WEBVTT
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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.