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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.

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So any state, as regards its spin, its orientation, should be expandable as a linear combination of the state plus,

3
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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.

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There will be these coefficients here and a complex number here and a complex number there.

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The amplitude to measure plus a half or less said or the amplitude to measure minus a half.

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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

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we want to be able to do is write the result of using some spin operator on this arbitrary state of sci fi.

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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,

9
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a half particle spin off system.

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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.

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If on the right, we put in the two numbers that characterise up sy on the right we get out on the left,

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the two numbers that characterise PHI after we've multiplied by this matrix of four complex numbers being the expectation value of the,

13
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of the relevant of whatever operator we're trying to use between the plus states,

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the minus states and then these non-classical of diagonal bits on each side.

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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,

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then this matrix is very simple because asset on plus is simply a half of plus.

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So we get a half appearing here, we get minus a half appearing here because our set of minus is minus one half times

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minus and we get nothing appearing here and here because plus and minus are orthogonal.

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So we have this diagonal matrix, which is no accident.

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It is simply the matrix that contains the eigenvalues of Z down its diagonal.

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Because we use this basis vectors the I can cancel offset.

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We made that choice and the result is that s the matrix representing said is diagonal with its eigenvalues two on the diagonal.

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And this matrix is conventionally written as a half times this matrix which is called Sigma

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Z and it's called apparently Matrix because Wolfgang Pauli introduced it into physics,

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although it was known to mathematicians matrices like this.

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Okay. So more interesting is if we ask ourselves, what is the matrix for X?

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So the matrix for my ex is going to involve things.

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Well, we're going to have, for example, plus s, x plus.

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This is a complex number. We want to know which complex number and the secret is to radius of calculating

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this is to right s x is a half of s plus plus minus where s plus minus.

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So the matrices, sorry, are the operators that we already introduced in the context of J and L to to

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reorient the angular momentum either towards the Z axis or away from the Z axis.

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So as these are X plus and minus I times s y, right?

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So this, this operator was, was introduced in the form of j plus minus.

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But remember, spin and total angle momentum have the same commutation relations the same the same behaviour in every way.

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So this letter, this letter operator is this.

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And obviously if you add X plus two S minus, you get to X because the Y terms cancel.

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So this is definitely the case. So this, this thing here can be written as a half of plus s plus plus plus plus.

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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.

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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.

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The total spin is only a half, so it kills it in the process.

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Therefore this one is zero. S minus successfully lowers this to minus, but minus is orthogonal to plus.

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So this is zero. So this element here is zero and that's the top left corner of the matrix for x is zero.

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Similarly,

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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.

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So if we look at plus s x minus, we're looking at a.

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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.

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So this number here is equal to one and minus tries to lower this and kills it in the process.

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And therefore, this is equal to zero. So this element, this off diagonal element, is in fact equal to a half.

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We know that the bottom right hand element is the complex conjugate of the top right hand element, because this is a mission operator.

51
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So we know now that the matrix is s x is represented by the Matrix, half of nothing,

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one one nothing, also known as a half of Sigma X, the Pauli Matrix.

53
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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?

54
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This is a half of plus one over two I of s plus minus this minus.

55
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Right. Because if you take the difference of X plus, i, y and s x minus I y, you will end up with 2isy.

56
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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.

57
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So therefore this is equal to one over two.

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I also known as a half minus a half minus I over two.

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So the matrix representing s y is going to be it's going to be a half.

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Of one minus I. I saw.

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I want to have one. Nothing. Nothing.

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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.

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So that's where the pounding matrices come from. They're simply the matrix representations of the spin operators in a basis.

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In the when you choose as your basis the eigenvectors the eigen kits of Sigma Z.

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So let's use these. Use this apparatus to.

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To do something slightly interesting.

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It's it's an excellent exercise both in in in in practising getting experimental predictions out of this abstract apparatus.

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And also we learn something interesting about how how the orientation of atomic scale things behave the somewhat counterintuitive arrangements.

69
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Okay. I don't think this computer is going to this system, projection system is going to work today for some reason.

70
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So. Okay.

71
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So the point is that, uh, so the point is that a spinning charged body.

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Is a magnetic is a magnetic dipole.

73
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I think that's kind of plausible. So that.

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So electrons, neutrons, protons, except what's right.

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Neutrons, electrons, protons, being spinning.

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Charge bodies have little magnetic moments. They are little magnets.

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So if you put a magnet in a beam field, you have this is the energy.

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Of a magnetic. Dipole in a mag field.

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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.

80
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Right. So when this dot product is positive, the energy is lowest.

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So that's why magnets, compass, needles and whatever align with the magnetic field.

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That also means that if a magnetic dipole is aligned with the field,

83
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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,

84
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whereas if it's anti aligned with a magnetic field,

85
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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,

86
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we have a more positive energy. So since things tend to move in the direction that minimises that potential energy we have,

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that magnets aligned with B will be sucked into a region of stronger B, so a magnet, a dipole.

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Aligned. With B so that means that mu dot be greater than nought is sucked.

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Into a field. If the field strength varies spatially, which it too often does,

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the what particles which have the fields dipoles aligned will be will be sucked into b and similarly the other ones will be repelled.

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So the anti aligned loops aligned.

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Dipoles will be repelled from a region of high P.

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So that was the physics that Stern and Gerlach exploited, exploited in 1922, in experiments which astounded the world.

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They found themselves. They made themselves a magnet. Or should we call this north and we'll call this south?

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They made themselves a magnet which had puzzle pieces, one of which was pointy and the other of which was flat or even.

96
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Well, I think it was flat, but it could also be concave like this.

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And then you can imagine how the field lines run. The field lines run like this somehow.

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I'm not doing a very good job of it. My diagrams are usually rather rubbish.

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The point is that here we have a crowding of field lines, which means we have high B.

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Near knife edge.

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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.

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So this is like the point of a knife, right? We're looking end on the point of a knife.

103
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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.

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So that they're heading for this, well, they're heading a bit below this region of high magnetic field like this.

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Then the ones that have their spin aligned this way into B are going to be sucked into the region,

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drawn, attracted by the region of high B near the point of the knife and move on up here.

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So this is the particles which have moved on b greater than nought and particles

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with and anti aligned with mew dot be less than nought will come down here.

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Of course, this is all grotesquely exaggerated.

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In fact, you'll have very you'll have a very subtle curvature and then you'll have a straight line in front.

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Right. So we we get we get the particles deflected either way.

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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.

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Then what they found, which surprised them and everybody else, that half of their particles,

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half the silver atoms went off this way and half of this little breath was went off that way.

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So when they they detected the atoms on a screen over here, they got two blobs distinctly separated.

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The quantum mechanical interpretation of this is that is these atoms.

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When the atoms are in here, they are.

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Sorry I haven't said that. Mu. The magnetic moment is equal to some number.

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The gyro magnetic ratio times the spin operator.

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So when they're in here, they're.

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The magnetic field is, as it were, measuring the component of of spin in the direction of the magnetic field.

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That's what you you say to yourself. So and there are only two answers possible.

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Either you'll get plus a half or you'll get minus a half for the for the value of this.

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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.

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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.

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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.

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You've made the silver vent, you've made you heat it up.

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Some silver in an oven, made some silver vapour, allowed it to diffuse out of some holes, culminating slits and that kind of stuff.

129
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So it's coming along here with some thermal velocities and out out of your filter.

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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.

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So it's a machine for for it's a practical device for creating silver atoms which are in this state.

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Now you can play some entertaining games by installing another Stern Gerlach filter.

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So let's just block these off something being a nuisance stick in another stern girl like filter here and now.

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Let's measure the. Let's measure up rn.

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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.

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We're going to have this to be the Z direction and what the Y direction will have to be out of the board.

137
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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.

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So n is going to be a vector which if thi is nothing, is just in the z direction.

139
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And if theatre is pi by two, it's in the y direction and it can be allowed to scan between these directions.

140
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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.

141
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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.

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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.

143
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So the probability that you pass F2 given that you passed.

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F one in quantum mechanical language is is plus a half on.

145
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Given that your well let's just we'll just say plus on n given that you were plus on Z.

146
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So this is the state that you're in. Up there is just called.

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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.

148
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This is nyan cat of s of n with eigenvalue a half.

149
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And this this pair of things makes me the amplitude for by the basic dogma of the subject for the probability of this outcome.

150
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So I need to mod square this and I've got the probability that I want.

151
00:19:38,210 --> 00:19:39,490
So we can work this out.

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00:19:39,500 --> 00:19:51,020
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.

153
00:19:51,020 --> 00:19:56,000
Right. Because so if we get this number and this number,

154
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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.

155
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So right to get out of this catch, you could get the bra you want up there by complex conjugate,

156
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get you to have a nice star bang in with plus on Z and you pick out a star.

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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.

158
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How to find A and B? Well, what's the point about? What's the point?

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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.

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And.

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It's totally characteristic of these sorts of calculations of a wide range of quantum mechanical calculations that this the sequence of arguments.

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I want a certain complex number. It will involve some cat.

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Ask yourself, what is the defining characteristic of the cat? It will usually be because it is not in case of some operator.

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Now we have a well-defined mathematical problem.

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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

166
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z sort of a dot product between the unit vector and and the vector made up of the three pally matrices.

167
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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.

168
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So at the end of the day, it is a half of.

169
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Now Sigma Z, we've got up there, it's got one in the top left hand corner and minus one in the bottom.

170
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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.

171
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So we get a minus sign feature appearing there and it's complex conjugate has to appear down here.

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So this is the matrix that represents s n where theta is defining the direction of.

173
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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

174
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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,

175
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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.

176
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So I'm looking for the eigen catch of this operator with eigenvalue one notice I don't

177
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waste my time finding out what the eigenvalues of this operator are of this matrix are.

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I know that the because this is a is a is a is a matrix that represents a spin operator.

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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.

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So we don't waste time finding out what the eigenvalues are. We just get on and solve these equations.

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What? There are two equations here, but because we're looking at an eigenvalue problem, only one of them.

182
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These two equations are linearly dependent upon one another.

183
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Only one of them contains useful information.

184
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The other one repeats that information. So we may need to look at the top equation and it says that a a minus one.

185
00:23:48,510 --> 00:23:53,610
Sorry, sorry, sorry. Eight times brackets, one minus cost theta.

186
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So if I'm going to get a costly to equals a on the right hand side.

187
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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.

188
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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.

189
00:24:18,960 --> 00:24:27,390
The absolute values have to be determined from a normalisation condition or are equal to.

190
00:24:33,090 --> 00:24:37,860
V override is equal to one minus cost to over.

191
00:24:38,890 --> 00:24:42,300
Minus I. Sine theta.

192
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And we can clean this up a bit if we use some half angle formulae,

193
00:24:47,820 --> 00:24:58,140
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.

194
00:25:04,700 --> 00:25:06,649
So we can cancel a number of things.

195
00:25:06,650 --> 00:25:18,590
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.

196
00:25:20,170 --> 00:25:28,070
So I can right now that a B is equal to costs.

197
00:25:28,090 --> 00:25:34,090
These are over to I signed the teacher over to.

198
00:25:35,110 --> 00:25:42,070
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.

199
00:25:43,060 --> 00:25:46,660
And moreover, this thing is correctly normalised. It just happens.

200
00:25:46,990 --> 00:25:52,420
So in principle, I would now need to deal with the normalisation. I've only been calculating the ratio of the components.

201
00:25:52,870 --> 00:25:59,290
I want more squared plus plus more B squared to come to one, but it jolly well does by good fortune.

202
00:25:59,290 --> 00:26:05,349
Right? So this is the, this is the complete bottom line. This gives you OC, right?

203
00:26:05,350 --> 00:26:09,790
So, so the probability that we pass.

204
00:26:11,940 --> 00:26:25,890
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.

205
00:26:29,510 --> 00:26:40,280
Does that make sense? If theatre is equal to nothing, then the second filter is also measuring the Z component of angular momentum.

206
00:26:41,000 --> 00:26:50,719
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.

207
00:26:50,720 --> 00:26:53,780
So this probability must be one and indeed col squared of nothing is one.

208
00:26:54,440 --> 00:27:03,890
If the theatre is pi then then the second one is plus a half.

209
00:27:04,120 --> 00:27:11,960
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.

210
00:27:13,070 --> 00:27:20,000
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.

211
00:27:20,690 --> 00:27:24,860
If I put features equal to pi, I'm going at cost squared pi upon two, which is nothing.

212
00:27:25,160 --> 00:27:32,390
So that makes sense. If I put to equal to pi upon two, then we're measuring.

213
00:27:32,930 --> 00:27:42,110
Then the end direction becomes the y direction and we're measuring in a direction which is orthogonal to the Z direction.

214
00:27:42,470 --> 00:27:48,080
And, and then you would think that knowing what components of the anchor mentum in the

215
00:27:48,080 --> 00:27:52,880
Z direction was couldn't possibly affect the angular momentum in the Y direction.

216
00:27:53,210 --> 00:28:05,480
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

217
00:28:05,480 --> 00:28:13,130
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,

218
00:28:13,400 --> 00:28:24,170
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.

219
00:28:24,170 --> 00:28:29,930
And that makes perfect sense as well. So this formula predicts the kind of thing that you would expect.

220
00:28:38,200 --> 00:28:43,330
Okay. Suppose we now have a we won't do this in all detail, but let's just sketch it out.

221
00:28:43,840 --> 00:28:48,909
Suppose we have now another filter.

222
00:28:48,910 --> 00:28:53,440
So we have f one as before. We have f two as we've just calculated.

223
00:28:53,440 --> 00:28:57,280
Now suppose on the output of f two, we include f three.

224
00:28:59,100 --> 00:29:03,940
Right. So this one is going to measure in the theatre direction.

225
00:29:04,210 --> 00:29:10,770
I said this one, let's say this one has its axis in the five direction, also in the x y plane.

226
00:29:10,780 --> 00:29:21,190
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.

227
00:29:21,220 --> 00:29:28,650
Then you measure and then and then those that return plus a half in that direction you measure in the direction.

228
00:29:29,200 --> 00:29:32,890
Nothing. Sine theta. Sine phi crossfire.

229
00:29:33,730 --> 00:29:37,120
Suppose we do that. So.

230
00:29:45,840 --> 00:30:00,780
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.

231
00:30:00,790 --> 00:30:09,090
We'll just use this notation. This will be a half on PHI.

232
00:30:10,580 --> 00:30:12,380
A half on feature.

233
00:30:13,860 --> 00:30:23,610
So the output from this filter definitely has particles with with plus a half component of angular momentum in the direction defined by theta.

234
00:30:23,880 --> 00:30:36,810
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.

235
00:30:37,290 --> 00:30:40,170
The answer to that is according to the dogma of the theory.

236
00:30:40,410 --> 00:30:49,350
It's that and I can expand that into here I can slide the identity operator taking the form of plus.

237
00:30:49,620 --> 00:30:53,820
On Z plus. On Z plus.

238
00:30:55,450 --> 00:30:59,049
Plus minus on Z minus on that.

239
00:30:59,050 --> 00:31:04,930
We've slid identity operators in many times before in more complicated contexts.

240
00:31:05,380 --> 00:31:10,830
So this thing that we're doing here is going to be a half by.

241
00:31:12,620 --> 00:31:23,570
Sorry. That's a blunt and a half. I plus said plus a half theatre plus.

242
00:31:26,690 --> 00:31:29,990
The half fy minus said.

243
00:31:31,280 --> 00:31:34,740
Minus said. The half sister.

244
00:31:36,480 --> 00:31:39,840
Now, these complex numbers we already know, we just calculated them.

245
00:31:39,840 --> 00:31:44,670
Right. This was a which we used. This was B which we didn't use.

246
00:31:44,680 --> 00:31:49,580
But we've got it written down up there. It's I's science teacher upon to. So this one here is cost.

247
00:31:49,590 --> 00:31:54,300
These are on two. This one here is I sign.

248
00:31:56,280 --> 00:32:03,810
Theatre over too. But we also know what this is, because this is going to be the same.

249
00:32:07,810 --> 00:32:13,000
Excuse me. Excuse me. We have a complex. Let's just ask ourselves carefully exactly what is b?

250
00:32:13,450 --> 00:32:17,130
B is actually a.

251
00:32:18,300 --> 00:32:23,370
The complex conjugate of this. Sorry. These need complex conjugate signs.

252
00:32:35,240 --> 00:32:38,990
Can we remind ourselves actually where we where we are on this?

253
00:32:39,860 --> 00:32:45,100
I'm not worried about whether I'm dealing with a complex problem. Some of these need complex and complex conjugate sites.

254
00:32:45,110 --> 00:32:52,990
What exactly are A and B? They were defined. Okay.

255
00:32:53,000 --> 00:33:04,100
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.

256
00:33:04,400 --> 00:33:08,720
That's what we said. That was the definition of A and B. So what is this?

257
00:33:08,750 --> 00:33:15,770
This thing here is, uh, is plus z, a half on theta.

258
00:33:22,220 --> 00:33:28,270
Yeah. So. So what I said originally was correct.

259
00:33:28,270 --> 00:33:31,820
There are no stars here. Okay.

260
00:33:32,120 --> 00:33:37,460
So that's just for note. All right.

261
00:33:37,490 --> 00:33:46,040
Now back to this. This is the complex conjugate of this is essentially the same as that with C2 replaced by PHI.

262
00:33:47,060 --> 00:33:53,300
So we know that this will be the complex conjugate of this with Theta replaced by PHI.

263
00:33:53,330 --> 00:33:57,470
This is in fact real. So this is going to be cos phi over two.

264
00:33:58,700 --> 00:34:05,870
Similarly this the complex conjugate of this is the same as that with theta replaced by Phi.

265
00:34:05,990 --> 00:34:13,790
So I now have to write down the complex conjugative that which is minus I signed phi over to.

266
00:34:19,000 --> 00:34:25,000
So that's what that comes to. So the probability that we get through F3, F3,

267
00:34:25,000 --> 00:34:35,350
given that we got through F2 is going to be cost squared phi over two plus because

268
00:34:35,350 --> 00:34:39,970
that minus sign and that I and the pair of eyes make a plus sign sign squared.

269
00:34:39,980 --> 00:34:59,030
These are. Which is also known as cos fi over to minus three to over two.

270
00:35:02,070 --> 00:35:06,350
I think if. By trick formula.

271
00:35:09,440 --> 00:35:16,790
So does this make sense? It tells me that I will if a fire over two.

272
00:35:16,790 --> 00:35:20,150
If fire is the same as theatre, I'm certain to get through. That's good.

273
00:35:20,540 --> 00:35:25,370
If. If. If the difference in the angles is.

274
00:35:26,030 --> 00:35:29,600
Is pi upon two, then I have a chance.

275
00:35:30,380 --> 00:35:35,870
Sorry, we ruled. We have failed to mod square the whole thing.

276
00:35:35,870 --> 00:35:37,910
That's what's gone wrong there. Maybe there's muttering about that.

277
00:35:38,180 --> 00:35:44,900
So this got expanded to this and this whole thing needed a mod square and this needed a mod square.

278
00:35:46,400 --> 00:35:53,180
And we were doing various calculations and that this needed a mod square and this needed a mod square.

279
00:35:53,180 --> 00:35:54,589
So it just became cost square, right?

280
00:35:54,590 --> 00:36:04,220
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,

281
00:36:04,580 --> 00:36:07,760
that the probability of getting through the third filter,

282
00:36:08,180 --> 00:36:09,890
given that we got through the second filter,

283
00:36:10,100 --> 00:36:15,050
should depend only on the difference in the two angles and indeed should go like the difference divided by two.

284
00:36:16,490 --> 00:36:20,870
Yep. Exactly. Costs five to coast to coast square fires.

285
00:36:23,520 --> 00:36:28,380
Oh, gosh. Yeah. Sorry. You're completely right. Right.

286
00:36:31,530 --> 00:36:51,540
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.

287
00:36:51,540 --> 00:36:55,290
And then we have to do a mod square of it. Yeah.

288
00:36:55,320 --> 00:36:58,820
Excuse me. And we have a formula in trick, right?

289
00:36:58,830 --> 00:37:01,200
Which says that this combination of.

290
00:37:01,230 --> 00:37:11,190
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.

291
00:37:12,090 --> 00:37:20,140
And then we have to square it. Sorry. Okay.

292
00:37:20,210 --> 00:37:23,690
Now we can learn something. We. We can. We can. We can make a little.

293
00:37:24,440 --> 00:37:34,519
Get a little. Physical result here by considering the case that theatre is equal to pi.

294
00:37:34,520 --> 00:37:38,330
On to pi is equal to pi.

295
00:37:38,360 --> 00:37:44,200
What does that mean? That means that n is equal to s of y.

296
00:37:45,380 --> 00:37:49,970
The the axis of the second filter is equal to e sub y.

297
00:37:49,970 --> 00:37:57,890
You're measuring the spin in the wider action. The axis of this one will call it m is then equal to minus.

298
00:37:58,280 --> 00:38:05,420
Is it? So what's the.

299
00:38:06,110 --> 00:38:12,950
So what's the probability of passing F3 given that you passed F1?

300
00:38:14,750 --> 00:38:16,879
And that's the same as the probability.

301
00:38:16,880 --> 00:38:23,870
What that is physically is the probability of eventually having your spin being measured to be in the minus Z direction.

302
00:38:24,140 --> 00:38:29,540
Given that as you emerged from F1, you had your spin in the plus Z direction, right?

303
00:38:32,200 --> 00:38:35,410
So we already had that. This. Okay. Right.

304
00:38:36,730 --> 00:38:44,590
What is that? Well, it's the probability of passing the second filter, given that you pass the first times,

305
00:38:44,590 --> 00:38:51,879
the probability of passing the third filter, given that you pass the second, and therefore it's equal.

306
00:38:51,880 --> 00:38:54,940
This probability was a half in this.

307
00:38:54,970 --> 00:38:57,280
We already discussed that in the case that.

308
00:38:58,540 --> 00:39:04,930
The theatre was pi up on two, so we were measuring in a perpendicular direction to the direction associated with the first filter.

309
00:39:05,290 --> 00:39:11,979
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

310
00:39:11,980 --> 00:39:16,540
on the difference of the two angles and the difference in the two angles. Here is pi upon two.

311
00:39:16,870 --> 00:39:20,590
So it's times a half. So it's a quarter.

312
00:39:21,730 --> 00:39:26,560
So a quarter of the particles which emerge with their spin,

313
00:39:26,920 --> 00:39:32,000
quote unquote in the Z direction are found eventually to have their spin in the minus direction.

314
00:39:32,170 --> 00:39:37,840
This is concrete evidence that the second filter hasn't just measured the spin of this of the particle.

315
00:39:37,990 --> 00:39:45,030
It's changed the spin of the particle. It's redirected it.

316
00:39:45,330 --> 00:39:49,020
So this is this is a manifestation. This result is a matter if we had.

317
00:39:49,350 --> 00:39:59,580
So the. So the probability of just doing F3 given F1 a no second filter is zero.

318
00:40:00,420 --> 00:40:05,640
So putting in the second filter, the intermediate filter affects the result.

319
00:40:06,600 --> 00:40:13,050
And that's the re-alignment. So we should talk briefly.

320
00:40:14,880 --> 00:40:17,520
Spin off is far and away the most important case.

321
00:40:18,060 --> 00:40:26,790
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.

322
00:40:26,970 --> 00:40:30,450
There's nothing we've been doing here which is really peculiar to spin a half.

323
00:40:32,220 --> 00:40:39,240
So in the case of spin one, we have the upside can be written as an amount of one.

324
00:40:39,900 --> 00:40:43,650
So one, one if you like, plus an amount of one.

325
00:40:43,650 --> 00:40:47,490
Nothing plus an amount of one minus one.

326
00:40:50,030 --> 00:41:03,800
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.

327
00:41:05,140 --> 00:41:08,510
Are. Particles with spin.

328
00:41:08,510 --> 00:41:13,310
One Photons also have spin one, but they have certain pathologies because they have zero arrest mass.

329
00:41:13,670 --> 00:41:16,790
So it's as well not to include them in this discussion.

330
00:41:21,910 --> 00:41:23,590
So. So we have that.

331
00:41:24,010 --> 00:41:38,229
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,

332
00:41:38,230 --> 00:41:48,910
B, C, D times one one plus E times one, nothing plus F, times one minus one.

333
00:41:49,090 --> 00:41:53,800
Right? So this is represented by three complex numbers D, E and F.

334
00:41:54,040 --> 00:41:56,650
This is represented by A, B, and C.

335
00:41:56,890 --> 00:42:09,670
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.

336
00:42:10,300 --> 00:42:16,870
Sorry, we will have one one. Let's leave off the total and total agreement and quantum numbers.

337
00:42:16,870 --> 00:42:32,919
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.

338
00:42:32,920 --> 00:42:42,459
And here we will have nothing, Asi one, nothing, Asi nothing work.

339
00:42:42,460 --> 00:42:48,280
So we have a three by three matrix operating on A, B, C, this is how we would concretely do our computations.

340
00:42:51,820 --> 00:43:00,580
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,

341
00:43:00,850 --> 00:43:06,130
as I said, will be the matrix of the eigen of its eigenvalues down the diagonals.

342
00:43:06,140 --> 00:43:10,270
That'll be one. Nothing minus one and nothing everywhere else.

343
00:43:14,500 --> 00:43:18,909
Which follows immediately from the fact that s said on this produce is one times this,

344
00:43:18,910 --> 00:43:25,930
etc., etc., etc. and when we want to work out what we want to do for s x,

345
00:43:26,710 --> 00:43:29,680
so when we want to work out 1sx,

346
00:43:29,680 --> 00:43:40,209
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.

347
00:43:40,210 --> 00:43:43,780
Or we'll have a nought in this slot when we when, when this,

348
00:43:44,020 --> 00:43:52,209
when we put s x in here we have a half of a plus plus, minus, minus will lower this to minus one,

349
00:43:52,210 --> 00:43:57,880
which is orthogonal, but plus will raise it to this and it will produce in fact root two will have

350
00:43:57,880 --> 00:44:03,220
that s plus operating or nothing will turn out to be a root two times one.

351
00:44:06,010 --> 00:44:19,749
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.

352
00:44:19,750 --> 00:44:24,130
But he can't drag it all the way up to one. And it's minus, of course, kills minus one.

353
00:44:24,760 --> 00:44:27,040
So we get a matrix that looks like this.

354
00:44:35,560 --> 00:44:49,070
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.

355
00:44:49,720 --> 00:44:53,140
Of nothing. One nothing. One nothing. One nothing.

356
00:44:53,140 --> 00:44:57,190
One nothing. Just taking out the factor of two.

357
00:44:57,760 --> 00:45:02,649
And this one is most easily written. I mean, is handily written just the same way we derive it.

358
00:45:02,650 --> 00:45:06,910
Nothing minus I, nothing minus I.

359
00:45:07,180 --> 00:45:11,200
It's a permission matrix. Ah, here goes I, here goes I. Nothing, nothing, nothing.

360
00:45:11,920 --> 00:45:15,520
So these are sort of these are the generalisations of the pounding matrices

361
00:45:15,520 --> 00:45:22,510
for spin one problem and it's worth doing some stone girl type experiments,

362
00:45:23,230 --> 00:45:28,150
the thought experiments with the spin one systems to just see what the differences are.

363
00:45:36,280 --> 00:45:42,340
I did want to talk about the OC. Let's just briefly talk about this.

364
00:45:43,210 --> 00:45:49,060
Let's go all the way to Spain s which is much greater than one.

365
00:45:49,100 --> 00:45:57,729
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,

366
00:45:57,730 --> 00:46:01,570
it has a well-defined orientation. None of this probabilistic this thing and that thing.

367
00:46:01,570 --> 00:46:08,469
And the other thing you can see where the damn thing points, right? We have to recover this out of this probabilistic apparatus.

368
00:46:08,470 --> 00:46:16,150
And the way to do that is to imagine the what the spin matrices look like for spin.

369
00:46:16,150 --> 00:46:19,450
And it's absolutely straightforward to construct them.

370
00:46:20,650 --> 00:46:25,900
Everything we've done carries over absolutely straightforwardly.

371
00:46:26,080 --> 00:46:29,800
We have that n well.

372
00:46:32,170 --> 00:46:42,690
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.

373
00:46:44,340 --> 00:46:49,800
It's going to be the matrix of the eigen values of said and diagonal.

374
00:46:52,230 --> 00:46:56,639
S x is going to be here.

375
00:46:56,640 --> 00:47:00,300
We will have the state s. S.

376
00:47:00,540 --> 00:47:04,260
X. S. Then here we will have s.

377
00:47:04,590 --> 00:47:08,160
S. X. X minus one. And so on.

378
00:47:08,190 --> 00:47:11,610
S. S. X. X minus two.

379
00:47:11,610 --> 00:47:14,130
If you want and if you want to apply this in classical physics,

380
00:47:14,400 --> 00:47:20,670
this matrix will be on the order of ten to the 30 something by ten to the 30 something, and it will be enormous.

381
00:47:21,780 --> 00:47:26,670
But nearly all the numbers will vanish because, well, this number we already know is equal to zero.

382
00:47:27,120 --> 00:47:35,940
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.

383
00:47:36,360 --> 00:47:42,780
This will be non-zero because s plus will raise that to x, which will couple to that.

384
00:47:43,020 --> 00:47:48,570
And in fact this will turn out to be alpha of S minus one.

385
00:47:49,680 --> 00:47:58,290
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.

386
00:47:58,620 --> 00:48:02,609
This will come to nothing because we'll have S plus it'll raise this twist minus one,

387
00:48:02,610 --> 00:48:05,970
which is not good enough, and S minus will lower it, which is useless.

388
00:48:06,210 --> 00:48:08,820
So this is equal to zero and everything else is going to be equal to zero.

389
00:48:08,820 --> 00:48:14,760
So this matrix is going to consist of a line of non-zero numbers just above the diagonal.

390
00:48:15,150 --> 00:48:18,720
Nothing's down the diagonal. So let me write this out.

391
00:48:19,080 --> 00:48:22,530
This is going to be on the diagonal precisely nothing.

392
00:48:24,120 --> 00:48:28,859
Above the diagonal. We will have alpha of S minus one, which is easily worked out.

393
00:48:28,860 --> 00:48:32,700
It's the square root here. We will have alpha of s minus two.

394
00:48:33,000 --> 00:48:36,510
Here we will have alpha of S minus three, s minus three,

395
00:48:36,510 --> 00:48:42,600
and so on down the diagonal and just below the diagonal we have the complex conjugates of those.

396
00:48:42,600 --> 00:48:49,559
These are in fact real numbers and therefore we have the same numbers and nothing's everywhere else.

397
00:48:49,560 --> 00:48:51,360
So this is a very simple matrix.

398
00:48:51,360 --> 00:49:05,370
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.

399
00:49:05,910 --> 00:49:12,270
Oh, sorry, but I think it probably is worth just doing this and I'll finish it off tomorrow.

400
00:49:13,620 --> 00:49:13,890
Yep.