WEBVTT
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PROFESSOR: This was a
differential equation
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for the energy eigenstates phi.
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Supposed to be
normalizable functions.
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We looked at this equation
and decided we would first
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clean out the constant.
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We did that by replacing x
by a unit-free coordinate u.
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For that we needed a constant
that carries units of length,
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and that constant is
given by this combination
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of the constants of the problem.
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H-bar, m, and omega--
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the frequency of the oscillator.
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We also defined the
unit-free energy--
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calligraphic e.
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In terms of which
the real energies
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are given by multiples
of h omega over 2.
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So the problem has now become--
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and this whole
differential equation
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turns into this simple
differential equation.
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Simple looking--
let's say properly--
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differential equation for phi--
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as a function of u--
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which is the new
rescaled coordinate,
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and where the energy
shows up here.
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And for some reason,
this equation
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doesn't have
normalizable solutions.
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Unless those
energies are peculiar
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values that allow a
normalizable solution to exist.
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We looked at this equation
for u going to infinity,
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and realized that e to the
plus minus u squared over 2
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are the possible dependencies.
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So we said-- without
loss of generality--
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that phi could be written
as some function of u
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to be determined times
this exponential.
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And we hope for a function
that may be a polynomial.
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So that the dependence
at the infinity
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is governed by this factor.
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So with this for
the function phi,
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we substitute back into
the differential equation.
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Now the unknown is h.
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So you can take the derivatives
and find this differential
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equation for h-- a second
order differential equation.
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And that's the equation.
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It may look a little
more complicated
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than the equation
we started with,
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but it's much simpler, actually.
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There would be no polynomial
solution of this equation,
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but there may be a
polynomial solution
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of the second equation.
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So we have to solve
this equation now.
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And the way to do it is to
attempt a serious expansion.
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So we would try to write h
of u equals the sum over j.
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Equals 0 to infinity.
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Ak, u to the k.
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P equals 0 to infinity.
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Now, one way to
proceed with this
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is to plot this expansion into
the differential equation.
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You will get three sums.
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You will have to shift indices.
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It's kind of a
little complicated.
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Actually, there's a simpler
way to do this in which you
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think in the following way.
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You have this series
and you imagine
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there's a term aj, u to the j,
plus aj plus 1, u to the j plus
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1, plus aj plus 2.
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U to the j plus 2.
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And you say, let me
look at the terms with u
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to the j in the
differential equation.
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So just look at the terms
that have a u to the power j.
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So from this second--
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h vu squared-- what do we get?
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Well, to get a term that has
a u to the j, you must start--
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if you take two derivatives
and to end up with u to the j--
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you must have started with this.
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U to the j plus 2.
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So this gives you j
plus 2, j plus 1--
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taking the two derivatives--
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aj plus 2, u to the j.
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From the series, the term with
u to the j from the second hvu
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squared is this one.
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How about for minus 2u dh, du?
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Well, if I start with h and
differentiate and then multiply
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by u, I'm going to get u to
the j starting from u to the j.
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Because when I differentiate
I'll get u to the j minus 1,
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but the u will bring it back.
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So this time I get minus 2 a j--
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or minus 2.
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One derivative j.
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That's aj, u to the j.
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So it's from this
[? step. ?] Minus 2.
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You differentiate
and you get that.
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From the last term,
e minus 1 times h,
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it's clearly e minus
1 times aj u to the j.
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So these are my three terms that
we get from the differential
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equation.
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So at the end of the
day, what have we gotten?
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We've gotten j plus 2, j plus
1, aj plus 2, minus 2jaj,
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plus e minus 1 aj.
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All multiplied by u to the j.
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And that's what we
get for u to the j.
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So if you wish, for the
whole differential equation--
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all of the
differential equation--
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you get the sum from j equals
0 to infinity of these things.
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And that should
be equal to zero.
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So this is the
whole left-hand side
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of the differential equation.
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We calculated what is
the term u to the j.
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And there will be terms
from u to the zeroth
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to u to the infinity.
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So that's the whole thing.
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And we need this differential
equation to be solved.
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So this must be zero.
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And whenever you have a function
of u like a polynomial-- well,
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we don't know if it's a
polynomial-- and it stops.
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But if you have a
function of u like this,
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each coefficient must be 0.
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Therefore, we have
that j plus 2,
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times j plus 1, times aj plus
2, is equal to 2 j plus 1 minus
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e, aj.
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I set this whole combination
inside brackets to 0.
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So this term is equal to
this term and that term
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on the other side.
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You get a plus 2j.
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A plus 1 and minus e.
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So basically this is
a recursion relation.
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Aj plus 2 is equal
to 2j plus 1 minus e,
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over j plus 2, j
plus 1 times aj.
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And this is perfectly nice.
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This is what should
have happened
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for this kind of
differential equation--
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a second-order linear
differential equation.
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We get a recursion
that jumps one step.
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That's very nice.
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And this should hold
for j equals 0, 1, 2--
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all numbers.
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So when you start solving this,
there's two ways to solve it.
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You can decide, OK, let me
assume that you know a0--
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you give it.
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Give a0.
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Well, from this equation--
from a0-- you can calculate a2.
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And then from a2 you
can calculate a4.
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And successively.
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So you get a2, a4--
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all of those.
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And this corresponds to an even
solution of the differential
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equation for h.
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Even coefficients.
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Even solution for h.
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Or-- given this recursion--
you could also give a1--
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give it-- and then
calculate a3, a5--
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and those would be
an odd solution.
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So you need two
conditions to solve this.
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And those conditions
are a0 and a1,
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which is the same as specifying
the value of the function h
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at 0--
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because the value of the
function h at 0 is a0.
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And the value of the derivative
of the function at 0,
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which is a1.
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[INAUDIBLE] h of mu is a0
plus a1u plus a2u squared.
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So the derivative
at 0 is [? h1. ?]
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And that's what you must have
for solving a differential
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equation-- a second-order
differential equation for h.
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You need to know the value
of the function at zero
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and the value of the
function at the derivative
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of the function--
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at zero, as well.
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And then you can
start integrating it.
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So this first gives you a
solution a0 plus a2u squared,
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plus a4u to the fourth.
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And the second is an a1u plus
a3u cubed, plus these ones.
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So all looks pretty much OK.