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PROFESSOR: So I
want to demystify

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a little of this equation.

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We sometimes use
the basis to show

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everything that is happening.

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So we have a good basis.

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So let's look at it.

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So how does it all look
in an orthonormal basis?

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So which basis?

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Well, you have one.

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H 0 was supposed to be known.

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So we'll call these states n.

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These are the
eigenstates of H 0.

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We'll have E n.

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We could put a zero.

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We used to have that zero there.

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But since we're
never going to try

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to solve this equation in a
different way, as time goes on,

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the time-dependent
system, I said to you,

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doesn't have energy eigenstates.

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So perturbation
theory is not going

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to be about perturbing
these energies.

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So I will erase.

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E n, and there will
be no confusion.

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That's never going to change.

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Now I'm going to
write an equation that

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may seem a little strange.

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We want to solve for this psi ~.

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So let's write an
ansatz for psi ~ of t.

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It's going to be the
following, sum over n C n n.

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If C ns are constants, this
is definitely not right.

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This state must have
some time-dependence.

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So at least, I should put
a time-dependence here.

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But even that sounds a
little wrong, a priori.

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You knew that in
time-independent problems,

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you can always write a state as
a superposition of your energy

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eigenstate.

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So now, this is the
interacting picture

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we have not any more energy
eigenstates in any sense.

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Can I write this?

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Does this make sense still?

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Well, the answer
is yes, you can.

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And the reason is
that whatever you

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can say about this
energy eigenstate,

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they apply for early times.

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They apply for late times.

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They anyway exist.

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They're a basis.

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And what I said,
they are a basis.

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So any wave function can be
written in terms of them.

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So if I look psi ~
at time 1 second,

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I should be able to find
numbers that make this possible.

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And if I look at
it at 2 seconds,

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I will find another set of
numbers and make it possible.

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So by the fact that I've
included here a time-dependence

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for the coefficients, it
is possible to write this.

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These coefficients change
in time in strange ways,

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but that's our unknown.

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If I knew how they
change in time,

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I would have solved the problem.

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I don't know how
the change in time.

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But in general, we will try
to find for this coefficient.

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So if you have psi ~, you
know at the end of the day,

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your goal is psi of t.

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So what is psi of t?

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Psi of t, from this
blackboard, is the action

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of this operator on psi ~.

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That operator moves through the
constants and through the sum.

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And with the end states
eigenstates of H 0,

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this just gives me minus
I E n t over H bar n.

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And this expansion should
make you feel good.

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You say, OK, here it is.

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These are my states in my
time-dependent we function,

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let's say.

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And it's given by this formula.

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If I didn't have a delta H, we
know that without a delta H,

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psi ~ is constant.

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So they C ns would be constant
if there is no delta H.

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And if the C ns are
constant and there's

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no delta H, that's exactly
how energy eigenstates evolve

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in time.

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They are evolving
with exponentials

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of the action of each 0.

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So this is consistent
with all you know.

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If delta H turns on,
the C ns are going

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to acquire time-dependence.

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And we will know what
the state is doing.

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So this is good.

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So let's plug, into
the new Schrodinger

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equation, this expansion.

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And this is our new
Schrodinger equation.

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So what do we get?

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I get I H bar d/dt of this.

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I'll write it here with a sum.

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I'll change to letter m.

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I'll use dots.

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C m of t dot.

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For time derivative,
we'll many time use dots.

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So I'm taking the time
derivative of psi ~

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as in the Schrodinger equation.

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I go here, C m m.

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And this is supposed
to be equal to delta H

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~ times this same state.

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So it's going to be sum
over n C n of t delta H ~ n.

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So this is the Schrodinger
equation for this state.

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We're now looking at
the Schrodinger equation

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in a basis, because that may
be a good way to solve it.

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Well, one way to
solve it is to do,

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in the right-hand
side, a complete set,

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introduce a complete
set of states.

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So I'll put a sum over
m m m and this whole sum

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over n C n of t delta H n.

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So this will be
equal to sum over m.

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The m is here.

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And sum over n, the bra
can go all the way in.

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And you write C n
of t m delta H n.

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So this is our Schrodinger
equation still.

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And now it's kind of in a
nice way in which we have

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similar letters on both sides.

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A little bit of notation.

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I'll call this delta H ~ m n.

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We've done that in perturbation
theory many, many times.

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So what is our equation?

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Well, compare terms with
equal value of the function

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in front of the state m.

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So we gave I H bar
C m dot of t is

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equal to the sum over n C
n or delta H m n C n of t.

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That's it.

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It's a nice looking equation.

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There's a couple sets of
differential equations

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for an infinite set of functions
in which the derivatives are

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obtained in terms of the
Hamiltonian matrix elements

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for the transition Hamiltonian
for the perturbation

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times these functions there.

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A little more notation
here in the sense

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of just understanding the
structure of that matrix

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element.

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That's useful, because in
practice, all the tilde things,

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at the end of the day,
we don't want them.

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We want the original ones,
things without tilde.

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And we always look for them.

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So what is this matrix element
delta H m n ~ is m delta H ~ n.

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But remember what delta H ~ was.

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You have it here.

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So we have m E to the I
H 0 t over H bar delta H

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E to the minus I H
0 t over H bar n.

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That's no problem.

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Everything is good here,
because those are eigenstates.

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So we know how much
you get by letting

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this act on that state and
that other exponential act

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on this state on the right.

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So what do we get?

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E to the I e m t over H bar.

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And from the other one, E
n t over H bar like this.

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The two exponential
give you that.

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And then you have just delta
H, your original perturbation,

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between states of
the Hamiltonian.

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I'll write it this way.

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Delta H mn, just without the
tilde, which is m delta H n.

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Most people call this
the frequency m n.

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E over H bar is a frequency.

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So the harmonic
oscillator reminds you

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of that, E equal H bar omega.

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And for easier writing,
you write an omega mn.

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So this becomes E to the I
omega mn t times delta H mn.

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So this is kind of
a nice notation.

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So this matrix elements there--

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well, I'll write again the
equation I H bar C m dot of t

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would be the sum over n of E to
the I omega mn t delta H mn C

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n of t.

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That same equation
has been rewritten

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in the viewpoint of this,
where we simplified the tilde.

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And we can refer everything
to our original basis.

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So in a sense, I think this
should demystify things.

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What's the situation?

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What has happened is
that if you have a basis,

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you write an ansatz for the
wave function of this form.

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And the coefficients are
solutions of those differential

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equations.

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So we've translated the
problem to something doable.

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If you have lots of
resources, a computer,

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and you solve coupled
time-dependent differential

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equations first order.

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There this are not
partricularly--

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well, it all depends
how difficult

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is the time-dependence here.

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The exponential is not bad.

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This?

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Well, it all depends.

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But this can be
solved numerically.

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It can be solved
with many methods.

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This has made the
problem concrete.

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And we're going to try to
understand how to solve it

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in cases of interest.