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BARTON ZWIEBACH: Today's
subject is momentum space.
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We're going to kind of discover
the relevance of momentum
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space.
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We've been working
with wave functions
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that tell you the
probabilities for finding
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a particle in a given
position and that's sometimes
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called coordinate
space or position space
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representations of
quantum mechanics,
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and we just talked about
wave functions that
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tell you about probabilities
to find a particle in a given
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position.
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But as we've been
seeing with momentum,
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there's a very intimate relation
between momentum and position,
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and today we're going to
develop the ideas that
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lead you to think
about momentum space
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in a way that is quite
complimentary to coordinate
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space.
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Then we will be able to
talk about expectation
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values of operators
and we're going
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to be moved some steps into
what is called interpretation
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of quantum mechanics.
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So operators have expectations
values in quantum mechanism--
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they are defined
in a particular way
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and that will be
something we're going
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to be doing in the second
part of the lecture.
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In the final part
of the lecture,
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we will consider
the time dependence
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of those expectation
values, which is again,
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the idea of dynamics--
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if you want to understand how
your system evolves in time,
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the expectation value--
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the things that you measure--
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may change in time
and that's part
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of the physics of the
problem, so this will tie
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into the Schrodinger equation.
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So we're going to begin
with momentum space.
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So we'll call this
uncovering momentum space.
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And for much of what
I will be talking
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about in the first
half of the lecture,
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time will not be relevant--
so time will become relevant
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later.
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So I will be writing
wave functions that
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don't show the
time, but the time
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could be put there everywhere
and it would make no difference
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whatsoever.
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So you remember these
Fourier transform statements
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that we had that a wave
function of x we could put time,
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but they said let's
suppress time.
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It's a superposition
of plane waves.
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So there is a superposition
of plane waves
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and we used it last time to
evolve wave packets and things
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like that.
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And the other side of Fourier's
theorem is that phi of k
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can be written by a pretty
similar integral, in which you
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put psi of x, you change
the sign in the exponential,
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and of course,
integrate now over x.
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Now this is the wave
function you've always
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learned about first, and
then this wave function,
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as you can see, is encoded
by phi of k as well.
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If you know phi of
k, you know psi of x.
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And so this phi of k has the
same information in principle
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as psi of x--
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it tells you everything
that you need to know.
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We think of it as
saying, well, phi of k
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has the same
information as psi of x.
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And the other thing
we've said about phi of k
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is that it's the weight with
which you're superposing plane
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waves to reconstruct psi of x.
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The Fourier transform
theorem is a representation
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of the wave function in terms of
a superposition of plane waves,
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and here, it's the
coefficient of the wave that
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accompanies each exponential.
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So phi of k is the
weight of the plane
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waves in the superposition.
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So one thing we want to do is
to understand even deeper what
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phi of k can mean.
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And so in order to do that,
we need a technical tool.
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Based on these
equations, one can
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derive a way of representing
this object that we
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call the delta function.
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Delta functions are pretty
useful for manipulating objects
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and Fourier transforms,
so we need them.
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So let's try to obtain what
is called the delta function
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statement.
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And this is done by
trying to apply these two
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equations simultaneously.
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Like, you start
with psi of x, it's
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written in terms of
phi of k, but then what
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would happen if you
would substitute
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the value of phi of k in here?
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What kind of equation you get?
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What you get is an equation
for a delta function.
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So you begin with psi of x
over square root of 2 pi,
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and I'll write it
dk e to the ikx,
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and now, I want
to write phi of k.
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So I put phi of k--
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1 over square root
of 2 pi integral psi,
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and now I have to
be a little careful.
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In here in the second integral,
x is a variable of integration,
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it's a dummy variable.
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It doesn't have any
physical meaning, per se--
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it disappears after
the integration.
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Here, x represents a point
where we're evaluating the wave
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function, so I just cannot
copy the same formula here
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because it would be confusing,
it should be written with
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a different x and x-prime.
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Because that x certainly
has nothing to do with the x
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we're writing here.
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So here it is, we've
written now this integral.
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And let's rewrite it
still differently.
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We'll write as
integral dx-prime.
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We'll change the orders of
integration with impunity.
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If you're trying to be very
rigorous mathematically,
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this is something
you worry about.
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In physics problems
that we deal with,
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it doesn't make a difference.
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So here we have dx-prime,
psi of x-prime, then
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a 1 over 2 pi integral dk e
to the ik x minus x-prime.
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So this is what the
integral became.
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And you look at it and you
say, well, here is psi of x,
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and it's equal to the
integral over x-prime
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of psi of x-prime times
some other function
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of x and x-prime.
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This is a function of x minus
x-prime, or x and x-prime
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if you wish.
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It doesn't depend on
k, k is integrated.
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And this function,
if you recognize it,
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it's what we call
a delta function.
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It's a function that,
multiplied with an integral,
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evaluates the integrand
at a particular point.
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So this is a delta function,
delta of x-prime minus x.
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That is a way these
integrals then would work.
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That is, when you
integrate over x-prime--
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when you have x-prime
minus something,
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then the whole integral
is the integrand
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evaluated at the point where we
say the delta function fires.
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So the consistency of
these two equations
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means that for all
intents and purposes,
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this strange integral
is a representation
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of a delta function.
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So we will write it down.
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I can do one thing here, but--
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it's kind of here you see
the psi of x-prime minus x,
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but here you see
x minus x-prime.
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But that's sign, in
fact, doesn't matter.
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It's not there.
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You can get rid of it.
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Because if in this integral,
you can let K goes to minus k,
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and then the dk changes sign,
the order of integration
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changes sign, and they
cancel each other.
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And the effect is
that you change
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the sign in the exponents, so
if you let k goes to minus k,
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the integral just becomes
1 over 2 pi integral dk e
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to the ik x-prime minus x.
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So we'll say that
this delta function
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is equal to this thing,
exploiting this sign ambiguity
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that you can always have.
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So I'll write it
again in the way
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most people write the formula,
which is at this moment,
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switching x and x-prime.
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So you will write this
delta of x minus x-prime
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is 1 over 2 pi integral from
infinity to minus infinity dk
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e to the ik x minus x-prime.
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So this is a pretty
useful formula
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and we need it all the time
that we do Fourier transforms
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as you will see very soon.
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It's a strange integral, though.
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If you have x equal to x-prime,
this is 0 and you get infinity.
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So it's a function,
it's sort of 0--
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when x minus x-prime
is different from 0,
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somehow all these
waves superimpose to 0.
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But when x is equal to
x-prime, it blows up.
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So it's does the right thing,
it morally does the right thing,
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but it's a singular
kind of expression
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and we therefore
manipulate it with care
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and typically, we use
it inside of integrals.
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So it's a very nice formula,
we're going to need it,
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and it brings here for the
first time in our course,
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I guess, the delta functions.
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And this is something if
should-- if you have not
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ever play without
the functions, this
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may be something interesting
to ask in recitation
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or you can try to prove, for
example, just like we show
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that delta of minus x
is the same as delta
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of x, the delta of
a being a number
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times x is 1 over absolute
value of a times delta of x.
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Those are two simple
properties of delta functions
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and you could practice by
just trying to prove them--
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for example, use this integral
representation to show them.