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This video is about the de Broglie hypothesis which supposedly says matter is a wave as
well as a particle and this equation that tells us how to find the ‘wavelength’
of a particle.
The issue is, there is a very precise way to interpret the hypothesis that leads to
interesting questions about quantum mechanics, but instead it’s often stated in an almost
meaninglessly imprecise form.
Let’s fix that.
This video is a bit more involved then the previous ones but hopefully not too bad!
But if I fail, don’t worry, you don’t need to understand all of this video to get
the next.
If you do want to stick around.
I recommend a little revision since it’s been a while–in particular these videos.
Ok, we’ve got this particle, it’s got a wavefunction.
Remember the wavefunction of the particle contains everything we can know about that
For example, it might say that the particle is in a superposition of being here, here
and here, and so if you measure it’s position then it will turn up in one of those places.
There are these special wavefunctions called position eigenstates.
This means that the particle is really only in one spot- if you measure it, it will definitely
be there.
Wavefunctions doesn’t just tell us about the position of a particle, it also tells
us about the speed, or momentum, of the particle.
There is a way to convert a wavefunction from the position basis to the momentum one.
When you do it you could find your particle is in a superposition of several speeds.
Here’s the question we’ll focus on: if the particle is going at one speed, so it’s
wavefunction is a momentum eigenstate, then where is it?
How do we write that wavefunction is the position basis?
Is it in a superposition of many places, or is it in one spot?
If this was every day life, at any point in time a moving particle is still only in one
But what we’ll see is, even at one particular time, a momentum eigenstate is a superposition
of many positions- in fact, all of them, equally.
I’ll show how we motivate this.
As with Noether’s theorem, we do thought experiments about how a state could be different.
We say, suppose we have some superposition of locations.
We just imagine, what if these places had been a bit to the right instead?
Let’s call this offset amount delta.
As you can see, position wavefunctions are the same but delta to the right.
Remember the way we defined this shift, the position is changed, but the speeds the particle
is going at is not.
Ok so imagine our original wavefunction was a momentum eigenstate, what happens to it
when it’s shifted?
We’d expect this wavefunction is exactly the same right?
Because the shift doesn’t affect the speed, so it should still be the same momentum wavefunction.
Well actually, that doesn’t mean the wavefunction has to be exactly the same.
We definitely want it to be going at the same speed if we measure it.
But there’s still some freedom here.
Remember the probability of a particular speed is the length of the coefficient in front
Imagine if the state had a negative one in front of it.
Negative one still has length one.
That’s means the probability of getting this speed is still 1.
What about if there was an i here That still has length 1.
In fact, if we go back to thinking of complex numbers as arrows, all complex number on this
circle has length one.
In physics we call these numbers ‘phases’.
There’s really convenient way to represent a phases.
Suppose this phase has angle theta.
Using our trigonometry, we could write that as this complex number.
So we’re just going to define this expression to be a complex exponential.
Multiplying an complex number by a phase is the same as just rotating it by that angle.
That’ll be useful.
Getting back to the point,
we require that, when you shift a momentum eigenstate in position, it doesn’t change
momentum, but it may pick up a phase.
This phase obviously needs to depend on the amount of shifting delta and it also depends
on the momentum, so those two things should go into the angle, but then we need a number
that scales how much that the angle changes as you change delta and p.
We’re going to call it oh I don’t know, how about... h bar.
Yes the same constant in all those famous quantum mechanics equations.
This is it’s grand entry- it’s just a constant there that decides how much a shift
changes the phase.
We can measure it some other ways, and it turns out it is very very small.
We’re almost there.
But first we’ve got to talk a little bit more about position wavefunctions.
I’ve talked about about the wavefunction when it’s a superposition of a few different
But the thing is, a particle could be in superposition of a whole infinite range of positions.
What if the particle is in a superposition of all these places but it’s more likely
to turn up in this area?
How do we represent that?
Before we put a complex number in front of each position.
Now we’re basically going to do the same, we’ll assign a complex number to each point
the particle could be, in other words along the whole line.
Assigning a complex number to every point on the line- that’s just a complex function
-we’ll call it f(x).
The bigger the complex number, the more likely the particle will turn up around there.
There are two ways we can draw this complex function, either, we just draw the real and
imaginary part separately, or draw it 3D, with the real bit coming out toward you, and
the imaginary bit going up, so all complex numbers are just arrows.
Both are useful.
For people who’ve seen calculus before, all I’m saying is that the wavefunction
can be written as an integral of all positions, where each position is weighted by this complex
If that made no sense to you, don’t worry, it’s not important.
Anyway, we have this momentum eigenstate, and we want to write it in the position basis.
We know the position wavefunction is going to be represented one of these complex functions,
we just need to find it.
Remember that when you translate the momentum eigenstate, it stays the same, except it picks
up an overall phase.
Then since this function is the momentum eigenstate, it must do the same.
Well there is a function that fits that bill: it is this complex squiggly function: where
the blue is the real bit and the red is the imaginary.
In other words this function is a complex exponential with this wavelength.
Why is this the function we’re looking for?
Well you can already see it almost has translation symmetry, the function is the same if you
shift it over a bit... only it’s a bit rotated.
But that’s exactly what would have happened if we multiplied by a phase!
Remember, multiplying any complex number by a phase just rotates it by that angle.
So translating is this function is the same as just multiplying it by a phase.
We’ve now actually derived the de Broglie hypothesis!
See, all it actually says is, the wavefunction of a particle with precise momentum is the
complex exponential with this wavelength.
Why the confusion then about the wave particle duality, where people say a particle acts
like a wave sometimes and like a particle sometimes.
I think the problem is that for a confusing time before quantum mechanics, people where
doing things like the double slit experiment and for understandable reasons, interpreted
them using the wave/ particle duality.
Let me show you an alternative explanation using all of our current quantum mechnaics.
The way the double or single slit experiment is done is by firing particles, let’s say
Those atoms are set up to have one precise speed.
In other word’s each atom’s wavefunction is a momentum eigenstate.
We’ll just look at the real bit, and so that looks a wave.
If we have a single slit, that wave, after it passes through the slit, spreads out- just
like a water wave would.
Then we measure where the particles end up.
As you can see, the wavefunction is biggest in the middle, and so that’s where each
particle is most likely to land.
But that’s what we would have expected even if we were throwing tennis balls you know?
So people thought, this is purely particle like behaviour.
The double slit experiment then should be similar- each particle goes through one slit
and lands most likely right behind that slit.
Instead they got this.
This is how you explain it with quantum mechanics.
Since we don’t measure which door the particle goes through, each particle’s wavefunction
is a superposition of going through both doors.
This means we add the separate wavefunctions together.
But interesting things happen when you add waves.
If the waves were in sync, like they are at these points because the crests meet other
crests and the toughs meet other troughs here, the result is a really big wave.
But here, the waves are totally out of sync, and so there’s no wave left there at all-
they cancel.
The result is that there are these alternating spots were the wavefunction is big then small
then big again etc etc, to get the right result.
People thought now the particle is acting as a wave, but wasn’t before.
But as you can see the current interpretation is much more subtle, but covers both cases
in the same framework.
People also imagined that the particle itself smeared out into a physical wave, but collected
itself back up as a point particle when measured at the wall.
Now we say that the particle itself is doing who knows what while it’s not measured,
if it exists at all, but it’s wavefunction is the one that spreads out.
Honestly, I’m not sure why people who know quantum mechanics still talk about the wave/
particle duality at all- maybe because that seems easier to explain than the wavefunction?
But it annoys me because it makes the world sound so
paradoxical so we couldn’t possibly hope to comprehend it.
In the end, maybe that’s the case, but we won’t know until we try to understand quantum
Anyway, thanks for watching this video- I know I’ve been away for far far too long.
As a reward for your loyalty, here’s some homework.
First, What did you think of my supposed derivation of the momentum wavefunction?
What assumptions did I make that you think need more justification (there are a few keys
one I saw) Second, Nils Bohr was a huge proponent of
the wave/particle duality even long after the modern form of quantum mechanics.
Try and understand his argument for it and report back.
Ok, see you very soon for the Heisenberg uncertainty principle.