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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