1
00:00:00,000 --> 00:00:05,010
PROFESSOR: Let's do
E less than V not.
2
00:00:05,010 --> 00:00:06,660
So we're back here.
3
00:00:06,660 --> 00:00:14,470
And now of the energy e
here is v not is x equal 0.
4
00:00:14,470 --> 00:00:14,970
X-axis.
5
00:00:17,560 --> 00:00:18,710
And that's the situation.
6
00:00:18,710 --> 00:00:22,985
Now you could solve this again.
7
00:00:26,276 --> 00:00:29,310
And do your
calculations once more.
8
00:00:29,310 --> 00:00:35,530
But we can do this
in an easier way
9
00:00:35,530 --> 00:00:41,320
by trusting the principle
of analytic continuation.
10
00:00:41,320 --> 00:00:44,920
In this case, it's very
clear and very unambiguous.
11
00:00:44,920 --> 00:00:48,250
So the big words,
analytic continuation,
12
00:00:48,250 --> 00:00:51,610
don't carry all the
mathematical depth.
13
00:00:51,610 --> 00:00:54,460
But it's a nice, simple thing.
14
00:00:54,460 --> 00:01:00,420
We first say that the solution
is the same for x less than 0.
15
00:01:00,420 --> 00:01:06,010
So for x less than 0, we
write the same solution.
16
00:01:10,000 --> 00:01:13,450
Because the energy
is greater than 0,
17
00:01:13,450 --> 00:01:16,540
or all what we said
here, the value of k
18
00:01:16,540 --> 00:01:20,590
squared, a into the ikhd
e to the minus i k x.
19
00:01:20,590 --> 00:01:22,450
It's all good.
20
00:01:22,450 --> 00:01:28,570
And k squared is still
2 m e over h squared.
21
00:01:28,570 --> 00:01:34,360
The problem is the region
where x is greater than 0.
22
00:01:34,360 --> 00:01:36,490
Because there you
have an exponential.
23
00:01:36,490 --> 00:01:39,940
But now you must have
a decaying exponential.
24
00:01:39,940 --> 00:01:43,450
But we know how
that should work.
25
00:01:43,450 --> 00:01:48,850
It should really be an e
to the minus some kappa x.
26
00:01:48,850 --> 00:01:51,400
So how could I achieve that?
27
00:01:51,400 --> 00:01:57,120
If I let k bar replace--
28
00:01:57,120 --> 00:02:03,060
everywhere you see k bar,
replace it by i kappa.
29
00:02:03,060 --> 00:02:13,160
Park then one thing
that happens is that--
30
00:02:13,160 --> 00:02:16,130
you learn from here,
from this k bar squared,
31
00:02:16,130 --> 00:02:20,250
would be minus kappa
squared equal to that.
32
00:02:20,250 --> 00:02:29,300
So kappa squared would be 2 m,
v not minus e over h squared.
33
00:02:29,300 --> 00:02:33,860
The sign is just the
opposite from this equation.
34
00:02:33,860 --> 00:02:38,310
That's what that equation
becomes upon that substitution.
35
00:02:38,310 --> 00:02:42,860
Now that substitution would
not make any difference, how
36
00:02:42,860 --> 00:02:47,390
they put a plus i or minus
i, I wouldn't have gotten
37
00:02:47,390 --> 00:02:51,060
my sine change, and this.
38
00:02:51,060 --> 00:02:54,350
But if I look at
the solution there,
39
00:02:54,350 --> 00:02:58,490
the solution psi becomes--
40
00:02:58,490 --> 00:03:06,240
on the region x greater than
0, turns into c, e to the i.
41
00:03:06,240 --> 00:03:12,340
kappa bar is a i kappa x.
42
00:03:12,340 --> 00:03:16,870
Therefore, it's equal to
c, e to the minus kappa
43
00:03:16,870 --> 00:03:18,880
x, which is the right thing.
44
00:03:18,880 --> 00:03:23,500
And that sine that I
chose, of letting k equal i
45
00:03:23,500 --> 00:03:28,480
kappa proves necessary
to get the right thing.
46
00:03:28,480 --> 00:03:32,240
So it's clear that to get the
right thing, you have that.
47
00:03:32,240 --> 00:03:34,170
And now you know
that, of course,
48
00:03:34,170 --> 00:03:38,310
if you would have written the
equation from the beginning,
49
00:03:38,310 --> 00:03:40,540
you would have said,
yes, in this region,
50
00:03:40,540 --> 00:03:41,880
there is a decaying thing.
51
00:03:41,880 --> 00:03:43,670
And looking at the
Schrodinger equation,
52
00:03:43,670 --> 00:03:47,610
you have concluded that
kappa is given by [? that. ?]
53
00:03:47,610 --> 00:03:50,610
But the place where
you now save the time
54
00:03:50,610 --> 00:03:56,670
is that, since I just must do
this change in the equations,
55
00:03:56,670 --> 00:04:00,580
I can do that change in
the solutions as well.
56
00:04:00,580 --> 00:04:04,920
And I don't have to write the
continuity equations again,
57
00:04:04,920 --> 00:04:06,960
nor solve them.
58
00:04:06,960 --> 00:04:10,020
I can take the solutions
and let everywhere
59
00:04:10,020 --> 00:04:16,200
that was a kappa bar replaced by
i, that was a k bar, replace it
60
00:04:16,200 --> 00:04:18,930
by i kappa bar.
61
00:04:18,930 --> 00:04:22,364
So what do we get?
62
00:04:22,364 --> 00:04:27,020
It should go here and
believe those circuits.
63
00:04:30,080 --> 00:04:37,900
OK, so b over a,
that used to be.
64
00:04:37,900 --> 00:04:40,790
Top blackboard there.
65
00:04:40,790 --> 00:04:48,180
Middle, k minus k bar
becomes k plus i--
66
00:04:48,180 --> 00:04:51,260
no, minus I k bar,
minus i kappa.
67
00:04:54,110 --> 00:04:57,960
And k plus i kappa.
68
00:04:57,960 --> 00:04:59,880
so it has changed.
69
00:04:59,880 --> 00:05:03,900
Suddenly this ratio
has become complex.
70
00:05:03,900 --> 00:05:07,330
It's kind of interesting.
71
00:05:07,330 --> 00:05:12,910
Well, let's make it clearer
by factoring a minus i here.
72
00:05:12,910 --> 00:05:16,120
So this becomes kappa.
73
00:05:16,120 --> 00:05:23,200
And you need plus k, so this
must be plus i k over i.
74
00:05:23,200 --> 00:05:28,660
This would be kappa minus i k.
75
00:05:28,660 --> 00:05:36,460
So this is just minus kappa
plus i k over kappa minus i k.
76
00:05:43,555 --> 00:05:46,730
But when you see
that ratio, you're
77
00:05:46,730 --> 00:05:53,370
seeing the ratio of two complex
numbers of equal length.
78
00:05:53,370 --> 00:05:57,900
And therefore, that
ratio is just a phase.
79
00:05:57,900 --> 00:06:00,580
It's not any magnitude.
80
00:06:00,580 --> 00:06:06,240
So this is just a phase,
and it deserves a new name.
81
00:06:06,240 --> 00:06:10,320
There is a phase shift
between the b coefficient
82
00:06:10,320 --> 00:06:12,270
and the a coefficient.
83
00:06:12,270 --> 00:06:16,170
And we'll write it as e minus--
84
00:06:16,170 --> 00:06:18,160
the minus I'll keep.
85
00:06:18,160 --> 00:06:20,930
e to the 2 i delta.
86
00:06:23,680 --> 00:06:25,180
That depends on the energy.
87
00:06:25,180 --> 00:06:29,050
I'll put delta of the energy,
because after all, kappa, k,
88
00:06:29,050 --> 00:06:31,030
everybody depends on the energy.
89
00:06:31,030 --> 00:06:38,380
So let's call it 2 i delta of e.
90
00:06:38,380 --> 00:06:40,235
And what is delta of e?
91
00:06:44,200 --> 00:06:50,500
Well, think of the
number kappa plus i k.
92
00:06:50,500 --> 00:06:52,750
This is the k, i k here.
93
00:06:55,552 --> 00:06:58,800
The angle-- this complex
number has an angle
94
00:06:58,800 --> 00:07:01,700
that, in fact, is delta.
95
00:07:01,700 --> 00:07:05,000
e to i delta is that phase.
96
00:07:05,000 --> 00:07:10,070
And delta is the arc
tangent, k over kappa.
97
00:07:10,070 --> 00:07:16,250
So I'll write it like that.
98
00:07:16,250 --> 00:07:17,150
Delta.
99
00:07:17,150 --> 00:07:21,495
Now you get a delta from the
numerator, a minus delta,
100
00:07:21,495 --> 00:07:24,740
as you can imagine,
from the denominator.
101
00:07:24,740 --> 00:07:29,080
And that's why you get
a total 2 i delta here.
102
00:07:29,080 --> 00:07:36,010
So delta of e is 10
minus 1, k over kappa.
103
00:07:38,820 --> 00:07:45,060
And if you look at what k
and kappa were, k over kappa
104
00:07:45,060 --> 00:07:48,120
is like the ratio of the
square root of the energy
105
00:07:48,120 --> 00:07:51,390
over v not minus the energy.
106
00:07:51,390 --> 00:08:01,780
So delta of e is equal to 10
minus 1 square root of energy
107
00:08:01,780 --> 00:08:05,040
over e v not minus and energy.
108
00:08:15,500 --> 00:08:19,940
Now I got the question
about current conservation.
109
00:08:19,940 --> 00:08:23,450
What happens to current
conservation this time?
110
00:08:23,450 --> 00:08:28,510
Well, you have all
these waves here.
111
00:08:28,510 --> 00:08:33,919
But on the region x greater
than 0, the solution is real.
112
00:08:33,919 --> 00:08:38,720
If the solution is real, there
is no probability current
113
00:08:38,720 --> 00:08:41,289
on the right.
114
00:08:41,289 --> 00:08:47,300
There's really no probability
that you get this thing,
115
00:08:47,300 --> 00:08:49,550
and you get current
flowing there.
116
00:08:49,550 --> 00:08:52,010
And you get this
pulse, or whatever
117
00:08:52,010 --> 00:08:55,400
you send to keep moving and
moving and moving to the right.
118
00:08:55,400 --> 00:08:58,070
Indeed, the solution decays.
119
00:08:58,070 --> 00:09:02,810
And it looks like the one of
the bound state in this region.
120
00:09:02,810 --> 00:09:05,167
So eventually there's
no current here,
121
00:09:05,167 --> 00:09:06,500
because there's no current here.
122
00:09:06,500 --> 00:09:07,250
No current there.
123
00:09:07,250 --> 00:09:07,958
No current there.
124
00:09:07,958 --> 00:09:09,920
Because the solution drops down.
125
00:09:09,920 --> 00:09:12,760
But it's a real solution anyway.
126
00:09:12,760 --> 00:09:14,780
So there's no current there.
127
00:09:14,780 --> 00:09:20,390
So Ja-- Jc is equal to 0.
128
00:09:25,120 --> 00:09:27,090
0.
129
00:09:27,090 --> 00:09:32,980
Solution is real for
x greater than 0,
130
00:09:32,980 --> 00:09:36,990
and any way goes
to 0 at infinity.
131
00:09:36,990 --> 00:09:41,970
So the fact that it's real
is a mathematical nicety
132
00:09:41,970 --> 00:09:45,270
that help us realize
that it must be 0.
133
00:09:45,270 --> 00:09:49,470
But the fact that there's no
current far away essentially
134
00:09:49,470 --> 00:09:51,780
telling you better be 0.
135
00:09:51,780 --> 00:09:57,835
So if the current Jc is
0, Ja must be equal to Jb.
136
00:10:01,570 --> 00:10:06,580
And therefore that means a
squared is equal to b squared.
137
00:10:06,580 --> 00:10:10,885
And happily that's what
happened because b over a
138
00:10:10,885 --> 00:10:14,410
is a complex number
of magnitude one.
139
00:10:14,410 --> 00:10:19,180
So the fact that b and a
differ by just the phase
140
00:10:19,180 --> 00:10:21,790
was required by
current conservation.
141
00:10:24,360 --> 00:10:30,510
A over b is a number
that has norm equal to 1.
142
00:10:30,510 --> 00:10:34,350
So that's a consistent picture.
143
00:10:34,350 --> 00:10:37,300
This phase is very important.
144
00:10:37,300 --> 00:10:41,500
So what happens for
resolution for x less than 0?
145
00:10:41,500 --> 00:10:48,960
Well, psi of x would
be a into the i k x,
146
00:10:48,960 --> 00:10:55,100
plus b all the way
to the left there.
147
00:10:55,100 --> 00:10:59,430
Your solution is a plus
b equal minus i k x.
148
00:10:59,430 --> 00:11:07,980
Of course, we now know what
the b is, so this is minus a
149
00:11:07,980 --> 00:11:10,425
from the ratio over there.
150
00:11:10,425 --> 00:11:14,822
e to the 2 i delta of e.
151
00:11:14,822 --> 00:11:17,625
E to the minus i k x.
152
00:11:21,090 --> 00:11:22,470
For x less than 0.
153
00:11:22,470 --> 00:11:28,090
And for x greater than 0,
psi of x is going to c e
154
00:11:28,090 --> 00:11:30,150
to the minus kappa x.
155
00:11:30,150 --> 00:11:33,380
And I'm not bothering
to write the coefficient
156
00:11:33,380 --> 00:11:40,630
c in Terms of a.
157
00:11:40,630 --> 00:11:45,060
Now this expression
for x less than 0
158
00:11:45,060 --> 00:11:48,040
can be simplified a little.
159
00:11:48,040 --> 00:11:53,200
You can factor an a.
160
00:11:53,200 --> 00:11:55,930
But it's very nice,
and you should
161
00:11:55,930 --> 00:11:58,790
have an eye for those
kind of simplifications.
162
00:11:58,790 --> 00:12:01,840
It's very nice to
factor more than an a
163
00:12:01,840 --> 00:12:07,770
and to factor one
phase like an i delta.
164
00:12:07,770 --> 00:12:11,840
I delta of e,
because in that way,
165
00:12:11,840 --> 00:12:19,040
you get e to the i k x minus
delta of e from the first term,
166
00:12:19,040 --> 00:12:22,500
where the two delta
appearances cancel each other.
167
00:12:22,500 --> 00:12:26,450
Because the first term
didn't have a delta.
168
00:12:26,450 --> 00:12:29,120
But then the second
term will have
169
00:12:29,120 --> 00:12:32,170
the same argument here,
of the exponential
170
00:12:32,170 --> 00:12:33,650
but with a minus sign.
171
00:12:33,650 --> 00:12:37,310
Minus i k x.
172
00:12:37,310 --> 00:12:40,970
And I claim also a
minus delta of e.
173
00:12:40,970 --> 00:12:43,790
And this time minus and
minus gives you plus.
174
00:12:43,790 --> 00:12:47,210
And the other e to the i
delta gives you back the 2.
175
00:12:47,210 --> 00:12:50,300
But now you've created the
trigonometric function,
176
00:12:50,300 --> 00:12:52,760
which is simpler to work with.
177
00:12:52,760 --> 00:13:01,580
So psi effects is equal to
2 i a e to i delta of e,
178
00:13:01,580 --> 00:13:07,710
sine of k x minus delta of e.
179
00:13:15,660 --> 00:13:23,750
And if you wish psi squared,
the probability density
180
00:13:23,750 --> 00:13:30,980
is proportional to 4 a
squared, sine squared
181
00:13:30,980 --> 00:13:36,450
of k x minus delta of e.
182
00:13:36,450 --> 00:13:39,980
So this can be plotted.
183
00:13:39,980 --> 00:13:47,580
Sine squared is like that.
184
00:13:50,190 --> 00:13:53,710
And where is x equal 0.
185
00:13:53,710 --> 00:13:58,545
OK, I'll say x equal
0, say is here.
186
00:14:02,120 --> 00:14:05,280
So this is not
really true anymore.
187
00:14:09,800 --> 00:14:15,710
But this point here is x 0.
188
00:14:15,710 --> 00:14:16,760
It vanishes.
189
00:14:16,760 --> 00:14:24,440
Would be the point at which
k x 0 is equal to delta of e.
190
00:14:24,440 --> 00:14:26,370
And the sine squared vanishes.
191
00:14:26,370 --> 00:14:28,700
So this is not a
solution either.
192
00:14:32,120 --> 00:14:33,290
Solution is like that.
193
00:14:33,290 --> 00:14:37,580
And then this is
for psi squared.
194
00:14:37,580 --> 00:14:41,640
And then it must couple to the
k exponential on this side.
195
00:14:41,640 --> 00:14:45,940
So that the true solution must
somehow be like this and well,
196
00:14:45,940 --> 00:14:46,570
whatever.
197
00:14:46,570 --> 00:14:49,060
I don't know how it looks.
198
00:14:49,060 --> 00:14:56,880
That is the e to the minus 2
kappa x decay and exponential.
199
00:14:56,880 --> 00:14:59,110
It must decay.
200
00:14:59,110 --> 00:15:02,160
There's continuity of the
derivative and continuity
201
00:15:02,160 --> 00:15:03,340
of the wave function.
202
00:15:03,340 --> 00:15:05,505
So that's how this should look.
203
00:15:14,530 --> 00:15:20,110
A couple more things we can
say about this solution that
204
00:15:20,110 --> 00:15:21,720
will play a role later.
205
00:15:28,180 --> 00:15:33,610
I want to get just a little
intuition about this phase,
206
00:15:33,610 --> 00:15:35,700
delta of e, this phase shift.
207
00:15:46,700 --> 00:15:52,100
So we have it there.
208
00:15:52,100 --> 00:15:57,240
Delta of e, I'll write it
here, so you won't have to--
209
00:15:57,240 --> 00:16:03,790
10 minus 1, square root
of e over v not minus e.
210
00:16:03,790 --> 00:16:05,700
So this is interesting.
211
00:16:05,700 --> 00:16:11,700
This phase shift just applies
for energies up to v not.
212
00:16:11,700 --> 00:16:14,550
And that corresponds to the
fact that we've been solving
213
00:16:14,550 --> 00:16:18,640
for energies under the barrier.
214
00:16:18,640 --> 00:16:20,930
And if we solve for
energies under the barrier,
215
00:16:20,930 --> 00:16:23,360
well, the solutions
as we're writing
216
00:16:23,360 --> 00:16:25,460
with these complex
numbers, apply up
217
00:16:25,460 --> 00:16:28,550
to energies equal to the
barrier, but no more.
218
00:16:28,550 --> 00:16:32,400
So we shouldn't plot
beyond this place.
219
00:16:32,400 --> 00:16:36,715
And here is delta of e.
220
00:16:36,715 --> 00:16:38,410
The phase shift.
221
00:16:38,410 --> 00:16:41,880
And when the energy is
0, when your particle
222
00:16:41,880 --> 00:16:45,580
you're sending in, or
the packet eventually
223
00:16:45,580 --> 00:16:48,340
is very low energy here.
224
00:16:48,340 --> 00:16:54,720
Then the phase shift is the
arctangent of 0, which is 0.
225
00:16:54,720 --> 00:17:00,060
As the energy goes
to the value v not,
226
00:17:00,060 --> 00:17:02,350
then the denominator goes to 0.
227
00:17:02,350 --> 00:17:04,770
The ratio goes to infinity.
228
00:17:04,770 --> 00:17:10,010
And the arctangent is pi over 2.
229
00:17:10,010 --> 00:17:15,810
So it's a curve that
goes from here to here.
230
00:17:15,810 --> 00:17:18,910
And it's not quite
like a straight line.
231
00:17:18,910 --> 00:17:22,109
But because of the
square roots, it sort of
232
00:17:22,109 --> 00:17:30,140
begins kind of vertical,
then goes like this.
233
00:17:30,140 --> 00:17:33,150
It's not flat either,
in the middle.
234
00:17:33,150 --> 00:17:35,820
So maybe my curve
doesn't look too good.
235
00:17:40,230 --> 00:17:42,665
Those more vertical here.
236
00:17:42,665 --> 00:17:46,570
Wow, I'm having a
hard time with this.
237
00:17:46,570 --> 00:17:47,530
Something like this.
238
00:17:51,050 --> 00:17:53,620
In fact, it's kind of
interesting to plug
239
00:17:53,620 --> 00:17:59,810
the derivative, d
delta, d energy.
240
00:17:59,810 --> 00:18:03,430
A little calculation will
give you this expression.
241
00:18:03,430 --> 00:18:09,180
You can do this with
mathematica or v not minus e.
242
00:18:09,180 --> 00:18:15,470
And shows, in fact, that here
is v not, and here is v delta,
243
00:18:15,470 --> 00:18:16,240
v e.
244
00:18:16,240 --> 00:18:18,810
We could call it
delta prime of e,
245
00:18:18,810 --> 00:18:24,420
because we wrote the phase shift
as a function of the energy.
246
00:18:24,420 --> 00:18:30,960
So that the delta, d e is
really delta prime of e.
247
00:18:30,960 --> 00:18:34,690
And it sort of infinite--
248
00:18:34,690 --> 00:18:38,300
goes to a minimum and infinite
again, in that direction.
249
00:18:38,300 --> 00:18:40,390
That's how it behaves.