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PROFESSOR: This is a wonderful
differential equation,

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because it carries a
lot of information.

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If you put this
psi, it's certainly

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going to be a solution.

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But more than that,
it's going to tell you

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the relation
between k and omega.

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So if you try your--

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we seem to have gone
around in circles.

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But you've obtained
something very nice.

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First, we claim that that's
a solution of that equation

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and has the deep
information about it.

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So if you try again, psi equal
e to the ikx minus i omega t,

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what do we get?

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On the left hand side, we
get ih minus i omega psi.

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And on the right
hand side, we get

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minus h squared over 2m and two
derivatives with respect to x.

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And that gives you an
ik times and other ik.

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So ik squared times psi.

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And the psis cancel from the
two sides of the equation.

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And what do we get
here? h bar omega.

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It's equal to h squared
k squared over 2m, which

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is e equal p squared over 2m.

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So it does the
whole job for you.

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That differential
equation is quite smart.

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It admits these as solutions.

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Then, this will have
definite momentum.

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It will have definite energy.

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But even more, when you
try to see if you solve it,

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you find the proper
relation between the energy

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and the momentum that tells
you you have a particle.

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So this is an infinitely
superior version

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of that claim that that is
a plane wave that exists.

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Because for example, another
thing that you have here

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is that this equation is linear.

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Psi appears linearly, so
you can form solutions

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by superposition.

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So the general solution,
now, is not just this.

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This is a free particle
Schrodinger equation.

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And you might say, well, the
most general solution must

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be that, those plane waves.

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But linearity means that you
can compose those plane waves

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and add them.

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And if you can add plane
waves by Fourier theorem,

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you can create pretty much
all the things you want.

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And if you have
this equation, you

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know how to evolve
free particles.

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Now, you can construct a
wave packet of a particle

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and evolve it with the
Schrodinger equation

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and see how the wave packet
moves and does its thing.

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All that is now
possible, which was not

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possible by just saying,
oh, here is another wave.

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You've worked back
to get an equation.

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And this is something
that happens in physics

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all the time.

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And we'll emphasize it
again in a few minutes.

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You use little
pieces of evidence

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that lead you-- perhaps not
in a perfectly logical way,

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but in a reasonable
way-- to an equation.

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And that equation
is a lot smarter

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than you and all the
information that you put in.

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That equation has
all kinds of physics.

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Maxwell's equations were found
after doing a few experiments.

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Maxwell's equation
has everything in it,

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all kinds of phenomena that
took years and years to find.

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So it's the same
with this thing.

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And the general solution
of this equation

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would be a psi of x
and t, which would be

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a superposition of those waves.

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So you would put an e to
the ikx minus i omega t.

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I will put omega of k
because that's what it is.

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Omega is a function of k.

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And that's what represents our
free particles-- omega of kt.

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And this is a solution.

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But so will be any superposition
of those solutions.

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And the solutions are
parametrized by k.

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You can choose different
momenta and add them.

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So I can put a wave
with one momentum

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plus another wave
with another momentum,

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and that's perfectly OK.

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But more generally,
we can integrate.

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And therefore, we'll write
dk maybe from minus infinity

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to infinity.

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And we'll put a phi of k, which
can be anything that's not part

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of the differential equation.

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Now, this is the
general solution.

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You might probably say,
wow, how do we know that?

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Well, I suggest you try it.

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If you come here,
the ddt will come in.

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We'll ignore the k.

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Ignore this.

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And just gives you
the omega factor here.

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That ddx squared-- we'll
ignore, again, all these things,

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and give you that.

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From the relation omega minus
k equals 0, you'll get the 0.

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And therefore, this
whole thing solves

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the Schrodinger equation--
solves the Schrodinger

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equation.

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So this is very general.

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And for this, applies
what we said yesterday,

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talking about the
velocity of the waves.

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And this wave, we
proved yesterday,

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that moves with
a group velocity,

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v group, which was equal
to d omega dk at some k0,

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if this is localized at k0.

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Otherwise, you can't
speak of the group

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velocity this thing will not
have a definite group velocity.

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And the omega dk--

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And you have this relation
between omega and k,

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such a to way that this is
the evp, as we said yesterday.

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And this is ddp of
b squared over 2m,

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which is p over m,
which is what we call

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the velocity of the particle.

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So it moves with the proper
velocity, the group velocity.

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That's actually a
very general solution.

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We'll exploit it to calculate
all kinds of things.

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A few remarks that come
from this equation.

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Remarks.

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1, psi cannot be a real.

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And you can see that
because if psi was real,

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the right hand
side would be real.

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This derivative would be
real because the relative

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of a real function
is a real function.

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Here you have an
imaginary number.

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So structurally, it is
forbidden to have full wave

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functions that are real.

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I call these full wave
functions because we'll

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talk sometime later about time
independent wave functions.

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But the full wave
function cannot be real.

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Another remark is that
this is not the wave

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equation of the usual type--

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not a usual wave equation.

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And what a usual wave
equation is something

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like d second phi dx
squared minus 1 over v

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squared d second phi
dt squared equals zero.

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That's a usual wave equation.

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And the problem with
that wave equation

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is that it has real solutions.

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Solutions, phi that go
like functions of x minus

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is vt, plus minus x over vt.

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And we cannot have
those real solutions.

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So we managed to get a wave, but
not from a usual wave equation.

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This, waves also all move
with some same velocity,

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velocity, v, of the wave.

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These waves don't do that.

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They have a group velocity.

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It's a little bit
different situation.

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And what has happened
is that we still

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kept the second derivative,
with respect to x.

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But in time, we replaced
it by first derivative.

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And we put an i.

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And somehow, it did
the right job for us.