1
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PROFESSOR: All right.

2
00:00:01,880 --> 00:00:04,850
So let's get now the state n1.

3
00:00:04,850 --> 00:00:08,570
So I want to make
a general remark.

4
00:00:08,570 --> 00:00:10,655
You have an equation like this.

5
00:00:14,940 --> 00:00:17,980
And you want to solve it.

6
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It's a vector equation.

7
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Operator and a vector
equal number and a vector--

8
00:00:24,430 --> 00:00:27,450
a more operator and a vector.

9
00:00:27,450 --> 00:00:31,320
To make sure you
have solved it, when

10
00:00:31,320 --> 00:00:33,480
you have a vector equation
you must make sure

11
00:00:33,480 --> 00:00:36,720
that every component-- you
can write a vector equation

12
00:00:36,720 --> 00:00:38,670
in the form vector equals zero.

13
00:00:38,670 --> 00:00:42,480
And then you must make sure that
every component of the vector

14
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is zero.

15
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What we did here
is we found what

16
00:00:46,890 --> 00:00:52,680
happens when I look at
the component along n0.

17
00:00:52,680 --> 00:00:55,620
And I figure out that,
whoops, this equation,

18
00:00:55,620 --> 00:00:58,530
when I look at the
component along n0,

19
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tells me what the energy is.

20
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So the rest of the
information of this equation

21
00:01:04,560 --> 00:01:09,090
arises when I look at
it along the components

22
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on the other states, not n0 but
the k states that we introduced

23
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from the beginning, the k0's
that run from one to infinity.

24
00:01:20,950 --> 00:01:24,660
So what we're
going to do is take

25
00:01:24,660 --> 00:01:27,960
that original--
this second equation

26
00:01:27,960 --> 00:01:54,040
and form k0 h0 minus Em0 m1 1
is equal to k0 Em1 minus delta H

27
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n0.

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So I now took the same equation
and I put it in a problem

29
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with k0.

30
00:02:04,790 --> 00:02:07,400
And I say, look, k
will be different

31
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from n, because
when we put k equal

32
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to n that already we've done.

33
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And we've learned all about it.

34
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And, in fact, n1 didn't appear.

35
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The state that we wanted
didn't appear at all.

36
00:02:23,490 --> 00:02:28,340
So now we do this
with arbitrary k.

37
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And we need to figure
out what this gives.

38
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So you have to look
at these things

39
00:02:38,170 --> 00:02:41,410
and try to remember a little
of the definitions with both.

40
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So h0, we know what
it gives from k0.

41
00:02:46,090 --> 00:02:49,040
It gives you a number,
the energy of that state.

42
00:02:49,040 --> 00:02:51,410
So this is another number.

43
00:02:51,410 --> 00:02:52,300
So that's great.

44
00:02:52,300 --> 00:03:03,810
This simplifies this
to ek0 minus En0

45
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times the overlap of k0 with n1.

46
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That's the left hand side.

47
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How about the right hand side?

48
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All right.

49
00:03:21,160 --> 00:03:25,480
Let's see what this is.

50
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First term.

51
00:03:26,500 --> 00:03:29,620
The En1 is a number,
so I must ask

52
00:03:29,620 --> 00:03:34,930
myself is what happens
when k0 meets n0?

53
00:03:34,930 --> 00:03:38,980
Well, those are our
original orthonormal states.

54
00:03:38,980 --> 00:03:42,040
And we said that k
is different from n.

55
00:03:42,040 --> 00:03:45,850
So this term is 0 with an En1.

56
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This is a number.

57
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And these two states
are orthogonal.

58
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So this term gives you a 0,
not because this number is 0,

59
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but because the overlap is 0.

60
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And I get here
minus k0 delta H n0.

61
00:04:07,640 --> 00:04:13,030
And it's good
notation to call this,

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to save writing, delta Hkn
It's a good name for it.

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It's the matrix k--

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the kn-th element of the matrix
delta H. And this is a number

65
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so I can solve k0 n1
is equal to minus delta

66
00:04:44,520 --> 00:04:51,580
Hkn divided by Ek0 minus En0.

67
00:04:54,170 --> 00:04:59,990
And this is true for
every k different for n.

68
00:04:59,990 --> 00:05:04,275
And here we find, for the first
time, our energy denominators.

69
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These energy denominators
are the things

70
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that are going to make life
interesting and difficult.

71
00:05:11,080 --> 00:05:14,870
And it answers the
question already

72
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that if you had
degenerate states,

73
00:05:18,710 --> 00:05:21,890
there would be some
k state that have

74
00:05:21,890 --> 00:05:25,250
the same energy as this one.

75
00:05:25,250 --> 00:05:28,090
And this blows up.

76
00:05:28,090 --> 00:05:31,870
And this is unsolvable
for this component.

77
00:05:31,870 --> 00:05:33,810
So you start
getting difficulties

78
00:05:33,810 --> 00:05:36,210
if you have degeneracies.

79
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As long as every k state--

80
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all the other states
of the spectrum

81
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have different energy from En,
nevermind if the other states

82
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are degenerate.

83
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They're not degenerate
with the state you care.

84
00:05:50,820 --> 00:05:54,530
You care just about one
state now, the n-th state.

85
00:05:54,530 --> 00:05:59,840
And if that's nondegenerate, all
these denominators are non-zero

86
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and you're OK.

87
00:06:03,110 --> 00:06:05,660
So here is the solution
for this thing.

88
00:06:05,660 --> 00:06:18,470
Now I can write the expressions
for the state and the energy.

89
00:06:18,470 --> 00:06:20,500
So let me do it.

90
00:06:20,500 --> 00:06:23,730
So I have this n1 like that.

91
00:06:23,730 --> 00:06:25,820
Now you can say the following.

92
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Let me do this very
deliberately first.

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n1 is equal to the sum
over all k of k0 k0 n1.

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This is the resolution
of the identity formula.

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That's the unit operator.

96
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You can always do that.

97
00:06:56,220 --> 00:07:03,780
And now you know that the
state n1 is orthogonal to n0.

98
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So this becomes the sum
over k different from n,

99
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because for k equal to n, these
are orthogonal of k0 k0 n1.

100
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And that's what we
calculated here.

101
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So what did we get?

102
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Therefore, the state n1, I can
substitute what we had there.

103
00:07:45,760 --> 00:08:02,570
It's the sum from k different
from n of k0 delta Hnk over Ek0

104
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minus En0.

105
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That's n1.

106
00:08:07,970 --> 00:08:09,550
I should have a minus sign.

107
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The minus sign is
there at the state n1.

108
00:08:20,720 --> 00:08:25,460
So the state n1 is a
complicated correction.

109
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It gets a little component
from every other state

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of the spectrum.

111
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And the coefficient depends
on the matrix element

112
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of your state with the state
you're contributing with.

113
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So you have the state n and
all the other states here.

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The amount of this state k
that enters into the correction

115
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is proportional to the matrix
element between n and k.

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If the matrix element is 0, that
state does not contribute here.

117
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And then there is the
energy denominator as well.

118
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So we're getting to the
end of this calculation.

119
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There's one more thing one
can do, which is to find--

120
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so I'm starting to
wrap up this, but still

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an important step
what we have to do.

122
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I'll get the second
order energy correction.

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What is our second
order energy correction?

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Our second order
energy correction

125
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can be found from the formula
on that blackboard, En2.

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We already found the first
order energy correction,

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which I happened to have
erased it right now.

128
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It was there.

129
00:09:51,550 --> 00:10:00,500
En2 is obtained by doing
n0 delta H times n1, which

130
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we already know.

131
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So I must do n0 delta H on that.

132
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So look what you get.

133
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You get minus the sum
over k different from n.

134
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Think of putting the n0 and the
delta H, they're all together.

135
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It's a [INAUDIBLE] so far.

136
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It's a delta H and n0.

137
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IT should go into n1.

138
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But the only state in n1 is k0.

139
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So here we have k0.

140
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And then we have delta
Hnk over Ek0 minus En0.

141
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OK.

142
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A little bit of work.

143
00:10:50,290 --> 00:10:52,436
So what is this?

144
00:10:52,436 --> 00:10:58,930
This is another matrix element.

145
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This is the matrix--

146
00:11:03,370 --> 00:11:04,300
OK.

147
00:11:04,300 --> 00:11:05,080
I'm sorry.

148
00:11:05,080 --> 00:11:08,540
Here do I have a mistake?

149
00:11:08,540 --> 00:11:10,860
Oh, yes, I have kn.

150
00:11:10,860 --> 00:11:12,460
I copied it wrong.

151
00:11:12,460 --> 00:11:13,650
It's kn.

152
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Yes.

153
00:11:16,110 --> 00:11:18,370
Yes.

154
00:11:18,370 --> 00:11:24,910
So here I have delta Hnk.

155
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but delta Hnk is this.

156
00:11:32,740 --> 00:11:36,220
If you complex conjugate--

157
00:11:36,220 --> 00:11:40,690
if you complex
conjugate delta Hkn,

158
00:11:40,690 --> 00:11:49,810
complex conjugate is k
delta H n complex conjugate,

159
00:11:49,810 --> 00:11:51,790
which changes the order.

160
00:11:51,790 --> 00:11:57,600
n delta H, which
is her mission k.

161
00:11:57,600 --> 00:11:59,480
And that's delta Hnk.

162
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So delta Hnk is equal
to delta Hkn star.

163
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And therefore the second
order energy correction

164
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has a nice formula.

165
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En2 is equal to minus the
sum over k different from n.

166
00:12:25,730 --> 00:12:31,160
Delta Hnk, which is the
star of that times this one,

167
00:12:31,160 --> 00:12:39,825
so you get delta Hnk absolute
value squared divided by Ekn.

168
00:12:43,228 --> 00:12:45,510
Ek0 minus En0.

169
00:12:53,470 --> 00:12:58,090
So we've done a lot of work.

170
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We've written the perturbation.

171
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Here is the answer.

172
00:13:02,730 --> 00:13:09,220
So far we have n of lambda
equal n0 plus lambda n1.

173
00:13:09,220 --> 00:13:12,520
n1 has been calculated.

174
00:13:12,520 --> 00:13:16,360
Energy is En0 plus lambda En1.

175
00:13:16,360 --> 00:13:18,640
That was calculated
what was just

176
00:13:18,640 --> 00:13:26,320
delta H in this state
plus lambda squared

177
00:13:26,320 --> 00:13:29,840
En2, which we have calculated.

178
00:13:29,840 --> 00:13:34,800
So this is as far as we will do
for nondegenerate perturbation

179
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theory.

180
00:13:35,300 --> 00:13:38,750
But we have found rather
interesting formulas.

181
00:13:38,750 --> 00:13:41,870
And we're going to spend
half of its lecture trying

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to understand them better.