 I'm going to continue with the start effect where we left off last night. The point was we were looking at the ground state of the hydrogen atom in the presence of an external electric field. And we found that the perforative shift in the original was zero, meaning that there was a permanent electric cycle moment. However, there isn't a perforative shift in the wave function, which is given by the standard formula here. Here's the ground state wave function, here's the correction in the presence of the field. It's a linear combination of all states other than the ground state. These are all states of a particle in the ground state with coefficients that come out of perturbation theory. Nymphomaphany establishes psi around the z-component of the dipole operator, which is minus zz. And the result was that we got a non-zero answer showing that in the presence of the electric field, there actually is a dipole moment to the ground state of the hydrogen atom. Not a permanent dipole moment, but as we say it's an induced dipole moment. An induced dipole moment is proportional to the electric field. Now this is a common situation in physics in which the dipole moment is proportional to the applied field. F here is an electric field, and the proportionality factor in alpha is just part of polarizability, and the ratio between those two. So one of the results of this is that it's possible to obtain the polarizability alpha with the ground state of hydrogen from alpha's principle. That's what we're looking for in quantum mechanics. And it turns out that two weeks were at S, where S is the sum, which appears here. S was faster than sum, but we arranged it slightly so that the denominator is positive and the sum S is positive. Okay, so this polarizability of the hydrogen atom can be used to compute the dielectric constant. Well, first of all, the electric susceptibility and then the dielectric constant. The polarization vector advantage is defined as the net dipole moment per unit volume. And so if we have hydrogen atoms and each one has a vector moment D, which is alpha times F, then the vector moment per unit volume is equal to the number of atoms per unit volume, which I'll call N here, just multiply times alpha F. So N here is the number of atoms per unit volume. And on the other hand, the proportionality between the polarization and the electric field is usually denoted by chi E, that's the electric susceptibility. And so what you see is, as you would expect, the electric susceptibility is proportional to the density of the gas, but the proportionality factor is the polarizability of the single atom. And in this manner, you see it would compute the electric susceptibility and then from that the dielectric constant, that's 1 plus 4 pi chi E, like this. Now, this is not a very realistic calculation of the dielectric constant for first principle because normally speaking, we don't have a gas and hydrogen atoms, you have a gas and hydrogen molecules. So this is actually rather unrealistic. But at least we can actually analyze the idea of what you'd use in a more realistic case. You'd have to analyze a molecule or if it were helium or you'd have single atoms, you'd have to analyze a two electron atom. And the basic procedure would follow just what we've done here. It's a case where they start, in fact. All right. And similarly, we want to compute the magnetic susceptibility in a similar manner. All right. As long as we're still talking about the ground state of hydrogen, there's one other thing we can do here, which is that since we've discovered that the first order energy ship vanishes because of parity, let's go on the second order perturbation theory and see if we've got an energy ship in the second order. So let's write this energy ship in this way as delta E with 100 subscript, that refers to the ground state. And I'll put two of the parentheses out here that indicate second order perturbation theory. If you apply the standard results from the second order perturbation theory for this problem, it becomes a sum over all states in NLM, which are not equal to the ground state because that's the one you're perturbing. And then you've got a matrix sum, a pair of matrix elements. It's 1, 0, 0. And then I'll have the perturbing Hamiltonian in the middle of NLM on the right. And then the same matrix element repeated backwards is complex conjugate, in other words, 1, 0, 0. These matrix elements are really real, so the complex conjugate doesn't matter, but I'll write it in this order anyway because it makes it look nice. And then it's divided by an energy denominator, which is E1 minus En, which is the unperturbed energy levels. They depend only on your principal quantum number N. Now, if you allow me to fix this up slightly, the H1 here is the same as EFZ, where F is the electric field strength and Z is the coordinate electron. The other H1 is the same as EFZ. And as far as the denominator goes, if the denominator is negative, so let me write it as minus En minus C1 like this. The energies are an increasing function of the principal quantum number, so En minus C1 is the positive number. Notice that it's number zero because we're excluding the state in question that's being perturbed. That was the point of this resultant operator introduced in balance state perturbation theory was to avoid the small denominators. All right, anyway, this is just a standard expression from a plot of second order non-degenerate perturbation theory where the 100 level is non-degenerate applied to this particular problem. Now, the same sum that appeared in the construction of the dipole moment, and hence the polarizability, what I'm calling S here, is now appearing again in the second order energy shift. The only difference is that I've got Z's here in the sum, and here I've got extra factors. In fact, this whole thing of transition would factor a Df taken twice, so it's E squared S squared, and then a minus sign from the denominator. What's left over is the same sum S like this. So this is the expression for the second order energy shift, and you see it's non-zero. The energy shift at first order is zero, and the second order is to find a non-zero energy shift. This energy shift is negative, by the way. That's a general feature that whenever you do a perturbation, it affects the perturbation on the ground state of the system if it's a non-degenerate ground state. And if the first order energy shift vanishes, and the second order energy shift is always negative, that's because the denominator here is the square of the matrix settlement, and it's always positive, and the numerator is always positive, and the denominator is always negative, so the overall shift is negative. As far as this sum is concerned, you can re-express it in terms of the polarizability. S is just L over 2 e squared, and if you put those together, what you get is that the second order energy shift, L22 of the ground state of hydrogen is equal to minus 1 half the polarizability times F squared. Like this, excuse me. So this is the first non-managing connection to the energy of the ground state. And as you see it goes as the square of the applied electric field, and for that reason this is called a chromatic stark effect. A chromatic stark effect exists when the linear stark effect doesn't, and the ground state of hydrogen is one of the easiest cases to analyze if you get this. All right, you know. There's a couple more things that I want to say about this calculation. One of them is that it's obvious in order to evaluate numerically the polarizability, the hydrogen atom, it's necessary to do this sum, which is an infinite sum. The sum's over an infinite number of terms, so it looks pretty hard. In fact, it's even worse than it looks because the NLM here, this is really a schematic notation. The NLM sum here, as you see, it kind of comes from a resolution of the identity, which is a pair of the numerator. That's where it came from, this Resolven operator, although one term is excluded. But if you want a resolution of the identity, you have to include the continuum states in hydrogen as well as the bound states. So this is really a schematic expression and it should be understood to include an integral over the continuous spectrum of the unbound states as well as their element of quantum numbers. So it's overall actually a rather complicated expression. We're going to have to know the unbound energy eigenfunctions for hydrogen as well as the bound ones just to do this by brute force method. It turns out that there is, however, an approximation to this and we can get a bound on this that's not so hard to work out. It works in a following way. I'll do what I can right down here below here and then I'll have to switch to another board and cover this up. It works in the following way, is that the denominator En minus E1 depends on N and that's what makes this sum hard, is the denominator. However, since the energies are an increasing function of N, this is always greater than or equal to E2 minus E1. This is for the case that N equals 2, 3, 4, and so on. Those are the only Ns that occur in the sum. And so the denominator is always larger than a certain quantity which is independent in a constant. Or if we take the reciprocal of this we get 1 over En minus E1 is less than 1 over E2 minus E1. And so I'm going to have to cover this up in just a moment, but it means that if I put this in and I can get a bound on S, S will be less than the sum we get if we just replace this denominator by E2 minus E1. So there's the formula and now I have to cover it up and here's what we're going to get. We're going to get that S, which is less than 1 over E2 minus E1 times the sum over these states. And then the matrix element is 1, 0, 0 over the C in here and then NLM and then NLM they make themselves backwards 1, 0, 0 like this and then the denominator is going to be removed. Now in this sum we excluded the case that NLM equals 1 is to exclude the ground state that's because we had to back here because otherwise you'd have a version of the denominator be 0 denominator. However, once I pulled out this constant denominator then once left over it doesn't have any problems with that but we can actually now include the 1, 0, 0 term because that term is 0 anyway if we vanish just by parity. So in other words we can now take this sum over all values of NLM and if we do this including the denominator that we continue with then what you have here is a nice resolution of the identity and the whole thing becomes 1 over E2 minus E1 times the matrix element 1, 0, 0 sandwiched around the operator C squared because there's a Z here and a Z there. Well this is a non-zero matrix element and if you plug in the wave functions you can evaluate it and it turns into the squared before radius. As far as E2 minus E1 goes E2 minus E1 is minus a half and then it's E squared over A0 which is the characteristic energy of hydrogen and then we have 1 over 2 squared minus 1 over minus 1 which is 1 over 1 squared like this and if you log in and bring the minus sign inside it becomes minus a quarter plus a 1 that makes 3 quarters times a half is 3 eighths the whole thing becomes 3 eighths and E squared over A0 and so plugging this in with the A0 squared here reciprocal of this what we find is that our sum is less than 8 thirds times A0 cubed by the way E squared and so it's an upper bound and this is sum S and if we look at our polarizability which is 2 E squared times plus the polarizability we find that alpha is less than 16 thirds A0 cubed which is a rather simple result to obtain for the polarizability of the Gonsted hydrogen 16 thirds is about 5.33 it turns out there's an exact analysis of the sum it's possible to do the sum exactly I won't take you through it but if you do what you find 9 halves A0 cubed and 9 halves is about 4.5 so this upper bound is not too bad but the point is is that by doing some work we can actually get a simple formula for the polarizability of hydrogen to the ground state and then from this we get the shift of energy which is alpha over minus 5 minus alpha over 2 the square of the electric field useful result for certain applications ok so this is the what about the stark effect now there's a couple more things that I'd like to say about so this seems to be what I want to say about the ground state of hydrogen in the stark effect there's a couple more things I want to say about this whole business however let me remind you that we did some searching around for cases where there would be a linear stark effect and therefore a permanent electric cycle moment and first of all we showed that such a thing did not exist in any non-degenerate eigenstate of hydrogen or the alkalis because then the perturbation sandwiched between states of definite parity so then we looked at excited states of the alkalis where you have degenerate states and we found that this still is a permanent electric cycle moment because all of those states all of those excited states of the alkalis are again eigenstates of parity now the question arises we finally found an example where it does work in hydrogen because we had a degeneracy because of this explicit symmetry in hydrogen we have a degeneracy between states of different L values and in central force problems the parity in states minus 1 to be L so the central force of fixed L all in definite parity unless you mix together different L's which is what happens in hydrogen but not elsewhere you won't get a permanent electric cycle moment I'd like to make a few comments now about how this generalizes to other systems this was only an electrostatic model of hydrogen or the alkalis it didn't include relativistic effects of spin, hyperfine all kinds of other small effects were emitted the question might be what of those effects would we find would it be possible to find a permanent electric cycle moment or maybe turn it into a multi-particle system there's a lot of generalizations so I want to generalize this question because it has to do with the question about parity as a good quantum number and more complex systems so here's a general idea let's suppose we have a system so you can think of an atom a molecule or a nucleus it could be quite complicated it doesn't have a tonic at all called H0 and we'll assume that it commutes with parity which in fact is what it means, which in fact is always true if you neglect the weak interactions which is an extremely good approximation in most cases so we'll just assume that it commutes with parity but otherwise we won't say much about H0 well we'll also assume that it's isolated but that means that it also commutes with J squared and Jz so that means that it's a collection of operators all the statement H is instead of H0 the collection of operators H, J squared and Jz are commuting sets and it means that we can first of all diagonalize J squared and Jz and then inside each of those eigenspaces we can diagonalize H and if we do what we get is a set of states that we can label by three quantum numbers like this without the Hamiltonian act on this this brings out an energy which is basically just a sequencing number of energy levels this will bring out an energy in Uj on this state Ujm like this the energy does not depend on him because of the rotational invariance of the system this is just a generalization of what happens in the central force function whereas we would use the simple we can in the principal quantum number instead of nu the basic structure of these eigenstates applies for any of these systems no one question is whether there might be any energy to generalities for different values of nu and J that is to say for different values of nu and J is it possible that different energies would coincide that's what they do in hydrogen where you've got for example the 2s and 2p have the same energy that however is a very exceptional case a general rule about the generalities is they don't happen unless there is some symmetry that makes them happen these systems don't have any symmetry beyond the obvious symmetry that's what's being used here as a rotational symmetry to write down these space we'll talk about parity in a moment but there isn't any most systems there isn't any extra symmetry around and so the result is that in the most realistic systems the energies in Uj in fact are non-degenerate well they have the degeneracy in M that's always the case but they don't have any extra degeneracy and that means you've got 2j plus 1 states let me remind you of the iron 57 nucleus which has a brown state of i equals 1 half this is 57 iron like this there's an exciting state called star then as i equals 3 has this is the 14kV transition that's often times used with the lost power effect and then there's another doubly decided state then as i equals 5 has then the point is that these states of iron are characterized with transitions like this and also like this as if these states of iron are characterized by an angular momentum form of number that's like the j here and they have a degeneracy where the evidence from minus i to plus i i here is the standard symbol for the nuclear spin but there isn't any degeneracy between these states these different states with a new here with stars 1 and 2 just labels these states and there's no degeneracy amongst them alright now the question is what about parity are these energy eigenstates also going to be eigenstates of parity because if they are then they won't get any linear stark effect so let's analyze this question let's consider what parity does to one of these states let's say parity act on one of these energy eigenstates well this is also an energy eigenstate itself because if I let h act on it since h can be used with parity I bring it through and it's going to bring on an e to j like this and to an i to j m in fact it's a eigenstate of energy with the same eigenvalue but there's no degeneracy amongst different e to j values and they're only degeneracy what remains is the m values so the energy eigenstates with this given eigenvalue are just the 2j plus 1 that come from varying m and therefore pi nu j m since it has an energy dj e nu j must be a linear combination sum over m prime when it states nu j m prime it must be a linear combination of vectors that lie in the same or useful subspace which is the energy eigenspace with some coefficients all called c sub m prime here and you do the coefficients that's a j and another e to prime this actually these coefficients will also depend on the n value of the state that you let pi act on and this is as far as we know that's when we first write this equation down now to find out more about these coefficients let's sort of sandwich on the other side with the same nu in the same j but in n prime what we get is the matrix elements of the parity operator inside the single irreducible subspace and this just turns into these coefficients c and prime m like this that's just what the coefficients are the matrix elements of the parity operator the parity operator how is a scalar operator it can get to all rotations and as a result it's matrix elements inside a single irreducible subspace are just some number of c that's independent of m times a chronicle delta in this prime this comes from the finger f arc and what that means is the matrix of the parity operator inside this irreducible subspace is multiple depending on the constant c factor and that means that all of the nu jms for a nu and j fixed an invariable where all eigenstates of parity with some eigenvalues don't see here but the eigenvalues of parity are plus or minus one or the other but whichever it is a plus one or a minus one it's the same value for all of the all of the different n values and the conclusion of this is is that these energy eigenstates which have this 2j plus one-fold degeneracy unless by some miracle they say there's a degeneracy for different values of nu and j which usually doesn't happen this will also be an eigenstate of parity states of definite parity and so for example if you look at energy level tables of energy levels nuclear energy levels will not only give the spin but it will also give the parity of the states these are all definite eigenstates of parity the same is true of very supplementary particles excitations vary on excitations and so on and then we land and melt the particles and things of that sort so all the eigenstates of parity as well as angular momentum these are the basic quantum numbers that they characterize in energy eigenstates with almost any isolated system I say almost any because the one exception is hydrogen where you have this accidental with X for degeneracy alright now and then of course since there are eigenstates of parity there's no linear stark effect hence no permanent dipole moment for these states even if they are degenerate okay now having said all that then if you've ever taken a course in chemistry this raises a question which I don't like to compress because in chemistry they tell you that there are certain molecules they call polar molecules which carry a certain amount of electric dipole moment and I see so many stones like erasers again so I'll have to be right back to see if I find they're as strong as they'll have to do the erasers they stole the good erasers like the bad ones the so they tell you in chemistry class that molecules are typically molecules like hydrogen chloride this is an example but in fact it applies to any heteronuclear diatomic molecule they tell you that these are polar molecules so they carry a there's a charge separation in them so they carry a dipole moment factor like this permanent dipole moment and then you might have done problems in stat neck where you take something that has a dipole moment vector and you have an energy delta e which is the dipole moment vector in the electric field and then you carry out the stat neck of this to find the orientation and say that it's in a permanent equilibrium and the orientation and so on energy and orientation and so on so how can I reconcile this permanent electric dipole moment with this analysis I gave up above which says that energy levels of realistic isolated systems almost never have an electric dipole or a dipole moment the exception being the excited states of hydrogen how do we reconcile this the answer is this is that the analysis I gave above was true it's correct that as long as there's no degeneracies amongst the different e new j sort of different knowledge of new a j then you then each of these states or eigenstates of parity however it is possible that e new j's become very close together in energy being nearly degenerate but then if you apply the electric field which is strong enough to sort of overwhelm the small splitting between these levels they will then start to behave as if they are practically speaking they become degenerate and then start to achieve a dipole moment that is effectively a permanent dipole moment involves mixing of states of opposite parity and so this is what happens in the case of diatomic molecules is that it's true that the electron distribution around the two atoms has a charge separation think of this as it would be useful to think of this as a dumbbell with a positive and negative two ends of the dumbbell this is an example of the diatomic molecule almost rigid body which I gave a lecture on earlier in the semester however if you look at the total wave function the total wave function is multiplied by a ylm and this gives a parity mistake which is minus 1 to the l now what happens is that in other words it's not just pointing in some direction in space but it's effectively averaged over all angles in a manner that's indicated by the ylms now if you then take the expectation value of the dipole moment vector over the ylm what you find is that it's 0 it is 0 just like I've just like you've been saying on the other hand if going from 1l to the next l the parity changes because it goes like minus 1 to the l this is the different rotational energy states of the molecule and these rotational energy states are separated by as I also explained this are separated by the energies that are down in the microwave the current for red and microwave regime so what you've got is a is a series of rotational energy levels as a function of l and they alternate plus, minus, plus, minus, plus, minus and so on in parity and if you now put them in a magnetic field which is strong enough to overwhelm this small splitting between neighboring energy levels you'll get significant mixing of the positive and negative parity states the minimum start to behave is if it really does have a permanent electric dipole moment and so this is how this is how this would justify these kind of statements that make in chemistry classes but to be rigorous about it in weak electric fields you don't get a lot you don't get a permanent dipole moment they really have a quadratic start effect for weak fields it becomes linear later on and the field becomes strong after overall splitting by the way even in hydrogen this applies because this degeneracy, this accidental degeneracy that happens in hydrogen where the 2s and the 2p are exactly degenerate they're exactly degenerate only in the electrostatic model of hydrogen if you start adding very small effects in hydrogen you find that this degeneracy is lifted as well and so even in hydrogen it's not true that you have exactly degeneracy between states between different irreducible multiples in rotations okay well that's all I want to say about the start effect it was partly an excuse to make some general points about parity as well as talking about the start effect and I'd like to turn now to a new subject which is the fund structure and Chris's listing will once again talk about the case of hydrogen and alkali atoms so the physics of fund structure is that it's the concern of the effects of spin and relativity on the structure of the atom so we'll be dealing with a non-returned Hamiltonian which is piece-ware of a 2m plus v of r so this is the electrostatic model and the v of r will be we'll not just do hydrogen we'll go hydrogen like atoms as well we'll put it in a C dependence so this is for H like atoms and as usual there's no simple formula for the case of the alkali and you should know what roughly the potential looks like anyway so this is the electrostatic model so as I said in the perturbation we want to consider here is the fund structure perturbation which concerns the effects of relativity and spin it turns out that there are three terms that occur here and they're all of the same order of magnitude so for a realistic treatment you have to take them all together and these are the first one is what I'll call the Noveltyvistic Kinetic Energy Connect Correction HRKB the second one is called the Darwin term and the third one is the spin orbit term that we talked about already so I don't need to say too much about that the fact that all three of these are the same order of magnitude is an indication that spin in a certain sense is a relativistic effect it's a relativistic effect that sometimes that is often times incorporated into non-noveltyvistic quantum mechanics by an ad-hoc manner by adding extra terms to the Schrodinger equations what's normally done, for example in advanced matter physics but from a more fundamental standpoint it really reflects relativistic effects and the deeper understanding has to come out of a relativistic theory I might have missed this point but when you were talking about when you said that HCL has a permanent to behave as a permanent electric diode but but that as opposed to induced so if you are talking about it happening only in the principle of an external electric field then why would it be permanent it isn't strictly speaking it's not permanent because in the absence of the electric in the external electric field there is no dipole load these YLMs average out the direction of the I mean it's true the electronic wave function it's a charge separation but it's like a charged dumbbell but the YLMs average this out over all angles so that the average which is what you call the dipole load is zero it's an induced energy eigenstate for the isolated molecule and what that means in turn is that if you apply a weak electric field you see only a quadratic start effect the energy only increases dramatically in the field however when the field strength gets large enough it will start producing strong mixing between neighboring states and neighboring L values they are pretty close together in the case of molecules they are pretty low energies and so beyond a certain field strength which you can calculate I mean some 100 volts per centimeter it's actually not all that small it's all states of opposite pairs they are neighboring L values in the rotational spectrum and beyond that the dependence of the energy on the applied field becomes linear rather than quadratic it's transitioned from a quadratic to a linear regime and so beyond that it actually behaves as if there is a first order start effect does that help? alright so then in effect it becomes like a permanent electrical moment and then the energy in the field becomes minus d dot e where the d is a permanent electrical moment alright so to go back to the fine structure then there are three fine structure terms Rolivistic kinetic energy Darwin terms spin orbit terms what I'd like to indicate is as best I can within the framework of non-rolivistic mechanics or non-rolivistic or Schrodinger equation what these terms mean physically non-rolivistic kinetic energy is probably the easiest one to work with in relativity the total energy of the system which we are trying to think of as being rest mass plus kinetic energy is given by the square root of m squared c with 4 plus c squared p squared this is a function of momentum and this is the same thing factoring out mc squared times 1 plus the momentum p divided by mc square and if we expand this out the theory is this becomes mc squared times 1 plus 1 half p over mc quantity squared minus 1 a if I carry out the 4th order of momentum p over mc to the 4th like this and I'll multiply this together the first term is the rest mass the next term becomes p squared over 2m which is the usual non-rolivistic expression for the kinetic energy the next term is p to the 4th divided by 8 mqc c squared like this and we stop there and so if we call the kinetic energy the contribution of the energy beyond the rest mass then it's everything all the terms of the series beyond the second one in other words it's a non-rolivistic kinetic energy term but there's also this first direction in fact this is the expression for the non-rolivistic kinetic energy hrkv is equal to minus p to the 4th divided by 8 mq times c squared this is an actual right this in a slightly different form it's minus 1 over 2 mc squared times p squared over 2m quantity squared so it's proportional to the square which is actually divided by twice mc squared so that term is fairly easy to understand and it's plausible in something like this should appear if you're interested in relativistic corrections to make a slight tangent let me say some things about relativity and atomic physics I pointed out earlier in my lecture on hydrogen atom that the velocity of the electron in the ground state of the hydrogen atoms characteristic velocity you find structure constant alpha times the speed of light so it's approximately c over 137 this is pretty fast but it's still also pretty substantial non-rolivistic when you compute differences in energies they typically go like the velocity squared and so v squared you see it goes like or mv squared well v squared goes like something like let's say v over c squared to make a conventional list goes like alpha squared which is something like 10 to the minus 4 so in hydrogen you'd expect the corrections due to relativity to be something like 10 to the minus 4 compared to the basic energy energy scale of the electrostatic Coulomb modeling on the other hand if the hydrogen like atoms in which the nuclear charge is increased by a factor of c then the characteristic velocity goes up to z also to get z times c over 137 if you go all the way to the uranium this is something like about 2 thirds the speed of light and so in the case of uranium non-rolivistic corrections are actually quite important at least for the electrons that are down in the case shell it's certainly true so anyway this gives you some idea of rolivistic corrections they're small in the order of 10 to the minus 4 hydrogen but this is something that grows as z squared and becomes more important as you move to heavier atoms and this will certainly be true of this rolivistic magnetic energy correction here alright now as far as the many two terms this is a Darwin term and the spin orbit term let me just deal with the spin orbit term first and all I'm going to do is write it down earlier what it means and here's what it is spin orbit term makes spin orbit is equal to it's a 1 over 2 twice m squared c squared times 1 over the radius r times dv dr times l dot s and just to remind you this physically represents the magnetic energy of interaction of the electron spin with the magnetic field is seen in the electron rest frame there is no magnetic field in the lab frame the frame of the nucleus there's only the Coulomb field in the nucleus but that's an electric field by Lorenz transformation you get a magnetic field when you go to the rest frame of the electron I went through this before so I won't elaborate on it but this is the spin orbit correction this turns out to be the same order of magnitude as the rolivistic magnetic energy finally there's the Darwin correction the Darwin term the Darwin term is very difficult to explain even a hand waving manner within a non rolivistic framework it really comes out of the Dirac equation I think the best I can do is to, I'll do the best I can it turns out that in the rolivistic Dirac theory of the electron that if you if you get a course of the Dirac equation into the mole of the Schrodinger equation what you find is that the electron in effect interacts with the electric field the Coulomb field in the nucleus in this case it interacts with the electric field in somewhat not a local manner it is as if the electron is smeared out over a certain radius which has a scaling which is of the order of the Compton wavelength which I call lambda C let me say something about the Compton wavelength because it's a basic it's a basic quantity which is relevant for rolivistic quantum mechanics the physics of it is really very simple if we had a particle and we put it in a box of size L, like this as you know there is a uncertainty in the momentum of the delta P which is the order of H part divided by L and so as you if you imagine squeezing the box in and making it narrower then the delta P increases and there is greater uncertainty in the momentum now let's ask ourselves how much do we have to squeeze the box before the momentum starts to take on rolivistic values well if that's a typical rolivistic value for the momentum it would be the mass times the speed of light so this gives us an estimate then for L and what we find is that L is equal to H part divided by mc this in fact is the definition of the Compton wavelength lambda C as you see it depends on the mass of the particle the lighter particles have larger Compton wavelengths in the case of the electron which is what we are mainly interested in here the Compton wavelength H bar over mc for the electron is in fact equal to the more radius A which I call A naught multiplied times the fine structure constant of H bar C which is about 100 more than around 837 you know there is a Compton wavelength for the electron the factor of the alpha down is the wavelength of the size of the standard size of the hydrogen atom it's done by a factor of 137 and so on this scaling less than a percentage of 1% or so of the size of the hydrogen atom rolivistic effects start to be important in the case of the electron and this smearing out that I was speaking about which lies in the Darwin effect takes place over the scaling which is precisely this constant wavelength because of this smearing out it turns out the electron interacts with the potential not only by the potential and the evaluated position of the electron the center of this block but also a correction term a secondary correction term which is in the order of magnitude of the Compton wavelength squared times the Laplace electron things like this occur in classical electrostatics too you may have seen averaging things out of the potentials but anyway it goes back to Laplace and this is how it turns out and so except for a factor of 1 to 8 that's exactly what the Darwin term is it's the square of the Compton wavelength which is h bar over mc h bar squared over m squared c squared multiplied times the Laplace of the potential and this is what you got for this okay so this is at best just panning the derivations of these three terms these terms all will come out automatically from the Dirac equation without the unique hand way of things at all and we'll do that in the next semester but for now what I'd like to do is just to use these three perturbation terms basically is practice and perturbation theory and also for gaining some knowledge about the fine structure I mentioned that we were talking about both hydrogen like atoms and alkalis actually what I'd really like to do is to concentrate almost entirely on the hydrogen like atoms first and we would never have come back to tell you about the alkalis so for the case of the hydrogen atom let's talk about the let's specialize to the case where the potential has this h like form i.e. zp squared over r as far as the the potential is concerned the potential v is the energy of the particle which is less equal to the charge which is minus e times the electrostatic potential and the the velocity of electrostatic potential by Gauss's laws minus 4 pi times the charge density this is the charge density which produces the potential of i and the charge density in turn for a hydrogen like atom is the nuclear charge z times the the electron charge z times the charge of the proton times the racked delta function of the origin treating the nucleus as if it were a point particle and so putting this together what we get is a del squared p for the case of a hydrogen like atom is equal to with a plus 4 pi zp squared the racked delta function of the the three dimensional racked delta function of the origin and so to combine this and for the case of the hydrogen like atom the yarn term becomes equal to pi over 2 times z b squared h bar squared divided by n squared c squared times racked delta function of r also for the spin orbit term it's easy enough to differentiate the potential dv dr here and divide by r and plug that in and if you do then here's what you get the term that becomes z times e squared divided by twice n squared c squared times one of the r cubed times delta dot s and so here they are here's the three terms which will take for perturbations on a hydrogen like atom so what would be the sum of what they do to the energy levels now as usual in perturbation theory before you start you ought to understand the unperturbed system the unperturbed system here is just an h like atom so we know a lot about that already it's not too much to say if we the unperturbed Hamiltonian is given up there it's a purely orbital operator and if you ignored the spin completely then you write the eigen space in ket language like this this is what we did in the case of the stark effect allow me to put an L subscript on the atom to indicate that it's the magnetic quantum number associated with orbital and angular momentum the reason is I bring this up is because the fine structure terms at least the spin or the term does anyway it involves the spin explicitly so to take these terms into account we need to include the spin in the electron so to do that the basis states then for the we need to size the Hilbert space to include the spin as well and the basis states or Hilbert space or the product of the basis states the orbital basis states times the spin basis states which will write as s and s like this and with a shorthand notation let's write these basis spaces n L, m sub L, m sub s where I suppressed the quantum number s because that's the spin of the electron and it's just constant of one half this is what we call the uncoupled basis when we were doing the coupling of angular momentum this is the uncoupled basis so the uncoupled basis is the is n or b or n eigenbasis of the unperturbed Hamiltonian and since the unperturbed Hamiltonian doesn't involve the spin at all the addition of the spin here didn't change the energies the unperturbed Hamiltonian is still the standard electrostatic model energies but what it did do is it doubled it to the degeneracy because the energy doesn't depend on the orientation of the spin and s-hopper and s-down and so now the degeneracies of the unperturbed system are 2 n squared instead of just n squared except for that there's not much change there's not much else to say about the unperturbed system now the 2 n squared is never equal to 1 so these energy levels are degenerate they're always degenerate and if we want to add up these three fine structure terms and find their effect on the energy levels we're going to have to use degenerate perturbation theory so if I call if I call the H-fine structure here's kind of what we need to think about let's at least think about this at first we want to form the matrix elements of the perturbing Hamiltonian inside one of the degenerate eigenspaces the degenerate eigenspaces are determined since the energy the temperature of the energy is only dependent on n they're determined by the principle of quantum number alone so we'll have states n l, m l and m s on one side and n l, m l and m s on the other side except I'll put primes on the l, m l and m s I don't put primes on the ns because that labels the eigenspace of the unperturbed system the primes on everything else because they label the basis vectors inside the degenerate eigenspace of the unperturbed system this is the basic strategy of degenerate perturbation theory before we get carried away with it well so this is going to be as you see is going to be 2 n squared by 2 n squared dimensional matrix and it might be intimidated by the diagonalizing such a big matrix before we get carried away with doing that though there's some rules that I'd like to tell you about in effect of choosing a good basis that saves labor allow me to instead of talking about h-fine structure allow me to just write h1 here where h1 is perhaps 21 of these 3-fine structure terms that we're talking about here thinking about this kind of matrix I'd like to illustrate a certain principle let's suppose that h1 commutes with lz let's suppose this is true this actually is true for these three cases here let's suppose this is true then if I take these two states and sandwich them on the commutator I'm going to get 0 I get 0 is equal to nl nl ns with primes on it and then I've got lz times h1 minus h1 times lz and then we have nl nl ns on the other side like this which just follows through the commutator now the first lc I can allow to act to the left and the second I can allow to act to the right so this brings out nl prime minus nl and then what's left over is is the matrix element of h1 alone nl nl ns the total of this subject is it takes so long to write everything out and so what you see is is that either the matrix element of the Hamiltonian vanishes or else the magnetic quantum numbers are equal and what that means is if the perturbing Hamiltonian here commutes with lc then this is actually a diagonal in the ML quantum this whole thing is proportional to nl prime and this actually applies obviously not only for lc but for the other quantumers like l squared and ns of z which are in here also and in all states of general rule we usually use basis states which are eigenstates of complete sets of commuting observables that's what's being done here yet the operator whose matrix element you're taking commutes with one of the members of the complete set then it means the matrix elements are diagonal in the corresponding quantum number well if you're diagonal in the corresponding quantum number you've already done some of the work of diagonalizing the matrix if you're smart you may be able to find a complete set of commuting observables such that all of the observables in the list can use with the operator in the middle and if that's so the whole matrix is diagonal and then there's nothing to diagonalize all you need to do is just compute the diagonal elements and those will be the energy shifts so that effectively converts degenerate perturbation theory into non-degenerate perturbation theory which is much easier to diagonal make consultants okay with that I'll let's go have a happy Thanksgiving and I'll see you in a minute oh