going to ignore it.

57
00:04:26,840 --> 00:04:28,160
We didn't go that far.

58
00:04:28,160 --> 00:04:31,970
In fact, nobody goes higher
in the WKB approximation.

59
00:04:34,850 --> 00:04:36,535
So what do we have here?

60
00:04:40,700 --> 00:04:42,160
I'll write it.

61
00:04:42,160 --> 00:04:46,310
This term is kind
of interesting.

62
00:04:46,310 --> 00:04:58,860
We have e to the i h bar
of x times e to the i s1.

63
00:04:58,860 --> 00:05:08,190
And s1 was i over 2 log
of p of x plus a constant.

64
00:05:13,280 --> 00:05:21,450
So look at this items
i is minus 1, 2.

65
00:05:21,450 --> 00:05:25,140
So you have 1/2.

66
00:05:25,140 --> 00:05:35,130
This becomes e to the minus
1/2 logarithm of b of x.

67
00:05:35,130 --> 00:05:37,950
And 1/2 the logarithm
of b of x is

68
00:05:37,950 --> 00:05:44,400
e to the minus log of square
root of b of x, and when you go

69
00:05:44,400 --> 00:05:49,410
like that e to minus that
is 1 over the function.

70
00:05:49,410 --> 00:06:01,680
So p of x like that, and then
we have e to the i over h

71
00:06:01,680 --> 00:06:09,920
bar integral from x0 to
x p of x prime dx prime.

72
00:06:09,920 --> 00:06:28,860
This is the classic WKB
approximation, classic result.

73
00:06:28,860 --> 00:06:37,110
So as promised, this is of the
form of a scale factor here,

74
00:06:37,110 --> 00:06:43,050
a rho, the square root
of rho times a phase.

75
00:06:43,050 --> 00:06:48,060
So we did begin with what
looked like a pure phase,

76
00:06:48,060 --> 00:06:51,510
but then we said s of
x is complicit in fact.

77
00:06:51,510 --> 00:06:56,390
s0 was real, but
s1 was imaginary.

78
00:06:56,390 --> 00:07:04,140
With s1 imaginary, the rho of
s1 was to provide the magnitude.

79
00:07:04,140 --> 00:07:09,480
And this is an intuition,
this approximation scheme.

80
00:07:09,480 --> 00:07:13,830
The first thing you have
to get right is the phase.

81
00:07:13,830 --> 00:07:16,920
Once you get the phase
right, the next order,

82
00:07:16,920 --> 00:07:19,740
you get the amplitude
of the wave right.

83
00:07:19,740 --> 00:07:22,250
That comes to second order.

84
00:07:22,250 --> 00:07:27,340
That's a next effect.

85
00:07:27,340 --> 00:07:29,910
So this is our solution.

86
00:07:29,910 --> 00:07:32,640
When I began today,
I reminded you

87
00:07:32,640 --> 00:07:37,350
that we have WKB solutions of
the form square root of rho

88
00:07:37,350 --> 00:07:43,900
e to the is, and we calculated
some things for that.

89
00:07:43,900 --> 00:07:47,580
So because of the
signs, s0, I dropped

90
00:07:47,580 --> 00:07:49,620
the sign, this plus or minus.

91
00:07:49,620 --> 00:07:59,180
Let me write the
general solutions of WKB

92
00:07:59,180 --> 00:08:01,250
slightly more complete.

93
00:08:04,740 --> 00:08:07,420
Let's be more complete.

94
00:08:07,420 --> 00:08:10,980
It's important to see
the whole freedom here.

95
00:08:10,980 --> 00:08:17,240
So if we have e greater
than v, remember

96
00:08:17,240 --> 00:08:25,790
when e is greater than v, the
p of x is a real quantity.

97
00:08:25,790 --> 00:08:31,200
And we wrote, and
we said that p of x

98
00:08:31,200 --> 00:08:34,190
we would write as h bar k of x.

99
00:08:38,710 --> 00:08:43,332
You know, I think I should have
probably, for convenience here,

100
00:08:43,332 --> 00:08:44,290
let's put the constant.

101
00:08:47,430 --> 00:08:50,970
We're not attempting to
normalize these wave functions.

102
00:08:50,970 --> 00:08:52,980
We could not attempt
to do it, because we

103
00:08:52,980 --> 00:08:55,350
don't know what p of x is.

104
00:08:55,350 --> 00:08:58,150
And this function may
have limited validity

105
00:08:58,150 --> 00:08:59,700
as we've spoken.

106
00:08:59,700 --> 00:09:02,250
But I had the constants here.

107
00:09:02,250 --> 00:09:05,180
This constant could have
a real or imaginary part.

108
00:09:05,180 --> 00:09:07,110
It would affect this a.

109
00:09:07,110 --> 00:09:08,360
So let's put it there.

110
00:09:11,150 --> 00:09:12,480
OK.

111
00:09:12,480 --> 00:09:18,370
So if p of x is hk of x, we can
have the following solutions,

112
00:09:18,370 --> 00:09:24,930
psi of x and t equal a, another
constant, square root of p--

113
00:09:24,930 --> 00:09:28,140
let's go simpler--
square root of k.

114
00:09:28,140 --> 00:09:30,456
It's a different a.

115
00:09:30,456 --> 00:09:38,310
And here, e to the i, since
p is h bar, it cancels here.

116
00:09:38,310 --> 00:09:40,920
So we have a simpler
integral as well,

117
00:09:40,920 --> 00:09:46,920
x0 to x, k of x prime, dx prime.

118
00:09:46,920 --> 00:09:49,900
So that's that term.

119
00:09:49,900 --> 00:09:54,670
I just use the
opportunity to replace p

120
00:09:54,670 --> 00:09:57,460
for k, which
simplifies your life,

121
00:09:57,460 --> 00:09:59,100
simplifies all these constants.

122
00:09:59,100 --> 00:10:02,560
So the other solution
is the wave moving

123
00:10:02,560 --> 00:10:04,030
in a different direction.

124
00:10:04,030 --> 00:10:10,870
So k of x, e to the
minus i x0 to x,

125
00:10:10,870 --> 00:10:15,100
k of x prime v of x prime.

126
00:10:15,100 --> 00:10:24,190
So that is your solution when
you have e greater than v.

127
00:10:24,190 --> 00:10:31,990
If we have e less than v,
we still have a solution,

128
00:10:31,990 --> 00:10:39,190
and we said that p
of x, in that case,

129
00:10:39,190 --> 00:10:47,120
would be equal to
i h bar kappa of x.

130
00:10:47,120 --> 00:10:49,280
We use that notation.

131
00:10:49,280 --> 00:10:56,460
If e is less than v, this
is a negative number.

132
00:10:56,460 --> 00:11:00,580
So p of x is i times
some positive number

133
00:11:00,580 --> 00:11:03,200
and square root of
a positive number.

134
00:11:03,200 --> 00:11:06,170
And we called it
kappa last time.

135
00:11:06,170 --> 00:11:11,450
So this is the
letter we usually use

136
00:11:11,450 --> 00:11:13,610
for spatial
dependence in regions

137
00:11:13,610 --> 00:11:16,680
where the wave function
decays exponentially,

138
00:11:16,680 --> 00:11:19,410
which is what it's
going to happen here.

139
00:11:19,410 --> 00:11:27,730
So what is the psi of x and
t is equal to a constant c

140
00:11:27,730 --> 00:11:35,061
over square root of
kappa of x e to the--

141
00:11:35,061 --> 00:11:37,440
the i will disappear,
and there will

142
00:11:37,440 --> 00:11:41,650
be two solutions, one
with plus, one with minus.

143
00:11:41,650 --> 00:11:44,640
That's the reason I don't have
to be very careful in saying

144
00:11:44,640 --> 00:11:46,810
whether this is i or minus i.

145
00:11:46,810 --> 00:11:48,570
There's, anyway, two solutions.

146
00:11:48,570 --> 00:11:51,120
At this stage, we
don't need to worry.

147
00:11:51,120 --> 00:12:00,060
So this is from x0 to x,
kappa of x prime, dx prime,

148
00:12:00,060 --> 00:12:08,590
plus d over square root of
kappa of x, e to the minus x0

149
00:12:08,590 --> 00:12:13,005
to x, kappa of x
prime, dx prime.

150
00:12:16,810 --> 00:12:20,315
So this is the complete
solution of WKB.

151
00:12:23,110 --> 00:12:27,250
If you are in the classically
allowed region top

152
00:12:27,250 --> 00:12:30,910
or in the classically
forbidden region,

153
00:12:30,910 --> 00:12:36,580
it's important to realize that
this function, the second term

154
00:12:36,580 --> 00:12:40,300
is the decaying exponential.

155
00:12:40,300 --> 00:12:45,580
As x increases, the integral
accumulates more and more

156
00:12:45,580 --> 00:12:51,340
value, and the wave function
gets more and more suppressed.

157
00:12:51,340 --> 00:12:56,890
This is a growing
kind of exponential.

158
00:12:56,890 --> 00:12:59,200
In the previous
iteration in your life,

159
00:12:59,200 --> 00:13:04,150
kappa was a constant, if
you had constant potentials.

160
00:13:04,150 --> 00:13:08,440
And this would be e to
the kappa x basically.

161
00:13:08,440 --> 00:13:10,810
But here, you must
think of kappa

162
00:13:10,810 --> 00:13:15,160
being some positive
number, positive function,

163
00:13:15,160 --> 00:13:18,700
as you integrate and x grows,
your cube-- the integral

164
00:13:18,700 --> 00:13:19,940
becomes bigger.

165
00:13:19,940 --> 00:13:22,000
And this is a
growing exponential.

166
00:13:22,000 --> 00:13:25,720
So the sign tells you that,
especially because we've

167
00:13:25,720 --> 00:13:30,790
ordered the limits properly.

168
00:13:30,790 --> 00:13:33,700
So we have a decaying
and growing exponential.

169
00:13:38,300 --> 00:13:40,850
At this moment,
we're pretty much

170
00:13:40,850 --> 00:13:48,290
done with what WKB does for you.

171
00:13:48,290 --> 00:13:51,740
Although, we have a few
things still to say.

172
00:13:55,120 --> 00:13:58,330
So this will be in
terms of comments

173
00:13:58,330 --> 00:14:05,680
about the general validity
of such approximation,

174
00:14:05,680 --> 00:14:10,180
but first even some comments
about the current and charge

175
00:14:10,180 --> 00:14:10,850
density.

176
00:14:10,850 --> 00:14:15,760
So let's consider
this equation I.

177
00:14:15,760 --> 00:14:25,090
Let's just make the comments,
comments for equation I on one.

178
00:14:27,760 --> 00:14:32,260
What is the charge density
or the probability density

179
00:14:32,260 --> 00:14:39,170
in this case, rho
would be psi squared,

180
00:14:39,170 --> 00:14:46,525
and in case one, is equal
to a squared over k of x?

181
00:14:50,540 --> 00:14:53,770
You could, if you wish, this
is a perfectly nice formula,

182
00:14:53,770 --> 00:14:59,730
multiply it by h
bar up and down.

183
00:14:59,730 --> 00:15:01,300
And that's the momentum.

184
00:15:01,300 --> 00:15:08,240
So it's h bar a
squared over p of x.

185
00:15:08,240 --> 00:15:14,146
And you could say
this is h bar over m

186
00:15:14,146 --> 00:15:20,210
a squared over v of x,
a local velocity, p of x

187
00:15:20,210 --> 00:15:23,720
is m over a local velocity.

188
00:15:23,720 --> 00:15:28,180
And this is an intuition
you've had for a long time.

189
00:15:28,180 --> 00:15:32,830
The probability density or
the amplitude of the wave

190
00:15:32,830 --> 00:15:36,190
is going to become
bigger in the regions

191
00:15:36,190 --> 00:15:41,540
where the particle
has smaller velocity.

192
00:15:41,540 --> 00:15:43,510
That's the regions of
the potential where

193
00:15:43,510 --> 00:15:46,630
the particle spends
more time, and it's

194
00:15:46,630 --> 00:15:49,990
an intuition that almost
immediately comes here.

195
00:15:49,990 --> 00:15:52,600
This k is essentially
the momentum.

196
00:15:52,600 --> 00:15:56,750
So that's essentially the
square root of the velocity.

197
00:15:56,750 --> 00:16:00,460
And this coefficient,
therefore, becomes bigger,

198
00:16:00,460 --> 00:16:03,670
as the velocity is smaller.

199
00:16:03,670 --> 00:16:07,240
That's part of the intuition
you've had for a long time,

200
00:16:07,240 --> 00:16:10,810
regarding these quantities.

201
00:16:10,810 --> 00:16:13,240
The other piece
is the computation

202
00:16:13,240 --> 00:16:18,430
of the current from this
equation I. Remember,

203
00:16:18,430 --> 00:16:23,800
the current is h bar over
m times the imaginary part

204
00:16:23,800 --> 00:16:27,160
of psi star gradient psi.

205
00:16:27,160 --> 00:16:28,730
So it's a long computation.

206
00:16:28,730 --> 00:16:33,760
But we did it for the
case we had before.

207
00:16:33,760 --> 00:16:40,030
We said that the current
is rho times gradient of s

208
00:16:40,030 --> 00:16:43,060
over m for the case when
the wave function is

209
00:16:43,060 --> 00:16:47,690
written in rho e to the s form.

210
00:16:47,690 --> 00:16:52,240
So in here, rho is
already determined

211
00:16:52,240 --> 00:16:57,700
is a squared over k of x.

212
00:16:57,700 --> 00:17:04,576
We have the 1 over m,
and the gradient of s--

213
00:17:04,576 --> 00:17:10,880
s is this quantity, e
to the i h bar times s.

214
00:17:10,880 --> 00:17:19,009
So the gradient of s is just
p of x, so h bar k of x.

215
00:17:22,720 --> 00:17:25,014
Now, they cancel.

216
00:17:29,860 --> 00:17:32,520
And it's very fast.

217
00:17:32,520 --> 00:17:33,390
They cancel.

218
00:17:33,390 --> 00:17:38,470
Would have been a major
disaster if they didn't.

219
00:17:38,470 --> 00:17:40,090
This is a number.

220
00:17:40,090 --> 00:17:43,210
It's a squared over m.

221
00:17:50,840 --> 00:17:55,520
The reason it cancels is
that it would have failed

222
00:17:55,520 --> 00:17:57,920
the conservation law otherwise.

223
00:17:57,920 --> 00:18:04,910
The rho dt plus the
divergence of j should be 0.

224
00:18:04,910 --> 00:18:12,920
In our case, rho has
no time dependence.

225
00:18:12,920 --> 00:18:15,620
The wave function that
we are considering,

226
00:18:15,620 --> 00:18:18,520
our time independent
Schrodinger equations,

227
00:18:18,520 --> 00:18:21,680
we're considering
energy eigenstates.

228
00:18:21,680 --> 00:18:26,030
And the current
must be a constant.

229
00:18:26,030 --> 00:18:30,260
In an energy eigenstate, the
current cannot be a spatially

230
00:18:30,260 --> 00:18:33,800
varying constant, because then
the current would accumulate

231
00:18:33,800 --> 00:18:38,180
in some place, and that's
inconsistent with stationary

232
00:18:38,180 --> 00:18:38,990
states.

233
00:18:38,990 --> 00:18:45,560
And in fact, this is 0, and
the versions of j in this case

234
00:18:45,560 --> 00:18:52,370
would be dj dx, and if it would
have had some x dependence,

235
00:18:52,370 --> 00:18:54,340
it would have destroyed
this equation.

236
00:18:54,340 --> 00:19:00,690
So dj dx is also 0, and
that's all consistent.