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