1
00:00:01,480 --> 00:00:11,410
PROFESSOR: This also solves the
exchange degeneracy problem.

2
00:00:11,410 --> 00:00:14,710
It kind of seems clear
that it solves it.

3
00:00:14,710 --> 00:00:17,680
But since we made
such a fuss about it,

4
00:00:17,680 --> 00:00:19,630
let's see it in more detail.

5
00:00:23,720 --> 00:00:26,280
So in what sense it solves it?

6
00:00:28,790 --> 00:00:33,980
It solves it in the
sense that you basically

7
00:00:33,980 --> 00:00:39,740
don't get degeneracy once
you have this postulate.

8
00:00:39,740 --> 00:00:41,930
Before this postulate,
for example,

9
00:00:41,930 --> 00:00:44,810
you had the two electron state.

10
00:00:44,810 --> 00:00:48,500
You said, well, it could be
this, or it could be that.

11
00:00:48,500 --> 00:00:50,480
And which should I choose?

12
00:00:50,480 --> 00:00:53,840
If they're equivalent,
we get into mistakes.

13
00:00:53,840 --> 00:00:56,780
So which one is the state
of the two electrons?

14
00:00:56,780 --> 00:00:59,130
Now we seem to have an answer.

15
00:00:59,130 --> 00:01:02,690
It's the antisymmetric
combination.

16
00:01:02,690 --> 00:01:05,810
The last part of the
answer is to say, well,

17
00:01:05,810 --> 00:01:10,270
there should be just one
antisymmetric combination.

18
00:01:10,270 --> 00:01:13,700
And when you have two
particles, it's kind of obvious.

19
00:01:13,700 --> 00:01:16,940
But if you have more,
you can think a little.

20
00:01:16,940 --> 00:01:21,670
And you will conclude that
it's kind of obvious as well.

21
00:01:21,670 --> 00:01:23,460
But let's just check it.

22
00:01:23,460 --> 00:01:26,790
So this would be
comment number 5,

23
00:01:26,790 --> 00:01:35,995
which is solving the
exchange degeneracy problem.

24
00:01:42,390 --> 00:01:46,860
So this is the way we solve
this problem, we discuss it.

25
00:01:46,860 --> 00:01:54,600
Suppose we have a state u
that belongs to V tensor N.

26
00:01:54,600 --> 00:02:00,540
Now I should be a little tongue
tied when I say this thing,

27
00:02:00,540 --> 00:02:03,030
a state u belonging to this.

28
00:02:03,030 --> 00:02:05,430
This is not a physical state.

29
00:02:05,430 --> 00:02:09,509
We already declared that
arbitrary states in V N

30
00:02:09,509 --> 00:02:11,880
are not physical states.

31
00:02:11,880 --> 00:02:13,850
What should they call this?

32
00:02:13,850 --> 00:02:18,360
Maybe a mathematical state or
a state, and then the others,

33
00:02:18,360 --> 00:02:20,670
call them physical states.

34
00:02:20,670 --> 00:02:22,270
You get the point.

35
00:02:22,270 --> 00:02:25,710
Assume you have this
u that belongs here.

36
00:02:25,710 --> 00:02:31,590
Then you can define V u,
a vector space associated

37
00:02:31,590 --> 00:02:36,720
with u, which is
the span, all you

38
00:02:36,720 --> 00:02:45,160
get by applying permutations,
the span of P alpha u for all P

39
00:02:45,160 --> 00:02:52,330
alpha in the permutation
group S N. You see,

40
00:02:52,330 --> 00:02:56,500
that was the idea of
the degeneracy exchange.

41
00:02:56,500 --> 00:02:58,570
You got one state.

42
00:02:58,570 --> 00:03:01,720
And you produce all these
other ones by permutation.

43
00:03:01,720 --> 00:03:04,340
And then you don't
know which one to use.

44
00:03:04,340 --> 00:03:07,090
So we're going back
to that situation.

45
00:03:07,090 --> 00:03:09,610
You have one state.

46
00:03:09,610 --> 00:03:12,880
You produce all you
can by perturbations.

47
00:03:12,880 --> 00:03:16,180
And now you've got
a big vector space.

48
00:03:16,180 --> 00:03:18,040
And which one to use?

49
00:03:18,040 --> 00:03:24,460
Well, you know that you should
use the states that are totally

50
00:03:24,460 --> 00:03:26,050
symmetric or antisymmetric.

51
00:03:26,050 --> 00:03:30,100
But let's first
just get the idea

52
00:03:30,100 --> 00:03:33,430
clear of what we have here.

53
00:03:33,430 --> 00:03:38,260
Notice, that depending
on u, which u you choose,

54
00:03:38,260 --> 00:03:42,510
this vector space may be
1-dimensional, 2-dimensional,,

55
00:03:42,510 --> 00:03:46,180
3-dimensional, very
large dimension.

56
00:03:46,180 --> 00:03:51,350
Could be of arbitrarily high
dimension, up to n factorial.

57
00:03:51,350 --> 00:03:55,210
So let's take an example.

58
00:03:55,210 --> 00:04:10,000
Suppose u is the state a
a b with tensor N equal 3.

59
00:04:10,000 --> 00:04:11,620
And you choose this.

60
00:04:11,620 --> 00:04:15,010
Well, if you apply
permutation operators,

61
00:04:15,010 --> 00:04:22,330
you can also get
the state b a a.

62
00:04:22,330 --> 00:04:25,420
You flip the first
and the third.

63
00:04:25,420 --> 00:04:29,791
Or you can get a b a.

64
00:04:29,791 --> 00:04:30,880
And that's it.

65
00:04:30,880 --> 00:04:36,370
So in this case, you get three
states in V u, for example.

66
00:04:38,920 --> 00:04:39,920
And here is the claim.

67
00:04:48,900 --> 00:04:52,020
The claim is that
there's just one

68
00:04:52,020 --> 00:04:54,600
state that is symmetric in V u.

69
00:04:57,900 --> 00:05:09,650
Up claim, up to
scale V u contains

70
00:05:09,650 --> 00:05:39,440
at most a single ket in Sym N
V and a single ket in Anti N V.

71
00:05:39,440 --> 00:05:42,530
That is the claim that we
have solved the problem.

72
00:05:42,530 --> 00:05:44,980
Now there's one state.

73
00:05:44,980 --> 00:05:49,410
You produce the whole list
of states by permutation.

74
00:05:49,410 --> 00:05:50,450
There's just one.

75
00:05:50,450 --> 00:05:55,745
I said at most, because
sometimes in the case of Anti--

76
00:06:01,150 --> 00:06:05,140
I should say here
actually, up the scale

77
00:06:05,140 --> 00:06:19,760
V u contains a single ket and at
most a single ket in Anti N V.

78
00:06:19,760 --> 00:06:24,990
For example, the one we
did up there, that example.

79
00:06:24,990 --> 00:06:28,460
If you wanted to produce a
totally antisymmetric state,

80
00:06:28,460 --> 00:06:34,070
you would not be able to do
it because of the a here.

81
00:06:34,070 --> 00:06:40,760
You cannot and antisymmetrize
with states a and a.

82
00:06:40,760 --> 00:06:43,640
If you have a state
of three fermions,

83
00:06:43,640 --> 00:06:45,760
they must each be different.

84
00:06:45,760 --> 00:06:47,570
And then you antisymmetrize.

85
00:06:47,570 --> 00:06:51,950
You cannot antisymmetrize
states that are the same.

86
00:06:51,950 --> 00:06:55,610
So that's the statement here.

87
00:06:55,610 --> 00:06:59,600
Let's say a couple
of words about it.

88
00:07:06,360 --> 00:07:15,980
So the first thing we can say
is that, yes, this is clear.

89
00:07:15,980 --> 00:07:27,840
You can get a state
in Sym N V by doing

90
00:07:27,840 --> 00:07:33,060
the symmetrizer acting in u.

91
00:07:33,060 --> 00:07:35,780
So it's clear you
can get one state.

92
00:07:35,780 --> 00:07:40,850
And you can get a
state in Anti N on V

93
00:07:40,850 --> 00:07:45,350
by applying the
antisymmetrizer on u.

94
00:07:45,350 --> 00:07:48,410
And you know.

95
00:07:48,410 --> 00:07:51,440
You can get those
states directly.

96
00:07:51,440 --> 00:07:54,470
Now, it may happen
that Anti of V

97
00:07:54,470 --> 00:07:59,480
vanishes, because in
Anti, when you apply a,

98
00:07:59,480 --> 00:08:03,140
you have some permutations
with plus, some with minus,

99
00:08:03,140 --> 00:08:05,180
and there could
be a cancellation.

100
00:08:05,180 --> 00:08:09,110
In this case, there
would be a cancellation.

101
00:08:09,110 --> 00:08:12,050
In the symmetrize case,
there's no cancellation .

102
00:08:12,050 --> 00:08:16,620
Possible So we'll see that.

103
00:08:16,620 --> 00:08:18,470
So here it is.

104
00:08:18,470 --> 00:08:24,690
The argument for not getting
more is kind of simple.

105
00:08:24,690 --> 00:08:26,900
Let me go through it.

106
00:08:26,900 --> 00:08:29,820
It's almost kind of obvious.

107
00:08:29,820 --> 00:08:34,080
You could say, OK, I got one
state applying the projector.

108
00:08:34,080 --> 00:08:35,929
So that's it, there's one state.

109
00:08:35,929 --> 00:08:38,320
Well, you could say,
well, maybe somebody

110
00:08:38,320 --> 00:08:40,909
hands you another state.

111
00:08:40,909 --> 00:08:45,210
And then you would claim, oh,
it must be the same as this one.

112
00:08:45,210 --> 00:08:47,450
So let's see that
that is the case.

113
00:08:47,450 --> 00:09:05,650
Suppose psi is also in Sym
N V and belongs to V u.

114
00:09:05,650 --> 00:09:11,380
So it would be a claim that
there's another state that

115
00:09:11,380 --> 00:09:12,530
is symmetric.

116
00:09:12,530 --> 00:09:16,810
So since it belongs
to V u, psi would

117
00:09:16,810 --> 00:09:23,710
be the sum of C alphas
times P alphas on u.

118
00:09:23,710 --> 00:09:27,610
What was V u was the
span of this state.

119
00:09:27,610 --> 00:09:31,300
So if it's in that
vector space, it

120
00:09:31,300 --> 00:09:35,660
would have to be in that space.

121
00:09:35,660 --> 00:09:40,150
But since psi is
symmetric, psi is also

122
00:09:40,150 --> 00:09:46,360
equal to S psi by
assumption or S

123
00:09:46,360 --> 00:09:55,670
acting on C alpha P
alpha u, which is psi.

124
00:09:55,670 --> 00:10:01,910
But the S comes in and
multiplies the P alpha.

125
00:10:01,910 --> 00:10:10,310
And you'll recall that S times
P alpha is S. So you get this.

126
00:10:10,310 --> 00:10:12,050
How nice.

127
00:10:12,050 --> 00:10:17,720
If you get that, this is a state
that this independent of alpha.

128
00:10:17,720 --> 00:10:21,230
So you get it out
here, S u times

129
00:10:21,230 --> 00:10:24,830
sum over alpha of C alpha.

130
00:10:24,830 --> 00:10:26,390
So what have you proven?

131
00:10:26,390 --> 00:10:29,750
If you hand out
another state that you

132
00:10:29,750 --> 00:10:33,350
claim is in the
symmetric subspace,

133
00:10:33,350 --> 00:10:37,670
it is proportional to the one
you obtain with the projector.

134
00:10:37,670 --> 00:10:40,055
There's no other
state you can find.

135
00:10:42,790 --> 00:10:45,980
You're done with these things.

136
00:10:45,980 --> 00:10:52,630
So I want to make
yet another comment,

137
00:10:52,630 --> 00:11:00,550
comment 6 and comment
7, and we'll be done.

138
00:11:00,550 --> 00:11:04,810
Comment 6 is a
construction for fermions.

139
00:11:04,810 --> 00:11:06,850
It's a famous construction.

140
00:11:12,590 --> 00:11:20,400
6, for fermions,
suppose you have

141
00:11:20,400 --> 00:11:27,440
a state u this time,
which is phi on 1,

142
00:11:27,440 --> 00:11:33,475
chi on the second, and
omega on the third.

143
00:11:41,160 --> 00:11:48,540
This time, we have to build
an antisymmetric state.

144
00:11:48,540 --> 00:11:55,450
So the state A on u, the
antisymmetric projector,

145
00:11:55,450 --> 00:12:07,800
ti is 1 over 6 sum over alpha
E alpha times the permutation P

146
00:12:07,800 --> 00:12:11,430
alpha acting on this state.

147
00:12:21,680 --> 00:12:27,200
So this will be a sum
of six terms in which

148
00:12:27,200 --> 00:12:31,490
this these states are
scrambled, sometimes with pluses

149
00:12:31,490 --> 00:12:32,990
and sometimes with minuses.

150
00:12:36,590 --> 00:12:41,840
There is a nice construction
in mathematics that does that.

151
00:12:41,840 --> 00:12:46,000
Moreover, you find that if
you have two states that are

152
00:12:46,000 --> 00:12:48,400
the same, if phi
would be equal to chi,

153
00:12:48,400 --> 00:12:50,590
you cannot antisymmetrize here.

154
00:12:50,590 --> 00:12:52,360
You will get 0.

155
00:12:52,360 --> 00:12:57,070
This is kind of like a
statement that the two columns

156
00:12:57,070 --> 00:13:00,910
of a matrix have 0
determinant if they're

157
00:13:00,910 --> 00:13:03,810
proportional to each other.

158
00:13:03,810 --> 00:13:08,090
So the claim is that
this construction

159
00:13:08,090 --> 00:13:12,860
is nothing else than
1 over 3 factorial,

160
00:13:12,860 --> 00:13:17,570
the following determinant in
which the three states are

161
00:13:17,570 --> 00:13:28,750
written like that, phi,
chi, omega, phi, chi, omega,

162
00:13:28,750 --> 00:13:35,360
and phi, chi, omega
into the determinant.

163
00:13:35,360 --> 00:13:49,245
And then you'll put 1, 2,
3, 1, 2, 3, and 1, 2, 3.

164
00:13:52,160 --> 00:13:56,830
The determinant of the
states written in that way--

165
00:14:03,690 --> 00:14:07,050
I would recommend any
one to just write it out.

166
00:14:07,050 --> 00:14:10,620
For three particles, you
have all the six elements.

167
00:14:10,620 --> 00:14:11,560
Write them out.

168
00:14:11,560 --> 00:14:13,080
Do the permutations.

169
00:14:13,080 --> 00:14:16,350
And check that this
product produces that.

170
00:14:16,350 --> 00:14:19,110
You know, for example,
that the determinant

171
00:14:19,110 --> 00:14:24,670
will produce this product along
the diagonal with a plus sign,

172
00:14:24,670 --> 00:14:27,870
this and that with a
plus sign, this and that

173
00:14:27,870 --> 00:14:30,720
with a plus sign, then the
other ones with a minus

174
00:14:30,720 --> 00:14:32,680
sign, but each one 3.

175
00:14:32,680 --> 00:14:34,980
So for example,
this, this, that is

176
00:14:34,980 --> 00:14:39,900
the one in which you have the
identity acting on the states.

177
00:14:39,900 --> 00:14:44,670
And the reason this works
is the general formula.

178
00:14:44,670 --> 00:14:46,740
I will put it in the notes.

179
00:14:46,740 --> 00:14:51,270
But here, I think I don't
want to get into more of it.

180
00:14:56,030 --> 00:15:00,070
It's a general formula
for determinants.

181
00:15:00,070 --> 00:15:02,710
It's a well known formula.

182
00:15:02,710 --> 00:15:05,540
If you have a
permutation P alpha,

183
00:15:05,540 --> 00:15:08,770
remember permutation
P alpha, alpha

184
00:15:08,770 --> 00:15:15,130
was supposed to be a list, alpha
of 1, alpha of 2, up to alpha

185
00:15:15,130 --> 00:15:24,170
of n, which is a permutation
of the list 1, 1, up to n.

186
00:15:24,170 --> 00:15:30,710
So if you have a matrix, the
determinant of a matrix B,

187
00:15:30,710 --> 00:15:36,530
of any matrix, can be written
as the sum over permutations

188
00:15:36,530 --> 00:15:40,850
of the sign of the
permutation times,

189
00:15:40,850 --> 00:15:43,310
essentially, the
permutation acting

190
00:15:43,310 --> 00:15:49,280
on the elements of the
matrix, that is B of alpha 1

191
00:15:49,280 --> 00:15:59,230
1, B of alpha 2 2,
up to B of alpha n n.

192
00:16:06,010 --> 00:16:09,520
This is like saying
in the matrix,

193
00:16:09,520 --> 00:16:12,370
you will select
the alpha 1 element

194
00:16:12,370 --> 00:16:16,360
from the first column,
the alpha 2 element

195
00:16:16,360 --> 00:16:19,600
from the second column, all
of those, and multiply them.

196
00:16:19,600 --> 00:16:21,340
And that's the determinant.

197
00:16:21,340 --> 00:16:29,430
And that formula is the
reason this is true,

198
00:16:29,430 --> 00:16:33,960
because this sum of
epsilons times Ps

199
00:16:33,960 --> 00:16:38,370
really is the formula that
calculates the determinant.

200
00:16:38,370 --> 00:16:40,590
You almost see the P here.

201
00:16:40,590 --> 00:16:46,350
The P is the thing that took
the element 1, 1, to 2, 2

202
00:16:46,350 --> 00:16:51,900
and and n, n and replaced
the 1 by alpha of 1, P of 2

203
00:16:51,900 --> 00:16:54,250
and replaced them that way.

204
00:16:54,250 --> 00:16:56,310
So I almost made the connection.

205
00:16:56,310 --> 00:16:58,380
But we probably
need a few minutes

206
00:16:58,380 --> 00:17:03,560
to make sure you
understand why it happened.

207
00:17:03,560 --> 00:17:07,504
Last comment, seventh,
is occupation numbers.

208
00:17:13,980 --> 00:17:19,740
If we're trying to describe
states in V tensor n,

209
00:17:19,740 --> 00:17:24,720
and suppose you have a
state with n particles,

210
00:17:24,720 --> 00:17:27,329
and that is V tensor n.

211
00:17:27,329 --> 00:17:34,830
But suppose that this vector
space V has a basis of basis

212
00:17:34,830 --> 00:17:41,250
vectors u 1, u 2,
up to u infinity.

213
00:17:41,250 --> 00:17:42,360
It just goes on forever.

214
00:17:47,600 --> 00:17:52,620
Then if you have
that, then there

215
00:17:52,620 --> 00:17:56,205
is a way to think of all the
states that you can build.

216
00:18:00,020 --> 00:18:03,220
And the way to think
of them is follows.

217
00:18:08,240 --> 00:18:13,460
List the vectors here,
u 2 up to all those.

218
00:18:13,460 --> 00:18:20,360
And you imagine that little n
1 particles are in this state.

219
00:18:20,360 --> 00:18:24,200
Little n 2 particles
are in this state.

220
00:18:24,200 --> 00:18:26,120
And it goes on like that.

221
00:18:26,120 --> 00:18:32,420
So you write your state with
occupation numbers, n 1, n 2,

222
00:18:32,420 --> 00:18:34,010
n 3.

223
00:18:34,010 --> 00:18:35,840
And what is that state?

224
00:18:35,840 --> 00:18:38,630
Well, there's n 1 of this one.

225
00:18:38,630 --> 00:18:43,580
So u 1 for the first
particle up to u 1

226
00:18:43,580 --> 00:18:47,060
for the and n 1 particle.

227
00:18:47,060 --> 00:18:52,700
And then you have u 2 for the
next particle up to a u 2.

228
00:18:52,700 --> 00:18:55,520
And you have n 2 of those.

229
00:18:55,520 --> 00:18:57,700
And you have all the others.

230
00:18:57,700 --> 00:19:04,190
And then you must apply into
it the symmetrization operator.

231
00:19:04,190 --> 00:19:07,070
But this is the list of states.

232
00:19:07,070 --> 00:19:10,040
This describes what you have.

233
00:19:10,040 --> 00:19:12,320
Symmetrizing is a lot of work.

234
00:19:12,320 --> 00:19:15,950
But listing the
occupation numbers

235
00:19:15,950 --> 00:19:20,150
is really all there is
to it, because the rest

236
00:19:20,150 --> 00:19:24,980
is a complicated algebraic
manipulation of symmetrization.

237
00:19:24,980 --> 00:19:27,170
But now you know what
you must symmetrize.

238
00:19:27,170 --> 00:19:32,390
You have n 1 times the state
1, n 2 times the state 2.

239
00:19:32,390 --> 00:19:34,550
When you have fermions,
this is the same,

240
00:19:34,550 --> 00:19:39,950
except that the n 1s, n
2s, n 3s can be 0 or 1.

241
00:19:39,950 --> 00:19:43,820
And finally here, if it's
a state of n particles,

242
00:19:43,820 --> 00:19:49,460
then and n 1 plus n 2 plus
all of those must add up to n,

243
00:19:49,460 --> 00:19:52,010
because you have n particles.

244
00:19:52,010 --> 00:19:55,670
So that brings us
to the end of 8.06.

245
00:19:55,670 --> 00:20:00,050
Occupation numbers is where
quantum field theory begins.

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You start constructing operators
that add particles and subtract

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00:20:05,330 --> 00:20:06,060
particles.

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00:20:06,060 --> 00:20:11,360
So it's a fitting place to end
your undergrad with quantum

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00:20:11,360 --> 00:20:14,420
mechanics education.

250
00:20:14,420 --> 00:20:15,940
I wish you all the best.

251
00:20:15,940 --> 00:20:19,900
I hope the course lived
up to your expectations.

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00:20:19,900 --> 00:20:23,030
It's an awesome
responsibility to teach

253
00:20:23,030 --> 00:20:25,010
such an interesting course.

254
00:20:25,010 --> 00:20:27,620
And we try to do it well.

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00:20:27,620 --> 00:20:33,200
And we wish that all of
you have enjoyed this

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00:20:33,200 --> 00:20:34,510
and have been motivated.

257
00:20:34,510 --> 00:20:38,420
And many of you will understand
this subject, one day,

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00:20:38,420 --> 00:20:43,190
better than your teachers and
will help move it forward.

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00:20:43,190 --> 00:20:45,050
So take care.

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00:20:45,050 --> 00:20:45,710
Good luck.

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00:20:45,710 --> 00:20:46,730
And see you soon.

262
00:20:46,730 --> 00:20:47,230
Bye.

263
00:20:47,230 --> 00:20:49,980
[APPLAUSE]