 Just go over some of that material. It's not really very much to say to finish it off. And then be done with the same amount of that. So in particular, it's fairly easy. It's quite easy actually to find the energy shifts in the case of a weak field. Weak magnetic field. And the form given here, it's a gourmet, a ton of tons of magnetic field, multiplied by some dimensionless numbers. And the hard part of evaluating this is to make yourself a spin operator, as you can call it a spin. These quantum numbers here represent the quantum numbers of an atomic state of an atom with a single electron. So hydrogen or an alkali atom is the idea. This is the usual quantum numbers for that. So it's evaluating this final matrix element that we use in the projection theorem. Now, I think last time I did an all right job of explaining how the projection theorem is derived from Dirac's identity. I didn't prove the identity, but if you believe it, then it's fairly easy to derive this result here from the projection theorem. This is written in general terminology in which gamma, J, and M are elements of a standard angular minimum basis for an arbitrary system where J is a total angular momentum. And then, so we have to change the notation to identify whether the quantum numbers of the same above are specific to the same artifact. But what I'd like to do now, something I didn't have time for last Wednesday, is to explain why it's called the projection theorem, because that also helps you understand the theorem you need to remember in case you're on the decimal line. There's rather some time it can't remember it. So it works like this. The first thing to notice about the projection theorem is it's actually a fairly special result in the way I presented it, because it's not an arbitrary care of standard angular momentum basis factors in the two sides. Rather, it's required that the J value is in the two sides to be equal. However, as far as the gamma and the M's are concerned, those are allowed to be different. So, as I say, it's rather a special result. There is a generalization of this theorem that allows also the J value to be distinct, but we won't need it for this course, and it's more complicated than so I won't go into it. You can find it and cut it in shortly, if you want to look in that book. In any case, the J then is the same on both sides, and the other quantum numbers are different. So here's a classical interpretation of this that helps you remember it, and also helps you understand why it's called the projection theorem. The operator A, vector operator A, and the other one is J, as if they were just classical vectors. So A is just a vector that sticks off from like this in some direction, and J is another vector that goes off from another direction. Now, if you wanted to project A on the J, what you usually do is you've got the perpendicular down like this, and let's call that projection, let's call it A parallel, meaning the component of A is in the same direction as J. Geometry, A parallel is the same thing as A dotted into J, multiplied times J, divided by J squared. J over J squared, J over negative J is just the unit vector in this direction, and all you're doing is taking the component of that unit vector and then multiplying by the unit vector itself. That's just what this means. So this is just a simple vector geometry to see this. Now, one mechanical projection theorem says that the matrix elements of the original vector between angular momentum states where the J value is the same on both sides, so it's a subspace that fixes angular momentum is the same thing as the matrix elements of the projection. The only thing you have to remember is that the classical J squared that appears here needs to be replaced by J times J plus 1, which is the usual quantum mechanical eigenvalue of the operator of J squared. And then you see this is what you get at the right hand side. Often, it's much easier to find matrix elements of J than it is of the original vector than this usual trick you can do to take care of the dot product. So, just as an example for the S and Z here, what this says is that what the projection theorem says, if I apply it, is that this matrix element, and now what we need to do is to identify the gamma here, and if we get an L, the J is the same as the J here and the N is the same as the NJ. It's a change of notation. So, by the rejection theorem, this is the same thing as NLJ MJ on one side, and then we have S divided into J, and then we've got JZ. I've taken the Z component of this general formula and identified A with a spin, that's what I'm doing. And then NLJ MJ on the other side, and then this is divided by J times J plus 1. In this particular used application of projection theorem, the NL of one side is the same as the NL on the other side. Although the formula of NLJ again, we have a problem with a lot of these things. It's the same thing about the Emsen imprints. Emsen imprints, by the way, are the same in this formula here because the JZ commutes with the perturbation in Hamiltonian so it's diagonally in Emsen-J's interpretation. But once you've got it to this stage, then the JZ part is easy because that just brings out an MJ that's acting like a eigenstate in itself. So as far as S dot J is concerned, you use a trick. You write the orbital angle when L is equal to J minus S like this. But then you square both sides and you get L squared is equal to J squared plus S squared minus twice S dotted into J. So you concoct it so that you get a particular dot product you're interested in. And then so solving for S dot J is 1 half J squared plus S squared minus L squared. And so this S dot J can be written this way because it says this operator 1 half J squared plus S squared minus L squared. However, this operator is now sandwiched between states which are eigenstates of J squared L squared and S squared is a constant operator. So it just turns into a number which is the same thing as 1 half of J times J plus 1 plus S times S plus 1 minus L times L plus 1. And S times S plus 1 turns the same thing as 3 quarters because of the force of the spin of the electron so it's 1 half times 3 halves. So in any case, you now just put the algebra together and you get the landed G factor. This is the outline of the of the application of the projection theorem for these weak fields that I like back. Sorry, can I ask this question? Yes, sorry. I've lost track of the overarching theme from one lecture to the next. So I have no... Yeah, I'm sorry. Maybe you wouldn't have remembered it if I'd come on a Friday. No, it's just that so I have no recollection of like a business student. Yeah, so this isn't the end of the same artifact but if you look at the notes on the same artifact this is like 9 to the end of the notes. Now I think it's actually all explained in the notes but this is the weak field same artifact which in some sense is the most difficult of them all. And basically you have the energy the eigenstates are degenerate. They're basically they're just characterized by the angular momentum value or the mj is what the degeneracy is it's 2j plus 1j. And so this is an example of the degenerate perturbation theory but as it always seems to happen to get by with not the degenerate perturbation theory because the matrix is a this case a styrofoam of instant j. So I'm indeed I skipped steps leading up to this but those were done in the last Wednesday's lecture. Right. We'll take a look at the notes because you'll see it's I think that part of the notes explains it. I don't recall that but then now we're saying okay so now we have somehow from that we have like a theorem that you generate the variables and things more. Yes. In fact I'll show you in just a moment if we have time. I'll show you several more applications of the projection theorem but the this matrix element is difficult to evaluate because the j and j involve Clepschordian coefficients combining ordinal and spin random momentum and s and z only refers to one part of that. So if I were to expand the amount of Clepschordian coefficients I'd get a linear combination of terms for each one of these and then indeed it would be the easy to let s see act on them because each term would be an eigenstate of s and c but then you get a whole bunch of terms of Clepschordian coefficients multiplied by things so that's a lot of algebra to do it. So this is actually much easier so do this way this allows you to transfer the z component from s over to j which helps because you have eigenstates of j and z. Okay. Now, so if you got that I'm going to cover it up because I want to study the new topic. The new topic is going to use the projection theorem in a fair amount. I want to turn to the subject of the deuteron and do just a little bit of nuclear physics. It's really quite interesting. So to begin with, the deuteron is a balanced state of the proton of the neutron if I color the neutron dark it looks like this so it's a proton of the neutron each of these, of course, is a spin on half particle and the deuteron is a balanced state of that. Now, the proton and neutron are held together by the strong forces and so let me say something first without the strong forces and how they contrast with the electromagnetic forces. One of the main differences is that the electromagnetic forces are considered long range that's because the Coulomb potential falls off as whatever are it doesn't have any natural scale in the sense the scale is infinity and it's also considered a rather slow slow decay of the force. And even the nuclear forces are roughly speaking fall off exponentially. I have to say roughly this is only a crude statement because it's the idea of the potential falls off exponentially and in fact as we'll see the interaction between the proton and neutron is only roughly described by potential if you push the accuracy it's not even a potential. Unless it's useful to think that way about them. So it has an exponential decay and therefore a scale length which is the scale length of the decay. What is that scale length of the nuclear forces? It's roughly about one Fermi. One Fermi is defined to be 10 to the minus 13 centimeters and the so anything involving the range of the nuclear forces comes out as a small multiple of one Fermi. Notice that this is about 10 to the fifth time smaller than the size of the matter and so the nucleus is of that relative size its volume of the force is that Q so it's 10 to the minus 15 which is a really very small volume in the atom. Fortunately the higher density the density is 10 to the 15th higher for the density of the atom which is approximately the density of ordinary matter. Nuclear matter is much much more dense than ordinary matter. So that's actually what the forces tell. For example, just to give you an example of what this means. By studying experiments that have studied electron scattering of protons to probe the internal structure of protons it determines what's called a form factor and allows you to get information about the charge distribution inside the proton. When this comes out with an RMS charge radius for the proton it's something like about 0 for the Fermi. The RMS value so the sense of charge continues out beyond that out to something above 1 Fermi but it gives you an idea of the order of magnitude. So those are some facts about the strong forces. Here's some more facts. The strong forces are strong. That means they're stronger than the electromagnetic forces. One can see this for example in the case of the alpha particle which of course is a nucleus of helium that has two protons and two electrons. That's two neutrons. Neutrons are neutral but the protons are positive so they repel each other by the Coulomb force. Now the electric force that goes as the inverse square of the distance and since this distance is 10 to the fifth times smaller in the size of an atom it means the force which repels pushing those two protons apart is something like 10 to the tenth times as large as the attractive force of the electron and the hydrogen atom. It's really quite a strong force. It just remains as a short distance and nevertheless the alpha particle is a valid state and it is so because despite that strong repulsion the nuclear forces are even stronger and hold the proton and the neutrons together than the alpha particle. So this is the sense in which they're considered strong. Alright. Now specifically about the oh yes one more thing I should say is that the strong forces as it turns out are actually independent of the distinction between the proton and the neutron is sometimes referred to as this charge state which is a neutral one and the other one is positive but the strong forces between these two particles are in fact independent of the charge state. So the force between two protons is the same as the force between a proton and a neutron the strong forces don't care about the difference between those two particles. Of course the electromagnetic forces do and in fact the analysis of such a phenomenon electromagnetic forces are often taken as a small perturbation on top of the strong forces. Alright. Anyway the quality of the nuclear forces between protons and neutrons is the beginning of isospin symmetry which I won't have time to go into when it's a basic a structural fact about nuclear business. Alright. Now to turn to the deuterons specifically let me give you some experimental facts about it since it's a balanced state of the two we can picture as the two kind of stuff together like this it obviously has a charge of proton which is one half heavy isotope of hydrogen the we can guess that the average separation between it is on the border of a faring as it turns out it's actually somewhat larger than that because it's only in the classical region. There's some more experimental facts it turns out that the spin of the deuteron which is normally denoted, the operator is normally denoted by I the spin of the deuteron is actually physically speaking it's the sum of the spins of the proton and the neutron plus the orbital angle mentioned between them by the way the masses of the proton and the neutron are almost equal so what you have is a picture of the classical of the two nearly equal mass objects rotating around each other like a dumbbell turning about its common center of mass which is halfway between the two the two particles so in any case in general there's both orbital angle and the spin and we can write the total angle of the deuteron in this way as L plus let's call it S1 plus S2 or 1 and 2 refer to the proton of the neutron this is the spin operator let's denote the eigenvalue around the quantum number of the operator I squared by I so that the value of I squared is taken as I times I plus 1 now it's an experimental result that the spin of the deuteron is 1 as to say that the quantum number I is equal to 1 but the operator I squared goes into 1 times 2 which is 2 it has a numerical value of 2 alright this means this means so if the numerical the value of the total spin is equal to 2 we can now use the rules for addition of angular momentum to get some idea on the restrictions of the quantum numbers for L squared S1 squared and S2 squared by the way let's do this let's write S to the total spin is equal to S1 plus S2 it's the total spin of the proton plus the neutron so that we have I is just equal to L plus S like this in simple equation so the total values of the quantum number S I'll call it lower case S I'll probably make it small because the lower case S can either equal 0 or 1 because you're combining it half of the half the 0 is the singlet state the 0 is the singlet state and the 1 is the triplet state if you count the number of magnetic substates this is in concerns of the total spin so the total spin quantum is therefore either 0 or 1 and then L takes on the usual orbital angular momentum values so if I make a table here of the different allowed L and S values and the one I've read is quantum numbers of all times S and first of all S could be equal to 0 just based on the rules for addition of angular momentum S could be 0 and if S is 0 then we must have L equals 1 in order to get a value of I which is equal to 1 that's the only way you could get 1 out of the spin 0 however if the spin were equal to 1 if you're in the triplet state then the orbital angular momentum could either be 0, 1 or 2 in so far as the rules for addition of angular momentum are concerned because that's what you have to need in order to reach the final value of 1 for the total angular momentum so this is a table of possible L and S values based purely on the rules for addition of angular momentum see in a moment these get restricted by further considerations and it ends up there's a fewer set of possibilities in the other table here's that so this is as a result of the consequence of the experimental value of the spin of the neuron here's some other experimental facts the ground state of the neuron has an energy of about minus 2.2 mV this is actually a rather small binding energy for a nuclei that's considerably more energy to remove a nucleon from an alpha particle so in a sense from the standpoint of nuclear physics the neuron is trying to weaken bound another experimental fact is that this is the only bound state that the neuron has it has no excited states if you try to no bound excited states if you try to excite it all you do is break it apart and it turns into an unbound proton and neutron which is a negative state for the system so unlike hydrogen because of infinite number of bound states of the neuron the most simple nuclear problem has over one alright and finally here's another experimental fact it has the neuron has a g factor g sub e which is equal to 0.857 determined nicely by magnetic resonance experiments where high accuracy is possible what this means is that the magnetic moment operator for the neuron is equal to g sub e times the nuclear magneton times the total spin operator total angular momentum operator for the neuron divided by h bar whereas the nuclear magneton as usual is equal to e h bar divided by twice the chrome time mass times the speed of light this is the standard unit for magnetic moments of nuclei okay so what I'd like to do now is to take these experimental facts and then try to gather as much information as we can about the wave function of the neuron in the quantum mechanics using the rules of quantum mechanics alright so in the first place it would be a simple guess that the interaction between the proton and the neutron is described by a potential that would have to be a central force potential in order for the system to be vocational to vary it and if that's so we'd have a simple central force Hamiltonian e squared over 2m plus potential v of r this is really a Hamiltonian for the relative motion so the linear is really nearly as the reduced mass of the proton neutron system but anyway let's take this as a guess as a starting point for what the interaction might be we don't know what the potential of v of r is let's see what some of the consequences are well here's one of the consequences if you look at the radial wave equation as you know the potential the effective potential is the sum of the true plus the centrifugal the true potential must be some kind of a well in order to do the bound states so let's just sketch a well here the centrifugal potential depends on the angular momentum but it's always positive it's L times L plus 1 divided by 2 mu r squared so it's always a positive potential that goes plus infinity as r goes to 0 if L is equal to 0 the centrifugal potential vanishes and then you just have the true potential that's all but in any case you can see that when you add the true potential to the centrifugal potential we've got a curve that comes down like this and it's made up below 0 and then it's been an asymptote to 0 like so create a well like this in that well it's going to be some bound state that looks like that, some energy but the fact that the centrifugal potential is an increasing function of L means that the absolute ground state of the system is always going to be an S-way it's going to be one force L a k 0 this is true for a central force Hamiltonian of course we know it's an S-way for hydrogen but you see the general argument applies for a central force Hamilton and since a deuteron has only one bound state this is the ground state that ground state must be an S-way that follows inevitably from this assumption of a central force Hamilton actually we'll see in a minute that this is really not adamant to describe the deuteron there's more to it but if we accepted this then it must follow that we have an S-way and so what we conclude is that L must be equal to 0 but if L is equal to 0 then this pin must be 1 and the total angle of member has to come from just the spin part because the L is not contributing since I is equal to 1 and S must be equal to 1 and we're in a spin triplet state that's what that means so in particular we need a wave function side to side to look like this it would be a radial wave function that's called r of r and then a y 0 0 of omega because that's L equal to 0 and then what you have is the total spin where I could operate this S-square with an S and C you'd have part of numbers 1 and m sub s we look like this the m sub s is the same thing here as m sub i since L is 0 and this gives us a state of total total angle of member 1 which is the same spin part now the next question I'd like to turn it to most of what I'll be talking about in the next 45 minutes is magnetic moments so let me turn to the magnetic question of magnetic moments the magnetic moments of the proton and neutron can be measured separately as separate particles and for the proton the answer is g is g factor is 5.588 and the g factor for the neutron is minus 3.823 negative this means that the magnetic moment for the proton is equal to the g proton times the nuclear magneton times the s1 over h bar over s1 I'm going to start setting h bar equals to 1 so I'll just put an s1 here s1 refers to the proton and likewise the magnetic moment of the neutron is the g factor of the neutron times the nuclear magneton times s2 yes the question is can we understand the experimentally measured value of the g factor of the deuteron 0.857 in terms of the g factors of the proton and the neutron to make it up this is the magnetic moment of the deuteron coming purely from the combined magnetic moments of the two constituent particles so how can we analyze that? well if we have our proton and neutron in orbit around each other and we put the system in an external magnetic field and there's going to be a delta H perturbing Hamiltonian involved in the field and it's going to be this it's going to have two terms it's going to be minus the mu dot v but one from the one from the proton one from the neutron right here going into minus mu sub n this is probably the factor of mu n and then there's g proton s1 plus g neutron s2 dotted in the magnetic field v or if we put the magnetic field in the z direction then it becomes the magnitude of v times mu n which of course is an energy and then we get gp times s1z plus gn times s2z and now if we wish to find the the the energy shifts in the magnetic field these will be the perturbing Hamiltonian sandwiched around the eigenstates of the Deuteron, the ground state eigenstates well those ground state eigenstates under the conclusion we reached earlier they've got a fixed radial wave function in any given one and the only thing it changes is the s squared sc and s squared by the way is the same thing as i here because l is equal to zero I think I'll call these ground state wave functions i in sub i but i would say that i is the same thing as i would say the total s so this is really the same thing as s, s and z referring to total spin so this is now you see i is just equal to s1 plus s2 since l is equal to zero so what is this matrix settlement you see it's going to be minus the use of nuclear random times the magnetic field and then we've got i mi on one side and then gp times s1z plus gn times s2z times i mi on the other side so we need to evaluate this matrix now this is straightforward to evaluate by spanning up the i mi in terms of classical equations and then just view force charging through it but it's easier to use the projection theorem so let me do that for you let's focus on the first of all the matrix settlement of s1z which is the z component of this home time so let's look at this matrix i mi sam from s1z now by the projection theorem this is used to i mi sandwiched around the total operator i dotted into s s1 times i i cz divided by i times i plus 1 using the projection theorem the i z part is now easy because it just becomes mi brings out the magnetic field of number the ground state of the deuteron is 3-fold in general because of the magnetic quantum numbers as far as the i di s1 is concerned here's what we do let's take the i there this formula applies in the case of an L is equal to 0 let's solve for s2 so we get s2 is equal to i minus s1 and if we square both sides we get s2 squared is i squared plus s1 squared minus twice i di and thereby you see we get the dot product v1 up here that's what s2 squared and the squares and the spin operators so the protons and the neutrons those are constant that both of them are 3-quarters because it's a spin focused spin those two terms cancel and what we get to something is that i dotted into s1 is the same thing as the one half of i squared and so this thing goes over into one half of i squared which because it's acting like an i-space of i squared turns out to be one half of i times i plus 1 and the ii plus 1 cancels the denominator and the overall result of this whole thing is it's just become one half of i so that's the expectation of how you get s1z and by very similar analysis with s2z you get a good factor and a half there as well and so the result is is that this becomes minus-neutral and unique times one half of gp plus gn times s of i and what this shows you is is that the g-factor of the neuron if it is indeed coming purely from the spins and the protons and the neutrons it would just be the average of the g-factor for the proton and the neutron so where are my numbers here's the numbers for the proton and the neutron if you do them you add them up and divide by 2 here's what you get and you get 0.883 which as you see is close to the experimental value 0.857 but it's too large it's off by certainly no and so what this indicates is that the this conclusion follows inevitably from the assumption that we have a simple force Hamiltonian because that led to the conclusion that the neuron state is an s-wave and then once you've got that then the magnetic moment has to come to just the spins and it turns into that value rather than the experimental value so this calls in the question whether this is the correct Hamiltonian in fact it turns out it's not there's more to it it's not just the simple force Hamiltonian by the way there's another wide of argument that shows you that it cannot be just a simple force Hamiltonian and here's the reason it's because there's no spin dependent terms in this system it's like dealing with hydrogen but leaving out the spin orbit terms so if we leave out the spin orbit terms so we just have a simple force Hamiltonian like this then the energy does not depend on a spin state but we know that the ground state to do wrong is a spin triple s equals it is if it's an s-wave the fact that these two numbers are close together means that it's not too bad to make this assumption but it's not perfect and it means that it's a crude model we can indeed think of the way function as being an s-wave but there's more to it than that so based on this crude model we can say that it's mostly an s-wave but with some kinds of corrections well if it was purely an s-wave and if there's no spin terms of the Hamiltonian the triplet state and the singlet state spin states would have the same energy this is just like in hydrogen if you ignore the spin-spin interaction between the proton and the electron in the spin-triple and spin-singlet state would have the same energy in fact in hydrogen they very nearly do this is called a hyperfine splitting it's a very small splitting it comes from the orientation of the proton and electron spins it's responsible for the this will be my first lecture next semester but this is responsible for a splitting of the ground state this is the 21 centimeter line and so important in astrophysics but now here it's more of the 21 centimeters which is actually quite a small energy in the scale of hydrogen energies because what we can see is that by flipping the spin by going to spin singlet state it must raise the energy by at least 2.2 MeV because otherwise there would be another bound state that would spin singlet of the ground state so in fact if there really were no spin orbit turns or any spin dependent turns then you'd have a degeneracy here because of the ground state because of the 2-spost spin values ok so this is all a bunch of indications that this simple potential is not completely added it's a second force model for taking the new one it's not completely accurate now so what do we do what I do now is I'll tell you about I'll tell you about another experimental fact which appeared in 1939 and this is the fact that the denom possesses a quadruple moment like a quadruple moment let me remind you of the quadruple moment tensor you have a single particle it's the coordinates of that tensile integral the coordinates of that particle let me call it x i x j the coordinates minus double i j r squared in the case of the Deuteron there's only the neutron doesn't contribute to the quadruple moment because it's neutral so these coordinates are really the proton coordinates let me remind you also that there's a dipole operator this is for the single charge particle when we were looking at the start we considered the question whether energy eigenstates could have a parallel electric dipole moment let's talk about dipole moments for simplicity so in size of the energy eigenstates the position operator are multiplied by the electron if you're talking about a single electron pattern this is the average of an electric dipole moment operator and as we saw this is all equal to zero because energy eigenstates are eigenstates of parity and the position operator is odd of a parity now however if I go from a dipole operator to a quadruple moment operator this story changes because this is quadratic in the position of that force even in the parity and so the quadruple moment operator is a parity and so there's no parity in conservation between exclude electric dipole moments of energy eigenstates now here's another issue which arises is as I mentioned the deuteron has a i equal to 1 as the total angle of the system that means that the ground state is a irreplaceable subspace under rotations of dimension 3 and negative one number still 0 minus 1 plus 1 and it means that on that subspace the ground state subspace in the deuteron system it means that the possible irreplaceable tensor operators or if I write them as tkq they're a useful notation they allow values of k and only be 0 1 and 2 that comes from adding a cross of 1 cross 1 which is a 0 plus 1 plus 2 and the one here goes from the state of the deuteron so this is part of the homework problem the maximum k that is allowed is going to give you a subspace like this it's actually twice the state so in particular your useful tensor operators are going to make this too far allow any spin limit a spin limit system they're not allowing a spin limit f system that means that for example the proton or the triton which is the third, the heaviest oxygen of hydrogen the triton has got 2 neutrons and 1 protons that's kind of like this one step up through the deuteron this actually has a spin of 0.5 not spin 1 SPI half integer it turns out to be 1 half and since it's 1 half the maximum irreplaceable tensor operator maximum regular irreplaceable tensor operator for the triton system is 1 the triton kind of can and does have a magnetic moment but it does not have an electric quadruple moment deuteron however is allowed to have an electric quadruple moment and it was discovered experimentally by Robbie in his room in the late 1930s that in fact it does they found this by looking at the energy levels and saw that it was necessary to have have a quadruple term included alright now given the fact then that the deuteron has an electric quadruple moment we can say about the wave function what this means is if I take the wave function whatever it is the sandwich is one of the quadruple moment operators we have to get another 0 answer well the quadruple moment operator is a T2Q operator as we know P equals Q and if psi were purely an S wave so that this had L equals 0 also it's a purely orbital operator it's the only involved orbital degrees of freedom then by the figure of that card there we see the matrix so it has to be 0 this has to be equal to 0 in an S wave because you can't probably get 0 with 2 you get an answer with 0 with 0 see 2 cross with 0 just equals 2 so the answer the answer is 0 therefore in order to get a non-vanishing quadruple moment it's necessary to have a higher values of L and so the ground state wave function of the neuron must include not only L equals 0 but also higher values of L so let's do this let's take the ground state let me call it psi as the ground state of the neuron let me write it as an expansion over L L equals 0 to infinity of some coefficients I'll call A L and then psi is L the idea here is that we take this state psi we project it onto the subspaces of angular momentum L and so like this you project it onto the subspace like this and psi L is a normalize factor inside that subspace and the coefficient gives us the magnitude of the projection so this is a really understanding of psi L psi L is equal to 1 then the ace of Ls are the expansion coefficients by probability conservation by the sum of L A L adds to the value of square less to be equal to 1 now we turns out we can get some restrictions on which L values are allowed by considering further symmetry principles let's look at parity it's quite easy to look at if I take parity and allow it to act on the ground state the ground state must be an eigenstate of parity so the answer has to turn out to be plus or minus 1 but it has to be one of those two choices of whole-plated psi on the other hand we take this expansion of the parity the states of the angular momentum L are parity of minus 1 to the L so this now becomes a sum on L of minus 1 to the L times the same coefficient of A L times these projected vectors psi L like this we know that A0 is not equal to 0 and this is our original model which we had central force Hamiltonian led us to the conclusion that it was an S wave and even if it didn't work perfectly it worked pretty well so there must be a strong component of an L equal to an S wave in the dunon wave function even if that's not all there is so in particular the A L equal to 0 term ends up in a plus sign here and the only way that can equal so we find the dunon as a parity plus 1 is even parity if there is something else that happens too if we look at the L equals 1 wave then the minus 1 to the L is a minus 1 and that's not consistent with this expansion unless those terms vanish so what we conclude by a conservation of parity is that A L is equal to 0 as well as I now, boom, so some has to be only over even an S wave and then a D wave and doesn't go on well the answer is it does not go on because we can't have L greater than 2 if we had L equals 4 for example then the maximum spin we can get is 1 and if you combine them together the minimum value of I would be I equals 3 but this is wrong because we know I is equal to 1 and so in fact the sum terminates at L equals 2 so what we conclude and I'll put it right here is that the ground state of the Deuteron must be some coefficient let's call it a 0 times a normalized vector of the L equals 0 subspace plus a set of 2 times a normalized vector of the L equals 2 subspace it's a combination of an S wave and a D wave and that's all now the question is what about these 2 coefficients if you don't have time to finish this I'll just go ahead and get on it I'll show you how by using this magnetic moment data there will be this disagreement here between the simpler model where it's a pure S wave not what we're going to do is improve on this and consider this linear combination of S waves and D waves and see if we can make the magnetic moment by using the experimental which is not going to erase this was the theoretical result on that simple model having a more sophisticated model that includes a D wave to try to determine what the coefficients are turns out you can't get the coefficients themselves but you can get the squares you can get the squares of the whereas the a0 and the a2 of these on the 2 are no longer a mystery all right so let's work on this now so yes it's actually a relative wave function of course it's a semi-nasa wave function too that's true so the we'll do it like this we'll again use magnetic moment information let me remind you that in the same effect in the single electron single electron atom we had a perturbing Hamiltonian we call it delta H for the interaction of the magnetic field I think previously I called it the same I called it Hc which is which was which was this, it was use of dv and then this was multiplied times else is the plus plus s is the the orbital spin angular momentum momentum contributed unequal proportions to the magnetic moment using this there's an effective g-factor of one for the orbital motion and a g-factor of two for the spin as this is too early as this is the g-factor of the electron the electron spin the direct this is for the this is for the case of a hydrogen atom now in the case of the deuteron it's going to look like this it's going to be first of all nuclear magneton times the magnetic field there's also a minus sign which is particles of positive charge and it turns out to be the following in the first place there's an effective g-factor not of one but rather of one half and the reason it's one half instead of one is because the hydrogen and proton is massive so the electron is moving around and the proton is mostly stationary so in the deuteron both protons and neutrons are orbiting around each other and the distance from either particle to the center of mass is one half of the distance between the particles so there's an analysis one can go through in which you transform from the laboratory to the center of mass variables the lack of time that will go through this but when the smoke clears you find that there's a g-factor of one half for the orbital angular momentum in the case of the deuteron and as far as the spin is concerned these are the same things we wrote down earlier it's g proton times s1z plus g neutron times s2z and now what we want to do is to sandwich this this Xamon Hamiltonian between the psi here which refers to the ground state wave function which refers to this linear combination up here to check for the two angular momentum states to be a little more specific about those two angular momentum states psi 0 and psi 2 let me write them out so psi 0 allow me to allow me to write states this way in terms of the following quantum numbers l, s, i and m sub i let's let those be the quantum numbers then psi 0 is a quantum number in which the orbital angular momentum is 0 that's what the 0 here refers to the spin has to equal 1 because that's the only way you can get the total total experimental spin of the deuteron to come out when l is equal to 0 i is equal to 1 that's the value of the spin of the deuteron as far as psi 2 is concerned this is a d-wave so l is equal to 2 the spin is still equal to 1 you can buy 2 with 1 or you can get a normal total value of 1 like this if you prefer to think about this in terms of wave functions the first one is the same wave function I wrote down earlier which is not disappearing this is let's call it r 0 r as a regular wave function times y 0 0 even for the spins what we've got is s where m sub s is equal to m sub i the same thing of the 2 sides and the s is equal to 1 so s and e sub s are the triplet states that are made out of this the 2 spins are the particles added together now then in psi 2 what we have is we've got a different wave function r 2 of r and now we need to use clutch Gordon coefficients to combine orbital and spin angular momentum with orbital 2 and the spin 1 to create a total final argument of y so I'm going to write this this way as the sum on l l and s of y sub 2 l of the sum on angle k to the phi when the spin state will be the spin of 1 and then m sub s so this is equal to the spin and then we have clutch Gordon coefficients good grief a lot of time we have clutch Gordon coefficients which is going to be 2 1 l m s in the final shape i values are i or i here is equal to 1 like this so explicitly in terms of adding orbital and spinning angular momentum these 2 states we're taking probably another half hour we're going to take this all the way through let me just say that when you finally do this you can actually calculate we need to use the projection there about 3 or 4 times and then we're going to evaluate the citizenship of the deuter on in the presence of the external magnetic field including the orbital wave of their effects we're taking to a complex linear combination of 2 different roll values and then you're going to have an equation when you solve the first which you've got 2 squared is equal to 0.956 and you get 2 squared is equal to 0.044 so in general the wave function has about a 4% probability of being in a d-wave about 96% of the wave this is the ok, well work i have a couple of announcements i have here a memory statement that is