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PROFESSOR: All right.
So, this homework
that is due on Friday
contains some questions on
the harmonic oscillator.
And the harmonic oscillator
is awfully important.
I gave you notes on that.
And I want to use about
half of the lecture,
perhaps a little
less, to go over
some of those points
in the notes concerning
the harmonic oscillator.
After that, we're going
to begin, essentially,
our study of dynamics.
And we will give the revision,
today, of the Schrodinger
equation.
It's the way Dirac, in his
textbook on quantum mechanics,
presents the
Schrodinger equation.
I think it's actually,
extremely insightful.
It's probably not
the way you should
see it the first
time in your life.
But it's a good way
to think about it.
And it will give
you a nice feeling
that this Schrodinger
equation is something
so fundamental and
so basic that it
would be very hard to
change or do anything to it
and tinker with it.
It's a rather complete theory
and quite beautiful [? idea. ?]
So we begin with the
harmonic oscillator.
And this will be a bit quick.
I won't go over every detail.
You have the notes.
I think that's pretty
much all you need to know.
So we'll leave it at that.
So the harmonic oscillator
is a quantum system.
And as quantum
systems go, they're
inspired by classical systems.
And the classical system
is very famous here.
It's the system in
which, for example, you
have a mass and a spring.
And it does an oscillation for
which the energy is written
as p squared over 2m plus 1/2
m, omega squared, x squared.
And m omega squared is
sometimes called k squared,
the spring constant.
And you are supposed to do
quantum mechanics with this.
So nobody can tell you this is
what the harmonic oscillators
in quantum mechanics.
You have to define it.
But since there's
only one logical way
to define the quantum
system, everybody
agrees on what the harmonic
oscillator quantum system is.
Basically, you use
the inspiration
of the classical system
and declare, well,
energy will be the
Hamiltonian operator.
p will be the momentum operator.
And x will be the
position operator.
And given that
these are operators,
will have a basic commutation
relation between x and p
being equal to i h-bar.
And that's it.
This is your quantum system.
Hamiltonian is--
the set of operators
that are relevant for this are
the x the p, and the energy
operator that will
control the dynamics.
You know also you should specify
a vector space, the vector
space where this acts.
And this will be complex
functions on the real line.
So this will act
in wave functions
that define the vector space,
sometimes called Hilbert space.
It will be the set of integrable
functions on the real line,
so complex functions
on the real line.
These are your wave functions,
a set of states of the theory.
All these complex functions
on the real line work.
I won't try to be more precise.
You could say they're
square integrable.
That for sure is necessary.
And we'll leave it at that.
Now you have to
solve this problem.
And in 804, we discussed this by
using the differential equation
and then through the creation
annihilation operators.
And we're going
do it, today, just
through creation and
annihilation operators.
But we want to
emphasize something
about this Hamiltonian and
something very general, which
is that you can right
the Hamiltonian as say
1/2m, omega squared, x squared.
And then you have plus
p squared, m squared,
omega squared.
And a great solution
to the problem
of solving the Hamiltonian--
and it's the best you could ever
hope-- is what is
called the factorization
of the Hamiltonian, in
which you would manage
to write this Hamiltonian as
some operator times the dagger
operator.
So this is the ideal situation.
It's just wonderful,
as you will see,
if you can manage to do that.
If you could manage to
do this factorization,
you would know immediately
what is the ground state
energy, how low can
it go, something
about the Hamiltonian.
You're way on your way
of solving the problem.
If you could just factorize it.
Yes?
AUDIENCE: [INAUDIBLE] if
you could just factorize it
in terms of v and v
instead of v dagger and v?
PROFESSOR: You want to factorize
in which way instead of that?
AUDIENCE: Would it be
helpful, if it were possible,
to factor it in terms of v
times v instead of v dagger?
PROFESSOR: No, no, I
want, really, v dagger.
I don't want v v. That
that's not so good.
I want that this factorization
has a v dagger there.
It will make things
much, much better.
So how can you achieve that?
Well, it almost looks possible.
If you have something like this,
like a squared plus b squared,
you write it as a minus
ib times a plus ib.
And that works out.
So you try here, 1/2
m, omega squared,
x minus ip over m omega,
x plus ip over m omega.
And beware that's
not quite right.
Because here, you have
cross terms that cancel.
You have aib b and minus iba.
And they would only
cancel if a and b commute.
And here they don't commute.
So it's almost perfect.
But if you expand this out,
you get the x squared for sure.
You get this term.
But then you get an extra term
coming from the cross terms.
And please calculate it.
Happily, it's just a number,
because the commutator
of x and b is just a number.
So the answer for
this thing is that you
get, here, x
squared plus this is
equal to this, plus h-bar over m
omega, times the unit operator.
So here is what you
could call v dagger.
And this is what we'd call v.
So what is your Hamiltonian?
Your Hamiltonian has become
1/2 m, omega squared, v dagger
v, plus, if you multiply out,
H omega times the identity.
So we basically succeeded.
And it's as good as what we
could hope or want, actually.
I multiply this
out, so h-bar omega
was the only thing
that was left.
And there's your Hamiltonian.
Now, in order to see
what this tells you,
just sandwich it
between any two states.
Well, this is 1/2 m, omega
squared, psi, v dagger, v,
psi, plus 1/2 half h, omega.
And assume it's a
normalized state,
so it just gives you that.
So this thing is the
norm of the state, v psi.
You'd think it's
dagger and it's this.
So this is the norm
squared of v psi.
And therefore that's positive.
So H, between any
normalized state,
is greater than or equal
to 1/2 h-bar omega.
In particular, if psi
is an energy eigenstate,
so that H psi is equal to E psi.
If psi is an energy
eigenstate, then you have this.
And back here, you
get that the energy
must be greater than or
equal to 1/2 h omega,
because H and psi gives
you an E. The E goes out.
And you're left with
psi, psi, which is 1.
So you already know
that the energy
is at least greater than
or equal to 1/2 h omega.
So this factorization
has been very powerful.
It has taught you
something extremely
nontrivial about the
spectrum of the Hamiltonian.
All energy eigenstates
must be greater than
or equal to 1/2 h omega.
In fact, this is
so good that people
try to do this for
almost any problem.
Any Hamiltonian, probably
the first thing you can try
is to establish a
factorization of this kind.
For the hydrogen atom, that
factorization is also possible.
There will be some
homework sometime later on.
It's less well known and
doesn't lead to useful creation
and annihilation operators.
But you can get the ground
state energy in a proof
that you kind of go below
that energy very quickly.
So a few things are done
now to clean up this system.
And basically, here I have the
definition of v and v dagger.
Then you define a to be square
root of m omega over 2 h-bar,
v. And a dagger must be m
omega over 2 h-bar v dagger.
And I have not written for
you the commutator of v and v
dagger.
We might as well do the
commutator of a and a dagger.
And that commutator turns
out to be extremely simple.
a with a dagger is
just equal to 1.
Now things that are useful,
relations that are useful
is-- just write
what v is in here
so that you have a
formula for a and a dagger
in terms of x and p.
So I will not bother writing it.
But it's here already.
Maybe I'll do the first one.
m omega over 2 h-bar.
v is here would be x,
plus ip over m omega.
And you can write
the other one there.
So you have an expression
for a and a dagger
in terms of x and p.
And that can be
inverted as well.
And it's pretty useful.
And it's an example
of formulas that you
don't need to know by heart.
And they would be in
any formula sheet.
And the units and
all those constants
make it hard to remember.
But here they are.
So you should know that x is a
plus a dagger up to a constant.
And p is a dagger minus a.
Now p is Hermitian, that's
why there is an i here.
So that this, this
anti-Hermitian,
the i becomes a
Hermitian operator.
x is manifestly Hermitian,
because a plus a dagger is.
Finally, you want to
write the Hamiltonian.
And the Hamiltonian is given
by the following formula.
You know you just have to put
the v and v dagger, what they
are in terms of the creation,
annihilation operators.
So v dagger, you
substitute a dagger.
v, you go back here
and just calculate it.
And these calculations
really should be done.
It's something that is
good practice and make sure
you don't make silly mistakes.
So this operator is so important
it has been given a name.
It's called the
number operator, N.
And its eigenvalues are numbers,
0, 1, 2, 3, all these things.
And the good thing about
it is that, once you
are with a's and a
daggers, all this m omega,
h-bar are all gone.
This is all that
is happening here.
The basic energy is h-bar omega.
Ground state energies, what
we'll see is 1/2 h-bar omega.
And this is the number operator.
So this is written as h-bar
omega, number operator--
probably with a hat-- like that.
So when you're talking
about eigenvalues,
as we will talk soon, or
states for which these thing's
are numbers, saying that
you have a state that
is an eigenstate
of the Hamiltonian
is exactly the same
thing as saying
that it's an eigenstate
of the number operator.
Because that's the only thing
that is an operator here.
There's this plus this number.
So this number
causes no problem.
Any state multiplied by a number
is proportional to itself.
But it's not true that every
state multiplied by a dagger a
is proportional to itself.
So being an
eigenstate of N means
that acting on a state,
N, gives you a number.
But then H is just
N times the number.
So H is also an eigenstate.
So eigenstates of N
or eigenstates of H
are exactly the same thing.
Now there's a couple
more properties
that maybe need to be mentioned.
So I wanted to talk in
terms of eigenvalues.
I would just simply write
the energy eigenvalue
is therefore equal h-bar
omega, the number eigenvalue--
so the operator is
with a hat-- plus 1/2.
So in terms of
eigenvalues, you have that.
From here, the energy is
greater than 1/2 h omega.
So the number must be greater
or equal than 0 on any state.
And that's also clear from the
definition of this operator.
On any state, the expectation
value of this operator
has to be positive.
And therefore, you have this.
So two more properties
that are crucial here
are that the
Hamiltonian commuted
with a is equal
to minus h omega a
and that the Hamiltonian
committed with a dagger
is plus h omega a dagger.
Now there is a
reasonably precise way
of going through
the whole spectrum
of the harmonic oscillator
without solving differential
equations, almost to
any degree, and trying
to be just very
logical about it.
It's possible to deduce the
properties of the spectrum.
So I will do that right now.
And we begin with the
following statement.
We assume there is
some energy eigenstate.
So assume there is a state E
such that the Hamiltonian--
for some reason in
the notes apparently
I put hats on the Hamiltonian,
so I'll start putting hats
here-- so that the states
are labeled by the energy.
And this begins a tiny bit of
confusion about the notation.
Many times you want to label
the states by the energy.
We'll end up labeling them
with the number operator.
And then, I said,
it will turn out,
when the number operator is
0, we'll put a 0 in here.
And that doesn't mean 0 energy.
It means energy equal
1/2 h-bar omega.
So if you assume there
is an energy eigenstate,
that's the first step
in the construction.
You assume there is one.
And what does that mean?
It means that this
is a good state.
So it may be normalized.
It may not be normalized.
In any case, it
should be positive.
I put first the equal, but
I shouldn't put the equal.
Because we know in a
complex vector space,
if a state has 0 norm, it's 0.
And I want to say
that there's really
some state that is non-0,
that has this energy.
If the state would be 0, this
would become a triviality.
So this state is good.
It's all good.
Now with this state, you can
define, now, two other states,
acting with the creation,
annihilation operators.
I didn't mention that name.
But a dagger is going to be
called the creation operator.
And this is the destruction
or annihilation operator.
And we built two states, E
plus is a dagger acting on E.
And E minus is a
acting on E. Now
you could fairly
ask a this moment
and say, well, how do you
know these states are good?
How do you know they even exist?
How do you know that
if you act with this,
don't you get an
inconsistent state?
How do you know
this makes sense?
And these are perfectly
good questions.
And in fact, this is exactly
what you have to understand.
This procedure can
give some funny things.
And we want to
discuss algebraically
why some things are safe and
why some things may not quite
be safe.
And adding an a dagger,
we will see it's safe.
While adding a's to the
state could be fairly unsafe.
So what can be bad
about the state?
It could be a 0 state, or it
could be an inconsistent state.
And what this an
inconsistent state?
Well, all our states are
represented by wave functions.
And they should be normalizable.
And therefore they have
norms that are positive,
norms squared that are positive.
Well you may find,
here, that you
have states that have
norms that are negative,
norm squareds that are negative.
So this thing that
should be positive,
algebraically you may
show that actually you
can get into trouble.
And trouble, of course,
is very interesting.
So I want to skip this
calculation and state something
that you probably checked
in 804, several times,
that this state has
more energy than E
and, in fact, has as much
energy as E plus h-bar omega.
Because a dagger, the
creation operator,
adds energy, h-bar omega.
And this subtracts
energy, h-bar omega.
This state has an
energy, E plus,
which is equal to
E plus h-bar omega.
And E minus is equal
to E minus h-bar omega.
Now how do you check that?
You're supposed to act
with a Hamiltonian on this,
use the commutation relation
that we wrote up there,
and prove that those are
the energy eigenvalues.
So at this moment, you
can do the following.
So these states have energies,
they have number operators,
they have number eigenvalues.
So we can test, if
these states are good,
by computing their norms.
So let's compute the
norm, a dagger on E,
a dagger on E for the first one.
And we'll compute a E, a E.
We'll do this computation.
We just want to
see what this is.
Now remember how you do this.
An operator acting
here goes with a dagger
into the other side.
So this is equal to
E a, a dagger, E.
Now a, a dagger is
not quite perfect.
It differs from the
one that we know
is an eigenvalue for this state,
which is the number operator.
So what is a, a
dagger in terms of N?
Well, a, a dagger--
it's something
you will use many,
many times-- is
equal to a commutator with
a dagger plus a dagger a.
So that's 1 plus
the number operator.
So this thing is E
1 plus the number
operator acting on the state E.
Well, the 1 is clear what it is.
And the number operate is clear.
If this has some
energy E, well, I
can now what is the eigenvalue
of the number operator
because the energy on
the number eigenvalues
are related that way.
So I will simply call
it the number of E
and leave it at that.
Times EE.
So in here, the
computation is easier
because it's just E a dagger
a E. That's the number,
so that's just NE times EE.
OK, so these are
the key equations
we're going to be
using to understand
the spectrum quickly.
And let me say a couple
of things about them.
So I'll repeat what we
have there, a dagger
E a dagger E is equal
to 1 plus NE EE.
On the other hand, 888
aE aE is equal to NE EE.
OK, so here it goes.
Here is the main thing that
you have to think about.
Suppose this state
was good, which
means this state has
a good norm here.
And moreover, we've
already learned
that the energy is
greater than some value.
So the number
operator of this state
could be 0-- could
take eigenvalue 0.
But it could be bigger
than 0, so that's all good.
Now, at this stage, we have
that-- for example, this state,
a dagger E has number
one higher than this one,
than the state E because it has
an extra factor of the a dagger
which adds an energy of h omega.
Which means that it
adds number of 1,
So if this state
has some number,
this state has a
number which is bigger.
So suppose you keep adding.
Now, look at the
norm of this state.
The norm of this state is pretty
good because this is positive
and this is positive.
If you keep adding
a daggers here,
you always have that this state,
the state with two a daggers,
you could use that
to find its norm.
You could use this formula,
put in the states with one
a dagger here.
But the states with one a
dagger already has a good norm.
So this state with two a daggers
would have also good norm.
So you can go on step by
step using this equation
to show that as long as
you keep adding a daggers,
all these states will
have positive norms.
And they have positive norms
because their number eigenvalue
is bigger and bigger.
And therefore,
the recursion says
that when you add
one a dagger, you
don't change the sign of this
norm because this is positive
and this is positive,
and this keeps happening.
On the other hand,
this is an equation
that's a lot more dangerous.
Because this says that in this
equation, a lowers the number.
So if this has some number,
NE, this has NE minus 1.
And if you added
another a here, you
would use this
equation again and try
to find, what is the norm
of things with two a's here?
And put in the one
with one a here
and the number of that state.
But eventually, the number can
turn into a negative number.
And as soon as the number turns
negative, you run into trouble.
So this is the equation
that is problematic
and the equation that
you need to understand.
So let me do it in two stages.
Here are the numbers.
And here is 5 4, 3, 2, 1, 0.
Possibly minus 1, minus
2, and all these numbers.
Now, suppose you start with
a number that is an integer.
Well, you go with this equation.
This has number 4.
Well, you put an a.
Now it's a state with
number 3, but its norm
is given 4 times that.
So it's good.
Now you go down another 1, you
have a state with number 3,
with number 2, with
number 1, with number 0.
And then if you keep lowering,
you will get minus 1,
which is not so good.
We'll see what happens.
Well, here you go
on and you start
producing the states--
the state with number 4,
state with number 3, state with
number 2, state with number 1.
And state here, let's call
it has an energy E prime.
And it has number equal 0.
Number of E prime equals 0.
So you look at this
equation and it
says aE prime times aE prime is
equal N E prime times E prime E
prime.
Well, you obtain this
state at E prime,
and it was a good
state because it
came from a state
that was good before.
And therefore, when
you did the last step,
you had the state at 1
here, with n equals to 1,
and then that was the
norm of this state.
So this E E prime is a
fine number positive.
But the number E prime is 0.
So this equation says that aE
prime aE prime is equal to 0.
And if that's equal
to 0, the state
aE prime must be equal to 0.
And 0 doesn't mean the
vacuum state or anything.
It's just not there.
There's no such state.
You can't create it.
You see, aE prime would be a
state here with number minus 1.
And everything suggests to
us that that's not possible.
It's an inconsistent state.
The number must be less than 1.
And we avoided the inconsistency
because this procedure
said that as you go ahead
and do these things,
you eventually run into this
state E prime at 0 number.
But then, you get that
the next state is 0
and there's no inconsistency.
Now, that's one possibility.
The other possibility
that could happen
is that there are
energy eigenstates that
have numbers which are not--
well, I'll put it here.
That are not integer.
So maybe you have a state
here with some number E
which is not an integer.
It doesn't belong
to the integers.
OK, so what happens now?
Well, this number is positive.
So you can lower it and you can
put another state with number
1 less.
Also, not integer
and it has good norm.
And this thing has
number 2.5, say.
Well, if I use the
equation again,
I put the 2.5 state
with its number 2.5
and now I get the
state with number 1.5
and it still has positive norm.
Do it again, you find
the state with 0.5 number
and still positive norm.
And looking at this,
you start with a state
with 0.5, with 0.5 here.
And oops, you get a
state that minus 0.5.
And it seems to be
good, positive norm.
But then, if this
is possible, you
could also build another
state acting with another a.
And this state is now very bad
because the N for this state
was minus 1/2.
And therefore, if
you put that state,
that state at the minus
1/2, you get the norm
of the next one
that has one less.
And this state now
is inconsistent.
So you run into a difficulty.
So what are the ways in
which this difficulty
could be avoided?
What are the escape hatches?
There are two possibilities.
Well, the simplest one would
be that the assumption is bad.
There's no state with
fractional number
because it leads to
inconsistent states.
You can build them and
they should be good,
but they're bad.
The other possibility
is that just
like this one sort
of terminated,
and when you hit 0--
boom, the state became 0.
Maybe this one with
a fractional one,
before you run into trouble
you hit a 0 and the state
becomes 0.
So basically, what you
really need to know now
on the algebraic method cannot
tell you is how many states are
killed by a.
If maybe the state of
1/2 is also killed by a,
then we would have trouble.
Now, as we will see now,
that's a simple problem.
And it's the only
place where it's
interesting to
solve some equation.
So the equation that
we want to solve
is the equation a on
some state is equal to 0.
Now, that equation already says
that this possibility is not
going to happen.
Why?
Because from this equation, you
can put an a dagger on this.
And therefore, you get
that NE is equal to 0.
This is the number operator,
so the eigenvalue of the number
operator, we call it NE.
So in order to be killed by a,
you have to have NE equals 0.
So in the fractional case,
no state will be killed
and you would arrive
to an inconsistency.
So the only possibility is that
there's no fractional states.
So it's still interesting to
figure out this differential
equation, what it gives you.
And why do we call it a
differential equation?
Because a is this
operator over there.
It has x and ip.
So the equation
is x a E equals 0,
which is square root of
m omega over 2 h bar x
x plus ip over m
omega on E equals 0.
And you've translated
these kind of things.
The first term is an x
multiplying the wave function.
We can call it psi E of x.
The next term, the
coefficient in front
is something you don't
have to worry, of course.
It's just multiplying
everything,
so it's just irrelevant.
So have i over m omega.
And p, as you remember, is h bar
over i d dx of psi E of x zero.
So it's so simple
differential equation,
x plus h bar over m omega d dx
on psi E of x is equal to 0.
Just one solution
up to a constant
is the Gaussian that
you know represents
a simple harmonic oscillator.
So that's pretty
much the end of it.
This ground state wave
function is a number
times the exponential of minus
m omega over 2 h bar x squared.
And that's that.
This is called the ground state.
It has N equals 0
represented as a state.
We say this number
is N equals 0.
So this state is the thing
that represents this psi
E. In other words, psi
E of x is x with 0.
And that 0 is a
little confusing.
Some people think
it's the 0 vector.
That's not good.
This is not the 0 vector.
The 0 vector is not a state.
It's not in the Hilbert space.
This is the ground state.
Then, the worst confusion is
to think it's the 0 vector.
The next confusion is
to think it's 0 energy.
That's not 0 energy,
it's number equals 0.
The energy is, therefore,
1/2 h bar omega.
And now, given
our discussion, we
can start building states
with more oscillators.
So we build a state
with number equal 1,
which is constructed by
an a dagger on the vacuum.
This has energy 1
h bar omega more.
It has number equal to 1.
And that's sometimes
useful to just
make sure you understand why
N on a dagger on the vacuum
is a dagger a a
dagger on the vacuum.
Now, a kills the
vacuum, so this can
be replaced by the
commutator, which is 1.
And therefore, you're left
with a dagger on the vacuum.
And that means that
the eigenvalue of n hat
is 1 for this state.
Moreover, this state is
where normalized 1 with 1
actually gives you
a good normalization
if 0 is well-normalized.
So we'll take 0 with 0
to be 1, the number 1.
And that requires fixing
that N0 over here.
Now, these are things
that you've mostly seen,
so I don't want to say
much more about them.
I'd rather go through
the Schrodinger thing
that we have later.
So let me conclude by just
listing the general states,
and then leaving for you to read
what is left there in the notes
so that you can just get an
appreciation of how you use it.
And with the practice
problems, you'll be done.
So here it is.
Here is the answer.
The n state is given by 1 over
square root of n factorial
a dagger to the n
acting on the vacuum.
And these n states are such
that m with n is delta mn.
So here we're using
all kinds of things.
First, you should check
this is well normalized,
or read it and do
the calculations.
And these are, in
fact, orthogonal
unless they have the same
number of creation operators
are the same number.
Now, that had to be expected.
These are eigenstates
of a Hermitian operator.
The N operator is Hermitian.
Eigenstates of a
Hermitian operator
with different
eigenvalues are always
orthogonal to each other.
If you have eigenstates
of a Hermitian operator
with the same eigenvalue,
if you have a degeneracy,
you can always arrange them
to make them orthogonal.
But if the eigenvalues are
different, they are orthogonal.
And there's no degeneracies
in this spectrum whatsoever.
You will, in fact, argue that
because there's no degeneracy
in the ground state, there
cannot be degeneracy anywhere
else.
So this result,
this orthonormality
is really a consequence of
all the theorems we've proven.
And you could check it
by doing the algebra
and you would start
moving a and a daggers.
And you would be left
with either some a's or
some a daggers.
If you're left
with some a's, they
would kill the
thing on the right.
If you're left with
some a daggers,
it would kill the
thing on the left.
So this can be proven.
But this is just a consequence
that these are eigenstates
of the Hermitian operator n
that have different eigenvalues.
And therefore, you've
succeeded in constructing
a full decomposition
of the state
space of the
harmonic oscillator.
We spoke about
the Hilbert space.
Are now very
precisely, see we can
say this is u0 plus
u1 plus u2 where
uk is the states of the form
alpha k, where N on k-- maybe
I should put n here.
It looks nicer.
n.
Where N n equal n n.
So every
one-dimensional subspace
is spanned by that
state of number n.
So you have the states of
number 0, states of number 1,
states of number 2.
These are all
orthogonal subspaces.
They add up to form everything.
It's a nice description.
So the general
state in this system
is a complex number times
the state with number 0
plus the complex number states
of number 1, complex number,
and that.
Things couldn't have
been easier in a sense.
The other thing that you
already know from 804
is that if you try to compute
expectation values, most
of the times you want to
use a's and a daggers.
So the typical thing
that one wants to compete
is on the state n, what is the
uncertainty in x on the state
n?
How much is it?
What is the
uncertainty of momentum
on the energy
eigenstate of number n?
These are relatively
straightforward calculations.
If you have to do the integrals,
each one-- by the time you
organize all your constants--
half an hour, maybe 20 minutes.
If you do it with
a and a daggers,
this computation
should be five minutes,
or something like that.
We'll see that
done on the notes.
You can also do them yourselves.
You probably have
played with them a bit.
So this was a brief review and
discussion of them spectrum.
It was a little detailed.
We had to argue things
carefully to make
sure we don't assume things.
And this is the
way we'll do also
with angular momentum
in a few weeks from now.
But now I want to leave that,
so I'm going to take questions.
If there are any questions
on this logic, please ask.
Yes.
AUDIENCE: [INAUDIBLE] for
how you got a dagger, a,
a dagger, 0, 2 dagger, 0?
PROFESSOR: Yes,
that calculation.
So let me do at the step
that I did in words.
So at this place--
so the question was,
how did I do this computation?
Here I just copied what N is.
So I just copied that.
Then, the next step was to
say, since a kills this,
this is equal to a dagger times
a a dagger minus a dagger a.
Because a kills it.
And I can add this, it
doesn't cost me anything.
Now, I added something
that is convenient,
so that this is a
dagger commutator of a
with a dagger on 0.
This is 1, so you get that.
It's a little more
interesting when
you have, for
example, the state 2,
which is 1 over square root
of 2 a dagger a dagger on 0.
I advise you to try to
calculate n on that.
And in general,
convince yourselves
that n is a number
operator, which means
counts the number of a daggers.
You'll have to use that
property if you have N with AB.
It's N with A B and then A N
with B. The derivative property
of the bracket has to
be used all the time.
So Schrodinger dynamics,
let's spend the last 20
minutes of our lecture on this.
So basically, it's
a postulate of how
evolution occurs in
quantum mechanics.
So we'll state it as follows.
What is time in
quantum mechanics?
Well, you have a state space.
And you see the
state space, you've
seen it in the
harmonic oscillator
is this sum of vectors.
And these vectors were wave
functions, if you wish.
There's no time anywhere there.
There's no time on
this vector space.
This vector space is an
abstract vector space
of functions or states, but time
comes because you have clocks.
And then you can ask,
where is my state?
And that's that vector
on that state space.
And you ask the
question a littler later
and the state has moved.
It's another vector.
So these are vectors and
the vectors change in time.
And that's all the dynamics
is in quantum mechanics.
The time is sort of
auxiliary to all this.
So we must have a
picture of that.
And the way we do this is to
imagine that we have a vector
space H. And here is a vector.
And that H is for Hilbert space.
We used to call it in our
math part of the course
V, the complex vector space.
And this state is the
state of the system.
And we sometimes
put the time here
to indicate that's what it is.
At time t0, that's it.
Well, at time t, some arbitrary
later time, it could be here.
And the state moves.
But one thing is clear.
If it's a state of a
system, if we normalize it,
it should be of unit length.
And we can think of a sphere
in which this unit sphere is
the set of all the tips of the
vectors that have unit norm.
And this vector will move here
in time, trace a trajectory,
and reach this one.
And it should do it preserving
the length of the vector.
And in fact, if you don't
use a normalized vector,
it has a norm of 3.
Well, it should preserve
that 3 because you'd
normalize the state
once and forever.
So we proved in our
math part of the subject
that an operator
that always preserves
the length of all vectors
is a unitary operator.
So this is the fundamental
thing that we want.
And the idea of quantum
mechanics is that psi at time t
is obtained by the action
of a unitary operator
from the state psi at time t0.
And this is for all t and t0.
And this being unitary.
Now, I want to make
sure this is clear.
It can be misinterpreted,
this equation.
Here, psi at t0 is
an arbitrary state.
If you had another
state, psi prime of t0,
it would also evolve
with this formula.
And this U is the same.
So the postulate of
unitary time evolution
is that there is this
magical U operator that
can evolve any state.
Any state that you
give me at time
equal 0, any possible
state in the Hilbert space,
you plug it in here.
And by acting with
this unitary operator,
you get the state
at the later time.
Now, you've slipped an
extraordinary amount of physics
into that statement.
If you've bought it, you've
bought the Schrodinger equation
already.
That is going to
come out by just
doing a little
calculation from this.
So the Schrodinger equation
is really fundamentally,
at the end of the
day, the statement
that this unitary
time evolution, which
is to mean there's a
unitary operator that
evolves any physical state.
So let's try to discuss this.
Are there any questions?
Yes.
AUDIENCE: So you
mentioned at first that
in the current
formulation [INAUDIBLE]?
PROFESSOR: A little louder.
We do what in our
current formulation?
AUDIENCE: So if you don't
include time [INAUDIBLE].
PROFESSOR: That's right.
There's no start of
the vector space.
AUDIENCE: Right.
So is it possible to consider
a vector space with time?
PROFESSOR: Unclear.
I don't think so.
It's just nowhere there.
What would it mean, even, to
add time to the vector space?
I think you would
have a hard time even
imagining what it means.
Now, people try to
change quantum mechanics
in all kinds of ways.
Nobody has succeeded in
changing quantum mechanics.
That should not be a
deterrent for you to try,
but should give you
a little caution
that is not likely to be easy.
So we'll not try to do that.
Now, let me follow on this
and see what it gives us.
Well, a few things.
This operator is unique.
If it exists, it's unique.
If there's another operator that
evolves states the same way,
it must be the same as that one.
Easy to prove.
Two operators that attack
the same way on every state
are the same, so that's it.
Unitary, what does it mean
that u t, t0 dagger times u t,
t0 is equal to 1?
Now, here these parentheses
are a little cumbersome.
This is very clear,
you take this operator
and you dagger it.
But it's cumbersome, so
we write it like this.
This means the dagger
of the whole operator.
So this is just the same thing.
OK, what else?
u of t0, t0, it's
the unit operator.
If the times are the same,
you get the unit operator
for all t0 because you're
getting psi of t0 here
and psi of t0 here.
And the only operator that
leaves all states the same
is the unit operator.
So this unitary operator must
become the unit operator,
in fact, for the two
arguments being equal.
Composition.
If you have psi t2, that can be
obtained as U of t2, t1 times
the psi of t1.
And it can be obtained as u of
t2, t1, u of t1, t0, psi of t0.
So what do we learn from here?
That this state itself is u of
t2, t0 on the original state.
So u of t2, t0 is u of
t2, t1 times u of t1, t0.
It's like time composition is
like matrix multiplication.
You go from t0 to t1,
then from t1 to t2.
It's like the second
index of this matrix.
In the first index
of this matrix,
you are multiplying them
and you get this thing.
So that's composition.
And then, you have
inverses as well.
And here are the inverses.
In that equation, you
take t2 equal to t0.
So the left-hand side becomes 1.
And t1 equal to t, so you get
u of t0, t be times u of t,
t0 is equal to 1,
which makes sense.
You propagate from t0 to t.
And then from t to
t0, you get nothing.
Or if it's to say that the
inverse of an operator--
the inverse of this
operator is this one.
So to take the inverse of a
u, you flip the arguments.
So I'll write it like that,
the inverse minus 1 of t, t0.
You just flip the arguments.
It's u of t0, t.
And since the
operator is Hermitian,
the dagger is equal
to the inverse.
So the inverse of an operator
is equal to the dagger.
so t, t0 as well.
So this one we got here.
And Hermiticity says that the
dagger is equal to the inverse.
Inverse and dagger are the same.
So basically, you can
delete the word "inverse"
by flipping the order
of the arguments.
And since dagger is
the same as inverse,
you can delete the
dagger by flipping
the order of the arguments.
All right, so let's try to
find the Schrodinger equation.
So how c we c the
Schrodinger equation?
Well, we try obtaining
the differential equation
using that time
evolution over there.
So the time evolution
is over there.
Let's try to find
what is d dt of psi t.
So d dt of psi of
t is just the d dt
of this operator u
of t, t0 psi of t0.
And I should only
differentiate that operate.
Now, I want an
equation for psi of t.
So I have here psi of t0.
So I can write this
as du of t, t0 dt.
And now put a psi at t.
And then, I could
put a u from t to t0.
Now, this u of t and t0 just
brings it back to time t0.
And this is all good now, I have
this complicated operator here.
But there's nothing too
complicated about it.
Especially if I
reverse the order here,
I'll have du dt of t, t0
and u dagger of t, t0.
And I reverse the order there
in order that this operator
is the same as that, the one
that is being [INAUDIBLE] that
has the same order of
arguments, t and t0.
So I've got something now.
And I'll call this
lambda of t and t0.
So what have I learned?
That d dt of psi and t is equal
to lambda of t, t0 psi of t.
Questions?
I don't want to loose
you in their derivation.
Look at it.
Anything-- you got lost,
notation, anything.
It's a good time to ask.
Yes.
AUDIENCE: Just to make sure when
you differentiated the state
by t, the reason that you don't
put that in the derivative
because it doesn't have a
time [INAUDIBLE] necessarily,
or because-- oh, because
you're using the value at t0.
PROFESSOR: Right.
Here I looked at that
equation and the only part
that has anything
to do with time t
is the operator, not the state.
Any other comments or questions?
OK, so what have we learned?
We want to know some important
things about this operator
lambda because
somehow, it's almost
looking like a
Schrodinger equation.
So we want to see a
couple of things about it.
So the first thing
that I will show to you
is that lambda is, in
fact, anti-Hermitian.
Here is lambda.
I could figure out,
what is lambda dagger?
Well, lambda dagger is you
take the dagger of this.
You have to think when you
take the dagger of this thing.
It looks a little worrisome,
but this is an operator.
This is another operator,
which is a time derivative.
So you take the dagger by
doing the reverse operators
and daggers.
So the first factor
is clearly u of t, t0.
And then the dagger of this.
Now, dagger doesn't interfere
at all with time derivatives.
Think of the time derivative--
operator at one time,
operator at another
slightly different time.
Subtract it.
You take the dagger
and the dagger
goes through the derivative.
So this is d u dagger t, t0 dt.
So I wrote here what
lambda dagger is.
You have here what lambda is.
And the claim is that one
is minus the other one.
It doesn't look
obvious because it's
supposed to be anti-Hermitian.
But you can show it is true by
doing the following-- u of t,
t0 u dagger of t, t0
is a unitary operator.
So this is 1.
And now you differentiate
with respect to t.
If you differentiate
with respect to t,
you get du dt of t, t0 u dagger
of t, t0 plus u of t, t0 du
dagger of t, t0 equals 0 because
the right-hand side is 1.
And this term is lambda.
And the second term
is lambda dagger.
And they add up to 0, so
lambda dagger is minus lambda.
Lambda is, therefore,
anti-Hermitian as claimed.
Now, look.
This is starting to
look pretty good.
This lambda depends on t and t0.
That's a little nasty though.
Why?
Here is t.
What is t0 doing here?
It better not be there.
So what I want to show to
you is that even though this
looks like it has a t0
in there, there's no t0.
So we want to show this operator
is actually independent of t0.
So I will show that if
you have lambda of t, t0,
it's actually equal to
lambda of t, t1 for any t1.
We'll show that.
Sorry.
[LAUGHTER]
PROFESSOR: So this will
show that you could take t1
to be t0 plus epsilon.
And take the limit
and say the derivative
of this with respect of t0 is 0.
Or take this to
mean that it's just
absolutely independent of t0
and t0 is really not there.
So if you take t1 equal
t dot plus epsilon,
you could just
conclude from these
that this lambda with
respect to t0 is 0.
No dependence on t0.
So how do we do that?
Let's go a little quick.
This is du t, t0 dt
times u dagger of t, t0.
Complete set of states
said add something.
We want to put the t1 here.
So let's add something
that will help us do that.
So let's add t, t0 and
put here a u of t0,
t1 and a u dagger of t0, t1.
This thing is 1, and I've put
the u dagger of t, t0 here.
OK, look at this.
T0 and t1 here and t
dot t1 there like that.
So actually, we'll do
it the following way.
Think of this whole
thing, this d dt
is acting just on this factor.
But since it's time,
it might as well
be acting on all of this factor
because this has no time.
So this is d dt on
u t, t0 u t0, t1.
And this thing is u of t1m t0.
The dagger can be
compensated by this.
And this dagger is u of t0, t.
This at a t and that's a comma.
t0, t.
Yes.
OK, so should I go there?
Yes.
We're almost there.
You see that the first
derivative is already
d dt of u of t, t1.
And the second operator
by compensation
is u of t1, t, which is the
same as u dagger of t, t1.
And then, du of t, t1 u dagger
of t, t1 is lambda of t, t1.
So it's a little
sneaky, the proof,
but it's totally rigorous.
And I don't think
there's any step
you should be worried there.
They're all very
logical and reasonable.
So we have two things.
First of all, that
this quantity,
even though it looks
like it depends on t0,
we finally realized that
it does not depend on t0.
So I will rewrite this
equation as lambda of t.
And lambda of t
is anti-Hermitian,
so we will multiply by an
i to make it Hermitian.
And in fact, lambda has
units of 1 over time.
Unitary operators have no units.
They're like numbers,
like 1 or e to the i phi,
or something like
that-- have no units.
So this has units
of 1 over time.
So if I take i h
bar lambda of t,
this goes from lambda being
anti-Hermitian-- this operator
is now Hermitian.
This goes from lambda
having units of 1 over time
to this thing having
units of energy.
So this is a Hermitian
operator with units of energy.
Well, I guess not much
more needs to be said.
If that's a Hermitian
operator with units of energy,
we will give it a name
called H, or Hamiltonian.
i h bar lambda of t.
Take this equation and multiply
by i h bar to get i h bar
d dt of psi is equal
to this i h bar
lambda, which is
h of t psi of t.
Schrodinger equation.
So we really got it.
That's the Schrodinger equation.
That's the question
that must be satisfied
by any system governed by
unitary time evolution.
There's not more information
in the Schrodinger equation
than unitary time evolution.
But it allows you to
turn the problem around.
You see, when you went to
invent a quantum system,
you don't quite know how
to find this operator u.
If you knew u, you know
how to evolve anything.
And you don't have
any more questions.
All your questions in life
have been answered by that.
You know how to find the future.
You can invest in
the stock market.
You can do anything now.
Anyway, but the unitary operator
then gives you the Hamiltonian.
So if somebody tells you,
here's my unitary operator.
And they ask you, what
is the Hamiltonian?
You go here and calculate
I h bar lambda, where
lambda is this derivative.
And that's the Hamiltonian.
And we conversely,
if you are lucky--
and that's what we're
going to do next time.
If you have a
Hamiltonian, you try
to find the unitary
time evolution.
That's all you want to know.
But that's a harder problem
because you have a differential
equation.
You have h, which is here
, and you are to find u.
So it's a first-order matrix
differential equation.
So it's not a simple problem.
But why do we like Hamiltonians?
Because Hamiltonians
have to do with energy.
And we can get inspired
and write quantum systems
because we know the energy
functional of systems.
So we invent a Hamiltonian
and typically try
to find the unitary
time operator.
But logically
speaking, there's not
more and no less in the
Schrodinger equation
than the postulate of
unitary time evolution.
All right, we'll
see you next week.
In fact--
[APPLAUSE]
Thank you.
