1
00:00:00,500 --> 00:00:04,170
PROFESSOR: I'll begin
by reviewing quickly
2
00:00:04,170 --> 00:00:05,460
what we did last time.
3
00:00:05,460 --> 00:00:12,350
We considered what are called
finite range potentials,
4
00:00:12,350 --> 00:00:18,330
in which over a distance
R, in the x-axis,
5
00:00:18,330 --> 00:00:20,830
there's a non-zero potential.
6
00:00:20,830 --> 00:00:29,430
So the potential is some v of x
for x between capital R and 0,
7
00:00:29,430 --> 00:00:35,310
is equal to 0 for x
larger than capital R,
8
00:00:35,310 --> 00:00:40,270
and it's infinity
for x negative.
9
00:00:40,270 --> 00:00:46,170
So there's a wall at x equals 0.
10
00:00:46,170 --> 00:00:48,900
And there can be some
potential, but this
11
00:00:48,900 --> 00:00:52,740
is called a finite range
potential, because nothing
12
00:00:52,740 --> 00:00:56,940
happens after distance R.
13
00:00:56,940 --> 00:01:02,580
As usual, we considered
scattering solutions, solutions
14
00:01:02,580 --> 00:01:06,320
that are unnormalizable
with energies,
15
00:01:06,320 --> 00:01:12,300
h squared k squared over 2m,
for a particle with mass m.
16
00:01:12,300 --> 00:01:20,030
And if we had no potential,
we wrote the solution phi
17
00:01:20,030 --> 00:01:23,420
of x, the wave function,
which was sine of kx.
18
00:01:26,150 --> 00:01:28,880
And we also wrote it
as a superposition
19
00:01:28,880 --> 00:01:31,250
of an incoming wave.
20
00:01:31,250 --> 00:01:34,070
Now, an incoming
wave in this set up
21
00:01:34,070 --> 00:01:38,660
is a wave that propagates
from plus infinity towards 0.
22
00:01:38,660 --> 00:01:43,400
And a reflected wave is
a wave that bounces back
23
00:01:43,400 --> 00:01:47,300
and propagates towards
more positive x.
24
00:01:47,300 --> 00:01:51,080
So here we'll
write this as minus
25
00:01:51,080 --> 00:02:00,980
e to the minus ikx over 2i,
plus e to the ikx over 2i.
26
00:02:06,550 --> 00:02:11,440
This is the sine
function rewritten
27
00:02:11,440 --> 00:02:13,990
in terms of exponential
in such a way that
28
00:02:13,990 --> 00:02:16,960
here is the incoming wave.
29
00:02:16,960 --> 00:02:23,480
Remember the time dependence
is minus iet over h bar.
30
00:02:23,480 --> 00:02:26,470
So this wave
combined with a time
31
00:02:26,470 --> 00:02:29,570
is a wave that is moving
towards the origin.
32
00:02:29,570 --> 00:02:32,500
This wave is moving outwards.
33
00:02:32,500 --> 00:02:39,940
Then we said that there would
be, in general, with potential.
34
00:02:39,940 --> 00:02:44,680
With a potential, you would
have a solution psi effects,
35
00:02:44,680 --> 00:02:47,620
which we wrote after some
tinkering in the farm
36
00:02:47,620 --> 00:02:53,905
i delta sine of kx plus delta.
37
00:02:58,040 --> 00:03:02,660
And if you look at the
part of the phase that
38
00:03:02,660 --> 00:03:08,150
has the minus ikx would have a
minus delta and a delta here.
39
00:03:08,150 --> 00:03:09,350
So they would cancel.
40
00:03:09,350 --> 00:03:17,390
So this solution has
the same incoming wave
41
00:03:17,390 --> 00:03:21,960
as the no potential solution.
42
00:03:21,960 --> 00:03:27,420
On the other hand here, you
would have e to the 2i delta,
43
00:03:27,420 --> 00:03:34,750
e to the ikx over
2i, and this solution
44
00:03:34,750 --> 00:03:39,910
is only valid for x
greater than R. You see,
45
00:03:39,910 --> 00:03:43,120
this is just a plane
wave after all.
46
00:03:43,120 --> 00:03:47,320
There's nothing more than a
plane wave and a phase shift.
47
00:03:47,320 --> 00:03:50,230
The phase shift,
of course, doesn't
48
00:03:50,230 --> 00:03:56,920
make the solution any more
complicated or subtle,
49
00:03:56,920 --> 00:04:01,580
but what it does is, by
depending on the energy,
50
00:04:01,580 --> 00:04:07,840
this phase shift delta depends
on the energy, and we're on k.
51
00:04:07,840 --> 00:04:10,570
Then, it produces
interesting phenomena
52
00:04:10,570 --> 00:04:13,250
when you send in wave packets.
53
00:04:13,250 --> 00:04:20,860
So if we write psi,
we usually write
54
00:04:20,860 --> 00:04:27,430
psi is equal to
the phi plus psi s,
55
00:04:27,430 --> 00:04:30,640
where psi s is called
the scattered wave.
56
00:04:30,640 --> 00:04:35,680
You see, the full wave that
you get, for x greater than R,
57
00:04:35,680 --> 00:04:39,910
we would have to solve and work
very carefully to figure out
58
00:04:39,910 --> 00:04:43,370
what is the wave function
in the region 0 to R.
59
00:04:43,370 --> 00:04:46,900
But for x greater
than R is simple,
60
00:04:46,900 --> 00:04:50,710
and for x greater than
R the wave function psi
61
00:04:50,710 --> 00:04:55,180
is the free wave function,
in the case of no potential,
62
00:04:55,180 --> 00:04:58,270
plus the scattered wave.
63
00:04:58,270 --> 00:05:02,140
Quick calculation with this,
things [? give to ?] you
64
00:05:02,140 --> 00:05:13,960
the scatter wave is e to the i
delta sine delta e to the ikx
65
00:05:13,960 --> 00:05:17,320
is an outgoing wave.
66
00:05:17,320 --> 00:05:20,830
And this coefficient is called
the scattering amplitude.
67
00:05:20,830 --> 00:05:24,920
It's the amplitude of
the scattered wave.
68
00:05:24,920 --> 00:05:29,030
This is a wave that is going
out, and this is its amplitude.
69
00:05:29,030 --> 00:05:34,810
So it has something to do with
the strength of the scattering,
70
00:05:34,810 --> 00:05:37,450
because if there
was no scattering,
71
00:05:37,450 --> 00:05:41,680
the wave function would
just behave like the no
72
00:05:41,680 --> 00:05:43,260
potential wave function.
73
00:05:43,260 --> 00:05:46,860
But due to the potential,
there is an extra piece,
74
00:05:46,860 --> 00:05:49,750
and that represents
an outgoing wave
75
00:05:49,750 --> 00:05:55,910
beyond what you get outgoing
with a free no potential wave
76
00:05:55,910 --> 00:05:57,070
function.
77
00:05:57,070 --> 00:06:08,170
So it's the
scattering amplitude,
78
00:06:08,170 --> 00:06:12,830
and therefore sometimes we
are interested in as squared,
79
00:06:12,830 --> 00:06:14,725
which is just sine
squared delta.
80
00:06:18,290 --> 00:06:23,260
Anyway, those are the
things we did last time.
81
00:06:23,260 --> 00:06:30,630
And we can connect to some
ideas that we were talking about
82
00:06:30,630 --> 00:06:34,080
in the past, having to
do with time delays,
83
00:06:34,080 --> 00:06:36,410
by constructing a wave packet.
84
00:06:36,410 --> 00:06:39,030
That's what's usually done.
85
00:06:39,030 --> 00:06:43,740
Consider the process
of time delay, which
86
00:06:43,740 --> 00:06:50,080
is a phenomenon
that we've observed
87
00:06:50,080 --> 00:06:54,550
happens in several
circumstances.
88
00:06:54,550 --> 00:06:58,150
If you have an incident
wave, how do you
89
00:06:58,150 --> 00:07:00,080
construct an incident wave?
90
00:07:00,080 --> 00:07:02,620
Well, it has to be
a superposition of e
91
00:07:02,620 --> 00:07:07,670
to the minus ikx, for sure.
92
00:07:07,670 --> 00:07:14,600
So we'll put the function in
front, we'll integrate over k,
93
00:07:14,600 --> 00:07:16,910
and we'll go from 0 to infinity.
94
00:07:20,520 --> 00:07:24,310
I will actually add the
time dependence as well.
95
00:07:24,310 --> 00:07:27,740
So let's do phi of x and t.
96
00:07:27,740 --> 00:07:36,150
Then, we would have e to the
minus i, e of k, t over h bar,
97
00:07:36,150 --> 00:07:44,500
and this would be valid
for x greater than R.
98
00:07:44,500 --> 00:07:47,405
Again, as a solution of
the Schrodinger equation.
99
00:07:51,760 --> 00:07:53,900
You see, it's a free wave.
100
00:07:53,900 --> 00:07:59,530
There's nothing extra from what
you know from the de Broglie
101
00:07:59,530 --> 00:08:02,990
waves we started
a long time ago.
102
00:08:02,990 --> 00:08:06,550
So if this is your
incident wave,
103
00:08:06,550 --> 00:08:16,870
you have to now realize that
you have this equation over here
104
00:08:16,870 --> 00:08:22,790
telling you about the general
solution of the Schrodinger
105
00:08:22,790 --> 00:08:23,470
equation.
106
00:08:23,470 --> 00:08:26,260
The general solution of
the Schrodinger equation,
107
00:08:26,260 --> 00:08:29,490
in this simple region,
the outside region,
108
00:08:29,490 --> 00:08:34,230
is of this form, and it
depends on this delta
109
00:08:34,230 --> 00:08:36,950
that must be calculated.
110
00:08:36,950 --> 00:08:40,600
This is the incoming wave,
this is the reflected wave,
111
00:08:40,600 --> 00:08:42,070
and this is a solution.
112
00:08:42,070 --> 00:08:50,050
So by superposition, I construct
the reflected wave of x and t.
113
00:08:50,050 --> 00:08:54,010
So for each e to
the minus ikx wave,
114
00:08:54,010 --> 00:08:58,870
I must put down
one e to the ikx,
115
00:08:58,870 --> 00:09:05,470
but I must also put an e to
the 2i delta of the energy,
116
00:09:05,470 --> 00:09:08,540
or delta of k.
117
00:09:08,540 --> 00:09:11,860
And I must put an
extra minus sign,
118
00:09:11,860 --> 00:09:14,500
because these two
have opposite signs,
119
00:09:14,500 --> 00:09:24,400
so I should put a minus
0 to infinity dk f of k.
120
00:09:24,400 --> 00:09:31,880
And we'll have the e to the
minus i, e of k, e over h bar.
121
00:09:31,880 --> 00:09:38,500
And just for reference, f of
k is some real function that
122
00:09:38,500 --> 00:09:40,690
picks at some value k naught.
123
00:09:46,610 --> 00:09:51,880
So you see, just
like what we did
124
00:09:51,880 --> 00:09:54,790
in the case of the step
potential, in which we
125
00:09:54,790 --> 00:09:57,280
had an incident wave,
a reflected wave
126
00:09:57,280 --> 00:10:01,430
packet, a transmitted wave
packet, the wave packets go
127
00:10:01,430 --> 00:10:05,400
along with the basic solution.
128
00:10:05,400 --> 00:10:08,670
The basic solution had
coefficients A, B, and C,
129
00:10:08,670 --> 00:10:12,030
and you knew what B was
in terms of A and C.
130
00:10:12,030 --> 00:10:16,920
Therefore, you constructed the
incoming wave with A e to ikx,
131
00:10:16,920 --> 00:10:21,330
and then the reflected wave
with B e to the minus ikx.
132
00:10:21,330 --> 00:10:28,270
The same thing we're doing
here inspired by this solution,
133
00:10:28,270 --> 00:10:32,697
the psi affects we
superpose many of those,
134
00:10:32,697 --> 00:10:34,030
and that's what we've done here.
135
00:10:37,310 --> 00:10:42,850
Now of course, we can do the
stationary phase calculations
136
00:10:42,850 --> 00:10:45,100
that we've done several
times to figure out
137
00:10:45,100 --> 00:10:49,210
how the peak of the
wave packet moves.
138
00:10:49,210 --> 00:11:02,602
So a stationary phase
at k equal k naught.
139
00:11:02,602 --> 00:11:06,000
As you remember, the
only contribution
140
00:11:06,000 --> 00:11:10,380
can really come when k is near
k naught, and at that point,
141
00:11:10,380 --> 00:11:14,880
you want the phase to be
stationary as a function of k.
142
00:11:14,880 --> 00:11:18,870
I will not do here
the computation again
143
00:11:18,870 --> 00:11:21,540
for psi incident.
144
00:11:21,540 --> 00:11:25,620
You've done this computation
a few times already.
145
00:11:25,620 --> 00:11:30,830
For psi incident, you find
the relation between x and t,
146
00:11:30,830 --> 00:11:32,490
and I will just write it.
147
00:11:32,490 --> 00:11:35,860
It's simple.
148
00:11:35,860 --> 00:11:39,550
You find that x is
equal to minus h
149
00:11:39,550 --> 00:11:50,520
bar k naught over mt, or
minus some v velocity, group
150
00:11:50,520 --> 00:11:52,080
velocity, times t.
151
00:11:54,660 --> 00:11:58,590
That is the condition
for a peak to exist.
152
00:11:58,590 --> 00:12:02,940
The peak satisfies
that equation,
153
00:12:02,940 --> 00:12:08,160
and this makes sense
when t is negative.
154
00:12:08,160 --> 00:12:12,480
This solution for psi incident
only makes sense for x positive
155
00:12:12,480 --> 00:12:19,500
if in fact x greater than R. So
this solution needs x positive.
156
00:12:19,500 --> 00:12:21,660
So it needs t
negative [? indeed. ?]
157
00:12:21,660 --> 00:12:25,900
This is a wave that is
coming from plus infinity,
158
00:12:25,900 --> 00:12:30,450
x equal plus infinity,
at time minus infinity,
159
00:12:30,450 --> 00:12:35,870
and it's going in
with this velocity.
160
00:12:35,870 --> 00:12:40,280
For psi reflected,
the derivative
161
00:12:40,280 --> 00:12:44,890
now has to take the derivative
of delta, with respect to e,
162
00:12:44,890 --> 00:12:50,480
and then the derivative of e
with respect to the energy.
163
00:12:50,480 --> 00:12:52,345
And the answer, in this case--
164
00:12:55,160 --> 00:12:56,480
you've done this before--
165
00:12:56,480 --> 00:13:11,940
it's v group times t minus
2 h bar delta prime of E.
166
00:13:11,940 --> 00:13:20,970
So yes, in the reflected
wave, x grows as t grows
167
00:13:20,970 --> 00:13:22,880
and it's positive.
168
00:13:22,880 --> 00:13:26,810
t must now be
positive, but in fact,
169
00:13:26,810 --> 00:13:30,870
if you would have a
just x equal v group t,
170
00:13:30,870 --> 00:13:32,810
this would correspond
to a particle that
171
00:13:32,810 --> 00:13:37,610
seems to start at the origin
at time equals 0 and goes out.
172
00:13:37,610 --> 00:13:42,560
But this actually there is
an extra term subtracted.
173
00:13:42,560 --> 00:13:46,910
So only for t greater
than this number
174
00:13:46,910 --> 00:13:48,590
the particle begins to appear.
175
00:13:48,590 --> 00:13:53,120
So this is a delay, t
minus some t naught,
176
00:13:53,120 --> 00:13:59,000
the packet gets delayed
by this potential.
177
00:13:59,000 --> 00:14:03,620
Now, this delay can
really get delayed.
178
00:14:03,620 --> 00:14:07,040
Sometimes it might
even accelerated,
179
00:14:07,040 --> 00:14:13,240
but in general, the delay
is given by this quantity
180
00:14:13,240 --> 00:14:16,060
So I'll write it here.
181
00:14:16,060 --> 00:14:24,580
The delay, delta t, is 2
h bar delta prime of E.
182
00:14:24,580 --> 00:14:28,120
And let's write it
in a way that you
183
00:14:28,120 --> 00:14:31,420
can see maybe the units better
and get a little intuition
184
00:14:31,420 --> 00:14:34,420
about what this
computation gives.
185
00:14:34,420 --> 00:14:39,910
For that, let's differentiate
this with respect to k,
186
00:14:39,910 --> 00:14:43,670
and then k with
respect to energy.
187
00:14:43,670 --> 00:14:51,845
So v delta with respect to k,
and dk with respect to energy.
188
00:14:56,120 --> 00:15:08,460
This is 2 over 1 over h
bar dE with respect to k.
189
00:15:08,460 --> 00:15:12,660
I do a little rearrangement
of this derivative
190
00:15:12,660 --> 00:15:16,490
is one function
of one variable k
191
00:15:16,490 --> 00:15:18,150
and neither is a
single relation.
192
00:15:18,150 --> 00:15:21,110
So you can just invert it.
193
00:15:21,110 --> 00:15:23,745
This is more dangerous when
you have partial derivatives.
194
00:15:23,745 --> 00:15:28,950
This is not necessarily true but
for this ordinary derivatives
195
00:15:28,950 --> 00:15:34,200
is true, and then you have
this 2 to the left here.
196
00:15:34,200 --> 00:15:39,960
The h bar went all the way
down, and I have d delta dk.
197
00:15:42,570 --> 00:15:46,680
And here, we recognize
that this is 2,
198
00:15:46,680 --> 00:15:50,280
and this is nothing else
than the group velocity
199
00:15:50,280 --> 00:15:51,740
we were talking before.
200
00:15:51,740 --> 00:15:57,510
The E, the energy, is h
squared k squared over 2m.
201
00:15:57,510 --> 00:16:00,300
You differentiate,
divide by h bar,
202
00:16:00,300 --> 00:16:05,220
and it gives you the group
velocity hk naught over m.
203
00:16:05,220 --> 00:16:07,680
Because these
derivatives all have
204
00:16:07,680 --> 00:16:11,460
to be evaluated at k naught.
205
00:16:11,460 --> 00:16:15,930
So this derivative is really
evaluated at k naught.
206
00:16:15,930 --> 00:16:18,490
This is also
evaluated at k naught.
207
00:16:21,640 --> 00:16:30,380
So this is the group velocity,
d delta, dk, and finally,
208
00:16:30,380 --> 00:16:35,100
let me rewrite it in a
slightly different way.
209
00:16:35,100 --> 00:16:41,320
I multiply by 1 over R. Why?
210
00:16:41,320 --> 00:16:46,360
Because d delta dk, k has
units of 1 over length.
211
00:16:46,360 --> 00:16:50,860
So if I multiply by 1 over
R, this will have no units.
212
00:16:50,860 --> 00:17:04,869
So I claim that one over R d
delta dk is equal to delta t,
213
00:17:04,869 --> 00:17:16,250
and you'll have 2 over vg
and R. So I did a few steps.
214
00:17:16,250 --> 00:17:22,190
I moved the 2 over
vg down to the left,
215
00:17:22,190 --> 00:17:25,880
and I multiplied by
1 over R, and now we
216
00:17:25,880 --> 00:17:29,430
have a nice expression.
217
00:17:29,430 --> 00:17:31,220
This is the delay.
218
00:17:31,220 --> 00:17:34,820
Delta t is the
delay, but you now
219
00:17:34,820 --> 00:17:41,690
have divided it by 2R divided
by the velocity, which
220
00:17:41,690 --> 00:17:46,010
is the time it takes the
particle with the group
221
00:17:46,010 --> 00:17:55,190
velocity to travel back and
forth in the finite range
222
00:17:55,190 --> 00:17:55,980
potential.
223
00:17:55,980 --> 00:17:57,920
So that gives you an idea.
224
00:17:57,920 --> 00:18:00,680
So if you compute
the time delay,
225
00:18:00,680 --> 00:18:03,530
again, it will have
units of microseconds,
226
00:18:03,530 --> 00:18:08,240
and you may not know if
that's little or much.
227
00:18:08,240 --> 00:18:11,210
But here, by computing
this quantity,
228
00:18:11,210 --> 00:18:15,800
not exactly delta prime
of v but this quantity.
229
00:18:15,800 --> 00:18:17,720
You get an [? insight,
?] because this is
230
00:18:17,720 --> 00:18:24,510
the delay divided by
the free transit time.
231
00:18:34,680 --> 00:18:37,330
It's kind of a nice quantity.
232
00:18:37,330 --> 00:18:43,060
You're dividing your delay
and comparing it with the time
233
00:18:43,060 --> 00:18:45,410
that it takes a particle,
with a velocity that
234
00:18:45,410 --> 00:18:49,940
is coming in, to do the
bouncing across the finite range
235
00:18:49,940 --> 00:18:51,790
potential.