WEBVTT

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PROFESSOR: We have to
ask what happens here?

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This series for h of u
doesn't seem to stop.

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You go a 0, a 2, a 4.

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Well, it could go on forever.

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And what would happen
if it goes on forever?

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So if it goes on
forever, let's calculate

00:00:28.350 --> 00:00:38.670
what this aj plus 2 over
aj as j goes to infinity.

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Let's see how the
coefficients vary

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as you go higher and higher
up in the polynomial.

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That should be an
interesting thing.

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So I pass the aj that is on
the right side and divide it,

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and now on the
right-hand side there's

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just this product of factors.

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

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

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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.

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

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is this sum from n
equals 0 to infinity 1

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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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So now, since we have j's
and they jump by twos,

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these exponents also
here jump by two.

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So that's about right.

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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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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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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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In which case, if
that is true, let's

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try to see what this
cj plus 2 over cj--

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the ratio of two consecutive
coefficients in this series.

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Well, cj plus 2 would be j plus
2 over 2 factorial, like this.

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That's the numerator,
because of that formula.

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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.

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

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These are integers.

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These are ordinary factorials.

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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.

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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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over j, which is exactly
what we have here.

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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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will diverge.

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And it will diverge like so.

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If the series does
not truncate, h of u

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will diverge like
e to the u squared.

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Needless to say,
that's a disaster.

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Because, first, it's kind of
interesting to see that here,

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yes, you have a safety factor,
e to the minus u squared over 2.

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But if h of u diverges
like e to the u squared,

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you're still in trouble.

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e to the u squared
minus u squared over 2

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is e to the plus
u squared over 2.

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And it actually coincides
with what we learned before,

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that any solution goes like
either plus or minus u squared

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over 2.

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So if h of u doesn't
truncate and doesn't

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become a polynomial,
it will diverge

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like e to the u squared, and
this solution will diverge

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like e to the plus
u squared over 2,

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which was a possibility.

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And it will not be normalizable.

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So that's basically the
gist of the argument.

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This differential
equation-- whenever

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you work with
arbitrary energies,

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there's no reason why
the series will stop.

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Because e there will have to
be equal to 2j plus 1, which

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is an integer.

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So unless je is an
integer, it will not stop,

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and then you'll
have a divergent--

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well, not divergent; unbounded--

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far of u that is
impossible to normalize.

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So the requirement that the
solution be normalizable

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quantizes the energy.

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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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without doing
numerical experiments,

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that what's going on here
is an absolute requirement

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that this series terminates.

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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.

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So if the series must
terminate, the numerator

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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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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.

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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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We call j equals n, and then
the energy is 2n plus 1.

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The h is an u to the
n plus an minus 2.

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

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And the energy e, remember,
was h omega over 2 times e--

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so 2n plus 1.

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So we'll move the
2 in, and e will

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be equal to h omega n plus 1/2.

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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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So here you see
another well-known,

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famous fact that energy
levels are all evenly spaced,

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h omega over 2,
one by one by one--

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except that there's
even an offset for n

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equals 0, which is supposed
to be the lowest energy

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state of the oscillator.

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You still have a
1/2 h bar omega.

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This is just saying that
if you have the potential,

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the ground state is
already a little bit up.

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You would expect that--

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

00:12:30.880 --> 00:12:32.270
point of the potential.

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But the first solution
has to be a little bit up.

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So it's here and then
they're all evenly spaced.

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And this begins with
E0; for n equals 0, e1.

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And there's a little bit
of notational issues.

00:12:53.800 --> 00:12:55.930
We used to call the
ground state energy

00:12:55.930 --> 00:12:59.060
sometimes e1, e2, e3,
going up, but this time

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it is very natural to call it
E0 because it corresponds to n

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equals 0.

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Sorry.

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Those things happen.

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No, it's not an approximation.

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It's really, in a sense,
the following statement.

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Let me remind everybody
of that statement.

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When you have even
or odd solutions,

00:13:36.630 --> 00:13:39.780
you can produce a
solution that you

00:13:39.780 --> 00:13:44.220
may say it's a superposition,
but it will not be an energy

00:13:44.220 --> 00:13:46.270
eigenstate anymore.

00:13:46.270 --> 00:13:51.010
Because the even solution
that stops, say, at u to the 6

00:13:51.010 --> 00:13:53.990
has some energy,
and the odd solution

00:13:53.990 --> 00:13:55.430
has a different energy.

00:13:55.430 --> 00:13:58.200
So these are different
energy eigenstates.

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So the energy
eigenstates, we prove

00:14:00.855 --> 00:14:04.080
for one-dimensional
potentials, are not

00:14:04.080 --> 00:14:07.920
chosen to be even or
odd for bound states.

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They are either even or odd.

00:14:11.490 --> 00:14:15.156
You see, a superposition--

00:14:15.156 --> 00:14:17.790
how do we say like that?

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Here we have it.

00:14:19.260 --> 00:14:23.280
If this coefficient is
even, the energy sum value--

00:14:23.280 --> 00:14:27.340
if this coefficient is odd,
the energy will be different.

00:14:27.340 --> 00:14:33.110
And two energy eigenstates with
different energies, the sum

00:14:33.110 --> 00:14:35.580
is not an energy eigenstate.

00:14:35.580 --> 00:14:38.700
You can construct the general
solution by superimposing,

00:14:38.700 --> 00:14:40.710
but that would be
general solutions

00:14:40.710 --> 00:14:44.970
of the full time-dependent
Schrodinger equation, not of

00:14:44.970 --> 00:14:46.470
the energy eigenstates.

00:14:46.470 --> 00:14:50.010
The equation we're
aiming to solve there

00:14:50.010 --> 00:14:53.340
is a solution for
energy eigenstates.

00:14:53.340 --> 00:14:57.590
And although this concept I
can see now from the questions

00:14:57.590 --> 00:15:02.100
where you're getting,
it's a subtle statement.

00:15:02.100 --> 00:15:04.840
Our statement was,
from quantum mechanics,

00:15:04.840 --> 00:15:08.610
that when we would solve
a symmetric potential,

00:15:08.610 --> 00:15:13.960
the bound states would turn
out to be either even or odd.

00:15:13.960 --> 00:15:15.620
It's not an approximation.

00:15:15.620 --> 00:15:16.940
It's not a choice.

00:15:16.940 --> 00:15:19.010
It's something forced on you.

00:15:19.010 --> 00:15:22.470
Each time you find the bound
state, it's either even

00:15:22.470 --> 00:15:26.520
or it's odd, and this
turned out to be this case.

00:15:26.520 --> 00:15:30.800
You would have said the general
solution is a superposition,

00:15:30.800 --> 00:15:32.280
but that's not true.

00:15:32.280 --> 00:15:34.960
Because if you put
a superposition,

00:15:34.960 --> 00:15:38.640
the energy will
truncate one of them

00:15:38.640 --> 00:15:41.440
but will not truncate
the other series.

00:15:41.440 --> 00:15:43.680
So one will be bad.

00:15:43.680 --> 00:15:46.140
It will do nothing.

00:15:46.140 --> 00:15:49.980
So if this point is
not completely clear,

00:15:49.980 --> 00:15:53.130
please insist later,
insist in recitation.

00:15:53.130 --> 00:15:55.650
Come back to me office hours.

00:15:55.650 --> 00:16:01.398
This point should
be eventually clear.

00:16:01.398 --> 00:16:01.898
Good.

00:16:05.460 --> 00:16:08.820
So what are the names
of these things?

00:16:08.820 --> 00:16:12.970
These are called
Hermite polynomials.

00:16:12.970 --> 00:16:18.550
And so back to the
differential equation,

00:16:18.550 --> 00:16:20.780
let's look at the
differential equations

00:16:20.780 --> 00:16:23.910
when e is equal to 2n plus 1.

00:16:29.550 --> 00:16:32.480
Go back to the
differential equation,

00:16:32.480 --> 00:16:40.225
and we'll write d second
du squared Hn of u.

00:16:40.225 --> 00:16:49.170
That will be called the Hermite
polynomial, n minus 2udHn du

00:16:49.170 --> 00:16:51.740
plus e minus 1.

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.

00:17:00.900 --> 00:17:04.360
This is the Hermite's
differential equation.

00:17:10.069 --> 00:17:14.010
And the Hn's are
Hermite polynomials,

00:17:14.010 --> 00:17:20.180
which, conventionally,
for purposes

00:17:20.180 --> 00:17:27.560
of doing your algebra nicely,
people figured out that Hn of u

00:17:27.560 --> 00:17:30.290
is convenient if--

00:17:30.290 --> 00:17:32.645
it begins with u
to the n and then

00:17:32.645 --> 00:17:36.460
it continues down
u to the n minus 2

00:17:36.460 --> 00:17:39.080
and all these ones here.

00:17:39.080 --> 00:17:44.010
But here people like it when
it's 2 to the n, u to the n--

00:17:44.010 --> 00:17:48.060
a normalization.

00:17:48.060 --> 00:17:54.890
So we know the leading
term must be u to the n.

00:17:54.890 --> 00:17:58.965
If you truncate with j,
you've got u to the j.

00:17:58.965 --> 00:18:02.670
You truncate with n,
you get u to the n.

00:18:02.670 --> 00:18:05.065
Since this is a linear
differential equation,

00:18:05.065 --> 00:18:09.490
the coefficient in
front is your choice.

00:18:09.490 --> 00:18:16.520
And people's choice has been
that one and has been followed.

00:18:16.520 --> 00:18:20.150
A few Hermite
polynomials, just a list.

00:18:20.150 --> 00:18:24.120
H0 is just 1.

00:18:24.120 --> 00:18:27.680
H1 is 2u.

00:18:27.680 --> 00:18:32.960
H2 is 4u squared minus 2.

00:18:32.960 --> 00:18:40.115
H3 is our last one, 8u
cubed minus 12u, I think.

00:18:40.115 --> 00:18:42.590
I have a little typo here.

00:18:42.590 --> 00:18:43.460
Maybe it's wrong.

00:18:48.740 --> 00:18:52.730
So you want to generate
more Hermite polynomials,

00:18:52.730 --> 00:18:58.952
here is a neat way
that is used sometimes.

00:19:07.076 --> 00:19:16.350
And these, too, are
generating functional.

00:19:16.350 --> 00:19:18.180
It's very nice actually.

00:19:18.180 --> 00:19:22.140
You will have in some
homework a little discussion.

00:19:24.810 --> 00:19:27.610
Look, you put the
variable z over there.

00:19:27.610 --> 00:19:31.420
What is z having to
do with anything?

00:19:31.420 --> 00:19:34.840
u we know, but z, why?

00:19:34.840 --> 00:19:38.110
Well, z is that formal
variable for what is

00:19:38.110 --> 00:19:39.820
called the generating function.

00:19:39.820 --> 00:19:45.310
So it's equal to the sum
from n equals 0 to infinity.

00:19:48.240 --> 00:19:51.980
And you expand it kind
of like an exponential,

00:19:51.980 --> 00:19:55.510
zn over n factorial.

00:19:55.510 --> 00:20:01.800
But there will be functions
of u all over there.

00:20:01.800 --> 00:20:07.090
If you expand this exponential,
you have an infinite series,

00:20:07.090 --> 00:20:12.260
and then you have to collect
terms by powers of z.

00:20:12.260 --> 00:20:14.720
And if you have
a z to the 8, you

00:20:14.720 --> 00:20:17.860
might have gotten from
this to the fourth,

00:20:17.860 --> 00:20:22.510
but you might have gotten
it from this to the 3

00:20:22.510 --> 00:20:26.240
and then two
factors of this term

00:20:26.240 --> 00:20:29.330
squared or a cross-product.

00:20:29.330 --> 00:20:34.040
So after all here, there
will be some function of u,

00:20:34.040 --> 00:20:37.970
and that function is called
the Hermite polynomial.

00:20:37.970 --> 00:20:40.160
So if you expand this
with Mathematica,

00:20:40.160 --> 00:20:43.250
say, and collect
in terms of u, you

00:20:43.250 --> 00:20:45.632
will generate the
Hermite polynomials.

00:20:49.400 --> 00:20:52.520
With this formula,
it's kind of not

00:20:52.520 --> 00:20:57.375
that difficult to see that
the Hermite polynomial begins

00:20:57.375 --> 00:20:59.140
in this way.

00:20:59.140 --> 00:21:01.470
And how do you
check this is true?

00:21:01.470 --> 00:21:06.080
Well, you would have to show
that such polynomials satisfy

00:21:06.080 --> 00:21:08.510
that differential
equation, and that's

00:21:08.510 --> 00:21:10.806
easier than what it seems.

00:21:10.806 --> 00:21:15.740
It might seem difficult,
but it's just a few lines.

00:21:15.740 --> 00:21:20.690
Now, I want you to feel
comfortable enough with this,

00:21:20.690 --> 00:21:24.380
so let me wrap it
up, the solutions,

00:21:24.380 --> 00:21:28.460
and remind you, well, you had
always u but you cared about x.

00:21:28.460 --> 00:21:35.300
So u was x over a.

00:21:35.300 --> 00:21:38.600
So let's look at
our wave functions.

00:21:38.600 --> 00:21:45.630
Our wave functions phi n of x
will be the Hermite polynomial

00:21:45.630 --> 00:21:52.190
n of u, which is
of x over a, times

00:21:52.190 --> 00:21:58.610
e to the minus u squared over
2, which is minus x squared

00:21:58.610 --> 00:22:00.443
over 2a squared.

00:22:00.443 --> 00:22:07.880
And you should remember that a
squared is h bar over m omega.

00:22:07.880 --> 00:22:11.690
So all kinds of funny factors--

00:22:11.690 --> 00:22:15.660
in particular,
this exponential is

00:22:15.660 --> 00:22:24.776
e to the minus x squared m
omega over h squared over 2.

00:22:24.776 --> 00:22:29.090
I think so-- m
omega over 2h bar.

00:22:29.090 --> 00:22:32.369
Let me write it differently--

00:22:32.369 --> 00:22:38.600
m omega over 2h bar x squared.

00:22:38.600 --> 00:22:42.360
That's that exponential, and
those are the coefficients.

00:22:42.360 --> 00:22:49.300
And here there should be
a normalization constant,

00:22:49.300 --> 00:22:50.312
which I will not write.

00:22:50.312 --> 00:22:51.440
It's a little messy.

00:22:55.088 --> 00:22:59.010
And those are the solutions.

00:22:59.010 --> 00:23:10.310
And the energies en were h
bar omega over 2 n plus 1/2,

00:23:10.310 --> 00:23:15.524
so E0 is equal to
h bar omega over 2.

00:23:15.524 --> 00:23:23.600
E1 is 3/2 of h bar omega, and
it just goes on like that.