Okay. So let's get underway. We were we were talking about spin a half, the most important type of spin yesterday and we got this far.
So any state, as regards its spin, its orientation, should be expandable as a linear combination of the state plus,
which means you are certain to get a plus a half if you measure the spin along the Z axis and minus, and there will be some coefficients.
There will be these coefficients here and a complex number here and a complex number there.
The amplitude to measure plus a half or less said or the amplitude to measure minus a half.
And I said, and we're calling these we it's obviously handy in notation to call that thing A on this thing B and then what
we want to be able to do is write the result of using some spin operator on this arbitrary state of sci fi.
We can also expand the linear combination of this and this because there are a complete set of states for the orientation of this spin,
a half particle spin off system.
And we, uh, I hope I persuaded you yesterday that these these numbers, these amplitude, C and D can be obtained as the vector on the left.
If on the right, we put in the two numbers that characterise up sy on the right we get out on the left,
the two numbers that characterise PHI after we've multiplied by this matrix of four complex numbers being the expectation value of the,
of the relevant of whatever operator we're trying to use between the plus states,
the minus states and then these non-classical of diagonal bits on each side.
And we said, I think we finished by saying that if, if I is Zed, in other words, if we're interested in the result of using a Z on ABC,
then this matrix is very simple because asset on plus is simply a half of plus.
So we get a half appearing here, we get minus a half appearing here because our set of minus is minus one half times
minus and we get nothing appearing here and here because plus and minus are orthogonal.
So we have this diagonal matrix, which is no accident.
It is simply the matrix that contains the eigenvalues of Z down its diagonal.
Because we use this basis vectors the I can cancel offset.
We made that choice and the result is that s the matrix representing said is diagonal with its eigenvalues two on the diagonal.
And this matrix is conventionally written as a half times this matrix which is called Sigma
Z and it's called apparently Matrix because Wolfgang Pauli introduced it into physics,
although it was known to mathematicians matrices like this.
Okay. So more interesting is if we ask ourselves, what is the matrix for X?
So the matrix for my ex is going to involve things.
Well, we're going to have, for example, plus s, x plus.
This is a complex number. We want to know which complex number and the secret is to radius of calculating
this is to right s x is a half of s plus plus minus where s plus minus.
So the matrices, sorry, are the operators that we already introduced in the context of J and L to to
reorient the angular momentum either towards the Z axis or away from the Z axis.
So as these are X plus and minus I times s y, right?
So this, this operator was, was introduced in the form of j plus minus.
But remember, spin and total angle momentum have the same commutation relations the same the same behaviour in every way.
So this letter, this letter operator is this.
And obviously if you add X plus two S minus, you get to X because the Y terms cancel.
So this is definitely the case. So this, this thing here can be written as a half of plus s plus plus plus plus.
Sorry, sorry. Yeah, well, this is what I'm trying to calculate. Yeah. S plus s minus plus s plus tries to raise this to an even larger value.
This is plus a half. It'll try and raise it to two plus three halves, but no such value is allowed because of spin.
The total spin is only a half, so it kills it in the process.
Therefore this one is zero. S minus successfully lowers this to minus, but minus is orthogonal to plus.
So this is zero. So this element here is zero and that's the top left corner of the matrix for x is zero.
Similarly,
exactly the same reasoning would lead you to conclude that the bottom right hand corner is zero and the non zero elements occur off diagonal.
So if we look at plus s x minus, we're looking at a.
Half of plus s plus minus plus plus plus s minus minus s plus races minus two plus successfully s plus or minus is exactly one times plus.
So this number here is equal to one and minus tries to lower this and kills it in the process.
And therefore, this is equal to zero. So this element, this off diagonal element, is in fact equal to a half.
We know that the bottom right hand element is the complex conjugate of the top right hand element, because this is a mission operator.
So we know now that the matrix is s x is represented by the Matrix, half of nothing,
one one nothing, also known as a half of Sigma X, the Pauli Matrix.
This is this is the Pauli matrix. Sigma X. And when we do the same thing to find out what s y is we right?
This is a half of plus one over two I of s plus minus this minus.
Right. Because if you take the difference of X plus, i, y and s x minus I y, you will end up with 2isy.
So, so we have this. And what do we get this as plus raises this two minus two plus so so plus s plus minus the gain equals one.
So therefore this is equal to one over two.
I also known as a half minus a half minus I over two.
So the matrix representing s y is going to be it's going to be a half.
Of one minus I. I saw.
I want to have one. Nothing. Nothing.
The dog, the elements will be nothing for the same reason that they were with X, also known as a half of powerless matrix sigma one.
So that's where the pounding matrices come from. They're simply the matrix representations of the spin operators in a basis.
In the when you choose as your basis the eigenvectors the eigen kits of Sigma Z.
So let's use these. Use this apparatus to.
To do something slightly interesting.
It's it's an excellent exercise both in in in in practising getting experimental predictions out of this abstract apparatus.
And also we learn something interesting about how how the orientation of atomic scale things behave the somewhat counterintuitive arrangements.
Okay. I don't think this computer is going to this system, projection system is going to work today for some reason.
So. Okay.
So the point is that, uh, so the point is that a spinning charged body.
Is a magnetic is a magnetic dipole.
I think that's kind of plausible. So that.
So electrons, neutrons, protons, except what's right.
Neutrons, electrons, protons, being spinning.
Charge bodies have little magnetic moments. They are little magnets.
So if you put a magnet in a beam field, you have this is the energy.
Of a magnetic. Dipole in a mag field.
So there's a minus sign here which says that the energy is lowest when the magnetic when the dipole is aligned with the magnetic field.
Right. So when this dot product is positive, the energy is lowest.
So that's why magnets, compass, needles and whatever align with the magnetic field.
That also means that if a magnetic dipole is aligned with the field,
it the its energy will drop as it moves into a region of bigger fields because it'll this will become a more negative number,
whereas if it's anti aligned with a magnetic field,
then its energy will increase if it moves into the magnetic field because this will become this will be negative and the two minuses will cancel,
we have a more positive energy. So since things tend to move in the direction that minimises that potential energy we have,
that magnets aligned with B will be sucked into a region of stronger B, so a magnet, a dipole.
Aligned. With B so that means that mu dot be greater than nought is sucked.
Into a field. If the field strength varies spatially, which it too often does,
the what particles which have the fields dipoles aligned will be will be sucked into b and similarly the other ones will be repelled.
So the anti aligned loops aligned.
Dipoles will be repelled from a region of high P.
So that was the physics that Stern and Gerlach exploited, exploited in 1922, in experiments which astounded the world.
They found themselves. They made themselves a magnet. Or should we call this north and we'll call this south?
They made themselves a magnet which had puzzle pieces, one of which was pointy and the other of which was flat or even.
Well, I think it was flat, but it could also be concave like this.
And then you can imagine how the field lines run. The field lines run like this somehow.
I'm not doing a very good job of it. My diagrams are usually rather rubbish.
The point is that here we have a crowding of field lines, which means we have high B.
Near knife edge.
So I have a nice picture of this, but the computer isn't willing to show it because this is the end view of a of a long of a long thing.
So this is like the point of a knife, right? We're looking end on the point of a knife.
And this is just this is a table somehow. So if you if you have some particles with some spin coming in here and aim it right.
So that they're heading for this, well, they're heading a bit below this region of high magnetic field like this.
Then the ones that have their spin aligned this way into B are going to be sucked into the region,
drawn, attracted by the region of high B near the point of the knife and move on up here.
So this is the particles which have moved on b greater than nought and particles
with and anti aligned with mew dot be less than nought will come down here.
Of course, this is all grotesquely exaggerated.
In fact, you'll have very you'll have a very subtle curvature and then you'll have a straight line in front.
Right. So we we get we get the particles deflected either way.
So if you have. So what they did was they took silver atoms because silver atoms turn out to be spin a half particles coming in here.
Then what they found, which surprised them and everybody else, that half of their particles,
half the silver atoms went off this way and half of this little breath was went off that way.
So when they they detected the atoms on a screen over here, they got two blobs distinctly separated.
The quantum mechanical interpretation of this is that is these atoms.
When the atoms are in here, they are.
Sorry I haven't said that. Mu. The magnetic moment is equal to some number.
The gyro magnetic ratio times the spin operator.
So when they're in here, they're.
The magnetic field is, as it were, measuring the component of of spin in the direction of the magnetic field.
That's what you you say to yourself. So and there are only two answers possible.
Either you'll get plus a half or you'll get minus a half for the for the value of this.
And therefore, mew will be either a half g in the direction of B or it'll be minus a half G in the direction of B.
If it's so and the half widgets it's up it's plus a half g will be deflected that way and the other lot will be deflected down here and.
There you go. So at the end of the day, you have to stern girl our filter you put in the particles with they've just come out of some oven.
You've made the silver vent, you've made you heat it up.
Some silver in an oven, made some silver vapour, allowed it to diffuse out of some holes, culminating slits and that kind of stuff.
So it's coming along here with some thermal velocities and out out of your filter.
You have a load of you have atoms which have their spins in this case up on Z and the ones that come out here are in this state.
So it's a machine for for it's a practical device for creating silver atoms which are in this state.
Now you can play some entertaining games by installing another Stern Gerlach filter.
So let's just block these off something being a nuisance stick in another stern girl like filter here and now.
Let's measure the. Let's measure up rn.
So let's measure the spin along some unit vector n and let's take so so we're going to have this to be the X direction.
We're going to have this to be the Z direction and what the Y direction will have to be out of the board.
All right. And what we're going to do is we're going to take n is equal to nothing, comma sine theta, comma cost theta.
So n is going to be a vector which if thi is nothing, is just in the z direction.
And if theatre is pi by two, it's in the y direction and it can be allowed to scan between these directions.
As we vary theatre and what we want to do is calculate which fraction of the uh of the atoms will survive, will get through the second filter.
So this is the filter f one, this is the filter f two and you want to calculate the probability that an atom gets through both filters.
So let's focus for the moment on the probability that an atom that has gone through the first filter gets through the second filter.
So the probability that you pass F2 given that you passed.
F one in quantum mechanical language is is plus a half on.
Given that your well let's just we'll just say plus on n given that you were plus on Z.
So this is the state that you're in. Up there is just called.
Plus when I put in a Z just to distinguish it from this which is in the direction of N that this is an icon of s z with eigenvalue a half.
This is nyan cat of s of n with eigenvalue a half.
And this this pair of things makes me the amplitude for by the basic dogma of the subject for the probability of this outcome.
So I need to mod square this and I've got the probability that I want.
So we can work this out.
We can get this complex number as soon as we know how to write a class on in as an amount of plus on Z plus an amount of minus on Z.
Right. Because so if we get this number and this number,
then we have the probability that we want is going to be model squared because because a star is going to be exactly that number.
So right to get out of this catch, you could get the bra you want up there by complex conjugate,
get you to have a nice star bang in with plus on Z and you pick out a star.
So the probability we want is just model squared. So that's our exercise to find A and B and we'll be all done.
How to find A and B? Well, what's the point about? What's the point?
What's the defining characteristic of that cat? It is that it is an I can catch of this operator with eigenvalue of this defines.
And.
It's totally characteristic of these sorts of calculations of a wide range of quantum mechanical calculations that this the sequence of arguments.
I want a certain complex number. It will involve some cat.
Ask yourself, what is the defining characteristic of the cat? It will usually be because it is not in case of some operator.
Now we have a well-defined mathematical problem.
Find it because what is s n s n or sorry s n is equal to a half of an x sigma x plus ny sigma y plus nz sigma
z sort of a dot product between the unit vector and and the vector made up of the three pally matrices.
What I need is zero. So basically we've got and then why we agreed was going to be sine theta and this we agreed was going to be cost theatre.
So at the end of the day, it is a half of.
Now Sigma Z, we've got up there, it's got one in the top left hand corner and minus one in the bottom.
So I get a cost theatre and a minus cost theatre appearing on the diagonal because of Sigma Z and this has got a minus I in the top right hand corner.
So we get a minus sign feature appearing there and it's complex conjugate has to appear down here.
So this is the matrix that represents s n where theta is defining the direction of.
And now all we have to do is say that is say that this matrix cost the two minus I sign c to I sine theta cost the two on a b is equal to a b this I
can get this this vector has to be A and I can head of this matrix with eigenvalue one in order that it's and I can count of n with eigenvalue a half,
right. Because the original expression was s n on this equals a half of that, but here is a half I can cancel on the two sides.
So I'm looking for the eigen catch of this operator with eigenvalue one notice I don't
waste my time finding out what the eigenvalues of this operator are of this matrix are.
I know that the because this is a is a is a is a matrix that represents a spin operator.
And I know before I start that the eigenvalues are plus and minus a well of this one plus minus one half of this one plus or minus one.
So we don't waste time finding out what the eigenvalues are. We just get on and solve these equations.
What? There are two equations here, but because we're looking at an eigenvalue problem, only one of them.
These two equations are linearly dependent upon one another.
Only one of them contains useful information.
The other one repeats that information. So we may need to look at the top equation and it says that a a minus one.
Sorry, sorry, sorry. Eight times brackets, one minus cost theta.
So if I'm going to get a costly to equals a on the right hand side.
So if I go on the right hand side will have a costly to a into one minus cost theta is equal to minus i b sine theta.
In other words, we're going to have that b over a, which is all that I can say the ratio of A to B that I can determine out of this.
The absolute values have to be determined from a normalisation condition or are equal to.
V override is equal to one minus cost to over.
Minus I. Sine theta.
And we can clean this up a bit if we use some half angle formulae,
because this on the top is twice the sine squared of Caesar over two sine theta is twice sine theta upon two cos these are upon two.
So we can cancel a number of things.
The Tus cancel one of the sine features cancel and we end up with sine feature of a two over minus I cos these are over two.
So I can right now that a B is equal to costs.
These are over to I signed the teacher over to.
So if you work out the ratio be over a of these two I think you will get that because this minus I could be puts an eye on the top.
And moreover, this thing is correctly normalised. It just happens.
So in principle, I would now need to deal with the normalisation. I've only been calculating the ratio of the components.
I want more squared plus plus more B squared to come to one, but it jolly well does by good fortune.
Right? So this is the, this is the complete bottom line. This gives you OC, right?
So, so the probability that we pass.
F2 given that we passed. If one is actually equal to we said it was gonna be more squared is therefore cos squared feature upon to.
Does that make sense? If theatre is equal to nothing, then the second filter is also measuring the Z component of angular momentum.
And we are. We are. The output from the first filter is guaranteed to return plus a half for the Z component of anchor momentum.
So this probability must be one and indeed col squared of nothing is one.
If the theatre is pi then then the second one is plus a half.
Then N is pointing in the minus z direction. So getting plus a half in the direction n is equivalent to getting minus a half in the direction Z.
But we know for certain that we're going to get plus a half in the direction Z so the probability of this happening is zero and indeed cost squared.
If I put features equal to pi, I'm going at cost squared pi upon two, which is nothing.
So that makes sense. If I put to equal to pi upon two, then we're measuring.
Then the end direction becomes the y direction and we're measuring in a direction which is orthogonal to the Z direction.
And, and then you would think that knowing what components of the anchor mentum in the
Z direction was couldn't possibly affect the angular momentum in the Y direction.
So you would expect that there was equal probability that half the probability of passing the second filter that I say of getting plus a
half for for the spin along y that plus one plus a half on Y and minus one half on y be equally likely by the symmetry of the situation,
and indeed cost squared of pi upon four across apartment pi upon four is is one upon root to two cost cost squared of of pi upon four is a half.
And that makes perfect sense as well. So this formula predicts the kind of thing that you would expect.
Okay. Suppose we now have a we won't do this in all detail, but let's just sketch it out.
Suppose we have now another filter.
So we have f one as before. We have f two as we've just calculated.
Now suppose on the output of f two, we include f three.
Right. So this one is going to measure in the theatre direction.
I said this one, let's say this one has its axis in the five direction, also in the x y plane.
Right. So you measure, first of all, the spin on Z, then you measure on the unit vector costly to nothing sine theta, costly to sorry.
Then you measure and then and then those that return plus a half in that direction you measure in the direction.
Nothing. Sine theta. Sine phi crossfire.
Suppose we do that. So.
So the probability of passing F3, given that you passed F2 is going to be we'll call this vector N and we'll call this vector M, say, no, no, no, no.
We'll just use this notation. This will be a half on PHI.
A half on feature.
So the output from this filter definitely has particles with with plus a half component of angular momentum in the direction defined by theta.
And I want to know the amplitude that those particles have will definitely give me a plus a half if I measure in the direction defined by PHI.
The answer to that is according to the dogma of the theory.
It's that and I can expand that into here I can slide the identity operator taking the form of plus.
On Z plus. On Z plus.
Plus minus on Z minus on that.
We've slid identity operators in many times before in more complicated contexts.
So this thing that we're doing here is going to be a half by.
Sorry. That's a blunt and a half. I plus said plus a half theatre plus.
The half fy minus said.
Minus said. The half sister.
Now, these complex numbers we already know, we just calculated them.
Right. This was a which we used. This was B which we didn't use.
But we've got it written down up there. It's I's science teacher upon to. So this one here is cost.
These are on two. This one here is I sign.
Theatre over too. But we also know what this is, because this is going to be the same.
Excuse me. Excuse me. We have a complex. Let's just ask ourselves carefully exactly what is b?
B is actually a.
The complex conjugate of this. Sorry. These need complex conjugate signs.
Can we remind ourselves actually where we where we are on this?
I'm not worried about whether I'm dealing with a complex problem. Some of these need complex and complex conjugate sites.
What exactly are A and B? They were defined. Okay.
Just, just to get this right. Um, what we said was that a half on feet was equal to A plus Z plus B minus said.
That's what we said. That was the definition of A and B. So what is this?
This thing here is, uh, is plus z, a half on theta.
Yeah. So. So what I said originally was correct.
There are no stars here. Okay.
So that's just for note. All right.
Now back to this. This is the complex conjugate of this is essentially the same as that with C2 replaced by PHI.
So we know that this will be the complex conjugate of this with Theta replaced by PHI.
This is in fact real. So this is going to be cos phi over two.
Similarly this the complex conjugate of this is the same as that with theta replaced by Phi.
So I now have to write down the complex conjugative that which is minus I signed phi over to.
So that's what that comes to. So the probability that we get through F3, F3,
given that we got through F2 is going to be cost squared phi over two plus because
that minus sign and that I and the pair of eyes make a plus sign sign squared.
These are. Which is also known as cos fi over to minus three to over two.
I think if. By trick formula.
So does this make sense? It tells me that I will if a fire over two.
If fire is the same as theatre, I'm certain to get through. That's good.
If. If. If the difference in the angles is.
Is pi upon two, then I have a chance.
Sorry, we ruled. We have failed to mod square the whole thing.
That's what's gone wrong there. Maybe there's muttering about that.
So this got expanded to this and this whole thing needed a mod square and this needed a mod square.
And we were doing various calculations and that this needed a mod square and this needed a mod square.
So it just became cost square, right?
So when the angle is so what all this tells us is that which is it had to tell us we would have been worried if we hadn't discovered this,
that the probability of getting through the third filter,
given that we got through the second filter,
should depend only on the difference in the two angles and indeed should go like the difference divided by two.
Yep. Exactly. Costs five to coast to coast square fires.
Oh, gosh. Yeah. Sorry. You're completely right. Right.
So let's let's go back to this line here. This was cos pi over two, cos theta over two plus sign phi over two, sine theta over two, that's what it is.
And then we have to do a mod square of it. Yeah.
Excuse me. And we have a formula in trick, right?
Which says that this combination of.
Of, uh, cosines and sines is the, is the cosine, that what's in here is actually the costs of five on two, minus these upon two.
And then we have to square it. Sorry. Okay.
Now we can learn something. We. We can. We can. We can make a little.
Get a little. Physical result here by considering the case that theatre is equal to pi.
On to pi is equal to pi.
What does that mean? That means that n is equal to s of y.
The the axis of the second filter is equal to e sub y.
You're measuring the spin in the wider action. The axis of this one will call it m is then equal to minus.
Is it? So what's the.
So what's the probability of passing F3 given that you passed F1?
And that's the same as the probability.
What that is physically is the probability of eventually having your spin being measured to be in the minus Z direction.
Given that as you emerged from F1, you had your spin in the plus Z direction, right?
So we already had that. This. Okay. Right.
What is that? Well, it's the probability of passing the second filter, given that you pass the first times,
the probability of passing the third filter, given that you pass the second, and therefore it's equal.
This probability was a half in this.
We already discussed that in the case that.
The theatre was pi up on two, so we were measuring in a perpendicular direction to the direction associated with the first filter.
This probability came out to be a half. We felt that was natural. This probability is going to be a half as well because we've seen that it depends
on the difference of the two angles and the difference in the two angles. Here is pi upon two.
So it's times a half. So it's a quarter.
So a quarter of the particles which emerge with their spin,
quote unquote in the Z direction are found eventually to have their spin in the minus direction.
This is concrete evidence that the second filter hasn't just measured the spin of this of the particle.
It's changed the spin of the particle. It's redirected it.
So this is this is a manifestation. This result is a matter if we had.
So the. So the probability of just doing F3 given F1 a no second filter is zero.
So putting in the second filter, the intermediate filter affects the result.
And that's the re-alignment. So we should talk briefly.
Spin off is far and away the most important case.
But let's just briefly talk about spin one and make make the point that everything that we've been doing here generalises to arbitrary spin.
There's nothing we've been doing here which is really peculiar to spin a half.
So in the case of spin one, we have the upside can be written as an amount of one.
So one, one if you like, plus an amount of one.
Nothing plus an amount of one minus one.
So there are three complex numbers needed to define the orientation of the spin of a spin one particle, and, for example, a W boson or a Z boson.
Are. Particles with spin.
One Photons also have spin one, but they have certain pathologies because they have zero arrest mass.
So it's as well not to include them in this discussion.
So. So we have that.
The consequence of that is that if I have a bin operator as I working on a upside that maps to a matrix problem where we we write this as a,
B, C, D times one one plus E times one, nothing plus F, times one minus one.
Right? So this is represented by three complex numbers D, E and F.
This is represented by A, B, and C.
And there will be a matrix relation between these and we will have that D, E and F are equal to a matrix which we will make with plus.
Sorry, we will have one one. Let's leave off the total and total agreement and quantum numbers.
So let's just call it one, si1 and then we'll have one, s I nothing and then I'll have one, s I minus one and so on and so forth.
And here we will have nothing, Asi one, nothing, Asi nothing work.
So we have a three by three matrix operating on A, B, C, this is how we would concretely do our computations.
And and we need to know. So we'll have three matrices, one for S X, one for S Y and one for Z, as I said, will be the easy one to do,
as I said, will be the matrix of the eigen of its eigenvalues down the diagonals.
That'll be one. Nothing minus one and nothing everywhere else.
Which follows immediately from the fact that s said on this produce is one times this,
etc., etc., etc. and when we want to work out what we want to do for s x,
so when we want to work out 1sx,
one will replace that s x by a half as plus plus s minus s plus will kill this x minus will lower this to nothing which is orthogonal to this.
Or we'll have a nought in this slot when we when, when this,
when we put s x in here we have a half of a plus plus, minus, minus will lower this to minus one,
which is orthogonal, but plus will raise it to this and it will produce in fact root two will have
that s plus operating or nothing will turn out to be a root two times one.
So as X will be a half of nothing route to nothing, we'll get nothing in the right thing because because as plus can raise minus one to nothing.
But he can't drag it all the way up to one. And it's minus, of course, kills minus one.
So we get a matrix that looks like this.
And we will get for why a matrix that's most handily written as one of of a route to this is more easily written as one of a route to.
Of nothing. One nothing. One nothing. One nothing.
One nothing. Just taking out the factor of two.
And this one is most easily written. I mean, is handily written just the same way we derive it.
Nothing minus I, nothing minus I.
It's a permission matrix. Ah, here goes I, here goes I. Nothing, nothing, nothing.
So these are sort of these are the generalisations of the pounding matrices
for spin one problem and it's worth doing some stone girl type experiments,
the thought experiments with the spin one systems to just see what the differences are.
I did want to talk about the OC. Let's just briefly talk about this.
Let's go all the way to Spain s which is much greater than one.
Right. So in the classical regime, we want to understand how out of this can we recover the classical situation that if I hold of a piece of chalk,
it has a well-defined orientation. None of this probabilistic this thing and that thing.
And the other thing you can see where the damn thing points, right? We have to recover this out of this probabilistic apparatus.
And the way to do that is to imagine the what the spin matrices look like for spin.
And it's absolutely straightforward to construct them.
Everything we've done carries over absolutely straightforwardly.
We have that n well.
For. A Z in this case is going to be s s minus one s minus to rip down to minus s along the diagonal.
It's going to be the matrix of the eigen values of said and diagonal.
S x is going to be here.
We will have the state s. S.
X. S. Then here we will have s.
S. X. X minus one. And so on.
S. S. X. X minus two.
If you want and if you want to apply this in classical physics,
this matrix will be on the order of ten to the 30 something by ten to the 30 something, and it will be enormous.
But nearly all the numbers will vanish because, well, this number we already know is equal to zero.
This number vanishes because this you replaced by a half of x plus plus x minus x plus kills this x minus loads it something orthogonal to this.
This will be non-zero because s plus will raise that to x, which will couple to that.
And in fact this will turn out to be alpha of S minus one.
So there's a when sex works on this, we get a horrible square root, which I'm calling alpha of S minus one times S So that's what this will come to.
This will come to nothing because we'll have S plus it'll raise this twist minus one,
which is not good enough, and S minus will lower it, which is useless.
So this is equal to zero and everything else is going to be equal to zero.
So this matrix is going to consist of a line of non-zero numbers just above the diagonal.
Nothing's down the diagonal. So let me write this out.
This is going to be on the diagonal precisely nothing.
Above the diagonal. We will have alpha of S minus one, which is easily worked out.
It's the square root here. We will have alpha of s minus two.
Here we will have alpha of S minus three, s minus three,
and so on down the diagonal and just below the diagonal we have the complex conjugates of those.
These are in fact real numbers and therefore we have the same numbers and nothing's everywhere else.
So this is a very simple matrix.
It just has to non non-zero diagonals and we can work with it and we can now do things like suppose we have no time, so it's time to stop.
Oh, sorry, but I think it probably is worth just doing this and I'll finish it off tomorrow.
Yep.