WEBVTT
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PROFESSOR: We ask,
is psi of x, 0 real?
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And I told you the answer is no.
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And how would I know
that this is not real?
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Well, we can take the
complex conjugate.
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And at the end of the
day, this will boil down
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to some property of phi of k.
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You see, you have an expression
phi of x in terms of phi of k.
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So it would not be surprising
that the requirement
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that psi is real means
something about phi of k.
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So let's just say, suppose
psi is given by that,
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then psi x, 0 star,
the complex conjugate
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would be 1 over square root
of 2 pi integral phi of k star
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e to the minus ikx dk.
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I conjugated everything in that
equation for psi of x and 0.
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Now, you want to compare
this with psi of x and 0
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to see if it's real.
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Or let's consider what is the
condition that this be real.
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So I want to simplify
here a little more.
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So what I'm going to do is
going to change variables,
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by changing k to minus k.
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If you prefer to go
a little more slowly,
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you could say you're
going to change k prime--
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you're going to be a
new k, called k prime,
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equals to minus k.
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But it's possible
to do it this way.
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Now, there's going to
be a couple of changes.
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Wherever you see k, you're
now going to see minus k--
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so 1 over 2 pi integral
phi of minus k.
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And I'll just put this star
here, not so many parentheses--
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e to the minus ikx becomes ikx.
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And the dk will go to minus dk,
but the order of integration,
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that was from minus infinity
to plus infinity, would switch.
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So those two signs cancel.
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So there's a sign from
doing dk to minus dk,
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and 1 from the limit
of integration--
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so at the end of the
day, you have dk,
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and you still have this--
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minus infinity to infinity.
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And you can say, well,
is this equal to--
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or what is the condition
of for psi to be real?
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Well, is this equal to 1
over 2 pi minus phi of k, e
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to the ikx dk--
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is that-- question mark--
is that equal to it?
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That would mean
that the psi of x, 0
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is real, because this thing
is just psi of x and 0.
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So this is a question
mark-- this is a condition.
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So here you could say, exploring
the reality condition--
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condition-- when
is a psi of x real?
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So what must be true
is that these two terms
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must equal each other.
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So, in fact, this requires--
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reality requires that 1 over
square root of 2 pi integral
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from minus infinity
to infinity phi of k
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minus phi star of minus k, e
to the ikx dx is equal to 0.
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I brought the two
terms to one side.
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Both are of the same type--
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they're integrated
against an e to the ikx.
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And therefore, we
can combine them,
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and that's what must be true
in order for the function
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to be real.
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And now you can
say, so what is it?
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What's the answer?
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Well, this integral
should vanish.
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Now, this integral
should vanish--
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it should vanish
for all values of k.
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So actually, what
you want to conclude
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is that this thing
is identically 0.
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AUDIENCE: Excuse me, shouldn't
that be dk and not dx?
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PROFESSOR: Yes, thank you.
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Thanks very much.
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So this property that this whole
integral be equal to 0, you
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were tempted to
conclude that it means
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that this thing is equal to 0.
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And that is correct--
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that is a perfectly
legal argument.
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And it basically-- if you want
to express it more precisely,
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you could base it on the
Fourier theorem, again.
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These two sets of equalities
here are Fourier's theorem.
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And look what this is saying--
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this is saying that this
quantity has a 0 Fourier
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transform.
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Because how do you do
the Fourier transform
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of a function of k?
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You multiply by e to
the ikx and integrate.
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And therefore, this function
has a 0 Fourier transform.
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So, but if a function has
a 0 Fourier transform,
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the function must be 0.
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Because already this is
0, and the integral is 0--
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0.
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So this is absolutely rigorous.
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And therefore, you
get the conclusion
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that phi of minus k star
must be equal to phi of k,
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and that's the
condition for reality.
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So if a phi of k satisfies
this property, that psi of x
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will be real, and our phi of k
doesn't satisfy this property,
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what do you see
in this property?
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Basically, if you have phi that
exists for some value of k,
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it should also exist for
the value of minus k.
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And in fact, should be
the complex conjugate
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of the other value.
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But here, you have some phis
of k, and no phis at minus k.
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So the phi that we wrote above
doesn't satisfy this condition.
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And therefore psi is not
real, and it all makes sense.
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OK, so basically,
if you were plotting
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not the absolute value,
but the real part
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and the imaginary
parts of psi, you
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would see some sort
of funny waves.
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I think if you were plotting
the real part, for example,
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you would see a wave like that.
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And if you were plotting
the imaginary part,
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you'd presumably see some
other wave like that.
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And the absolute value,
it's much nicer and simpler.