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PROFESSOR: We start with
the stationary states.
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In fact, stationary
states are going
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to keep us quite busy for
probably a couple of weeks.
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Because it's a place where you
get the intuition about solving
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Schrodinger's equation.
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So the stationary states are
simple and useful solutions
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of the Schrodinger equation,
very nice and simple.
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So what are they by definition?
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Well, we begin
with a definition.
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And the intuition of a
stationary state will follow.
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See the word stationary
is not the same as static.
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Stationary is something that
maybe it's kind of moving,
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but things don't change.
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Static is something
that's just not moving.
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Stationary states
have time dependence.
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It is very simple,
as we will see.
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So, your definition
of a stationary state
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has a factorized space
and time dependencies.
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So this psi of x and t
is a stationary state.
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If you can write it as a
product of a function of time
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times a function of position.
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And now, I will try to
be careful about this.
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Wave functions that
have position and time
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will have this
bar at the bottom.
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Wave functions that don't
have x will not have it.
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If I slip on that,
please let me know.
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So this is a stationary
state, but a stationary state
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has factorized space
and time dependencies
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and solves the Schrodinger
equation-- the solution
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of Schrodinger's equation.
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So what we need to understand is
what this factorized dependence
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tell us for the
Schrodinger equation.
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So this stationary state
has time dependence.
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But the thing that
makes them stationary
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is that if you look
at some observable,
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a Hermitian operator,
and you say,
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OK, the state has
time dependence,
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so maybe my observable
will have time dependence.
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No.
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The observables don't
have time dependence.
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That is the nice thing
about stationary states.
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So, what we call time
independent observables which
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are all observables that
are familiar [INAUDIBLE]--
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Hamiltonian, the
momentum, the precision,
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the angular momentum.
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Time independent observables
have no time dependence.
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And it kind of looks simple
when you write it like that.
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Time independent means
don't have time dependence.
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But you've seen that d
dt of the expectation
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value of x is equal to p
over m, or the sum of p
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over m, the velocity.
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And here it is-- a time
independent observable
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that does have time dependence.
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So the observable
is time independent,
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but expectation value
have no time dependence
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in their expectation values.
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The time dependence
comes from the state--
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the state, the psi of x
and t have time dependence,
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and sometimes it just
doesn't drop out.
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But for stationary states, it
will drop, as you will see.
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So, time independent observables
have no time dependence
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in their expectation values.
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OK.
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So enough of saying things.
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And let's just get to them.
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So we look at the
Schrodinger equation,
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i h-bar d dt of
psi of x and t is
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equal to h-bar psi of x and t.
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And just to remind, this minus
h squared over 2m d second d x
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squared plus V of x.
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And I will consider states that
have just that at this moment.
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Otherwise, it's not so easy
to get time-dependent--
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to get stationary states.
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If you have a potential
that has time dependence,
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we kind of do the nice thing
that we're going to do.
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So we're going to look only at
time independent potentials.
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So, V of x, like this,
times psi of x and t.
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OK.
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So what we do next is
to simply substitute
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the value of the wave function
into the differential equation
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and see what we get.
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So on the left
hand side, we will
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get i h-bar The psi of x goes
out but you have d dt, Now
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a normal derivative of g of t.
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And now, this factor, H of
psi acts on these two things.
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Imagine the function
of time times
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the function of x sitting here.
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Well, the function of
time can be moved out.
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So the function of time can be
moved across the Hamiltonian
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operator.
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It doesn't do anything to it.
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So we'll have g of t times
H-hat of the psi of x.
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This is H-hat.
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And because we had
no time dependence
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in the potential,
our assumption,
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this whole thing
is a function of x.
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All right.
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Next step.
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Divide this whole equation
by the total wave function.
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Divide by psi.
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Well, if you divide by psi,
you cancel the middle psi here,
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and you get the 1 over g.
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So you get i h-bar 1
over g dg dt is equal--
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on the right side, you
cancel the g and you get a 1
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over psi of x H-hat psi of x.
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And now you look
at this equation.
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And this equation is saying
something very strange.
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The left hand side is a
function of time only.
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The right hand side is a
function of space only.
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How can a function of time be
equal to a function of space?
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The only way this can
be is if both are not
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a function of what they
were supposed to be.
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They're just numbers.
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Any function of time cannot be
equal to a function of space,
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in generality.
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It just doesn't make sense.
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So each side must be
equal to a constant,
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and it's the same constant.
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So each side, this is
all equal to a constant.
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And we'll call the constant E.
And this E has units of energy.
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E equal to be a real constant
with units of energy.
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You can see the units
because the Hamiltonian
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has units of energy.
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And whatever psi units it has--
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whatever every unit
psi has, they cancel.
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Here, whatever units
g has, they cancel.
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And h-bar over time
is units of energy,
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like in energies
equal h-bar omega.
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So it has units of energy.
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The only thing that you
may be could say, why real.
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Quantum mechanics
loves complex numbers.
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So why don't we
put the complex E?
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We'll see what trouble
we get if you choose
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something that is complex.
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So here we go.
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It's a real quantity
to be-- let's
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do it real for the time being.
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And let's solve
the first equation.
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The left hand side, i h-bar dg
dt is now equal to gE, or E,
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where E is a number and
g is a function of time
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from where g of p is
equal to constant E
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to the minus iEt over h-bar.
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Let's just check
it works correctly.
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It's a first order
differential equation.
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Just one function
of integration.
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If you guess the answer,
must be the answer.
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And that's the time dependence
of a stationary state.
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It's exponential
minus iEt over h-bar.
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What about the other equation?
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The other equation has become
H psi of x equals E psi of x.
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Or, we should write at least
once, minus h squared over 2m--
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did I make a mistake?
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No, I didn't-- d
second dx squared--
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I got this normal
derivatives here
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because this is just
a function of x--
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plus V of x psi is
equal to E psi of x.
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This is the same equation
that I'm boxing twice,
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because it's written
in those two ways.
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And both ways are
very important.
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And this is part of solving
for stationary state.
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You've solved for g of t.
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The time dependence
was easy to solve for,
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but the x dependence is
complicated, in general.
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There, you have to do some work.
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You have to solve a
differential equation.
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It's not that easy.
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So many people-- most people--
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call this the time independent
Schrodinger equation.
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So that's the time independent
Schrodinger equation,
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where H psi equal E psi.
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And as you can imagine, solving
this differential equation can
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be challenging, or
sometimes very interesting
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because it may be that, as
far as the first equation
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is concerned, of
what we did here,
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we don't know what
this number E is.
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But it may be that the
only reasonable solutions
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that this equation has
are for some values of E.
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The analogy with
matrices should tell you
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that's probably what's
going to happen.
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Because you remember eigenstates
and eigenvalues of matrices
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are peculiar numbers.
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If you have a matrix,
they're peculiar eigenvalues.
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So this equation is an
eigenfunction equation.
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And it's possible that
it has the solution
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for some particular
values of the energy.
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Let me write the
whole solution then.
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If you've solved
these two things,
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the whole solution
psi of x and t
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is now a constant
times psi of x times
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e to the minus iEt over
h-bar, where this psi of x
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solves this equation.
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So this is the stationary state.
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How about normalizing
the stationary state?
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Can we do that?
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Well, if we try
to normalize it--
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psi star of x and t
and psi of x and t dx,
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and you set this equal to 1.
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This should be the
case, because this
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should be interesting solutions
of the Schrodinger equation.
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We expect that we could
do particles with them.
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And we can start wave packets
or peculiar states with them.
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And let's see what we get here.
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Maybe I should know.
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I'm really [INAUDIBLE].
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I'm going to erase
that constant C here.
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Since we want to
normalize this, we
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will think of this as
a normalization of psi.
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When we try to
normalize psi, we'll
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be normalizing middle
psi, as you will see here.
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There's no need to put
that constant there.
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So what do we get here?
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We get integral dx
psi star of x and t,
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so you have psi of x star.
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And now you could say it's
E to the iEt over h-bar.
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That's the complex conjugate.
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Now, on the other
hand, suppose--
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I'll do this this way.
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[INAUDIBLE] of
this other term is
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psi of x into the
minus iEt over h-bar.
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And now the good
thing about this, you
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see this integral should be
normalized to 1 to make sense.
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And it's a great thing that
the time dependence drops out.
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And it would not have dropped
out if the energy had not
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been real.
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If the energy was
not real, I would
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have had to put here E start.
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And here I would have
had E star minus E
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and some function of time.
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And how can a function
of time be equal to 1?
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Would be a problem.
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We would not be able to
normalize this wave function.
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So E must be real
because otherwise we
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don't cancel this time
dependence, which happily,
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when it cancels,
it just tells you
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that the integral dx of
psi star of x psi of x
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must be 1, which is
a very nice thing.
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So in a stationary state,
the normalization condition
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for a full time dependent
stationary state
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is that the spacial
part is normalized.