PROFESSOR: SHO algebraically.
And we go back to the
Hamiltonian, p squared over 2m
plus 1/2 m omega
squared x hat squared.
And what we do is observe
that this some sort
of sum of squares plus
p squared over m--
p squared over m
squared omega squared.
So the sum of two
things squared.
Now, the idea that
we have now is
to try to vectorize
the Hamiltonian.
And what we call vectorizing is
when you write your Hamiltonian
as the product of two vectors,
V times W. Well actually,
that's not quite
the vectorization.
You want kind of the same
vector, and not even that.
You sort of want this to be the
Hermitian conjugate of that.
And if there is a
number here, that's OK.
Adding numbers to a Hamiltonian
doesn't change the problem
at all.
The energies are all
shifted, and it's
just how you're defining
the zero of your potential,
is doing nothing but that.
So vectorizing the Hamiltonian
is writing it in this way,
as V dagger V. And you
would say, why V dagger V?
Why not VV dagger or VV
or V dagger V dagger?
Well, you want the
Hamiltonian to be Hermitian.
And this thing is Hermitian.
You may recall that AB dagger.
The Hermitian conjugate of AB
dagger is B dagger A dagger.
So the Hermitian
conjugate of this product
is V dagger times the
dagger of V dagger.
A dagger of a dagger is the same
operator, when you dagger it
twice, you get the same.
So this is Hermitian.
V dagger times V is
a Hermitian operator,
and that's a very good thing.
And there will be
great simplifications.
If you ever succeed in writing
a Hamiltonian this way,
you've gone 90% of the way
to solving the whole problem.
It has become infinitely easier,
as you will see in a second,
if you could just write
this vectorization.
So if you had x minus--
x squared minus this, you
would say, oh, clearly that's--
A squared minus B squared
is A minus B time A plus B,
but there's no such thing here.
It's almost like A
squared plus B squared.
And how do you sort
of factorize it?
Well, actually, since
we have complex numbers,
this could be A minus
IB times A plus IB.
That is correctly A
squared plus B squared,
and complex numbers are supposed
to be friends in quantum
mechanics, so having Is, there's
probably no complication there.
So let's try that.
I'll write it.
So here we have x
squared plus p squared
over m squared omega squared.
And I will try to write it as
x minus i p hat over m omega
times x plus I p
hat over m omega.
Let's put the question
mark before we
are so sure that this works.
Well, some things work.
The only danger here is
that these are operators
and they don't commute.
And when we do this, in one
case, in the cross-terms,
the A is to the left of B,
but the other problem the B
is to the left of A. So we
may run into some trouble.
This may not be exactly true.
So what is this?
This x with x, fine.
x squared.
This term, p with ps, correct.
Plus p squared over m
squared omega squared.
But then we get
plus i over m omega,
x with p minus p with x,
so that x, p commutator.
So vectorization of operators
in quantum mechanics
can miss a few concepts
because things don't commute.
So the cross-terms
give you that,
and this x, p is I h bar, so
this whole term will give us
the following statement.
What we've learned is
that what we wanted,
x squared plus p
squared over m squared
omega squared is equal to--
so I'm equating this
line to the top line--
is equal to x hat minus i
p hat over m omega times
x hat plus i p hat over m omega.
And then, from this whole
term, i with i is minus,
so it's h bar over m omega.
So I'll put it in--
it's a minus h bar
over m, [INAUDIBLE].
So here is plus h bar
over m omega times
a unit vector, if you wish.
OK.
So this is very good.
In fact, we can call this
V dagger and this V. Better
call this V first
and then ask, what
is the dagger of this operator?
Now, you may remember that,
how did we define daggers?
If you have phi with psi
and the inner product--
with an integral
of five star psi--
if you have an A
psi here, that's
equal to A dagger phi psi.
So an operator is acting
on the second wave
function, moves as A dagger
into the first wave function.
And you know that x moves
without any problem.
x is Hermitian.
We've discussed that p
is Hermitian as well,
moves to the other side.
So the Hermitian
conjugate of this operator
is x, the p remains means p,
but the i becomes minus i.
So this is correct.
If this second
operator is called V,
the first operator should
be called V dagger.
That is a correct statement.
One is the dagger
of the other one.
So the Hamiltonian
is 1/2 m omega
squared times this
sum of squares,
which is now equal to V dagger
V plus h bar over m omega.
So h hat is now
1/2 m omega squared
V dagger V plus a sum, which
is plus 1/2 h bar omega.
So we did it.
We vectorized the
Hamiltonian V dagger V,
and this is quite useful.
So the Vs, however, have units.
And you probably are aware that
we like things without units,
so that we can see
the units better.
This curve is perfectly nice.
It's a number added
to the Hamiltonian.
It's h omega, it
has units of energy,
but this is still
a little messy.
So let's try to clean up those
Vs, and the way I'll do it
is by computing their
commutator, to begin with.
So let's compute the
commutator of V and V dagger
and see how much
is that commutator.
It's a simple commutator,
because it involves vectors
of x and V. So
it's the commutator
of x plus ip over
m omega, that's V,
with x minus ip over m omega.
So the first x talks
only to the second piece,
so it's minus i
over m omega x, p.
And for the second
case, you have plus
i over am omega p with x.
This is i h bar, and
this is minus i h bar.
Each term will contribute
the same, i times minus i
is plus, so h bar over
and, omega times the 2.
That is V dagger V.
VV dagger, I'm sorry.
2 h bar over m omega.
So time to change
names a little bit.
Let's do the following.
Let's put square root of
m omega over 2 h bar V.
Have a square root of m
omega over 2 h bar V dagger,
commute to give you 1.
That's a nice commutator.
It's one number-- or an
operator is the same thing.
So I brought the square
root into each one.
And we'll call the first term--
because of reasons
we'll see very soon--
the destruction operator, A
square root of m omega over 2 h
bar V. It's called the
destruction operator.
And the dagger is
going to be A dagger.
Some people put hats on them.
I sometimes do too,
unless I'm too tired.
2h bar V dagger.
And those A and A daggers
are now unit-free--
and you can check That--
Because they have
the same units.
And A with A dagger is
the nicest commutator, 1.
Is A a Hermitian operator?
Is it?
No.
A is not Hermitian.
A dagger is different from
A. A is basically this thing,
A dagger is this thing.
So not Hermitian.
So we're going to work
with these operators.
They're non-Hermitian.
I need to write the
following equations.
It's very-- takes a
little bit of writing,
but they should be
recorded, they will always
make it to the formula sheet.
And it's the basic relation
between A, A dagger, and x
and p.
A is this, A
dagger, as you know,
is x minus ip hat over m omega.
Since I'm copying, I'd
better copy them right.
x, on other hand,
is the square root
of h bar over 2m
omega A plus A dagger,
and p is equal to i square
root of m omega h bar over 2
A dagger minus A.
So these four equations,
A and A dagger
in terms of x and p and
vice versa, are important.
They will show up all the time.
Here are the things to notice.
A and A dagger is
visibly clear that
on is the Hermitian
conjugate of the other.
Here, x is Hermitian.
And indeed, A plus A
dagger is Hermitian.
When you do the Hermitian
conjugate of A plus A dagger,
the first A becomes an A dagger.
The second A, with another
Hermitian conjugation,
becomes A. So this is Hermitian.
But p is Hermitian, and here we
have A dagger minus A. This is
not Hermitian, it changes sign.
Well, the i is there for that
reason, and makes it Hermition.
So there they are, they're
Hermitian, they're good.
