PROFESSOR: ih bar
d psi dt equal E
psi where E hat is equal to p
squared over 2m, the operator.
That is the
Schrodinger equation.
The free particle
Schrodinger equation--
you should realize it's
the same thing as this.
Because p is h bar over i ddx.
And now Schrodinger did the
kind of obvious thing to do.
He said, well, suppose I have a
particle moving in a potential,
a potential V of x and t--
potential.
Then the total energy is kinetic
energy plus potential energy.
So how about if we think of
the total energy operator.
And here is a guess.
We'll put the just p squared
over 2m, what we had before.
That's the kinetic
energy of a particle.
But now add plus V of
x and t, the potential.
That is reasonable from
your classical intuition.
The total energy
is the sum of them.
But it's going to change
the Schrodinger equation
quite substantially.
Now, most people,
instead of calling this
the energy operator,
which is a good name,
have decided to call
this the Hamiltonian.
So that's the most popular
name for this thing.
This is called
the Hamiltonian H.
And in classical
mechanics, the Hamiltonian
represents the energy
expressed in terms
of position and momenta.
That's what the
Hamiltonian is, and that's
roughly what we have here.
The energy is [? in ?]
[? terms ?] [? of ?] momenta
and position.
And we're going to soon
be getting to the position
operator, therefore.
So this is going to
be the Hamiltonian.
And we'll put the hat as well.
So Schrodinger's
inspiration is to say, well,
this is going to be H hat.
And I'm going to say that ih bar
d psi dt is equal to H hat psi.
Or equivalently, ih bar ddt
of psi of x and t is equal
to minus h squared over
2m, [? v ?] [? second ?] dx
squared--
that's the p squared over 2m--
plus V of x and t,
all multiplying psi.
This is it.
This is the full
Schrodinger equation.
So it's a very simple departure.
You see, when you
discover the show
that the equation for a free
particle, adding the energy
was not that difficult. Adding
the potential energy was OK.
We just have to interpret this.
And maybe it sounds to
you a little surprising
that you multiply this by psi.
But that's the only way it could
be to be a linear equation.
It cannot be that psi is
acted by this derivative,
but then you add v.
It would not be a
linear equation.
And we've realize that the
structure of the Schrodinger
equation is d psi dt is equal
to an energy operator times psi.
The whole game of
quantum mechanics
is inventing energy
operators, and then solving
these equations, then
see what they are.
So in particular, you
could invent a potential
and find the equations.
And, you see, it looks funny.
You've made a very
simple generalization.
And now you have an equation.
And now you can put the
potential for the hydrogen atom
and calculate, and
see if it works.
And it does.
So it's rather unbelievable
how very simple generalizations
suddenly produce
an equation that
has the full spectrum
of the hydrogen atom.
It has square wells, barrier
penetration, everything.
All kinds of dynamics
is in that equation.
So we're going to say a few more
things about this equation now.
And I want you to understand
that the V, at this moment,
can be thought as an operator.
This is an operator,
acts on a wave
function to give you a function.
This is a simpler operator.
It's a function of x and t.
And multiplying by a
function of x and t gives you
a function of x and t.
So it is an operator.
Multiplying by a given
function is an operator.
It changes all the functions.
But it's a very simple one.
And that's OK, but
V of x and t should
be thought as an operator.
So, in fact, numbers
can be operator.
Multiplication by a
number is an operator.
It adds on every function and
multiplies it by a number,
so it's also an operator.
But x has showed up.
So it's a good time to
try to figure out what
x has to do with these things.
So that's what we're
going to do now.
Let's see what's x
have to do with things.
OK, so functions
of x, V of x and t
multiplied by wave functions,
and you think of it
as an operator.
So let's make this formal.
Introduce an operator,
X hat, which,
acting on functions of
x, multiplies them by x.
So the idea is that if you
have the operator X hat acting
on the function f
of x, it gives you
another function, which is
the function x times f of x--
multiplies by x.
And you say, wow,
well, why do you
have to be so careful in
writing something so obvious?
Well, it's a good
idea to do that,
because otherwise
you may not quite
realize there's something
very interesting happening
with momentum and
position at the same time,
as we will discover now.
So we have already
found some operators.
We have operators
P, x, Hamiltonian,
which is p squared over 2m.
And now you could
put V of x hat t.
You know, if here you
put V of x hat, anyway,
whatever x hat does
is multiplied by x.
So putting V of x hat here--
you may want to do
it, but it's optional.
I think we all know
what we mean by this.
We're just multiplying
by a function of x.
Now when you have operators,
operators act on wave functions
and give you things.
And we mentioned that
operators are associated
or analogs of matrices.
And there's one fundamental
property of matrices.
The order in which you multiply
them makes a difference.
So we've introduced
two operators, p and x.
And we could ask whether
the order of multiplication
matters or not.
And this is the way Heisenberg
was lead to quantum mechanics.
Schrodinger wrote
the wave equation.
Heisenberg looked at operators
and commutation relations
between them.
And it's another way of
thinking of quantum mechanics
that we'll use.
So I want to ask the question,
that if you have p and x
and you have two operators
acting on a wave function,
does the order matter,
or it doesn't matter?
We need to know that.
This is the basic
relation between p and x.
So what is the question?
The question is, if I have--
I'll show it like that--
x and p acting on a
wave function, phi,
minus px acting on a wave
function, do I get 0?
Do I get the same result or not?
This is our question.
We need to understand
these two operators
and see how they are related.
So this is a very good question.
So let's do that computation.
It's, again, one of
those computation that
is straightforward.
But you have to be careful,
because at every stage,
you have to know very
well what you're doing.
So if you have two
operators like a and b
acting on a function,
the meaning of this
is that you have a acting on
what b acting on phi gives you.
That's what it means to
have two things acting.
Your first act with
the thing on the right.
You then act on the other one.
So let's look at this thing--
xp phi minus px phi.
So for the first
one, you would have
x times p hat on phi minus p hat
times x on phi, phi of x and t
maybe--
phi of x and t.
OK, now what do we have?
We have x hat acting on this.
And this thing, we
already know what it is--
h over i ddx of phi of x and t--
minus p hat and
x, acting on phi,
is little x phi of x and t.
Now this is already a
function of x and t.
So an x on it will
multiply it by x.
So this will be h
over i x ddx of phi.
It just multiplies it
by x at this moment--
minus here we have h
bar over i ddx of x phi.
And now you see that when
this derivative acts on phi,
you get a term
that cancels this.
But when it acts on x, it
gives you an extra term. ddx
of x is minus h over i phi--
or ih phi.
So the derivative
acts on x or an phi.
When it acts on phi,
gives you this term.
When it acts on x, gives you
the thing that is left over.
So actually, let me write
this in a more clear way.
If you have an operator,
a linear operator A
plus B acting on a function
phi, that's A phi plus B phi.
You have linear
operators like that.
And we have these things here.
So this is actually equal
to x hat p hat minus p hat
x hat on phi.
That's what it means when
you have operators here.
So look what we got, a
very surprising thing.
xp minus px is an operator.
It wants to act on function.
So we put a function
here to evaluate it.
And that was good.
And when we evaluate it, we got
a number times this function.
So I could say--
I could forget about the phi.
I'm simply right that xp
minus px is equal to ih bar.
And although it looks a little
funny, it's perfectly correct.
This is an equality
between operators--
equality between operators.
On the left-hand side, it's
clear that it's an operator.
On the right-hand side,
it's also an operator,
because a number acts as an
operator on any function it
multiplies by it.
So look what you've
discovered, this commutator.
And that's a notation
that we're going
to use throughout this semester,
the notation of the commutator.
Let's introduce it here.
So if you have two
operators, linear operators,
we define the commutator
to be the product
in the first direction minus the
product in the other direction.
This is called the
commutator of A and B.
So it's an operator,
again, but it shows you
how they are non-trivial, one
with respect to the other.
This is the basis, eventually,
of the uncertainty principle.
x and p having a
commutator of this type
leads to the
uncertainty principle.
So what did we learn?
We learned this
rather famous result,
that the commutator of x and p
in quantum mechanics is ih bar.
