1
00:00:01,090 --> 00:00:03,940
PROFESSOR: We have already
begun with the degenerate

2
00:00:03,940 --> 00:00:06,430
perturbation theory.

3
00:00:06,430 --> 00:00:09,730
And we obtained
the first result--

4
00:00:09,730 --> 00:00:12,010
perhaps the most
important result--

5
00:00:12,010 --> 00:00:12,640
already.

6
00:00:12,640 --> 00:00:15,700
It came rather quick.

7
00:00:15,700 --> 00:00:18,340
And let me review what we had.

8
00:00:18,340 --> 00:00:21,820
So we were talking
about the Hamiltonian

9
00:00:21,820 --> 00:00:26,750
H0 that had degeneracies.

10
00:00:26,750 --> 00:00:31,640
And we were now interested to
see what the perturbation does

11
00:00:31,640 --> 00:00:33,380
to those degenerate states.

12
00:00:33,380 --> 00:00:35,330
What does it do to its energies?

13
00:00:35,330 --> 00:00:38,030
What does it do to
the states themselves?

14
00:00:38,030 --> 00:00:46,310
So we called that degenerate
subspace the space V capital N.

15
00:00:46,310 --> 00:00:51,230
And the rest of the state
space of the theory was V hat.

16
00:00:51,230 --> 00:00:54,830
For those states in
the degenerate space,

17
00:00:54,830 --> 00:01:01,970
we use a label n for being an
n-th state or having energy En.

18
00:01:01,970 --> 00:01:07,850
They all have the same energy,
unperturbed energy, En0.

19
00:01:07,850 --> 00:01:10,100
These are the states.

20
00:01:10,100 --> 00:01:13,430
And with k running
from 1 to n-- these

21
00:01:13,430 --> 00:01:17,450
are a basis of linearly
independent states

22
00:01:17,450 --> 00:01:21,410
that span the
degenerate subspace.

23
00:01:21,410 --> 00:01:23,780
So that's our
degenerate subspace.

24
00:01:23,780 --> 00:01:28,940
And then, the rest of the
space is the space V hat.

25
00:01:28,940 --> 00:01:32,630
And it's spanned by states
that we call p0's, we

26
00:01:32,630 --> 00:01:33,540
used to call them.

27
00:01:33,540 --> 00:01:36,120
And the notation
is good that allows

28
00:01:36,120 --> 00:01:39,260
you to distinguish,
with two labels, states

29
00:01:39,260 --> 00:01:42,710
in the degenerate subspace
and, with one label, a state

30
00:01:42,710 --> 00:01:45,380
in the rest of the space.

31
00:01:45,380 --> 00:01:48,470
The rest of the space can
also have degeneracies

32
00:01:48,470 --> 00:01:50,520
that wouldn't matter.

33
00:01:50,520 --> 00:01:52,600
You see, when you do
non-degenerate perturbations

34
00:01:52,600 --> 00:01:54,980
theories, because
you are focusing

35
00:01:54,980 --> 00:01:57,980
on a single,
non-degenerate state,

36
00:01:57,980 --> 00:02:01,100
it's not that you're
focusing on a theory in which

37
00:02:01,100 --> 00:02:04,720
every state is non-degenerate.

38
00:02:04,720 --> 00:02:06,360
Non-degenerate
perturbation theory

39
00:02:06,360 --> 00:02:09,440
means focusing on a
non-degenerate state

40
00:02:09,440 --> 00:02:11,990
of a general spectrum.

41
00:02:11,990 --> 00:02:14,400
Same thing with degenerate
perturbation theory.

42
00:02:14,400 --> 00:02:17,930
You focus on a particular
non-degenerate space.

43
00:02:17,930 --> 00:02:21,200
Whether there are other
non-degenerate spaces

44
00:02:21,200 --> 00:02:22,670
doesn't matter.

45
00:02:22,670 --> 00:02:26,090
They will be
treated as the rest.

46
00:02:26,090 --> 00:02:29,420
And that makes your life simple.

47
00:02:29,420 --> 00:02:33,530
So we had our
perturbative equations

48
00:02:33,530 --> 00:02:39,800
assume that we were going
to find states nk that

49
00:02:39,800 --> 00:02:42,020
depend on lambda.

50
00:02:42,020 --> 00:02:48,460
And they would have energies
Enk that depend on lambda.

51
00:02:48,460 --> 00:02:50,450
And looking at the
Schrodinger equation,

52
00:02:50,450 --> 00:02:53,300
we found an expansion
in which this state

53
00:02:53,300 --> 00:02:58,490
is the state n0 plus lambda
n1 plus lambda squared n2.

54
00:02:58,490 --> 00:03:04,430
And this energy is En0
plus the first corrections

55
00:03:04,430 --> 00:03:08,390
plus the second corrections
and those things.

56
00:03:08,390 --> 00:03:12,710
So these were our equations
that probably by now, they're

57
00:03:12,710 --> 00:03:15,560
starting to look
pretty familiar.

58
00:03:15,560 --> 00:03:23,160
And as usual, we say that
a correction to a state

59
00:03:23,160 --> 00:03:25,620
doesn't receive corrections.

60
00:03:25,620 --> 00:03:28,110
We can work in a
convention where

61
00:03:28,110 --> 00:03:31,170
it doesn't receive
corrections proportional

62
00:03:31,170 --> 00:03:33,300
to the original state.

63
00:03:33,300 --> 00:03:36,870
This could be reabsorbed
into normalization.

64
00:03:39,520 --> 00:03:42,730
So that was the setup.

65
00:03:42,730 --> 00:03:48,100
And we said we would calculate
things using three steps.

66
00:03:48,100 --> 00:03:54,820
And the first step was to act
with those unperturbed states

67
00:03:54,820 --> 00:04:00,670
on the order lambda equation,
so lambda to the 1 there.

68
00:04:00,670 --> 00:04:03,680
The first equation
is always trivial.

69
00:04:03,680 --> 00:04:06,100
So we acted in this equation.

70
00:04:06,100 --> 00:04:11,710
And these states are
killed by this operator

71
00:04:11,710 --> 00:04:15,730
because they are
states of energy En0.

72
00:04:15,730 --> 00:04:20,350
And therefore, you get the
right-hand side equal to 0

73
00:04:20,350 --> 00:04:22,720
when you have the n0l.

74
00:04:22,720 --> 00:04:27,460
And we discovered that the
consistency of our expansion

75
00:04:27,460 --> 00:04:32,710
requires that the chosen basis--

76
00:04:32,710 --> 00:04:35,500
because, after all,
you chose that basis.

77
00:04:35,500 --> 00:04:37,180
Nobody gave it to you.

78
00:04:37,180 --> 00:04:39,790
If you have a
degenerate subspace,

79
00:04:39,790 --> 00:04:43,630
you can take any set of
linearly independent vectors

80
00:04:43,630 --> 00:04:47,800
that are orthonormal,
and that's a good basis.

81
00:04:47,800 --> 00:04:49,760
It's not a uniquely
defined basis.

82
00:04:49,760 --> 00:04:53,050
But in the chosen basis
that you're working with,

83
00:04:53,050 --> 00:04:56,830
it must happen that
the perturbation

84
00:04:56,830 --> 00:05:01,235
is diagonal-- delta lk.

85
00:05:07,630 --> 00:05:12,700
And therefore, the first
order perturbations,

86
00:05:12,700 --> 00:05:18,280
when you take l equal to k, is
given by this matrix element

87
00:05:18,280 --> 00:05:21,640
between nk and nk.

88
00:05:21,640 --> 00:05:26,650
So this is a key result,
and our first result

89
00:05:26,650 --> 00:05:29,800
that in the degenerate
subspace, either you

90
00:05:29,800 --> 00:05:34,600
choose a basis that diagonalizes
the perturbation to begin with,

91
00:05:34,600 --> 00:05:42,610
or you compute the matrix delta
H in that degenerate subspace

92
00:05:42,610 --> 00:05:44,770
and then work and
diagonalize it.

93
00:05:44,770 --> 00:05:47,630
So either way, you do it.

94
00:05:47,630 --> 00:05:53,590
So I want to make a couple
of remarks on this thing.

95
00:05:53,590 --> 00:06:02,440
First remark is a terminology.

96
00:06:02,440 --> 00:06:09,610
We can say that the degeneracy
was lifted by the perturbation.

97
00:06:09,610 --> 00:06:11,770
And that's a great
thing if it happens.

98
00:06:16,460 --> 00:06:18,280
We'll comment more.

99
00:06:18,280 --> 00:06:27,350
It's a very nice thing if it
happens, the degeneracy lifted.

100
00:06:27,350 --> 00:06:29,450
And you have the
intuition what happens.

101
00:06:29,450 --> 00:06:34,200
You have these states are
all here of the same energy.

102
00:06:34,200 --> 00:06:36,720
And then, if the
degeneracy is lifted--

103
00:06:36,720 --> 00:06:40,370
these are four states-- then,
suddenly, one, two, three,

104
00:06:40,370 --> 00:06:45,710
four, as lambda
changes, they split.

105
00:06:45,710 --> 00:06:49,650
So their energies start
to become different.

106
00:06:49,650 --> 00:06:51,680
And that depends
on the first energy

107
00:06:51,680 --> 00:06:57,170
corrections being different for
every single one of the states.

108
00:06:57,170 --> 00:07:07,910
So this means that Enk1
is different from Enl1

109
00:07:07,910 --> 00:07:09,410
for the l-th state.

110
00:07:09,410 --> 00:07:12,440
And for the k-th state, the
first energy corrections

111
00:07:12,440 --> 00:07:17,930
are different for all
k different for l.

112
00:07:21,840 --> 00:07:26,970
So for any two states that you
pick, from the list of your N--

113
00:07:26,970 --> 00:07:29,940
capital N states there--

114
00:07:29,940 --> 00:07:32,760
for any two states
that you pick,

115
00:07:32,760 --> 00:07:34,440
their energies
must be different.

116
00:07:34,440 --> 00:07:38,610
And then, the
degeneracy is lifted.

117
00:07:38,610 --> 00:07:42,360
And that means that
you've found what

118
00:07:42,360 --> 00:07:48,530
we will call a good
basis, a basis of states

119
00:07:48,530 --> 00:07:52,160
that diagonalizes delta H.

120
00:07:52,160 --> 00:07:56,990
And already, the
states have split.

121
00:07:56,990 --> 00:08:01,970
You can now follow each of
them and see what happens.

122
00:08:01,970 --> 00:08:06,320
You know what is
a preferred basis.

123
00:08:06,320 --> 00:08:10,850
So this is nice.

124
00:08:10,850 --> 00:08:14,510
And so this is our first remark.

125
00:08:14,510 --> 00:08:17,786
Degeneracy is lifted
when this happens.

126
00:08:17,786 --> 00:08:19,160
And this is the
case, we're going

127
00:08:19,160 --> 00:08:23,690
to assume, for the
rest of the discussion,

128
00:08:23,690 --> 00:08:25,730
for the first part
of the lecture,

129
00:08:25,730 --> 00:08:27,740
that the degeneracy is lifted.

130
00:08:27,740 --> 00:08:30,950
And then, we'll try to find
the correction to the state.

131
00:08:30,950 --> 00:08:36,880
Let me comment that our first
order shifts that we have

132
00:08:36,880 --> 00:08:39,400
found-- these eigenvalues--

133
00:08:39,400 --> 00:08:49,240
are valid even if the
degeneracy is not lifted.

134
00:08:56,500 --> 00:09:02,650
The second remark
has to do with a rule

135
00:09:02,650 --> 00:09:06,040
to help you do this stuff.

136
00:09:06,040 --> 00:09:09,760
This will be very useful
in about a week, when

137
00:09:09,760 --> 00:09:13,670
we apply some of these
things to the hydrogen atom.

138
00:09:13,670 --> 00:09:17,440
So first matter of notation--
some people use this,

139
00:09:17,440 --> 00:09:20,230
I kind of like it--

140
00:09:20,230 --> 00:09:37,770
a basis in VN that diagonalizes
delta H is called a good basis.

141
00:09:43,210 --> 00:09:49,590
So good, bad-- it's kind of
easy to relate to those things.

142
00:09:49,590 --> 00:09:50,450
So it's good.

143
00:09:50,450 --> 00:09:52,450
It gives you the answer.

144
00:09:52,450 --> 00:09:55,090
So we'll call it a good basis.

145
00:09:55,090 --> 00:10:02,140
So here comes a question
that you want to answer.

146
00:10:02,140 --> 00:10:07,000
Typically, you have some
basis that you like.

147
00:10:07,000 --> 00:10:10,830
And you have a delta H.

148
00:10:10,830 --> 00:10:14,100
Now, you can say, OK, do
I know that delta H is

149
00:10:14,100 --> 00:10:16,620
diagonal in this basis?

150
00:10:16,620 --> 00:10:18,240
Maybe I don't.

151
00:10:18,240 --> 00:10:19,840
So what can I do?

152
00:10:19,840 --> 00:10:22,740
I can diagonalize it and check.

153
00:10:22,740 --> 00:10:25,650
But sometimes, there
is a better way,

154
00:10:25,650 --> 00:10:31,510
and it helps you explain why
the matrix delta H is diagonal.

155
00:10:31,510 --> 00:10:33,280
So this is a shortcut.

156
00:10:33,280 --> 00:10:36,120
It's a conceptual
idea that you can

157
00:10:36,120 --> 00:10:41,700
check that delta H is diagonal
without diagonalizing it.

158
00:10:41,700 --> 00:10:43,380
So here it is.

159
00:10:47,520 --> 00:10:49,060
So here's the rule--

160
00:10:49,060 --> 00:10:49,560
"rule."

161
00:10:52,810 --> 00:11:12,240
If the basis vectors are
eigenstates of a Hermitian

162
00:11:12,240 --> 00:11:19,870
K with different eigenvalues--

163
00:11:25,500 --> 00:11:28,150
so so far, what am I saying?

164
00:11:28,150 --> 00:11:30,270
I'm bringing up a new operator.

165
00:11:30,270 --> 00:11:34,780
Somebody gives you
another operator K--

166
00:11:34,780 --> 00:11:36,210
Hermitian.

167
00:11:36,210 --> 00:11:38,430
And then, you check
that, after all,

168
00:11:38,430 --> 00:11:43,980
for this Hermitian operator,
the basis, or the basis vectors,

169
00:11:43,980 --> 00:11:50,240
are eigenstates of that operator
and with different eigenvalues.

170
00:11:50,240 --> 00:11:54,460
All of them have
different eigenvalues.

171
00:11:54,460 --> 00:11:56,070
[MUMBLING]

172
00:11:57,080 --> 00:12:10,470
And K commutes with delta
H, then the basis is good.

173
00:12:13,760 --> 00:12:16,770
Basis is good.

174
00:12:19,800 --> 00:12:23,530
Now, first time I heard this--

175
00:12:23,530 --> 00:12:25,990
I still remember--
it seemed to me

176
00:12:25,990 --> 00:12:27,820
like a very complicated rule.

177
00:12:27,820 --> 00:12:30,681
Just diagonalize
and forget about it.

178
00:12:30,681 --> 00:12:31,180
No.

179
00:12:31,180 --> 00:12:34,520
But it actually is very helpful.

180
00:12:34,520 --> 00:12:38,040
So let's explain this rule.

181
00:12:42,030 --> 00:12:46,380
So K with delta H is equal to 0.

182
00:12:46,380 --> 00:12:50,190
So let's assume we
have two eigenstates.

183
00:12:50,190 --> 00:12:53,540
So "proof" here.

184
00:12:53,540 --> 00:12:59,180
Two eigenstates-- n0p and n0q.

185
00:13:04,490 --> 00:13:16,410
And the K eigenvalues are
lambda p and lambda q that

186
00:13:16,410 --> 00:13:19,140
are different from each other.

187
00:13:19,140 --> 00:13:22,890
So we said that the
basis states are supposed

188
00:13:22,890 --> 00:13:28,590
to be eigenstates of
this Hermitian operator K

189
00:13:28,590 --> 00:13:31,090
and all of them with
different eigenvalues.

190
00:13:31,090 --> 00:13:35,490
So here, I wrote two states
that are part of our basis.

191
00:13:35,490 --> 00:13:39,330
And I say that K eigenvalues
are lambda p and lambda q,

192
00:13:39,330 --> 00:13:42,600
and they are different
from each other.

193
00:13:42,600 --> 00:13:44,350
So far, so good.

194
00:13:44,350 --> 00:13:52,820
So then we can form
the following--

195
00:13:52,820 --> 00:14:01,120
n0p, the commutator of K
with delta H with K or K

196
00:14:01,120 --> 00:14:07,640
with delta H n0q.

197
00:14:11,890 --> 00:14:16,170
Consider that matrix element.

198
00:14:16,170 --> 00:14:22,800
Since we say that K commutes
with delta H, this must be 0.

199
00:14:22,800 --> 00:14:25,950
That operator that we've put
in between the states is 0.

200
00:14:25,950 --> 00:14:33,220
So this is 0 because this is 0.

201
00:14:33,220 --> 00:14:36,060
But on the other hand,
let's expand this.

202
00:14:36,060 --> 00:14:38,050
This is a commutator.

203
00:14:38,050 --> 00:14:43,290
So you have K times delta
H minus delta H times K.

204
00:14:43,290 --> 00:14:46,410
Since it's a Hermitian
operator, the first term

205
00:14:46,410 --> 00:14:49,620
K is near this state.

206
00:14:49,620 --> 00:14:53,070
And being Hermitian,
you get the eigenvalue

207
00:14:53,070 --> 00:15:06,230
of 8, which is lambda
p n0p delta H n0q.

208
00:15:09,370 --> 00:15:12,100
That's from the first term.

209
00:15:12,100 --> 00:15:16,000
And the second term is a
minus the delta H on the left.

210
00:15:16,000 --> 00:15:27,570
So you get the
minus n0qp delta H.

211
00:15:27,570 --> 00:15:31,980
And the K acting on the right
this time gives you lambda q.

212
00:15:39,680 --> 00:15:48,700
So this is equal to lambda
p minus lambda q times n--

213
00:15:48,700 --> 00:15:52,360
it's the same factor
for both terms--

214
00:15:52,360 --> 00:15:55,060
delta H n0q.

215
00:15:59,851 --> 00:16:00,350
OK.

216
00:16:00,350 --> 00:16:03,710
So that matrix element
has been evaluated.

217
00:16:03,710 --> 00:16:07,640
And it's now equal to the
product of the difference

218
00:16:07,640 --> 00:16:12,530
of eigenvalues times the
matrix element of delta H.

219
00:16:12,530 --> 00:16:16,430
But here, we know that
this product must be 0.

220
00:16:16,430 --> 00:16:22,550
But since the two
eigenvalues are different,

221
00:16:22,550 --> 00:16:24,680
this factor is not 0.

222
00:16:24,680 --> 00:16:29,060
So the only possibility is
that the second factor is 0.

223
00:16:29,060 --> 00:16:31,820
And you have shown,
therefore, at this moment,

224
00:16:31,820 --> 00:16:42,140
since this is 0, this
implies that the n0p delta H

225
00:16:42,140 --> 00:16:50,120
n0q is equal to 0 whenever
p is different from q.

226
00:16:50,120 --> 00:16:53,900
And that's exactly
the statement that all

227
00:16:53,900 --> 00:16:58,760
the off-diagonal elements
of delta H are 0.

228
00:16:58,760 --> 00:17:02,630
So indeed, it has
been diagonalized.

229
00:17:06,569 --> 00:17:11,180
Or you can, without
diagonalizing the delta H,

230
00:17:11,180 --> 00:17:13,130
you can say, oh,
look at this basis.

231
00:17:13,130 --> 00:17:14,510
This basis is good.

232
00:17:14,510 --> 00:17:18,619
Because there is this operator--
many times it will be angular

233
00:17:18,619 --> 00:17:21,680
momentum, lz, it
may be other thing--

234
00:17:21,680 --> 00:17:24,589
for which all the states
have different eigenvalues.

235
00:17:24,589 --> 00:17:26,839
And it commutes
with a perturbation.

236
00:17:26,839 --> 00:17:29,600
So you look for an
operator that commutes

237
00:17:29,600 --> 00:17:35,930
with a perturbation for
which your basis vectors

238
00:17:35,930 --> 00:17:38,650
may be eigenstates.