WEBVTT
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PROFESSOR: Let me do a
little exercise using still
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this manipulation.
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And I'll confirm
the way we think
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about expectations values.
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So, suppose exercise.
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Suppose you have indeed that
psi is equal to alpha i psi i.
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Compute the expectation value
of Q in the state of psi.
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Precisely, the expectation
value of this operator
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we've been talking
about on the state.
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So this is equal to the
integral dx psi star Q psi.
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And now I have to
put two sums before.
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And go a little fast here.
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dx sum over i alpha i psi i
star Q sum over j alpha j psi j.
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No star.
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This is equal to sum over i
sum over j alpha i star alpha j
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integral dx psi i star Q psi j.
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But Q psi j is
equal to qj psi j.
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Therefore, this whole
thing is equal to qj
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times the integral dx
of psi i star psi j,
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which is qj delta ij.
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So here we go.
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It's equal to sum
over i, sum over j,
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alpha i star alpha
j, qj delta ij, which
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is equal to the sum over i.
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The j's disappear.
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And this is alpha i squared qi.
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That's it.
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OK.
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Now you're supposed to
look at this and say, yay.
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Now why is that?
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Look.
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How did we define
expectation values?
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We defined it as the sum of
the value times the probability
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that this value have.
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It's for a random variable.
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So here our random variable is
the result of the measurement.
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And what are the
possible values?
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qi's.
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And what are the probabilities
that they have Pi?
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OK.
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So the expectation value
of q should be that,
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should be the sum of
the possible values
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times their probabilities, and
that's what the system gives.
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This is how we defined
expectation value of x.
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Even though it's
expectation value of P.
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And it all comes from
the measurement postulate
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and the definition.
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Now, this definition and
the measurement postulate
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just shows that this
is what we expect.
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This is the result of
the expectation value.
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OK.
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I think I have a nice example.
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I don't know if I
want to go into all
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the detail of these
things, but they illustrate
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things in a nice way.
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So let's try to do it.
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So here it is.
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It's a physical example.
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This is a nice concrete example
because things work out.
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So I think we'll
actually illustrate
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some physical points.
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Example.
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Particle on a circle.
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x 0 to L. Maybe you haven't
seen a circle described by that,
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but you take the x-axis,
and you say yes, the circle
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is 0 to L. L and 0.
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And the way you think
of it is that this point
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is identified with this point.
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If you have a line and you
identify the two endpoints,
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that's called a circle.
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It's in the sense of topology.
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A circle as the set of points
equidistant to a center
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is a geometric description
of a round circle.
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But this, topologically
speaking, anything
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that is closed is
topologically a circle.
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We think of a circle
as this, physically,
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or it could be a curved line
that makes it into a circle.
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But it's not important.
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Let's consider a free
particle on a circle,
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and suppose the circle has
an end L. So x belongs here.
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And here is the wave
function, psi equals 2 over L,
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1 over square root
of 3 sine of 2 pi
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x over L, plus 2 over square
root of 3 cosine 6 pi x over L.
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This is the wave function of
your particle on a circle.
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At some time, time equals
0, it's a free particle.
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No potential.
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And it lives in the
circle, and these functions
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are kind of interesting.
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You see, if you
live on the circle
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you would want to emphasize
the fact that this point 0 is
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the same as the point L,
so you should have that psi
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and L must be equal to psi at 0.
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It's a circle, after
all, it's the same point.
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And therefore for 0 or
for L, the difference here
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is 0 or 2 pi, and the
sine is the same thing.
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And 0, when x equals 0, and 6
pi, so that's also periodic,
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and it's fine.
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It's a good wave
function result.
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The question is,
for this problem,
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what are, if you measure
momentum, measure momentum,
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what are the possible values
and their probabilities?
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Probabilities.
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So you decide to measure
momentum of this particle.
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What can you get?
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OK.
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It looks a little nontrivial,
and it is a little nontrivial.
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Momentum.
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So I must sort of find
the momentum eigenstates.
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Momentum eigenstates, they are
those infinite plane waves,
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e to the ikx, that we
could never normalize.
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Because you square it, it's 1,
and the integral over all space
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is infinite.
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So are we heading
for disaster here?
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No.
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Because it lives
in a finite space.
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Yes, you have a question?
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STUDENT: Should it be a wave
function [INAUDIBLE] complex?
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Because right now, it just
looks like it's a real value.
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And we can't [INAUDIBLE]
real wave functions, can we?
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PROFESSOR: Well, it is the
wave function at time equals 0.
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So the time
derivative would have
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to bring in complex things.
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So you can have a
wave function that
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is 0, that is real at
some particular time.
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Like, any wave function psi of
x e to the minus iEt over h bar
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is a typical wave function.
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And then at time equal
0 it may be real.
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It cannot be real forever.
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So you cannot assume it's real.
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But at some particular
times it could be real.
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Very good question.
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The other thing you
might say, look, this
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is too real to have momentum.
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Momentum has to do with waves.
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That's probably not
a reliable argument.
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OK, so, where do
we go from here?
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Well, let's try to find
the momentum eigenstates.
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They should be things
like that, exponentials.
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So how could they look?
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Well, e to the 2 pi i, maybe.
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What else?
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x, there should be an
x for a momentum thing.
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Now there should
be no units here,
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so there better be an L here.
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And now I could put, maybe,
well the 2 maybe was--
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why did I think of
the 2 or the pi?
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Well, for convenience.
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But let's see what.
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Suppose you have
a number m here.
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Then the good thing about this
is that when x is equal to 0,
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there is some number here,
but when x is equal to L,
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it's a multiple of e to the
2 pi i, so that's periodic.
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So this does satisfy, I
claim, it's the only way
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if m is any integer.
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So it goes from minus
infinity to infinity.
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Those things are periodic.
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They satisfy psi.
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Actually they satisfy psi of
x plus L is equal to psi of x.
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OK.
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That seems to be something that
could be a momentum eigenstate.
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And then I have to normalize it.
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Well, if I square
it and integrate it.
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If I square it then the
phase cancels, so you get 1.
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If you integrate it
you get L. If you put 1
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over the square root of L, when
you square it and integrate,
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you will get 1.
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So here it is.
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Psi m's of x are going to
be defined to be this thing.
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And I claim these things
are momentum eigenstates.
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In fact, what is the
value of the momentum?
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Well, you calculate h
bar over i d dx on psi m.
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And you get what?
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You get 2 pi m over L
times h bar times psi.
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The h bar is there, the i
cancels, and everything then
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multiplies, the x falls down.
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So this is the
state with momentum
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P equals to h bar 2 pi m over L.
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OK.
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Actually, doing that, we've
done the most difficult part
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of the problem.
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You've found the
momentum eigenfunctions.
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So now the rest of the thing
is to rewrite this in terms
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of this kind of objects.
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I'll do it in a second.
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Maybe I'll leave a
little space there
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and you can check
the algebra, and you
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can see it in the notes.
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But you know what
you're supposed to do.
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A sine of x is e to the ix minus
is e to the minus ix over 2i.
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So you'd get these things
converted to exponentials.
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The cosine of x is
equal to e to the ix
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plus e to the minus ix over 2.
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So if you do that with
those things, look.
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What the sine of 2 pi
x going to give you?
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It's going to give you some
exponentials of 2 pi ix over L.
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So suppose that m equals 1.
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And m Equals minus 1.
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And this will give you m
equals 3, 3 times 2 is 6.
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And m equal minus 3.
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So I claim, after some work,
and you could try to do it.
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I think it would
be a nice exercise.
00:14:47.410 --> 00:14:50.560
Psi is equal square
root of 2 over 3,
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1 over 2 i psi 1 minus square
root of 2 over 3, 1 over 2i psi
00:14:58.860 --> 00:15:04.095
minus 1 plus 1 over
square root of 3 psi 3,
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plus 1 over square
root of 3 psi minus 3.
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And it should give
you some satisfaction
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to see something like that.
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You're now seeing
the wave function
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written as a superposition
of momentum eigenstates.
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This theorem came through.
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In this case, as a
particle in the circle,
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the statement is that
the eigenfunctions
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are the exponentials, and
it's Fourier's theorem.
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Again, for a series.
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So finally, here is the answer.
00:15:40.420 --> 00:15:44.510
So psi 1, we can measure psi 1.
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What is the momentum of psi 1?
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So here are p values.
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And probabilities.
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The first value, psi 1, the
momentum is 2 pi h bar over L.
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So 2 pi h bar over L. And
what is its probability?
00:16:16.750 --> 00:16:19.570
It's this whole number squared.
00:16:19.570 --> 00:16:25.060
So square root of 2/3,
1 over 2i squared.
00:16:25.060 --> 00:16:28.510
So how much is that?
00:16:28.510 --> 00:16:34.200
It's 2/3 times 1/4.
00:16:34.200 --> 00:16:39.520
2/3 times 1/4, which is 1/6.
00:16:39.520 --> 00:16:43.930
And the other value that you
can get is minus this one,
00:16:43.930 --> 00:16:49.830
so minus 2 pi h bar over L.
This minus doesn't matter,
00:16:49.830 --> 00:16:53.550
probability also 1/6.
00:16:53.550 --> 00:16:55.570
The next one is with 3.
00:16:55.570 --> 00:17:03.410
So you can get 2, 6
pi, 6 pi h bar over L,
00:17:03.410 --> 00:17:06.405
with probability
square of this, 1/3.
00:17:09.900 --> 00:17:16.800
And minus 6 pi h bar over
L with probability 1/3.
00:17:16.800 --> 00:17:19.329
Happily our
probabilities add up.
00:17:22.310 --> 00:17:23.099
So there you go.
00:17:23.099 --> 00:17:28.230
That's the theorem expressed
in a very clear example.
00:17:28.230 --> 00:17:30.160
We had a wave function.
00:17:30.160 --> 00:17:34.300
You wrote it as a sum of
four momentum eigenstates.
00:17:34.300 --> 00:17:37.240
And now you know, if
you do a measurement,
00:17:37.240 --> 00:17:40.570
what are the possible
values of the momentum.
00:17:40.570 --> 00:17:44.140
This should have
been probably 1/6.
00:17:44.140 --> 00:17:46.820
You can do anything you want.