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ROBERT FIELD: Last lecture, I did a fairly standard treatment
of the harmonic oscillator, which is not
supposed to make you excited, but just
to see that you can do this.
And the path to the solution was to define
dimensionless position coordinate,
and then you get a dimensionless Schrodinger equation.
And the solution of that involves two steps.
One is to insist that the solutions to the Schrodinger
equation have an exponentially damped form.
And then the Schrodinger equation
is transformed into a new equation called the Hermite
equation.
You shouldn't get all excited about that-- the mathematicians
take care of it.
And so the solutions to this Hermite differential equation
gives you a set of orthogonal and normalized wave functions.
They give you the energy levels expressed as a quantum number
plus 1/2, times constant.
The energy levels with the quantum number v
are even or odd in v, and they're even or odd in psi.
Even v corresponds to even psi.
V is the number of internal nodes,
and there are all sorts of things you can do to say,
well, I expect if I know how to do certain things
in classical mechanics, they're are
going to come out pretty much the same way in quantum
mechanics.
So that's the standard structure,
but more importantly, I want you to have in your head
the pictures of the wave functions, the idea that there
is a zero point energy, and that there's a reason for that--
that the wave functions have tails extending out
into the non-classical, or the classically forbidden region.
And this turns out to be the beginning of tunneling.
You'll be looking at tunneling more specifically.
I also want you to know how the spacing of nodes is,
and that involves generalization of the bright idea
that the wavelength is h over p.
But if the potential is not constant,
then p is a function of x.
This is not the quantum mechanical operator,
this is a function which provides you
with a lot of intuition.
And then if you know where the node spacings are, and you
know the shape of the envelope, you
have basically everything you need
to have a classical sense of what's going on.
And then-- I guess it's supposed to be hidden--
I did a little bit of semiclassical theory,
and I showed that if you integrate from the left turning
point to the right turning point at a given
energy of this momentum function,
you get h over 2 times the number of nodes.
And this is the semiclassical quantization--
it's incredibly important, and it's
useful either as an exact or approximate result
for all one-dimensional problems.
And so it tells you how to begin.
Now, before I start talking about what
we're going to do today, I want to stress where we're going.
So we're going to be looking at some exactly solved problems.
And so we have a particle in a box, the harmonic oscillator,
the hydrogen atoms-- you have to count them--
and the rigid rotor.
Now, all of these problems have an infinite number
eigenfunctions, an infinite number of energy levels,
and that's intimidating, but it's true.
Now, these infinite number of functions
are explicit functions of the quantum number.
And so we have an infinite number,
but in order to describe systems,
we're going to be calculating integrals.
We're going to be calculating a lot of integrals
between these infinite number of functions.
So we have an infinity squared of integrals.
Well, that shouldn't scare you because what
I'm going to show you is that all of the integrals that we
are going to encounter are explicit functions
of the quantum numbers, and they have
relatively selection rules.
In other words, which integrals are non-zero
based on the difference in quantum
numbers between the left-hand side and the right-hand side?
So we're collecting these things in order
to calculate a whole bunch of stuff.
Now, I told you that this is a course for use
rather than philosophy or history.
And so if you encounter any quantum mechanical problem,
you'd like to be able to describe
what you could observe with it.
And so if you're armed with the infinite number of energy
levels and eigen solutions for our problem,
you can calculate any property.
So you define some property you're interested in--
there is a quantum mechanical operator that
corresponds to that property.
And in order to be able to describe observations
of that property, you need to calculate
integrals of that operator.
Well, la dee dah.
That should be intimidating, but it's not
because almost all of these integrals
can be expressed as a simple constant times
a function of the quantum numbers
or the difference of quantum numbers,
and that's a fantastic thing.
So we have any operator--
suppose the Hamiltonian is an exactly solved problem
plus something else, which we'll call h1.
And this is a complexity in the--
or it's the reality in the problem.
And in order to deal with this, again, you're
going to need to calculate integrals of this operator.
And the last thing that's really going to be exciting
is once we look at the time dependent Schrodinger equation,
we're going to get wave packets.
And these are functions of position and time,
and these wave packets are classical-like, localized
objects that move following the Newton's equations
of motion with the center of the wave packet.
And again, there are a whole bunch
of integrals you're going to need in order
to do these things.
And so right now, we're starting with the best problem
for these integrals, because a harmonic oscillator
has some special properties.
And the lecture notes are incredibly tedious,
and they're mostly proofs.
And I'm going to try to go fast over the tedious stuff,
and give you the important ideas,
but since there is some important logic,
you should really look at these notes.
So what we're going to be doing today
is we start with the coordinate momentum operators,
we're going to get these operators
in dimensionless form, and then we're
going to get these a and a-dagger guys.
So this step is reminiscent of what
I did at the beginning of the previous lecture,
and then this is magic because this magic enables
you to just look at integrals and say,
I know that integral is zero, or I know that area is not zero.
And with a little bit more effort--
maybe something that you'd put on the back of a postage
stamp--
you can evaluate that integral, not by knowing integral tables,
but by knowing the properties of these simple little a
and a-dagger.
And that's a fantastic thing.
And it's so fantastic that this is
one of the reasons why almost all problems in quantum
mechanics start with a harmonic oscillator approximation,
because there is so much you can do with this a and a-dagger
formalism.
Now, at the beginning I also told you
that in quantum mechanics, the important thing that
contains everything we're allowed to know about a system
is the wave function.
But I also told you we can never measure the wave function.
We can never experimentally determine it,
and so we need to be able to calculate what this wave
function does as far as what we can observe,
and these a's and a-daggers are really important in being
able to do that.
So I'm going to start with covering
what I did in the notes, but I'm going to jump to final results
at some point--
governed by the clock.
And so the first thing we're going to do is these.
And so what we do is we define the relationship
between the ordinary position coordinate.
And this little twiddle means it's dimensionless,
and so we can write the inverse of that.
And that's the one we are going to want to--
well, actually, we go both ways.
And we do the same thing for the momentum--
p is equal to h-bar mu omega square
root, p twiddle and the inverse which I don't need to write.
And finally, we get the Hamiltonian,
which is p squared over 2 mu plus 1/2 kx squared.
And we'll put that into these new units.
So we have h-bar mu omega over 2 mu, p twiddle squared
plus k over 2.
This is all very tedious, but the payoff is very soon.
k over 2 times hr mu omega, x twiddle squared.
Oh, isn't that interesting?
We can combine-- we can absorb a k over mu in omega,
and so we get, actually, a big simplification.
We get h bar omega over 2 times p twiddle
squared plus x twiddle squared.
Well, that looks simpler.
And so the next thing we do is-- it looks like a simple problem
from algebra, let's factor this.
Now, it's a little tricky because you
know you can factor something in real terms
if this is a minus sign.
But we are allowed to talk about complex quantities,
so we can factor that.
And so this term, p twiddle squared plus x twiddle squared
is equal to ip twiddle, plus x twiddle times minus ip twiddle,
plus x.
And you can work that out-- that ip times minus ip is p squared,
and x times x is x squared.
And then we have these cross-terms, ip times
x and x times minus ip.
Whoops, they don't commute.
If this were algebra--
well, they would go away, but they don't.
And so what you end up getting is
p twiddle squared plus x twiddle squared, plus i times p--
I'm going to stop writing the twiddles.
So we have this.
I want to make sure that I haven't sabotaged myself--
that's going to be-- yeah, that's right.
So we have something here that isn't zero.
And it looks like i times the commutator of p twiddle
with x twiddle.
But we can work that out because we
know the commutator of p ordinary with x ordinary.
And so I did that.
And so we have this commutator, p twiddle, x twiddle.
After some algebra, we get plus 1.
A number-- pure number--
no?
I want you to check my algebra.
So you just substitute in what this
is in terms of ordinary p and the ordinary x.
Use a commutator for ordinary xp, which is ih-bar,
and magically, you get plus 1.
So this very strange and boring derivation says, OK--
well, let's now give these two things names.
This guy, we're going to call as the square root of 2 times a,
and this one is going to be the square root
of 2 times a-dagger.
So H is going to be h-bar omega over 2,
times square root of 2a-hat times
the square root of 2a-dagger-hat minus 1.
Remember, when we factored it, we
got this extra term which was 1.
And in order to make it correct, we have to subtract it away.
And so this becomes h-bar omega, a-dagger-hat minus 1/2.
Well, isn't that nice?
Now, we have the Hamiltonian expressed as a constant, which
we know is important because it's related to the energy
levels, and times these two little things,
which turn out to be the gift from God.
It's an incredible thing, what these do.
So we have gone through some algebra,
and we know the relationship between a, and the x and p
twiddles, and similarly for a-dagger.
And we can go in the other direction,
and we know the commutator, and now we're
going to start doing some really great stuff.
Well, one thing we're going to want to know about is a-hat.
a-hat, a-dagger, that commutator-hat.
And that turns out to be--
well, I already derived it-- it turns out to be plus 1.
And as a result, we can say things like this-- a a-dagger--
So using this trick, we can show we can always replace something
like a, a-dagger by a-dagger a plus this commutator, which
is 1.
And so we have this really neat way
of reversing the order of the a's and a-daggers.
So with this, we're going to soon discover
the a operating on the eigenfunction gives square root
of v times psi v minus 1.
And a-dagger operating on this wave function
gives v plus 1 square root of psi v plus 1.
Which is the reason these things are valuable,
because if you have any eigenfunction,
you can get all the others.
So suppose you have the lowest eigenfunction--
you apply a-dagger on it as many times
as you need to get to, say vth function.
So you don't actually--
you're not going to be evaluating integrals, you're
going to be counting a's and a-daggers,
and permuting them around, and getting 1's, and stuff
like that.
Yes?
AUDIENCE: In this line with the commutator,
you didn't move the dagger.
ROBERT FIELD: I didn't what?
AUDIENCE: For a, a times the square root of a a-dagger plus
a-dagger a, it should be a a-dagger.
And on the right-hand side, you need to move the dagger.
ROBERT FIELD: OK, so this is to switch the order,
and I've done that.
And that then is--
no, I think-- wait a second.
So we have a a-dagger--
so that's a a-dagger minus a-dagger a, and that's--
oh, yeah.
Thank you.
It's very, very easy to get lost, and once you're lost,
you can never be found because you've
made a mistake that's built into your logic,
and you're never going to see it.
You see it took me a couple of minutes to even accept--
that the insight from my TA who's sitting there calmly
thinking, and I'm trying to do several things in addition
to the thinking.
So we can do things like this.
Suppose we have psi--
I can't use this notation yet.
So suppose we have psi-star v, and we
have a-dagger, a-dagger, a-dagger, psi, v prime, dx.
These are raising operators, so this
is going to take v prime to v prime plus 3.
That's the only integral that's not 0.
And we get v prime plus 1, v prime plus 2, v prime plus 3,
square what?
It has the constants.
And this would be v prime plus 3.
So instead of evaluating an integral,
looking at what the x's and p's are,
we just have a little game we play.
So now, we have to prove some of the things I've said.
So we have h, and we're going to operate on a-dagger psi v.
So what does the Hamiltonian do to this thing?
So what we're going to want to do
is to show that this thing is an eigenvalue--
eigenfunction of v plus 1, and that's
what we are going to get.
So let's just go through this.
So we have h-bar omega, a-dagger a plus 1/2, times psi v.
So what I did is--
where did I-- yeah, I showed that the Hamiltonian--
or did I not do that yet?
Oh, yeah-- I did it right here.
The Hamiltonian is h-bar omega, a a-dagger minus 1/2,
or if we reverse these, it's equal to h-bar omega, a-dagger,
a plus 1/2.
So we can use either one, so I'm using that one--
except I wanted an a-dagger here,
because we want to show what the Hamiltonian does to this.
Now, we can pull in a-dagger out to the right, because this--
if it's 1/2 times a-dagger, well, that doesn't matter.
This a-dagger, a, a-dagger-- well,
we can pull this a-dagger out.
So we have h-bar omega, a-dagger is equal to a a-dagger plus
1/2--
so iv.
Now, we use our magic commutator trick
to replace this by a-dagger a plus 1.
So now, we have h-bar omega, a-dagger,
and we have a-dagger a plus three halves is iv.
Well, this is Ev plus h-bar omega.
We've increased the number that started here.
Here is Ev-- that was Ev plus 1.
And so now, we have no operators in here,
and we can stick the a-dagger back here.
And so we have h-bar omega, Ev plus 1, a-dagger
Well, what do we have here?
We have an operator, we have this function,
we have some constant times the same function.
So what we've shown is that this thing
is an eigenfunction of the Hamiltonian
that belongs to the eigenvalue Ev plus 1.
We've increased the energy by 1.
So what we have--
so we can show that we apply a-dagger
to any function-- we increase its energy,
and we can do this forever.
We could also do a similar thing if we apply a to psi v. We
can go down, but at some point, we run out of steam
because we've gone to the lowest energy, and if we go lower,
we get 0.
So a operating on psi min gives 0.
So we have this stack of energy levels and wave functions,
and we have the same stack being repeated as we go down,
but this one has an end.
We bring back what a is, and so a psi min is 0--
that's the equation.
We bring in what a is, and it's ip twiddle x plus x twiddle.
So we do some algebra, and what we end up
with is a differential equation, psi min.
dxx twiddle is equal to--
again, a little bit more algebra--
minus mu omega over h-bar times psi min.
So what function gives--
there's an x in here too.
So what function has a derivative,
which is the function you had started
with, times the variable, times a constant?
And so the answer to that is that psi min
has to have the form--
some normalization factor times e to the minus m omega--
or mu omega-- sorry--
over 2 h-bar x squared--
a Gaussian.
Well, it had to be a Gaussian, right?
We know when we did the algebra that we're
going to get some function times a Gaussian.
But for the lowest function, the Hermite polynomial is 1,
and all there is is the Gaussian.
And so we found the lowest level, and we can normalize it.
So let's start over here.
So what we have found is psi min of x
is equal to mu omega over pi h-bar to the 1/4 power, e
to the minus mu omega over 2 h-bar squared.
Well, that's useful.
We knew that, but this time we got it out
of a completely different path.
And now, we can get all higher v by
a-dagger, a-dagger, et cetera.
So remember, we don't care anything
about what the function is, we just
know that we can bring it in and get rid of it at will,
because what we want is the values of integrals involving
that function and some operator.
So yeah, we can have all of those functions,
and this is a way of generating all of the functions.
And so if we wanted psi v, we would do a-dagger to the vth
power divided by v-dagger--
v-- what do you call this with an exclamation point?
Factorial-- ha!
So we apply this operator that raises us to whatever level
we want starting from this Gaussian at the bottom,
and we have this normalization factor which cancels out
the fact the stuff that you get by applying av.
Now, there is some more logic in my notes,
and I don't want to do that, but what
we'd like to be able to show is that a-dagger on psi v
gives some constant, and that this constant has
some value-- we're going to evaluate what it is.
And similarly, a psi v gives dv, and some constant v minus that.
We can derive those things, and I'm not going
to waste time deriving them--
I'm going to just give you the values.
But we already know that cv is square of v plus 1/2--
e plus 1 and dv--
and you can see the derivation in my notes.
I don't think going through them is going to be instructive.
And that's just going to be v and 1/2.
So now, we have something that's wonderful,
because everything you need to know about getting numbers
concerning harmonic oscillator is obtained from these five
equations.
a-dagger on psi v is v plus 1 square root psi v plus 1.
a on psi v is v square root psi, v minus 1.
I've said this before, but these are the most useful things
you'll ever encounter.
We have this thing called the number operator,
and that number operator is a-dagger-hat,
and the number operator operating on psi v
gives v psi v. And so that's a kind of benign operator
that can suck up all sorts of factors of a-dagger a,
because it just gives a useful thing.
And then we have a, a-dagger, and this is 1.
Well, you sort of know it's going
to be 1, because a a-dagger gives an increase-- it gives v
plus 1, and a gives v minus 1.
So it's plus 1, not minus 1-- you know that it's hardwired.
Well, I did it--
I got to the point where it starts to get interesting.
So we're going to be using this notation, a-dagger and a,
for all sorts of stuff.
And one sort of thing is transition intensities
and selection rules.
So you have a harmonic oscillator.
A harmonic oscillator is, say a diatomic molecule
which is heteronuclear.
And so as the molecule vibrates, you
have a dipole moment which is oscillating.
And so any oscillating electric field
will grab a hold of that dipole moment
and stretch or compress it, especially
if that field is in resonance with h-bar omega.
And I've got some beautiful animation showing this,
but we can't do that until we have time dependent quantum
mechanics.
So we have a time dependent radiation
field, which is going to interact
with the dipole associated with the vibrating molecule,
and it's going to cause transitions.
And so we can write the quantum mechanical operator
that causes the transitions--
this is the electric dipole moment operator--
as a function of coordinate.
And we can do a power series expansion of this,
and we have--
so we have mu 0-- the constant term--
the first derivative of the dipole
with respect to x, and the second derivative of the dipole
with respect to x.
And we have the x cofactor and the x squared cofactor.
And so this guy doesn't have any x on it-- it's a constant.
The only integrals involving--
the only integrals are delta v, v prime following the selection
rule delta v v prime.
So these integrals are 0 unless v and v prime are the same.
And that says, well, an isolating field
isn't going to do anything key to it,
it's just going to leave it in the same vibration level.
But it might have an electric Stark effect,
but that's something else.
So this term does nothing as far as vibration is concerned.
This guy, which is a plus a-dagger,
has a selection rule, delta v of plus and minus 1,
and this guy has a selection rule, delta v
of plus and minus 2 and 0.
So if we're interested in the intensities
of vibrational transitions, it says,
well, this is the important term and it causes transitions,
changing the vibrational quantum number by one, which
is called the fundamental.
This gives rise to overtones.
So all of a sudden, we're in real problem land, where
if we're looking at vibrational transitions in a molecule,
that this enables us to calculate what's important,
or to say these are the intensities I measure,
and these are the first and second derivative
of the dipole moment operator as a function of internuclear
distance.
Isn't that neat?
I've gone so fast, I'm more or less at the end of my notes,
but I can blather on for a while.
So suppose you have some integral involving
an operator and a vibrational wave function.
So we have psi v star, some operator, psi v prime dx.
And we'd like to know how to focus our energies.
We're very busy people--
we don't want any value integrals
that come out to be 0, we'd like to just know.
So if this operator is some function of x or function of p,
we'd have a power series expansion of the operator,
and we then know what the selection rules are.
So usually, you look at the operator
and you find that it's a linear quadratic cubic function of x--
the leading term is usually linear.
Bang-- you have a delta v of plus 1--
selection rule.
Or if someone has bothered to actually convert
the operator to some form--
oh, it's the operator--
this might have some form, a-dagger cubed times
some constant.
So if the operator looks like a-dagger cubed,
we know that the selection rule is v to v plus 3,
and we know that the matrix of the integral is v plus 1,
times v plus 2, times v plus 3, square root of that,
times the constant.
So there is a huge number of problems
that, instead of being pages and pages of algebra,
are just reduced something that you can tell by inspection.
So one of the tricks is we have an operator
like x squared or x cubed--
what we want to do is write this in terms
of a squared, a-dagger squared, and maybe
some combination of a-dagger a.
So we want to take the a-dagger a's with the a a-daggers
and combine them using the commutation rule.
And then we have expressed this in this maximally simple form,
and then you just apply a squared,
apply a-dagger squared, and you apply this,
then you've got the value of the integral.
So if you're a busy person and you
want to actually calculate stuff,
you want to know how to reduce operators--
usually expressed as some power of the coordinate
or the momentum--
into a sum of terms involving these organized
products of a's and a-daggers.
And you're going to be absolutely shocked at how
perturbation theory--
which leads to basically all of the formulas and spectroscopy--
it's an ugly theory, but it reduces everything to things
that you can just write down at the speed of your pen
or pencil, and that's a fantastic thing.
So you can't do this with the particle in a box,
you can't do this with a hydrogen atom,
you can't do this with the rigid--
well, you can do some of this with a rigid rotor.
But the harmonic oscillator is so ubiquitous
because every one-dimensional problem
is harmonic at the bottom.
And so you can use it and then you can put in the corrections.
But also because you want to describe dynamics,
you almost always use the harmonic oscillator,
because not only do you know the integrals,
but you know there's only a few.
Normally, you're going to be summing
over an infinite number of quantum numbers,
and that takes time, and it takes judgment to say,
well, only certain of these are important.
But for the harmonic oscillator, the sums are finite.
All of these things are wonderful,
and that's why whenever you look at a theory,
you're going to discover that hidden in there
is the harmonic oscillator approximation,
because everything is doable in no effort.
And sometimes when you look at a paper like that,
it doesn't show you the intermediate steps,
because everybody knows what a harmonic oscillator does.
And there's also a lot of insight
because something like this--
this is an odd symmetry term, this is an even symmetry term,
and there are all sorts of things that have to do with,
are you're conserving symmetry or changing symmetry?
And sometimes, the issue is how does the molecule spontaneously
change symmetry by doing something
interacting with a field, or interacting
with some feature of the potential surface.
So this is a place where it's labor saving and insight
generating, and it's really amazing.
So maybe I've bored you with this,
but this is the beginning of almost every theory
that you encounter just because of the simplicity of the a's
and a-daggers.
OK, I'm done.
Thank you.