1
00:00:00,770 --> 00:00:03,440
PROFESSOR: It's
time for step two.

2
00:00:03,440 --> 00:00:05,180
What don't we know here?

3
00:00:05,180 --> 00:00:09,170
Well, we don't know anything
about the state correction.

4
00:00:09,170 --> 00:00:13,150
So n1, what is n1?

5
00:00:13,150 --> 00:00:28,200
So step two is find the
piece of n1 k, the piece that

6
00:00:28,200 --> 00:00:31,860
is in the space v hat.

7
00:00:31,860 --> 00:00:35,400
So I will use this
notation, a bar here,

8
00:00:35,400 --> 00:00:41,460
saying, of this vector,
the piece in that subspace.

9
00:00:41,460 --> 00:00:44,250
Then we, of course,
need to find the piece

10
00:00:44,250 --> 00:00:47,160
in the degenerate subspace.

11
00:00:47,160 --> 00:00:52,350
Remember, the corrections
are orthogonal

12
00:00:52,350 --> 00:00:56,220
to the original state,
but the corrections

13
00:00:56,220 --> 00:00:59,910
can have a piece on vn.

14
00:00:59,910 --> 00:01:01,940
The vn has more states.

15
00:01:01,940 --> 00:01:05,810
So it can have a piece on vn,
and that's typically the part

16
00:01:05,810 --> 00:01:08,490
that is a little hard to do.

17
00:01:08,490 --> 00:01:10,590
So what do we do here?

18
00:01:10,590 --> 00:01:16,700
We take the order one
equation, lambda equal 1,

19
00:01:16,700 --> 00:01:21,380
and we hit it with p0.

20
00:01:21,380 --> 00:01:27,815
So we'll hit with p0 on
the order lambda equation.

21
00:01:34,890 --> 00:01:41,590
So think of the p0 vector,
bra, appearing here.

22
00:01:41,590 --> 00:01:47,040
We know the energy of it, so
this will become a number.

23
00:01:47,040 --> 00:02:01,245
So we get Ep0 minus En0 p0 n1 k.

24
00:02:06,090 --> 00:02:08,560
And indeed, when you
see here-- you say,

25
00:02:08,560 --> 00:02:10,350
oh, this is my unknown.

26
00:02:10,350 --> 00:02:12,450
I don't know the first
order correction,

27
00:02:12,450 --> 00:02:16,320
and I'm getting the components
of this unknown correction

28
00:02:16,320 --> 00:02:18,630
along the v hat space.

29
00:02:18,630 --> 00:02:21,600
That's why this calculation
can only give me

30
00:02:21,600 --> 00:02:27,080
the piece of n1 k along v hat.

31
00:02:27,080 --> 00:02:29,090
And then, what do we have here?

32
00:02:29,090 --> 00:02:44,190
We have p0 Enk1
minus delta H n0 k.

33
00:02:48,370 --> 00:02:55,120
It's just that order one
equation calculated here.

34
00:02:55,120 --> 00:02:56,190
So what is it?

35
00:02:56,190 --> 00:03:00,120
You recognize the
challenge here is always

36
00:03:00,120 --> 00:03:05,321
sort of trying to remember what
the symbols mean and simplify

37
00:03:05,321 --> 00:03:05,820
things.

38
00:03:05,820 --> 00:03:08,750
So for example, here.

39
00:03:08,750 --> 00:03:09,990
This is a number.

40
00:03:09,990 --> 00:03:12,440
So what you must ask
is whether this vector

41
00:03:12,440 --> 00:03:14,690
is orthogonal to this
one because you're

42
00:03:14,690 --> 00:03:16,490
going to get an inner product.

43
00:03:16,490 --> 00:03:18,920
And this is a vector
in v hat, and this

44
00:03:18,920 --> 00:03:24,210
is a vector in v capital N. So
therefore, they are orthogonal.

45
00:03:24,210 --> 00:03:32,830
So this term vanishes by
orthogonality of these two.

46
00:03:32,830 --> 00:03:36,730
On the other hand, we are
left with this part here.

47
00:03:36,730 --> 00:03:41,500
Now, that's a matrix
element of delta H

48
00:03:41,500 --> 00:03:46,430
between a degenerate state
and the rest of the states.

49
00:03:46,430 --> 00:03:48,650
So that's something
you have to calculate.

50
00:03:48,650 --> 00:03:50,990
There's no way around it.

51
00:03:50,990 --> 00:03:53,630
So we use our notation.

52
00:03:53,630 --> 00:03:56,190
And in our notation,
this will be delta Hpnk.

53
00:04:04,180 --> 00:04:04,870
All right.

54
00:04:04,870 --> 00:04:06,730
So this gives us--

55
00:04:06,730 --> 00:04:08,620
we're done with this equation.

56
00:04:08,620 --> 00:04:10,240
It was simple.

57
00:04:10,240 --> 00:04:11,680
No big sweat.

58
00:04:11,680 --> 00:04:14,710
We now have these coefficients,
because these are numbers,

59
00:04:14,710 --> 00:04:16,839
and we can just
divide over here.

60
00:04:20,680 --> 00:04:24,980
And this factor is
nonzero because all the p

61
00:04:24,980 --> 00:04:29,190
states that are in v hat
have different energy

62
00:04:29,190 --> 00:04:30,720
from our degenerate ones.

63
00:04:30,720 --> 00:04:36,660
So there are the
components of this.

64
00:04:36,660 --> 00:04:49,440
So we have p0 n1 k is equal
to minus delta Hpnk divided

65
00:04:49,440 --> 00:04:55,680
by this quantity, Ep0 minus En0.

66
00:04:55,680 --> 00:04:57,960
And therefore, we
can write n1 k.

67
00:05:05,540 --> 00:05:11,540
If you have the inner products
of a state with p arbitrary,

68
00:05:11,540 --> 00:05:17,510
the state is the sum of other
p's times that inner product.

69
00:05:17,510 --> 00:05:23,030
So I'll write it as minus
because of that minus here.

70
00:05:23,030 --> 00:05:40,050
The sum over p of delta
Hpnk over Ep0 minus En0 p0

71
00:05:40,050 --> 00:05:42,540
written like that.

72
00:05:42,540 --> 00:05:46,680
This is something-- maybe
it sounds a little fast,

73
00:05:46,680 --> 00:05:53,550
but if you have a basis
vector, alpha i with psi

74
00:05:53,550 --> 00:05:57,180
being the number
beta i, the state

75
00:05:57,180 --> 00:06:05,600
psi is equal to the
sum of beta i alpha i.

76
00:06:05,600 --> 00:06:09,020
Once you have the
components, the vector

77
00:06:09,020 --> 00:06:12,320
can be reconstructed as the
components times the basis

78
00:06:12,320 --> 00:06:13,310
vectors.

79
00:06:13,310 --> 00:06:14,920
So that's what we did here.

80
00:06:14,920 --> 00:06:18,480
We have the components here.

81
00:06:18,480 --> 00:06:22,180
And therefore, the
vector is the components

82
00:06:22,180 --> 00:06:24,560
times the basis vectors.

83
00:06:24,560 --> 00:06:25,760
You can check that.

84
00:06:25,760 --> 00:06:28,030
If that makes you
a little uneasy,

85
00:06:28,030 --> 00:06:33,110
just put the p0 here,
and recalculate that,

86
00:06:33,110 --> 00:06:35,120
and you will get that answer.

87
00:06:35,120 --> 00:06:39,440
Now, as written it there,
it's really not precise.

88
00:06:39,440 --> 00:06:41,690
You're making,
technically, a mistake.

89
00:06:41,690 --> 00:06:47,660
This equation is technically
wrong because that's not n1 k.

90
00:06:47,660 --> 00:06:53,390
That is only the components
of n1 k along v hat.

91
00:06:53,390 --> 00:06:57,070
So this is v hat.

92
00:06:57,070 --> 00:07:01,240
If n1 k has something along
the degenerate subspace,

93
00:07:01,240 --> 00:07:03,250
we haven't found it.

94
00:07:03,250 --> 00:07:05,550
But this much we found.

95
00:07:05,550 --> 00:07:10,240
We've found part
of the correction.

96
00:07:10,240 --> 00:07:15,070
So degeneracy is always
a complicated thing.

97
00:07:15,070 --> 00:07:18,130
So we're going to try to find--

98
00:07:18,130 --> 00:07:19,890
we're going to
find, in fact, now

99
00:07:19,890 --> 00:07:25,470
the component of the correction
around the degenerate space.

100
00:07:25,470 --> 00:07:30,460
Now, in nondegenerate
perturbation theory

101
00:07:30,460 --> 00:07:36,340
we manage to use this equation
to calculate fully the state.

102
00:07:36,340 --> 00:07:39,580
But now we've used this
equation all the way.

103
00:07:39,580 --> 00:07:44,480
We put states in the
degenerate subspace,

104
00:07:44,480 --> 00:07:46,340
and we found the energies.

105
00:07:46,340 --> 00:07:49,460
We put, in the other
part of the space, v hat,

106
00:07:49,460 --> 00:07:55,070
and we found the part
of n1 along v hat.

107
00:07:55,070 --> 00:07:59,990
And we ran out of things
from that order one equation.

108
00:07:59,990 --> 00:08:03,940
It has no more information.

109
00:08:03,940 --> 00:08:09,700
So somehow it must be that the
next equation that we usually

110
00:08:09,700 --> 00:08:13,150
need to go to second
order in energy

111
00:08:13,150 --> 00:08:18,470
will tell us something about
the missing part of n1.

112
00:08:18,470 --> 00:08:21,550
So that's the surprising thing.

113
00:08:21,550 --> 00:08:27,040
You have to go to order lambda
square to find the first order

114
00:08:27,040 --> 00:08:29,830
correction to the degenerate
part of the state.

115
00:08:29,830 --> 00:08:33,220
That's why degenerate
perturbation theory

116
00:08:33,220 --> 00:08:36,700
is famous for its complication.

117
00:08:36,700 --> 00:08:40,970
You really need to go pretty
high to find the things.

118
00:08:40,970 --> 00:08:44,660
So let's do it.

119
00:08:44,660 --> 00:08:45,745
So this is step 3.

120
00:08:49,300 --> 00:08:58,155
You hit the order lambda
squared equation with n0 l.

121
00:09:02,130 --> 00:09:04,680
And therefore,
the left-hand side

122
00:09:04,680 --> 00:09:09,090
is going to be 0 because that
operator on the left-hand side

123
00:09:09,090 --> 00:09:11,370
always kills those states.

124
00:09:11,370 --> 00:09:17,010
Now, you could have been
a little sloppy here.

125
00:09:17,010 --> 00:09:20,790
I was sloppy when I first
solved those equations.

126
00:09:20,790 --> 00:09:23,870
I didn't say, OK, this is v hat.

127
00:09:23,870 --> 00:09:25,800
I said, that's the answer.

128
00:09:25,800 --> 00:09:27,540
That's the whole state.

129
00:09:27,540 --> 00:09:31,720
There's nothing along
the degenerate subspace.

130
00:09:31,720 --> 00:09:35,710
But then, if you stop
there, you can live happily

131
00:09:35,710 --> 00:09:37,160
the rest of your life.

132
00:09:37,160 --> 00:09:38,950
But if you go to
the next equation

133
00:09:38,950 --> 00:09:42,190
and find that it's
wrong, there was a piece

134
00:09:42,190 --> 00:09:43,710
along the degenerate subspace.

135
00:09:43,710 --> 00:09:45,430
So let's see that.

136
00:09:45,430 --> 00:09:49,420
So we'll also write,
as we've started to do,

137
00:09:49,420 --> 00:10:04,170
n1 k being equal to n1 k along
v hat plus the piece of n1 k

138
00:10:04,170 --> 00:10:08,220
along the vn.

139
00:10:08,220 --> 00:10:11,590
And this one we got.

140
00:10:11,590 --> 00:10:15,080
And this one, well, is it there?

141
00:10:15,080 --> 00:10:17,060
Do we need it?

142
00:10:17,060 --> 00:10:18,760
Does the state
receive a correction

143
00:10:18,760 --> 00:10:22,500
in that space or not?

144
00:10:22,500 --> 00:10:24,540
Well, let's do the work.

145
00:10:24,540 --> 00:10:29,820
Let's hit this
with those states.

146
00:10:29,820 --> 00:10:32,780
So what happens?

147
00:10:32,780 --> 00:10:48,904
We have n0 l Enk 1
minus delta H n1 k.

148
00:10:55,390 --> 00:10:57,820
That's our term there.

149
00:10:57,820 --> 00:11:02,180
Actually, I will write it twice.

150
00:11:02,180 --> 00:11:05,280
So I'll copy it again.

151
00:11:05,280 --> 00:11:19,580
n0 k l Enk 1 minus delta H n1 k.

152
00:11:19,580 --> 00:11:22,090
Why do I copy it again?

153
00:11:22,090 --> 00:11:27,610
Because I'll just put
here on v hat and on vn.

154
00:11:30,290 --> 00:11:33,610
So it was one term with an n1 k.

155
00:11:33,610 --> 00:11:36,880
But you know, the
n1 k vector has

156
00:11:36,880 --> 00:11:40,300
components along two subspaces.

157
00:11:40,300 --> 00:11:46,670
And therefore, we might as
well put each one separately

158
00:11:46,670 --> 00:11:52,750
so we can think about
them in a clear way.

159
00:11:52,750 --> 00:11:54,710
Then, what else?

160
00:11:54,710 --> 00:11:59,610
There's one more
term, the En2 k.

161
00:11:59,610 --> 00:12:09,540
And that is relatively
simple because that's Enk 2.

162
00:12:09,540 --> 00:12:14,400
And then we have the overlap
of an n0 l with an n0 k which

163
00:12:14,400 --> 00:12:18,410
is a delta lk.

164
00:12:18,410 --> 00:12:20,330
So this whole thing must be 0.

165
00:12:20,330 --> 00:12:23,240
That was the right-hand
side of the equation.

166
00:12:23,240 --> 00:12:29,560
But the left-hand side of the
equation was 0, so that's 0.

167
00:12:29,560 --> 00:12:30,920
OK.

168
00:12:30,920 --> 00:12:35,300
So now we have to think
about these terms.

169
00:12:35,300 --> 00:12:36,380
What do we know?

170
00:12:36,380 --> 00:12:39,060
What is 0 to begin with?

171
00:12:39,060 --> 00:12:40,030
What is not 0?

172
00:12:42,700 --> 00:12:47,550
This term in here,
we know n1 k already.

173
00:12:47,550 --> 00:12:48,800
We found it there.

174
00:12:48,800 --> 00:12:50,530
So that's nice.

175
00:12:50,530 --> 00:12:53,440
We know n1 k.

176
00:12:53,440 --> 00:12:54,940
We know this corrections.

177
00:12:54,940 --> 00:12:59,800
But this is simpler because n0
l is in the degenerate subspace.

178
00:12:59,800 --> 00:13:03,220
This is a number, and this is
in b hat, so that's 0 again.

179
00:13:11,000 --> 00:13:11,500
OK.

180
00:13:11,500 --> 00:13:13,230
So that's a simplification.

181
00:13:13,230 --> 00:13:18,880
Here we have a very
interesting situation.

182
00:13:18,880 --> 00:13:26,280
We have the delta H, and
we have this basis state

183
00:13:26,280 --> 00:13:29,800
of the degenerate subspace.

184
00:13:29,800 --> 00:13:35,290
We know that delta H is diagonal
in the degenerate subspace,

185
00:13:35,290 --> 00:13:43,540
but is this n0 l an
eigenvalue of delta H?

186
00:13:43,540 --> 00:13:48,220
Not quite because
being diagonal,

187
00:13:48,220 --> 00:13:52,990
these basis vectors
make delta H diagonal.

188
00:13:52,990 --> 00:13:56,020
So delta H is diagonal
in this subspace,

189
00:13:56,020 --> 00:13:59,380
but doesn't mean this thing
is an eigenvector because it

190
00:13:59,380 --> 00:14:04,060
can give you something outside
the degenerate subspace.

191
00:14:04,060 --> 00:14:09,100
So we cannot quite just say
that the eigenvalue of this is

192
00:14:09,100 --> 00:14:12,700
the first energy correction.

193
00:14:12,700 --> 00:14:16,240
But actually, we can.

194
00:14:16,240 --> 00:14:18,230
Let me explain that.

195
00:14:18,230 --> 00:14:22,090
So let's look at
this term alone.

196
00:14:22,090 --> 00:14:28,510
With this term alone-- now
this time I cannot kill this

197
00:14:28,510 --> 00:14:32,770
constant because this is
in the degenerate subspace,

198
00:14:32,770 --> 00:14:34,840
and this is in the
degenerate subspace.

199
00:14:34,840 --> 00:14:38,780
So we will have
to deal with that.

200
00:14:41,480 --> 00:14:44,015
So let's do this.

201
00:14:48,550 --> 00:14:58,360
So output here.
n0 l delta H n1 k.

202
00:15:01,570 --> 00:15:04,960
We're going to try
to simplify this.

203
00:15:04,960 --> 00:15:06,760
And the way to do it--

204
00:15:06,760 --> 00:15:12,030
this is n1 k, very
important in vn--

205
00:15:12,030 --> 00:15:17,430
is to insert the resolution
of the identity here.

206
00:15:17,430 --> 00:15:18,950
So I'll do it.

207
00:15:18,950 --> 00:15:26,410
We'll put n0 l delta
H, and now we'll

208
00:15:26,410 --> 00:15:29,800
put the whole resolution
of the identity.

209
00:15:29,800 --> 00:15:51,350
So we'll put the sum over q n0
q n0 q plus a sum over p p0 p0,

210
00:15:51,350 --> 00:15:59,660
all acting on n1 k of vn.

211
00:15:59,660 --> 00:16:02,570
I actually want
to remark that all

212
00:16:02,570 --> 00:16:04,500
what I'm doing in this line--

213
00:16:04,500 --> 00:16:06,880
so let's break this.

214
00:16:06,880 --> 00:16:11,800
The equation was up to
here, and we decided

215
00:16:11,800 --> 00:16:15,800
to try to understand this term.

216
00:16:15,800 --> 00:16:18,420
So we're trying to
understand this term.

217
00:16:18,420 --> 00:16:20,790
Just that term.

218
00:16:20,790 --> 00:16:27,890
Now, delta H here is
acting between degenerate

219
00:16:27,890 --> 00:16:32,715
eigenstate here and here,
some arbitrary state

220
00:16:32,715 --> 00:16:34,320
in the degenerate subspace.

221
00:16:37,100 --> 00:16:43,090
So we go here and we
say, OK, since this

222
00:16:43,090 --> 00:16:47,410
is in the degenerate subspace,
this inner product with vector

223
00:16:47,410 --> 00:16:49,210
in v hat, this is 0.

224
00:16:49,210 --> 00:16:52,510
So that part of the
resolution of the identity

225
00:16:52,510 --> 00:16:53,320
is not relevant.

226
00:17:00,140 --> 00:17:02,450
If that part of the
resolution of the identity

227
00:17:02,450 --> 00:17:12,020
is not relevant, we are
left with n sum over q n0 l

228
00:17:12,020 --> 00:17:30,220
delta H n0 q n0 q times n1 k vn.

229
00:17:33,710 --> 00:17:39,620
And now you can use that
this matrix is, indeed,

230
00:17:39,620 --> 00:17:42,830
diagonal in the
degenerate subspace.

231
00:17:42,830 --> 00:17:45,440
And therefore,
this matrix element

232
00:17:45,440 --> 00:17:54,830
is the energy Enl 1, the
first order correction,

233
00:17:54,830 --> 00:17:57,380
times a delta lq.

234
00:18:03,270 --> 00:18:06,150
And then we can
do the sum over q

235
00:18:06,150 --> 00:18:09,330
because there is a
delta function here.

236
00:18:09,330 --> 00:18:30,900
And this becomes
Enl 1 n0 l n1 k vn.

237
00:18:34,660 --> 00:18:37,510
So this is very nice.

238
00:18:37,510 --> 00:18:41,020
It took us a little bit of work
but look what has happened.

239
00:18:41,020 --> 00:18:45,610
This delta H term is going
to become the same structure,

240
00:18:45,610 --> 00:18:49,150
n0 l n1 k overlap.

241
00:18:49,150 --> 00:18:53,330
So it's great progress.

242
00:18:53,330 --> 00:18:57,970
So what does our
equation become?

243
00:18:57,970 --> 00:19:02,350
Well, from the first
term, this is the rest

244
00:19:02,350 --> 00:19:04,420
of the end of the aside.

245
00:19:04,420 --> 00:19:10,540
From the first term
we have minus n0 l

246
00:19:10,540 --> 00:19:18,450
delta H n1 k in v hat.

247
00:19:18,450 --> 00:19:22,870
That's all that was left
from that first term.

248
00:19:22,870 --> 00:19:27,000
And this term is
known because n1

249
00:19:27,000 --> 00:19:32,530
is known along the rest
of the Hilbert space.

250
00:19:32,530 --> 00:19:36,630
The second term over there
that was giving us trouble

251
00:19:36,630 --> 00:19:39,010
has become something
very simple.

252
00:19:39,010 --> 00:19:49,500
It has become Enk
minus Enl, both 1,

253
00:19:49,500 --> 00:19:56,654
multiplied by n0 l n1 k.

254
00:20:02,380 --> 00:20:08,050
I could put this vn, or
now I may not put it either

255
00:20:08,050 --> 00:20:12,830
because I am already
projecting to a state

256
00:20:12,830 --> 00:20:14,450
in the degenerate subspace.

257
00:20:14,450 --> 00:20:18,140
So I am finding the component
along the degenerate subspace.

258
00:20:18,140 --> 00:20:23,120
So even if n1 k had
a piece along v hat

259
00:20:23,120 --> 00:20:24,350
it would drop out here.

260
00:20:24,350 --> 00:20:29,350
So it's completely legal, and
it's simpler to erase the vn.

261
00:20:32,280 --> 00:20:40,830
And then the last term is still
there, plus Enk 2 delta lk

262
00:20:40,830 --> 00:20:41,565
equals 0.

263
00:20:44,750 --> 00:20:45,270
OK.

264
00:20:45,270 --> 00:20:47,250
This is our master equation.

265
00:20:47,250 --> 00:20:50,880
After thinking of this
equation for a while

266
00:20:50,880 --> 00:20:53,490
and using our properties,
we got this far.

267
00:20:58,810 --> 00:21:01,210
It's a very nice equation.

268
00:21:01,210 --> 00:21:04,810
It does give you the
second order energies.

269
00:21:04,810 --> 00:21:11,880
We were looking for
the part of the state

270
00:21:11,880 --> 00:21:13,800
along the degenerate subspace.

271
00:21:13,800 --> 00:21:15,415
This was our main unknown.

272
00:21:20,070 --> 00:21:22,810
But still, we can
get the energies.

273
00:21:22,810 --> 00:21:23,310
Why?

274
00:21:23,310 --> 00:21:31,050
Because we can take l
equals to k, in which case

275
00:21:31,050 --> 00:21:37,140
the term that we don't know
drops out, because nl equals

276
00:21:37,140 --> 00:21:40,530
to k, these things cancel.

277
00:21:40,530 --> 00:21:49,580
So when l is equal to k we get
that Enk 2, the second order

278
00:21:49,580 --> 00:22:07,980
correction to the energy, is
n0 k delta H n1 k in v hat.

279
00:22:07,980 --> 00:22:08,670
That's it.

280
00:22:08,670 --> 00:22:11,260
That's your second
correction to the energy,

281
00:22:11,260 --> 00:22:12,945
and that's very nice.

282
00:22:16,490 --> 00:22:17,900
Why do we say that's it?

283
00:22:17,900 --> 00:22:22,470
Because we actually
have the answer for n1.

284
00:22:22,470 --> 00:22:26,600
So we can find the
complete formula for this.

285
00:22:26,600 --> 00:22:28,490
I'll write it here.

286
00:22:28,490 --> 00:22:32,090
After a couple of
steps of algebra

287
00:22:32,090 --> 00:22:37,580
this gives minus the
sum over p delta H

288
00:22:37,580 --> 00:22:51,220
nkp over Ep0 minus En0 squared.

289
00:22:51,220 --> 00:22:52,185
Look at that formula.

290
00:22:55,260 --> 00:22:59,720
It has the same form as
the second order correction

291
00:22:59,720 --> 00:23:03,980
to nondegenerate states.

292
00:23:03,980 --> 00:23:08,630
Same look, except
that you only sum over

293
00:23:08,630 --> 00:23:12,680
the states that are outside
the degenerate space.

294
00:23:12,680 --> 00:23:17,820
It was exactly of this form the
second order energy correction.

295
00:23:17,820 --> 00:23:20,375
So it's kind of nice and simple.

296
00:23:23,400 --> 00:23:28,020
But let's look at
this equation again.

297
00:23:28,020 --> 00:23:31,560
And here is the thing that
would have been shocking

298
00:23:31,560 --> 00:23:34,050
if we didn't do this right.

299
00:23:34,050 --> 00:23:43,700
If we had set l different from k
here, if l is different from k,

300
00:23:43,700 --> 00:23:48,200
then this term is 0.

301
00:23:48,200 --> 00:23:49,610
If we didn't think--

302
00:23:49,610 --> 00:23:53,560
if we hadn't suspected
that the state n1 k had

303
00:23:53,560 --> 00:23:56,180
a piece along with
the generate subspace,

304
00:23:56,180 --> 00:23:57,980
we would have not
introduced it, and we

305
00:23:57,980 --> 00:24:00,260
would have had not this term.

306
00:24:00,260 --> 00:24:01,700
And the equation
would have become

307
00:24:01,700 --> 00:24:05,330
this equal to 0, which is 0.

308
00:24:05,330 --> 00:24:07,610
You have there n1 of k.

309
00:24:07,610 --> 00:24:10,190
And you can look at it,
and look at it for hours,

310
00:24:10,190 --> 00:24:12,170
and it's just not 0.

311
00:24:12,170 --> 00:24:16,220
So unless we have this
piece, the equation

312
00:24:16,220 --> 00:24:18,170
doesn't make sense.

313
00:24:18,170 --> 00:24:23,300
So this proves that the state
must develop a component

314
00:24:23,300 --> 00:24:26,480
along the degenerate subspace.

315
00:24:26,480 --> 00:24:30,080
And let's now finally get it.

316
00:24:30,080 --> 00:24:32,130
It's just one line
at this point.

317
00:24:32,130 --> 00:24:33,470
We don't have to do much.

318
00:24:42,680 --> 00:24:57,310
So when l is different from
k, l different from k--

319
00:24:57,310 --> 00:24:59,170
so what do we get
for the equation?

320
00:25:02,280 --> 00:25:04,700
Look at it there.

321
00:25:04,700 --> 00:25:10,700
We can solve directly
for the piece n0 l

322
00:25:10,700 --> 00:25:17,710
n1 k is equal to-- solve
for the second term.

323
00:25:17,710 --> 00:25:20,500
The other term goes on
the right-hand side.

324
00:25:20,500 --> 00:25:38,440
So n0 l delta H n1 k
over Enk 1 minus Enl 1.

325
00:25:41,760 --> 00:25:47,650
And here is another thing
that ended up working well.

326
00:25:47,650 --> 00:25:51,750
If the degeneracy
had not been broken,

327
00:25:51,750 --> 00:25:54,240
we're doing l different from k.

328
00:25:54,240 --> 00:25:58,200
If some of these
would have been 0,

329
00:25:58,200 --> 00:26:00,630
this would give you
0 in the denominator.

330
00:26:00,630 --> 00:26:04,470
This wouldn't work out.

331
00:26:04,470 --> 00:26:08,610
So it was urgent here
that all the degeneracy

332
00:26:08,610 --> 00:26:10,980
had been broken to first order.

333
00:26:10,980 --> 00:26:13,830
Otherwise, you couldn't have
computed this state this way.

334
00:26:16,990 --> 00:26:28,570
This means that n1 k, to wrap
it up here, in the subspace vn,

335
00:26:28,570 --> 00:26:33,660
is the sum over l
different from k

336
00:26:33,660 --> 00:26:35,940
because this inner
product only makes

337
00:26:35,940 --> 00:26:44,100
sense for l different from k of
n0 l times this coefficient, 1

338
00:26:44,100 --> 00:27:00,375
over Enk 1 minus Enl
1 n0 l delta H n1 k.

339
00:27:13,630 --> 00:27:15,520
It's a lot of work to get here.

340
00:27:15,520 --> 00:27:18,910
But actually, it's good
that we could get there.

341
00:27:18,910 --> 00:27:21,600
If you even want to
write it more explicitly,

342
00:27:21,600 --> 00:27:25,120
you substitute what n1 k is.

343
00:27:25,120 --> 00:27:27,270
And it goes there.

344
00:27:27,270 --> 00:27:32,060
Now, there is something a
little strange at first sight.

345
00:27:32,060 --> 00:27:38,750
If you look at it, you
say, OK, let's see.

346
00:27:38,750 --> 00:27:42,670
This is first order in
perturbation theory.

347
00:27:42,670 --> 00:27:45,740
We had to go to the
second order equation.

348
00:27:45,740 --> 00:27:49,910
And that's why we seem to
be counting orders wrong,

349
00:27:49,910 --> 00:27:53,390
because this is first
order in perturbation.

350
00:27:53,390 --> 00:27:55,310
This is another order
in perturbation.

351
00:27:55,310 --> 00:27:57,560
That's second order
in perturbation.

352
00:27:57,560 --> 00:28:01,430
That's, indeed, delta
H n1 is second order.

353
00:28:01,430 --> 00:28:04,290
Delta H n1 appear
to second order.

354
00:28:04,290 --> 00:28:07,910
So how come we get a
first order term that

355
00:28:07,910 --> 00:28:12,380
seems to depend on
second order stuff?

356
00:28:12,380 --> 00:28:15,500
Maybe it's wrong
what we did in the--

357
00:28:15,500 --> 00:28:19,280
we thought we were going to
get the degenerate first order

358
00:28:19,280 --> 00:28:21,860
piece from the second
equation, but here this

359
00:28:21,860 --> 00:28:25,824
is hitting us back.

360
00:28:25,824 --> 00:28:28,710
AUDIENCE: [INAUDIBLE]

361
00:28:28,710 --> 00:28:30,700
PROFESSOR: Yes.

362
00:28:30,700 --> 00:28:34,390
The denominator is
also first order.

363
00:28:34,390 --> 00:28:36,760
So this is second order.

364
00:28:36,760 --> 00:28:37,990
This is first order.

365
00:28:37,990 --> 00:28:40,000
The ratio is first order.

366
00:28:40,000 --> 00:28:41,350
All is good here.

367
00:28:41,350 --> 00:28:43,510
This is a good formula.

368
00:28:43,510 --> 00:28:46,100
This is a real result.

369
00:28:46,100 --> 00:28:50,440
So this completes our analysis
of degenerate perturbation

370
00:28:50,440 --> 00:28:55,900
theory when the perturbation,
the first order perturbation,

371
00:28:55,900 --> 00:28:59,040
splits all the levels.

372
00:28:59,040 --> 00:29:03,610
What we're going to do now is
degenerate perturbation theory

373
00:29:03,610 --> 00:29:05,640
when the first order
correction doesn't

374
00:29:05,640 --> 00:29:07,650
split any of the levels.

375
00:29:07,650 --> 00:29:09,090
Doesn't split them at all.

376
00:29:09,090 --> 00:29:10,220
What happens?

377
00:29:10,220 --> 00:29:12,290
What can we do?