1
00:00:00,500 --> 00:00:03,410
PROFESSOR: E less than v0.

2
00:00:03,410 --> 00:00:11,650
So you have an incident
wave, e to the ikx, incident,

3
00:00:11,650 --> 00:00:17,240
and a reflected wave that you
have, e to the minus ikx--

4
00:00:17,240 --> 00:00:27,840
remember minus the other face--
and e to the 2i delta of E.

5
00:00:27,840 --> 00:00:32,880
So this is the incident wave,
and this is the reflected wave.

6
00:00:36,120 --> 00:00:39,690
They correspond to
your Ae to the ikx

7
00:00:39,690 --> 00:00:43,230
plus Be to the minus ikx.

8
00:00:43,230 --> 00:00:46,410
Remember, when the
energy was less than v0,

9
00:00:46,410 --> 00:00:50,870
the range of B over A was
minus e to the 2i delta.

10
00:00:50,870 --> 00:00:55,680
And since I take A equals
1, you get this thing.

11
00:00:55,680 --> 00:01:04,830
So suppose I construct a psi
incident of x less than 0

12
00:01:04,830 --> 00:01:15,085
and t as the sum 0 to infinity
dk f of k e to the ikx

13
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e to the minus iEt over h bar.

14
00:01:19,394 --> 00:01:20,690
Whew.

15
00:01:20,690 --> 00:01:24,640
So I superimpose the
incident thing here.

16
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Then the reflected one
should be superimposed too

17
00:01:29,964 --> 00:01:33,895
and would be 0 to
infinity, a minus in front

18
00:01:33,895 --> 00:01:43,431
because there's a minus, dk
f of k e to the minus ikx

19
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e to the 2i delta of E e to
the minus iEt over h bar.

20
00:01:52,342 --> 00:01:52,842
Whew.

21
00:01:55,800 --> 00:01:58,890
That's the reflected
wave superimposed.

22
00:01:58,890 --> 00:02:01,120
So now you've
constructed everything.

23
00:02:01,120 --> 00:02:04,230
Here the reflected wave
is more interesting

24
00:02:04,230 --> 00:02:06,630
than the transmitted
wave, because there's

25
00:02:06,630 --> 00:02:08,699
no real big transmitted wave.

26
00:02:08,699 --> 00:02:10,919
It just whistles out.

27
00:02:10,919 --> 00:02:12,960
But the reflective
thing is interesting.

28
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If you're doing the experiment,
you send in a particle,

29
00:02:15,660 --> 00:02:17,677
you want to see
what you get back.

30
00:02:17,677 --> 00:02:19,635
That's going to tell you
what kind of potential

31
00:02:19,635 --> 00:02:23,960
you can expect it encountered.

32
00:02:23,960 --> 00:02:28,520
So let's do the stationary
phase for this one,

33
00:02:28,520 --> 00:02:29,570
for the reflected.

34
00:02:29,570 --> 00:02:31,260
Let's see how it moves.

35
00:02:31,260 --> 00:02:33,670
We know how the incident moves.

36
00:02:33,670 --> 00:02:38,060
The incident moves
with x equals--

37
00:02:38,060 --> 00:02:43,860
we've done it
there-- hk0 over m t.

38
00:02:43,860 --> 00:02:45,450
But how about this one?

39
00:02:45,450 --> 00:02:53,820
Well, this one you would
have to do d dk of minus kx

40
00:02:53,820 --> 00:03:03,990
plus 2 delta of E minus Et over
h bar, all that at k0 equals 0.

41
00:03:03,990 --> 00:03:07,140
And you'll probably
remember that this thing

42
00:03:07,140 --> 00:03:10,590
was in your midterm
and your first test.

43
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You had this wave, and
you had to analyze,

44
00:03:13,590 --> 00:03:16,820
what did stationary phase do?

45
00:03:16,820 --> 00:03:17,570
And it does that.

46
00:03:20,300 --> 00:03:21,715
So what do you get?

47
00:03:26,202 --> 00:03:28,110
Well, when you take
the derivative,

48
00:03:28,110 --> 00:03:31,110
you have to take the derivative
of delta with respect

49
00:03:31,110 --> 00:03:35,610
to energy, that's delta prime,
and then derivative of energy

50
00:03:35,610 --> 00:03:37,980
with respect to t.

51
00:03:37,980 --> 00:03:40,440
Let me save you a little time.

52
00:03:40,440 --> 00:03:50,610
The answer is minus h bar
k0 m t minus 2h bar delta

53
00:03:50,610 --> 00:03:56,160
prime of E. OK.

54
00:03:56,160 --> 00:03:59,670
That's what you get.

55
00:03:59,670 --> 00:04:02,850
That's how this packet moves.

56
00:04:02,850 --> 00:04:05,970
And what does it do really?

57
00:04:05,970 --> 00:04:10,890
Remember, this is defined
for x less than 0.

58
00:04:10,890 --> 00:04:12,420
So this is valid--

59
00:04:12,420 --> 00:04:16,920
forget about this
little term here--

60
00:04:16,920 --> 00:04:19,890
this is valid for t positive.

61
00:04:19,890 --> 00:04:23,200
For t positive, you're going
to get this to satisfy.

62
00:04:23,200 --> 00:04:26,980
So this is a big wave
packet for t positive.

63
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It's the reflected wave.

64
00:04:29,470 --> 00:04:30,750
That's what you would expect.

65
00:04:30,750 --> 00:04:31,880
This is the reflected.

66
00:04:35,030 --> 00:04:38,630
Now, if this factor
was not here,

67
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it is as if, well, the incoming
packet hit the origin at t

68
00:04:45,085 --> 00:04:46,550
equals 0.

69
00:04:46,550 --> 00:04:49,340
And this will be perfect
bouncing in which

70
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the packet gets reflected.

71
00:04:51,380 --> 00:04:55,520
And at t equals 0, it
starts to move to the left.

72
00:04:55,520 --> 00:04:58,970
And as t increases, it's moved
more and more to the left.

73
00:04:58,970 --> 00:05:03,420
You see it there, because
x must be negative.

74
00:05:03,420 --> 00:05:06,080
But if there is
this term, it really

75
00:05:06,080 --> 00:05:08,750
doesn't start to
move to the left

76
00:05:08,750 --> 00:05:12,870
until t is bigger than
that so that x is negative.

77
00:05:12,870 --> 00:05:17,720
So only at t equal to
this amount of time

78
00:05:17,720 --> 00:05:18,950
the packet reflects.

79
00:05:18,950 --> 00:05:23,660
So there's a delay,
and the delay

80
00:05:23,660 --> 00:05:29,240
is 2h bar delta prime of E.

81
00:05:29,240 --> 00:05:32,420
So this is a
technology people use

82
00:05:32,420 --> 00:05:35,690
in scattering theory to figure
out what kind of potential

83
00:05:35,690 --> 00:05:38,500
you have, figure out
how much things get

84
00:05:38,500 --> 00:05:41,840
delayed from the bouncing.

85
00:05:41,840 --> 00:05:44,870
Now, this derivative-- we've
plotted it there-- delta

86
00:05:44,870 --> 00:05:50,090
prime of E. You get a
big delay for low energy,

87
00:05:50,090 --> 00:05:55,000
for energies near v0, and in
the middle it's not so big.