Okay. So now we've investigated the states are a well-defined momentum, which as you recall with these plane waves with a wave number.
Pierpont Momentum upon HPR,
we can go back to this problem of two split interference and ask ourselves why it is that quantum interference isn't observed
in macroscopic objects and also assess what experimental setup will be required to see quantum interference with electrons.
So we put some numbers into this experiment.
So as you recall, we had some gun here that was symmetrically placed with respect to a couple of slits here in an obscuring screen screen,
it fired out particles.
Some of the particles got through the holes and they came to here and we said that the amplitude up here would be the sum of the quantum.
The quantum amplitude to arrive at this point would be the sum of the amplitude to go via this route or to go for this route.
These routes. Well, the the the distance from the gun to the two slits by set up is symmetrical is equal.
So any difference in the amplitude that they come there is to do with the change in the amplitude when it goes along this route as
against along this route so that D D plus b the distance from the upper slit to this place and D minus the distance to the lower slit.
Then we can write a formula that D plus by Pythagoras theorem is going to be L squared plus x minus x squared square root,
and correspondingly D minus is going to be the square root of l squared plus x plus s squared.
And we make the reasonable conjecture.
I mean, the right way to think about this is these particles.
Let's imagine that we've got our gun here, has been tuned to emit particles with some well-defined energy.
That means that as they go along here,
the particles will have some reasonably well-defined momentum because they the energy will consist of that kinetic energy.
So along here, we will have that, that the amplitude is going to be on the order of e to the i t upon h bar times x.
So that's a reasonable model for what the how the amplitude varies with position.
That's the new information that we bring to bear on this. So what will be the difference?
So the probability to arrive at X, as you recalled, was equal to the amplitude to arrive by the top slot,
plus the amplitude to arrive at the bottom slot mod squared.
And the argument that this was a plus mod squared plus a minus mod squared plus twice the real part of a plus a minus.
And these were about equal and well, these were the classical probabilities.
So these were P plus plus, P minus.
And then we have this quantum interference term, which we want to assess which.
So so what is this quantum interference term?
So the interference term. Is going to be.
Mod plus mod minus neither.
Not very interesting.
And the crucial thing is that this one is going to be e to the eye p upon bar d plus and the other one's going to be e to the minus i pierpont bar.
Sorry. Plus. Sorry.
One of these needs to have a star in it, which means that one of these requires a minus sign.
So if we're interested. Excuse me in the real in the real part of this and the real part so this thing
can be written as cos Pierpont bar de plus de minus minus D plus etc. etc. etc.
So this is looking like to the probability to go either way times through
through each hole separately times the cosine of p over each bar d plus minus,
d minus. Actually it's easier to do it the other way around it.
Minus plus. So what is this difference of distances?
This difference of distances from up there?
Well, let's binomial expand this and because we can argue that el is going to be big experimentally compared to x, x will be mm.
L will be a metre or so.
So we can binomial expand this and say that this is l it's about half of l brackets one plus x minus x squared of a l squared plus a dot, dot, dot.
And the other one's going to be about half L one plus x plus s.
Squared over l squared plus dot, dot, dot. So when we tell you the difference of those two.
So this is going to be two p plus minus this probability times the cosine of p upon bar of the
difference is going to be x plus x squared over l minus x minus x squared over two l in fact.
So when you take the difference of those two, you will be looking at two.
X. S of l. Yes to excessive l is what that difference will be.
So. So now let's put it now. We need to put in some numbers.
All right. Suppose we take the energy, which is P squared over two.
What do we want to do? We want to know. So what does this do?
This gives us a probability of arrival as a function of X,
which is going to consist of a of twice the sum of these two, which are going to be about equal.
And so about twice this plus twice. This thing times this cosine.
So it's going to be an oscillating commodity which will be doing this.
And there's some characteristic distance between these, between these minima which so this will call this delta X say no, let's call it big x.
That's what the so we call this big X the distance between the places where it's a
minimum because the quantum interference term is cancelling the classical term.
And this difference is what causes that.
The argument is that cosine to become two pi so we can write the two pi is equal to p over four times two s of l times x.
In other words, we have a formula for the distance between the minima, which is two pi to pi h bar, but h bar is h planck's constant over two pi.
So that's going to be h over p h l sorry.
Yes. Of S.
Yeah. And we could also write this.
I guess we we could also say that.
Oh, no, it's not, Paula. So let's put some numbers in.
Let's say that e the energy is.
So in order to get a big value of X, we want to take a big value of L.
Needless to say, we want to take a small value of P and of course a small value of S two on the bottom of.
I've lost a two. Yeah, you're quite right. So, so let's take L to be a metre.
Let's take S to be what got I think.
What did I do. I think I think it's Yeah.
A micron because you want to make it as small as you can.
But if you make it much smaller than a micron, you'll find it difficult to make the hole using ordinary materials.
And let's and we also in order to get a small value of P, we want to take a small value of the energy,
but you can't take too small a value of the energy or your particles will be deflected by stray electromagnetic fields and stuff,
and it'll be difficult to keep any kind of coherence.
So let's take 100 ev say as a, as a sort of convenient low speed if you plug all this stuff into there.
So that gives you what does it give you a 20th of a millimetre or something.
I think it's is it 0.06 millimetres which is obviously perfectly, perfectly observable.
Yeah. So this this such an interference experiment is, is, is possible but hard using electrons if you do the same thing with bullets.
When we're not expecting anything to happen, what could we do?
We would we would take the velocity for a gun is say 300 metres a second might be a bit faster these days.
I'm not sure. But that's a classical that's, you know, faster than sound.
So that's sort of a reasonable ballpark figure.
Suppose we took L to be one kilometre a thousand yards, pretty reasonable shooting distance for a rifle.
And if we took the mass to be ten grams, put it into the same formula and we discover that X is some ridiculous figure ten to the -29 metres.
So it's obvious that you cannot observe this interference using anything like a bullet, any kind of macroscopic, any,
any kind of macroscopic object, because it's going to be vastly bigger itself than the than the size of the interference pattern.
Obviously, an absolutely basic requirement for this experiment to work is the physical size of
your particle has to be smaller than the value of X that you derive out of this.
So you haven't a hope of measuring this interference. So that's why classically we don't we are unaware of this interference term.
But I would remind you that in the last lecture we recovered classical results,
which which explain why cricket balls move as they do, why satellites and so on move as they do by interpreting.
We calculated this.
We obtained results which recovered classical physics by decomposing the amplitude to arrive into a sum of contributions from states of different,
well-defined momentum. And these were all interfering with each other.
And the classical physics came back as a result of quantum interference.
So this quantum interference, on the one hand, is something which is very hard to observe with classical objects.
On the other hand, our entire picture of the classical world, a classical world, is only recovered through quantum interference.
It's not is it's not some esoteric corner of the subject, but it is hard to.
It's hard to have it happened in a controlled way. Okay.
So we. Yeah.
We should just. So we've done, we've done the position representation in just one dimension.
Everything has been a one dimensional motion motion long x.
We obviously need to generalise the position representation to three dimensions because we live in a three dimensional world for whatever reason,
and the generalisation is nice and trivial. We don't need to worry about it.
We have we now have three position operators X, Y and Z, also known as X, Y.
All right. And we have, of course, three momentum operators, three more operators, X, Y and Z, also known as P.
And we have that every one of these operators commutes with the other one.
So we have that exi comma sj.
It's nothing. And every one of the momentum operators commutes P-I, comma, PGA.
Equals nothing. So it is possible to simultaneously know your x, coordinate,
your y coordinate and your z coordinate as a complete set of eigen function of can states.
Of well-defined states. States. Where you know all those strings simultaneously.
Or you can know all three components of momentum, but you can't know there's not a complete set of states for knowing and so on.
And the only other interesting thing we have to have is XY commuted with PJ.
He's h baa delta. So it is possible to know the exposition and the Y momentum, but it's not possible to know the exposition and the momentum.
So most of these operators commute with well, each operator commutes with five of the of the sorry,
four of the remaining five operators, but it does not commute with its own momentum.
That's what each of these position operators. So that's the generalisation there.
What else do we have to say? Well, we used to have a wave function of PSI being a function of scalar x.
We now it's trivial the argument of the wave.
We we can now label a complete set of states by X,
Y and Z so we can write that there's a we have states of well-defined position
which are labelled by a vector now vector position x because there are three.
This is an eigen state of the x operator. It's an elegant state of the Y operator.
And then I can say to the Z operator, so we need three eigenvalues written inside here to describe what this is.
It is that's the mathematical level of the physical level. This is the state of being at the location position vector x correspondingly,
our wave functions become functions of X, Y and Z because they become these complex numbers.
Right? That's still a complex number, this complex number, but it's a function of X and Y and Z, the locations of the particles.
Similarly, we have states of well-defined momentum up.
We have states. Yeah. Yeah. You. You p.
Of X, which is x. P.
So now we have. Here we have p, p, p, z.
Because we have a state of well-defined momentum which is labelled with all three components of momentum.
So we have this function of a complex of three components, right?
This complex function of three variables X, Y and Z labelled by the momentum.
This is just an identical notation. Whereas in single, when we were doing this in one dimension, we found that this was e to the.
P over h bar times. X not vectors.
Now that's a vector. That's a vector.
And whereas on the bottom we used to have h bar to the one half, now we have h bar two,
the three halves reflecting the fact that there are there's an X component to this, y component to this and Z component to this.
So. So this the way this wave function of a state of well-defined momentum has now become a plain wave.
Whose wave whose wave surfaces are normal to the vector p.
And that's and that's what it is. It's easy to check that that stuff works.
It's a it's a very straightforward generalisation of what we did before. And I think that's all we have to say.
Oh, no, not quite. We also want to say what the momentum operator P looks like.
So previously we had that x. P xp.
Sorry. Not right. Talking about. Yeah. X p ci was minus h, but it was introduced by this formula here.
DPD x of x ci. All right.
That was what we did in one dimension. That generalises in three dimensions, very straightforwardly to x p psi.
So. So that's become a vector.
That's become a vector because we have to write down what it is for P and p z.
This is really going to be a shorthand for three formulae and it's going to be
minus H bar gradient operator on the function of three variables functional space.
This one here. All right. The wave function. So this is a vector reflecting the fact that that's a vector.
This is just a label which appears on both sides of the equation. That's what this this formula generates.
Generalise this to that formula. I don't think we need to be detained about that any longer before we leave the position representation.
It's good to do. And a useful result.
Which falls into our laps now because of what we've already done call the variable theorem, which is a a theorem.
It's a result in classical physics, which you may not have met, I don't know, but you in a way, should have met.
Did you cover the variable theorem in classical mechanics anyway?
No. Anyway, so it's there's nothing quantum mechanical about the variable theorem, but it has a classical counterpart,
but it's going to fall into our laps because we've got this powerful machinery.
So do you recall if we are in a stationary state, that is to say, a state in which the result of measuring energy is certain then.
All expectation values for such a state are constants.
That's why we call it a stationary state. It's going nowhere.
So every expectation value for a stationary state for a state of well-defined energy is independent of time.
So we want to exploit that result. So for a stationary state.
Is this just recalling what we already had? It was it was a consequence of our first theorem for a stationary state.
We have that DVD time or even deep DVD time of E Q E equals nought for all for all operators.
Q It doesn't matter what observable you stuff in there, as long as the observable doesn't is defined in a way that is independent of time.
So it's something like position, momentum, angular momentum, whatever it has,
it has a vanishing rate of change or with respect to time, it's a constant.
So we now apply this result to CU is equal to x dot p.
So then we have that nought is equal to.
T. Sorry.
Subtle. Some moment of doubt. Yeah.
Yeah. So I want to apply this to export p e and let's divide.
No stinking H-bomb. That by Ehrenfest theorem is x dot p comma h.
Whoops. So Aaron, First Theorem tells us that this rate of change which vanishes is equal to this here and now.
Let's take suppose we're dealing with a particle which has.
Which has kinetic energy and potential energy. So we'll take the Hamiltonian to be of that form, which is.
Pretty useful form. And stuff it in there and we're going to have that nought is equal to E!
X dot p comma p squared over two m plus v close square brackets.
So now we need to work out what this comitato is.
And this is where a little bit of so this is where we get a bit of practice in using the three dimensional generalisation.
We obviously have two things to work out. We've got a combination of X come up with P squared, so let's work that out.
X come up with P squared.
Now we write that in components. We write x x dot piece, x dot peters I say comma exactly comma squared.
This x dot can be written as a sum over j equals 1 to 3 of x i p i sorry sj pj so that's just a way of writing that.
And now I have a some peak. Well.
P-square. Okay, so I'm something of a K as well.
All right, p squared is p squared plus p y squared plus p said squared.
Now we can work out this using our rules for a commentator.
We had that rule that a comma B sorry, ap comma C was equal to a comma c.
B plus a, B, C, P.
We know that peak commutes with PGA that's been written down up there.
So that commentator vanishes. That's this one here in some sense.
Sorry, this this one here in some sense. And so what we're left with is so we have this double sum, we're going to have X Jay Peak, PJ.
Sorry. That squared. Squared. Comma.
No, no, no comma. So that's what we get. So.
So this has to be commuted with that? That's what I've written down, I hope.
And then there should in principle, be another term, this commuting with this.
But that vanishes because commutes would peak for all for all jank.
So we have to work out what this one is now and we can use the same rule.
If we're being pedantic, we would say this is X on peak peak.
So we would say that this is x j peak, peak plus peak.
The commentator of x and peak. I'm using the same rule and that all has to be multiplied by p j.
The same because. Because I'm not writing p squared is peak. Peak.
But this is h bar. This is bar.
So these two terms actually contribute the same thing. This becomes to h bar p p times, delta j k times, peak times.
PJ And I'm sorry, I've lost track of the sum sign.
Here we have a sum sine. With something over J and with something over K.
Some have a J and you get nothing because of this delta j k unless j is equal to K, so this becomes PCP.
K Some do have a k, but PCP k some of AK is the same thing as p squared.
So this is 2ih bar p squared.
That's what the commentator is of x dot p with p squared.
Now let's let's right now, let's do the exact P commentator with V, which is itself a function of X, of course.
These things ought to have hats, really. But one gets difficult to write down enough things.
Well. What we want to do is is write this thoroughly in the position.
Representation in the position representation x dot p is minus i h bar x dot gradient.
Right? That's what this becomes in the position representation on v, which becomes a function of x just so this is in the position representation.
So what does that mean. That means I ball minus age brackets x dot gradient.
Uh, working on v minus the x dot gradient.
And this is an operator statement. So it's waiting for you to put in the function of your choice of PSI on the right.
Right. There's a virtual function there for it to work on.
That's what that's the meaning of this vehicle gradient.
And this exact gradient V doesn't mean extra gradient only V means of everything that is the right of it, including Europe PSI.
So when you use this exact v on v times alone, you'll get a term and then you will still have to use the V on the up side.
But the result of using the V on the up side will be killed by here and next v on up say.
So what this is equal to is minus h bar x dot gradient of v.
That's all it survives. This is the action of of the nebula, the gradient operator on the potential itself,
the operation of the action of the gradient operator on the wave function that's virtually sitting here is cancelled by this contribution here.
So we now have we we can put these results back into what we had up there.
So what we had was nought is equal to yeah.
Is equal to the sum of these of these commentators is equal to e x dot p comma p squared.
E plus e x dot p v.
That's just summarising where we stand. This we've discovered to be.
This is to i h bar. This commentator turned out to be p squared.
So it becomes the expectation. Oops, there should have been over two m on this, shouldn't it?
Because it was the Hamiltonian p squared over to him. Yeah.
This P came from the Hamiltonian where it was P squared over two m this v came from the Hamiltonian where it was just V.
So we have over two M Uh, no, let's leave it alone of p squared over to m e plus we figured out that this one was a minus h bar.
So we want to cancel what we.
This is the expectation value of the kinetic energy. Clearly.
Right. P squared over two m is the kinetic energy.
That is the expectation value of it.
So cancelling the ball, we can say that two times the expectation value of the kinetic energy is equal to this stuff.
That's as far as we can go in general. But consider now very important cases have that V of X is proportional to model X to the alpha.
So, for example, for a simple harmonic oscillator we're about to discuss, the potential energy goes like x squared alphas two.
If we were dealing with a dealing with a Coulomb interaction, the potential and potential energy goes like one over radius.
So it would be V of all is proportional to one overall.
Well, this model X is R, so alpha would be minus one.
So we can say that alpha equals two in simple harmonic motion.
Alpha equals minus one is coulomb.
There are, you know, you can think of other power laws which are relevant in this case.
So then we ask ourselves, what is X dot?
Gradient of. Evie.
Well, that's going to be so sad. We'll say that this is equal to some constant A times, X to the alpha.
What is this going to be? It's going to be alpha model X to the alpha minus one times X dot the gradient of model X and the gradient of Model X is.
The gradient of Model X is. X the unit vector x, so it's the vector x over model x.
So this is equal to. Sorry, this is a we have a X to the alpha minus one times x dot x over monarch's.
So here this mod x is going to make this an alpha to the minus two.
But from this x dot x we're going to get mod x squared.
So this is going to be and I've lost sorry this was an alpha.
That was also an A unfortunately. Yeah. Sorry. We need an A and an alpha.
This is going to be alpha times a x to the alpha, which is alpha times V.
So if V has a power lower dependence on distance from the origin, then x dot grad v is simply alpha times v.
So when we put this result back into that formula, back into this statement,
here we have that twice the k e expectation value is equal to alpha times the expectation value of the potential energy.
So that's our Kepler formula. In the case of simple harmonic motion, alpha is two, and kinetic energy is equal potential energy.
In the case of Coulomb interaction, where alpha is minus one, you have that the potential energy is minus twice the kinetic energy,
which is to say that the particle has lost two units of energy, radiate in falling in from infinity into a bound orbit.
It's lost two units of energy, one units being sent off to infinity and radiation or something, and one unit is used as kinetic energy of its orbit.
So that's the this is a very theorem. So now we open a new chapter, as it were, by talking about harmonic motion.
The harmonic oscillator is the single most important dynamical system in physics.
Most of field theory, most of contents of quantum field theory,
most of condensed matter physics is fiddling with more or less with harmonic oscillators, which which are which are decorated in some way.
So the basic physics is that of the harmonic oscillator.
And it's worth just taking a moment to understand why harmonic oscillators are all over the place, the universe, the physicists.
A fundamental position of physicists. Well, physicists like to represent the universe is a collection of harmonic oscillators.
And this is partly because physicists may be brighter than some other people, but they're still pretty stupid.
We have a quite a small bag of tricks and harmonic oscillators is a trick that we we have.
And it's an incredibly useful trick for this reason that if you plot force in some direction versus displacement from a point of equilibrium,
you will get a curve which does something like this, the force.
Vanishes at an equilibrium at the point of equilibrium of a system.
The force on it obviously vanishes. So if you do a plot of force versus distance, you'll get a curve.
Something like this passing through zero at the point of equilibrium, which I happen to put at the origin of X.
But, you know, that's by construction, clearly.
And the general idea is that most of the time you can destroy.
That's meant to go through the through the origin. Most of the time, you can represent this to some, to a good approximation,
you can say that F of X is about equal to x plus order of x squared or whatever,
so to lowest order approximation, because f has to vanish to the point of equilibrium in the neighbourhood of the point of equilibrium.
F is going to be proportional to x and if is if we neglect this, if this is small,
then we have harmonic motion for displacements, these small displacements around here.
So this is why harmonic oscillators are ubiquitous, a very credibly, an incredibly valuable model.
We can apply. We can use to understand many, many systems because many systems for small displacements,
almost all systems for small displacements look like a harmonic oscillator.
Okay, so let's agree what the Hamiltonian of this thing should be.
The Hamiltonian of our harmonic oscillator should be P squared over two M plus a half K X squared.
Right. Because if the force is going like that, you integrated up, this becomes the potential energy.
And this is of course the kinetic energy. We're familiar with that already. It's better, though, to write this in a different way,
to anticipate results that are to come and to write this is p squared plus omega x squared over to M.
So, uh, and of course omega squared is k over m so it's easy defining omega squared to be okay over him.
It's easy to write this formula like that, and that's how I want to write it.
So you want to reproduce this formula? Just think about dimensional analysis.
We want to have piece going over to him because it's the kinetic energy. We're always saying that.
And here I want something that's proportional to x squared and has the dimensions of momentum and obviously omega x has dimensions of speed.
So M omega X's dimensions momentum. So that's why, you know, that enables you to recover that quickly from this.
And that's the way to go for practical purposes. So that's our Hamiltonian and we're trying, of course, to solve.
So these stationary states are the key to understanding dynamics because they have this
trivial time evolution and by by decomposing any initial condition into with some of,
of stationary states, into a linear superposition of stationary states,
then evolving the stationary states, we find out how any arbitrary initial condition evolves in time.
So that's why we want these stationary states. I've said that before and I'll say that again.
So we want to find states of well-defined energy.
This is the problem we want to solve. And this is a completely generic situation in physics.
First of all, you think about your physical system on the grounds of physics, you write down the Hamiltonian.
Then the next thing you do is you find the down stationary states, because once you've got those, you can do anything you want, pretty much.
So that's what we're trying to solve. The the the way to do this is the proper way to, to find these states.
So we need to find the energies that that are possible and we need to find the corresponding states.
And the way to do this is to introduce some new operators.
Let's introduce a which is m omega x plus i p over the square root of two m omega or h bar omega.
Why do I write that down? Well, basically because I know where I'm going.
But just to give you some sense of direction, the general idea here is that we want to factories, that that's the general idea.
We want the factories, the Hamiltonian into.
So it's a quadratic expression. It seems kind of reasonable to factories.
It if these were if these weren't operators because these are operators, sorry.
In future I'm not going to even attempt to put hats on operators.
Right. These are operators.
Despite the absence of hats, it's just too difficult to remember to put the hats on and takes too much time and grown ups never do.
But these are operators now, if they but if they weren't operators this in its complex conjugate would factories that.
So that's the drift.
Okay let's write down it's it's a bit complex this is this of course is an operator and it's not an observable it's not a mission operator.
It's what is its dagger? A dagger is not joint is this thing dagger, which is itself because X is a mission operation.
It's own dagger. So it's M omega x plus this thing dagger p is its own dagger.
But I has the the dagger of I the joint of eyes minus II.
So this is minus I p over.
Of course, this on the bottom is a real number. So it's own it's its own complex conjugate.
So here we have two operators and the general idea is they're going to factories h or
they almost do whatever because the plan and this is called in annihilation operator.
And this is a correctional operator and. Well, the reason they have these names will emerge.
But it is that if you use this on a state,
this operator increases the excitation of our harmonic oscillator and this oscillator reduces the expectation of our harmonic oscillator.
And since in quantum field theory, particles are excitations of the vacuum.
This thing creates a particle because it creates an excitation, which is a particle,
and this thing destroys a particle because it destroys the excitation.
So what we next do is work out what a a dagger A is, because the idea was that this product would be more or less the Hamiltonian.
So what exactly is it? Let's get this right. M omega x minus IP and omega x plus i p over to m bar omega.
Now, when we write this out, we have the obvious terms.
We have p squared and we have m omega x squared.
So let's write those down. That's P squared plus m omega x squared all over two bar into m omega bar, whatever.
And. And then we have some additional terms which would class that would cancel in classical algebra but don't now because we have an X,
we have an M Omega X times IP and here we have an m omega x on the so we have an I minus IP times name and we are x.
So the additional term is an m omega x.
Am Omega I x comma p and it's a gain of a two and h bar omega.
Right. So this is the Hamiltonian overreach for Omega and this is an H bar which and the I's make a minus one with this.
So this is going to be minus a half and everything else will cancel.
Right. Because we'll we've got an Omega here.
We're going to get an H bar from there. So the rest cancels. So I should have I should have explained.
Sorry, what is the factories, this and this on the bottom, this normalising factor on the bottom is put in.
It's not really essential, but it's very convenient and it's put in in order to make this dimensionless.
So just to check that that's true. H Bar has dimensions of position, times momentum.
Right. So it has the dimensions of position, times momentum.
So what we have here is m x sorry, m omega x which we've agreed has dimensions of momentum, times p which has dimensions of momentum.
Then we take the square root.
So this on the bottom has dimensions of the whole square root of momentum and therefore cancel the dimensions of what's on the top.
So it's dimensionless. That's the purpose of the that's the purpose of the horrible square root.
So we find that this product, which is dimensionless, is equal to the Hamiltonian divided by H Bar Omega,
which has the dimensions of energy because H Bar also has the dimensions of energy,
times, time, omega, of course, has dimensions of one over time it's a free is the frequency of the oscillator.
So this is has dimensions of energy minus a half, which is obviously dimensionless.
So we have indeed almost fact arised.
We have the statement now that H can be written as H Bar Omega, which carries the dimensions time z dagger a plus a half.
We've almost factorisation just this that there. The next thing we want to do is calculate the commentator a dagger, a comma, a yes.
We just got time to do this, a dagger of these two operations.
So of course we will have a one over two H Bar Omega as a factor on the bottom because each of these A's brings in its own square root.
And then we will have the commentator of m omega x minus i p on m omega x plus i p now.
We and we have this breaks down into four commentators in principal.
There's the commentator of this with this and the commentator of this with this the commentator of this
with this obviously vanishes because excuse with itself and the commentator of this with this is.
So we're going to have an m omega AI times X comma p, that's the computation of this with this.
And now we have to deal with these with these terms. This produces a non-negotiable commentator.
With that, we're going to have minus M omega ai times p, comma x.
And then we'll have the commentator repeat with self, which will vanish.
If I swap those two over, then clearly I change the sign in front and then this becomes a plus x comma p.
It becomes this thing all over.
So that cancels this and this whole caboodle is going to equal i x comma p over h bar that because we're going to get a two.
These two terms are going to add together to make us a two, which cancels with this.
And the moment is clearly go x, come a p is itself equal to h bar.
So the i's make a minus one, the bars cancel and this is equal to minus one.
So these two operators have non vanishing comitato actually equal to minus one.
Yeah. Well, we seem to still have time to to nail this this problem, I think.
So let us suppose we have got a state of a stationary state.
Let us now apply the operator a dagger to both sides of this equation.
Right then, this is just an eigenvalue. It's only a number.
So I can then write E a dagger.
E is equal to a dagger.
H e that's obvious.
I would like to swap these over so I jolly well do. I say this is equal to AJ Dagger plus a dagger.
Commentator H. So this commentator puts in what I'm supposed to have and takes away what I'm not supposed to have but have previously written down.
But we know what age is in terms we have that age is equal to there it is age bar, omega eight Dagger.
So let's use that. So this is H, a dagger plus commentator of dagger.
And H turns out to be a dagger. A plus a half.
Close brackets. H bar omega to carry the dimensions.
Close that, close that and stick in our E that we first thought of.
So all I have done is replace H by an expression we already derived.
Yeah. Now I have to take the commentator of a dagger with this.
And with this. The commentator of Dagen with a half clearly vanishes because a half is just a number, not an operator.
The commentator of a dagger with itself then vanishes.
So. So when we do the commentator with this product, there should in principle be two terms, but only one of them survives.
And that term is. That term is this sticks.
This stands idly by while the dagger works on that.
And then I have an omega. Omega.
Close brackets. Close brackets. He.
But we just worked this thing out and found that it's minus one, right?
A dagger turned out to be minus one. So this is equal to.
H a dagger e minus H bar omega a dagger e.
Just to remind you, remind us what we had on the left. What we had on the left was e a dagger.
E is just a restatement of what's been at the top.
So we take this a dagger E and we obviously join it on to that degree and we discover that h on a dagger e is equal to well, h working on this.
The cap that you get by using a dagger on E is equal to E plus h bar omega of dagger working on E, what does this tell us?
It tells us that we have out of a state which had energy e we have constructed a state by multiplying by a dagger which has energy e plus h by omega.
So this means that a dagger E is equal to a constant normalising constant not discussed times e plus h bar omega, a new stationary state.
This is an incredibly powerful result because it immediately follows that we have states,
if we can find a state e we can immediately generate E plus h bar omega by using this a dagger based.
And also if we use a dagger on this, it follows we're going to get E plus two H Bar Omega and we're going to get another.
If we use a dagger on this, we're going to get E plus three H Bar Omega.
So we're going to get a whole infinite series of states of ever increasing energy simply by applying a dagger again and again and again.
So what remains is to find what e what number E is, and that we will do first thing tomorrow.