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PROFESSOR: Simple
harmonic oscillator.
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So what is there about this
simple harmonic oscillator?
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Well, it's a classical
system that you
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understand perfectly well.
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An oscillator, a
spring with a mass
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oscillates and has
an energy, which
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is the kinetic energy
plus the potential energy,
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and that's p
squared over 2m plus
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1/2 m omega squared x squared.
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And this may be a
tiny bit unfamiliar,
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this way of writing it.
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But you may recall that omega
is equal to the square root of k
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over m, the so-called
spring constant,
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in which the potential in terms
of k would be 1/2 k x squared.
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And that's the potential
energy stored in a spring
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that you stretch at distance x.
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That's the total energy
of a harmonic oscillator.
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So when physicists starting
with quantum mechanics
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in the '20s decided, let's
do a harmonic oscillator,
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a quantum harmonic
oscillator, they
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had to invent the Hamiltonian.
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And the Hamiltonian they
invented was a simple one.
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They looked at
that and said, h is
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going to be p hat
squared over 2m
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plus 1/2 m omega
squared x hat squared.
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And now the difference is
going to be that x with p
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are operators, and
this is our h bar,
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and that's going to
be my quantum system.
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So this is a quantum
system that is
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inspired by classical mechanics
in the purest and simple way.
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Anyone could have invented
this quantum system.
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It was very natural.
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Still, the result
of the quantization
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is very surprising because while
this mechanical oscillator can
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oscillate with any amplitude,
the quantum oscillator
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has quantized amplitudes and
quantized energies therefore.
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So all kinds of
interesting things
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happen with this oscillator.
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Now the reason this is
also very ubiquitous is
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that this potential is exactly
a quadratic potential v of x.
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v of x is 1/2 m omega
squared x squared.
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We have x without the hat.
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This is a good approximation
to almost any system
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we consider in nature,
any oscillating system.
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Because for any potential
that has a minimum,
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there is some parabolic
approximation at the bottom.
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At the bottom the
derivative vanishes,
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so the Taylor series says that
approximately at the bottom
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is a quadratic potential.
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And therefore this
quadratic potential
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will govern the
quantum oscillations
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of a diatomic molecule,
the quantum oscillations
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of a periodic system, all
kind of quantum oscillations
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will be approximately governed
by a harmonic oscillator.
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Light has a harmonic oscillator
description for its photons.
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This Hamiltonian is the most
famous Hamiltonian there is.
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In fact, when you have
electrons in a magnetic field,
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somehow this shows up.
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And this becomes
some sort of problem
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that you solve very well,
understand very well,
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and suddenly it pops up
in all kinds of contexts.
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So we need to understand it.
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And let's go directly to the
issue of solving this problem,
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because it has many
important lessons.
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So there's two ways to solve
this problem of finding
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the bound states of
the energy eigenstates
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of the harmonic oscillator.
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This is a very
interesting potential
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because all its energy
eigenstates are bound states.
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That's not the case for the
delta function potential.
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In the delta function potential
we found one bound state,
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but they're all kind of unbound
states with positive energy.
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But this potential grows
forever, never stops growing.
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So whatever energy you
have, it is a bound state.
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It will decay.
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It will be localized.
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So you just have bound states.
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It's marvelously nice
because of that property.
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Much simpler than
anything you can imagine.
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So what do we have to do?
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We want to find the
energy eigenstates.
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So we'll write h, and
I will write phis.
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People write sometimes phis.
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Phi n of x is equal
to E phi n of x.
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And we don't know the
energy eigenstates.
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You know it's a
symmetric potential.
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It's a real potential.
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This we used to go psi
n's, but I will write them
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as phi n's as many
people do, because they
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are the harmonic oscillator
ones that are very famous.
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And we don't know--
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this could be En, energy
of the n-th state.
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And we don't know what
are the energies nor how
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the wave functions look.
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And we have to solve a
differential equation.
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As I was saying,
there's two ways
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of solving this
differential equation.
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One, treating this as
a differential equation
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and understanding
why the energy is
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quantized from the
differential equation.
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This actually gives you a lot of
insight as to what's going on,
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and it will relate to the
kind of things you're doing
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in the homework this week.
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The other way is
to be very clever
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and invent what are called
raising and lowering operators,
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and sort of solve this
whole system without solving
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the general
differential equation,
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by solving a first order very
simple differential equation,
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and then doing everything
else algebraically
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with creation and
annihilation operators.
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We will also do that.
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But we will not
develop that too far.
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That's The applications of
that method are mostly for 805.
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So we'll introduce creation and
annihilation operators, which
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are very nice and very useful.
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But we leave some
of the applications
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to coherent states of
harmonic oscillators,
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to squeezed states of harmonic
oscillators, for later.
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So this is what
we want to solve.
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So we have minus h squared over
2m d second phi n dx squared.
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And I will probably
forget about the labels.
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The labels will come later
as we solve the equation.
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Plus v of x 1/2 m omega
squared x squared phi of x,
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is equal to E phi of x.
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OK.
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This is the question
we want to solve.
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Now we're going to do one thing
first with this differential
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equation.
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We don't like all these
dimensionful constants.
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If you had to put it
in the computer, what?
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Are you going to put 6 times
10 to the minus 23 over m?
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And you won't solve a
differential equation.
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We have to clean this up.
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To clean this up,
there is a procedure
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that is guaranteed to
do the job for you.
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And the procedure is to change
the x variable into a variable
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that has no units.
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This is guaranteed to
lead you to the way
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to solve this
differential equation.
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You will be using this
throughout the semester.
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Cleaning differential
equations is a nice skill.
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And the fact is that
there's a method,
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and the method always
proceeds by first
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writing x equals au, where
this will be unit-free,
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this quantity u.
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So this will become a
differential equation
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on a unit-free
constant, which is
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ideal for your numerical
solution and is much nicer.
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But then this a to
have units of length.
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So the first thing you have to
do is look up in your problem.
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What are your constants?
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And you have a mass m, a
frequency omega, an h bar.
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And you need to find a
constant with units of length.
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Wasn't that in your test?
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I think so.
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How do you find a constant
with units of length here?
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Well, energy is equal to--
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you can write it in two ways.
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The energy can be written
as p squared over m,
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so it will be h squared over
m times a length squared.
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But from the potential,
it also has the units
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of m omega squared a squared.
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These are units,
equations for units.
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The units of energy are
these, and the units
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of energy from the second term
in the Hamiltonian are those.
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From where you get
that a squared is
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equal to h over m omega.
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So that's the constant
that you need.
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If you have that constant,
your differential equation
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becomes what?
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Well, it becomes the following.
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Let's write it out.
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It's very simple,
because x is equal to au,
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and therefore you get
minus h squared over
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2m a squared d second
phi du squared.
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x is equal to au, so it
basically just shows up here.
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Plus 1/2 m omega squared
a squared u squared phi,
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is equal to E phi.
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Now things have to work nicely.
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If you did the job well,
they have to work nicely.
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Let's think of the
units of this equation.
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Phi is here, phi is here, phi
is here, so phi is not an issue.
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The units of phi are irrelevant.
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Here is units of energy,
but u has no units.
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This has no units.
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So this must have
units of energy.
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And since u has no units,
this derivative has no units,
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and this must have
units of energy.
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And these two numbers
must be something
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nice if you
substitute a squared.
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And indeed, if you
substitute a squared here,
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this whole number
becomes h bar omega.
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And this number
becomes h bar omega
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as well, which is very nice.
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So this whole
equation has become
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minus 1/2 times h bar omega
d second phi du squared
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plus 1/2 h bar omega u
squared phi equal E phi.
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The next step is
to say, you know,
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I don't even want
these energy units.
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Even though they
don't look that bad,
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this equation looks much nicer
than the original equation
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which had all kinds
of strange units.
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So I will multiply this equation
by 2 over h omega to cancel it.
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So I get minus d second
phi du squared plus u
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squared phi is equal to
2E over h bar omega phi.
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So look at this.
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The equation is now
almost in perfect form.
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And in order to
make it perfect, I
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would say that the
right hand side--
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now, see again.
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Phi, whatever units it
has, it doesn't matter.
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It's all over the place.
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But this has no units,
this derivative,
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and this has no units,
this multiplication.
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So this must have no units.
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And indeed, you know that h
omega has units of energy,
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and that's energy.
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So it suggests that you define
a unit-free energy, free energy,
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which is 2E over h omega.
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And calculating curly
E is the same thing
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as calculating the energy,
because if you know this number
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you know the energy.
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The energy is h bar
omega over 2 times
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curly E. The advantage is that
curly E will be either 1, 2,
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1/2, a nice number, while E
is some 0.87 Ev, or things
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like that.
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So all of this equation
has been reduced
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to this very nice equation.
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Minus d second phi
du squared plus u
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squared phi is equal to E phi.
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Or, d second phi du squared is
equal to u squared minus E phi.