1
00:00:00,500 --> 00:00:03,830
PROFESSOR: We have to
ask what happens here?

2
00:00:03,830 --> 00:00:12,514
This series for h of u
doesn't seem to stop.

3
00:00:12,514 --> 00:00:15,930
You go a 0, a 2, a 4.

4
00:00:15,930 --> 00:00:17,610
Well, it could go on forever.

5
00:00:17,610 --> 00:00:20,610
And what would happen
if it goes on forever?

6
00:00:25,350 --> 00:00:28,350
So if it goes on
forever, let's calculate

7
00:00:28,350 --> 00:00:38,670
what this aj plus 2 over
aj as j goes to infinity.

8
00:00:38,670 --> 00:00:43,760
Let's see how the
coefficients vary

9
00:00:43,760 --> 00:00:48,050
as you go higher and higher
up in the polynomial.

10
00:00:48,050 --> 00:00:49,550
That should be an
interesting thing.

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00:00:49,550 --> 00:00:55,360
So I pass the aj that is on
the right side and divide it,

12
00:00:55,360 --> 00:00:57,380
and now on the
right-hand side there's

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just this product of factors.

14
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And as j goes to
infinity, it's much larger

15
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than 1 or e, whatever
it is, and the 2

16
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and the 1 in the denominator.

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So this goes like
2j over j squared.

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And this goes roughly
like 2 over j.

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So as you go higher and higher
up, by the time j is a billion,

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the next term is 2
divided by a billion.

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And they are decaying,
which is good,

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but they're not
decaying fast enough.

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That's a problem.

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So let's try to figure out
if we know of a function that

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decays in a similar way.

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So you could do
it some other way.

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I'll do it this way.

28
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e to the u squared-- let's
look at this function--

29
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is this sum from n
equals 0 to infinity 1

30
00:02:12,890 --> 00:02:17,000
over n u squared to the n.

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So it's u to the 2n--

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1 over n factorial, sorry.

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00:02:27,350 --> 00:02:32,570
So now, since we have j's
and they jump by twos,

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these exponents also
here jump by two.

35
00:02:36,440 --> 00:02:38,720
So that's about right.

36
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So let's think of 2n being
j, and therefore this

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becomes a sum where j equals 0,
2, 4, and all that of 1 over--

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so even j's.

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00:02:55,410 --> 00:02:58,090
2n is equal to j--

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so j over 2 factorial over 1,
and then you have u to the j.

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So if I think of this as some
coefficient c sub j times u

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00:03:19,650 --> 00:03:27,960
to the j, we've learned
that c sub j is equal to 1

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over j divided by two factorial.

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00:03:33,690 --> 00:03:38,410
In which case, if
that is true, let's

45
00:03:38,410 --> 00:03:43,890
try to see what this
cj plus 2 over cj--

46
00:03:43,890 --> 00:03:48,980
the ratio of two consecutive
coefficients in this series.

47
00:03:48,980 --> 00:03:57,000
Well, cj plus 2 would be j plus
2 over 2 factorial, like this.

48
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That's the numerator,
because of that formula.

49
00:04:00,910 --> 00:04:05,180
And the denominator would
have just j over 2 factorial.

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Now, these factorials
make sense.

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You don't have to worry that
they are factorials of halves,

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because j is even.

53
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And therefore, the numerators
are even-- divided by 2.

54
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These are integers.

55
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These are ordinary factorials.

56
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There are factorials
of fractional numbers.

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You've seen them probably
in statistical physics

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and other fields, but we
don't have those here.

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This is another thing.

60
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So this cancels.

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If you have a number and the
number plus 1, which is here,

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you get j plus 2 over 2,
which is 2 over j plus 2.

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And that's when j
is largest, just 2

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00:04:56,970 --> 00:05:04,740
over j, which is exactly
what we have here.

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00:05:04,740 --> 00:05:10,580
So this supposedly
nice, innocent function,

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polynomial here--

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if it doesn't truncate, if
this recursive relation keeps

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producing more and more
and more terms forever--

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00:05:23,480 --> 00:05:24,950
will diverge.

70
00:05:24,950 --> 00:05:28,490
And it will diverge like so.

71
00:05:28,490 --> 00:05:39,370
If the series does
not truncate, h of u

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00:05:39,370 --> 00:05:43,505
will diverge like
e to the u squared.

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00:05:48,080 --> 00:05:51,500
Needless to say,
that's a disaster.

74
00:05:51,500 --> 00:05:56,780
Because, first, it's kind of
interesting to see that here,

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00:05:56,780 --> 00:06:02,870
yes, you have a safety factor,
e to the minus u squared over 2.

76
00:06:02,870 --> 00:06:07,490
But if h of u diverges
like e to the u squared,

77
00:06:07,490 --> 00:06:09,285
you're still in trouble.

78
00:06:09,285 --> 00:06:12,700
e to the u squared
minus u squared over 2

79
00:06:12,700 --> 00:06:15,250
is e to the plus
u squared over 2.

80
00:06:15,250 --> 00:06:19,450
And it actually coincides
with what we learned before,

81
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that any solution goes like
either plus or minus u squared

82
00:06:24,710 --> 00:06:25,280
over 2.

83
00:06:25,280 --> 00:06:30,560
So if h of u doesn't
truncate and doesn't

84
00:06:30,560 --> 00:06:32,390
become a polynomial,
it will diverge

85
00:06:32,390 --> 00:06:36,515
like e to the u squared, and
this solution will diverge

86
00:06:36,515 --> 00:06:39,700
like e to the plus
u squared over 2,

87
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which was a possibility.

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And it will not be normalizable.

89
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So that's basically the
gist of the argument.

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00:06:50,990 --> 00:06:54,020
This differential
equation-- whenever

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you work with
arbitrary energies,

92
00:06:56,570 --> 00:07:01,250
there's no reason why
the series will stop.

93
00:07:01,250 --> 00:07:06,040
Because e there will have to
be equal to 2j plus 1, which

94
00:07:06,040 --> 00:07:07,010
is an integer.

95
00:07:07,010 --> 00:07:11,530
So unless je is an
integer, it will not stop,

96
00:07:11,530 --> 00:07:14,810
and then you'll
have a divergent--

97
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well, not divergent; unbounded--

98
00:07:18,590 --> 00:07:23,720
far of u that is
impossible to normalize.

99
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So the requirement that the
solution be normalizable

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quantizes the energy.

101
00:07:31,460 --> 00:07:35,290
It's a very nice effect of
a differential equation.

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It's very nice
that you can see it

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00:07:36,770 --> 00:07:39,190
without doing
numerical experiments,

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00:07:39,190 --> 00:07:42,520
that what's going on here
is an absolute requirement

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00:07:42,520 --> 00:07:44,120
that this series terminates.

106
00:07:46,640 --> 00:07:51,210
So here, phi of u
would go like e to

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the u squared over 2,
what we mentioned there,

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and it's not a solution.

109
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So if the series must
terminate, the numerator

110
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on that box equation must
be 0 for some value of j,

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and therefore there must
exist a j such that 2j plus 1

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is equal to the energy.

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So basically, what this means
is that these unit-free energies

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00:08:43,240 --> 00:08:46,450
must be an odd integer.

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So in this case, this can be
true for j equals 0, 1, 2, 3.

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In each case, it will
terminate the series.

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With j equals 0, 1, 2, or 3
there, you get some values of e

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that the series will terminate.

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And when this series terminates,
aj plus 2 is equal to 0.

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Because look at
your box equation.

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aj, you got her number,
and then suddenly you

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get this 2j plus 1 minus e.

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And if that's 0, the
next one is zero.

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So, yes, you get
something interesting

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even for j equals 0.

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Because in that case,
you can have a0,

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but you will have no
a2, just the constant.

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So I will write it.

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So if aj plus 2 is
equal to zero, h of u

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will be aj u to the j plus
aj minus 2u to the j minus 2.

131
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And it goes down.

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The last coefficient
that exists is aj,

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and then you go down by two's.

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So let's use the
typical notation.

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00:10:28,630 --> 00:10:41,620
We call j equals n, and then
the energy is 2n plus 1.

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00:10:44,710 --> 00:10:56,100
The h is an u to the
n plus an minus 2.

137
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You do the n minus
2, and it goes on.

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If n is even, it's
an even solution.

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00:11:04,170 --> 00:11:07,990
If n is odd, it's
an odd solution.

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00:11:07,990 --> 00:11:15,690
And the energy e, remember,
was h omega over 2 times e--

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so 2n plus 1.

142
00:11:20,880 --> 00:11:26,620
So we'll move the
2 in, and e will

143
00:11:26,620 --> 00:11:31,580
be equal to h omega n plus 1/2.

144
00:11:34,160 --> 00:11:41,710
And n in all these solutions
goes from 0, 1, 2, 3.

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We can call this the energy en.

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00:11:46,890 --> 00:11:51,130
So here you see
another well-known,

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00:11:51,130 --> 00:11:57,760
famous fact that energy
levels are all evenly spaced,

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00:11:57,760 --> 00:12:01,990
h omega over 2,
one by one by one--

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00:12:01,990 --> 00:12:06,160
except that there's
even an offset for n

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00:12:06,160 --> 00:12:08,970
equals 0, which is supposed
to be the lowest energy

151
00:12:08,970 --> 00:12:10,420
state of the oscillator.

152
00:12:10,420 --> 00:12:14,530
You still have a
1/2 h bar omega.

153
00:12:14,530 --> 00:12:19,270
This is just saying that
if you have the potential,

154
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the ground state is
already a little bit up.

155
00:12:22,870 --> 00:12:25,360
You would expect that--

156
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you know there's no solutions
with energy below the lowest

157
00:12:30,880 --> 00:12:32,270
point of the potential.

158
00:12:32,270 --> 00:12:34,430
But the first solution
has to be a little bit up.

159
00:12:34,430 --> 00:12:38,000
So it's here and then
they're all evenly spaced.

160
00:12:42,214 --> 00:12:47,430
And this begins with
E0; for n equals 0, e1.

161
00:12:50,485 --> 00:12:53,800
And there's a little bit
of notational issues.

162
00:12:53,800 --> 00:12:55,930
We used to call the
ground state energy

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00:12:55,930 --> 00:12:59,060
sometimes e1, e2, e3,
going up, but this time

164
00:12:59,060 --> 00:13:03,800
it is very natural to call it
E0 because it corresponds to n

165
00:13:03,800 --> 00:13:04,540
equals 0.

166
00:13:07,460 --> 00:13:08,890
Sorry.

167
00:13:08,890 --> 00:13:09,890
Those things happen.

168
00:13:15,820 --> 00:13:17,820
No, it's not an approximation.

169
00:13:22,770 --> 00:13:29,050
It's really, in a sense,
the following statement.

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Let me remind everybody
of that statement.

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00:13:32,430 --> 00:13:36,630
When you have even
or odd solutions,

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00:13:36,630 --> 00:13:39,780
you can produce a
solution that you

173
00:13:39,780 --> 00:13:44,220
may say it's a superposition,
but it will not be an energy

174
00:13:44,220 --> 00:13:46,270
eigenstate anymore.

175
00:13:46,270 --> 00:13:51,010
Because the even solution
that stops, say, at u to the 6

176
00:13:51,010 --> 00:13:53,990
has some energy,
and the odd solution

177
00:13:53,990 --> 00:13:55,430
has a different energy.

178
00:13:55,430 --> 00:13:58,200
So these are different
energy eigenstates.

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00:13:58,200 --> 00:14:00,855
So the energy
eigenstates, we prove

180
00:14:00,855 --> 00:14:04,080
for one-dimensional
potentials, are not

181
00:14:04,080 --> 00:14:07,920
chosen to be even or
odd for bound states.

182
00:14:07,920 --> 00:14:11,490
They are either even or odd.

183
00:14:11,490 --> 00:14:15,156
You see, a superposition--

184
00:14:15,156 --> 00:14:17,790
how do we say like that?

185
00:14:17,790 --> 00:14:19,260
Here we have it.

186
00:14:19,260 --> 00:14:23,280
If this coefficient is
even, the energy sum value--

187
00:14:23,280 --> 00:14:27,340
if this coefficient is odd,
the energy will be different.

188
00:14:27,340 --> 00:14:33,110
And two energy eigenstates with
different energies, the sum

189
00:14:33,110 --> 00:14:35,580
is not an energy eigenstate.

190
00:14:35,580 --> 00:14:38,700
You can construct the general
solution by superimposing,

191
00:14:38,700 --> 00:14:40,710
but that would be
general solutions

192
00:14:40,710 --> 00:14:44,970
of the full time-dependent
Schrodinger equation, not of

193
00:14:44,970 --> 00:14:46,470
the energy eigenstates.

194
00:14:46,470 --> 00:14:50,010
The equation we're
aiming to solve there

195
00:14:50,010 --> 00:14:53,340
is a solution for
energy eigenstates.

196
00:14:53,340 --> 00:14:57,590
And although this concept I
can see now from the questions

197
00:14:57,590 --> 00:15:02,100
where you're getting,
it's a subtle statement.

198
00:15:02,100 --> 00:15:04,840
Our statement was,
from quantum mechanics,

199
00:15:04,840 --> 00:15:08,610
that when we would solve
a symmetric potential,

200
00:15:08,610 --> 00:15:13,960
the bound states would turn
out to be either even or odd.

201
00:15:13,960 --> 00:15:15,620
It's not an approximation.

202
00:15:15,620 --> 00:15:16,940
It's not a choice.

203
00:15:16,940 --> 00:15:19,010
It's something forced on you.

204
00:15:19,010 --> 00:15:22,470
Each time you find the bound
state, it's either even

205
00:15:22,470 --> 00:15:26,520
or it's odd, and this
turned out to be this case.

206
00:15:26,520 --> 00:15:30,800
You would have said the general
solution is a superposition,

207
00:15:30,800 --> 00:15:32,280
but that's not true.

208
00:15:32,280 --> 00:15:34,960
Because if you put
a superposition,

209
00:15:34,960 --> 00:15:38,640
the energy will
truncate one of them

210
00:15:38,640 --> 00:15:41,440
but will not truncate
the other series.

211
00:15:41,440 --> 00:15:43,680
So one will be bad.

212
00:15:43,680 --> 00:15:46,140
It will do nothing.

213
00:15:46,140 --> 00:15:49,980
So if this point is
not completely clear,

214
00:15:49,980 --> 00:15:53,130
please insist later,
insist in recitation.

215
00:15:53,130 --> 00:15:55,650
Come back to me office hours.

216
00:15:55,650 --> 00:16:01,398
This point should
be eventually clear.

217
00:16:01,398 --> 00:16:01,898
Good.

218
00:16:05,460 --> 00:16:08,820
So what are the names
of these things?

219
00:16:08,820 --> 00:16:12,970
These are called
Hermite polynomials.

220
00:16:12,970 --> 00:16:18,550
And so back to the
differential equation,

221
00:16:18,550 --> 00:16:20,780
let's look at the
differential equations

222
00:16:20,780 --> 00:16:23,910
when e is equal to 2n plus 1.

223
00:16:29,550 --> 00:16:32,480
Go back to the
differential equation,

224
00:16:32,480 --> 00:16:40,225
and we'll write d second
du squared Hn of u.

225
00:16:40,225 --> 00:16:49,170
That will be called the Hermite
polynomial, n minus 2udHn du

226
00:16:49,170 --> 00:16:51,740
plus e minus 1.

227
00:16:51,740 --> 00:17:00,900
But e is 2n plus 1 minus 1
is 2n Hn of u is equal to 0.

228
00:17:00,900 --> 00:17:04,360
This is the Hermite's
differential equation.

229
00:17:10,069 --> 00:17:14,010
And the Hn's are
Hermite polynomials,

230
00:17:14,010 --> 00:17:20,180
which, conventionally,
for purposes

231
00:17:20,180 --> 00:17:27,560
of doing your algebra nicely,
people figured out that Hn of u

232
00:17:27,560 --> 00:17:30,290
is convenient if--

233
00:17:30,290 --> 00:17:32,645
it begins with u
to the n and then

234
00:17:32,645 --> 00:17:36,460
it continues down
u to the n minus 2

235
00:17:36,460 --> 00:17:39,080
and all these ones here.

236
00:17:39,080 --> 00:17:44,010
But here people like it when
it's 2 to the n, u to the n--

237
00:17:44,010 --> 00:17:48,060
a normalization.

238
00:17:48,060 --> 00:17:54,890
So we know the leading
term must be u to the n.

239
00:17:54,890 --> 00:17:58,965
If you truncate with j,
you've got u to the j.

240
00:17:58,965 --> 00:18:02,670
You truncate with n,
you get u to the n.

241
00:18:02,670 --> 00:18:05,065
Since this is a linear
differential equation,

242
00:18:05,065 --> 00:18:09,490
the coefficient in
front is your choice.

243
00:18:09,490 --> 00:18:16,520
And people's choice has been
that one and has been followed.

244
00:18:16,520 --> 00:18:20,150
A few Hermite
polynomials, just a list.

245
00:18:20,150 --> 00:18:24,120
H0 is just 1.

246
00:18:24,120 --> 00:18:27,680
H1 is 2u.

247
00:18:27,680 --> 00:18:32,960
H2 is 4u squared minus 2.

248
00:18:32,960 --> 00:18:40,115
H3 is our last one, 8u
cubed minus 12u, I think.

249
00:18:40,115 --> 00:18:42,590
I have a little typo here.

250
00:18:42,590 --> 00:18:43,460
Maybe it's wrong.

251
00:18:48,740 --> 00:18:52,730
So you want to generate
more Hermite polynomials,

252
00:18:52,730 --> 00:18:58,952
here is a neat way
that is used sometimes.

253
00:19:07,076 --> 00:19:16,350
And these, too, are
generating functional.

254
00:19:16,350 --> 00:19:18,180
It's very nice actually.

255
00:19:18,180 --> 00:19:22,140
You will have in some
homework a little discussion.

256
00:19:24,810 --> 00:19:27,610
Look, you put the
variable z over there.

257
00:19:27,610 --> 00:19:31,420
What is z having to
do with anything?

258
00:19:31,420 --> 00:19:34,840
u we know, but z, why?

259
00:19:34,840 --> 00:19:38,110
Well, z is that formal
variable for what is

260
00:19:38,110 --> 00:19:39,820
called the generating function.

261
00:19:39,820 --> 00:19:45,310
So it's equal to the sum
from n equals 0 to infinity.

262
00:19:48,240 --> 00:19:51,980
And you expand it kind
of like an exponential,

263
00:19:51,980 --> 00:19:55,510
zn over n factorial.

264
00:19:55,510 --> 00:20:01,800
But there will be functions
of u all over there.

265
00:20:01,800 --> 00:20:07,090
If you expand this exponential,
you have an infinite series,

266
00:20:07,090 --> 00:20:12,260
and then you have to collect
terms by powers of z.

267
00:20:12,260 --> 00:20:14,720
And if you have
a z to the 8, you

268
00:20:14,720 --> 00:20:17,860
might have gotten from
this to the fourth,

269
00:20:17,860 --> 00:20:22,510
but you might have gotten
it from this to the 3

270
00:20:22,510 --> 00:20:26,240
and then two
factors of this term

271
00:20:26,240 --> 00:20:29,330
squared or a cross-product.

272
00:20:29,330 --> 00:20:34,040
So after all here, there
will be some function of u,

273
00:20:34,040 --> 00:20:37,970
and that function is called
the Hermite polynomial.

274
00:20:37,970 --> 00:20:40,160
So if you expand this
with Mathematica,

275
00:20:40,160 --> 00:20:43,250
say, and collect
in terms of u, you

276
00:20:43,250 --> 00:20:45,632
will generate the
Hermite polynomials.

277
00:20:49,400 --> 00:20:52,520
With this formula,
it's kind of not

278
00:20:52,520 --> 00:20:57,375
that difficult to see that
the Hermite polynomial begins

279
00:20:57,375 --> 00:20:59,140
in this way.

280
00:20:59,140 --> 00:21:01,470
And how do you
check this is true?

281
00:21:01,470 --> 00:21:06,080
Well, you would have to show
that such polynomials satisfy

282
00:21:06,080 --> 00:21:08,510
that differential
equation, and that's

283
00:21:08,510 --> 00:21:10,806
easier than what it seems.

284
00:21:10,806 --> 00:21:15,740
It might seem difficult,
but it's just a few lines.

285
00:21:15,740 --> 00:21:20,690
Now, I want you to feel
comfortable enough with this,

286
00:21:20,690 --> 00:21:24,380
so let me wrap it
up, the solutions,

287
00:21:24,380 --> 00:21:28,460
and remind you, well, you had
always u but you cared about x.

288
00:21:28,460 --> 00:21:35,300
So u was x over a.

289
00:21:35,300 --> 00:21:38,600
So let's look at
our wave functions.

290
00:21:38,600 --> 00:21:45,630
Our wave functions phi n of x
will be the Hermite polynomial

291
00:21:45,630 --> 00:21:52,190
n of u, which is
of x over a, times

292
00:21:52,190 --> 00:21:58,610
e to the minus u squared over
2, which is minus x squared

293
00:21:58,610 --> 00:22:00,443
over 2a squared.

294
00:22:00,443 --> 00:22:07,880
And you should remember that a
squared is h bar over m omega.

295
00:22:07,880 --> 00:22:11,690
So all kinds of funny factors--

296
00:22:11,690 --> 00:22:15,660
in particular,
this exponential is

297
00:22:15,660 --> 00:22:24,776
e to the minus x squared m
omega over h squared over 2.

298
00:22:24,776 --> 00:22:29,090
I think so-- m
omega over 2h bar.

299
00:22:29,090 --> 00:22:32,369
Let me write it differently--

300
00:22:32,369 --> 00:22:38,600
m omega over 2h bar x squared.

301
00:22:38,600 --> 00:22:42,360
That's that exponential, and
those are the coefficients.

302
00:22:42,360 --> 00:22:49,300
And here there should be
a normalization constant,

303
00:22:49,300 --> 00:22:50,312
which I will not write.

304
00:22:50,312 --> 00:22:51,440
It's a little messy.

305
00:22:55,088 --> 00:22:59,010
And those are the solutions.

306
00:22:59,010 --> 00:23:10,310
And the energies en were h
bar omega over 2 n plus 1/2,

307
00:23:10,310 --> 00:23:15,524
so E0 is equal to
h bar omega over 2.

308
00:23:15,524 --> 00:23:23,600
E1 is 3/2 of h bar omega, and
it just goes on like that.