 Okay, glad you all found the room. I wanted to begin today by discussing these terms, the two of them here. The first one I was telling you about from the end of the hour last time, it says that if you have two observables, I'll remind you that an observable is a complete remission operator. If you have two observables, A and B, they possess a simultaneous eigenbasis, if and only if they commute. Simultaneous eigenbasis means, first of all, a basis. Second of all, the basis vectors are eigenvectors. And third of all, the eigenvectors are both A and D at the same time. I wanted to prove to you this, the first part of this theorem, that if they indeed commute and they possess a simultaneous eigenbasis at the end of the hour last time, the controversy is that if they possess a simultaneous eigenbasis, then they commute. And that's fairly easy to discuss all your notes. So I won't go over the class. I'll just remark that this is quite an effective quantum mechanics, as we'll see as we go on. By the way, in the matrix language, the equivalent of this first theorem is that if you have two remission matrices, then they can be diagonalized by the same military transformation if and only if they commute. In that case, the military transformation takes you from some original basis you start with into a final one of which both matrices are diagonal. And that final basis is one whose existence is guaranteed by this theorem is the simultaneous eigenbasis. In that case, the eigenvectors are the eigenvectors of the vectors of the eigenvectors that go into matrices. Now there's a special case of this theorem, which is part B here, which is really a trivial extension of it, but it happens so often and it's so useful, and of course, that I want to make a particular mention of it. It says that if you have two observables A and B which commute, then any non-degenerate eigenstate of A is also an eigenstate of B. I'll just outline the proof of this for you. Let's suppose we've got A, and let's say that U is an eigenstate of A. I'm really thinking of a non-degenerate case here. Excuse me, excuse me. I'm thinking of the discrete case here. So this is the discrete and discrete spectrum. So the view is actually a vector of a, a normalizing a vector that belongs to open space. Now let's suppose that this is non-degenerate. Let's suppose that, let's suppose that also, we have a number of observable A which commutes with B. Then the statement is that if I take B and let it act on U, I get it, I get that, I get something that's proportional to U. That's to say that U, the same vector U, is also an eigenvector of B. To prove this this way, let's let psi be equal to, just to find it to be acting on U. And if we do this and let's let A act on psi, this is the same thing as A and B acting on U, but that's the same thing as B A acting on U which has A and B commutes. Now the A is adjacent to U, so it brings out the eigenvalue A, more case A, but more case A is just a number, so we can move it left to the B. And the result is this becomes A multiplying B times U, which is the same thing as A multiplying psi. And the result of this is that psi is also an eigenvector of A with the same eigenvalue lower case A as the original state U that we started with. Now what does that mean that U is an eigenvector of B? Well psi is equal to BU. The reason is that the reason is that we're assuming this is a non-degenerate eigenvalue, so the eigenspace corresponding to this eigenvalue lower case A is one-dimensional. And therefore any two vectors that lie in that space, which is to say any two eigenvectors of A with that eigenvalue have to be proportional to one another. So B acting on U has to be proportional to U. Let's call the proportionality factor more case B, and thus you see that it's also an eigenstate of B. Notice this argument doesn't tell you what the eigenvalue B is, it just tells you that it can't exist, that this is an eigenstate of B. To give you an example of this, let's suppose we have a Hamiltonian which commutes with parities. This is a typical example of this bear which occurs in quantum mechanics. Such Hamiltonians are very common in practice. Now let's suppose I've got an A-actor that has some state in it like this, and this is the eigenvalue B, which is not a degenerate. And the statement is that B is also an eigenstate of parity. The eigenvalue is a parity of plus or minus one, so it's got to be one of the two choices plus or minus one. As an example of this, you might think of the ordinary harmonic oscillator. This eigenstate is a non-degenerate because it's one dimensional, and it also commutes with parity. And according to this theorem, all of the harmonic oscillator eigenstates are also eigenstates of parity. The fact that you know the even ones are even, and the odd ones are odd. Now that fact doesn't follow to this theorem here. You'd have to work with the harmonic oscillator to figure out which ones are even or odd. But the fact that they're eigenstates of parity does follow to this. All right, so those are some simple theorems. Now let's turn to some projection operators. Projection operator, this is the definition, projection operator, or it's also called a projector, is first of all an observable, it's a computer emission oscillator, and such that its square is equal to itself, at least what it would need. An operator that satisfies this property is said to be item potent. It just means that if you raise it to a higher power, you'll always get the same thing as the first power. The way to visualize a projection operator is to think about the floor here like this, to think about light coming down from the sun and straight overhead. And the projection is equivalent to the shadow of the vector. So if I have a vertical vector like this, it rejects to a point, gets annihilated, and just turns into zero. But if you have a horizontal vector, its shadow is equal to itself. So in this case, the projector applied to a vertical vector as zero, and the projector applied to a horizontal vector as the horizontal vector itself, like this. And you see the vertical vector as an eigenvector of p with eigenvalue zero, because I can multiply by p here if I want, and it's zero. And the horizontal vector is said to have an eigenvalue of one. In fact, it's easy to show that the eigenvalues of projectors are always zero or one, because we'll hear capital P as the operator, but if I let lowercase p be the eigenvalue, then the eigenvalues have to satisfy the same algebraic equation as the operators. So p squared equals p, and this implies that p is either equal to zero or one. There's only two eigenvalues, and it's obviously a discrete spectrum. And so, in the case of a projection operator, there are only two eigenvalues, and therefore only two eigenspaces. And this means that the moment space decomposes into the direct sum of the zero space and the one space. When this shadow image here, the zero, the one space is the horizontal space, and the zero space is the vertical space, the one space are the ones where the eigenvalue is one, so the vector, if nothing happens to the vector, if you have to be on it, then the zero space is the one taken by a vector. Anyway, this is all this holds for projection operators. Now, projection operators occur frequently. They occur always, as a matter of fact, in resolutions of the identity. Let's suppose we have an observable A, and for simplicity, let's suppose it has a discrete spectrum. So let's call this eigenstates n and r here, eigenvalues a, n, nr like this, where r is the index which is used to resolve the generalities that's to say, it specifies the basis of the eigenspace in the end of the eigenvalue a, so then, like this. So r here runs from one up to the convention of the even eigenspace which will probably be n like this, the generality index. Then we know there's a resolute flow, and then the fourth normality relations, by the way, are the nr-squared part of n, r prime, or delta n, n prime, delta r prime, as I mentioned last time, that's all. Now, the resolution of the identity in this case is that the identity operators, the sum, are n and r, and the other part of n are nr. It's something like this, whenever you have a discrete spectrum. What we can do here is to define an operator that's actually a projection operator pn, which is the sum on r and the other part of n are nr. Here we're summing only on the degeneracy index but not also on n, so this depends on n. Now, this is the projection operator on to the eigenspace dn. Projects on to, projects on to, this projects on to the eigenspace dn. How do I know that? I know it because if I let this operator act on any vector that's orthogonal to the eigenspace dn, I get 0 because all these vectors seem to lie in that space and the various eigenspaces are orthogonal to one another. Likewise, if I allow this to act on a vector which is in this space, I just read through this vector because this is a resolution of the identity inside the eigenspace. In fact, if we take this pn and just restrict it to the eigenspace dn, it becomes the identity operator in that space. That's really what a projection operator is. In any case, that's the projector on to dn. And the result is that this resolution of the identity here by separating it in the r-sum is going to be right this way. It's still in the r-sum of r. The resolution of the identity becomes this. In one, this is the sum of the eigenvalues of the original operator of the projectors or the different spaces. This is actually a statement of the completeness of the operator written in terms of the projected language. Another interesting thing is the operator itself can be written in terms of this projector instead of its eigenvalues. It looks like this. The operator a is the sum of dn times dn. Which I think is fairly obvious if you think about it because if you let a act on one of its own eigenvectors that lies in the space dn and the pn doesn't do anything to it and the result is it just gets multiplied by an. All the other vectors that are affordable will not get annihilated by that pn. And so this has the effect on the basis of eigenvectors that the right-hand side is the same effect on that basis as does the left-hand side. So the two sides are in fact equal operators. So in the case of the discrete spectrum this is how you express both the identity operator and the original operator itself in terms of the projected sum of the eigenspaces. Now, let me say something about the continuous spectrum. Here it might be easiest if we just concentrate on an example which we used before which is the case of the momentum operator, the p operator or p hat as we were talking about in wave mechanics. So I explained last time the eigenfunctions of this operator are plane waves which don't belong to the moment space because they're not normalizable and they do form a continuous basis. Now, let's suppose, so give us a moment of access like this and let's suppose we choose a moment of value of p here and this of course corresponds to an eigenfunction which is a given plane wave. Now, the first question is it doesn't make sense to talk about the subspace, the eigenspaces each e corresponding to a single moment of value. And the answer is no because the eigenfunction in this case is not a subspace in the moment space because the eigenfunction doesn't even belong to the moment space. It's not normalized enough. And so this actually has no meaning. On the other hand, it's possible to talk about an interval let's say going from p0 to p1 like this let's call it interval i and we'll have space taking this from p0 to p1 and then what one can talk about is a projector onto the projector corresponding to the interval let's call it pi. What is this projector? For example, it's the interval of momentum from p0 to p1 in bounds of the interval and the outer product of p with itself. Notice that if these waves were extended to infinity you'd just get the identity operator. Since it's only a finite range of momentum what you get is a projector onto a range of momentum. This kind of projector should be familiar to you from electronics and signal analysis if you're working in a time domain extended to a space domain. The idea is that you design a circuit you call it a filter which would take a signal and it would reject all frequencies outside of a certain frequency range while on the spatial domain that's equivalent to momentum. And so the idea is you'll have an acceptance function and what's the step function that looks like this and that's an input signal. And in fact this is what this so in fact this is what that circuit does to time dependent signals because this really is a projection operator but something similar to that when applying quantum mechanics is this projection operator here we move all momentum components that lie outside the interval you can see it's a projection operator because once you're inside the interval you apply it twice you get the same result all over again no change in anything if you apply it twice. So in a continuous spectrum you do have projection operators corresponding to the interval so we actually also have corresponding subspaces of the Hilbert space corresponding to the intervals of the continuous spectrum. Let me tell you a little bit about the resolution of the identity and the resolution of the operator is in the case of a continuous spectrum in fact I'll even do more than that in the case of a fixed spectrum where you have both the screen and continuous eigenvalues this is the dimension last time this is what you get in the case of a hydrogen atom so let's suppose we've got A acting on N and R this is the discrete spectrum is A and N and R so this is R where it's more than the dimension of the eigen space again the discrete part of the spectrum and let's suppose we've got A acting on let's call it A R it turns out the value of A as A R this is in the continuous spectrum so if A here belongs to some interval of the sum interval which represents the continuous spectrum pretty well let's normalize these vectors in the following way in our scalar product N prime R prime is delta K N prime delta R R prime this is the discrete normalization for the continuous normalization we have A R A prime R prime is a direct delta function in A is A minus A prime and in prime they're delta in the R's the R here is the R scale of the resolution the R here is just an extra index which is introduced in order to resolve degeneracies if you have more than one plane in the independent writing function for a given value of A this actually occurs in a hydrogen atom with positive energies where you've had, I think I mentioned last time mistakes have been written this way D L N so the positive energies E is continuous and L N is still discrete these are essentially scattering wave functions in a hydrogen atom with given values of L N but in a positive energy in any case these are the relative relative relations and they continue and then finally we have the order of the relative relations connected with the continuum of the discrete which is that those are zero those are orthogonal states with some kind of bound states that are orthogonal to the unbound states alright this is a complete set of order of the relative relations then in this case the resolution of the identity looks like this one is the sum of A and R the discrete state of the outer product of N R and N R plus an integral over the range of the continuous eigenvalues of D A and then the sum over R of A R of the outer product of A R likewise the operator A itself can be represented by such a decomposition it's the sum of N R of A N times the outer product of N R and N R it's just only locally these terms by the corresponding eigenvalue and we're going to continue to respect the considerable D A of the multiple of A with the sum of R of A R of the outer product of A R like this this last formula can be generalized to give us a representation not just of the operator A but in fact a function of the operator A when I want to function of the operator A I just take a function of the eigenvalues just replace the eigenvalue by a function of the eigenvalues what this means is if you have an operator A and you want to find a function of it you just go to the eigenspaces and on each eigenstation just use the function you're interested in with the eigenvalue we'll see interesting functions non-trivial functions for operators that appear later on in the course and clearly things like delta of A minus A not even delta functions which are kind of desired functions of operators but in fact these are moving forward to give us this representation of a function of an operator now that's all I want to say about the mathematical formula of quantum mechanics for now but what I'd like to do now is to turn to the physical possibilities of quantum mechanics there's a lot of ways to connect the mathematical formula with these experimental results we're actually not ready to present the physical possibilities in a final form because we haven't talked about mixed states so these are in an incomplete form I'll say in a minute in what sense they're incomplete but here's what they are these possibilities, by the way are the distillation of a formatted mystery of the development of concepts of quantum mechanics they're more or less this form already by the late 1920s so we're not exactly recent but the process of getting there was a very long and painful process so they're not intended to be obvious and in fact they're certainly not obvious but we're going to take them as giving here as a starting point and see where they go so Poshlet's run like this first of all Poshlet says that every physical system is associated with certain Hilbert states called script D Hilbert's space is in a complex vector space with a scalar product on it it has certain properties such as its dimension and the dimension of Hilbert's space is actually determined by the physical system I'll give you an illustration of that in just a moment how we know that, how we know the dimension is secondly, every pure state of the physical system corresponds to a ray in the Hilbert space I haven't explained yet what a pure state is and that's why these Poshlets are incomplete I'll have to come back later to fill that in so let's just leave that handling for now this is going to be a neat incomplete from a logical standpoint we should learn a lot anyway by going through this at this stage let me just say that a pure state roughly speaking is a state which corresponds to a weight function that would give you an idea of what we mean by that a ray I'll remind you of is a set of vectors in the Hilbert space which are proportional to one another in other words it's one dimensional subspace this is usually represented by some nonzero cat which lies in the ray the state the real correct thing to say is the ray itself corresponds to the physical state not any particular vector in the ray nevertheless we're pretty sloppy about this in language we frequently refer to a nonzero vector in the ray as the quote unquote state even though it's the ray that's the state and not the vector this is because the normalization states don't have any physical surface so we just choose some dimensional nonzero vector in there frequently we remain side normalized but in most of this I'm going to assume that it's not necessarily side normalized the third possibility is that any measurement process you can carry out in the physical system corresponds to an observable which is an operator that acts on the cat space the observable means that it has say of course a condition operator yes the correspondence always one to one and is everything in the Hilbert space a physical state is every physical state an element in the Hilbert space if you're talking about pure states yes by state you mean a ray yes if you mean vectors then there's a set of global vectors corresponding to the state so the measurement process corresponds to observables fourth possibility says that the results of the measurements are the eigenvalues so to be observable there's two possibilities that could be discrete or they could be continuous discrete eigenvalues or continuous both possibilities exist depending on the measurement you're making the fifth possibility of measuring a discrete eigenvalue called it equals a n or maybe I should say capital A equals a n is given by the ratio which is on the top is the projection operator pn on the eigenspace corresponding to the eigenvalue which you measured sandwiched between the state vector psi divided by the norm of the state vector psi notice we're not requiring psi to be normalized and in fact this answer is independent of the phase of normalization of psi which as I say they're not visible anyway in the continuous case we need to we can't talk about the problem of measurement of the value for a continuous variable which has to be zero we're talking about the probability of measuring a on some interval let's say a0 to a1 in fact let's call it interval i like this capital i the probability is given by the expectation value projection operator on that interval again divided by the norm of psi finally there's the last postulate which is sometimes called the collapsed postulate which says that after the measurement the state psi, the original state psi is replaced by the psi in which the psi with the projection upcoming to the strict case replaced by the original state psi with the projection operator pn which projects onto the eigenspace of the eigenvalue which was observed in other words psi becomes an eigenstate now of the original operator a with eigenvalue a n which is projected onto the eigenspace likewise in the continuous case if we measured a0 to i as a continuous interval then the projection operator is the one that corresponds to the continuous interval alright so those are the postulates now they're able to reform now what I'd like to do is to illustrate these postulates in the case of the Stern-Gerlach experiment to show how these postulates can be connected to a real example it isn't quite a real example we'll be talking about the predominance experiments there was a real Stern-Gerlach experiment which has been repeated many times but certain variations of it have never been done as far as I know so part of this is the predominance experiments but nevertheless you can take the results that I call as being experimental results I don't think anybody doubts that they would be true so let's go through the Stern-Gerlach experiment just using these postulates now in this process we play a certain game which is if we continue we know nothing about quantum mechanics it's a postulate particularly yes so where to continue the Stern-Gerlach I mean is it an interval kind of arbitrary we can make an interval as small as we want as long as it closes A so in what way is this consistent to say that it changes to the P of i of psi when P of i is some arbitrary interval I don't understand the question the probability measurement depends on the interval because if you narrow if you just think about a columnator that measures the position of a beam that's really what the i is in that case if you narrow the columnator you're narrowing the interval the probability goes down of course because you're reducing the number of particles you're accepting does that answer your question well I mean you still get a certain answer A yeah but no you get an answer that why is it inside the interval probability of having the answer why is it inside the interval is giving right this explanation but you narrow the interval probability goes down yes so in the case of of a degenerate if you make a measurement corresponding to an eigen if you get a degenerate eigenvalue then does that no yes it's a unique state and the eigen subspace however the eigen subspace is multidimensional there's other states in that subspace that have the same eigenvalue right so in other words yeah so if the eigen space is multidimensional we might get different output states depending on the input state because if I'm projecting on the table and nothing is expressed this way then both the eigenvectors would give an eigenvalue but it wouldn't be proportional to each other it would not be the same state afterwards on the other hand if the eigen space were multidimensional then all of these would represent the same state because they'd all be proportional to one another alright so yeah so I'm going to talk about the Scheren-Gerlach so as I was saying in this we're going to play a game here in which we pretend the way the functions always spin matrices or any of that stuff Scheren-Gerlach didn't know about any of that stuff themselves and here are the purposes to understand how these postulates can be how they believe to an understanding of the space that we're using unfortunately we have to cover up the postulates in order to get a format for right on so first let me say some things about the Scheren-Gerlach experiment the purpose of this was done in 1923 the purpose of it was to measure the magnetic moments of atoms magnetic moment class of magnetic moment is a vector near like this it's determined by any localized current distribution and magnetic moments so we could in principle you could measure magnetic moments of small dust particles of little iron bar magnets attached to them if you wanted to do it they had definitely magnetic moment so you could measure them that was the idea of the experiment now a magnetic moment is measured by placing in an inhomogeneous magnetic field in the magnetic field the magnetic moment requires an energy which is minus mu dot b and if you take minus the gradient of this in order to get a force the two minus signs cancel and we've got the force acting on it so this is the force on a magnetic dot pole that requires the force requires a magnetic field to have a spatial dependence the mu is characteristic of the particle and doesn't depend on the space so the dell operator really only acts on the b in the experiment which Janet and Gerlach used two magnetic poles that looked like this one was at a sharp point and the other was flat so this is the north pole or the south pole the purpose of this is to create a strong magnetic field gradient near the pole tip of the north pole here because you need strong radians to get a reasonable force so the idea is the this pole piece would go down like this and so would the bottom one too going across and you'd run a beam through and the beam would run this goes so this way you actually want to make this a meter or more it's a very long magnet like this and you have a beam that comes in like this and you run the beam through here this is an atomic beam of beautiful atoms the beam by the way they use the solar atoms the beam by the way conceptually is created by having a actually two we have a lump of silver in an oven which would heat to a high temperature they use silver atoms other atoms can be used too so that you get some silver vapor in here the silver vapor passes through a small hole in the oven and from there it passes through a column here to create a beam of silver atoms all right and then this beam is passed next to the magnetic pole so I'll say that the z direction is the direction of the vertical direction in this diagram like this then along the axis the magnetic field is largely in the z direction so that v is approximately equal to the magnitude of v times z hat and thus this means that the force of the atom is given by u sub z the z component of the magnetic moment of the atom is the gradient of the magnetic field which itself is largely in the z direction so the force is essentially in the z direction now and so while the atom passes through this inhomogeneous magnetic field over a certain period of time over that time this force of time gives us momentum transfer momentum transfer is in the vertical direction so the beam will move will tan out in the vertical direction like this with an amount which is proportional of u c the z component of the magnetic moment when it entered the beam we can assume that the original magnetic moment the assumption is the magnetic moment u as a definite magnitude of the atom is a definite magnitude let's call it v is 0 and if so the z component ranges between plus and minus v is 0 depending on the random angle of orientation and so what you would expect classically is you would expect a continuous distribution of a deposit of the final beam on the screen ranging between a plus and a zero and you could figure out what v is 0 is by looking at those extremes of course with Stern and Gerlach found that something completely different instead of getting a continuous smear on this screen it actually got two spots at plus and minus v is 0 they were able to measure the value of v is 0 and it turned out to be a good agreement with the Bohr magneton that was actually known at that time the zh bar between 2 and 3 the momentum value came out to about right alright so this was the big surprise in this regard it's one of the crazy aspects of quantum mechanics that you should only have two spots here it seems crazy because how can you and z have only two components plus and minus v is 0 there's nothing special about the z direction you can orient in the x or the y direction and you get the same two spots the classical vector can't possibly have just two different distributions of particles that you possibly have just plus and minus components in all three directions that makes no sense so how do you interpret this okay so let's try to do this from the standpoint of measurements in quantum mechanics in the first place let's take this z from the oven and let's schematically indicate Stern and Gerlach apparatus by measuring here let's say a component of z here but let's say we measure the component of x in the x direction first and as I say what comes out of this beam is two outcomes which I'll just note by plus and minus but it stands for plus and minus before magneton of the values that are measured and so what this means is if you go to the postulates in quantum mechanics it tells us that ux the intermission operator actually can't space in the system which is the can't space in the atoms and since there are two outcomes it means that there are two eigenvalues of the operator ux plus and minus mu zero and therefore the Hilbert space decomposes into two orthogonal subspaces corresponding to these two eigenvalues so we don't know what the dimensionality is of those subspaces that's to say we don't know what the degeneracy is but there's at least two subspaces so the Hilbert space is at least two dimensional also we get the same gen answers with the same mu naught no matter how we worry about the beam x, y, or z so as far as making measurements of components of mu's concern there's no evidence that the Hilbert space has any more than two dimensions the easiest way to proceed with this discussion is just to assume at this point that these two eigenspaces are in fact non-degenerate so they're one dimensional and later on come back to the question of degeneracies so allow me to do that let's just make an assumption let's say the Hilbert space is two dimensional if that's so when the eigenspaces correspond to these plus and minus outcomes are both one dimensional and non-degenerate let's suppose in this experiment we take the minus outcome and we just throw that beam away and we take the plus beam and carry it out to do further experiments on it in accordance with the possible mechanics this is now represented by one dimensional sub-space of the Hilbert space let's introduce a vector all called ux with a plus sign ux is the operator plus is the eigenvalue this will take us to be a normalized vector which lies in this one dimensional space this is not specified by the phase of this vector which is still arbitrary so this is not uniquely determined we have to remember that this is the eigenspace and call it ux plus now let's take this this resulting beam let's feed it into a second let's make this a tandem let's feed it into a mu y where we measure mu y in a different direction the experimental or economic experimental results once again we get two values plus and minus would be a zero and they each come up with 50% probability same thing would actually happen that's a matter of fact so, oh excuse me that's not mu y let's make this mu z let's make this mu z and so let's call the states by the, okay so I have to measure mu z then according to the projection or number six the last hypothesis is after measuring mu z the state is projected on the eigenstates of the operator mu z therefore the state of the system within these two beams is represented by two vectors let's call them plus and minus here I'm using the notation that mu z plus or minus is just the same thing as plus or minus although we meant the mu z in the case of the z component because we're going to take that standard basis but in any case the state of the system is now represented by either of these two states depending on which beam you look at coming out following the cost of this all right we're going to take these to the globalized states now as a result of this we can take this mu s plus these two states now are giving us a basis on the entire Norbert space and so we can expand this with a linear combination of the mu z states like this with coefficient c plus and c minus also as I said the probability is 0 or 50 to 70 also the probability of measuring mu z equals plus mu zero let's say coming off the top side here why are possible let's say mu x plus sandwiched around the projection operator p plus like this I don't need to divide by denominator because we're assuming this was normalized where the p plus was the projector onto the the plus by the state of the of the mu z operator so this probability becomes mu x plus scalar product of plus times the scalar product of plus mu x plus which is equal to the absolute value of mu x plus scalar product of plus squared that's the same thing as the expansion coefficient c plus in this expansion so this becomes c plus squared which is discriminately equal to one half thus we conclude that c plus is equal to one over the square root of two times the phase which will write as e to the i alpha one where alpha one is unknown at this point similarly if I do this for the minus state we find that c minus is equal to two times e to the i beta one where beta one excuse me, yeah alpha one beta one where beta one is another unknown phase by applying these postulates this is one over the square root of two that alpha one plus e to the i beta one one of these allow me to in effect absorb this arbitrary phase into the phase convention of mu x plus I'll do this by writing this as mu x plus for the prime one times e to the i alpha one and then we'll multiply both sides by e to the minus i alpha one and what we get in is that mu x plus prime is one over the square root of two where the space is going to eliminate it plus e to the i beta one prime minus beta one times alpha one now if you allow me to do a very trick for lecturers which is to use the eraser I'm just going to erase the primes and I'll put a check next to the mu x plus and what the check means is we've now established phase convention for this state and we can't change its phases anymore so when there's phase convention for the mu x plus state we can write it as linear combination of the eigenstates of mu z in this form now obviously we could have done something with mu x minus the other the other being coming out of the first dashed area in the upper lattice and then we did it with this mu x minus is equal to one over the square root of two plus let's call it e to the i beta two minus like this and I'll put a check next to that one because that means the phase convention now these phases can be nailed down further however because we know that mu x plus must be orthogonal to mu x minus because they belong to because these are lecturers that belong to eigenstates within distinct eigenvalues in this case of the operator mu x so this is equal to zero and if you do a little bit of algebra here what you find is if you do the i beta two is equal to minus e to the i beta one so we can now combine these two statements together in this form to state that mu x plus or minus is one over the square root of two as the state plus plus or minus e to the i beta one times the state minus like this it's a condensed version of the two lines above there's only one independent phase and the phase conventions on the left side are determinative similarly instead of mu x first we measure mu y first we did the same thing we would obviously end up with similar conclusions we'd get mu y plus or minus could be represented this way one over the square root of two times plus plus or minus of phase e to the i gamma one minus all specifying phase conventions for the mu y plus or minus phase so now there's two phases again and what about the relation between them well before I do let me point out that one need to say that this point is convenient to actually talk about the operators remember the operator can be represented by a sum over the projectors which is weighted by the eigenvalues so in particular the operator mu x itself there's two eigenvalues plus and minus mu zero so it looks like this it's equal to mu zero times the object product of mu x plus mu itself minus the object product of mu x minus mu itself like this because plus mu zero is that eigenvalue minus mu zero is that eigenvalue mu x plus expression down here plug it in here and do the algebra you'll find that mu x is equal to mu zero times the following is e to the minus i theta one times the outer product of plus with minus plus e to the plus i theta one times the outer product of minus with plus similarly for the mu y you'll find this is mu zero times e to the minus i theta one plus minus plus e to the i theta one minus plus and finally mu z is using this mode right down this plus plus out of the product minus minus out of the product here what I've done is I've written in three operators in terms of other products the eigenstakes of the operator mu z like this and I've sold these two unknown faces that are going to occur here alright now we can not receive these phases by imagining yet another tandem experiment instead of feeding mu z and mu x and mu z they're going to feed a mu x into a mu y and if you do that again we get two outcomes when probability is 50% that means that a mu x instead of plus sign of mu y with either plus or minus sign in the square that's probability the answer is one half and if you take these states here the plus sign on the other side and the plus and minus sign on the lower one for your scalar product 70 to the one half through the algebra this gives you a relationship between the phase one and gamma one in fact what you apply is e to the i of gamma one is with a sign plus or minus sign e to the i of gamma one so again if I may use my eraser e to the i of gamma one here I'll replace by plus i times e to the i of beta one and e to the minus i of gamma one I'll make it minus i e to the i of beta one and as far as this plus or minus sign here which is not determined by this formula we're going to plus or minus out front e to the y so by using these various combinations of 10 experiment in making phase conventions we can reduce these 3 operators to this form now there's still this final unknown phase of beta one and we've pinned down the phase conventions for the mu x and mu y plus and minus we can't change the phase convention for the mu z plus state because it would mess up this equation you see it would have changed the alpha one of the sort but we can change the phase convention for the mu z minus state in fact let's use it to absorb e to the i of beta one in other words we'll say a minus prime is equal to e to the i of beta one times the minus like that and if we do this I'll again use my eraser and it just makes the e to the i of beta one go away everywhere and now what you see here in here is the three problem matrices because if I take the operator mu x and I write it in this plus or minus basis it's mu zero one zero one zero one one zero the mu y is equal to plus or minus mu zero times a zero minus i i zero and the e to the c is equal to mu zero times one zero zero minus one one remark here is there's still remaining unknown sign and plus or minus sign it hasn't been determined by the possible so they experegrable data but apart from that you see the standard poly matrices appearing the y matrix is turning out to be purely imaginary and the x matrix is purely real we could have reversed that it's done a different phase convention for the z minus state we put a factor i in here and the result would be that the x matrix would be real excuse me x matrix to the imaginary would be real so one of the rules of this is you can make one of the other of these two matrices real but if you do the other one has to be remaining one has to be imaginary this is by the way an indication of proof that in quantum mechanics it's necessary to use complex vector spaces with complex numbers the possibilities of quantum mechanics cannot be reconciled with the experimental results of sharing all our type experiments now what about this final plus or minus sign that's something that I will let you think about there's still remaining on this sign here but we're pretty close to the standard results okay so this is this is an indication of how jolbert's spaces are constructed out of experimental data and I notice what the operators are based on experimental results and how one can use space conventions and other things to obtain standard representations for the operators the quantum mechanics literature is just crawling with various space conventions that have been established over the years and through practical calculations you have to know how to use them but there's no physics in them and the real physics is contained back in the postulates I'll be the return of the question how do we know that these operators are non-generated? that was an important step here because it's allowed us to use a two-dimensional corporate space how do we know those eigenstaces of the new x-operator for example are really only one-dimensional the answer to this is connected with the concept of compatible or commuting insurmables let's consider an idealized measurement process in which we measure an observable A so we've got some kind of an input state here let's call it size 0 like this there's a whole bunch of outcomes that come out these correspond to what's called the discrete eigenvalues of A let's say one of them is A n and let's say all the others we throw away and we take the A n state out here and call this resulting state after measuring the observable A on the initial state continuing the value of A to the n let's say the resulting state is side 1 well according to our postulate side 1 is the projection operator corresponding to operator A with eigenvalue n applied in size 0 now let's take this side 1 and let's feed it into a second measurement of an observable that's called it B let's say this is a bunch of outcomes let's say we throw all of them away except for one of them all called B to the n and what comes out of here is another state we call side 2 which according to the postulates of quantum mechanics side 2 must be the projection operator onto the eigen space of the operator B with eigenvalue of B to the n like this this is acting on side 1 so this is the same thing as piece of B to the n acting on piece of A to the n acting on the original state size 0 by the postulates of quantum mechanics now let's look at the probability measuring let's say first A n as we've indicated here and then secondly B to the n on these two measurements this is a conditional probability probability getting B to the n given that we got A to the n on the first measurement so the probability of getting A to the n on the first measurement which is side 0 sandwiched around P to the n divided by the normal side 0 times the probability of getting B to the n given that first times the probability of B to the n with side 1 coming in so this is the probability of the side 1 this is the scalar product of side 1 projection operator P B m side 1 divided by the normal side 1 side 1 denominator here now this numerator up here can be written this way because side 1 is P A n acting on side 0 this is supposed to cancel that and it's not going to occur ok I'm going to stop it's a little over time anyway I'll finish filling the rest of this next time