1
00:00:00,500 --> 00:00:02,580
PROFESSOR: Important
thing to do is to just try
2
00:00:02,580 --> 00:00:05,410
to understand one more thing.
3
00:00:05,410 --> 00:00:08,800
The creation and annihilation
operators-- what do they
4
00:00:08,800 --> 00:00:12,030
do to those states?
5
00:00:12,030 --> 00:00:17,490
You see, a creation operator
will I add one more a dagger,
6
00:00:17,490 --> 00:00:24,750
so somehow must change
phi n into phi n plus 1.
7
00:00:24,750 --> 00:00:31,920
A destruction operator with an a
will kill one of these factors,
8
00:00:31,920 --> 00:00:36,950
and therefore it will give
you a state with lower number
9
00:00:36,950 --> 00:00:38,600
of phi n minus 1.
10
00:00:38,600 --> 00:00:45,290
And we would like to know
the precise relations.
11
00:00:45,290 --> 00:00:46,310
So look at this.
12
00:00:46,310 --> 00:00:50,270
Let's do with an A on phi n.
13
00:00:50,270 --> 00:00:56,171
And we know it should be
roughly phi n minus 1.
14
00:00:56,171 --> 00:01:01,080
This is one destruction
operator, but we can do it.
15
00:01:01,080 --> 00:01:08,620
Look-- this is 1 over
square root of n.
16
00:01:08,620 --> 00:01:16,340
A times a dagger to the n phi 0.
17
00:01:20,570 --> 00:01:25,041
A with a dagger to the n phi 0,
we can replace by a commutator
18
00:01:25,041 --> 00:01:25,540
again.
19
00:01:28,050 --> 00:01:35,450
Commutator of a, a
dagger to the n phi 0.
20
00:01:35,450 --> 00:01:38,630
This is 1 over square
root of n factorial,
21
00:01:38,630 --> 00:01:43,700
and here we get a
factor of n times
22
00:01:43,700 --> 00:01:49,300
a dagger to the n minus 1 phi 0.
23
00:01:51,830 --> 00:01:54,700
You know, it's all a matter
of those commutators we
24
00:01:54,700 --> 00:01:55,960
on the left blackboard.
25
00:02:00,870 --> 00:02:10,190
But this state--
by definition, we
26
00:02:10,190 --> 00:02:13,350
have n square root
of n factorial.
27
00:02:13,350 --> 00:02:19,280
That's state, by definition,
is phi n minus 1 times
28
00:02:19,280 --> 00:02:22,160
square root of n
minus 1 factorial.
29
00:02:29,400 --> 00:02:33,360
See, by looking at this
definition and saying,
30
00:02:33,360 --> 00:02:38,230
suppose I have n minus 1, n
minus 1, this is phi n minus 1.
31
00:02:38,230 --> 00:02:42,980
So n minus 1 a
daggers on phi 0 is
32
00:02:42,980 --> 00:02:49,610
n minus 1 factorial square
root multiplied phi n minus 1.
33
00:02:49,610 --> 00:02:52,750
And now we can simplify this--
34
00:02:52,750 --> 00:02:55,660
square root of n
factorial and square root
35
00:02:55,660 --> 00:02:58,450
of n minus 1 factorial
gives you just a factor
36
00:02:58,450 --> 00:03:01,090
of square root of
n that with this n
37
00:03:01,090 --> 00:03:07,206
here, this square root
of n, phi n minus 1.
38
00:03:07,206 --> 00:03:09,220
Sop there we go--
39
00:03:09,220 --> 00:03:11,830
here is the first relation.
40
00:03:11,830 --> 00:03:17,320
A is really a lowering operator.
41
00:03:17,320 --> 00:03:22,600
It gives you an
eigenstate 1 less energy,
42
00:03:22,600 --> 00:03:28,510
but it gives it with a
factor of square root of n,
43
00:03:28,510 --> 00:03:31,180
that if you care
about normalizations,
44
00:03:31,180 --> 00:03:33,490
you better keep it.
45
00:03:33,490 --> 00:03:39,640
That factor is there because
the overall normalization
46
00:03:39,640 --> 00:03:44,050
of this equation was designed
to make the states normalized.
47
00:03:48,020 --> 00:03:54,530
Similarly, we can do
the other operation,
48
00:03:54,530 --> 00:04:00,470
which is what is a
dagger acting on phi n.
49
00:04:00,470 --> 00:04:03,590
This would be 1 over
square root of n factorial,
50
00:04:03,590 --> 00:04:10,045
but this time a dagger to
the n plus 1 on phi n, phi 0.
51
00:04:12,790 --> 00:04:15,730
Because you had already
a dagger to the n,
52
00:04:15,730 --> 00:04:18,220
and you put one more a dagger.
53
00:04:18,220 --> 00:04:23,790
But this thing is equal to what?
54
00:04:23,790 --> 00:04:29,490
This is equal to square root
of n plus 1 factorial times
55
00:04:29,490 --> 00:04:31,270
phi n plus 1.
56
00:04:34,200 --> 00:04:35,640
From the definition--
57
00:04:41,310 --> 00:04:45,210
I hope you're not getting dizzy.
58
00:04:45,210 --> 00:04:50,120
Lots of factors here.
59
00:04:50,120 --> 00:04:57,400
But now you see that the n
part of the factorial cancels,
60
00:04:57,400 --> 00:05:01,640
and you get that
a hat dagger phi
61
00:05:01,640 --> 00:05:07,810
n is equal to square root
of n plus 1 phi n plus 1.
62
00:05:12,530 --> 00:05:16,310
OK let's do an application.
63
00:05:16,310 --> 00:05:21,020
Suppose somebody asks
you to calculate example.
64
00:05:23,830 --> 00:05:26,770
The expectation
value of the operator
65
00:05:26,770 --> 00:05:34,880
x on phi n, the expectation
value of p on phi n.
66
00:05:34,880 --> 00:05:35,960
How much are they?
67
00:05:40,580 --> 00:05:43,330
OK.
68
00:05:43,330 --> 00:05:47,440
This, of course, in conventional
language, at first sight
69
00:05:47,440 --> 00:05:48,790
looks prohibitive.
70
00:05:52,480 --> 00:05:58,060
I would have to get those phi n
[? in ?] some Hermit polynomial
71
00:05:58,060 --> 00:06:03,050
hn, for which I don't know
the closed form expression.
72
00:06:03,050 --> 00:06:06,000
It's a very large
polynomial, jumps 2 by 2,
73
00:06:06,000 --> 00:06:10,900
there is exponentials, I
will have to do an integral.
74
00:06:10,900 --> 00:06:14,090
That's something that
we don't want to do.
75
00:06:16,990 --> 00:06:21,710
So how can we do it
without doing integrals?
76
00:06:21,710 --> 00:06:27,030
Well, this one's-- actually, you
can do without doing anything.
77
00:06:27,030 --> 00:06:29,950
You don't have to do integrals,
you don't have to calculate.
78
00:06:29,950 --> 00:06:34,940
The answers are kind
of obvious, if you
79
00:06:34,940 --> 00:06:38,000
think about it the right way.
80
00:06:38,000 --> 00:06:42,620
That's not the obvious part,
to think about the right way.
81
00:06:42,620 --> 00:06:43,840
But here it is.
82
00:06:43,840 --> 00:06:45,830
Look, what is this integral?
83
00:06:45,830 --> 00:06:53,780
This is the integral of x times
phi and of x, those are real,
84
00:06:53,780 --> 00:06:54,800
quantity squared.
85
00:06:58,978 --> 00:07:05,330
And the phi n's are either even
or odd, but the fight n squared
86
00:07:05,330 --> 00:07:07,950
are even.
87
00:07:07,950 --> 00:07:10,220
And x is odd.
88
00:07:10,220 --> 00:07:15,182
So this integral should be 0,
and we shouldn't even bother.
89
00:07:15,182 --> 00:07:17,500
That's it.
90
00:07:17,500 --> 00:07:18,970
Momentum.
91
00:07:18,970 --> 00:07:22,570
Expectation value
of the momentum.
92
00:07:22,570 --> 00:07:25,480
All these are stationary states.
93
00:07:25,480 --> 00:07:26,920
Cannot have momentum.
94
00:07:26,920 --> 00:07:30,340
If it had momentum, here
is the harmonic oscillator,
95
00:07:30,340 --> 00:07:31,660
here is the wave function.
96
00:07:31,660 --> 00:07:34,980
If it has momentum, half
an hour later it's here.
97
00:07:34,980 --> 00:07:36,310
It's impossible.
98
00:07:36,310 --> 00:07:38,050
This thing cannot have momentum.
99
00:07:38,050 --> 00:07:40,510
This must be 0 as well.
100
00:07:40,510 --> 00:07:42,250
OK.
101
00:07:42,250 --> 00:07:44,590
Now this one is
something you actually
102
00:07:44,590 --> 00:07:46,990
proved in the first test--
103
00:07:46,990 --> 00:07:51,000
the expectation value
of the momentum operator
104
00:07:51,000 --> 00:07:55,340
on a bound state with a
real wave function was 0.
105
00:07:55,340 --> 00:07:58,840
And you did it by integration--
but in fact you proved it
106
00:07:58,840 --> 00:08:02,530
in two ways, in momentum
space, in coordinate space,
107
00:08:02,530 --> 00:08:04,970
is [? back ?] the same thing.
108
00:08:04,970 --> 00:08:06,680
OK.
109
00:08:06,680 --> 00:08:10,410
So these ones were too easy.
110
00:08:10,410 --> 00:08:12,110
So let's try to
see if we can find
111
00:08:12,110 --> 00:08:16,130
something more difficult to do.
112
00:08:16,130 --> 00:08:21,020
Well, actually, before doing
that I will do them anyway
113
00:08:21,020 --> 00:08:22,500
with this notation.
114
00:08:22,500 --> 00:08:24,830
So what would I have here?
115
00:08:24,830 --> 00:08:34,235
I would have phi n x phi n.
116
00:08:34,235 --> 00:08:37,330
And I say, oh, I don't know
how to do things with x.
117
00:08:37,330 --> 00:08:38,720
That's a terrible thing.
118
00:08:38,720 --> 00:08:40,539
I would have to do integrals.
119
00:08:40,539 --> 00:08:43,309
But then you say, no.
120
00:08:43,309 --> 00:08:48,630
X-- I can write in terms
of a and a daggers.
121
00:08:48,630 --> 00:08:51,520
And a and a daggers you
know how to manipulate.
122
00:08:51,520 --> 00:08:55,740
So this is a formula
we wrote last time,
123
00:08:55,740 --> 00:09:03,410
and it's that x is equal to
square root of h over 2m omega,
124
00:09:03,410 --> 00:09:07,200
a plus a dagger.
125
00:09:07,200 --> 00:09:11,860
So x is proportional
to a plus a dagger.
126
00:09:11,860 --> 00:09:18,910
So here is a square root
of h, 2m omega, phi n,
127
00:09:18,910 --> 00:09:23,135
a plus a dagger, on phi n.
128
00:09:30,260 --> 00:09:38,210
Now, this is 0, and why is that?
129
00:09:38,210 --> 00:09:46,270
Because this term is
a acting on phi n.
130
00:09:46,270 --> 00:09:47,860
Well, we have it there--
131
00:09:47,860 --> 00:09:52,930
is square root of
n, phi n minus 1.
132
00:09:52,930 --> 00:09:59,270
And a dagger acting on
phi n is square root of n
133
00:09:59,270 --> 00:10:02,450
plus 1, phi n plus 1.
134
00:10:05,680 --> 00:10:14,790
But the overlap of phi
n minus 1 with phi n
135
00:10:14,790 --> 00:10:19,410
is 0, because all these
states with different energies
136
00:10:19,410 --> 00:10:21,210
are orthogonal.
137
00:10:21,210 --> 00:10:23,160
It's probably a
property I should
138
00:10:23,160 --> 00:10:26,370
have written somewhere here.
139
00:10:26,370 --> 00:10:30,510
Which is-- not only
they're well-normalized,
140
00:10:30,510 --> 00:10:36,735
but phi n phi m is delta nm.
141
00:10:43,920 --> 00:10:48,560
If the numbers are
different, it's zero.
142
00:10:48,560 --> 00:10:52,880
And you see this is
something intuitively clear.
143
00:10:52,880 --> 00:10:57,110
If you wish, I'll just
say here-- these are 0,
144
00:10:57,110 --> 00:11:00,650
and this is 0 because the
numbers are different.
145
00:11:00,650 --> 00:11:05,600
If you have, for
example, a phi 2
146
00:11:05,600 --> 00:11:13,960
and phi 3, or let's do
the phi 3 and a phi 2,
147
00:11:13,960 --> 00:11:21,730
then you have roughly a dagger,
a dagger, a dagger, phi 0,
148
00:11:21,730 --> 00:11:26,430
a dagger, a dagger, phi 0.
149
00:11:26,430 --> 00:11:33,005
And then is equal to phi 0,
three a's, and two a daggers.
150
00:11:36,690 --> 00:11:39,050
Correct?
151
00:11:39,050 --> 00:11:43,840
And now you say, OK,
this a is ready to kill
152
00:11:43,840 --> 00:11:46,880
what is on the right hand side.
153
00:11:46,880 --> 00:11:49,090
On the right side to it.
154
00:11:49,090 --> 00:11:51,190
But it can't because
there are a daggers.
155
00:11:51,190 --> 00:11:55,250
But that a is going to kill
at least one of the a daggers.
156
00:11:55,250 --> 00:11:58,150
So an a kills an a dagger.
157
00:11:58,150 --> 00:12:03,810
The second a will kill the
only a dagger that is left.
158
00:12:03,810 --> 00:12:07,250
And now you have an a
that is ready to go here,
159
00:12:07,250 --> 00:12:10,690
no obstacle whatsoever,
and kills the phi 0,
160
00:12:10,690 --> 00:12:12,310
so this is zero.
161
00:12:12,310 --> 00:12:16,120
So each time there are
some different number
162
00:12:16,120 --> 00:12:21,280
of eight daggers on the left
input and the right input,
163
00:12:21,280 --> 00:12:23,090
you get 0.
164
00:12:23,090 --> 00:12:29,020
If you have more a
daggers on the right,
165
00:12:29,020 --> 00:12:31,915
then move them to
the left, and now you
166
00:12:31,915 --> 00:12:34,990
will have more a's than a
daggers and the same problem
167
00:12:34,990 --> 00:12:36,100
will happen.
168
00:12:36,100 --> 00:12:40,060
The only way to get something
to work is they are the same.
169
00:12:40,060 --> 00:12:44,440
But this of course is
guaranteed by our older theorems
170
00:12:44,440 --> 00:12:46,840
that the--
171
00:12:46,840 --> 00:12:49,030
eigenstates, if
Hermitian operators
172
00:12:49,030 --> 00:12:51,890
with different eigenvalues
are orthogonal.
173
00:12:51,890 --> 00:12:58,420
So this is nice to check
things, but it's not something
174
00:12:58,420 --> 00:13:02,170
that you need to check.
175
00:13:02,170 --> 00:13:02,830
All right.
176
00:13:02,830 --> 00:13:07,210
So now let's say you
want to calculate
177
00:13:07,210 --> 00:13:16,780
the the uncertainty
of x in phi n.
178
00:13:16,780 --> 00:13:21,280
Well, the uncertainty
of x squared
179
00:13:21,280 --> 00:13:23,980
is the expectation
value of x squared
180
00:13:23,980 --> 00:13:29,840
and phi n minus the expectation
value of x on phi n.
181
00:13:29,840 --> 00:13:34,720
On this already we know is 0,
but now we have a computation
182
00:13:34,720 --> 00:13:37,300
worth our tools.
183
00:13:37,300 --> 00:13:45,160
Let's calculate the expectation
value of x squared in phi n.
184
00:13:45,160 --> 00:13:50,470
And if you had to do it
with Hermit polynomials,
185
00:13:50,470 --> 00:13:55,890
it's essentially
a whole days work.
186
00:13:55,890 --> 00:13:57,870
Maybe a little
less if you started
187
00:13:57,870 --> 00:14:02,160
using recursion relations and
invent all kinds of things
188
00:14:02,160 --> 00:14:03,330
to do it.
189
00:14:03,330 --> 00:14:05,220
It's a nightmare,
this calculation.
190
00:14:05,220 --> 00:14:07,440
But look how we do it here.
191
00:14:07,440 --> 00:14:19,110
We say, all right, this is
phi n x hat squared phi n.
192
00:14:19,110 --> 00:14:28,140
But x hat squared would be h
bar over 2m omega, phi n times
193
00:14:28,140 --> 00:14:34,620
a plus a dagger time
a plus a dagger phi n.
194
00:14:41,080 --> 00:14:50,890
Now I must decide what to
do, and one possibility
195
00:14:50,890 --> 00:14:56,540
is to try to be clever and
do all kinds of things.
196
00:14:56,540 --> 00:14:58,420
Now, you could do
several things here,
197
00:14:58,420 --> 00:15:02,540
and none is a lot
better than the other.
198
00:15:02,540 --> 00:15:06,120
And all of them
take little time.
199
00:15:06,120 --> 00:15:10,650
You have to develop
a strategy here,
200
00:15:10,650 --> 00:15:20,710
but this is sufficiently doable
that we can do it directly.
201
00:15:20,710 --> 00:15:22,830
So what does it
mean doing directly?
202
00:15:22,830 --> 00:15:25,200
Just multiply those operators.
203
00:15:25,200 --> 00:15:33,810
So you have phi n times a
a plus a dagger a dagger
204
00:15:33,810 --> 00:15:40,660
plus a a dagger plus a dagger a.
205
00:15:40,660 --> 00:15:44,676
All that on phi n.
206
00:15:44,676 --> 00:15:50,415
I just multiplied, and
now I try to think again.
207
00:15:53,460 --> 00:16:00,150
And I say oh, the first term
is to annihilation operators
208
00:16:00,150 --> 00:16:01,500
acting on phi n.
209
00:16:01,500 --> 00:16:05,440
The first is go give
you phi n minus 1.
210
00:16:05,440 --> 00:16:11,420
Second is going to give me a phi
n minus 2 by the time it acts.
211
00:16:11,420 --> 00:16:19,290
And a phi n minus 2 is
orthogonal to a phi n.
212
00:16:19,290 --> 00:16:23,730
So this term cannot contribute.
213
00:16:23,730 --> 00:16:29,740
You know, this term has
two more a's than this one.
214
00:16:29,740 --> 00:16:32,790
So as we just sort
of illustrated,
215
00:16:32,790 --> 00:16:35,190
but it just doesn't match.
216
00:16:35,190 --> 00:16:39,810
These two terms acting on phi n
would give you a phi n minus 2.
217
00:16:39,810 --> 00:16:42,250
And that's orthogonal.
218
00:16:42,250 --> 00:16:45,790
So this term cannot do anything.
219
00:16:45,790 --> 00:16:49,210
Nor can this,
because both raise.
220
00:16:49,210 --> 00:16:52,570
So this will end up
as phi n plus two,
221
00:16:52,570 --> 00:16:58,410
for example, using that
top property over there.
222
00:16:58,410 --> 00:17:00,900
Over there-- the
box equation there.
223
00:17:00,900 --> 00:17:04,290
If you have two a
daggers acting on phi n,
224
00:17:04,290 --> 00:17:07,050
you will end up
with a phi n plus 2.
225
00:17:07,050 --> 00:17:11,160
So this term also
doesn't contribute.
226
00:17:11,160 --> 00:17:14,609
And that's progress-- the
calculation became half as
227
00:17:14,609 --> 00:17:18,280
difficult.
228
00:17:18,280 --> 00:17:21,099
OK, that-- now we--
229
00:17:21,099 --> 00:17:24,740
maybe it's a little
more interesting.
230
00:17:24,740 --> 00:17:27,200
But again, you should
you should refuse
231
00:17:27,200 --> 00:17:29,030
to do a [? long ?] computation.
232
00:17:29,030 --> 00:17:31,160
Whenever you're looking
at those things,
233
00:17:31,160 --> 00:17:33,380
you have the temptation
to calculate--
234
00:17:33,380 --> 00:17:35,990
refuse that temptation.
235
00:17:35,990 --> 00:17:41,100
Look at things and let it
become clear what's going on.
236
00:17:41,100 --> 00:17:42,680
There are two terms here--
237
00:17:42,680 --> 00:17:45,320
a dagger and a dagger a.
238
00:17:45,320 --> 00:17:47,720
That's not even a
commutator, it's sort of
239
00:17:47,720 --> 00:17:49,380
like an anti-commutator.
240
00:17:49,380 --> 00:17:52,470
That's strange.
241
00:17:52,470 --> 00:17:57,460
But this, a dagger
a, is familiar.
242
00:17:57,460 --> 00:17:59,115
That's n.
243
00:17:59,115 --> 00:18:00,650
The operator n.
244
00:18:00,650 --> 00:18:05,620
And we know the n eigenvalue, so
this is going to be very easy.
245
00:18:05,620 --> 00:18:06,954
This is n hat.
246
00:18:09,800 --> 00:18:12,440
The other one is not
n hat, because it's
247
00:18:12,440 --> 00:18:14,720
in the wrong order.
248
00:18:14,720 --> 00:18:18,005
N hat has a dagger a.
249
00:18:22,350 --> 00:18:31,290
But this operator can be
written as the commutator
250
00:18:31,290 --> 00:18:33,660
plus the thing in
reverse order--
251
00:18:33,660 --> 00:18:36,990
that equation we
had on top-- ab is
252
00:18:36,990 --> 00:18:39,660
equal to ab commutator plus ba.
253
00:18:39,660 --> 00:18:47,310
So this is equal to a a
dagger plus a dagger a.
254
00:18:50,538 --> 00:18:55,640
And this is 1.
255
00:18:55,640 --> 00:18:58,800
Plus another n hat.
256
00:18:58,800 --> 00:19:03,150
So look-- when you have
a and a dagger multiply,
257
00:19:03,150 --> 00:19:07,561
it's either n hat or
it's 1 plus n hat.
258
00:19:10,267 --> 00:19:17,470
And Therefore x squared
expectation value has become
259
00:19:17,470 --> 00:19:24,970
h bar over 2mw phi m, and
this whole parenthesis
260
00:19:24,970 --> 00:19:31,280
is 1 plus 2 n hat phi n.
261
00:19:34,200 --> 00:19:42,210
And this is h bar over 2mw,
phi n, and this is a number.
262
00:19:42,210 --> 00:19:45,960
Because phi in is
an n hat eigenstate.
263
00:19:45,960 --> 00:19:52,300
So it's 1 plus two little
n, phi n phi times 1
264
00:19:52,300 --> 00:19:55,544
plus two [? little ?] n.
265
00:19:55,544 --> 00:19:58,290
And here is our final answer--
266
00:19:58,290 --> 00:20:02,130
expectation value
of x squared is
267
00:20:02,130 --> 00:20:11,980
equal to h bar over and m
omega, n plus 1/2 phi n.
268
00:20:15,730 --> 00:20:18,010
This is a fairly
non-trivial computation.
269
00:20:21,310 --> 00:20:28,110
And that is, of course, because
the expectation value of x
270
00:20:28,110 --> 00:20:33,420
is equal to zero, is the
uncertainty or x squared.
271
00:20:33,420 --> 00:20:41,800
It grows, the state is bigger,
as the quantum number n grows.
272
00:20:41,800 --> 00:20:44,710
By a similar computation,
you can calculate
273
00:20:44,710 --> 00:20:46,320
that you will do
in the homework,
274
00:20:46,320 --> 00:20:50,650
the expectation value
of b squared and phi n,
275
00:20:50,650 --> 00:20:56,980
and then you will see how
much is delta x, delta p,
276
00:20:56,980 --> 00:21:01,120
on the [INAUDIBLE] on phi n.
277
00:21:01,120 --> 00:21:03,240
How much it is.