Ok, so I promised a while ago to explain the
maths behind the heisenburg uncertainty principle
and I've finally going to do it now.
To understand this video you're going to need
to know about something called the Fourier
Transform.
If you don't and you have no intention of
knowing then feel free to skip this video.
Otherwise, I made 2 videos explaining it,
so watch those if you'd like.
Ok, I'm going to assume now we're all on the
same page.
Today's discussion is going to center around
one really important position wave function.
If you watched the last video, then you'll
already be well acquainted with it.
It is the complex exponential.
I.e. the function whose real part is a cos
wave of wavelength 2 on k, and its imaginary
part is a sin wave with the same wavelength.
Ok, now's probably a good time for me to make
an apology.
Yes, this wavefunction has imaginary numbers
in it.
In general all wavefunctions do.
I never mentioned it because it wasn't very
important up until now.
Trust me though; the ideas are all straight
forward.
Ok, let's get back to our exponential.
What's so special about having a position
wavefunction like this?
As you well know, usually if we try to measure
the momentum of a particle, we're uncertain
about what value we'll get.
However, the weird thing is, if a particle's
position wavefunction is a complex exponential
like this, then we know with out doubt exactly
what momentum we'll get is.
In fact the value we'll get is k times h bar.
This is called the De Broglie hypothesis,
and honestly this is the only mysterious part
of this episode.
As you'll see, the De Broglie hypothesis is
all you need to make the uncertainty principle
happen.
I don't know why it's true, for now we can
just take this as an experimental fact but
hopefully in the future we'll come back to
it.
For now, I'm going to show you how we can
use this fact to find the momentum wavefunction.
Ok so we now know that if a particle has this
kind of position wavefunction, it certainly
will have momentum h bar k.
What about if the wavefunction is some kind
of superposition like this?
Well then if we measure the momentum of a
particle, it will either be hbar k1 or hbar
k2.
The probability of measuring either depends
on the size of the coefficients.
The bigger the coefficients, the more likely
that value is.
A similar thing works no matter how big the
superposition.
Therefore, if we're given a random wavefunction,
and we want to know what the momentum is most
likely to be, our first task is to write the
function as a superposition of exponential
functions, and then look at the coefficients.
But wait, isn't this is exactly what the Fourier
transform is good at doing?
Remember, that if we to decompose the function
into exponentials, the Fourier transform is
the thing that basically tells us how much
of each of the exponentials we have.
Therefore, if we look at the Fourier transform
of a function and see it is very peaked at
a point, it means the weights on the exponentials
in that range are very high.
But that means the probability of getting
the corresponding momentums is also high.
Well then a more convenient version of this
graph is to multiply all these k values by
hbar, because that gives us the momentums
using the De Broglie hypothesis.
In this version of the graph, I can find out
how likely a certain range of momentum's are
by just looking at how big the graph is around
there.
But that means this graph is infact the momentum
wavefunction.
So all we have to do to get from the position
wavefunction to the momentum one is take the
Fourier transform, and multiply all these
k values by h bar.
Ok, so this is the link between momentum and
position in Quantum Mechanics, but what's
so special about it?
Remember the uncertainty in a variable is
measured by the width of the wavefunction,
or more precisely its standard deviation.
The uncertainty principle basically says,
you can't make the uncertainty in the position
and momentum arbitrarily small.
Well turns out that this is just a property
of the fourier transform.
Say we had some function, with uncertainty
delta x and its fourier transform has uncertainty
delta p.
Let's try to make the uncertainty in the function
smaller by squishing the function to make
its uncertainty go down by a factor of d.
Now surely, the product of the uncertainties
has gone down because delta x has been reduced.
Unfortunately, the fourier transform thwarts
us.
You see, the fourier transform now spreads
out so its uncertainty goes up by a factor
of d.
Therefore the product of uncertainty doesn't
decrease at all, despite our best efforts.
It's pretty straight forward to show this
will happen no matter what type of function
we have using the formulas for the fourier
transform but we'll look at some examples
instead.
This function is the hat function.
As you can see it's fourier transform is pretty
crazy.
As we squish the hat though, the fourier transform
gets more and more spread out.
A similar thing happens if you start out with
a Gaussian function.
So you see, its this property that if a function
is lean then its fourier transform is wide,
and vis versa, that causes the position/momentum
uncertainty.
Let's just try one last vain time to defy
the uncertainty principle.
Let's go back and look at the function e to
the ikx.
If this is a particle's position wavefunction,
then what is it's momentum wavefunction?
Well, we know with out a doubt that the particle
will have momentum h bar k, so its momentum
wavefunction is just a huge spike at h bar
k.
In other words, the uncertainty in the momentum
is zero.
But then if delta p is zero, then delta p
times delta x must be zero- so the uncertainty
principle is wrong right?
Hm, not quite.
If we actually calculate the position uncertainty,
we'll find that it is infinite.
But this is very weird, because this means
that if the momentum is perfectly known, and
then we try and measure its position, then
the particle is equally likely to turn up
anywhere in the universe.
I think this is utterly ridiculous.
So then, is it actually possible to get a
particle into a state where we know its momentum
perfectly?
Well last episode I told you that if you measure
the particles momentum, it collapses into
the state of only having that momentum.
So then my conclusion is that its impossible
to measure the momentum one hundred percent
precisely, even in theory, because that would
allow the particle to turn up anywhere.
We can use pretty similar logic to conclude
that we can't perfectly measure a particle's
position either.
So when we talk about measuring a particle
in a certain spot, or with a certain momentum,
we really mean that it approximately has that
value.
Well, that's it for our discussion of the
uncertainty principle for a bit.
I'll see you next time for some more quantum
mechanics though
