So we finished yesterday with this is an application of Ehrenfest Theorem,
which showed that on the understanding that the Hamiltonian is the Hamiltonian operator is P squared over two M plus V,
the potential energy inspired by classical physics.
And on the understanding that P is the operator that I claim that it is, which is defined by the relation p hat on psi is is equal to.
Right. So this is this is I'm claiming that this the opera is defined by this equation.
P hat is a momentum operator. It seems reasonable to take this to be the energy operator,
the Hamiltonian that being so when we use Aram first theorem to find the rate of change of the expectation value of X,
which in classical physics would be the actual value of X, we find that it's in fact equal to the expectation value of the momentum divided by M,
which is in a classical sense, what we would call the velocity.
So that's, that's one good thing. It's obvious that we should move forward and calculate the rate of change.
Use Aram First Theorem to calculate the rate of change, the momentum,
the expectation value live in hope that this becomes the this becomes the force anyway.
This is going to be maybe will lift the bar and put it over here similar.
So this is going to be p comma h expectation value of Ehrenfest theorem of h bar because I haven't put the bar here now in second thoughts.
So we need to calculate p kummer h p comma h is p comma p squared over two m plus v.
Com of their obviously commutes with itself.
So forget that. And so therefore this is P comma V and we when we discussed comitatus, we showed that if you take the commentator of.
Of an operator with a function of an operator. This is a function of x.
Then what you end up with is the derivative of this operator.
What do you do? End up with that in the event that the commentator.
So so you remember we expanded what we did was we expanded V of x as v nought plus v1x plus v2x squared over two factorial,
etc., etc., etc. And then when we calculated p hat v what did we get?
We got v nought plus v one, p hat, etc. plus and here we would have v two over two.
This would be because we're taking the commentator of P with x squared,
which is P with x because the other extending idly by plus x p comma the other x from our
basic rule for doing the commentator with products because this thing is only a number,
it's minus H bar. In fact, we can take this number outside.
It doesn't matter the fact that this number is in front of X and this number is behind here,
it's behind X because it's a number we can just pull it out.
This becomes to X hat which cancels this and at the end of the day we are looking at p, comma,
p, comma, x, the common factor in all these things plus sorry brackets v one plus v two,
x plus v three, x squared over three, etc. etc. etc. sorry over two, which is the tailor series for DV by the x and this one here is minus h bar.
So we have minus h bar DV by the x.
So our equation of motion. So putting this commentator back in up there, we discover that D But it's the rate of change of the expectation.
Value of the momentum is what we pick up a minus sign from here because we have a minus H bar here
and we want to find the commentator over H bar so we get minus the expectation value of DV by the x.
So lo and behold, we have Newton's Law of motion. We have the rate of change and momentum is equal to force.
But in this expectation, value, sense, expectation, value of the rate of change, the rate of change of the expectation value,
the momentum is equal to the expectation value of the force because in some sense the force has to be thought of as something.
That's what it is. It's something that has quantum uncertainty because it has uncertainty because the position is uncertain,
different positions will give rise to different forces, etc.
So I think that makes a pretty convincing case that we've that the momentum operator is as advertised because we're able to recover on that.
Understanding Newton's Laws of Motion. So now let's look at states.
Very important topic. Let me do it here. In fact, states.
Well-defined momentum. That is to say.
So we want we want to know what are the wave functions,
what are the states look like and which are certain to the measurement of of the momentum is certain to produce a given number.
Okay. So we're interested in the eigen states of the momentum operator.
Right. The operator p on an Oregon State of P on an Oregon State labelled by p.
This is a number is equal to that number times P so this is this is the definition this defines.
These states.
If we want to know what these things look like in terms of in real space, we want to browse through with an X and then we're looking at x.
P hat p is equal to p.
S P. This is the wave function of our state of well-defined momentum.
Let's, let's introduce a newfangled notation and declare that this is u sub p of x, right?
This is just the definition. The way function x p equals up of x.
And this left side, by the definition of the P operator is minus H bar D up by the X.
So here we have one of these trivial differential equations which we know how to solve.
It tells us that you p of x is equal to a constant times e to the i p over h bar x.
If we put each of the IP over h power x in for you, when we do this differentiation, we get down an IP average bar,
the bars, cancel the minus, sign the eye together, make a one and the p sticks around is what we want.
So that's it. So a state of well-defined momentum.
The states in which you are certain to measure a given value of the momentum is a plane wave is a wave like this with.
So we have. So it's a wave. And we have the wave number.
Usually cool. K Is the momentum divided by age bar because each bar is incredibly small.
Typically this wave number is extremely large and the wavelength of course, lambda being two pi over k is two pi h bar over.
P is h over. P is going to be very small.
The bigger and the bigger the momentum, the smaller the wavelength.
That's that's obviously crucial for for physical applications.
What else can we say? We can say that.
There's complete if if you know the momentum, then.
So if we're in a state of well defined momentum, the result of measuring momentum is certain.
So you do know the momentum then the way your wave function looks like this,
which means that the, uh, the probability density is independent of space.
So the probability density, which is u p squared is equal to some constant, which is independent X.
In other words, you know absolutely nothing about the location of your particle.
Absolutely nothing. It might be it's it's likely to be here is on the other side of the universe.
So from that it follows you already got. It's like it's like these states are well-defined energy.
These states are well-defined momentum. Do not in practice occur. They are mathematical idealisation.
Right, because no, you would never see a particle which had totally uncertain position because it would spend all of its time not in a laboratory.
Your boss research an actual part of the universe. Okay.
So it's it's that's something to be a bit clearer about.
But all I want to say about this. Oh yeah.
We should address this wavelength.
Yep. I should mention this, of course, is called the debris.
Wavelength. Roy was thinking about relativity curiously in 1924, whatever in his thesis for which he won the Nobel Prize in 1929.
And he came up with the idea that there was this relationship between the two, though, that the two particles would be associated with a wavelength.
So that's going to destroy your wavelength. In his honour.
And as regards numbers, well, we'll look at some numbers later on,
but the general idea is the general idea is that the size of an atom is
determined by the debris wavelength of the electrons that are in make up atoms.
So if you have a hydrogen atom and it's in its ground state,
its characteristic size is given by the debris wavelength of the electron that's in there.
And the electron that's in there is is is in orbit around the proton with a certain momentum.
Right. So this this deploy wavelength is setting the size of atoms.
I think that's a point worth a point worth making, but we'll look at some numbers later on.
So if you have an electron so an electron in hydrogen.
Right. Is moving around. It has the binding energy of hydrogen is 13.6 EV.
It has a kinetic energy which is half that because of the variable theorem which we'll,
we'll, we'll have all these results later on, but they're already in classical physics.
So it has a, it has a kinetic energy of all the electron volts.
And that gives you a de broglie wavelength, which is which is a 10th of a nanometre.
That gives you some kind of sense of scale. Okay.
What about normalisation? So we've deduced that these the wave function of a state of well-defined momentum should be some constant times.
This exponential. It's it's good to decide what this constant should be.
We usually normalise our wave functions,
so usually we want to have we like to have the integral d x of a psi mod squared is one because that's the total probability to find it somewhere.
But this normalisation isn't going to work because the if upside is proportional to each of the i k x upside mod squared is going to be one.
The integral from minus infinity to infinity of one is just infinite and no constant in front is going to normalise it successfully.
So we don't use that normalisation. The normalisation that we, we use is, is this normalisation that you remember yesterday we agreed that X primed X.
Should be delta of X minus x prime. So this thing here is the amplitude to be it x primed.
If you're certainly at x, which is why it's nothing. Unless X prime is equal to x and this amplitude becomes very large when x equals x prime.
So that when you integrate over this, you get you get one.
So that's what we should do in this case. P is an operator with the continuous spectrum, same as X.
So we want to we want to choose the normal to normal.
The normalisation constant. Choose the constant. Such that.
P primed. P equals one.
Sorry, not one. Delta p minus.
P prime. If I precise analogy with that.
So that's something that's fairly straightforward to do. We write this, we put out we put an identity operator into here, made up of Xs.
So this, this, this implies that.
Well, this thing here is equal to. It's equal to p, primed x x.
P. Right. That's just taking an identity operator.
This is this is our.
So we go we're going to say that x p is equal to some normalising constant times.
Easily I p on h bar x.
Right. And the name of the game is to find the value of this because we we know that this thing is this.
The nice thing is that this is the complex conjugate of that.
Right? So what we have is that this is equal to a mod squared because we get an A from here and a star from here,
the integral d x of E to the minus i p primed over by x.
That's from here the complex conjugate of that with p made into p primed.
And from this we simply have an e to the i p over h4x, and that can be written just to clean it up a little bit.
P minus p primed x over h bar.
D x over bar. H bar.
All right, so this age bar was always present.
This one I've put in, I've. I've divided the X by age bar and multiplied by compensating each bar here.
So the variable in the variable of integration is now x over each bar which is still running from minus infinity to plus infinity.
And now this is a standard integral which I hope you will recognise from Professor s loss course from for your analysis and for your analysis.
We know that this integral is to pi times-delta p minus p probes.
So what we're concluding is going right back up to the top that that original delta p
minus P primed right up there is equal through these integrals to more squared times,
times to pi h bar to pi h bar, delta p minus p primed.
And that clearly tells us that a mod squared is equal to two pi h bar is just h is equal to one over h and the phase of a is unimportant.
So we're entitled to take it to be real. So what we do is we choose a21 over the square root of H, not H bar, but H.
So that means that the correctly normalise thing x wave function XP is e to the i p over for x over the square root of H.
So this is an important result. It tells us something else that's of interest.
If we take it's complex conjugate because it's complex conjugate says that p x is equal to E to the minus p over h by x over root h.
What is it? What does this mean? This means the amplitude.
To find that you have momentum. P Given that you are definitely the place.
X So if you have an electron that's localised at the place x it's wave function is is a delta function essentially, right?
It's localised. At X you can ask what's the amplitude for this to have various momenta?
The answer is given by this complex number here. The modulus of this complex number here is independent of P.
So what does that mean? It implies that the probability of having P given X is some consequence.
All values of momentum are equally likely.
From from a momentum, which is nothing very much or zero even up to a momentum which is is associated with some relativistic gamma,
you know, some large value of gamma, all momentum equally likely, including for extremely high ones.
So that's clearly un physical. And what that tells you is you will never succeed in localising.
It will never succeed in localising a particle precisely to an exact x.
The state of being definitely at X is unreliable because it would it would imply that there was enough energy somehow in
the system that there was a non-negligible probability of finding the momentum to have some extraordinarily large values.
Right. So there we are.
That's the so what we've discovered so far is if you if if if X is certain.
P is totally uncertain. And conversely, if P is certain.
X is totally uncertain. Moves.
Let's therefore investigate these. Both of these situations are clearly and physical.
So let's try and discuss something which is physical. And let's let's suppose that we're dealing with a probability distribution, an x,
which is a calcium E to the minus x squared oops x squared over two sigma squared over the square root to pi sigma squared.
Right. So this is a Gaussian distribution of probability and x, which sort of is generically we have,
which is our generic model of well we've got this thing localised at the origin to within plus or minus sigma more or less.
Right. We can ask what wave function yields this probability, what the answer is.
Essentially it's a wave function which is the square root of this. So a suitable wave function.
There are many possible way functions because phase information isn't conveyed by the probability.
But let's, let's write down this wave function which is e to the minus x squared.
Over four sigma squared over two pi sigma squared to the quarter power.
So if you take the mod square of this wave function, you get the probability and the probability you get to that one there.
So I could multiply this by all kinds of complex, all kinds of numbers of modulus, one and arbitrary phase.
And I would still get that. But this is the this real wave function is the simplest one that we can write down.
And now let's calculate for this.
So this is a well-defined wave function which we know localises our particle two plus to the origin, plus or minus sigma.
Let's ask, so what is the probability distribution for this upside of measuring a particular value of P?
Right. So what we want to discover for this is what? So what's P upside?
Well, that's the integral the x of p x x cy.
We know what this is because we've just been working it out. This is a state of well-defined momentum.
So this is one this is the integral d x of E to the minus p upon h bar x I believe.
I hope I've got that minus sign right somewhere up there over the square root of H times this which is the wave function
we just wrote down e to the minus x squared over four sigma squared over two pi sigma squared to the quarter power.
So this this is a we have to get this from minus infinity to infinity.
Now, physics is full of integrals of this sort. And there's a box in the book explaining how to do them.
I don't want to take the time to to go into the sordid details now,
but all you do is you gather all these all these exponents of the exponential together.
And what we've got here is an integral de x of each the I quadratic.
In X. Right. If you gather this together, there's a linear term and there's a quadratic term.
So what? So you can you can express that.
I mean, it is each the quadratic expression in X.
And what you do is complete the square of the quadratic.
Change your variable of integration and use a standard result that the integral the x e to
the minus x squared from minus infinity to infinity is equal to the square root of pi.
We use the standard result, and that's how we evaluate these integrals here.
But it's I would recommend learning how.
Checking the box out, making sure you understand how that goes and doing this yourself as a as an example.
After the lecture, when I did want to take time to do it now because it's just algebra.
Let's just write down the answer, discusses the physical implications.
So this is this turns out to be that p upside is equal to E to the minus sigma squared, p squared over, h bar squared over.
And there's a normalising constant which is two pi h bar squared for over four sigma squared.
To the quarter power. So that if we square this up, we get the probability of measuring various momenta,
which is clearly going to be e to the minus two sigma squared p squared average
bar squared over two pi h bar squared for sigma squared to the quarter.
So our position of probability and position in real space, we have the particle localised in a Gaussian distribution with a width sigma.
It turns out from this calculation that the possible values of the momentum,
the probabilities associated different momenta is also a Gaussian distribution centred on zero in momentum and the width of this distribution,
the spread in momentum. So this in order to find what that is, you'd have to express this is e to the minus p squared over over two sigma p squared.
So the, the, the dispersion. And momentum is.
Is HPR. Over Two Sigma.
And so we write. So the dispersion in momentum is small when the uncertainty in real position is large.
And conversely, right. So this so we have a result for this particular model.
The dispersion in X times, the dispersion in momentum is h bar over to yes we ability to measure the hall.
This. This. You worried about this, too? To know that you got call ops.
Thank you. It should be a half. Yes, of course. Because I've squared the quarter. And thank you very much.
I've square the quarter and it's become a half. So this is the classical statement of the uncertainty principle.
It's really only an order of magnitude in this particular model.
This is an exact mathematical statement. It's a statement about about the width of two.
Gaussians But in, in a generic case, if you if you know,
your probability solution is sort of it's like this just some curve that's sort of natural width in a, in a location in X.
Then. The corresponding probability probability distribution in P will have a width
which is broadly related to the width in X here by a relationship of this type.
But it will be exactly h bar over two in the generic case.
It's exactly h bar over two just for these Gaussian distributions.
But the really key idea is that if is the product of the uncertainties in these two things.
That has to be will be on the order of HBO.
So there are two important points to make here.
We need to be clear what we're saying.
We are not saying that if you measure the position of an electron and then you measure its momentum, you will find results which scatter in this way.
This is not the uncertainty of a measurement in X and then the uncertainty of the following measurement, the following momentum.
Measurement. This is a statement about if I have a large supply of electrons,
different electrons set up so that they are pretty much in the same in the same wave function.
And I choose to measure the momenta of half of them and I'll get a dispersion sigma p and if I measure the positions of the other half of them,
I'll get an uncertainty sigma x which satisfies this relationship because we have this this
uncertainty in momentum is the uncertainty associated with the original wave function exercise.
I. And if I would measure the position of that electron,
I would change the wave function into some kind of something near to a delta function centred on whatever answer I got.
Right.
So, so when you make a measurement, you change the wave function and we've calculated the dispersions for measurements using the same wave function,
not the wave, not not an initial wave function.
And then the wave function that we get when we make the measurement.
And the reason we've done this partly is that we do not know what the wave function is.
We get when we make the measurement that's in the lap of the gods, you make a measurement.
So remember the basic dogma. We if we have to go back to the discrete case because it's simpler if I have my
wave function is some sum and then some linear combination of stationary states.
This is a well-defined wave function if I measured the energy.
Then this thing collapses. To absorb is equal to EAC for some K and which is in the lap of the gods.
The apparatus does not tell us that it's a wheel as the roulette wheel is spun and one of the one of the KS is chosen.
So it is up here. If you measure the position, you will find some value.
And after you've made that measurement, your wavefunction will be different.
It'll be more or less a delta function associated with that X and not the wave function we're working with here.
And the uncertainty on a subsequent measurement of P will be larger, will be large.
The other thing to say is how do we understand this physically, this uncertainty relationship?
We say to ourselves, well. If you.
If the wave function is highly localised in space.
If you think about that, wave function is made up as an interference pattern between states with twin plane waves,
which are states of well-defined momentum. Then in order to have the interference pattern highly localised so that the sum of
all these waves cancels to high precision everywhere except in some narrow region.
You will need to have you will need to use waves with a very large range in wave numbers.
And that's why the momentum is very uncertain. If the position is rather certain.
So it's it's because because of this basic quantum, the basic principle of of of adding amplitudes.
A highly localised lecture. We're entitled to think about a highly localised election as an interference pattern between states of of different
momenta and we will need to have a very large range of possible momenta if we want to have a highly localised electoral.
And tightly defined confined interference pattern. Let us now talk about the free the dynamics of a free particle.
So. So we've just got a particle whose energy it's not there's no potential energy.
It's just free to roam. So the Hamiltonian operator is going to be p squared over two m we dropped the plus v of x.
It's a free particle. But what we're going to do now is talk about the time, the time evolution of this particle.
So imagine that you've got the particle that tingles once you've got it localised around the origin.
And let's, let's set this up a little bit by saying let's localised around the origin,
but it's moving with some, you know we've got some, some idea of what its velocity is.
So we're going to say it's initially we're going to write down.
An appropriate expression for its momentum. So this is the this is the wave function in.
Well, it's the it's the. It's a complete set of amplitudes with respect to momentum of a particle which is localised at the origin and has no meet.
The mean momentum is nothing. Suppose we start from.
P ci is each of the minus sigma squared over bar squared.
P squared minus p nought squared. Sorry.
P minus P not. What do I want to do?
Yeah. P minus p not squared over this horrible normalising constant to pi.
H bar squared. Over four sigma squared.
A quarter. So it would be reasonable to conjecture that we'll find out whether this is true or not when we do the calculation.
But the conjecture is the reasonable conjecture is that this complete set of amplitudes characterises a state of the particle where it is,
it is moving with momentum.
P zero. P zero is a constant, right? This is the minimum. This is the this is the momentum.
Eigenvalue. This is just some constant.
So it has a velocity which is on the order of P zero over M and it's localised at T and it's localised at the origin to plus or minus sigma.
We'll find out whether this is true or not, but that's the conjecture. Okay.
Now let's ask ourselves, what is the wave function in real space that corresponds to that at different times as a function of time?
Why can we do this? Because we have a free particle.
The Hamiltonian is just p squared over M, which means that a state of well-defined energy is going to be a state of well-defined momentum.
The Hamiltonian is a function of the momentum, so it has the same eigen states as the momentum.
So a state of well-defined momentum is going to be an eigen state also of the energy.
Now we know how to we we know how to evolve in time states.
Once we so remember our basic equation,
which is the desire at time t is equal to the sum a and e to the each of the minus i e and t over h bar times e n nought.
Remember, this was why we were excited by the states, why these states are, well, well-defined energy.
The stationary states are so important is because they enable us to evolve in time a system where an is equal to end nought upside.
These things set the initial condition for the calculation and the time evolution is given by these exponentials.
So we want to use this formula in this other context here.
We know what this is. This is a state of well-defined momentum. We know what this is.
Right? This is just some exponential with the relevant energy going in there and this is the amplitude to have momentum.
P So this transforms, this is the discrete case, this transforms in our case into a possi is equal to an integral overall possible momenta.
That's the analogue of the summing over the energies.
When you sum have a momentum, you are summing over energy because different momentum or rate e to the minus.
I What's this? This is the energy. It says you amentum p I called it ep up there, but we can be more definite.
It's P squared over to m h bar t.
Sorry to bother. Excuse me. Tea over bar.
Right. That's the exponential thing either. What's this got to be?
This has got to be a state of well defined p.
And let's. We wanted to know what this looked like in real space.
So let's barrel through with X and then this.
Sorry. Sorry, sorry. I'm missing something altogether. Excuse me. Excuse me. Let's keep.
Let's leave it out. I miss something out. I missed out the awards, didn't I?
What are the ends? It's the amplitude.
To have at the time two equals nought is the amplitude to have energy n which in our case is the amplitude at equals nought.
To have a momentum. P.
Right. And then now we have the state.
P. And now if we want the wave function information, we should barrel through with X.
Then everything over here becomes a function of momentum and a known function of momentum.
This is a function of momentum. Also time. This is a function of momentum.
We just put it down by conjecture. It's that thing there. This is a function of momentum.
It's the. This is this is a plane way.
This is each the IP of punch bar X within a within a sign.
Now it is exactly that. So so let's just see what we get here.
So this is a dirty, great integral d p e to the minus i p squared t over to m h bar.
Let's put this one. No, let's keep to the right order.
E then here we have e what we said it was going to be eaten minus sigma squared over bar squared p minus
p nought squared over horrible two pi h bar squared over four sigma squared to the one quarter power.
If I've got that right. And this thing is our wave function for a state of well-defined momentum, which is e to the i p over h bar.
Sorry, x average bar. Over the square root of just H.
So what do we have here? We have an integral of an exponential of a quadratic expression in P, right.
Because here we have a P squared.
When you square this thing up, you're going to have a p squared and a minus two P and a linear part in P and here's a linear part in P.
So it's another of these integrals of a exponential of a quadratic expression in P,
which can be solved by the methods described in the box that we used just before.
Now, the algebra in this case is a little bit it's a little bit wearisome.
It's absolutely straightforward. It's absolutely straightforward, but it's just a bit wearisome.
And the answer, in fact, that this comes to is quite a complicated expression,
because what we're going to arrive at is something which has both phase information and amplitude information,
but we only want to know what the probability is of finding the particle at this place or the other place on that probability.
The mod square of the answer to this calculation is much simpler and I'm going to write it down.
So what follows now is a very straightforward calculation. I would I would urge you it's this box doing it in the book.
I would urge you afterwards to look through this and make sure you understand it.
But it is it is just algebra. And what's more what's interesting is to is to is to understand the physical implication of this.
So we're going to we're going to extract the mod square of the resulting of the answer when you've done all this integration.
And what apparently it is is sigma of a root to pi h bar squared mod b squared E to the minus x minus p zero.
T over m squared.
I need to tell you what B squared is, don't I? So. And here.
B squared is a complex animal. If sigma squared over h bar squared plus i t over to an h bar.
So what have we got? This is a Gaussian distribution in ex at any fixed time.
It's a Gaussian distribution in X.
The centre of the Gaussian is it pp0 of m times time, which means that it's centred on what one would call v times time, right?
Because P zero over m we said this was the mean momentum of the of the it was the,
it was the expectation value of the momentum of our original wave function.
So it's, it's this is the mean if you thought of this as many different particles,
it's really any one particle I thought of as many different particles would be the mean momentum.
So this is essentially the mean velocity. So that's what you would expect.
The the probability distribution is moving in space with a speed V nought equal to nought over M as we would expect.
And the dispersion associated with this Gaussian is determined by by that stuff.
So we have a we have a sigma as a function of time, which is going to be given by.
So. So what should this be? This should be two sigma squared.
So Sigma is going to be given by by the square root of those two, which is going to be from this sigma squared plus t squared.
I better write this down. It's too hard to do it in one's head.
Plus, page 40 over two sigma signatures in the book.
Yeah. Sorry. So this. Sorry. This should be another sigma. What should we call this?
This should be cool. Well, let's just call this the dispersion. Well, we call it sigma sub t, right.
Whereas this other sigma is the original sigma. So we've got a Gaussian distribution and, uh.
Right. That's. That has a dispersion given by this.
You know this story. Sorry. It's always right. There's something wrong here, isn't it?
Because on dimensional grounds. Yeah, you're right.
Did I write down the right integral? No, I didn't. That's exactly what's gone wrong.
Sorry we're missing from here. A sigma squared on top. It's crucial.
So when I. When I say what the dispersion should be, the dispersion we should arrange.
This is two pi dispersion squared. So dispersion squared is equal to this divided by that.
Sorry. Then I have to square root it so it's divided by that which makes it that and this I've copied out of my notes.
So I expect it's still right. But I was trying to do some of this in my head, which was dangerous.
So what do we got? We've got the the dispersion at time t is equal to the original dispersion at time t zero plus this extra bit here.
And what is this extra bit here? What was the original uncertainty?
The original uncertainty in momentum from the uncertainty principle here.
The uncertainty momentum was equal to a bar of a two sigma.
So the uncertainty and the velocity was equal to H bar over to M Sigma.
So what's this? This is equal to sigma. Plus the uncertainty in the velocity times time.
Uh. Squared and shouldn't be square in this, should I?
I think we might need to take a square root of a square, actually. Let's.
Let's let's not chase that down in the moment, because this is the basic idea.
The basic idea is the uncertainty in position is growing like the uncertainty in position times velocity.
But that's what you would expect, right? Because you have what do we have?
We have a bunch of particles originally at the origin and moving to the right with with the nought plus delta plus or minus delta v,
some are going faster, some are going slower. At some later time.
This is moved over by an amount v nought times time and this width of of of the this was there was a width sigma here.
But the ones that were going slower than the average will have slipped behind.
They were already there was some of them will already be sigma behind,
but then they sit behind extra by an amount delta V times T and some of the ones which were in front have got even more in front because they've they,
they have, they had bigger velocities by a delta V so that the total width is equal to the original width plus this extra width.
And I think probably we should be taking some squares and square routeing.
But you see that we are what we're getting from this calculation makes perfectly good sense physically.
Let me just remind you how we've done this calculation, because it's the it's the methodology, which is in many ways.
Well, it's good to say it's crucial to see that what emerges from this makes sense physically.
But it's also good to remind yourself, how can you how do you actually calculate these things in this damn theory?
The way we've done this is we've used this central expression.
We've said that states, I can I can evolve something in time so long as I can express my original state.
It's a linear combination of states of well-defined energy.
In this particular case, of a free particle, a state of well-defined energy is exactly the same as a state, a well-defined momentum.
So we wrote we we wrote that some expression in the integral form is appropriate because momentum has a continuous spectrum.
And then we just turned the handle and out came these perfectly sensible results.
I think we're probably pretty much ready to finish the.
Again, I want to stress I think I should stress that we've obtained this perfectly sensible physical picture,
but we've obtained this perfectly sensible physical picture through an orgy of quantum interference,
because we have in order to in order to get what we what we wanted,
we took a perfectly well defined spatial distribution and expressed it as as an interference pattern between states of well-defined momentum,
which we then evolved each state a well-defined momentum in time in its trivial way, with that just an exponential.
And then we allowed them to interfere at this later time in their evolved form to find out what the what the distribution was in real space.
So that's what I mean by it's an orgy of quantum interference. We've taken something, we've decomposed it into an infinite number of other things.
We've taken something physical, we've decomposed into an infinite number of things which are not really very physical,
namely states of well-defined momentum. We've evolved each one of those independently in time because they're states of well-defined energy.
And then we've interfered the evolved momentum states we've allowed by working
out this integral was working out the result of the corresponding interference,
right? We were adding up an infinite number of of amplitudes and allowing them to to interfere in outcomes,
something that makes sense, which is a wave packet that's travelling and spreading.
And behaves in a way which does make perfect sense from a physical point of view, from a classical physical point of view.
Okay, we'll finish with that.