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So we finished yesterday with this is an application of Ehrenfest Theorem,
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which showed that on the understanding that the Hamiltonian is the Hamiltonian operator is P squared over two M plus V,
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the potential energy inspired by classical physics.
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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.
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Right. So this is this is I'm claiming that this the opera is defined by this equation.
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P hat is a momentum operator. It seems reasonable to take this to be the energy operator,
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the Hamiltonian that being so when we use Aram first theorem to find the rate of change of the expectation value of X,
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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,
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which is in a classical sense, what we would call the velocity.
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So that's, that's one good thing. It's obvious that we should move forward and calculate the rate of change.
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Use Aram First Theorem to calculate the rate of change, the momentum,
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the expectation value live in hope that this becomes the this becomes the force anyway.
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This is going to be maybe will lift the bar and put it over here similar.
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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.
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So we need to calculate p kummer h p comma h is p comma p squared over two m plus v.
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Com of their obviously commutes with itself.
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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.
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Of an operator with a function of an operator. This is a function of x.
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Then what you end up with is the derivative of this operator.
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What do you do? End up with that in the event that the commentator.
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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,
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etc., etc., etc. And then when we calculated p hat v what did we get?
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We got v nought plus v one, p hat, etc. plus and here we would have v two over two.
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This would be because we're taking the commentator of P with x squared,
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which is P with x because the other extending idly by plus x p comma the other x from our
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basic rule for doing the commentator with products because this thing is only a number,
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it's minus H bar. In fact, we can take this number outside.
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It doesn't matter the fact that this number is in front of X and this number is behind here,
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it's behind X because it's a number we can just pull it out.
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This becomes to X hat which cancels this and at the end of the day we are looking at p, comma,
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p, comma, x, the common factor in all these things plus sorry brackets v one plus v two,
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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.
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So we have minus h bar DV by the x.
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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.
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Value of the momentum is what we pick up a minus sign from here because we have a minus H bar here
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and we want to find the commentator over H bar so we get minus the expectation value of DV by the x.
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So lo and behold, we have Newton's Law of motion. We have the rate of change and momentum is equal to force.
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But in this expectation, value, sense, expectation, value of the rate of change, the rate of change of the expectation value,
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the momentum is equal to the expectation value of the force because in some sense the force has to be thought of as something.
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That's what it is. It's something that has quantum uncertainty because it has uncertainty because the position is uncertain,
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different positions will give rise to different forces, etc.
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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.
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Understanding Newton's Laws of Motion. So now let's look at states.
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Very important topic. Let me do it here. In fact, states.
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Well-defined momentum. That is to say.
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So we want we want to know what are the wave functions,
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what are the states look like and which are certain to the measurement of of the momentum is certain to produce a given number.
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Okay. So we're interested in the eigen states of the momentum operator.
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Right. The operator p on an Oregon State of P on an Oregon State labelled by p.
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This is a number is equal to that number times P so this is this is the definition this defines.
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These states.
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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.
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P hat p is equal to p.
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S P. This is the wave function of our state of well-defined momentum.
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Let's, let's introduce a newfangled notation and declare that this is u sub p of x, right?
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This is just the definition. The way function x p equals up of x.
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And this left side, by the definition of the P operator is minus H bar D up by the X.
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So here we have one of these trivial differential equations which we know how to solve.
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It tells us that you p of x is equal to a constant times e to the i p over h bar x.
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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,
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the bars, cancel the minus, sign the eye together, make a one and the p sticks around is what we want.
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So that's it. So a state of well-defined momentum.
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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.
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So we have. So it's a wave. And we have the wave number.
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Usually cool. K Is the momentum divided by age bar because each bar is incredibly small.
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Typically this wave number is extremely large and the wavelength of course, lambda being two pi over k is two pi h bar over.
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P is h over. P is going to be very small.
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The bigger and the bigger the momentum, the smaller the wavelength.
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That's that's obviously crucial for for physical applications.
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What else can we say? We can say that.
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There's complete if if you know the momentum, then.
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So if we're in a state of well defined momentum, the result of measuring momentum is certain.
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So you do know the momentum then the way your wave function looks like this,
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which means that the, uh, the probability density is independent of space.
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So the probability density, which is u p squared is equal to some constant, which is independent X.
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In other words, you know absolutely nothing about the location of your particle.
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Absolutely nothing. It might be it's it's likely to be here is on the other side of the universe.
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So from that it follows you already got. It's like it's like these states are well-defined energy.
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These states are well-defined momentum. Do not in practice occur. They are mathematical idealisation.
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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.
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Your boss research an actual part of the universe. Okay.
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So it's it's that's something to be a bit clearer about.
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But all I want to say about this. Oh yeah.
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We should address this wavelength.
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Yep. I should mention this, of course, is called the debris.
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Wavelength. Roy was thinking about relativity curiously in 1924, whatever in his thesis for which he won the Nobel Prize in 1929.
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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.
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So that's going to destroy your wavelength. In his honour.
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And as regards numbers, well, we'll look at some numbers later on,
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but the general idea is the general idea is that the size of an atom is
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determined by the debris wavelength of the electrons that are in make up atoms.
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So if you have a hydrogen atom and it's in its ground state,
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its characteristic size is given by the debris wavelength of the electron that's in there.
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And the electron that's in there is is is in orbit around the proton with a certain momentum.
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Right. So this this deploy wavelength is setting the size of atoms.
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I think that's a point worth a point worth making, but we'll look at some numbers later on.
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So if you have an electron so an electron in hydrogen.
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Right. Is moving around. It has the binding energy of hydrogen is 13.6 EV.
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It has a kinetic energy which is half that because of the variable theorem which we'll,
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we'll, we'll have all these results later on, but they're already in classical physics.
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So it has a, it has a kinetic energy of all the electron volts.
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And that gives you a de broglie wavelength, which is which is a 10th of a nanometre.
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That gives you some kind of sense of scale. Okay.
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What about normalisation? So we've deduced that these the wave function of a state of well-defined momentum should be some constant times.
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This exponential. It's it's good to decide what this constant should be.
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We usually normalise our wave functions,
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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.
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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.
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The integral from minus infinity to infinity of one is just infinite and no constant in front is going to normalise it successfully.
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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.
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Should be delta of X minus x prime. So this thing here is the amplitude to be it x primed.
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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.
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So that when you integrate over this, you get you get one.
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So that's what we should do in this case. P is an operator with the continuous spectrum, same as X.
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So we want to we want to choose the normal to normal.
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The normalisation constant. Choose the constant. Such that.
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P primed. P equals one.
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Sorry, not one. Delta p minus.
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P prime. If I precise analogy with that.
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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.
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So this, this, this implies that.
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Well, this thing here is equal to. It's equal to p, primed x x.
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P. Right. That's just taking an identity operator.
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This is this is our.
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So we go we're going to say that x p is equal to some normalising constant times.
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Easily I p on h bar x.
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Right. And the name of the game is to find the value of this because we we know that this thing is this.
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The nice thing is that this is the complex conjugate of that.
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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,
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the integral d x of E to the minus i p primed over by x.
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That's from here the complex conjugate of that with p made into p primed.
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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.
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P minus p primed x over h bar.
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D x over bar. H bar.
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All right, so this age bar was always present.
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This one I've put in, I've. I've divided the X by age bar and multiplied by compensating each bar here.
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So the variable in the variable of integration is now x over each bar which is still running from minus infinity to plus infinity.
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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.
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We know that this integral is to pi times-delta p minus p probes.
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So what we're concluding is going right back up to the top that that original delta p
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minus P primed right up there is equal through these integrals to more squared times,
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times to pi h bar to pi h bar, delta p minus p primed.
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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.
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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.
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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.
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So this is an important result. It tells us something else that's of interest.
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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.
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What is it? What does this mean? This means the amplitude.
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To find that you have momentum. P Given that you are definitely the place.
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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?
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It's localised. At X you can ask what's the amplitude for this to have various momenta?
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The answer is given by this complex number here. The modulus of this complex number here is independent of P.
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So what does that mean? It implies that the probability of having P given X is some consequence.
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All values of momentum are equally likely.
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From from a momentum, which is nothing very much or zero even up to a momentum which is is associated with some relativistic gamma,
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you know, some large value of gamma, all momentum equally likely, including for extremely high ones.
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So that's clearly un physical. And what that tells you is you will never succeed in localising.
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It will never succeed in localising a particle precisely to an exact x.
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The state of being definitely at X is unreliable because it would it would imply that there was enough energy somehow in
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the system that there was a non-negligible probability of finding the momentum to have some extraordinarily large values.
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Right. So there we are.
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That's the so what we've discovered so far is if you if if if X is certain.
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P is totally uncertain. And conversely, if P is certain.
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X is totally uncertain. Moves.
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Let's therefore investigate these. Both of these situations are clearly and physical.
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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,
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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.
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Right. So this is a Gaussian distribution of probability and x, which sort of is generically we have,
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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.
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Right. We can ask what wave function yields this probability, what the answer is.
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Essentially it's a wave function which is the square root of this. So a suitable wave function.
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There are many possible way functions because phase information isn't conveyed by the probability.
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But let's, let's write down this wave function which is e to the minus x squared.
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Over four sigma squared over two pi sigma squared to the quarter power.
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So if you take the mod square of this wave function, you get the probability and the probability you get to that one there.
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So I could multiply this by all kinds of complex, all kinds of numbers of modulus, one and arbitrary phase.
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And I would still get that. But this is the this real wave function is the simplest one that we can write down.
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And now let's calculate for this.
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So this is a well-defined wave function which we know localises our particle two plus to the origin, plus or minus sigma.
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Let's ask, so what is the probability distribution for this upside of measuring a particular value of P?
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Right. So what we want to discover for this is what? So what's P upside?
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Well, that's the integral the x of p x x cy.
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We know what this is because we've just been working it out. This is a state of well-defined momentum.
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So this is one this is the integral d x of E to the minus p upon h bar x I believe.
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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
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we just wrote down e to the minus x squared over four sigma squared over two pi sigma squared to the quarter power.
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So this this is a we have to get this from minus infinity to infinity.
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Now, physics is full of integrals of this sort. And there's a box in the book explaining how to do them.
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I don't want to take the time to to go into the sordid details now,
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but all you do is you gather all these all these exponents of the exponential together.
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And what we've got here is an integral de x of each the I quadratic.
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In X. Right. If you gather this together, there's a linear term and there's a quadratic term.
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So what? So you can you can express that.
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I mean, it is each the quadratic expression in X.
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And what you do is complete the square of the quadratic.
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Change your variable of integration and use a standard result that the integral the x e to
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the minus x squared from minus infinity to infinity is equal to the square root of pi.
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We use the standard result, and that's how we evaluate these integrals here.
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But it's I would recommend learning how.
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Checking the box out, making sure you understand how that goes and doing this yourself as a as an example.
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After the lecture, when I did want to take time to do it now because it's just algebra.
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Let's just write down the answer, discusses the physical implications.
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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.
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And there's a normalising constant which is two pi h bar squared for over four sigma squared.
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To the quarter power. So that if we square this up, we get the probability of measuring various momenta,
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which is clearly going to be e to the minus two sigma squared p squared average
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bar squared over two pi h bar squared for sigma squared to the quarter.
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So our position of probability and position in real space, we have the particle localised in a Gaussian distribution with a width sigma.
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It turns out from this calculation that the possible values of the momentum,
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the probabilities associated different momenta is also a Gaussian distribution centred on zero in momentum and the width of this distribution,
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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.
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So the, the, the dispersion. And momentum is.
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Is HPR. Over Two Sigma.
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And so we write. So the dispersion in momentum is small when the uncertainty in real position is large.
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And conversely, right. So this so we have a result for this particular model.
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The dispersion in X times, the dispersion in momentum is h bar over to yes we ability to measure the hall.
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This. This. You worried about this, too? To know that you got call ops.
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Thank you. It should be a half. Yes, of course. Because I've squared the quarter. And thank you very much.
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I've square the quarter and it's become a half. So this is the classical statement of the uncertainty principle.
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It's really only an order of magnitude in this particular model.
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This is an exact mathematical statement. It's a statement about about the width of two.
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Gaussians But in, in a generic case, if you if you know,
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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.
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Then. The corresponding probability probability distribution in P will have a width
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which is broadly related to the width in X here by a relationship of this type.
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But it will be exactly h bar over two in the generic case.
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It's exactly h bar over two just for these Gaussian distributions.
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But the really key idea is that if is the product of the uncertainties in these two things.
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That has to be will be on the order of HBO.
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So there are two important points to make here.
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We need to be clear what we're saying.
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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.
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This is not the uncertainty of a measurement in X and then the uncertainty of the following measurement, the following momentum.
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Measurement. This is a statement about if I have a large supply of electrons,
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different electrons set up so that they are pretty much in the same in the same wave function.
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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,
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I'll get an uncertainty sigma x which satisfies this relationship because we have this this
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uncertainty in momentum is the uncertainty associated with the original wave function exercise.
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I. And if I would measure the position of that electron,
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I would change the wave function into some kind of something near to a delta function centred on whatever answer I got.
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Right.
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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,
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not the wave, not not an initial wave function.
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And then the wave function that we get when we make the measurement.
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And the reason we've done this partly is that we do not know what the wave function is.
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We get when we make the measurement that's in the lap of the gods, you make a measurement.
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So remember the basic dogma. We if we have to go back to the discrete case because it's simpler if I have my
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wave function is some sum and then some linear combination of stationary states.
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This is a well-defined wave function if I measured the energy.
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Then this thing collapses. To absorb is equal to EAC for some K and which is in the lap of the gods.
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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.
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So it is up here. If you measure the position, you will find some value.
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And after you've made that measurement, your wavefunction will be different.
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It'll be more or less a delta function associated with that X and not the wave function we're working with here.
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And the uncertainty on a subsequent measurement of P will be larger, will be large.
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The other thing to say is how do we understand this physically, this uncertainty relationship?
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We say to ourselves, well. If you.
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If the wave function is highly localised in space.
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If you think about that, wave function is made up as an interference pattern between states with twin plane waves,
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which are states of well-defined momentum. Then in order to have the interference pattern highly localised so that the sum of
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all these waves cancels to high precision everywhere except in some narrow region.
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You will need to have you will need to use waves with a very large range in wave numbers.
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And that's why the momentum is very uncertain. If the position is rather certain.
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So it's it's because because of this basic quantum, the basic principle of of of adding amplitudes.
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A highly localised lecture. We're entitled to think about a highly localised election as an interference pattern between states of of different
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momenta and we will need to have a very large range of possible momenta if we want to have a highly localised electoral.
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And tightly defined confined interference pattern. Let us now talk about the free the dynamics of a free particle.
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So. So we've just got a particle whose energy it's not there's no potential energy.
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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.
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It's a free particle. But what we're going to do now is talk about the time, the time evolution of this particle.
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So imagine that you've got the particle that tingles once you've got it localised around the origin.
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And let's, let's set this up a little bit by saying let's localised around the origin,
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but it's moving with some, you know we've got some, some idea of what its velocity is.
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So we're going to say it's initially we're going to write down.
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An appropriate expression for its momentum. So this is the this is the wave function in.
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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.
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The mean momentum is nothing. Suppose we start from.
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P ci is each of the minus sigma squared over bar squared.
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P squared minus p nought squared. Sorry.
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P minus P not. What do I want to do?
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Yeah. P minus p not squared over this horrible normalising constant to pi.
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H bar squared. Over four sigma squared.
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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.
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But the conjecture is the reasonable conjecture is that this complete set of amplitudes characterises a state of the particle where it is,
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it is moving with momentum.
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P zero. P zero is a constant, right? This is the minimum. This is the this is the momentum.
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Eigenvalue. This is just some constant.
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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.
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We'll find out whether this is true or not, but that's the conjecture. Okay.
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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?
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Why can we do this? Because we have a free particle.
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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.
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The Hamiltonian is a function of the momentum, so it has the same eigen states as the momentum.
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So a state of well-defined momentum is going to be an eigen state also of the energy.
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Now we know how to we we know how to evolve in time states.
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Once we so remember our basic equation,
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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.
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Remember, this was why we were excited by the states, why these states are, well, well-defined energy.
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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.
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These things set the initial condition for the calculation and the time evolution is given by these exponentials.
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So we want to use this formula in this other context here.
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We know what this is. This is a state of well-defined momentum. We know what this is.
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Right? This is just some exponential with the relevant energy going in there and this is the amplitude to have momentum.
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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.
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That's the analogue of the summing over the energies.
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When you sum have a momentum, you are summing over energy because different momentum or rate e to the minus.
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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.
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It's P squared over to m h bar t.
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Sorry to bother. Excuse me. Tea over bar.
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Right. That's the exponential thing either. What's this got to be?
311
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This has got to be a state of well defined p.
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And let's. We wanted to know what this looked like in real space.
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So let's barrel through with X and then this.
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Sorry. Sorry, sorry. I'm missing something altogether. Excuse me. Excuse me. Let's keep.
315
00:38:26,520 --> 00:38:32,030
Let's leave it out. I miss something out. I missed out the awards, didn't I?
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What are the ends? It's the amplitude.
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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.
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To have a momentum. P.
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Right. And then now we have the state.
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P. And now if we want the wave function information, we should barrel through with X.
321
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Then everything over here becomes a function of momentum and a known function of momentum.
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This is a function of momentum. Also time. This is a function of momentum.
323
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We just put it down by conjecture. It's that thing there. This is a function of momentum.
324
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It's the. This is this is a plane way.
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This is each the IP of punch bar X within a within a sign.
326
00:39:22,200 --> 00:39:26,210
Now it is exactly that. So so let's just see what we get here.
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So this is a dirty, great integral d p e to the minus i p squared t over to m h bar.
328
00:39:40,420 --> 00:39:43,200
Let's put this one. No, let's keep to the right order.
329
00:39:43,210 --> 00:39:52,270
E then here we have e what we said it was going to be eaten minus sigma squared over bar squared p minus
330
00:39:52,270 --> 00:40:02,139
p nought squared over horrible two pi h bar squared over four sigma squared to the one quarter power.
331
00:40:02,140 --> 00:40:13,240
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.
332
00:40:13,810 --> 00:40:20,370
Sorry, x average bar. Over the square root of just H.
333
00:40:22,320 --> 00:40:30,330
So what do we have here? We have an integral of an exponential of a quadratic expression in P, right.
334
00:40:30,330 --> 00:40:31,770
Because here we have a P squared.
335
00:40:31,950 --> 00:40:39,090
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.
336
00:40:39,420 --> 00:40:44,550
So it's another of these integrals of a exponential of a quadratic expression in P,
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which can be solved by the methods described in the box that we used just before.
338
00:40:48,420 --> 00:40:52,110
Now, the algebra in this case is a little bit it's a little bit wearisome.
339
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It's absolutely straightforward. It's absolutely straightforward, but it's just a bit wearisome.
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And the answer, in fact, that this comes to is quite a complicated expression,
341
00:41:01,350 --> 00:41:06,570
because what we're going to arrive at is something which has both phase information and amplitude information,
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00:41:06,690 --> 00:41:15,780
but we only want to know what the probability is of finding the particle at this place or the other place on that probability.
343
00:41:15,780 --> 00:41:21,269
The mod square of the answer to this calculation is much simpler and I'm going to write it down.
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So what follows now is a very straightforward calculation. I would I would urge you it's this box doing it in the book.
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00:41:27,180 --> 00:41:30,330
I would urge you afterwards to look through this and make sure you understand it.
346
00:41:30,720 --> 00:41:40,110
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.
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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.
348
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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.
349
00:42:02,820 --> 00:42:06,270
T over m squared.
350
00:42:13,810 --> 00:42:16,830
I need to tell you what B squared is, don't I? So. And here.
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00:42:18,890 --> 00:42:29,570
B squared is a complex animal. If sigma squared over h bar squared plus i t over to an h bar.
352
00:42:32,000 --> 00:42:38,799
So what have we got? This is a Gaussian distribution in ex at any fixed time.
353
00:42:38,800 --> 00:42:40,570
It's a Gaussian distribution in X.
354
00:42:43,450 --> 00:42:53,499
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?
355
00:42:53,500 --> 00:43:01,840
Because P zero over m we said this was the mean momentum of the of the it was the,
356
00:43:01,840 --> 00:43:05,560
it was the expectation value of the momentum of our original wave function.
357
00:43:07,000 --> 00:43:12,100
So it's, it's this is the mean if you thought of this as many different particles,
358
00:43:12,100 --> 00:43:15,760
it's really any one particle I thought of as many different particles would be the mean momentum.
359
00:43:15,970 --> 00:43:22,110
So this is essentially the mean velocity. So that's what you would expect.
360
00:43:22,130 --> 00:43:30,560
The the probability distribution is moving in space with a speed V nought equal to nought over M as we would expect.
361
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And the dispersion associated with this Gaussian is determined by by that stuff.
362
00:43:36,860 --> 00:43:43,490
So we have a we have a sigma as a function of time, which is going to be given by.
363
00:43:50,890 --> 00:43:54,930
So. So what should this be? This should be two sigma squared.
364
00:43:54,940 --> 00:44:11,830
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.
365
00:44:15,040 --> 00:44:18,190
I better write this down. It's too hard to do it in one's head.
366
00:44:19,120 --> 00:44:28,030
Plus, page 40 over two sigma signatures in the book.
367
00:44:28,570 --> 00:44:32,170
Yeah. Sorry. So this. Sorry. This should be another sigma. What should we call this?
368
00:44:32,200 --> 00:44:38,200
This should be cool. Well, let's just call this the dispersion. Well, we call it sigma sub t, right.
369
00:44:38,210 --> 00:44:44,840
Whereas this other sigma is the original sigma. So we've got a Gaussian distribution and, uh.
370
00:44:46,000 --> 00:44:49,900
Right. That's. That has a dispersion given by this.
371
00:45:04,420 --> 00:45:07,510
You know this story. Sorry. It's always right. There's something wrong here, isn't it?
372
00:45:07,510 --> 00:45:12,280
Because on dimensional grounds. Yeah, you're right.
373
00:45:12,580 --> 00:45:16,600
Did I write down the right integral? No, I didn't. That's exactly what's gone wrong.
374
00:45:17,050 --> 00:45:20,980
Sorry we're missing from here. A sigma squared on top. It's crucial.
375
00:45:21,580 --> 00:45:27,430
So when I. When I say what the dispersion should be, the dispersion we should arrange.
376
00:45:27,430 --> 00:45:32,320
This is two pi dispersion squared. So dispersion squared is equal to this divided by that.
377
00:45:35,890 --> 00:45:41,860
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.
378
00:45:41,860 --> 00:45:45,760
So I expect it's still right. But I was trying to do some of this in my head, which was dangerous.
379
00:45:47,880 --> 00:45:58,710
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.
380
00:46:00,190 --> 00:46:04,130
And what is this extra bit here? What was the original uncertainty?
381
00:46:04,150 --> 00:46:08,920
The original uncertainty in momentum from the uncertainty principle here.
382
00:46:10,990 --> 00:46:15,040
The uncertainty momentum was equal to a bar of a two sigma.
383
00:46:18,450 --> 00:46:26,190
So the uncertainty and the velocity was equal to H bar over to M Sigma.
384
00:46:27,030 --> 00:46:33,090
So what's this? This is equal to sigma. Plus the uncertainty in the velocity times time.
385
00:46:34,370 --> 00:46:43,250
Uh. Squared and shouldn't be square in this, should I?
386
00:46:46,040 --> 00:46:50,299
I think we might need to take a square root of a square, actually. Let's.
387
00:46:50,300 --> 00:46:55,570
Let's let's not chase that down in the moment, because this is the basic idea.
388
00:46:55,580 --> 00:47:02,930
The basic idea is the uncertainty in position is growing like the uncertainty in position times velocity.
389
00:47:02,930 --> 00:47:07,459
But that's what you would expect, right? Because you have what do we have?
390
00:47:07,460 --> 00:47:15,530
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,
391
00:47:16,010 --> 00:47:19,970
some are going faster, some are going slower. At some later time.
392
00:47:21,650 --> 00:47:34,970
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.
393
00:47:35,690 --> 00:47:40,310
But the ones that were going slower than the average will have slipped behind.
394
00:47:40,360 --> 00:47:44,479
They were already there was some of them will already be sigma behind,
395
00:47:44,480 --> 00:47:54,000
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,
396
00:47:54,020 --> 00:48:02,180
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.
397
00:48:02,510 --> 00:48:05,390
And I think probably we should be taking some squares and square routeing.
398
00:48:06,430 --> 00:48:12,040
But you see that we are what we're getting from this calculation makes perfectly good sense physically.
399
00:48:13,200 --> 00:48:21,909
Let me just remind you how we've done this calculation, because it's the it's the methodology, which is in many ways.
400
00:48:21,910 --> 00:48:27,790
Well, it's good to say it's crucial to see that what emerges from this makes sense physically.
401
00:48:28,000 --> 00:48:32,920
But it's also good to remind yourself, how can you how do you actually calculate these things in this damn theory?
402
00:48:33,760 --> 00:48:38,290
The way we've done this is we've used this central expression.
403
00:48:38,290 --> 00:48:46,659
We've said that states, I can I can evolve something in time so long as I can express my original state.
404
00:48:46,660 --> 00:48:48,969
It's a linear combination of states of well-defined energy.
405
00:48:48,970 --> 00:48:54,370
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.
406
00:48:55,600 --> 00:49:05,080
So we wrote we we wrote that some expression in the integral form is appropriate because momentum has a continuous spectrum.
407
00:49:06,250 --> 00:49:11,200
And then we just turned the handle and out came these perfectly sensible results.
408
00:49:22,950 --> 00:49:26,240
I think we're probably pretty much ready to finish the.
409
00:49:27,640 --> 00:49:35,470
Again, I want to stress I think I should stress that we've obtained this perfectly sensible physical picture,
410
00:49:35,920 --> 00:49:42,820
but we've obtained this perfectly sensible physical picture through an orgy of quantum interference,
411
00:49:43,420 --> 00:49:48,670
because we have in order to in order to get what we what we wanted,
412
00:49:48,670 --> 00:49:57,130
we took a perfectly well defined spatial distribution and expressed it as as an interference pattern between states of well-defined momentum,
413
00:49:57,910 --> 00:50:03,910
which we then evolved each state a well-defined momentum in time in its trivial way, with that just an exponential.
414
00:50:04,450 --> 00:50:13,480
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.
415
00:50:14,050 --> 00:50:20,740
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.
416
00:50:20,890 --> 00:50:26,200
We've taken something physical, we've decomposed into an infinite number of things which are not really very physical,
417
00:50:26,200 --> 00:50:32,800
namely states of well-defined momentum. We've evolved each one of those independently in time because they're states of well-defined energy.
418
00:50:33,100 --> 00:50:39,129
And then we've interfered the evolved momentum states we've allowed by working
419
00:50:39,130 --> 00:50:42,310
out this integral was working out the result of the corresponding interference,
420
00:50:42,310 --> 00:50:48,549
right? We were adding up an infinite number of of amplitudes and allowing them to to interfere in outcomes,
421
00:50:48,550 --> 00:50:51,880
something that makes sense, which is a wave packet that's travelling and spreading.
422
00:50:55,690 --> 00:51:02,740
And behaves in a way which does make perfect sense from a physical point of view, from a classical physical point of view.
423
00:51:02,780 --> 00:51:03,700
Okay, we'll finish with that.