PROFESSOR: So we go
back to the integral.
We think of k.
We'll write it as k
naught plus k tilde.
And then we have
psi of x0 equal 1
over square root of 2pi
e to the ik naught x--
that part goes out--
integral dk tilde phi
of k naught plus k tilde
e to the ik tilde x dk.
OK.
So we're doing this integral.
And now we're focusing
on the integration
near k naught, where the
contribution is large.
So we write k as k naught
plus a little fluctuation.
dk will be dk tilde.
Wherever you see a k, you must
put k naught plus k tilde.
And that's it.
And why do we have to worry?
Well, we basically have now
this peak over here, k naught.
And we're going
to be integrating
k tilde, which is
the fluctuation, all
over the width of this profile.
So the relevant region of
integration for k tilde
is the range from delta k over
2 to minus delta k over 2.
So maybe I'll make this
picture a little bigger.
Here is k naught.
And here we're going to be going
and integrate in this region.
And since this is delta
k, the relevant region
of integration--
integration-- for k tilde
is from minus delta k over 2
to delta k over 2.
That's where it's
going to range.
So all the integral has to
be localized in the hump.
Otherwise, you don't
get any contribution.
So the relevant
region of integration
for the only variable that
is there is just that one.
Now as you vary this k tilde,
you're going to vary the phase.
And as the phase changes, well,
there's some effect [? on ?]
[? it. ?] But if x is equal
to 0, the phase is stationary,
because k tilde is going to
very, but x is equal to 0.
No phase is stationary.
And therefore, you will
get a substantial answer.
And that's what we know already.
For x is going to
0 or x equal to 0,
we're going to get a
substantial answer.
But now think of the
phase in general.
So for any x that
you choose, the phase
will range over some value.
So for any x different from
0, the face in the integral
will range over minus delta k
over 2x and to delta k over 2x.
You see, x is here.
The phase is k tilde x.
Whatever x is, since k tilde
is going run in this range,
the phase is going to run in
that range multiplied by x.
So as you do the integral--
now think you're
doing this integral.
You have a nice, real,
smooth function here.
And now you have a running
phase that you don't
manage to make it stationary.
Because when x is
different from 0,
this is not going
to be stationary.
It's going to vary.
But it's going to vary from
this value to that value.
So the total, as you
integrate over that peak,
your phase excursion is
going to be delta k times x--
total phase excursion
is delta k times x.
But then that tells
you what can happen.
As long as this total phase
excursion is very small--
so if x is such
that delta k times
x is significantly less than 1--
or, in fact, I could
say less than 1--
there will be a
good contribution
if x is such that--
then you will get
a contribution.
And the reason is because the
phase is not changing much.
You are doing your integral,
and the phase is not killing it.
On the other hand,
if delta k times x--
delta k times x is
much bigger than 1,
then as you range over
the peak, the phase
has done many, many cycles and
is going to kill the integral.
So if k of x is greater than
1, the contribution goes to 0.
So let's then just extract
the final conclusion
from this thing.
So psi of x 0 will be sizable
in an interval x belonging
from minus x0 to x0.
So it's some value
here minus x0 to x0.
If, even for values
as long as x0,
this product is still about 1--
if for delta k times x0, roughly
say of value 1, we have this.
And therefore the uncertainty
in x would be given by 2x0.
So x0 or 2x0, this x0 is
basically the uncertainty in x.
And you would get
that delta k times
delta x is roughly equal to 1--
so delta k delta x
roughly equal to 1.
So I'm dropping factors of 2.
In principle here,
I should push a 2.
But the 2s, or 1s, or
pi's at this moment
are completely unreliable.
But we got to the
end of this argument.
We have a relation of
uncertainties is equal to 1.
And the thing that comes
to mind immediately
is, why didn't Fourier invent
the uncertainty principle?
Where did we use
quantum mechanics here?
The answer is nowhere.
We didn't use quantum mechanics.
We found the relation between
wave packets, known to Fourier,
known to electrical engineers.
The place where quantum
mechanics comes about
is when you realize that these
waves in quantum mechanics, e
to the ikx represent states
with some values of momentum.
So while this is fine and it's
a very important intuition,
the step that you
can follow with is--
it's interesting.
And you say that, well,
since p, the momentum,
is equal to h bar k and
that's quantum mechanical--
it involves h bar.
It's the whole discussion about
these waves of matter particles
carrying momentum.
You can say-- you can
multiply or take a delta here.
And you would say, delta p
is equal to h bar delta k.
So multiplying this
equation by an h bar,
you would find that delta
p, delta x is roughly h bar.
And that's quantum mechanical.
Now we will make the
definitions of delta p and delta
x precise and rigorous
with precise definitions.
Then there is a precise
result, which is very neat,
which is that delta
x times delta p
is always greater than or
equal than h bar over 2.
So this is really exact.
But for that, we need to
define precisely what we
mean by uncertainties, which
we will do soon, but not today.
So I think it's
probably a good idea
to do an example,
a simple example,
to illustrate these relations.
And here is one example.
You have a phi of k of
the form of a step that
goes from delta k over 2
to minus delta k over 2,
and height 1 over
square root of delta k.
That's phi of k.
It's 0 otherwise--
0 here, 0 there.
Here is 0.
Here is a function of k.
What do you think?
Is this psi of x, the
psi x corresponding
to this phi of k-- is it going
to be a real function or not?
Anybody?
AUDIENCE: This equation
[? is ?] [? true, ?] [? but-- ?]
PROFESSOR: Is it true or not?
AUDIENCE: I think it is.
PROFESSOR: OK.
Yes, you're right.
It is true.
This phi of k is real.
And whenever you have
a value at some k,
there is the same
value at minus k.
And therefore the star doesn't
matter, because it's real.
So phi is completely real.
So phi of k is equal
to phi of minus k.
And that should give
you a real psi of x--
correct.
So some psi of x-- have to
do the integral-- psi of x0
is 1 over square root of
2 pi minus delta k over 2
to delta k over 2.
The function, which is
1 over delta k in here--
that's the whole function.
And the integral was supposed
to be from minus infinity
to infinity.
But since the function only
extends from minus delta
k over 2 to plus delta k over
2, you restrict the integral
to those values.
So we've already got the phi
of k and then e to the ikx dx.
Well, the constants go out--
2 pi delta k.
And we have the integral
is an integral over x--
no, I'm sorry.
It's an integral over k.
What I'm writing here--
dk, of course.
And that gives you e
to the ikx over ix,
evaluated between delta k over
2 and minus delta k over 2.
OK, a little simplification
gives the final answer.
It's delta k over 2pi
sine of delta kx over 2
over delta kx over 2.
So it's a sine of x
over x type function.
It's a familiar looking curve.
It goes like this.
It has some value-- it goes
down, up, down, up like that--
symmetric.
And here is psi of x and 0.
Here is 2 pi over delta k, and
minus 2 pi over delta k here.
Sine of x over x
looks like that.
So this function already was
defined with the delta k.
And what is the delta x here?
Well, the delta x is
roughly 2 pi over delta k.
No, it's-- you could say it's
this much or half of that.
I took [? it half ?] of that.
It doesn't matter.
It's approximate
that at any rate now.
So delta x is this.
And therefore the
product delta x, delta k,
delta x is about 2 pi.
