 Are you ready? All right. I want to begin today with the subject of electric dipole transitions. Yeah, you could wave until the end of the hour for this. I want to start today with the subject of electric dipole transitions in the prime structure model of hydrogen. I doubt what I'm going to do in the very schematic form of what the energy levels look like, but the notation I explained last time. This is only a model of the prime structure model, and it doesn't get its own physics, but it's certainly more accurate than the electrostatic model. In any case, dipole transitions are governed by the basic sense of position operator, which is basically the same as the dipole operator. So it's position operator r vector, and it's sandwiched between the initial and final states, which in the prime structure model, as one of the numbers nlj and nj, like this, so we're running half these part of numbers on one side, and we want to prime the numbers on one side to indicate an upper state, and then we want to prime them on the lower side to indicate a lower state. All of this is one of the numbers on both sides of the difference. All right. So we used the finger record theorem for this, too. That is an exercise I'm using in determining selection rules by using the finger record theorem. With this, notice the position operator is a t1q operator, as that came with one irreversible tensor operator. This is in Cartesian form, and I won't worry about the q, but let's just concentrate on the k value. The position operator is, of course, purely spatial operator, and that means that the vector operator not only under rotations, but purely spatial rotations, or orbital rotations, which are generated by l, but also under total rotations of both order ones, spin degrees, which are generated by j. In fact, you can use the finger record theorem here twice. If you use it with l, there's a selection rule with l quantum numbers. It says that the delta l, which is l prime minus l is the difference between the initial and final ls here, is either equal to zero or plus or minus one. This is what comes from the finger record theorem, and it's just the addition. The rules of addition may, they remember the l here, combined with the one here, has to be able to get rid of each l prime from that. However, if you take a look at Perry, that Perry excludes the value delta l equals zero, and that goes out. And then we end up with the bifold selection rules, the delta l is equal to plus or minus one. These are the same rules we had when we were talking about the microstatic rule, no change. Now, although we also have rotations generated by j, these are overall rotations. And we've got j quantum numbers on the two sides. And again, this is an irreducible tensor operator with this batch of total rotations. So we get another selection rule that says that delta j is equal to j prime minus j is either equal to zero or plus or minus one. This is again coming from the finger record theorem. In the case of L, the finger record theorem plus Perry, Perry excludes the delta l equals zero. Well, Perry has no effect on delta j equals zero. Perry is not a function of j, it's a function of l. And the result is delta j equals zero, which is allowed by the vector record theorem, actually is allowed and can't happen. So this is a slight difference to the L and the j rules. Anyway, if we apply this, what we find is that, for example, in the 2p to the 1s transitions, there's actually two of them you can take place. You can go like this, and so the 2p to the 1s transition in hydrogen actually appears as a doublet. The same thing happens by the way in the 3p to 3s transition in sodium, that's the doublet line of sodium, the doublet of sodium b line, split by structure. Similarly, the 3p to 2s can go like this. The 3s to the 2p can go like this, there's again two lines moving down like this. If we look at the 3d to the 2p, the 3d to the 3s, we can go down to the 2p to the 1s, and we can also go down to the 2p to the 3s. However, the 3d to the 5s is only allowed to go down to the 2p to the 3s because to go to the 2p to the 1s would be delta j of two, and that's not allowed. So this 3d to the 2p transition is actually a triplet and not a driplet line. Anyway, these have been a lot of transitions, and if you improve higher stage, you can fill in a diagram like this. Okay, so that's an example of application of the Vingram-Hackard theorem for dipole transition slideshow rules. Alright, now, I'd like to go to the board and start a new topic. The new topic I'd like to address is the Zeeman Effect, if you can hear me well, I'll finish today. The Zeeman Effect, as you know, I'm sure you know, involves the response of an atom to an external magnetic field. And we'll be considering here valine hydrogen, but also some tension given to alkalis. These are both examples of single electron atoms, which is mainly what we're hoping to look at. Before we begin, I want to say something about units and what is the magnitude. It's continued in this calculation to use atomic units. In fact, atomic units are used a great deal in atomic physics. Atomic units takes the three physical constants which occur in the Schrodinger equation from the hydrogen atom, and this sets them equal to one. This is the electron charge, the electron mass, and the H bar are also equal to one. You can't set any more physical constants equal to one, because this takes care of everything. There's only three independent ones you can use. And this is the most you can set equal to one. The basic idea of atomic units is that it gives you an orderly units of charge, mass, etc. And also units of distance, time, and energy momentum, all the physical quantities. And what these units are is that they are the basic quantities which appear in the ground state of hydrogen. So for example, the atomic unit of velocity is the velocity of the electron in the ground state of hydrogen, which is alpha times the speed of light, 1 over 137 C. Give us some idea what goes on with atomic units if we take the fine structure constant, which is E squared over H bar C, which is 1 over 137. This is dimensionless, and so it has the same value in all the units. But if we restrict it to atomic units, then the E squared and H bar go away, and what's left over is just 4 over C. And so what we see is the C in the speed of light is equal to 137 in atomic units. That's its numerical value. This corresponds to the fact that V equals 1, the velocity of the electron in the ground state of the hydrogen atom is alpha times the speed of light. So the speed of light is 1 over alpha. This is what you get. Another thing to look at is the Bohr magneton. UB is equal to EH bar in the 2MC. And if we go to atomic units, the E, the H bar in the H bar go away, and it just turns into 1 over 2C, which is the view of the resultist above. This is sending this alpha over 2. That's the Bohr magneton. The plus and magnetic moment of the electron, U, which is equal to minus the chief factor of the electron, times the Bohr magneton, UB, times the spin divided by H bar. This puts it in its dimensions, right? The dimension comes from UB. Well, if we put this in atomic units, the H bar goes away, and UB becomes alpha over 2, and the GE, if we approximate this and we try to do this upside down, is equal to 2, a good approximation. Then it accounts as a 2 here. And the result is this becomes just simply minus alpha times the spin. For the magnetic moment. And thus, finally, the energy of interaction with the magnetic moment, where the external magnetic field, which is minus mu dot B, becomes plus alpha times V dotted into the spin at S. Okay. So this is some basics of atomic units. Now, in addition to atomic units for distance velocity and so on, we also need atomic units for magnetic field, because we're going to be doing the same line effect. Let's call the field B naught, which is equal to one atomic unit. And ask, what is B naught in terms of Gauss? Well, the answer is obtained by finding the unique magnetic field that can be constructed out of these three physical constants. By dimensional analysis, you can just work it out. Actually, since the magnetic field, the Gaussian units, as the same dimension as an electric field, it's also the same thing as E naught, measured in statinoles per centimeter. And the atomic unit of electric field is easy to see what that is. It's the electric field at the radius of the boron radius and the hydrogen atom. And so it's the same thing as the charge of the proton divided by the boron radius squared. And if you put that all together in the basic units, this becomes the mass squared times e to the fifth power divided by h bar two to the fourth, I believe. And then converting that into Gauss turns into 1.7 times by 10 to the 7th Gauss. So this is the unit of magnetic field in the Gaussian atomic units. The ordinary units, that's that much. Now, by way of reference, the strongest magnetic fields that can be created in laboratories in normal circumstances are created by using superconducting magnets. They need to get up to about 10 Tesla, which is, roughly speaking, which is the same as about 10 to the fifth Gauss. So by comparison, the largest laboratory magnetic field is about 10 to the fifth Gauss. And that means that if we're dealing with laboratory magnetic fields and then V over P naught, which is the same thing as the magnetic field measured in atomic units, is less than or comparable to something like 10 to the minus two. It's a small number. Coincidentally, that's also about the same thing as the fine structure constant alpha. So the atomic units for V zero is equal to 1. The V's that we'll be dealing with are generally small, less than about 1 to the 7th. I'm not going to add that in astronomical situations there are much larger, stronger magnetic fields that arise all the way up to 10 to the 12th or 10 to the 15th if you want to get speculative Gauss. This is in their neutron stars and magnetars that can be stronger fields and magnetars are somewhat speculative. But in any case, because of astrophysical applications of that sort, there's been some interest in recent years in the kind of exotic physics that arises with very strong magnetic fields. However, for this discussion with the same line, in fact, I'm going to assume that our magnetic fields are laboratory fields and so they're limited by this estimate of order of magnitude here. Magnetic fields. Now, the Hamiltonian is, let's do the following let's place our atom, our single electron atom in a magnetic field which is purely in the z direction and from this we can write down a vector potential a simple vector potential of this one is one hand of V cross into R which if you compute its curl you'll find that's equal to the V here the constant V and so if we put this into the Hamiltonian in ordinary units then the Hamiltonian is this it's going to be one over two M and as the momentum is plus U over C times the vector potential which is one half V cross R quantity squared there's a central point potential for the hydrogen atom or the alkali and then I'm also going to put in the fine structure turns because the fine structure turns I'll remind you of the fine structure turns in the case of hydrogen are down by a factor about alpha squared compared to the kinetic energy and the potential energy of the amplitude system as the order of the correction fine structure corrections is basically a view of the C squared correction and we'll have to in a moment we'll have to compare the size of the fine structure turns to the z bar turns which will come out from an expansion of the Hamiltonian and I forgot one more thing which is the interaction of the electrons then with the magnetic field now this is an ordinary unit if I convert this to Gaussian excuse me to convert this into atomic units it becomes one half of P vector the E goes away the C becomes one over alpha so we'll write this plus alpha over two times V cross R that whole thing squared and then the rest of it I've mostly copied V of R plus H fine structure and then for U dot V minus U dot V we work that out over here plus alpha times V of S V dot of V of the spin so this is what it looks like in atomic units allow me to draw your attention at first to the kinetic energy here which I'll call T the whole expression here is the kinetic energy if we expand out the square with three terms let me write this as T1 plus T2 plus T3 just to write out those three terms and let's look at them over here look at the orders of magnitude the T1 is just P squared over two the T2 is equal to there's two from a cross term but there's two twos in the denominator so it turns into alpha over two times P doted into V cross R but by rearranging the triple products so that it becomes R cross P we get orbital angular momentum so it becomes the same thing as alpha over two times P doted into L the orbital angular momentum but as far as T3 is concerned it's just the square of this last term divided by two so that's alpha squared over 8 times V cross R quantity squared if we put V in the Z direction as I was suggesting to do here this becomes the same thing as alpha squared V squared over 8 times X squared plus Y squared it's a term that depends on the position of the electron and in fact it sort of looks like potential, you could throw that potential if you wanted to it's not a second force potential it's got cylindrical symmetry not full rotational symmetry alright, now the question is what is the order of magnitude of these relative order of magnitude of these terms and just speaking in terms of order of magnitude let's look at the ratio of T2 over T1 to just make estimates we're working here in atomic units what that means for example is that if we're near the ground state of the atom but which the quantum numbers are small they're of order unity then it means that the values of quantities such as energy momentum angular momentum are all of order unity because that's what atomic units do so in particular the piece of T in the momentum is of order unity and T1 is just of order unity now what about T2 there's alpha times b times l this of course is the same thing as alpha over 2 the magnitude of b times lc let's write it that way if we assume that lc is of order unity because it's a quantum number then the magnitude is basically given by the product of alpha b so this has this has the ratio of alpha b now alpha is 1 over 137 and this is going to be not much more than 1 percent so this is very roughly something like 10 to the minus 4 the ratio of those two terms this is a small perturbation of momentum here this 10 to the minus 4 is coincidentally about the same as the relative magnitude of the fine structure term compared to that of alpha squared it's also about 10 to the minus 4 now if we take the ratio of T3 over T2 you'll find that this is also the same order of magnitude of alpha b which is about 10 to the minus 4 the result is that of these three terms they're decreasing the magnitude by the factor of 10 to the minus 4 every time you go down T3 is then the smallest term of all and you can't get the flow way on the other hand the T2 which as you see here is over 2 to the b dotted in the lc of the same order of magnitude is the lc v dot s which is of course the same thing as l times b times s is the if we put the magnetic field in the z direction and so the T2 and the mu dot v term in fact are of the same order of magnitude they're the only differences between the s and the l and the factor of 2 so throwing away the term T3 then we can rewrite the Hamiltonian in the following form if the Hamiltonian becomes this it becomes p squared over 2 plus v of r plus h find structure plus a term called h Zeimann h Zeimann is defined this way it's the sum of T2 which is up there involving v times lc and here a v times s e to the v dot b it becomes alpha over 2 times the magnetic field times lc plus twice s to the v like this and in fact since alpha over 2 is the same as the more magneton it's convenient to write it that way it's more magneton times v which is certainly okay because I just made the substitution in atomic units but this is convenient because in ordinary units BVB has dimensions of energy and Hamiltonian does also so if you just make the lc and s e you'll get the right answer anyway it's easy to see this as an energy okay so this is the Hamiltonian we're going to work with now we'll examine this Hamiltonian in several different limits in the first place as I mentioned already really partly by coincidence it has to do with the maximum value of magnetic fields that are obtainable in the laboratories is that the H. Zeeman and H. Klein structure are of the same order of magnitude if you use those magnetic fields 100 kilogalts or something like that so these two are of the same order of magnitude it's a question about how to treat them should we take them together or what by making the magnetic field weak you can of course make the Zeeman in terms small you may make it as small as you want so there is a limit where H. Zeeman is much less than H. Klein structure that's for weak magnetic fields on the other hand if you go to excited states where the quantum numbers suppress the importance of the fine structure but not the importance of H. Z then you can get the other one of H. Klein structure as much greater than H. Z this wouldn't apply at the Brown state unless you went to magnetic fields this H. Z much greater than H. Klein structure as much as possible if you go to excited states this wouldn't be possible in the Brown state unless you went to unrealistically large fields from a laboratory standpoint but still it's a useful limit to consider so let's take this in various cases the first case I want to consider is the second one in which the Zeeman term completely dominates the fine structure term overwhelming and so we'll just set the fine structure term to 0 and ignore it as I say this is unrealistic for laboratory fields but it gives us a useful reference case that we can use later on to your question ok so in that case in case one then will we ignore H. Klein structure let's set up the perturbation problem this way take H0 to be P squared over 2 plus V over R and then take H1 as equal to H Zeeman which we'll write this way as UBB times LZ plus twice S LZ plus twice S now this too which appears here in the spin is really the G factor of the electron and it's a way of indicating that the spin of the electron interacts with the magnetic field in a different proportion to its angular momentum than does the orbital angular momentum when the G factor is 1 so L plus 2S is of course not the same thing as L plus S which is a total angular momentum but the vector that interacts with the magnetic field is not the same as J it's a different vector because of the G factors alright so this is our perturbation problem now we'll take this as an exercise in perturbation theory how to analyze it the other perturbed system is when we were talking hydrogen it would have the 2N squared fold in generality of the unperturbed eigenstates factor 2 comes from the spin if you were talking about alkalis we have 2L plus 2 times 2L plus 1 fold in generality but in the case H0 is degenerate and so we'll have to think about the degenerate perturbation theory and so once again there's a question that we talked about before what is a good basis to choose for doing the unperturbed eigenstates for diagonalizing the matrix of the perturbing Hamiltonian and try to get by about diagonalizing anything by a clever choice of basis of the intelligence so let's make a list of operators as we did before and at the table we'll see which ones of them commute with a perturbing Hamiltonian H1 in order to choose a proper basis there's really only 2 choices of bases that are reasonable here one is the uncoupled basis which is NL Ns of L and Ns of S and the other is the coupled basis NLJ of J this is just the same as what we talked about in the case of the fund structure perturbation in the last lecture but anyway these are the choices so let's make a list of operators here let's start with Z component of L and then L squared and the S component of Z component of S and S squared and the Z component of J and J squared let's make these the operators because these are the operators that occur here in the two sets of two complete sets of commuting observables and we want to look at the commutation relations with H sigma now by the way before I get into that one thing that's fairly obvious is that the full rotational invariance of the NLJ Hamiltonian has been broken by the introduction of the external magnetic field H0 can be used in all rotations H1 however the magnetic field is in the C direction that's why you see LZ and SC here this means that the system still commutes with rotations about the Z axis as you can see the magnetic field is vertical and you've got rotational symmetry it's a two-dimensional rotation SO2 symmetry but the full three-dimensional rotational symmetry has been broken alright so anyway, let's look at commutators so first of all does H z line communicate with LC because LC communicates with itself that it also communicates with SC because LC and SC act on different degrees of freedom so they commute so the answer is yes, I'll put a check here what about L squared well L squared communicates with LZ because that's one of the angular momentum commutation relations and it communicates with SC because they act on different degrees of freedom again so the answer is a check here also when you look at SC and S squared it's the same logic in both of them now what about JC does JC communicate with this geometrically the answer is clear that it does communicate because JC generates rotations in all the variables in order to understand about the Z axis and this time I'm telling you there isn't very new rotations about the Z axis on the other hand if you wonder about how JC is equal to LC plus SC and just look at the commutators here you'll see that they're all zero because LC communicates with itself so anyway it does communicate with JC now what about J squared turns out it does not communicate with JC the reason is if JC is the same thing as L plus S squared as vectors this is L squared plus S squared plus twice L dot S now the L squared and S squared do commute with the same amount of Hamiltonian that's already here at the table but the L dot S doesn't because it involves the X and Y components of L which don't communicate with LC and involves the X and Y components of S which don't communicate with SC so the result is J squared does not communicate so here's our table of commutators now if we ask which of these two bases is going to be the better bases to choose it clearly must be the other couple of bases because L squared, L Z and S and Z are all here they are L squared, L Z and S and Z are all checked here and the other bases wouldn't work so well because it's kind of J and then J squared is not one of the members and so it just tells us then that we should use the uncomplicated basis for doing this perturbation calculation let's talk about hydrogen just to be specific and let's talk about the 1S the 1S to have the the unreturned Hamiltonian doesn't have the fine structure in it so let me just talk about 1S and then we'll have 2S and 2P and the 1S is 2-fold the generator if you count the spin and the 2S is 2-fold the generator and the 2P is 6-fold the generator this is the 2L plus 1 multiplied by 2 for the spin so we're going to go with 2 here and for 8 here for the states so the N equals 1 and N equals 2 levels are both in the generator 2 or 8 and we're going to use then the matrix elements of this perturbing Hamiltonian in the side of some general ligand spaces so here's what we need then we put the same on the middle we're going to have matrix elements of both sides now the N has to be the same on both sides because that identifies the unperturbed eigenstate but the other three indices indicate the basis vectors on the side of the unperturbed eigenstate so in principle they ought to allow them to be different however since Hz can be used with L squared Lz and SNC in fact the whole thing is actually diagonal in those matrix elements we can remove the crimes and so we only need to compute the diagonal matrix elements and once again in general perturbation theory we get away without having to diagonalize the matrices we just calculate the diagonal matrix and those will be the energy shifts so the energy shifts and both of these will be equal to this bank of solid well it's easy to evaluate that Lz just turns into a magnetic Lz and SNC turns into of course magnetic numbers so this becomes UBB times M sub L plus twice M sub S and if we write it this way you see it's actually correct also it's actually what needs to be mentioned so let's look at what this does to the energy levels first of all for the 1S let me take the 1S level let me take the 1S draw it like this the for an S level L is equal to 0 so ML is 0 plus or minus a half so what you get is times 2 you get plus or minus 1 times UBB so under the magnetic field with 1S levels let's say 2 that look like this where the the energies are plus 1 and minus 1 measured in the units of UBB relative to the temperature of energy the upper state if I indicate states by their magnetic quantum numbers ML and MS the upper state here is 0, 1 half and the lower state is 0 and minus 1 half you can see that it's splitting which is logical since there's no overlying momentum now let's go on to the N equals 2 levels here there's a 2S and 2P as far as 2S is concerned it splits in exactly the same way as the 1S for exactly the same reasons so you get a plus 1 and a minus 1 here and this is the quantum numbers 0 and a half and 0 and minus a half again this is the ML and the N equals 2S quantum numbers and they're right in here as far as the 2P state is concerned it's 6-fold degenerate in the underterm system but if you look at the different combinations that you can get by adding up the N and MS you'll find that there are actually 5 of them it pans out like this and they're measured in multiples of UBB they range from minus 2 up to plus 2 and one of the states the maximum one is 1 and MS is a half the next one now it's got ML equals 0 and MS is a half the next one now it's got ML minus 1 and MS is a half but it also has ML is equal to ML is equal to plus 1 and MS is equal to minus a half then down below that we've got 0 and minus a half and then down below that we've got minus 1 and minus a half so you see if you look at the 2P levels alone the 6-fold degeneracy has been resolved into 5 levels the center one is still 2-fold degenerate and all the others are now seamless as far as the 2S level the 2-fold degeneracy is completely resolved in the case of hydrogen however the 2S and 2P levels are degenerate themselves so the plus 1, plus 1, minus 1 and the 2S are actually degenerate plus 1, minus 1 and the 2P level all the degeneracies are one pair between those two one pair and the 2P levels and another for these two so some degeneracy remains after perturbation is turned on now that's the effect of very strong magnetic fields so it's so strong that the that the fine structure can be ignored now next let's take the limit in which in case 2 let's take the limit in which the Xamon term is still much larger than the fine structure term but we don't neglect the fine structure term it's not negligible what we do in this case is we take the previous Hamiltonian H0 plus H1 including the Xamon term as a new unperturbed Hamiltonian in regard to each fine structure is in perturbation so let's do it this way let's take H0 as P squared of 2M plus V of R plus H of Xamon let's take H1 is equal to each fine structure to a perturbation analysis of this form this will give us perturbations that are superimposed on top of the interchangeable diagram that's up in that quarter to 4 all right now the fine structure term consists of three terms, you know, all of the mystic kinetic energy garland and dust in orbit and they're all of the same order of magnitude so that you can realize that you have to include all of them however, for simplicity I'm just going to look at the the spin order term alone just to shape writing so even though this is not realistic this is an exercise to do let's do this perturbation calculation now in this case as you can see from those diagrams up there if we look at N equals two levels anyway there still remains some degeneracy so we have to think about doing degenerate perturbation theory however, there's no longer any choice of basis because we had to use the uncoupled basis to get the eigenstates of H0 so the uncoupled basis is all there is so now what we need to do then is to consider the matrix elements of the spin orbit Hamiltonian and the uncoupled basis the spin orbit Hamiltonian if we write it in atomic units in case of hydrogen well, for the same electron that we write it in atomic units it looks like this and I write it down for you somewhere I do write it down but I remember it's this it's alpha squared over two and then it's one of R, P, D, R where it gives the potential and then it's L dot S I mean the only question is the constants in the front but in atomic units they turn into alpha squared over two showing you quite clearly that the final structure terms are of order alpha squared compared to the under term Hamiltonian so we now need to take this in the sandwich of between vectors of the uncoupled basis so in other words in L in L and in S H spin orbit in the middle and then in L and in S now I might have to put some primes on some of these that is to say to indicate distinct vectors inside a degenerate hiding space to take the case of hydrogen there's actually let's take the N equals two levels just for example as I mentioned there's a two-fold degeneracy here you can see the M, L and M S are distinct there's a two-fold degeneracy between there and there the L values are distinct and between the lower the lower level here and the minus one level there is also the degeneracy there's actually three-two-fold degeneracies here and so it looks like what we need to do is to let the N and L and M S be different on the two sides this will give us a matrix a set of three two matrices to generate hiding spaces up there well however right away we can see the spin orbit Hamiltonian commutes with L squared that's because L squared commutes with L there it also commutes with S so the spin orbit Hamiltonian is rotationally invariant along the rotation so in fact it commutes with all functions of orbital angular momentum including L squared so that means in particular that this HSO is going to be diagonal in the L quantum number and I can erase the primes there now right away what that means is that there is no couple between these two degeneracies here or those two up there the S and P states are decoupled what about M L and M S? well this is L, Z and S in C and the L, Z and S in C I don't obviously commute with L, G, S and so I'll leave them as primes here for the moment but the only two by two degeneracy that remains is this one right here in which the M L and M S have these values so M L is either plus or minus along the M S is plus or minus half so now at this point what I'll do is use an identity it says that L dot S if you write it in terms of the operators we like it has this form it's L, Z, S of Z is the Z component the X and Y components you can write it this way as one half of L plus S minus plus L minus S plus so let's consider the off-diagonal matrix elements inside this two by two two-dimensional subspace here you can see that the L the L values are plus or minus one so that delta L is actually equal to two of those off-diagonal elements now L, Z here is an operator that has I'm sorry it's not delta L it's called delta M sub L it's the magnetic one that's what this minus one of the one is delta M sub L is equal to two so the L, Z is an operator that has delta M sub L equals zero it doesn't change the M L value whereas L plus and L minus change the M L by plus or minus one and so in either case do we get a delta M L plus or minus two which is needed to connect them to those two states and thus the result is the magnetic elements inside this degenerative space here and vanish and we will again end up with a diagonal matrix that means that again we just brought the primes here and just consider the diagonal matrix elements now if we do this then only the L, Z, S, C term will contribute because as I said the others if you have a delta M L plus or minus one you can also get two such more people and so this now just becomes delta S is concerned it can be replaced by L, C, S, and C and this will of course bring out a factor of M sub L times M sub S and then all that's left over is the expectation value of one over R dv dr and so this whole thing turns into alpha squared over two times the expectation value of one over R dv dr times M sub L times M sub S if you want to specialize in the hydrogen you'll find that this is proportional to one over R q and you can actually do that with the expectation value in any case this gives us the extra initial term which is due to the spin orbit term and we're not superimposed on the structure of the board here so you can see what happens when you turn on the spin orbit term on top of the earlier case in which the same on-term is dominant all right so that's the second case I don't want to consider now let's move on to the third case the third case is the one in which the same on-term is much smaller than the fine structure terms this of course is easily achieved by turning down the magnetic field a weak field in this case what we'll do is we'll take the temperature of Hamiltonian to be the same as the fine structure terms and we'll take page one to be equal to the Zaymar on-term now of course the temperature of Hamiltonian is degenerate this is the same as the fine structure levels which I proved in the beginning of the hour here they are some of the degeneracy is lifted by the structure itself if there's some that remains if 2 s is half and 2 t is half let's concentrate on the beginning it's two levels just to have a picture of the case we're working on so just to copy those levels again we've got the 2p1.5 which is degenerate by the 2 s and half which is degenerate by the 2p and half and then above that is the 2p3 house which as I explained last time the splitting here is about 10 gigahertz for hydrogen and this then is the basic level structure of the H equals 2 levels of the hundred-perchart system which is H zero which is right here alright and so now we need to certainly for the J equals a half levels well this too is also degenerate we have two degenerate levels of different degrees of degeneracy and we need to again we're checking Hamiltonian inside these two degenerate eigenspaces there's two of them what basis to use there is no choice of basis because the if we're going to restrict our choices to the couple and uncoupled basis we can only use the couple of basis because the uncoupled basis is not an eigenmeses of H zero not including the fine structure terms it must go to the couple of basis so the matrix elements of the question are going to be this it's going to be NLJMJ sandwiched from H same on NLJMJ and the N and the J on the two sides determine the energy level so those are the same on both sides but the L and the NJ determine the basis vectors inside the hundred-perchart eigenspaces so we allow them to be different from the two sides now however H same on commutes with LZ I had a table of it it's up there and so in fact the matrix elements are diagonal and the L quantum number moreover H same on commutes with J's as you can see up here as well and so I can remove the quantum numbers I can move the prime on the NJ and so once again we escape from having to diagonalize matrices and doing regenerative perturbation theory and we only need to calculate the diagonal elements so the energy shifts that are given by these diagonal elements you know now what are those energy shifts well where is my expression for H same on it's right up here it calls LZ plus twice SC why don't we to write this in this form H same on is equal to UBB or magneton times the magnetic field we write it as JZ plus SC LZ plus SC is JZ at least a single factor a single term of SC over so that's what this is this is convenient to do because the states on the two sides are eigenstates of JZ and this JZ will just bring up the magnetic quantum number given to J and so this matrix only then is equal to UBB times first of all m sub j from the JZ and then what's left over is the matrix element in all the mj sandwiched around tens of z like this that's a matrix element that requires some techniques in particular this is what projection theorem comes in projection theorem is a theorem which is useful for evaluating magnetic elements like this and it arises in quite a few places quite a few applications we'll see it again when we do hyperfine structure which will be next semester here's the I'll outline for you the derivation of the projection theorem it starts with an identity which is due to the wrap let's let A be a vector operator then we consider the double commutator of J squared with J squared commutation relations for a vector operator we can show that this is equal to twice vector of A times J squared plus J squared times vector of A minus 4 A dot of N to J times J and I might add I'm spending H more than 1 here and this is an identity which I won't prove in fact I'm going to flip a homework column on you to do this to prove this but I'll show you the consequences of this now allow me to use a general notation not the notation for the specific problem of the Zeeman effect let me use a general notation in which the standard angular momentum basis is again and again in the manner that we've same notation we used earlier general notation for this if you want to translate this notation into the Zeeman problem then again it is replaced by N and L and today is J and the N gets replaced by N and so these general basis vectors turn into rows up there from the Zeeman problem but we might as well treat the problem in generality because we'll use it in other contexts later as well so allow me to take both sides of this equation and sandwich it around the sandwich between gamma and J N so on the left-hand side let's take the left-hand side first let's write this as the commutator of J squared with a vector all called X where X is the same thing as the commutator of J squared with A so the left-hand side then so I want to sandwich the left-hand side between basis vectors like this so let's consider gamma J N and then the commutator of J squared with X gamma J N for the Zeeman problem the only thing we need is whether the vectors on both sides are the same vector however the projection there is more general and in fact we can allow the gamma from one side to be different from the gamma on the other side we will however require that the J value is going to be the same I know this is another special case but it's one that occurs if the only one will need for this course and it's simpler than the case where the J's are also different so let's take this case then let's make it so easy because it's J squared times X minus X times J squared and since J squared is being sandwiched on eigenvectors of J in the same eigenvalue on two sides each J produces J times J plus 1 and the difference just cancels to give you 0 so the left-hand side sandwiched between two vectors is 0 that means the right-hand side sandwiched between the same two vectors is 0 or in other words it means this is that gamma prime J M prime on the left J vector J squared plus J squared times A vector with gamma J M that's this term divided by 2 is equal to twice the same vector sandwiched around A dot J times J like this now on the left-hand side this J squared X on the right and this J squared X on the left and they both bring out J times J plus 1 to the factor of 2 which will cancel that too and all this left over is the vector operator A and so we ended up with this gamma prime J M prime greater than A gamma J M is equal to the same vector as gamma prime J M prime and then it's A dot of M of J times J times gamma J M and this is divided by J times J plus 1 and this is the this is the projection theorem in the form in which we will use it a lot of time okay well it's just a few short steps that leads you here to the landed G factor, unfortunately I don't do it but the result is that this becomes equal to UBB for the factor of M J that comes out there's a 1 here and then the rest of this is evaluated by the projection theorem and it turns into J times J plus 1 minus S times S plus S times S plus 1 minus L times L S1 divided by 2J and J plus 1 and this is the leadership to the weak same-on effect the quantity of the squared brackets is called the landed G factor you'll allow me to make one more comment which is I'm standing back in this population and looking at what it means to be the standpoint it's actually at the first sight it's remarkable that the energy shifts depend on the L sub J considering the factor of the energy itself it's because J you see is O plus S but the spin of the magnetic moments that are at the magnetic field by the vector of O plus 2S so why do the energy levels energy shifts depend only on the same the answer to this is is that we're working in a similar reducible subspace in the system and in any such space all vector operators are proportional if you could have that operator proportional to that one this is a result of the given red heart theorem that was the homework a couple of weeks ago and the proportionality factor here has been worked out explicitly using the projection theorem that's what this landed G factor is it gives you that proportionality factor in the right subspace so I think