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