 So, we began to get some feel about the fully vataisan transformations, and essentially the fully vataisan transformations achieve quite a lot, but one thing that I want to draw your attention to is this remark from Bjorken and Drell's book that they help us see different interactions of an electron with the electromagnetic field in a form that we can really recognize, we can interpret easily that is the kind of form that we see in various books. If you read Branstad and Joshen's book physics of atoms and molecules, you will see various terms you know which are referred to as relativistic terms, spin orbit interaction and so on. But those terms are not directly manifest in the Dirac equation and what the fully vataisan transformations help us achieve is to display them in a form that can be very easily interpreted. So, our basic problem is right here that you have the Hamiltonian which contains the odd operators and you want to transfer it to a different representation, transform it into a different representation using this transformation relation that we have discussed. And the objective is to make the transformed Hamiltonian relatively free from the odd operators, we cannot make it completely free from the odd operators, but as much as we can. So, we discussed the first fully vataisan transformations and then you have the leading odd terms which is of the order of 1 over m and then what we are going to do is to have another fully vataisan transformation and now you subject the first transformed Hamiltonian to the second fully vataisan transformation which is S 2 which is through this operator and now you get a transformed Hamiltonian which is H double prime and now you do it one more time following the same procedure. So, we have seen how to work out this algebra, it is a little laborious, it is fun to do and when you do it the third time you get a transformed Hamiltonian which is free from the odd operators to a very satisfactory extent and this is the form you get for the Hamiltonian in the Dirac Hamiltonian in the representation after subjecting the original Hamiltonian to three successive fully vataisan transformations. Now, you know what these operators are theta and epsilon, these are the short forms for the various terms that we introduced earlier and if you look at it you still need to evaluate all these commutators and there are a number of terms to work with and we will suggest to you how these terms are analyzed or at least you know the general trend in the analysis of this. So, that you can work out the details. So, look at this term this is theta square by twice m c square which is appearing here. So, let us see it is explicit form theta is the c alpha dot p minus e over c by a. So, now I am using the Gaussian units which is what I need to use and some of the earlier relations I think were in natural units where I had put c equal to 1 and you know h cross was also set equal to 1. So, here we have placed them and now you have this alpha dot this generalized momentum and then you have the square of it. So, now if you plug in the expression for alpha which is this matrix operator 0 sigma sigma 0 then you get sigma dot pi and sigma dot pi from these terms and you can write this as sigma dot pi sigma dot pi times the unit 4 by 4 matrix. And for sigma dot pi sigma dot pi you can use the poly identity which we have used a number of times as a matter of fact we are going to use it quite often it is something which gets very extensively used. So, this is our explicit expression for this term and we have to analyze this further. So, this is what we get for one of the terms in the transform Hamiltonian. Now, you need the pi cross pi over here which will involve this curl of a operator and a dot del operator, but then we have studied this that this is an operator and this will operate on an operand and we have done this analysis earlier we should always be very careful whenever we have the gradient operator. So, this is not just the magnetic field b, but it is the del cross a operator which would operate on an operand. So, you will end up taking the curl of a product of a and f. So, when you do that then these two terms cancel and you get the magnetic field coming from this. So, that is what we will go here and now you have the magnetic field and that tells you that this theta square over twice m c square operator is this pi square over 2 m and then with this i sigma dot pi cross pi term for which we use this relation we get h cross over c times sigma dot b. So, this is the term which will contribute to the beta theta square over twice m c square in the transform Hamiltonian, but then there are the terms there is this beta theta square over twice m c square and there are plenty of other terms to work with. So, let us have a look at this one this is theta comma e plus i theta dot and we have some indication of how this term is to be used. So, substitute the terms explicitly. So, theta is c times alpha dot the generalized momentum which is p minus e a then this epsilon is e phi. So, all I have done is to substitute these terms and I have a time derivative of the scalar product of these two operators. So, I get d alpha by d t dot pi plus alpha dot d pi by d t, but the rate of change of any operator d omega by d t you get it from the Heisenberg equation of motion. So, you can use that and insert it. So, first of all this term would go away this is not going to contribute anything this commutator would go to 0 then you have this a dot alpha dot gradient term and again we have to remind ourselves that whenever we see the gradient term we have to be careful with that operator because this is the commutator of alpha dot del with phi. So, this commutator is alpha dot del phi minus phi alpha dot del operating on f and if you follow the same reasoning as we did earlier you see that these two terms cancel and you are left with only alpha dot del phi. So, that is what you have got from this commutator. So, from this commutator there are two terms which reduce to a single equivalent operator which is alpha dot del phi and we will use this over here and then simplify this relation for combination of these two terms. So, this is the term we are looking at this alpha dot del comma phi commutator we have simplified and this del phi you know is related to the electric intensity, but there will in the expression for the electric intensity there will also be the rate of change of the vector potential. So, there is a minus 1 over C del a by del t and now you can put in this term in place of this commutator. Now, these are known things these things you know from electrodynamics you know them from elementary vector calculus you know them from quantum mechanics and what you do here is put it all together. So, that is what makes it really interesting. So, put it all together and now you have these terms systematically written as e plus 1 over say del a by del t you see where it is coming from and then you have the d alpha by d t and d phi by d t for these two terms and here you can use the Heisenberg equation of motion and use these to represent these derivatives. So, do not forget that this is an operator. So, which is why you have to use the equation of motion for the operator and now you have got from the equation of motion these terms you have collected all of them and since we have them written on this screen it saves me the time to write all these terms on the board, but this entire file is uploaded at the course web page. So, you can go through it carefully and make sure that you follow the derivation term by term, but here I want to show you in the class and this way we can go much faster than we would if I were to write this on the blackboard, but if you have any difficulty you should stop me. So, this entire file has been uploaded and you have these terms coming from the equation of motion. So, what have we got all the terms have been spelled out explicitly. Now, these commutators again if you make use of the matrix structure and look at the commutation properties you will be able to see that they cancel each other for that you have to work out the matrix operations and there is also a vector algebra which is involved you have to respect the vector algebra, the matrix algebra, the operator algebra everything comes together and by using that these terms will make no contribution these two terms cancel each other they are the same with opposite sign this one with a plus sign and this one with a minus sign. So, these two terms cancel each other and you are left you are led to a lot of simplification and this is what gives you one part of this commutator and that means that you have to get the commutation of theta with alpha dot e. So, here you had so many terms now all of them are replaced by just one commutator is now what you have to determine. So, your task is much easier now and when you do that again do it term by term it simplifies a lot because you get rather familiar expressions over here this alpha dot p alpha dot e this will suggest to you what kind of relations you can use because whenever you see terms of this kind you know that you will be able to use the poly identity. So, it is not that you have to remember what am I going to do next those suggestions are built into the structure of the equation and if you just use them you will be automatically led to do the right thing that is the idea. So, here you see the poly identity merging out of this matrix structure and you get the sigma dot p sigma dot e from this matrix multiplication of these two operators these are matrix operators. So, you have the term here in the first row in the first row of the equation. So, in first column will be the sigma dot p sigma dot e from these two right. So, that is how you get sigma dot p sigma dot e over here and also in the 2 2 position, but each sigma is a 2 by 2 matrix of course. So, you are really working with 4 by 4 matrices which is what you always do in the Dirac formalism. So, you got the sigma dot p sigma dot e and now you can use the poly identity which is which will give you p dot e plus i sigma dot p cross e and you have a similar expression over here, but this time you have got the e cross p. So, these are the 2 terms now let us have a look at this p dot e term p again has a gradient operator. So, you have to be careful with it and what you find from a careful handling of this gradient operator which now we have done a number of times you see that this p dot e will give you this scalar p dot e plus e dot p. So, this is not the same as this, this is an operator this is the result of all the operations carried out. So, now you can put all of these terms back in this expression for 1 over 8 m square c to the 4th and you write all of these terms coming from here and you find that this e dot p term cancels this e dot p, this e dot p is a descendant of this term here this is minus e dot p. So, that is the one which is coming here and these 2 cancel each other now you have this these 3 terms 1 2 and 3. So, these are the 3 terms 1 2 and 3 of which p cross e again with a careful handling of the gradient sitting in the p will give you these 2 terms p cross e minus e cross p we have done this. So, there is a gradient sitting over here. So, using this these 2 terms. So, instead of this p cross e term now you have this p cross e coming from here and then here you have a term in minus of e cross p, but then together with this minus sign it becomes a plus. So, you get a term in this plus e cross p and then you can combine this term and this term which is coming from these 2. Now, you notice that the last 2 terms these 2 terms are exactly the same this is e h cross over 8 m square c square sigma dot e cross p. So, these are the same term. So, instead of the 1 over 8 factor you will have the 1 over 4 because you can combine those 2 terms. So, that is what you have got you get the 1 over 4 here coming from these 2 terms and then from here you get this sigma dot curl of e and now you are perhaps beginning to see some familiarity with some of the terms you may have seen in perturbation theory, but they will become even more explicitly manifest after just a few more steps. So, this is what we have got these are the 3 terms. So, you get a divergence of e you get sigma dot curl of e and a sigma dot e cross p. So, let us write insert them in the transform Hamiltonian. Now, when you take care of all the terms and put them back into the transform Hamiltonian which is the transform Hamiltonian arrived at after 3 full d vitties and transformations. The transform Hamiltonian then has got so many terms 1, 2, 3 and then you have got another 1 over here and all of these terms are now amenable to easy interpretation rather easy interpretation. This was not the form in which we could see them in the Dirac Hamiltonian they were there it is not anything that we have added ad hoc. This is come straight out of Dirac Hamiltonian by subjecting it to appropriate transformations you immediately see that this electric intensity is nothing but the gradient of the potential and this e r is this position vector divided by r. So, you see the form minus 1 over r del v by del r and now you have got this r which will come here. So, you will get sigma dot r cross p which is the sigma dot l which is the spin orbit term that you would have seen in perturbation theory. It comes so nicely out of the full d vitties and transformations it was there in the Dirac equation but it was not visible and when you subject the full d vitties and transformations to 1, 2, 3 transformations then in the third result which is the h triple prime you see this form appearing. So, this is the e h cross p. So, over 4 m square c square 1 over r del v by del r sigma dot r cross p then you have got a term in p to the 4 this happens to be the relativistic kinetic energy correction I will discuss this in just a few minutes. Now, this is the relativistic kinetic energy which is e minus m c square and you remember we defined the energy mass equivalence very carefully at the very beginning without which we would not have this connection. So, there was a strong reason to spend some time discussing that and you will see that it is important. So, the relativistic kinetic energy is this difference which is the energy minus the rest mass energy what you take away after removing the rest energy is the kinetic part. So, that is the relativistic kinetic energy and now all you do is to expand this term to the power half and when you do that you get a power series and you have got c in the denominator whose powers keep increasing. So, you can truncate it at some level of approximation wherever you want and if you keep the leading term the most important contribution comes from the p to the 4 term. So, this is the relativistic kinetic energy term this is precisely the term that we saw in the foldivotison transformed Hamiltonian. So, there was a term in p to the 4 and we see its origin in the fact that the relativistic kinetic energy is different from the Galilean kinetic energy the two are different. So, its origins can be traced the origin of the p to the 4 term which is here this can be traced to the relativistic kinetic energy part this is what we just saw. This sigma dot b is a magnetic dipole moment term as is easily recognizable then you have got a term in the divergence of e and this was introduced in the pre foldivotison days as a perturbation and it was referred to as the Darwin correction. So, in some books on perturbation theory they will begin with a non relativistic Hamiltonian and add perturbations that this is the perturbation due to the relativistic kinetic energy because your original Hamiltonian is not relativistic let us take into account some of the relativistic features. And let us do it perturbatively because after all the non relativistic theory is not absurd it gives a good starting point it certainly does at the reason of course, is that the speed of light is extremely high even if finite it is extremely high. So, the non relativistic Schrodinger equation non relativistic quantum mechanics is not absurd, but not quite accurate. And we are now looking at those effects which are really important because an atomic spectroscopy and atomic processes it does become important to take into account relativistic effects and not just for heavy atoms, but even for very small atoms even for the hydrogen atom. It is absolutely important to take these effects into account and these could be introduced perturbatively in non relativistic mechanics. So, you can add the kinetic energy correction as a perturbation you can add this is the correction you can add the spin orbit correction actually these two terms together contribute to the spin orbit term rather than just the last one. The last one is more famous because curl of e for the hydrogen atom vanishes that is spherical symmetric potential. So, you know the curl of e vanishes. So, this one is not of important, but if you are dealing with the potential which is not strictly central field then you are going to have to take this into account and you cannot throw it away. So, relativistic quantum mechanics will give you not just the sigma dot l term not just the s dot l term, but an additional term which must be taken into account as an integral part of the spin orbit interaction that is a relativistic effect for a central field it is just the sigma dot l and of course, there is this d v by d r and so on. So, there are these terms and one can make corrections perturbatively and it is a good idea to see the correspondence. So, let us look at the perturbative correction due to the kinetic energy part this is the relativistic kinetic energy correction coming from the p to the 4 term. Now, you see that this is the square of p square and this is e n minus v whole square and you really need the expectation values of the potential energy and also the square of the potential energy. So, you need the average values of 1 over r because the potential energy goes as 1 over r and you also need the average energy of 1 over r square. Now, these problems you would have done in non relativistic hydrogen atom how to get the average expectation value of any operator. So, if you take the n th wave function for the hydrogen atom and determine the average values of 1 over r and 1 over r square you will see that 1 over r goes as 1 over n square 1 over r square goes as 1 over n cube a is a more constant and then there are other constants other quantum numbers which come in. So, you can use these relations put them in the expression for the average energy for the potential energy and the square of the potential energy and that will give you an estimate of the perturbative correction due to the relativistic kinetic energy and it turns out to go as alpha square where alpha is the fine structure constant. This is the reason it is called as a fine structure constant because it changes this non relativistic structure even if so in only a fine small detail and it contributes some details to the atomic structure. So, this is the fine structure constant, but this is not the only term which is coming because of relativity and this is the most important lesson from the Foldy-Vortaisan transformations because if you do perturbative corrections to non relativistic Hamiltonian, then which are the corrections you have to make means you could make the Darwin correction you could make this correction you could make this correction. So, there are three corrections that you can talk about and perturbatively you might say that this is the most important one I will do that I could do the second term or I could do the third term or I could do the second and the third, but not the first, but whenever you are making corrections it is extremely important that if you include one term of a given order then you consider all terms of the same order and you are certainly making an approximation. So, it is not an exact formalism that you are developing anyway whether it is the Foldy-Vortaisan scheme of doing things or the perturbative way of doing things you are not doing an exact analysis you are making approximations. So, make sure that your approximation is sensible because you when you make an approximation and say that argue that these terms I am not going to take into account because they are weak they are ignorable nothing wrong in making that argument, but you cannot ignore those terms which are of the same order as some of the other terms you have chosen to include. So, you have to take into account all of them and when you do not do that you will obviously, get wrong results. So, now this is the correction due to the relativistic kinetic energy and now let us see the correction due to the spin orbit term which is the sigma dot l term over here and here again you can go ahead and determine the expectation value of this operator this is the first order perturbation theory result which is quite familiar to you. So, you need the radial integral for 1 over r cube because del v by del r will give you the derivative of 1 over r which goes as 1 over r square. So, you need the expectation value of 1 over r cube because of that reason and then you get you need the expectation value of this operator in angular momentum states. So, when you evaluate this s dot l you know how to get it from j square because j is l plus s. So, j dot j will be l square plus twice l dot s minus s square and you can you know do the substitutions and get this result. So, you have a twice s dot l terms. So, you will get h cross square by 2 j into j plus 1 minus l into l plus 1 minus s into s plus 1 and then the average value of 1 over r cube will give you this term in 1 over n cube and some of the other quantum numbers is including the orbital angular quantum number. So, this is the correction to the energy because of the spin orbit term and it is as important as the relativistic kinetic energy term which is sometimes referred to as the relativistic mass term, but more appropriately it should be called as the relativistic kinetic energy term because we do not make a distinction between mass and energy that something that we have already taken care of. So, you can simplify this expression for the average value of the spin orbit interaction. These are the constants that you make use of the Bohr radius and the fine structure constant and you find that this is the same as what is some books call it as relativistic mass correction, but it is a relativistic kinetic energy correction as I prefer to call. So, this is what you get the spin orbit correction again goes as z alpha square. Now, this is the correction from relativistic kinetic energy which goes as z alpha square and now you are left with one more term because here is a simplification of this further the spin orbit interaction of course, is important only for l not equal to 0 states because for l equal to 0 the l plus s and l minus s will give you the same thing. So, this is when l is not equal to 0 you can explicitly put in the values and find these simplified expressions for the spin orbit splitting. What about the divergence term? This is the Darwin correction. Now, the Darwin correction requires you to take the divergence of e which is the divergence of 1 over r square which gives you a delta function. The divergence of 1 over r square gives you a Dirac delta and you will therefore, need the average value or the expectation value involving this delta function. So, when you evolve when you determine the expectation value of the operators you must use the delta function integration expectation value is an integral obviously. So, this is let us say you take the expectation value for some n s state this is important for l equal to 0. So, this is let us say for some non relativistic wave function psi n with l equal to 0 and m equal to 0 and you determine this integral and this will give you minus e n z alpha square over n. So, again you find that the Darwin correction is also of the same order of magnitude as the other two. So, it is important to take into account all the three if you want to make relativistic corrections at all or none at all. If you do only one or the other then it is not enough and that is the difficulty with perturbative approach to quantum mechanics. Because whenever you try to make improvisation and add terms which you even if they are in the right direction they can still give you completely wrong answer because you may have two contributors of the same order of magnitude. They could come with opposite signs and they could just kill each other or they could come with the same sign they can they could add to each other. So, these are various factors that you have to consider when you do perturbation theory. So, this of course is important only for l equal to 0 because it requires a finite amplitude of the wave function at the nucleus at r equal to 0. And you know that all the radial functions of the hydrogen atom they go as r to the power l as r goes to 0. So, as r goes to 0 the wave function goes as r to the power l and for l equal to 0 this will be finite at the origin at r equal to 0. But for l equal to 1, 2, 3 and everything else this would go to 0 and it would go to 0 faster for higher values of l which is a centrifugal barrier effect which I had discussed in unit 1 I believe. But this is of importance for the n s state. So, this is for the l equal to 0 state that this is important. And then you have these three corrections the relativistic kinetic energy the spin orbit correction the Darwin correction they are all of the same order z alpha square. And you therefore put them all together and add the net effect of these three terms to the correction. And these three terms when you add together you get a net relativistic correction for all orbital angular momentum quantum numbers. And this is the result that you get from relativistic quantum mechanics. But if you did not use all the three but only one or this this is not what you are going to get. So, please make a very major note of this point that all perturbations of the same importance must be included together. So, now our primary intention in doing this exercise was to solve the Dirac equation for the hydrogen atom. Now, we have seen what some of the terms mean and these are some of the references that I like very much for this topic Bjorken and Greiner and Triggs quantum mechanics messiah. So, these are some of the books that I have used and for the coulomb problem for the hydrogen atom you can solve the Dirac equation exactly. In all the forms that we used until now we had the potential v of r but we never actually specified it to be the coulomb potential. So, for the coulomb potential Dirac equation has got an exact analytical solution. So, now let us see how we are going to handle this and it is not at all trivial. The reason to discuss it is because it is not a trivial extension of the non relativistic methods. In non relativistic quantum mechanics you varies in a very simple manner you separate the radial part from the angular part because of the 1 over r potential. Now, that is not so obvious for the Dirac case because of the presence of the odd operators and if you look at the Dirac equation now. In fact, if you look at the Dirac equation even for a free particle let alone a particle in an electromagnetic field. So, even if you throw the terms in A and phi you find that the orbital angular momentum operator and the spin angular momentum operators do not commute with the Dirac Hamiltonian even for a free particle. And then of course, they will not commute for the Dirac Hamiltonian for the coulomb problem because those terms are also there. So, it is not going to commute which means that these will not give you good quantum numbers because our whole idea of good quantum numbers is to get them from those operators which commute with the Hamiltonian. So, that they correspond to simultaneous compatible observations. So, what constitute compatible observations to give you good quantum numbers is a question and we have to get the good quantum numbers for the Dirac Hamiltonian and it is not going to be n l m. L cannot be a good quantum number because l does not commute with the Hamiltonian. So, your quantum numbers are going to be different because of the spin orbit interaction, the spin orbit coupling and then parity also has to be defined in a very careful manner for the Dirac case because of the beta operator which is what is that 1 0 0 minus 1. So, you have to be careful with parity as well. So, this is our spherical symmetry Dirac Hamiltonian and now we want to exploit this spherical symmetry. So, somehow we want to separate the equation into a radial part and angular part and we have to extract angular features and radial features. We have to identify them separately and the main question is how are you going to handle this term this alpha dot p. So, let us discuss this. So, these are our operators I define two new operators. So, this operator which is got 1 in the off diagonal position is what I call as rho 1 I will use it alpha we have already used which is the Dirac operator 0 sigma sigma 0 and this is sometimes called as the Dirac spin. This is made up of the poly spin this is sigma sigma along the diagonal along the off diagonal is the Dirac alpha along the diagonal it is the sigma into the unit matrix operator or this is sometimes written with a sigma with a superscript d for Dirac or sometimes with this uppercase sigma. So, these are the various notations that you see sometimes it is written only as this sigma which is the same as the poly sigma, but from the context you know whether it is the poly sigma or the Dirac sigma. If it is the Dirac sigma it is a 4 by 4 vector operator likewise the spin angular momentum which is h cross over 2 sigma is now you extend this idea to the 4 by 4 operators and this is h cross over 2 this uppercase sigma. So, this is the notation I will be using. So, this is h cross over 2 sigma and you have got a unit operator you can also write this as h cross over 2 sigma or you can just not write this one operator in which case you will read it as just the usual poly h cross over 2 sigma, but from the context you should know what you are talking about. So, notice this rho 1 sigma is the same as alpha rho 1 alpha is the same as sigma and if you look at sigma dot p you can extract the sigma dot p and you have a unit operator. So, sometimes you will find books in which this sigma dot p is written only as this sigma dot p, but they are the same things they are the same yet different one has got a 4 by 4 structure and the other has got a 2 by 2 structure. So, which is why one is the poly sigma the other is the Dirac sigma, but you have to be careful while reading literature because many books do not alert you to this although I am sure that they would have said that somewhere, but you have to read it very carefully. So, this is sigma dot p now I have got sigma dot p here as well. So, obviously the rest of it over here is the unit operator which you can easily recognize because this is the projection of sigma in some direction because this r over r is the unit operator I do it twice and sigma square along any direction is equal to 1. So, this is just the unit operator. Everybody is comfortable this is just the projection of sigma on some direction sigma dot u and what is the square of sigma dot u it is like sigma z square it is equal to 1. Now, everybody will agree it is right indeed it is. So, it is just like sigma z square which is equal to 1. So, that is what you have got and now here you make use of the poly identity. So, this is r dot p plus i sigma dot r cross p here you would expect a 4 by 4 unit matrix sitting implicitly. So, this is sigma dot r this is r over r square. So, this is e r over r this is the unit vector in the direction of the position vector and this r dot p gives you if you represent the gradient operator in the spherical polar coordinate system you will need to work with only one of the components which is this and you know that the radial momentum operator is not just minus i h cross del by del r it is minus i h cross del by del r minus i h cross over r. So, be careful about it because you have to put that in the expression for r dot p which is minus i h cross r del by del r which is r p r plus i h cross. So, you have this extra term. So, now you write this expression for sigma dot p which is sigma dot e r and then you had r dot p which is here which you know is r p r plus i h cross. So, you can put that over here you have got a sigma dot l coming from here. So, you put all the terms together and you have essentially sigma dot e r, but then our interest is in alpha dot p and what is alpha it is rho 1 sigma and rho 1 is this operator this matrix which has got 0 1 1 0 with the elements 1 in the off diagonal position. So, when you pre multiply sigma by rho 1 you get the alpha. So, the rest of the expression is the same and you have alpha dot p which is alpha r p r i over r and then you have got h cross plus sigma dot l. Now, p r I have written explicitly in terms of this minus i h cross del over del r minus i h cross over r and now this term and this term will cancel. Now, this term now you see on the right side you have got expressions involving radial character and this is what we set out to do because our Dirac Hamiltonian for the spherical potential is the c alpha dot p plus beta m c square plus v and now we have succeeded in writing alpha dot p with radial features. It is not completely separated into radial and angular part and you will see that in the next class, but you are beginning to see that you are now able to see the radial features which is what you need to separate the radial part from the angular part and to be able to do the analysis completely we are going to introduce a new operator which is defined as k which is this h cross plus sigma dot l h cross plus sigma dot l why is this so important it is coming here h cross plus sigma dot l. This is the one which we think could create some trouble for us this is the operator which needs to be carefully handled. So, we in fact define a new operator which is beta k which is equal to h cross plus sigma dot l. So, that makes k itself go as beta h cross plus sigma dot l beta square is equal to 1 we know that. So, this is a new operator that we introduce and now in terms of this new operator your alpha dot p operator becomes alpha r p r plus i over r beta k where k is this new operator that we have defined. We can rewrite the Dirac Hamiltonian in this form in terms of this new operator and if you collect all of these terms together you have got the Dirac Hamiltonian for the spherical potential in terms of the k operator and this is where I will stop today and take this up from here in the next class. So, we need one more class to see how you actually work out the separation between the radial and the angular part and then see how you get the solutions to the Dirac equation the Coulomb functions and then use them in your relativistic applications. So, thank you very much.