 No one always thought of the lectures, but if you've been in this person, you're close. I'm probably a lecture on some business variations, of course. Okay, the last time at the end of the hour, we were in the process of examining this question about the possibility of degeneracies. In my research in the Stern-Gerlach experiment, there was no evidence in the basis of measurements of spin that the Hilbert space had any more than two dimensions. And so we're just as slim for the sake of simplicity that it had exactly two dimensions so that each of the ideal spaces of the various components of magnetic moment were one-dimensional. Now the question is that, and I promise we'll come back and examine the question of degeneracies later. So that's what we're doing now. The question we want to ask is, if there were degeneracies, how would we know? This is not an mathematical question. It's one of physics that has to start from the postulates of quantum mechanics. So to answer this question, I'd like to examine and still an idealized experimental situation in which we imagine that we've got some pure state, initial pure state, psi zero. We make a measurement of an observable A and we throw away all the outcomes except A sub n. Then according to the postulates of quantum mechanics, after this measurement, the state of the system, which I'm calling psi one, is given by the projection operator which projects on the new eigenstate of the observable A, which is what we mean by P a n, acting on psi zero, gives me the new state psi one. We then take this psi one state and we feed it into a second measurement of a different observable B, which also has a bunch of outcomes, and we throw it all away except for one of them, we call B sub n. And again, according to the postulates of quantum mechanics, the state that emerges after the measurement of B sub n with the eigenvalue of B sub n, is given by the projection operator on the eigenspace of the operator being with eigenvalue of Bm, acting on psi one, that's the state psi two. Okay, that's just following the postulates. Let's ask for the probability that we measure first A n and then secondly B sub n in a sequence of experiments, a sequence of measurements like this. This is a product probability that gives me to multiply the probability measured A n in the first place by the postulates of quantum mechanics given by taking the input state psi zero here and sandwiching it around the projection operator P A n and then divided by the norm of psi zero, the square norm of psi zero. That's the probability of getting A n and then we need to multiply by the conditional probability of getting B sub n given that we had A n in the first place. And that probability is again by the postulates of quantum mechanics given by taking the input state psi one sandwiching it around this projection operator piece of D here and dividing by the norm of psi one. So this is the compound probability. Now let's take this denominator psi one and psi one here and let's plug in this expression for psi one in terms of this projector. It gives us in terms of P A n acting on psi zero. I take this as a broad, I turn it around and I get a P A n dagger. So this is the same thing as psi zero piece of A n dagger times piece of A n times psi zero, that denominator. However, projection operators have emissions so I can drop the dagger if I want the left-hand factor and projection operators, I have to put them too. That means the square is equal to the operator itself. So we get the square of P A n in here but it can just be replaced by a single copy of P A n like that. And if we do this and we see that this denominator psi one psi one cancels with a numerator up above. Now likewise, this numerator that has psi one's in here we can substitute in and express them in terms of psi zero using this expression. And so if I copy this down, clean it up and copy it down, the probability of these two successive measurements giving two given successive outcomes as being equal to the state psi zero sends from a product of three projection operators, P B n, excuse me, that's P A n P V m P A n psi zero divided by the square order of psi zero, psi zero of the normalization. And I've got this one again. All right. Now, suppose we perform the experiments in the reverse order doing the A first, the B first and measuring P sub m and n A second and making the A sub n. And let's write down what the probability is for doing that in the reverse order. The formula is just the same as the one above except we swap the A's and the B's and the n's and the n's. So it's psi zero P V m P A n P V m P sub n psi zero divided by the word squared psi zero like this. Now the question is, are these probabilities equal? Do we get the same answer? And the answer is no because in general these projection operators, the A and the B projection operators still commute with one another. So the triple product in the first expression or the one in the second expression are the same. This is something that is one of the surprising aspects of quantum mechanics. In a sense, it's a saying that the measurement of one of the observables mixes up the statistics of the other one. We saw this already in measurements of magnetic moment if we measure the new x first it affects the outcome of the subsequent museum measurement, changes the statistics of the museums. And likewise here doing this in one order to the other makes a difference because measuring one observable affects the statistical outcome of the measurement of the second observable. Now, on the other hand, there are circumstances in which these two probabilities are equal. They're equal in particular if the operators A and B are the two observables commute with one another. This is because if the observables commute it implies that the projection operators commute P A n is P V m commute. This follows because projection operators are actually functions of the original operators themselves and functions of community operators are commuting. So in the present case it means, for example, this is the numerator of the above. I can take the P V m and move it to the right of the P A. We're talking of P A n squared. That's just P A because it's height of potent. So you get the product P A n times P V m in the upper expression. Likewise in the lower expression I can take this P V m and move it to the right of the P A and then you get a P V squared which is the same as the, without the square. So again you get the same product here. You can write this in either order, of course, because they commute. But the point is that these are the two expressions now become equal. So we can say that these probabilities are equal if the different operators commute with A and B. This is a consequence, as I've just explained, of the measurement postures on the canvas. Now there's a converse of this as well. By the way, the mu x and mu z, for example, in this triangular experiment, certainly don't commute. And so these probabilities are not equal for that kind of experiment. But they do commute when we have equal probabilities. Now there's a converse of this as well. We should have said if the measurement of the probability of the measurement of the compound measurements through observables like this are equal for all choices of the eigenvalues, A, N, and D, M, and also for all choices of the initial state size 0, then one can show that the operator is commutative. So it's really a definite only if is it necessary and sufficient conditions for the equality of these probabilities is the commutativity of the operators. This is what this means among other things is that the commutativity of operators has experimental consequences and, in fact, it can be determined in principle. It can be determined experimentally. Not that anyone actually does this because it would be very difficult to do these in full completeness. But the point here is that it's a principal experimental matter to decide whether operators commute. Remember here in the postulates of quantum mechanics, we're trying to build up the mathematics from the physics. We're not starting with an mathematical problem, which is what she's mostly done at this point in studying quantum mechanics. We're rather starting with the physics. All right. Now, so that's the meaning of commutative operators. Now, let's go back to the question about the possible degeneracies. This is related to how do we know the dimensionality of the Hilbert space when we were starting with the possibilities of quantum mechanics and building up the Hilbert space. How do we know, for example, those two-fold that the indication is thin operators or magnetic moments would decompose the Hilbert space into two subspaces. How do we know those subspaces are really old one-dimensional? The answer on something like this is that this state, side one, lies in the eigenspace A and the operator A according to postulates and quantum mechanics. Now, if the operators A and B commute, it then follows that the projection operators are B. There's more than one of them here. I'm just listing one of them for the single outcome. There's other outcomes here. There's a whole series of projection operators. It turns out that those B projectors leave this A eigenspace invariant. So they take any vector in that eigenspace such as this side one and they map it into another vector in the same eigenspace. So what that means is that the states have come out of all these different B outcomes. There are still eigenspaces of A with the same eigenvalue. This is a change. This is if they can move as I'm assuming that. On the other hand, the B values are different. So what that means is we've got a collection of states of what they now allow in our simultaneous eigenspaces of both A and B. And if there's more than one of these and the B values are different and the states don't vanish, then these different states have to be orthogonal to one another because the B values are distinct. And so what we've got now is a collection of several vectors that lie inside this eigenspace of A and N which therefore must be multidimensional. As we say, what we've done is we've resolved the genesis of the A operator by applying the B operator. And in fact, what we've discovered is simultaneous eigenspace by using commuting operators. We've discovered simultaneous eigenspaces. This is clearly related to this theorem we discussed earlier about commuting operators possessing a simultaneous eigenspace basis. Now, once we've got these simultaneous eigenspaces, how do we know that those are maybe degenerated? They're multidimensional. How do we know? Physically, how do we know? Well, the answer is we search for further observables that can be with both A and D and remember it in principle. That's an experimental matter to decide that. To see whether these measurements of the further observables will cause any further resolutions of these degeneracies. Now, what we've done with the A operator is we've reached the end of the game. These resulting simultaneous eigenspaces are one-dimensional, and we have what we call a complete set of commuting observables. A complete set of commuting observables means the simultaneous eigenspaces of all the observables taken together are one-dimensional. And so if you just choose normalized vectors in each of those spaces, you get an eigenbasis for the whole Hilbert space. These NLM states like this the first one is the eigen quantity of the Hamiltonian, the second one is the operator L squared and the third one is Lz. If you just talk about the Hamiltonian itself, it's degenerate. But the list taken together gives you a basis, a non-degenerative basis in this space. A question in principle in this procedure is how do you know if you've gone far enough? Maybe you can't find any observables that produce any more resolution but maybe you didn't look hard enough. Maybe they're actually there. The answer is that question is in principle you really don't know what you've gone far enough. What you do if you stop at a certain point is you're restricting yourself to a certain kind of physics. Here's an example of this. Frequently, as I'm sure you know in discussing quantum mechanics of electron, we frequently treat the electron as a property in, you know, in inductive courses. But we're just writing this psi of r or something that reminds us spin component. Well, what this amounts to is using position variables as a complete set of commuting observables but ignoring the spin variables. It's physically justifiable to do that if the spin variables are not important for the physics you're interested in. In fact, this happens frequently in practice because spin only interacts with magnetic fields and electric fields. So if you have electrons that are moving in electrostatic fields and they're not known to this state in low velocity, then it often happens that the spin is not important. And you can, in effect, stop. You can deliberately stop when you search for a complete set of commuting observables before you've taken into account all the degrees of freedom of the system even if you know about them because maybe you don't care about that physics. There's many examples of physics. Then what you'll find is that what you were previously speaking of as non-degenerate spaces like this one actually become degenerate. Because, for example, you need to say then you need to put in an extra quantum number that tells you the spin quantum number. This is a resolution of the P.D.C. but you go out of a non-degenerate eigenspace and now you have to keep going together with the magnetic quantum number the spin tells you which one you're talking about. Okay. So this is the story of the dimensionality of the coordinate space. What to do about the generalities. All right. Now, here's another important point which is that the... I didn't write out the possibilities of quantum mechanics once again today because I didn't get here in time. But the... as you'll recall the predictions of quantum mechanics refer to probabilities. That is, they are a statistical nature. So by the possibility of quantum mechanics as I presented them quantum mechanics makes predictions about the probabilities of various outcomes and measurements. These probabilities are measured in practice by taking an ensemble of identically prepared systems and subjecting them to the same measurement and statistical accounting of the results. The quantum mechanics doesn't tell you anything about what happens to an individual system. In the orthodox interpretation of quantum mechanics quantum mechanics only makes statistical predictions. So for example we go back to the Stern-Gerlach apparatus so where we first met with a new accident we send a plus key in through a measurement of a new z and then we got two outcomes and we flew away. We got two outcomes here, plus and minus and each one had a 15% probability. This is the Stern-Gerlach experiment. Now quantum mechanics tells you these probabilities. We can calculate them. But if you have an individual silver atom coming down to this beam quantum mechanics is not going to tell you and it's polarized in the x direction with plus, that's given. Quantum mechanics is not going to tell you what happens in the next magnet. It makes no statements about that. So it doesn't tell you what is an individual system. So in the orthodox interpretation of quantum mechanics it's probably safest to regard the wave function and later on we'll talk about the density operator as representing not an individual system but rather an ensemble of identically prepared systems. Although we oftentimes use this language such as the wave function of the electron the wave function is really describing the statistics of measurements on an ensemble of electron systems which are prepared to be in the same state such as this state here coming down to this first measurement. All right. Now this is a way if you keep this in mind it's a way of avoiding many of the paradoxes of quantum mechanics. All right. So people don't like this and they want to believe that quantum mechanics is not statistical and the reason it isn't the reason we think it is is because there are extra variables that we haven't measured so-called hidden variables and this leads into the subject of hidden variable theories which you can make some mention of in the notes you can read about them and you may also want to read the section in the soccerized book on Bell's theorem which deals with the same subject. Maybe if we have time later in the semester right now I'm just giving you the orthodox interpretation of quantum mechanics. As long as I'm talking about statistics let me make a few remarks, simple remarks I think you know about this already. If we're doing ordinaries well, if we have that observable A and we measure a certain value of A in and the probabilities let's call them depth in are given by taking the state in question assuming we did pure state and sandwiching it around the projection operator on the corresponding eigenspace this is from the postulates. If we were doing ordinary statistics and we had certain outcomes with certain probabilities we compute the average of the outcomes being the just the weighted sum of the probabilities times the values it's just a definition of the meaning of the statistics. However in quantum mechanics this meaning value can also be expressed in terms of nowhere to space operations and it turns out to be operator A sandwiched around the excuse me the state side sandwiched around operator A like this as I'm sure you well know. Similarly in statistics we can define a mean square deviation or dispersion which is the same thing as the sum of n of fn times a in the square minus the square of the meaning that's the meaning of statistics and it turns out this can also be written in terms of nowhere to space operations as we say the expectation value for operator A squared minus the expectation value of the operator itself 1 to be squared this can also be written in another way the size sandwiched around the operator A1 squared where the operator A1 is equal to the operator A minus its average value that's the definition of A1 and again this is straightforward to prove this and I'm sure you've seen it before so I won't go through it but it connects statistics with nowhere to space expectation values this first expectation value is often known as just abbreviated this way leaving out the psi if we do then the edge of the brackets has kind of a double meaning it's first of all a statistical average an expectation value in the sense of nowhere to space operations and likewise this is a similar thing this could be written as just A squared without the size if you use this notation then some state size understood you're writing this down alright now so that's all the simple stuff that I think you know let me carry along with this idea of statistics I'd like to address the question of generalized uncertainty relations the situation is the is the following let's suppose we've got two observables A and B which don't necessarily commute I'll put the result first it says that the dispersion in A squared times the dispersion in B squared the product of the two variants in some statistical sense is equal to the quantity which is one one-quarter of the absolute value of the expectation value of the commentator of A with B or in the square that's the identity so now allow me to prove this the delta A squared is the same thing as psi sandwiched around the operator A1 squared as I've indicated up here where A1 is between A1 and B in value let me write this as just A1 times A1 acting on psi like this A1 squared writing it up like that these are observable so they're permissioned so allow me to take back one of the left which is okay let me also define a ket alpha which I'll define A1 acting on the ket psi so that if I do this this becomes the same thing as the scalar product of alpha with itself similarly we'll say for B we'll say that this is psi sandwiched around B1 squared psi which is the same thing as the scalar product of beta with beta where beta ket is the same thing as of course defined to be B1 acting on psi like this and so the left-hand side which I want to prove here is the same thing as alpha alpha times beta beta is a product of two squares of these two vectors now by the Schwarz's inequality this is greater than or equal to the absolute value of the scalar product of alpha with beta squared so let's take a look at the scalar product of alpha with beta what does that equal to well it's the same thing as psi sandwiched around product of operators A1 and B and I'm using these equations here now to make a little regression if I had two operators x and y I can write them this way 1 half xy minus yx plus 1 half xy plus yx in an obvious identity and that's the same thing as 1 half of the commutator of x and y plus 1 half of the anti-commutator of x and y where I'm using curly brackets for the anti-commutator except that it's a plus sign now if x and y are commissioned then the commutator is actually anti-commission as you can easily show whereas the anti-commutator is commissioned so in this case A1 and B1 are both permission operators and I can write this therefore as 1 half the average value of the commutator of A1 with B1 what's 1 half the average value of the anti-commutator of A1 with B1 now since the commutator is anti-permission its expectation value is purely imaginary and since the anti-commutator is remission its expectation value is purely real so this scalar product over here which is a complex number is decomposed into so many real imaginary numbers which must be the real imaginary parts and therefore the absolute value of the square at the left hand side which is what appears in the Schwarz's inequality is the sum of the squares of these two terms and so therefore it's greater than or equal to the sum of the square just one of the terms separately which is the first term here is greater or equal to 1 quarter the square of the commutator of A1 with B1 expectation value of the square of A1 with B1 would be equal to a device if the anti-commutator managed when you see this is almost the result we need except there's just ones here instead of out there without the ones that's easy to take care of because the difference between the ones and the lack of ones is given up here it's just the average value which is the ordinary number of everything so it drops out of this commutator and it just drops the ones here who obviously use my fingers and that proves the inequality well so this is the generalized version of the of the uncertainty of relations and it puts limits on the possibilities of dispersions for measurements of two operators to be clear about the measurement of physical situation that lies behind this formula I'm not talking about making successive measurements of two different observables like I was a minute ago when we measured A1 and B2 and then swapped the order this is a different situation here we're talking about taking a single ensemble and then measuring the As on it and then taking the same ensemble and measuring the Bs on it imagine this is a beam we're going to put the beam through two different measurements A measurement and then a B measurement this is the kind of thing you'd like to do for example if you wanted to make your beams localized as possible in space and also as localized as possible let's say in the direction in which the beam goes this is an application of the uncertainty principle because there's limits on what you can do by minimizing those two quantities the most famous example of this is where we let A be the position and B be the momentum of the particle and if we take the square roots like this then what you get is of course then the commutator X of P is equal to my H bar so if you take the square roots you get delta X delta P is greater than the H bar of 2 which is the famous Heisenberg et cetera now I'd like now to return to a Stern-Gerlach after that I may have another point the the original Stern-Gerlach after that is the beam of the silver atoms it was obtained by taking silver and heating it to a high temperature in the oven make a small hole in the oven so the atoms come out you run it through a column here to make a narrow beam and this gives us what I'll call a thermal beam because it comes out of the oven and this goes into the way we describe it before so let's make the X component is 10 now instead of the magnetic moment it's really the same thing and there's two outcomes plus and minus let's say we throw away the minus and as I've explained in the basis of the possible upon mechanics and the assumption that the Norbert space is two-dimensional we know that the system is described by state vector which is normalized we'll assume it's normalized it's some eigenvector of i in the inside is one of the eigen space where s of x equals plus h bar over 2 h bar over 2 is going to be a knot because it's thin instead of magnetic moment alright so the point is is that the beam that emerges from the second affidavit is described by the definite state vector well definite to the normalization of the phase the question that arises is what about the beam that came out of the oven is it described by a state vector in other words, what's the wave function what's the thin wave function of the beam coming from the oven it doesn't even have one the answer is it does not have a wave function and here's how we can know this we know this because if we take this thermal beam and we make measurements of all three components of spin I'll call this s vector here which means s x, s y, s x s z s z and by an average what I mean is just what you do experimentally you just measure the results and weight them by the number of atoms that came out one way or another for example in the x component here, this is what this thing does it gives you 50% on either side and since the answer is plus and minus h bar over 2 the weighted average is 0 the same thing is true for s y and s z so the result of the thermal beam is that the average of the spin vector is 0 as we say the thermal beam is not polarized the reason for that is it's thermal there's nothing in the oven to pick out any preferred direction and because of thermal fluctuations and random collisions and so on the average spin vector is 0 at that point in any direction now again if the thermal beam were described by some state let's call it chi it wouldn't necessarily be possible to express that state as a linear combination of the z i in states which we're calling s of z i in states which are coming plus and minus with coefficients let's call them alpha and beta complex numbers if we assume that chi is normalized then these coefficients satisfy the normalization condition of one in the structure of the chi is equal to one we can assume the state is normalized here it's not also generality so what I point is to go back to the point is that if let's suppose the thermal beam were described by some definite state vector if it were we could do this expansion and now we'll just calculate what the components expectation value of spin is using the polynomial which we derived already in our previous exercise if you go through the algebra guess what you find is the answer is h bar over 2 and then what the polynomial gives you for the x component is twice the real part of alpha star beta in the imaginary part excuse me for the y component it's twice the imaginary part of alpha star beta and for the z component it's the absolute value of alpha squared minus the absolute value of beta squared that's what you get to do the algebra now it turns out that this thing as you find by showing that the impact dotted into itself is equal to 1 if that's some algebra to do that it also runs down to the normalization again all that should do that as an exercise but let me just write this there for h bar over 2 times the unit vector of impact and so what we see is that if the state of the beam is described by a state vector then it necessarily is polarized in some direction of impact whereas the beam that comes from the oven is not polarized it cannot be described by a state vector when you ask the question what is the wave function the spin wave function or the atoms coming out of the oven the answer is that it doesn't have one we need to understand this situation before proceeding let me remark that there's a kind of reverse of this or converse of this here I took an arbitrary two components spinner and I found out what direction it's pointing in reverse order two let's say we've got an in-hat distance in some given direction theta and phi so the usual coordinates this is sine theta cosine phi for the x sine theta sine phi for the y component and cosine theta for the z component so if there's a unit vector of the sphere of the coordinates I guess I'm going to make this a script phi here then one can show that if you choose these expansion coefficients all in beta would be equal to e to the i5 or 2 times the cosine of theta over 2 then e to the minus i5 or 2 times the sine of theta over 2 and you sandwich this around the three polymaking states you will find that you should get this unit vector so this is just a way of saying we can find pure states state vectors for a 10 and a half system which point in any direction that we decide alright now so the question remains how do we describe the quantum state of the beam that emerges from the other to answer that let's think of something of a slightly different experiment let's do it this way in the first place let's take instead of measuring the x components measured in some direction in hat and then pick out the beam which emerges with a plus sign and this is therefore described by a state vector that I don't know if I ask the impact plus some normalized vector lying in the eigen space of the impact components then with an eigenvalue of plus h part of 2 let's take this beam and let's feed it into some second measurement with some observable a we can imagine there's actually two observers doing this in the first apparatus in the second apparatus this one makes the measurements now to make the game interesting let's suppose that observable number 1 before passing the beam on to observable number 2 let me say that better let's suppose that observable number 1 before making the measurement of the skin suppose observable number 1 when he or she sees an atom coming out of the beam spins a dial and then she orients this apparatus to magnet in a random direction impact and then makes the measurement and if the atom passes on the plus beam it passes on to the second observer so what we might say is that the atom which emerges is polarized in a definite direction impact but it was one that was randomly chosen by the first observer now let's suppose the second observer doesn't have any information about the random direction of the first observer all the second observer can do is make measurements of different observables such as the components of spin what will the second observer get if he or she measures the average value of all three components of spin what do you think they're going to get the answer is zero because the polarization is in a random direction the answer is zero because of the randomness that was introduced by the random direction of polarization but this is physically equivalent to what you get by measuring a beam coming straight from the oven and if we're talking about the spin it's a spin Hilbert space the spin is the only operators there are then there is no measurement that the observer too can make that would distinguish between the thermal beam and this beam which is polarized but in a random direction there's no physical difference between the two so except the physical equivalence of those two situations then we can say that the thermal beam has to be a spin not by the choice of the definite state vector but rather by a random orientation of the state vectors so this leads us into consideration of statistical ensembles of state vectors so let me just say a few things about how the set is now there's two cases to consider a continuous case at a discrete one let's suppose that that state vector is a function of some continuous parameter which could stand for more than one parameter so for example in this bog experiment the bog lambda can stand for the spherical angles if they didn't find angles of orientation of the first polarization so lambda is a function of these parameters and let's suppose these parameters are described by probability density which has to be non-negative and which has to be normalized by the lambda we get one I'll remind you lambda here could stand for theta and five in the direction of n had above so this would give us an example of the statistical ensemble of state vectors another possibility is that it might have to be discrete set of state vectors described by discrete index i instead of a continuous parameter lambda with corresponding probabilities f i which are non-negative and which sum up to one to do that these are two cases a continuous or a discrete distribution of state vectors now I want to emphasize that these vectors of state vectors that I'm talking about are not necessarily orthonormal so for example in a discrete case like this we just have a collection of state vectors with certain probabilities assigned to them I'm not trying to say that there were a model of one of those to give you an example of that go back to the continuous case in this example up here n has 21 direction pointing in another direction those two state vectors are not orthogonal to each other they're just different state vectors that are assigned probabilities now if you have an observable a then we wish to compute the average of a as seen by an experimenter let's work this out in terms of both quantum mechanics and ordinary statistics in the continuous case if we knew the state with side lambda for a particular value of lambda the answer would be given by just taking side lambda and sandwiching around a that's for a known a known state state vector side lambda but since we don't know what lambda is we have to weight this by the probability density in lambda and then integrate over lambda and that is just the the average value as in the seen experimentally or in a discrete case what we get is this we take by some high standing to the around a for a particular state in the ensemble multiplied by the probability and some of the probabilities so in the discrete and continuous cases then this is the result of the measurement of the average value of an operator in these two cases now these two cases can be condensed into a single case by the introduction of the so-called probability density operator the density operator if you find two ways for the continuous and the discrete cases called rho in a continuous case it looks like this the integral of the parameters lambda times the probability density in lambda times the outer product of the of the state in side lambda with itself whereas in the discrete case the sum over i the probability is fi times the state side i outer product with themselves like this it's a weighted sum of projection operators as you see the advantage of this notation so this is the definition of density operator the advantage of this notation this is at least of expectation values here in both cases can be written this way as the trace of rho times a in fact I'll write that as the expectation value and put a box around it because that's the formula for the expectation value of a in the general sense in which the state of the system is only known in a statistical sense alright now the reason this works is because of the properties of the trace which are discussed in this first set of notes but I didn't even go over to your class so let me make a few remarks about the trace as you probably know the trace of a matrix the trace of a matrix is the sum of the diagonal matrix elements this has to be done on an orthonormal basis so if a is an operator and the trace of the operator is taken into the sum again of the diagonal elements a and n or n is a discrete basis and in fact the answer doesn't depend on which basis you choose so you can choose any orthonormal basis to do this the trace has the properties is that if you take the trace of the product of two operators that's the trace in the reverse order or if you take the trace of the product of three operators that's the same as the trace under cyclic permutations CAB BCA et cetera and the trace of the product of any number of operators can be subjected to a cyclic permutation and the trace doesn't change these are simple properties of the trace but I will not prove because you can do this exercise if you haven't done it before a similar property is that if you take a trace of an operator which is an object product let's say alpha and theta times the ordinary operator a like this again there's a kind of cyclic property you just take this ket alpha to the left and you just cycle it over to the right hand side and you get the answer this is equal to theta a alpha like that that's the number that comes out by taking the trace so these are the principal properties of the trace but then this formula follows because you see let's take an individual term right here this term right here this scalar product is the same thing is the trace of the projection operator psi i psi i times the ordinary operator a that's what that is and so pulling out the sum on half and defining where you get this expression for the expectation value but this is an important formula because this gives the this is the general formula in quantum mechanics for computing expectation times alright now okay so now I can tell you the difference between a cure and a mixed stage in particular I can find the pure state and then we're talking about it at this point the pure state is the state in which the statistical mixture only has one state in it if I take the discrete example the discrete case for example let's suppose all the probabilities are zero except for one of them which is equal to 100% if that were the case then the operator would just be the other product of the single state that happens to have 100% probability and in that case you see the trace of global times A is just the usual expectation value this is what you call a pure state on the other hand if there are different states which are that have probabilities that are not equal to 100% probabilities have to have up to 100% but if there's if there's not just a single probability at 100% then you have a general case of mixed state mixed state is mixed state is no f i is equal to 1 in discrete case actually I have to qualify this lightly because there's some solitudes about this definition of a pure state let me put a brackets around this because I'm going to make a qualification of it here's the qualification suppose I have a state let's call it sin1 and I decide I'm going to give this let's suppose I have another state which I'll call e to the i alpha sin1 it's the same state but it's got a different phase I'm assuming these states are normalized by the time and let's say I find that also a 50% probability is that a mixed state or a pure state? physically those two states are the same it should be a pure state shouldn't it because physically they're the same well once what happens is when you condense the operator it's going to be 1 half psi 0, psi 0 coming from the first term the outer product this with itself times the probability and then for the second term the probability is 1 half and I need to take the outer product of this thing with itself but when I do the fixed fact drops out and so what you get is just a single psi 0, psi 0 and in fact it's a pure state so what I actually said here isn't quite right because it's not just that no f i is equal to really to say that it's a pure state to say it's a mixed state to say that it's a mixed state means I'll say it in words in a while to write it out, to say that it's a mixed state means the row cannot be written in this form but this is nice you see because this is what you expect and it tells you the density of the operator is representing the physics and not some phase conventions notice that the phase conventions of the state psi cancel out of the density operator the density operator here's an important point it contains both a complete and minimal description of the physics it contains all the physics and also the minimum description it contains all the physics because every measurement process that you can carry out can be written in the form of the expectation value of some operator so if you know the density operator you know the outcome of all experiments and it contains the minimal description in the following sense too there are arbitrary conventions in it unlike weight functions which have normalization of all kinds of things involved density operators don't and this is reflected in the fact that it's possible to measure the density operator that's to say that if I have a system and I make enough measurements on different observables A to get these numbers out I can use those numbers to determine what the density operator actually is so the density operator is measurable and it's a method to measure the statistical state and what's the statistical quantum state of the ensemble you're dealing with here's a question that may have occurred to you I said a minute ago that if you ask what is the weight function of the spin of the particles coming out of this thermal beam the answer is it doesn't have a weight function it has density operator which we'll calculate when you make the next time if you're dealing with an ordinary accelerator of particles if you've ever done this and you took a course in quantum mechanics you surely asked yourself what is the weight function of the particles coming out then the course is that when the homework problems they'll talk about plane waves that can't be right because plane waves are not normalizable sometimes they talk about weight packets so they make them nice Gaussians are they Gaussian weight packets coming out of the beam well the answer is that the real beams are described that they don't have weight functions they've got density operators so the question is actually meaningless there's some further subtleties of this that I'll tell you about coming up in next lecture but I'll come further subtleties about this what is the density operator for example what is the density operator in this I've erased it now in this example of a thermal beam coming out well it's going to be this it's going to be side impact this is another product of side impact these are linear kind of spinor polarized in the impact direction but then we need to average this over the direction so I'll call it a solenoid beam and this will be the density operator but as far as the chi-inhead is concerned if I say in head it's the same thing as theta and phi to give you the coordinates of this as I explained earlier if we do this integral we get excuse me if we do this outer product here it's equivalent to the outer product is taking equivalent of taking a row after a row applied by a column vector we'll find the matrix in this plus and minus basis looks like this we'll find the matrix in this plus and minus basis looks like this it's cosine squared theta over 2 sine squared theta over 2 there's an e to the i phi here and an e to the minus i phi here which I'm supposed to stop and when we average over d omega what happens is that e to the i phi terms cancels they drop to zero because e to the i phi averages to zero whereas cosine squared and sine squared averages to a half and the result is that the density operator in matrix form is one half of the diagonal from zero to odd diagonals or if you like it's one half times the identity matrix and this is the density operator which is produced by this randomized polarization in this first in measurement that I was talking about a little while ago and accepting the physical equivalence of that with a thermal theme this is also the density operator alright now here's an interesting thing you can do let's suppose the first apparatus where the observer measured spin in the hand direction and then deliver the plus beam to the second observer the thermal beam came in let's change the game instead of polarizing this in a random direction suppose the first observer just polarizes it randomly between plus and minus z and a continuous set of choices but both with 50% probability so randomly half the time it's considered plus z direction and half the time it's in minus z direction and what's the density operator in that case well it's equal to one half times the trajectory of the z plus plus one half times the trajectory of the z minus direction like this but this is the same thing as one half times the identity operated because plus plus plus minus is a resolution of the identity so it's the same density operator that you get if you randomly polarize it in a random direction of the entire sphere there's two different ensembles but they get the same density operator now there's an important point in this which is that the second observer over here can't tell the difference between them they get the same statistics so they get the same results of all possible measurements now this is relevant because 5 times the city this is relevant because when you're dealing with a density operator and you're thinking of it as being weighted sum of pure states and if you grew up thinking about pure states and wave functions which we all do as being somehow not given in primary then you're inclined to think oh well the FIs are just the probabilities I don't know those because I'm missing some information but the real states are the psi i's that's what you're going to think and so if you see a density operator you're going to ask what are the real states the real wave functions and why we have a density operator what this example shows is there's no answer to that there's no unique answer to that because a density operator can be composed in the pure states and weights in more than one way so the reality is the density operator the question about what is the real wave functions in fact is meaningless and this applies also to this question about the wave functions as part of this of your game there's no meaning to the wave functions unless you can design a game so it produces a pure state then there are ways of doing that well I better stop and just mention that the issue of the density operator in quantum mechanics is very similar to polarizations in optics where you have partial polarizations in mathematical formulas okay that's all but before you go I want to say would you please prepare to have a make up lecture next Tuesday night 7 to 8 p.m I'll make an announcement by email what the room will be this will give us three lectures next week because we're not meeting on Monday so it's no worse than a regular week it'll give me ahead of the game we're going to have to make talent which is coming up very soon I'll send that email to you