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PROFESSOR: So it's a
beautiful recording.
OK, so to get
started, questions?
From last time?
Barton covered for me last time.
I fled.
I was out of town.
I was at a math conference.
It was pretty surreal.
Questions, yes.
AUDIENCE: [INAUDIBLE]
about the exam?
PROFESSOR: About--?
AUDIENCE: The exam.
PROFESSOR: The exam.
Yes, absolutely.
So the exam is, as you all
know, on Thursday, a week hence.
So on Tuesday we
will have a lecture.
The material Tuesday will
not be covered on the exam.
The exam will be a
review of everything
through today's lecture,
including the problems that,
which for some
technical reason I
don't know why
didn't get posted.
But it should be up
after lecture today.
The exam will be a
combination of short questions
and computations.
It will not focus on an
enormous number of computations.
It will focus more
on conceptual things.
But there will be a few
calculations on the exam.
And I will post some
practice problems
over the next couple of days.
AUDIENCE: Do we have a
problem set [INAUDIBLE]?
PROFESSOR: You do have a
problem set due Tuesday.
And that is part of your
preparation for the exam.
Here's a basic strategy
for exams for this class.
Anything that's on a
problem set is fair game.
Anything that's not
covered on a problem set
is not going to be fair game.
If you haven't seen
a new problem on it,
broadly construed, then you
won't-- I won't test you
on a topic you haven't
done problems on before.
But I will take
problems and ideas
that you've studied before and
spin them slightly differently
to make you think through
them in real time on the exam.
OK?
From my point of view,
the purpose of these exams
is not to give you a grade.
I don't care about the grade.
The purpose of these
exams is to give you
feedback on your understanding.
It's very easy to slip through
quantum mechanics and think,
oh yeah, I totally-- I got this.
This is fine.
But it's not always
an accurate read.
So that's the point.
Did that answer your question?
Other questions?
Exam or-- yeah.
AUDIENCE: About the harmonic
oscillator actually.
PROFESSOR: Excellent.
AUDIENCE: So when
we solved it Tuesday
using the series method,
so there are two solutions
technically, the even solution
and the odd term solution.
So did boundary conditions
force the other one
to be completely zero, like
the coefficient in front of it?
So there's like an A0 term which
determines all the other ones.
But there's an A0 term and an
A1 term for the evens and odds.
So did the other ones
just have to be 0?
PROFESSOR: This is a
really good question.
This is an excellent question.
Let me ask the question
slightly differently.
And tell me if this
is the same question.
When we wrote down our
differential equation--
so last time we did the
harmonic oscillator.
And Barton did give you
the brute force strategy
for the harmonic oscillator.
We want to find the
energy eigenstates,
because that's what we do to
solve the Schrodinger equation.
And we turn that into a
differential equation.
And we solve this
differential equation
by doing an asymptotic analysis
and then a series expansion.
Now, this is a second order
differential equation.
Everyone agree with that?
It's a second order
differential equation.
However, in our
series expansion we
ended up with one integration
constant, not two.
How does that work?
How can it be that there was
only one integration constant
and not two?
It's a second order
differential equation.
Is this he question?
AUDIENCE: Yeah.
PROFESSOR: OK, and this
is an excellent question.
Because it must be true,
that there are two solutions.
It cannot be that there
is just one solution.
It's a second order
differential equation.
Their existence in uniqueness
theorems, which tell us there
are two integration constants.
So how can it possibly be
that there was only one?
Well, we did something
rather subtle
in that series expansion.
For that series expansion
there was a critical moment,
which I'm not going
to go through but you
can come to my office
hours again, but just
look through the notes.
There's an important
moment in the notes when
we say, aha, these terms matter.
But what we did is we
suppressed a singular solution.
There's a solution of
that differential equation
which is not well-behaved,
which is not smooth,
and in particular
which diverges.
And we already did, from
the asymptotic analysis,
we already fixed that the
asymptotic behavior was
exponentially falling.
But there's a second solution
which is exponentially growing.
So what we did, remember
how we did this story?
We took our wave
function and we said,
OK, look, we're
going to pull off--
we're going to first
asymptotic analysis.
And asymptotic analysis
tells us that either we
have exponentially growing
or exponentially shrinking
solutions.
Let's pick the exponentially
shrinking solutions.
So phi e is equal
to e to the minus
x over 2a squared
squared, times some--
I don't remember what
Barton called it.
I'll call it u of x.
So we've extracted, because
we know that asymptotically it
takes this form.
Well, it could also
take the other form.
It could be e to the
plus, which would
be bad and not normalizable.
We've extracted that,
and then we write down
the differential equation for u.
And then we solve that
differential equation
by series analysis, yeah?
However, if I have a secondary
differential equation for phi,
this change of variables
doesn't change the fact
that it's a secondary
differential equation for u,
right?
There's still two
solutions for u.
One of those solutions
will be the solution
of the equation that has
this asymptotic form.
But the other solution
will be one that has an e
to the plus x squared
over a squared
so that it cancels off
this leading factor
and gives me the exponentially
growing solution.
Everyone cool with that?
So in that series
analysis there's
sort of a subtle
moment where you
impose that you have
the convergent solution.
So the answer of, why did we
get a first order relation,
is that we very carefully,
although it may not
have been totally obvious,
when doing this calculation
one carefully chooses the
convergent solution that
doesn't have this
function blowing up so
as to overwhelm the envelope.
That answer your question?
AUDIENCE: Yep.
PROFESSOR: Great.
It's a very good question.
This is an important
subtlety that
comes up all over the place
when you do asymptotic analysis.
I speak from my heart.
It's an important
thing in the research
that I'm doing
right now, getting
these sorts of subtleties right.
It can be very confusing.
It's important to think
carefully through them.
So it's a very good question.
Other questions
before we move on?
OK.
So I'm going to erase
this, because it's not
directly germane,
but it is great.
OK, so one of the lessons
of this brute force analysis
was that we constructed
the spectrum, i.e., the set
of energy eigenvalues allowed
for the quantum harmonic
oscillator, and we constructed
the wave functions.
We constructed
the wave functions
by solving the
differential equation
through asymptotic analysis,
which give us the Gaussian
envelope, and series
expansion, which
give us the Hermite polynomials.
And then there's some
normalization coefficient.
And then we got the energy
eigenvalues by asking,
when does this series
expansion converge?
When does it, in fact
truncate, terminate,
so that we can write
down an answer?
And that was what gave
us these discrete values.
But fine, we can see that
it would be discrete values.
We're cool with that.
In fact, Barton went through the
discussion of the node theorem
and the lack of degeneracy
in one dimensional quantum
mechanics.
So it's reasonable that we
get a bunch of discrete energy
eigenvalues, as we've talked
about now for two lectures.
However, there's
a surprise here,
which is that these
aren't just discrete,
they're evenly spaced.
We get a tower, starting with
the lowest possible energy
corresponding to a--
sorry, E0-- starting
with the lowest possible
energy, which is greater than 0,
and a corresponding ground
state wave function.
And then we have a whole bunch
of other states, phi 1, phi 2,
phi 3, phi 4, labeled
by their energies
where the energies
are evenly spaced.
They needed to be discrete,
because these are bound states.
But evenly spaced is a surprise.
So why are they evenly spaced?
Anyone, based on the
last lecture's analysis?
Yeah, you don't
have a good answer
to that from last
lecture's analysis.
It's one of the
mysteries that comes out
of the first analysis.
When you take a
differential equation,
you just beat the crap out of
it with a stick by solving it.
With differential equations
strategies like this
you don't necessarily get some
of the more subtle structure.
One of the goals
of today's lecture
is going to be to explain
why we get this structure.
Why just from the
physics of the problem,
the underlying
physics, should you
know that the system is going to
have evenly spaced eigenvalues?
What's the structure?
And secondly, I want to
show you a way of repeating
this calculation without doing
the brute force analysis that
reveals some of that more fine
grain structure of the problem.
And this is going
to turn out to be
one of the canonical
moves in the analysis
of quantum mechanical systems.
So from quantum mechanics
to quantum field theory
this is a basic
series of logic moves.
What I'm going to do today
also has an independent life
in mathematics, in algebra.
And that will be
something you'll
studying in more detail in
8.05, but I would encourage
you to ask your recitation
instructors about it,
or me in office hours.
So our goal is to understand
that even spacing and also
to re-derive these
results without the sort
of brutal direct assault
methods we used last time.
So what I'm going to
tell you about today
is something called
the operator method.
It usually goes under the
name of the operator method.
To get us started let's go back
to look at the energy operator
for the harmonic
oscillator, which
is what, at the end of
the day, we want to solve.
p squared over 2m plus m omega
squared upon 2, x squared.
And this is the operator that we
want to-- whose eigenvalues we
want to construct,
whose eigenfunctions
we want to construct.
Before we do anything else, we
should do dimensional analysis.
First thing you when
you look at a problem
is do some dimensional analysis.
Identify the salient
scales and make things,
to the degree possible,
dimensionless.
Your life will be better.
So what are the parameters
we have available to us?
We have h bar, because
it's quantum mechanics.
We have m, because we have
a particle of mass, m.
We have omega,
because this potential
has a characteristic
frequency of omega.
What other parameters do
we have available to us?
Well, we have c.
That's available to us.
But is it relevant?
No.
If you get an answer that
depends on the speed of light,
you made some horrible mistake.
So not there.
What about the number
of students in 8.04?
No.
There are an infinite
number of parameters
that don't matter
to this problem.
What you want to know is, when
you do dimensional analysis,
what parameters matter
for the problem.
What parameters could possibly
appear during the answer?
And that's it.
There are no other
parameters in this problem.
So that's a full set of
parameters available to us.
This has dimensions of
momentum times length.
This has dimensions of mass,
and this has dimensions of one
upon the time.
And so what
characteristic scales can
we build using these
three parameters?
Well, this is a
moment times a length.
If we multiply by a mass,
that's momentum times mass
times x, which is
almost momentum again.
We need a velocity
and not a position,
but we have 1 over time.
So if we take h bar
times and omega,
so that's px times
m over t, that
has units of momentum squared.
And similarly, this
is momentum which
is x, which is length
mass over time.
I can divide by mass
and divide by frequency
or multiply by time,
so h bar upon m omega.
And this is going to have
units of length squared.
And with a little bit of
foresight from factors of two
I'm going to use these to
define two link scales.
x0 is equal to h
bar-- I want to be
careful to get my coefficients.
I always put the two
in the wrong place.
So 2 h bar upon m omega.
Square root.
And I'm going to define p0 as
equal to square root of 2 h bar
times m omega.
So here's my claim.
My claim is at the end of
the day the salient link
scales for this
problem should be
integers or dimensionless
numbers times this link scale.
And salient momentum scales
should be this scale.
Just from dimensional analysis.
So if someone at
this point says,
what do you think is
the typical scale, what
is the typical size of
the ground state wave
function, the typical link scale
over which the wave function is
not 0?
Well, that can't possibly
be the size of Manhattan.
It's not the size of a proton.
There's only one link scale
associated with the system.
It should be of order x0.
Always start with
dimensional analysis.
Always.
OK, so with that we can
rewrite this energy.
Sorry, and there's
one last energy.
We can write an energy,
the thing with energy,
which is equal to h bar omega.
And this times a frequency
gives us an energy.
So we can rewrite
this energy operator
as h bar omega times
p squared over p0.
So this has units of energy.
So everything here
must be dimensionless.
And it turns out to be p
squared over p0 squared
plus x operator squared
over x0 squared.
So that's convenient.
So this has nothing to do
with the operator method.
This is just being reasonable.
Quick thing to note, x0 times p0
is just-- the m omegas cancel,
so we get root 2h
bar squared, 2 h bar.
Little tricks like
that are useful to keep
track of as you go.
So we're interested in
this energy operator.
And it has a nice form.
It's a sum of squares.
And we see the sum of squares,
a very tempting thing to do
is to factor it.
So for example, if I have
two classical numbers,
c squared plus d squared,
the mathematician in me
screams out to write c
minus id times c plus id.
I have factored this.
And that's usually a step
in the right direction.
And is this true?
Well yes, it's true. c
squared plus d squared
and the cross terms cancel.
OK, that's great.
Four complex numbers,
or four C numbers.
Now is this true for operators?
Can I do this for operators?
Here we have the energy
operator as a sum of squares.
Well, let's try it.
I'd like to write that
in terms of x and p.
So what about writing the
quantity x minus ip over x0
over p0, operator, times
x over x0 plus ip over p0.
We can compute this.
This is easy.
So the first term gives us
the x squared over x0 squared.
That last term gives
us-- the i's cancel,
so we get p squared
over p0, squared.
But then there are cross terms.
We have an xp and a minus
px, with an overall i.
So plus i times xp
over x0 times p0.
x0 times p0,
however, is 2 h bar.
So that's over 2 h bar.
And then we have the
other term, minus px.
Same thing.
So I could write that as
a commutator, xp minus px.
Everyone cool with what?
Unfortunately this is
not what we wanted.
We wanted just p
squared plus x squared.
And what we got
instead was p squared
plus x squared close
plus a commutator.
Happily this
commutator is simple.
What's the commutator
of x with p?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah, exactly.
Commit this to memory.
This is your friend.
So this is just i
h bar, so this is
equal to ditto plus
ditto plus i h bar.
Somewhere I got a minus sign.
Where did I get my minus
sign wrong? x with ip.
Oh now, good.
This is good.
So x with p is i--
so we get an i h bar.
No, I really did
screw up the sign.
How did I screw up the sign?
No I didn't.
Wait.
Oh!
Of course.
No, good.
Sorry, sorry.
Trust your calculation,
not your memory.
So the calculation gave us this.
So what does this give us?
It gives us i h bar.
So plus.
But the i h bar times i is
going to give me a minus.
And the h bar is
going to cancel,
because I've got an h bar
from here and an h bar
at the denominator minus 1/2.
So this quantity is equal to the
quantity we wanted minus 1/2.
And what is the
quantity we wanted,
x0 squared plus p0 squared?
This guy.
So putting that
all together we can
write that the energy
operator, which
was equal to h bar omega
times the quantity we wanted,
is equal to-- well,
the quantity we wanted
is this quantity plus 1/2.
h bar omega times--
I'll write this
as x over x0 plus ip over p0,
x over x0 plus ip over p0, hat,
hat, hat, plus 1/2.
Everyone cool with that?
So it almost worked.
We can almost factor.
So at this point it's
tempting to say, well
that isn't really
much an improvement.
You've just made it uglier.
But consider the following.
And just trust me on this one,
that this is not a stupid thing
to do.
That's a stupid symbol
to write, though.
So let's define
an operator called
a, which is equal to x
over x0 plus ip over p0,
and an operator, which
I will call a dagger.
"Is that a dagger
I see before me?"
Sorry. x over x0
minus ip over p0.
Hamlet quotes are harder.
So this is a dagger.
And we can now write
the energy operator
for the harmonic oscillator
is equal to h bar omega times
a dagger a plus 1/2.
Everyone cool with that?
Now, this should
look suggestive.
You should say, aha, this looks
like h bar omega something
plus 1/2.
That sure looks familiar from
our brute force calculation.
But, OK, that familiarity is
not an answer to the question.
Meanwhile you should
say something like this.
Look, this looks kind of
like the complex conjugate
of this guy.
Because there's an i and you
change the sign of the i.
But what is the complex
conjugate of an operator?
What does that mean?
An operator is like take
a vector and rotate.
What is the complex
conjugation of that?
I don't know.
So we have to define that.
So I'm now going to start
with this quick math aside.
And morally, this
is about what is
the complex conjugate
of an operator.
But before I move on, questions?
OK.
So here's a mathematical
series of a facts and claims.
I claim the following.
Given any linear operator we can
build-- there's a natural way
to build without making
any additional assumptions
or any additional ingredients.
We can build another
operator, o dagger, hat, hat,
in the following way.
Consider the inner
product of f with g,
or the bracket of f with g.
So integral dx of f
complex conjugate g.
Consider the function
we're taking here
is actually the
operator we have on g.
I'm going to define my-- so
this is a perfectly good thing.
What this expression says
is, take your function g.
Act on it with the operator o.
Multiply by the
complex conjugate.
Take the integral.
This is what we
would have done if we
had taken the inner product
of f with the function we get
by taking o and acting
on the function g.
So here's the thing.
What we want-- just an
aside-- what we want to do
is define a new operator.
And here's how I'm
going to define it.
We can define it by
choosing how it acts.
I'm going to tell you
exactly how it acts,
and then we'll
define the operator.
So this operator, o with a
dagger, called the adjoint,
is defined in the following way.
This is whatever
operator you need, such
that the integral
gives you-- such
that the following is true.
Integral dx o on f,
complex conjugate, g.
So this is the
definition of this dagger
action, the adjoint action.
OK so o dagger is the adjoint.
And sometimes it's called
the Hermetian adjoint.
I'll occasionally say
Hermetian and occasionally not,
with no particular order to it.
So what does this mean?
This means that
whatever o dagger is,
it's that operator
that when acting on g
and then taking the
inner product with f
gives me same answer as
taking my original operator
and acting on f and taking
the inner product with g.
Cool?
So we know how-- if we know
what our operator o is,
the challenge now is going to be
to figure out what must this o
dagger operator be such that
this expression is true.
That's going to be my
definition of the adjoint.
Cool?
So I'm going to do
a bunch of examples.
I'm going to walk through this.
So the mathematical
definition is
that an operator o
defined in this fashion
is the Hermetian adjoint of o.
So that's the
mathematical definition.
Well, that's our version of
the mathematical definition.
I just came back from
a math conference,
so I'm particularly chastened
at the moment to be careful.
So let's do some quick examples.
Example one.
Suppose c is a complex number.
I claim a number is
also an operator.
It acts by multiplication.
The number 7 is an operator
because it takes a vector
and it gives you 7
times that vector.
So this number is a particularly
simple kind of operator.
And what's the adjoint?
We can do that.
That's easy.
So c adjoint is going to be
defined in the following way.
It's integral dx f
star of c adjoint g
is equal to the integral dx of c
on f complex conjugate times g.
But what is this?
Well, c is just a number.
So when we take its
complex conjugate
we can just pull it out.
So this is equal to the
integral dx c complex conjugate,
f star g.
But I'm now going
to rewrite this,
using the awesome power of
reordering multiplication,
as c star.
And I'm going to put
parentheses around this
because it seems like fun.
So now we have this
nice expression.
The integral dx of
f c adjoint g is
equal to the integral dx of f c
star g, c complex conjugate g.
But notice that this
must be true for all f.
It's true for all.
Because I made no
assumption about what
f and g are, true
for all f and g.
And therefore the adjoint
of a complex number
is its complex conjugate.
And this is the basic
strategy for determining
the adjoint of any operator.
We're going to play
exactly this sort of game.
We'll put the adjoint in here.
We'll use the definition
of the adjoint.
And then we'll do
whatever machinations
are necessary to rewrite
this as some operator acting
on the first factor.
Cool?
Questions?
OK, let's do a more
interesting operator.
By the way, to
check at home, and I
think this might be
on your problem set--
but I don't remember
if it's on or not.
So if it's not, check
this for yourself.
Check that the
adjoin of the adjoint
is equal to the operator itself.
It's an easy thing to check.
So next example.
What is the adjoint of
the operator derivative
with respect to x?
Consider the operator, which
is just derivative with respect
to x.
And I want to know what is
the adjoint of this beast.
So how do we do this?
Same logic as before.
Whatever the operator is, it's
defined in the following way.
Integral dx, f complex
conjugate on dx dagger on g.
This is equal to the integral
dx of-- how we doing on time?
Good-- integral dx of dx,
f complex conjugate on g.
Now, what we want is we want
to turn this into an expression
where the operator
is acting on g, just
as our familiar operator ddx.
So how do I get
the ddx over here?
I need to do two things.
First, what's the
complex conjugate
of the derivative with respect
to x of a complex function?
MIT has indigestion.
So this is integral
dx, derivative
with respect to x of f
complex conjugate, g.
And now I want this operator.
I want derivative acting on g.
That's the definition.
Because I want to know
what is this operator.
And so I'm going to do
integration by parts.
So this is equal to
the integral, dx.
When I integrate by
parts I get an F complex
conjugate, and then an overall
minus sign from the integration
by parts minus f
complex conjugate dx g.
Was I telling you
the truth earlier?
Or did I lie to you?
OK, keep thinking about that.
And this is equal
to, well the integral
of dx, f complex conjugate
if minus the derivative
with respect to x acting on g.
Everyone cool with that?
But if you look at
these equalities,
dx adjoint acting on g is the
same as minus dx acting on g.
So this tells me that
the adjoint of dx
is equal to minus dx.
Yeah?
AUDIENCE: Are you assuming
that your surface terms vanish?
PROFESSOR: Thank you!
I lied to you.
So I assumed in this that
my surface terms vanished.
I did a variation by parts.
And that leaves me with
a total derivative.
And that total derivative
gives me a boundary term.
Remember how integration
by parts works.
Integration by parts
says the integral of AdxB
is equal to the
integral of-- well,
AdxB can be written as
derivative with respect
to x of AB minus B derivative
with respect to x of A.
Because this is A prime B plus B
prime A. Here we have AB prime.
So we just subtract off
the appropriate term.
But this is a total derivative.
So it only gives
us a boundary term.
So this integral is equal
to-- can move the integral
over here-- the integral
and the derivative,
because an integral is
nothing but an antiderivative.
The integral and the
derivative cancel,
leaving us with
the boundary terms.
And in this case, it's
from our boundaries
which are minus infinity
plus infinity, minus infinity
and plus infinity.
Now, this tells us
something very important.
And I'm not going to speak
about this in detail,
but I encourage the
recitation instructors
who might happen
to be here to think
to mention this in recitation.
And I encourage you
all to think about it.
If I ask you, what
is the adjoint
of the derivative
operator acting
on the space of functions
which are normalizable,
so that they vanish
at infinity, what
is the adjoint of the
derivative operator acting
on the space of functions which
is normalizable at infinity?
We just derive the answer.
Because we assume that
these surface terms vanish.
Because our wave functions,
f and g, vanish at infinity.
They're normalizable.
However, if I had asked you a
slightly different question,
if I had asked you,
what's the adjoint
of the derivative
operator acting
on a different set
of functions, the set
of functions that
don't necessarily
vanish at infinity, including
sinusoids that go off
to infinity and don't vanish.
Is this the correct answer?
No.
This would not be
the correct answer,
because there are
boundary terms.
So the point I'm making
here, first off, in physics
we're always going to be talking
about normalizable beasts.
At the end of the day, the
physical objects we care about
are in a room.
They're not off infinity.
So everything is going
to be normalizable.
That is just how
the world works.
However, you've got to
be careful in making
these sorts of
arguments and realize
that when I ask you, what is
the adjoint of this operator,
I need to tell you
something more precise.
I need to say,
what's the adjoint
of the derivative acting when
this operator's understood
as acting on some
particular set of functions,
acting on normalizable
functions?
Good.
So anyway, I'll leave that
aside as something to ponder.
But with that
technical detail aside,
as long as we're talking
about normalizable functions
so these boundary terms from
the integration by parts cancel,
the adjoint of the
derivative operator
is minus the
derivative operator.
Cool?
OK, let's do another example.
And where do I want to do this?
I'll do it here.
So another example.
Actually, no.
I will do it here.
So we have another
example, which is three.
What's the adjoint of
the position operator?
OK, take two minutes.
Do this on a piece of
paper in front of you.
I'm not going to call on you.
So you can raise you hand if
you-- OK, chat with the person
next to you.
I mean chat about
physics, right?
Just not-- [LAUGHS].
AUDIENCE: [CHATTING]
PROFESSOR: OK, so how do
we go about doing this?
We go about solving
this problem by using
the definition of the adjoint.
So what is x adjoint?
It's that operator
such that the following
is true, such that the integral
dx of f complex conjugate
with x dagger acting on g
is equal to the integral dx
of what I get by taking the
complex conjugate of taking
x and acting on f and then
integrating this against g.
But now we can use
the action of x
and say that this is equal
to the integral dx of x f
complex conjugate g.
But here's the nice thing.
What is the complex
conjugate of f
times the complex function of f?
x is real.
Positions are real.
So that's just x times the
complex conjugate of f.
So that was
essential move there.
And now we can
rewrite this as equal
the integral of f
complex conjugate xg.
And now, eyeballing
this, x dagger
is that operator
which acts by acting
by multiplying with little x.
Therefore, the adjoint
of the operator x
is equal to the same operator.
x is equal to its own adjoint.
OK?
Cool?
So we've just learned a
couple of really nice things.
So the first is-- where
we I want to do this?
Yeah, good.
So we've learned a
couple of nice things.
And I want to encode them
in the following definition.
Definition-- an
operator, which I
will call o, whose
adjoint is equal to o,
so an operator whose
adjoint is equal to itself
is called Hermetian.
So an operator which is
equal to its own adjoint
is called Hermetian.
And so I want to note a couple
of nice examples of that.
So note a number which
is Hermetian is what?
Real.
An operator-- we found
an operator which
is equal to its own adjoint.
x dagger is equal to x.
And what can you say about the
eigenvalues of this operator?
They're real.
We use that in the
proof, actually.
So this is real.
I will call an operator
real if it's Hermetian.
And here's a
mathematical fact, which
is that any operator
which is Hermetian
has all real eigenvalues.
So this is really-- I'll
state it as a theorem,
but it's just a fact for us.
o has all real eigenvalues.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah?
AUDIENCE: Is it if
and only [INAUDIBLE]?
PROFESSOR: No.
Let's see.
If you have all
real eigenvalues,
it does not imply
that you're Hermetian.
However, if you have
all real eigenvalues
and you can be
diagonalized, it does imply.
So let me give you an example.
So consider the
following operator.
We've done this
many times, rotation
in real three-dimensional
space of a vector
around the vertical axis.
It has one eigenvector,
which is the vertical vector.
And the eigenvalue
is 1, so it's real.
But that's not enough
to make it Hermetian.
Because there's another
fact that we haven't
got to yet with
Hermetian operators,
which is going to tell us
that a Hermetian operator has
as many eigenvectors
as there are dimensions
in the space, i.e., that the
eigenvectors form a basis.
But there's only one
eigenvector for this guy,
even though we're in a
three-dimensional vector space.
So this operator, rotation
by an angle theta,
is not Hermetian, even
though its only eigenvalue
isn't in fact real.
So it's not an only if.
If you are Hermetian, your
eigenvalues are all real.
And you'll prove this
on a problem set.
Yeah?
AUDIENCE: If you're Hermetian,
are your eigenfunctions normal?
PROFESSOR: Not necessarily.
But they can be made normal.
We'll talk about this
in more detail later.
OK.
Let's do a quick
check, last example.
And I'm not actually going
to go through this in detail,
but what about p?
What about the
momentum operator?
First off, do you think
the momentum is real?
It sure would be nice.
Because its eigenvalues are the
observable values of momentum.
And so its eigenvalues
should all be real.
Does that make it Hermetian?
Not necessarily,
but let's check.
So what is the adjoint of p?
Well, this actually
we can do very easily.
And I'm not going to go
through an elaborate argument.
I'm just going to
know the following.
p is equal to h bar upon i ddx.
And this is an operator.
This is an operator.
So what's the adjoint
of this operator?
Well, this under an adjoint
gets a minus sign, right?
It's itself up to a minus sign.
So is the derivative Hermetian?
No, it's in fact what
we anti-Hermetian.
Its adjoint is minus itself.
What about i?
What's its adjoint?
Minus i.
Sweet.
So this has an adjoint,
picks up a minus.
This has an adjoint,
picks up a minus.
The minuses cancel.
p adjoint is p.
So p is in fact Hermetian.
And here's a stronger
physical fact.
So now we've seen that each
of the operators we built
is x and p, true
of the operators
we've looked at so
far is Hermetian,
those that correspond
to physical observables.
Here's a physical fact.
All the observables you
measure with sticks are real.
And the corresponding
statement is
that all operators
corresponding to observables,
all operators must be Hermetian.
To the postulate that
says, "Observables
are represented by
operators," should
be adjoined the
word "Hermetian."
Observables are represented
in quantum mechanics
by Hermetian operators,
which are operators
that have a number
of nice properties,
including they have
all real eigenvalues.
Cool?
OK.
Questions?
Yeah.
AUDIENCE: If it has to be
Hermetian and not just have
real eigenvalues, does that
mean the eigenvalues always
need to form some kind of basis?
PROFESSOR: Yeah, the
eigenvectors will.
This is connected to the
fact we've already seen.
If you take an
arbitrary wave function
you can expand it in states
with definite momentum
as a superposition.
You can also expand it in a set
of states of definite energy
or of definite position.
Anytime you have a
Hermetian operator,
its eigenvectors suffice
to expand any function.
They provide a basis for
representing any function.
So that's the end of
the mathematical side.
Let's get back to
this physical point.
So we've defined this operator
a and this other operator
a dagger.
And here's my question first.
Is a Hermetian?
No.
That's Hermetian.
That's Hermetian.
But there's an i.
That i will pick
up the minus sign
when we do the
complex conjugation.
Oh, look.
Sure was fortuitous that
I called this a dagger,
since this is equal to a dagger.
So this is the adjoint of a.
So this immediately tells
you something interesting. x
and p are both observables.
Does a correspond
to an observable?
Is it Hermetian?
Every intervals is associated
to a Hermetian operator.
This is not Hermetian.
So a does not represent
an observable operator.
And I will post notes on
the web page, which give us
a somewhat lengthy
discussion-- or it
might be in one
of the solutions--
a somewhat lengthy
discussion of what
it means for a and a dagger
to not be observable.
You'll get more
discussion of that there.
Meanwhile, if a is not
observable, it's not Hermetian,
does it have real eigenvalues?
Well, here's an important thing.
I said if you're Hermetian,
all the eigenvalues are real.
If you're not Hermetian,
that doesn't tell you
you can't have any
real eigenvalues.
It just says that I
haven't guaranteed for you
that all the
eigenvalues are real.
So what we'll discover
towards the end of the course
when we talk about something
called coherent states is that
in fact, a does have a
nice set of eigenvectors.
They're very nice.
They're great.
We use them for lasers.
They're very useful.
And they're called
coherent states.
But their eigenvalues
are not in general real.
They're generically
complex numbers.
Are they things you can measure?
Not directly.
They're related to things
you can measure, though,
in some pretty nice ways.
So why are we bothering
with these guys
if they're not observable?
Yeah.
AUDIENCE: E [INAUDIBLE].
PROFESSOR: Yeah, good.
Excellent.
That's really good.
So two things about it.
So one thing is this form
for the energy operator
is particularly simple.
We see the 1/2.
This looks suggestive
from before.
But it makes it obvious
that E is Hermetian.
And that may not be
obvious to you guys.
So let's just check.
Here's something that you'll
show on the problem set.
AB adjoint is equal to
B adjoint A adjoint.
The order matters.
These are operators.
And so if we take the
adjective of this,
what's this going to give us?
Well we change the order.
So it's going to be
a dagger, and then we
take the dagger of
both of the a dagger.
So this is self-adjoint,
or Hermetian.
So that's good.
Of course, we already
knew that, because we
could have written it
in terms of x and p.
But this is somehow simpler.
And it in particular
emphasizes the form,
or recapitulates the form
of the energy eigenvalues.
Why else would we care
about a and a dagger?
OK, now this is a good moment.
Here's the second reason.
So the first reason
you care is this sort
of structural
similarity and the fact
that it's nicely Hermetian
in a different way.
Here's the key thing.
Key.
a and a dagger satisfy the
simplest commutation relation
in the world.
Well, the second simplest.
The simplest is that it's
0 on the right-hand side.
But the simplest not trivial
commutation relationship.
a with a dagger is equal to--
so what is a dagger equal to?
We just take the definition.
Let's put this in.
So this is x over x0
plus ip over p0, comma,
x over x0 minus i, p over p0,
hat, hat, hat, hat, bracket,
bracket.
Good.
So here there are
going to be four terms.
There's x commutator x.
What is that?
What is the commutator of
an operator with itself?
0.
Because remember the
definition of the commutator A,
B is AB minus BA.
So A with A is equal
to AA minus AA.
And you have no options there.
That's 0.
So x with x is 0. p with p is 0.
So the only terms that
matter are the cross terms.
We have an x with p.
And notice that's going to be
times a minus i with p0 and x0.
And then we have
another term which
is p with x, which
is i, p0 over x0.
So you change the order
and you change the sign.
But if you change the
order of a commutator,
you change the side.
So we can put them
both in the same order.
Let me just write this out.
So this is i over x0 p0.
So this guy, minus i over x0p0.
But x0p0 is equal to 2h
bar, as we checked before.
This was x with p.
And then the second
term was plus i, again
over x0p0, which is
2h bar, p with x.
This x with p is equal to?
i h bar.
So the h bar cancels.
The i gives me a plus 1.
And p with x gives
me minus i h bar.
So the h bar and the
minus i gives me plus 1.
Well that's nice.
This is equal to 1.
So plus 1/2, therefore a
with a dagger is equal to 1.
As advertised, that is
about as simple as it gets.
Notice a couple of
other commutators
that follow from this.
a dagger with a is
equal to minus 1.
We just changed the order.
And that's just an
overall minus sign.
And a with a is what?
0.
a dagger with a dagger?
Good.
OK.
So we are now going to use
this commutation relation
to totally crush the
problem into submission.
It's going to be weeping before
us like the Romans in front
of the Visigoths.
It's going to be dramatic.
OK, so let's check.
So let's combine the two things.
So we had the first thing
is that this form is simple.
The second is that the
commutator is simple.
Let's combine these
together and really milk
the system for what it's got.
And to do that, I need
two more commutators.
And the lesson of this
series of machinations,
it's very tempting
to look at this
and be like, why
are you doing this?
And the reason is,
I want to encourage
you to see the power of
these commutation relations.
They're telling you a tremendous
amount about the system.
So we're going through and
doing some relatively simple
calculations.
We're just computing
commutators.
We're following our nose.
And we're going to
derive something awesome.
So don't just bear with it.
Learn from this, that
there's something very useful
and powerful about
commutation relations.
You'll see that at the end.
But I want you to on to
the slight awkwardness
right now, that it's not
totally obvious beforehand where
this is going.
So what is E with a?
That's easy.
It's the h bar omega
a dagger a plus 1/2.
So the 1/2, what's 1/2
commutator with an operator?
0.
Because any number
commutes with an operator.
1/2 operator is operator 1/2.
It's just a constant.
That term is gone.
So the only thing
that's left over
is h bar omega, a
dagger a with a.
The h bar omega's
just a constant.
It's going to pull out no matter
which term we're looking at.
So I could just pull
that factor out.
So this is equal to h
bar omega times a dagger
a minus a a dagger a.
But this is equal to
h bar omega-- well,
that's a dagger a
a, a a dagger a.
You can just pull out
the a on the right.
a dagger a minus a a dagger a.
That's equal to h bar omega.
Well, a dagger with a is equal
to a dagger with a minus 1
is equal to minus h bar omega.
And we have this a leftover, a.
So E with a is equal to minus a.
Well, that's interesting.
Now, the second
commutator-- I'm not
going to do it--
E with a dagger is
going to be equal to--
let's just eyeball
what's going to happen.
They can be a dagger.
So we're going to
have a dagger a dagger
minus a dagger a dagger a.
So we're going to have
an a dagger in front
and then a dagger.
So all we're going
to get is a sign.
And it's going to be a
dagger plus a dagger.
I shouldn't written
that in the center.
Everyone cool with that?
Yeah.
AUDIENCE: The h bar where?
PROFESSOR: Oh shoot, thank you!
h bar here.
Thank you.
We would have
misruled the galaxy.
OK, good.
Other questions?
You don't notice-- you haven't
noticed yet, but we just won.
We just totally
solved the problem.
And here's why.
Once you see this, any
time you see this, anytime
you see this commutator,
an operator with an a
is equal to plus a times some
constant, anytime you see this,
cheer.
And here's why.
Yeah, right.
Exactly.
Now.
Whoo!
Here's why.
Here's why you should cheer.
Because you no longer have
to solve any problems.
You no longer have to solve
any differential equations.
You can simply write
down the problem.
And let's see that you can
just write down the answer.
Suppose that we already
happened to have access--
here in my sleeve I have
access to an eigenfunction
of the energy operator.
E on phi E is equal to E phi
E. Suppose I have this guy.
Cool?
Check this out.
Consider a new
state, psi, which is
equal to a-- which do
I want to do first?
Doesn't really matter,
but let's do a.
Consider psi is
equal to a on phi E.
What can you say
about this state?
Well, it's the state you get
by taking this wave function
and acting with a.
Not terribly illuminating.
However, E on psi
is equal to what?
Maybe this has some nice
property under acting
with E. This is equal
to E on a with pfi E.
Now, this is tantalizing.
Because at this point
it's very-- look, that E,
it really wants to hit this phi.
It just really wants to.
There's an E it
wants to pull out.
It'll be great.
The problem is it's not there.
There's an a in the way.
And so at this point we add 0.
And this is a very
powerful technique.
This is equal to Ea
minus aE plus aE, phi E.
But that has a nice expression.
This is equal to Ea minus aE.
That's the commutator
of E with a.
Plus a.
What's E acting on phi E?
Actually, let me just
leave this as aE.
So what have we done here
before we actually act?
What we've done is
something called
commuting an operator through.
So what do I mean by
commuting an operator through?
If we have an operator A and
an operator B and a state f,
and I want A to act
on f, I can always
write this as-- this is
equal to the commutator of A
would be plus BA acting on f.
So this lets me act A
on f directly without B.
But I have to know what
the commutator of these two
operators is.
So if I know what the
commutator is, I can do this.
I can simplify.
When one does this, when
one takes AB and replaces it
by the commutator of A with B,
plus BA, changing the order,
the phrase that one uses is
I have commuted A through B.
And commuting
operators through other
is an extraordinarily useful
tool, useful technique.
Now let's do y.
So here what's the
commutator of E with a?
We just did that.
It's minus h bar omega a.
And what's aE on phi E?
What's E on phi E?
E. Exactly.
Plus Ea on phi.
And now we're cooking with gas.
Because this is equal to minus
h bar omega a plus Ea, hat.
I'm going to pull out
this common factor of a.
So if I pull out that common
factor of a, plus E, a phi E,
and now I'm going to
just slightly write
this instead of minus
h bar omega plus E,
I'm going to write this
as E minus h bar omega.
I'm just literally changing
the order of the algebra.
E minus h bar omega.
And what is aE?
Psi.
That was the original
state we started with, psi.
Well, that's cool.
If I have a state with
energy E and I act on it
with the operator a, I
get a new state, psi,
which is also an eigenstate
of the energy operator,
but with a slightly
different energy eigenvalue.
The eigenvalue is now
decreased by h bar omega.
Cool?
And that is what we wanted.
Let's explore the
consequences of this.
So if we have a state
with eigenvalue E,
we have phi E such that E on
phi E is equal to E phi E.
Then the state a
phi E has eigenvalue
as energy, eigenvalue
E minus h bar omega.
So I could call this phi
sub E minus h bar omega.
It's an eigenfunction
of the energy operator,
the eigenvalue, E
minus h bar omega.
Agreed?
Do I know that this is in
fact properly normalized?
No, because 12
times it would also
be a perfectly
good eigenfunction
of the energy operator.
So this is proportional to
the properly normalized guy,
with some, at the moment,
unknown constant coefficient
normalization.
Everyone cool with that?
So now let's think about
what this tells us.
This tells us if
we have a state phi
E, which I will denote
its energy by this level,
then if I act on
it with a phi E I
get another state where the
energy, instead of being E,
is equal E minus h bar omega.
So this distance in
energy is h bar omega.
Cool?
Let me do it again.
We'll tack a on phi E. By
exactly the same argument, if I
make psi as equal to a on
a phi E, a squared phi E,
I get another state, again
separated by h bar omega, E
minus 2h bar omega.
Turtles all the way down.
Everyone cool with that?
Let's do a slightly
different calculation.
But before we do that,
I want to give a a name.
a does something really cool.
When you take the state phi
E that has definite energy E,
it's an energy eigenfunction,
and you act on it
with a, what happens?
It lowers the energy
by h bar omega.
So I'm going to call a
the lowering operator.
Because what it does is it
takes a state with phi E,
with energy
eigenvalue E to state
with energy E minus h bar omega.
And I can just keep doing
this as many times as I like
and I build a tower.
Yes?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Very good question.
Hold on to that for a second.
We'll come back to
that in just a second.
So this seems to build
for me a ladder downwards.
Everyone cool with that?
But we could have done the
same thing with a dagger.
And how does this story change?
What happens if we take
a dagger instead of a?
Well, let's go through
every step here.
So this is going to
be E on a dagger.
And now we have E a dagger,
a dagger, E, a dagger.
What's E with a dagger?
E with a dagger is equal to
same thing but with a plus.
And again, psi.
Same thing, because the
a dagger factors out.
Yeah?
So we go down by acting with a.
We go up by acting
with a dagger.
And again, the spacing
is h bar omega.
And we go up by acting
with a dagger again.
So a and a dagger are called the
raising and lowering operators.
a dagger, the raising operator.
a dagger phi E plus h bar omega.
So what that lets us do is
build a tower of states,
an infinite number of
states where, given a state,
we can walk up this ladder
with the raising operator,
and we can walk down it
by the lowering operator.
So now I ask you
the question, why
is this ladder evenly spaced?
There's one equation on the
board that you can point to-- I
guess two, technically--
there are two equations
on the board that you
could point to that
suffice to immediately
answer the question,
why is the tower of energy
eigenstates evenly spaced.
What is that equation?
AUDIENCE: [INAUDIBLE]
PROFESSOR: Yeah,
those commutators.
These commutators
are all we needed.
We didn't need to
know anything else.
We didn't even need to know
what the potential was.
If I just told you there's
an energy operator E
and there's an operator
a that you can build out
of the observables
of the system, such
that you have this
commutation relation,
what do you immediately know?
You immediately know that
you get a tower of operators.
Because you can act with
a and raise the energy
by a finite amount,
which is the coefficient
of that a in the commutator.
This didn't have to be
the quantum mechanics
of the harmonic
oscillator at this point.
We just needed this commutator
relation, E with a, E
with a dagger.
And one of the
totally awesome things
is how often it shows up.
If you take a bunch of
electrons and you put them
in a magnetic field,
bunch of electrons,
very strong magnetic
field, what you discover
is the quantum
mechanics of those guys
has nothing to do with the
harmonic oscillator on the face
if it's magnetic fields, Lorentz
force law, the whole thing.
What you discover is
there's an operator, which
isn't usually called a, but
it depends on which book
you use-- it's n
or m or l-- there's
an operator that
commutes with the energy
operator in precisely this
fashion, which tells you
that the energy eigenstates
live in a ladder.
They're called Landau levels.
This turns out to
be very useful.
Any of you who are
doing a [INAUDIBLE]
in the lab that has
graphene or any material,
really, with a magnetic
field, then this matters.
So this commutator
encodes an enormous amount
of the structure of
the energy eigenvalues.
And the trick for us
was showing that we
could write the harmonic
oscillator energy
operator in terms of operators
that commute in this fashion.
So we're going to run into this
structure over and over again.
This operator
commutes with this one
to the same operator times
a constant that tells you
have a ladder.
We're going to run
into that over and over
again when we talk about
Landau levels, if we get there.
When we talk about
angular momentum
we'll get the same thing.
When we talk about the
harmonic oscillator
we'll get the same thing.
Sorry, the hydrogen system.
We'll get the same thing.
So second question, does this
ladder extend infinitely up?
Yeah, why not?
Can it extend infinitely down?
AUDIENCE: Nope.
PROFESSOR: Why?
AUDIENCE: Ground state.
PROFESSOR: Well, people
are saying ground state.
Well, we know that from the
brute force calculation.
But without the brute
force calculation,
can this ladder extend
infinitely down?
AUDIENCE: [INAUDIBLE]
you can't go [INAUDIBLE].
PROFESSOR: Brilliant.
OK, good.
And as you'll prove
on the problems,
that you can't make the
energy arbitrarily negative.
But let me make that sharp.
I don't want to appeal to
something we haven't proven.
Let me show you that concretely.
In some state, in any state,
the energy expectation value
can be written as the
integral of phi complex
conjugate-- we'll say
in this state phi--
phi complex conjugate E phi.
But I can write this
as the integral,
and let's say dx, integral dx.
Let's just put in what the
energy operator looks like.
So psi tilda, we can
take the 4a transfer
and write the psi tilda p,
p squared upon 2m-- whoops,
dp-- for the kinetic energy
term, plus the integral--
and now I'm using the
harmonic oscillator--
plus the integral dx of psi of
x norm squared, norm squared,
m omega squared upon 2x squared.
Little bit of a
quick move there,
doing the 4a transfer
for the momentum term
and not doing the 4a
[INAUDIBLE] but it's OK.
They're separate integrals.
I can do this.
And the crucial thing here
is, this is positive definite.
This is positive definite,
positive definite,
positive definite.
All these terms are
strictly positive.
This must be greater
than or equal to 0.
It can never be negative.
Yeah?
So what that tells
us is there must
be a minimum E. There
must be a minimum energy.
And I will call it minimum E0.
We can't lower
the tower forever.
So how is this possible?
How is it possible that,
look, on the one hand,
if we want, if we have a state,
we can always build a lower
energy state by acting
with lowering operator a.
And yet this is telling
me that I can't.
There must be a last one where
I can't lower it anymore.
So what reaches out
of the chalkboard
and stops me from
acting with a again?
How can it possibly be
true that a always lowers
the eigenfunction but
there's at least one
that can't be
lowered any further.
Normalizable's a good guess.
Very good guess.
Not the case.
Because from this argument we
don't even use wave functions.
AUDIENCE: [INAUDIBLE]
PROFESSOR: That would be bad.
Yes, exactly.
So that would be
bad, but that's just
saying that there's
an inconsistency here.
So I'm going to come
back to your answer,
a non-normalizable.
It's correct, but
in a sneaky way.
Here's the way it's sneaky.
Consider a state
a on phi-- let's
say this is the lowest state,
the lowest possible state.
It must be true that the
resulting statement is not
phi minus h bar omega.
There can't be any such state.
And how can that be?
That can be true if it's 0.
So if the lowering operator acts
on some state and gives me 0,
well, OK, that's an eigenstate.
But it's a stupid eigenstate.
It's not normalizable.
It can't be used to describe
any real physical object.
Because where is it?
Well, it's nowhere.
The probability density,
you'd find it anywhere.
It's nowhere, nothing, zero.
So the way that this
tower terminates
is by having a last
state, which we'll
call phi 0, such that
lowering it gives me 0.
Not the state called 0,
which I would call this,
but actually the function called
0, which is not normalizable,
which is not a good state.
So there's a minimum E0.
Associated with that is a
lowest energy eigenstate
called the ground state.
Now, can the energy
get arbitrarily large?
Sure.
That's a positive
definite thing,
and this could get
as large as you like.
There's no problem with the
energy eigenvalues getting
arbitrarily large.
We can just keep raising
and raising and raising.
I mention that
because later on in
the semester we
will find a system
with exactly that commutation
relation, precisely
that commutation
relation, where there
will be a minimum and a maximum.
So the communication
relation is a good start,
but it doesn't
tell you anything.
We have to add in some
physics like the energy
operators bounded below for
the harmonic oscillator.
Questions at this point?
Yeah?
AUDIENCE: So you basically
[INAUDIBLE] this ladder
has to [INAUDIBLE] my
particular energy eigenstate
and I can kind of
construct a ladder.
How do I know that
I can't construct
other, intersecting ladders?
PROFESSOR: Yeah, that's
an excellent question.
I remember vividly when I
saw this lecture in 143A,
and that question plagued me.
And foolishly I didn't ask it.
So here's the question.
The question is, look, you
found a bunch of states.
How do you know
that's all of them?
How do you know
that's all of them?
So let's think through that.
That's a very good question.
I'm not going to worry
about normalization.
There's a discussion of
normalization in the notes.
How do we know
that's all of them?
That's a little bit tricky.
So let's think through it.
Imagine it's not all of them.
In particular, what
would that mean?
In order for there to be
more states than the ones
that we've written
down, there must
be states that are
not on that tower.
And how can we possi-- wow, this
thing is totally falling apart.
How do we do that?
How is that possible?
There are two ways to do it.
Here's my tower of states.
I'll call this one phi 0.
so I raise with a dagger
and I lower with a.
So how could it be that
I missed some states?
Well, there are ways to do it.
One is there could be extra
states that are in between.
So let's say that there's
one extra state that's
in between these two.
Just imagine that's true.
If there is such a state,
by that commutation relation
there must be another tower.
So there must be this state,
and there must be this state,
and there must be this state,
and there must be this state.
Yeah?
OK, so that's good so far.
But what happens?
Well, A on this guy gave me 0.
And this is going to
be some phi tilde 0.
Suppose that this tower ends.
And now you have to
ask the question,
can there be two
different states
with two different
energies with a0?
Can there be two
different states
that are annihilated by a0?
Well, let's check.
What must be true of any
state annihilated by a0?
Well, let's write the energy
operator acting on that state.
What's the energy of that state?
Energy on phi 0 is
equal to h bar omega.
This is a very good question,
so let's go through it.
So it's equal to h bar
omega times a dagger
a plus 1/2 on phi 0.
But what can you say about this?
Well, a annihilates phi 0.
It gives us 0.
So in addition to a being called
the lowering operator it's also
called the
annihilation operator,
because, I don't know, we're
a brutal and warlike species.
So this is equal to h bar
omega-- this term kills phi
0-- again with the kills-- and
gives me a 1/2 half leftover.
1/2 h bar omega phi 0.
So the ground state, any state--
any state annihilated by a
must have the same energy.
The only way you can
be annihilated by a
is if your energy is this.
Cool?
So what does that tell you
about the second ladder
of hidden seats that we missed?
It's got to be degenerate.
It's got to have
the same energies.
I drew that really
badly, didn't I?
Those are evenly spaced.
So it's got to be degenerate.
However, Barton proved for you
the node theorem last time,
right?
He gave you my spread
argument for the node theorem?
In particular, one of
the consequences of that
is it in a system
with bound states,
in a system with
potential that goes up,
you can never have
degeneracies in one dimension.
We're not going to prove
that carefully in here.
But it's relatively
easy to prove.
In fact, if you come
to my office hours
I'll prove it for you.
It takes three minutes.
But I don't want to set
up the math right now.
So how many people know
about the Wronskian?
That's awesome.
OK, so I leave it to you as an
exercise to use the Wronskian
to show that there cannot be
degeneracies in one dimension,
which is cool.
Anyway, so the Wronskian for
the differential equation,
which is the energy
eigenvalue equation.
There can be no degeneracies
in one dimensional potentials
with bound states.
So what we've just shown is
that the only way that there
can be extra states
that we missed
is if there's a tower with
exactly identical energies all
the way up.
But if they have exactly
identical energies,
that means there's a degenerate.
But we can prove that there
can't be degeneracies in 1D.
So can there be an extra
tower of states we missed?
No.
Can we have missed any states?
No.
Those are all the
states there are.
And we've done it without
ever solving a differential
equation, just by using
that commutation relation.
Now at this point it's
very tempting to say,
that was just sort of
magical mystery stuff.
But what we really did
last time was very honest.
We wrote down a
differential equation.
We found the solution.
And we got the wave functions.
So, Professor Adams,
you just monkeyed around
at the chalkboard with
commutators for a while,
but what are the
damn wave functions?
Right?
We already have the answer.
This is really quite nice.
Last time we solved that
differential equation.
And we had to solve that
differential equation
many, many times,
different levels.
But now we have a very
nice thing we can do.
What's true of the ground state?
Well, the ground
state is annihilated
by the lowering operator.
So that means that a acting
on phi 0 of x is equal to 0.
But a has a nice expression,
which unfortunately I erased.
Sorry about that.
So a has a nice expression.
a is equal to x over
x0 plus ip over p0.
And so if you write
that out and multiply
from appropriate
constants, this becomes
the following
differential equation.
The x is just multiplied by x.
And the p is take a
derivative with respect
to x, multiply by h bar upon i.
And multiplying by
i over h bar to get
that equation, this gives us dx
plus p over h bar x0-- sorry,
that shouldn't be an i.
That should be p0.
x on phi 0 is equal to 0.
And you solved
this last time when
you did the asymptotic analysis.
This is actually a
ridiculously easy equation.
It's a first order
differential equation.
There's one
integration constant.
That's going to be the
overall normalization.
And so the form is
completely fixed.
First order
differential equation.
So what's the
solution of this guy?
It's a Gaussian.
And what's the width
of that Gaussian?
Well, look at p0 over h bar x0.
We know that p0 times
x0 is twice h bar.
So if I multiply by x amount
on the top and bottom,
you get 2 h bar.
The h bars cancel.
So this gives me
two upon x0 squared.
Remember I said it would
be useful to remember
that p0 times x0 is 2h bar?
It's useful.
So it gives us this.
And so the result is
that phi 0 is equal to,
up to an overall normalization
coefficient, e to the minus
x squared over x0 squared.
Solid.
So there.
We've solved that
differential equation.
Is the easiest, second
easiest differential equation.
It's our first order
differential equation
with a linear term
rather than a constant.
We get a Gaussian.
And now that we've
got this guy-- look,
do you remember the
third Hermite polynomial?
Because we know the
third excited state
is given by h3
times this Gaussian.
Do you remember it off
the top of your head?
How do you solve what it is?
How do we get phi 3?
First off, how do we get phi 1?
How do we get the next
state in the ladder?
How do we get the wave function?
Raising operator.
But what is the
raising operator?
Oh, it's the differential
operator I take with-- OK,
but if I had a dagger, it's just
going to change the sign here.
So how do I get phi 1?
Phi 1 is equal to up
to some normalization.
dx minus 2 over x0
squared, x0, phi 0.
So now do I have to solve
the differential equation
to get the higher states?
No.
I take derivatives and
multiply by constants.
So to get the third Hermite
polynomial what do you do?
You do this three times.
This is actually an
extremely efficient way--
it's related to something
called the generating function,
and an extremely efficient
way to write down
the Hermite polynomials.
They're the things
that you get by acting
on this with this operator
as many times as you want.
That is a nice formal definition
of the Hermite polynomials.
The upshot of all of
this is the following.
The upshot of all this is that
we've derived that without ever
solving the differential
equation the spectrum just from
that commutation relation,
just from that commutation
relation-- I cannot emphasize
this strongly enough--
just from the
commutation relation,
Ea is minus a
times the constant,
and Ea dagger is a dagger
times the constant.
We derive that the energy
eigenstates come in a tower.
You can move along this tower
by raising with the raising
operator, lowering with
the lowering operator.
You can construct
the ground state
by building that simple
wave function, which
is annihilated by the
lowering operator.
You can build all
the other states
by raising them, which is
just taking derivatives
instead of solving differential
equations, which is hard.
And all of this came from
this commutation relation.
And since we are going to see
this over and over again--
and depending on how
far you take physics,
you will see this in 8.05.
You will see this in 8.06.
You will see this in
quantum field theory.
This shows up everywhere.
It's absolutely
at the core of how
we organize the
degrees of freedom.
This structure is something you
should see and declare victory
upon seeing.
Should see this and immediately
say, I know the answer,
and I can write it down.
OK?
In the next lecture we're
going to do a review which
is going to introduce a slightly
more formal presentation of all
these ideas.
That's not going to be
material covered on the exam,
but it's going to help
you with the exam, which
will be on Thursday.
See you Tuesday.
