 We're going to continue today with the fine structure in hydrogen, which we started last time. It reminds you very briefly, we're taking the under-term Hamiltonian, which is here at the top of the board. This governs the, this is the electrostatic model of the growth school on the effects of the atom. The perturbation H1, which we also call HFS, which stands for fine structure, consists of about three terms, relativistic to the energy in the garbant term and the spin order term, all of which are the same order of magnitude. These three terms, in the case of a hydrogen like atom, are listed out explicitly here, which we talked about that one last time, and I explained the physical meaning of them, so I will go over that again. Now, at the end of the hour last time, I was presenting you some strategies for dealing with the time dependent, excuse me, with the general perturbation theory, in which it's necessary to diagonalize, diagonalize the perturbing Hamiltonian inside the degenerate eigenspace of the under-term system, so that the matrix is diagonalized. And it's going to be a prime structure problem, the matrix can get pretty big. So, when you diagonalize the matrix, of course, the amount of work depends on the basis that you choose, and some bases are easier than others. So, at the end of the hour last time, I was addressing the question of strategy for the choice of a good basis for an analysis of the perturbing Hamiltonian. The basic rule is that, well, typically when we choose bases for calculations like this and quantum mechanics, they are simultaneous eigenbases of some set of complete set of community observables. And as I showed last time, if the perturbing Hamiltonian commutes with one of the members of the complete set, then the matrix elements are diagonal and, of course, a quantum number. So, in an extreme case, you might have a perturbing Hamiltonian which commutes with all of the members of the complete set of community observables, which are used to create the basis that you're using. If that's the case, then the matrix of the perturbing Hamiltonian will be strictly diagonal, and there won't be any matrix to diagonalize. The eigenvalues will be the diagonal elements. Any effect that reduces the generic perturbation theory to the non-degenerate case. All right. So, in the case of the fine structure calculation, there are two obvious bases that we can use. Well, first of all, the eigenspaces of the unperturbed Hamiltonian have an energy we call e n. They depend only on the quantum number n, and these are two n squared fold degenerate, coming with spin. So, the matrix in question is going to be two n squared by two n squared. It could be quite a big matrix. There are two fairly obvious choices for bases in the degenerate eigenspace. One of them is what we call the uncoupled basis that we were coupling angular momentum. So, the uncoupled basis consists of the central force eigenfunctions for the hydrogen-like Hamiltonian, which I'll denote in ket language by nl and imsebel, multiplied by the spin basis functions s and s, where ms is either over or down. That's the spin division factor of two here. And for sure hand for this, let me write this in the form of nl, imsebel, and imsebes, listing out all the quantum numbers here, except for the spin itself, which I'll make you list. It's a constant, which is one half, and it never changes. So, the others over here are the ones that are variable. These quantum numbers are listed out here. Of course, one of the certain operators is a complete set. The n is the Hamiltonian, and the l is l squared. The ml is l sub z, and the ms is s sub z. And here's the collection of four community operators, a complete set that gives us an eigenbasis. Now, in this perturbation calculation, the index n indicates the general subspace of the integer system that we're working in, with remaining three indices are labeling the vectors that lie inside that integer eigenspace. Another obvious choice for a basis is what we'll call a cut-index. This is obtained when we couple the orbital and spin angular momentum. We get the total angular momentum. And all the other vectors are the coupled basis by the minimum of space here, nlj and m sub j, in which the quantum numbers correspond to operators h, l squared, j squared, and j sub c. And these basis vectors are given by a much shorter than expansion in terms of the uncoupled basis vectors. So, for some of the magnetic quantum numbers, nl and ms, there is a vector that is the uncoupled basis nl, nl, and ms. And they're multiplied by a clutch working coefficient, which is ls, nl, ms, together product of j and mj. And you can kind of remember this formula because you're summing up nl and ms, and if you look at the first ket and bra here, it resembles a resolution of the identity where the nl and ms are the labels of the basis vectors that you're summing up. All right. So these are the two obvious bases that one might choose for this perturbation problem. And now what I'd like to do is to point out which one of them is the best. And we'll do that by making a table of operators that communicate with the perturbing Hamiltonian h1 for more exactly the three terms that are in it. So, for the three terms, I'll list them out for both for this kinetic energy, the garland term, and the spin order term, like this, to make the table. And then for operators, let's list, first of all, the orbital angular momentum, all three components l, the next is square, the next is spin angular momentum, and then the square of that, and then the total angular momentum, j, and the square of that for a grand total of of six operators here across the row. Now let's start with the relativistic kinetic energy half-tone. It's a square, and you see it's a bunch of constants multiplied by the fourth power of momentum. The fourth power of momentum means p squared squared, and p squared is the vector p dotted at p, so it's a dot product of two vectors. And as a result, it's a scalar operator. It's a scalar operator under purely orbital rotations because the momentum p is a purely orbital operator. It's a vector under orbital rotations. And as a result, it can use all three components of l, which are the generators of orbital rotations. Now if you want to check the commutators directly by computing using the commutator calculation, to find the commutator of l with p to the fourth, you're welcome to do that. We'll find the answer is zero. But the argument I'm showing you is an illustration of how going back to the theory of rotations, you can see that the p dot p is invariant in the rotations. It means that the operator must compute the generators of rotations, which are the l. Now if the operator commutes with all three components of l, it commutes also with any function of those, which includes l squared. The next we move on to the span. This operator commutes with a span because hrk is a purely orbital operator, and it doesn't act in the same spaces again. So it commutes with any function of the spin operators, including all three components of s squared. And then finally we come to j at the total length of momentum. j is the generator of total rotations, both orbital and spin rotations simultaneously. Well p dot p is invariant in total rotations just as it is under only orbital rotations. So it's a scalar under rotation generated by j just as well as those generated by l alone. And therefore the operator also commutes with j squared. And we see that hrk impacts commutes with all of the operators in this list. Now next let's look at the Darlin term. The Darlin term is a bunch of constants times the delta function of the origin. It's a functionality of the position vector. So this is once again a purely orbital operator. It doesn't involve the spin at all. The question is the singular operator because of the delta function here. The delta function can be at the origin, as you see here, the delta function can be regardless of the rotation invariant function of position. The reason is you can think of it as being the limit of a function, an ordinary function, which is zero outside a small sphere centered on the origin. And then as a non-zero value inside the sphere it gets larger and larger as you let the sphere shrink. But the point is that it's rotation invariant usually can happen to be in a concentrated in the sphere. And as a result of that the arguments we've used previously for the purely orbital operator hrke also apply to h Darlin. And it commutes with all six of these operators running across this row as well. Alright, then now let's turn to the spin orbit term. It's constants times 1 over r cube times l dot s. 1 over r cube is a purely orbital term and is certainly a scalar. What about the l dot s? Well, first of all, let's talk about the commutation relations of orbital angle momentum l, the generator of orbital rotations. If we apply orbital rotations to h spin orbit what it will do is it will rotate if we want to leave it on one of our cube invariant it will rotate the l because l is an orbital operator. It will not, however, rotate the spin because they have different spaces. It doesn't do anything to the spin. As far as orbital rotations are concerned the spin is a collection of three scalar operators. They don't change under orbital rotation. So you've got a dot product here if you think it was two vectors you're rotating one of them but not the other one including the dot product is non-invariant. And as a result, h spin orbit does not commute with the three components of l. It does, however, commute with the l squared because l squared commutes with the three components of l and also with s and puts a standard commutation relation in a momentum l squared with three components of l. So l squared actually is a diagonal quantity. Now what about the spin s? The spin s generates spin rotations but not orbital ones. So, again, what this will do is it's the opposite now. It rotates the spin vector but not the orbital vector. So, again, this is non-invariant under spin rotations and the operator does not commute with the three components of spin. But, however, for the same reason it commutes with s squared. Actually, s squared in this case is a constant operator and it will commute with everything. Now we come to j, the term rotations. j rotates both orbital and spin variables. And if it does so, then both sides of this dot product are rotated by the same rotation and the dot product is invariant. So, in fact, l dot s commutes with j and therefore with j squared. And so, here's what we get. There's our table of commentators. And you can see that the spin orbit term is the one that makes us think a little bit because it doesn't commute with everything. And in particular, you can see the spin orbit term does not commute with all of the operators that go to make up the complete set that specifies the unpopular basis. It doesn't commute with either l, z or s and c. However, looking at the operators in constrictive for the make up to the coupled basis, you see it does commute with j squared. And j c commutes with l squared also. So, in fact, what we see is the coupled basis is really the right one to use because it makes all three of these terms diagonal. And that means that in order to compute the energy shifts, we just need to compute the diagonal elements of all three of these terms in a coupled basis, and then just add them up. And that will be the fine structure energy shifts. All right. So, I'll outline for you the right couple of questions involved in doing that. Let's start with the relativistic kinetic energy term. We take a vector of the coupled basis in l j m sub j and send it around the relativistic kinetic energy like this. And the same value of m sub j of both sides because we just decided to go in and look at the diagonal elements. Same value as a whole three of quantum numbers on both sides. Now, h r k e is a purely orbital operator, but it's being sandwiched between the vectors of the coupled basis. So, let's use the clutch coordinate expansion which is up there for both the bra and the cat here to expand this back into the uncoupled basis. And if we do, we get a four-way sum on m l and m s at m l prime and m s prime like this. And then there's again to the clutch coordinate coefficient it's j m j. And then we get l s m l and m s. And then in the center we've got the vector of the uncoupled basis m l m l and m s on the left. We do perturbing Hamiltonian h r k e in the middle and then m l m l prime and m s prime on the right. And then the final clutch coordinate coefficient l s m l prime and s prime and then j m j. j m j and j m j and the two sides are the same. That's the same as these two corner vectors here and here. They're the same on the two sides. All right. Now, in the central matrix element, since h r k e is a purely normal operator, the spin parts of the uncoupled basis just in effect shine right through this operator. And they combine together and give you a chronic redulter of the spins. So the central matrix element is even a chronic redulter m s and m s prime. And then what's left over is a purely normal optical matrix element which is n l m l on one side h r k e in the middle and n l m s del prime on the other side of the fact. So now what we're doing is keeping this h r k e into sandwiching between the central force wave functions for the hydrogen atom. The h r k e is a scalar operator so it has the form of t 0 0 in the line between the Wigner-Eckhart theorem. And what that means is if you look at the if you look at the selection of the magnetic quantum numbers, it means the answer is 0 unless n l is equal to n l prime. So by the Wigner-Eckhart theorem, this matrix element, this time matrix element is chronic redulter m l m l prime times a reduced matrix element. But the reduced matrix element, I like to write it in the form of a reduced matrix element. But the reduced matrix element is independent of magnetic quantum numbers. In other words, regardless of matrix and m l m l prime, it's not only diagonal, diagonal elements are actually independent of the magnetic quantum numbers. So I could write it as a reduced matrix element, but instead what I'll do is just choose some value for the magnetic quantum numbers in 0 instead of convenient value to use because the answer doesn't depend on which value you choose. And so this gets multiplied times n l 0, let's write it this way, with h r k in the middle and then n l 0 like this, setting that magnetic quantum number equal to 0. And if we do this in the final matrix element, there's something that's independent of the variables, the indices of summation. And so this can be taken out and moved out there outside the sum. And then what's left what remains of this big mental matrix element is just these two chronic redelfas and the magnetic quantum numbers which allows us to kill off the prime sums here and replace these prime denials by unprimed ones. And then when we've got these two adjacent clutch coordinate coefficients but we're summing on the m l and m s and m l and m s at the final sum here that's just the resolution of the identity and the whole thing collapses into the scanner problem from j of j of itself which is gone. And so the result is that this whole thing turns into m l 0 h r k d m l 0 like this. It becomes a purely orbital matrix element which you could have expected since it's a normal operator. So how much between these two states? All right. Now to further revaluate this well as not even more let's look up at the other board that our h r k d is and if you look at that for a moment you'll see that I can write this this way as 1 over 2 minus i 1 over twice m c squared times the p squared over 2 m quantity squared it's the same expression but of course p squared over 2 m is the non-robastic kinetic energy so this factor here is the same thing as let's call it t squared if we write the temperature of Hamiltonian which is up to the top of the board there let's write it as t plus v v of c is the hydrogen-like coulomb potential then t squared is the same thing as h0 minus the potential v quantity squared and so our matrix element up here then becomes I'll write it up here it becomes nl0 sandwiched around h0 squared minus h0 times the potential minus the potential times h0 plus the potential squared like that but the potential well h0 as far as h1 is concerned it acts either to the right or to the left or to the right of its test and when it does it brings out the energy level e sub n of the temperature of the system so the first term is just e sub n squared the second two terms combined together to give you a factor of twice e sub n times the expectation value of the potential and the remaining term is the expectation value of the potential squared now as far as the expectation value of potential is concerned since the potential is minus cd squared over r this is the same thing as minus cd squared times the average value 1 over r in the right or to the left way functions and as far as the square average value of the square of the potential is concerned this is z squared into the fourth times the average value 1 over r squared then it will be the hydrogen atom way functions now over the top of the board here I've summarized the expectation values of various powers of r in the hydrogen atom I assume this is familiar to you because you have a homework problem on it and so we can take those results and plug them in here for these two things and then there's an algebra to do and I will mint the algebra but the result is is that we get an energy shift for the relativistic kinetic energy and I need to look at my notes because this can be somewhat messy formalists we get z alpha quantity squared I'll write this as minus times minus e and that's the temperature of the energy level times 1 over h is squared times the quantity three quarters minus minus n divided by l plus a half if you do the algebra that's what you get alright and by the way the minus e in here will be 1 over 2 2 in the square times z squared e squared over a naught if I'm going to put this and these are just the moral levels and I'm going to put it in the minus side to make the quantity positive so that this is the positive quantity here alright now before going on a few comments about this the result here of course the left hand side is an energy the minus n is the energy level of the unperturbed system the the four energy levels so all the rest of this stuff is giving us the energy shift due to the fight structure of perturbation as a fraction of the of the unperturbed energy level now the stuff that comes after the 1 over n squared and afterwards it depends just on the quantum numbers the n and the l in particular if we think about an atomic state for which the quantum numbers are not too large such as the one state or the first few excited states and the hydrogen atom and this entire expression after including 1 over n squared and the following is just some number of order unity so the actual magnitude of the perturbation is done by this factor z alpha squared relative to the to the unperturbed energy levels so the z alpha squared is what we really don't look at or understand the order of magnitude of this velocity over the speed of light or c times the alpha's velocity of the electron in the ground state of the hydrogen like that it's alpha times the speed of light in hydrogen but for heavier nuclear higher charge because the scales is charged so this is something which is 1 over 137 in the case of hydrogen it turns into something like 0.6 in the case of uranium in the reaction of the speed of light of the velocity of the electron so we're finding the energy shift goes to the square of the velocity v squared if you really to be over c squared this is just what you expected relativity theory for energy corrections they go like the square of v over c so in the case of hydrogen it's the alpha squared something like 10 to the minus 4 so these are that means that this shift is something like 1 part of separation between energy levels alright now next let's go on to the Darwin term the next one is this here if you'll allow me I'll it's very similar to the something that's still like something that's still like a big eraser look look at this something that's still a big eraser so to go to the Darwin term let me use my eraser and we'll just put the Darwin in here the Darwin term just like relativistic this is a purely ordinal operator and so all of this analysis of switching back to the uncoupled basis goes through it exactly the same way and so I'll just erase it in the same way as it did with all this kinetic energy so I'll just erase it and summarize the dissolve which is that this becomes equal to in L0 sandwiched around the Darwin term like this and it becomes a purely again a purely dissolve now this in turn if you examine the Darwin term it's a bunch of constants times a delta function if I copy the constants prior to C B squared H4 squared divided by M squared C squared and then the remaining making settlements going to look like this will be a integral over all space dQ of R and then we've got the wave function psi NL0 of R squared and then it's multiply times a delta function of R now I'll remind you that the wave functions in central force motion go like R to the L near the origin the delta function is going to evaluate this happy origin so what that means is that all the wave functions vanish at the origin except when L is equal to 0 and the result is that this darling term only has an effect on the L and on the L equal to 0 the S waves so this if I take this if I take the integral let me take the whole integral we take the whole integral then this is equal to delta L 0 unless L is equal to 0 and then the result is at square at the origin at R equal to 0 this becomes then multiply by side N 0 0 absolute value square and this square of the wave function in turn is the radial wave function R N 0 evaluated at 0 square times as far as the square of the wave function the radial wave function at the origin is concerned that was also something that appeared in the homework problem this is in the volume effect which I guess is this week's homework isn't it lost track anyway it's up there at the top of the board and one can plug that into into here and then pull all the pieces together and so once again there's some algebra here and it is equal to this it becomes it's arriving in the same form as the alpha square times the unperturbed energy level where minus on minus then what's left is 1 over N times delta L 0 okay and that leaves only this in orbit term the last one which goes as you can see constants times 1 over R Q times L as far as 1 over R Q goes that's of course a scalar operator under both purely orbital rotations as well as total rotations generated by J but L dot S is different because it involves the spin so we're going to be interested in the energy shifts which involve sandwiching the spin orbit term between vectors diagonal vectors of the spin this N L J M sub J it's going to be constant Z squared over twice M squared Z squared multiplied times the matrix solid which is 1 over R Q times L dot S and then N L J M sub J over R side and this is the energy shift delta each spin orbit equal to this the L dot S is the place to begin begin analyzing this we use a simple identity that says the square of the terminal angular momentum which is the orbital plus spin vector squared is the same thing as L squared plus twice L dot S plus S squared just expanding it out so we can solve for L dot S and this is then one half of J squared minus L squared minus S squared J squared L squared and S squared are members are operators which are members of the complete set which is used to specify these basis vectors the L the lower case L is the L squared the lower case J is J squared and J C is the operator corresponding to M sub J and S squared is suppressed but it's in there too it's a constant operator and so in sandwich between these these basis vectors this operator just takes on its eigenvalue which is one half of the quantity J times J plus one minus L times L plus one minus S times S plus one and S times S plus one is actually three quarters because it's a half times three half this is then its constant and so the L dot S then with this substitution is replaced by this constant value and then you're left with the expectation value of one more R cubed which can be handled in exactly the same way as the earlier two operators which were purely orbital operators and we end up needing to use this expectation value one over R cubed here now there's a slight problem which arises namely that the average value one over R cubed has a one over L in the denominator so it would appear that it's infinity for an S way it actually is it is infinity for an S way infinity clearly comes from the fact that the R cubed diverges at the origin so the integral doesn't give you an infinite value of that case so what does this mean about the spin orbit term well if you're dealing with an S way which is where L is equal to zero the L operator the orbital the L vector operator is restricted to that subspace becomes a zero operator so in a sense we have a singular formula that looks like zero over zero in a case of S way so there's a question about how to interpret it here's a way of handling that is take the Coulomb potential which is hiding hiding in the way functions here take the Coulomb potential which is hiding in the way functions and you smooth it out near the singularity but physically realistic way of doing this is just what you're doing in the volume effect you spread the charge over a sphere in a small radius if you do that then you find this divergence for the S ways of whatever R cubed goes away you get a finite answer however it's still true that the L operator is zero in the subspace of where lowercase L is equal to zero so not only that zero divided by something which is finite and the answer is never equal to zero so the correct limit that you get if you shrink the size of the charge a distribution is and the answer is zero in the case of L equals zero in other words we only use this expression given up here for the average value one over R cubed in the case where L is not equal to zero when you put that together then here's what you get for the energy shift through the fine structure terms it's the same C alpha square times minus the N and then it goes as one over two N and then it is the J times J plus one minus L times L plus one minus three quarters that came from our L dot S and then it's divided by the L times L plus a half times L plus one that came from the expectation value of one over R cubed and this is only in the case of L equals zero in fact if I do this right I need to put a bracket and say that the answer is zero and L is equal to zero all right and so this is the reason for the results then of the first order perturbation theory to generate perturbation theory using the three fine structure terms and so what we want to do is add these up now and to get the total energy shift through the fine structure so obviously there's some algebra involved in doing that and I'll just quote the results and skip the algebra and what we get is delta D fine structure and here's what it turns into we'll write it in the same form as C alpha quantity squared times minus piece of N and then there's a factor of one over N squared then you get three quarters minus N divided by J plus a half and this has to be considered as an important result a fundamental result of the theory of a hydrogen atom hydrogen like atoms should be shifting the energy levels due to spin in relativistic effects all right so to begin a discussion of this result probably the most striking feature about it is that the energy shift is not dependent on the L quantum number you'll recall that the energy levels of the electrostatic model of hydrogen are degenerating in the L quantum numbers the energies don't depend on L they depend on the principle quantum and that the fact that the energies are independent of L persists to at least the first order perturbation theory when we include the fine structure terms this is kind of a miracle because the three it turns individually do depend on L as you see there and here and there's L dependence here and there too now the energies do depend on the quantum number J which is the new quantum number that's been introduced in addition to N so I'm going from the electrostatic model to what we might call the fine structure model the energies which originally depended on N now have the form of these N and J they depend on both N and J but not L of course they don't depend on the LJ either because that's because it's a scalar operator but the fact that it's independent of L the extra symmetry that exists in the Coulomb model of hydrogen actually persists into the relativistic version of the hydrogen atom alright now the next thing to say is I've already went through the order of magnitude of this it's something like 10 to the minus 4 relative to unreturbed levels in the case of ordinary hydrogen if you look at this a little more detail you can show that this is always negative but you can show it suppresses the smaller values of a smaller J more than it does the values of higher J so that the energy levels are an increasing function of J when the fine structure terms are included and so now let me show you what that does qualitatively at least to the energy levels of the hydrogen atom so I think I can get rid of all this now so the ground state and hydrogen atom the microstatic model is of course the 1s level s of course means that L equals 0 and if we can combine 0 with a half the half of the spin the only possible value of J is a half so this is L, S, and J in August and so the ground state of hydrogen is necessarily J equals 1 half level it's customary to explicitly indicate the J value of the subscript underneath the letter for the angular momentum so this is called how to indicate fine structure at last similar to the 2s level and the unperturbed system becomes a 2s 1 half level because again J equals 1 half is the only J value allowed on the other hand for the 2p level what p state means that L is equal to 1 we're now combining orbital and spin with 1 cross a half which gives us two possible values of J which is 1 half and 3 halves and so the J values and you get a 2p 1 half and 2p 3 halves the 2p 1 half in this model is exactly in general 2s 1 half because the L values are the same or the J values are the same however the 2p 3 halves has a different J value and it is raised by a certain amount like this and so you get you see a splitting of the levels of the unperturbed system likewise if you go on with N equals 3 you have the 3s 1 half which is degenerate with the 3p 1 half somewhere above that is the 3p 3 halves that's degenerate with the 3d 3 halves and somewhere above that is the 3d 5 halves so including fine structure effects the N equals 3 level of the unperturbed system is now split into three different levels these two are degenerate these two are degenerate there's now a floating at the top the N equals 2 has two levels like this so this is qualitatively the effects of the fine structure this of course is not to scale because this splitting at least the hydrogen is 10,000 times smaller in the separation between these two now so that's the qualitative effects now there is a question that arises we did this by perturbation theory and we got the energy shifts in the first version of the perturbation theory and we thought whether that would persist in higher orders in perturbation theory is it exact or is it only resolved in first order perturbation theory the easiest way to analyze this question is not to go in higher order in perturbation theory but rather to use the Dirac equation which is something like a relativistic version of a Pauli equation for a formal electron it includes a spin but it's a fully relativistic equation and it turns out that it's possible in the Coulomb field and if you do what you find is you get energy levels even J which are depend only on the energy and they don't depend on L and this is so much complicated formula for it I'll write down like this this is square divided by a great big square root it's 1 plus a big bracket there's C alpha numerator here there's n minus J minus half where there's plus the square root of J plus half squared this whole thing this whole thing of the square brackets gets squared it's a fairly complicated formula like this now Z alpha is the quantity which is small in hydrogen not so small in uranium let's expand this result in the powers of Z alpha and if you do what you get is n minus square the rest of the electron is the leading term the first term then you get minus quantity Z alpha squared divided by twice n squared the next term is C alpha to the fourth divided by twice n to the fourth and this is multiply times this three quarters minus n divided by J plus a half and then you get extra terms out there again like this this is square brackets so they bounce this is the expansion of the of the energy eigenvalues of the Dirac equation in powers of Z alpha what you see is there's a leading term which is the rest of S of the electron which is normally included in the relativistic theories the next term is the energy levels of the electrostatic model these are the more energy levels they're a factor of Z alpha squared down to the rest of S the next the next correction is another factor of Z alpha squared down from that these are the fine structural levels which are just evaluated the same result comes from expanding that there is the eigenvalues of the Dirac equation and if you wanted to you could go on the next order to Z alpha to the sixth and get the next corrections down now actually there's no good reason to do that to go on the order to Z alpha to the sixth because there's other effects physical effects which are smaller than the fine structure effects but they're larger than the next Z alpha to the sixth term which are not incorporated into the Dirac equation and become important physically there's actually two effects that appear at that order one of these is the hyperfine effects which I will lecture on but not the semester and the other one is the land shift the hyperfine effects involve the magnetic interaction of the electron with the proton nucleus there's two spins they have magnetic dipole moments there's a dipole-dipole magnetic interaction I won't say more about that now but like that the land shift however I won't say something about that but because it's relevant to this picture here the land shift involves the interaction of the atomic electron with the quantized electromagnetic field as well as the quantized electron-positron field it involves diagrams like this there's an atomic electron it emits a photon so here's the climate diagram like this it emits and then we absorb the photon diagrams like this give rise to an energy shift which is the which is the land shift the main effect the most important effect of the land shift is that it suppresses the 2p1 half level relative to the 2 1 s 1 half level in other words it breaks the L degeneracy which is present all the way through the Dirac model so if I include the land shift more realistic picture here so if I draw a dotted line for the 2 s 1 half and the numbers are something like this is that the the land shift itself is something like about 1.0 gigahertz if I put it in frequency terms whereas the splitting between the 2p1 half and 2p3 half is about 10 gigahertz about 10 times as large and so this is a this is actually a realistic picture of the n equals 2 levels of hydrogen between all the physics and you see you've got a splitting of the 2 s 1 half and the 2 levels okay well I'm going to stop at this point because I have to hand out these these evaluations and I would like to ask you to call the tier to collect them and then take them to the office next time I'll