 So, we saw the quantization relation last time, we have to quantize this momentum operator, momentum in classical mechanics is a dynamical variable and in quantum mechanics it is an operator. So, we begin with the momentum scalar which is this p mu p mu which we found is equal to m square c square and now we know what mass we are talking about and you will recognize the importance of defining mass carefully, because you cannot define a relativistic mass separately since mass and energy are equivalent. The interpretation of mass it is usage in the relativistic equation is of importance it is at the very heart of relativistic quantum mechanics, we did this in our previous class and that is a mass which goes in over here and we see that this scalar p mu p mu is an invariant quantity, because it is equal to the square of m and the square of c and this m is as much invariant it is as much as scalar as c s it has to be the same in every frame of reference, if it was the other mass it would not be. So, we have this relationship this p mu p mu this is the Einstein summation convention as it is sometimes called that if you have an index which is repeated then you sum over it. So, this is the sum of four terms and one term gives you this e square over c square and the remaining three terms give you the conventional three dimensional scalar product p dot p. So, those are the four terms which are included in this summation. Now, we follow the same quantization prescription, because we have done this in non relativistic quantum mechanics that this operator is replaced by the gradient operator. Likewise this operator p 0 will also be replaced by the corresponding gradient like term which is the derivative essentially gradient is the derivative with respect to space, but then we are not making any distinction between space and time and here we take therefore, the derivative with respect to the fourth coordinate which is time and therefore, the e over c which is p 0 when quantized becomes the derivative operator with respect to time. So, this is our quantization condition and when you plug in this in this invariant relationship and now that you have an operator for completeness the operator to get physics out of it you would have the operator operate on an operand which the wave function is and then from this operation you would then develop the algebra further and directs tells us and then extract physical properties about it. So, this is the equation that you get by having the operator operate on the wave function psi which is the Klein Gordon equation sometimes you find it written in a different notation because this whole this operator which is the set of these operators gradient like or Laplacian like operators this is the second derivative with respect to time and second derivative with respect to space both. So, this is sometimes called as a D elimination operator written as a box it is also called as a box operator sometimes and this is your Klein Gordon equation the difficulty with that equation is that it leads to an indefinite probability density and there are other issues fairly complex issues some of which can be handled some of which cannot be handles. So, the Klein Gordon equation has it is own range of applicability it also has it is limitation and it is not the appropriate equation for electrons our focus of interest is the atomic structure and we want to describe the electron dynamics in an atom. So, we are going to look for a relativistic quantum equation for an electron which the Klein Gordon equation is not and therefore, we will not discuss this any further the only thing I will like to point out over here is that you should note that this is the second derivative with respect to time whereas, in the Schrodinger equation you have the first derivative. So, you would like to look for an equation which was one of the motivations to look for an alternative equation that was not the only equation, but again I am not going to be able to trace the historical development of the Dirac equation. So, that will take me off the target and that is not my intention, but those of you who are interested will find it very interesting to read some of the developments in relativistic quantum mechanics. So, you are going to look for an equation which involves the first derivative with respect to time and this has been achieved in Dirac's equation in two excellent papers in proceedings of the Royal Society of London in 1928. And there are excellent sources for Dirac's work including his book and this is the fundamental relation which is at the very foundation of atomic structure and atomic processes in which we are really interested. So, we know that P mu P mu this scalar is Lorentz invariant, this is a fundamental requirement of any relativistic theory because we are branching out from Galilean relativity. So, we have to look for a relationship which has the dynamical variables which can be quantized. So, it has to have the momentum operator. So, this is obviously the correct relationship to begin with, it has got the attractive features that it has got the momentum built into it, it has got Lorentz invariance built into it. We are interested in quantizing it, but we are also interested in looking for an operator which involves the first derivative with respect to time whereas, P mu P mu as we saw in the Klein Gordon equation obviously has the second derivative with respect to time. So, can we play with this relationship a little bit, so that we can look for how to extract the first derivative term rather than the second derivative which is manifest over there. So, these are the quantization conditions which we will continue to use, but we are going to look for a relationship which will have the first derivative in time rather than the second derivative which is manifest. So, what you can see is if you look at this relation the P mu P mu it this scalar expanded is P 0 square minus P dot P minus M square C square. So, this is our fundamental relation and if you now take a special case in which this term is 0, if the three momentum the traditional three momentum the three dimensional momentum that we normally use or use even in non-relativistic mechanics. If this momentum is 0 then you get only P 0 square minus M square C square equal to 0 and this is like a square minus b square equal to 0. So, you can factorize it as a plus b and a minus b. So, P 0 plus M c into P 0 minus M c equal to 0 you can factorize it and this relation is valid either when P 0 plus M c is equal to 0 or when P 0 minus M c is equal to 0 or both and now you have P 0 if you peel out one of these factors and set P 0 plus M c equal to 0 or P 0 minus M c equal to 0 then you get P 0 alone and not the quadratic momentum. The 0 th component of momentum is now no longer quadratic and you get the first derivative with respect to time. So, your requirement of finding the first order time derivative equation is satisfied your consistency with Lorentz invariance is also satisfied because it has come out of that basically and you can take either of these two relationships and continue to develop the algebra further. You can take either P 0 plus M c equal to 0 or P 0 minus M c equal to 0 you can take either of these and it turns out that it really does not matter which one you take because you are led to the same physics no matter which you take, but that is a matter of detail and essentially what we find is that this factorization is possible and we then ask that is factorization possible when P is not equal to 0 because we know that we cannot be dealing with special cases. If you do the algebra only with special cases then you can apply it only to special cases and that will limit our range. So, we are looking for a relationship in which you have a similar kind of factor, but now it cannot be based on assuming that this three vector scalar product P dot P is equal to 0 that is a tough one and that is where you need somebody with the intuition and intellect of Dirac. So, what Dirac did is to explore a factorization of this kind explore try it out set this quantity on the left hand side equal to a product of two factors and you know that it cannot be easily factorized. So, you insert some unknowns and you insert a beta over here and a gamma over here and then you ask is such a factorization possible because if it turns out to be possible then you can peel out one of these factors and set it equal to 0 and you will get the del by del t term. So, there is some motivation for it there is some hope, but then there is the query as to what will make such a factorization possible will some very peculiar properties of beta and gamma which are the unknowns over here will they make such a factorization possible may be not and inspired by the hope that it will be possible you then demand what properties of beta and gamma will enable such a factorization and then you include that in your condition. So, this is what you are going to look for which is to explore the possibility of factorization of this term and ask the question what properties of beta and gamma will enable this notice that beta kappa p kappa is actually a summation over kappa. So, there are four terms over there kappa goes from 0 1 2 3 that is the Einstein summation convention likewise you have four terms over here this is gamma lambda p lambda. So, lambda takes four values 0 1 2 and 3 and between these four terms and these four terms there are eight unknowns which are to be determined that is a part of your exploration process. So, you ask is such a factorization possible and then you just expand these products. So, beta kappa p kappa times gamma lambda p lambda gives you the first term and there are two terms in the first bracket another two in the second bracket. So, you get a set of four terms and you make sure that you write them in a constant consistent order because you have to be careful about commutation properties if any are involved and if they happen to commute you would not have to worry about it, but that is a question that is a matter of detail. So, you get these four terms. So, these are the four terms 1 2 3 and 4 and you find that this m square c square minus m square c square is common to both the left side and the right side. So, you can cancel it that makes life easy and you are left with fewer terms and since lambda is summed over it is a dummy index and instead of lambda you could use kappa as well it is a dummy index. So, instead of gamma lambda p lambda I use gamma kappa p kappa because now I find that I can actually combine these terms in which I have a summation over kappa and then I get the left hand side equal to this quadratic term in momentum and then minus m t m c times the linear terms in momentum. Now, this is an interesting relationship because you find that the left hand side is quadratic in momentum you find that the first term is quadratic in momentum, but this one is not and that suggests that to get rid of the linear term you can choose beta to be equal to gamma. If each beta kappa is equal to the corresponding gamma kappa beta 0 is equal to gamma 0, beta 1 is equal to gamma 1 then you can get rid of the linear term and then on both sides of the equation you have quadratic terms and you can really balance the equation. So, that is a good strategy that you can use. So, our query was what properties of beta and gamma would allow such a factorization we get a partial answer to it that whatever beta and gamma you discover or you hope to discover will need to be equal to each other. So, that is something we have made some progress beta must be equal to gamma now that we know that beta must be equal to gamma we can put that beta equal to gamma over here. So, let us do that. So, the left hand side which is p mu p mu scalar which is equal to beta kappa which is the same as gamma kappa because that is something that we have already learned and now you have got a summation over kappa and also a summation over lambda each taking 4 values. So, you get 16 terms of the right which should give you the 4 terms of the left and that will put some additional requirement on the right hand side which will lead us to what gamma must be like. We have already learned that beta must be equal to gamma and now we are going to find what this unknown gamma will turn out to be. So, this is the condition that must be satisfied. So, let us look at these 16 terms of the right and these 4 terms of the left carefully. So, this is the relationship that has to be satisfied. So, let us write this explicitly. So, I have summed over kappa for kappa equal to 0 I get gamma 0 and this kappa is also equal to 0. So, this is p 0 and then there is this gamma lambda p lambda. So, this is the double summation. So, there are actually 4 terms sitting in this single term likewise there are 4 terms sitting in the second another 4 over here and another 4 over here. These are the 16 terms that we had referred to. So, now you sum over lambda explicitly. So, these 4 terms you sum over lambda. So, lambda equal to 0 is the first term. So, this is gamma 0 and this lambda is 0. So, this is gamma 0 and this p 0 and then you take the next value of lambda and whatever I done or this would be p 0 p 1. This would be p 0 p 1. Now, it is all right this is actually a typographical error, but in our situation it really does not matter which is why it did not hit me. Because the components of momentum we know that they actually do commute with each other x does not commute with p x, but p x commute with p y p y commute with p z. So, it really does not matter and to look at these terms I have written the same 16 terms on this slide, but I have inserted some gaps. So, that I can show you what is going on with the terms what you will make use of is this commutation which I mentioned that p i p j is equal to p j p i which essentially means that these 2 terms p 1 p 0 and p 0 p 1 these are actually equal to each other and you can combine these 2 terms. So, these 2 terms can be combined then you do the same with the remaining ones. So, you have got p 2 p 0 over here and p 0 p 2 over here. So, essentially what you are going to find is terms I have written these 16 terms in such a manner that those terms which are equidistant from the diagonal can actually be combined that is how I have written them. So, terms which are equidistant from the diagonal can be combined. So, you have this term which is p 3 p 0 which is equal to p 0 p 3. So, these 2 terms can be combined then these 2 terms can be compiled which is p 2 p 1 and p 1 p 2 then these 2 terms p 3 p 1 with this p 1 p 3 and finally, these 2 terms p 3 p 2 and p 2 p 3. So, all of these terms which are equidistant from the diagonal can actually be combined and that leads to some simplification that we are looking for. So, this is how we have combined them these 4 are the diagonal terms with the indices 0 1 2 and 3. So, these are the terms in the i th row and i th column i going from 0 1 2 3 and then the off diagonal terms which we decided can be combined how do these coefficients gamma 0 gamma 1 and gamma 1 gamma 0. So, no approximation made as yet no postulate made as yet we have written them exactly and we know that this must correspond to the left hand side which has got only these terms only 4 terms. So, now what is it that we can demand on gamma. So, that the 16 terms on the right hand side will give you the 4 terms of the left can you make some demand on gamma. You see that if gamma 0 square is equal to 1 you get the first term happily you see that if gamma 1 square is equal to minus 1 you get the second term. So, you start making these demands and then you have to get rid of these terms. So, our question is if we could achieve that then the factorization of the Dirac equation not in the raw equation that we began with from the invariant momentum scalar, but by inserting unknowns betas and gammas and by making demands on betas and gammas. We find that such a factorization is possible and these are what the gammas must be because they have the right properties that we are looking for. Notice that these are 4 by 4 matrices these are not just numbers these are matrices. They have a block diagonal form as you can see right they have got a structure which is immediately manifest. They have got a block diagonal structure and you can see that this is a 2 by 2 unit matrix. This is a 2 by 2 negative unit matrix and what you find in the remaining positions are the poly matrices the poly 2 by 2 matrices which we have used earlier. So, you have a sigma 1 over here and a minus sigma 1 and then you have got sigma 2 and sigma 3. So, these matrices which are made up of the poly matrices, but the poly matrices are 2 by 2 matrices these are 4 by 4 matrices they appear in a block diagonal form and these are the poly 2 by 2 matrices which we are going to be using and with the use of these matrices we can actually factorize the Dirac which we can factorize the invariant momentum scalar P mu P mu which is what leads to the Dirac equation. So, together with the poly matrices the 4 by 4 matrices are called as Dirac matrices and this is the 0 th component which is which is written here as a 2 by 2 matrix, but each element is a 2 by 2 matrix. So, these are the 4 by 4 matrices. So, this is the structure of these 4 by 4 matrices these are called as Dirac matrices and there are 3 poly matrices gamma i i equal to 1 2 3 and these are made up of the 3 sigmas which are the 3 poly matrices i going from 1 2 3. So, we have in fact found that factorization of the 4 momentum scalar product is possible and it is possible by demanding that the gammas are 4 by 4 matrices. Now, this is wonderful because now you can peel out one of these factors and again it does not matter which factor you peel out you can take either this or that there are two of these factors and either one of them must be 0 or both of them could be 0, but you can take any one at this is the one that one normally takes and it does not matter which one you take this is the one that you take gamma kappa P kappa minus m c this is the factor equal to 0 and this is the summation over kappa going from 0 1 2 and 3. So, there are 4 terms in this summation the momentum is quantized there are 4 momentum operators the 0th component gives you the time derivative the remaining 3 components give you the space derivatives and you quantize this and keep track of the indices which is the superscript which is the subscript which is contra variant which is covariant you can lower the indices make sure that the operators are written in a consistent fashion. And then you have got an operator relation because this momentum is now replaced by the operator which is the derivative operator and this derivative operator this is just a matter of notation del kappa is derivative with respect to x superscript kappa. So, just use the covariant and contra variant indices carefully and this operator as we know from non relativistic quantum mechanics as well would operate on an operand which is what our wave function would be and the resulting equation would be the quantum relativistic equation of motion. The operator here however is now a 4 by 4 operator it has got an operator structure it also has a matrix structure and that demands that the wave function over here must have 4 components. Now, this equation can be solved for a few problems exactly for other problems you have to make certain approximations our interest will be in the hydrogen atom for the coulomb field for which exact solution is possible. And we will discuss that this is sometimes referred to as the Feynman's notation this gamma kappa p kappa is written as a p slash. So, this is sometimes called as a slash notation or Feynman notation but this is just a matter of notation basically this is what it is whenever you see a p slash you should recognize that it is gamma kappa p kappa and it is a set of 4 terms which are summed over. So, this is your Dirac equation now you can recognize these gammas to be the 4 by 4 matrices. So, this is gamma 0 p 0 this is gamma 1 p 1 look at the matrix structure. So, you can simplify some of these things by just doing matrix algebra and you will find that this p 0 will come here here here and here. So, you can write this relationship in a matrix form as well. So, this is just a matter of you know writing it in different forms that you might find in different books, but essentially what we have got is a relativistic quantum equation in which you have got first order time derivative operator which is nice because it has got something similar to the Schrodinger equation, but then you also have the first derivative operators with respect to space, but that is ok let us see what it leads us to, but we should certainly make a note of it. So, this is what we have got you can now you are dealing with very simple matrices you can use the poly matrices their properties are well known you have done some algebra and you know manipulations with these matrices. So, you can easily figure out that alpha is beta inverse gamma and you know you can write this in a representation which is called as a poly representation in which you use two operators beta and alpha, alpha is defined as beta inverse gamma where gamma is made up of these poly matrices and this is your alpha matrix. So, alpha is equal to 0 sigma sigma 0 this is a 4 by 4 matrix each element is made up of the poly matrices which are 2 by 2. So, in the poly representation instead of the 4 gammas you use 1 beta and 3 alphas, but this is just a matter of renaming them there is no new physics there is no new mathematics it is just a new nomenclature and that is the one which is commonly seen in a lot of literature. The 2 by 2 matrices the poly operators operate in the poly space the 4 by 4 Dirac operators and what is sometimes referred to as the Dirac space and you can see that the poly space is a subspace of the Dirac space. You can do a lot of very interesting mathematics with the Dirac matrices and this is a good exercise although I will not spend much time discussing the mathematics of the Dirac matrices, but these properties can be very easily verified and I will not spend any time on it because we will not be using some of these relationships directly in our development of the subject. But, I will certainly like to mention that you can actually build additional matrices like from the gammas you can build the gamma squares you can build the sigmas like defined as i gamma mu gamma nu you can build these additional matrices not all of them will be linearly independent, but 16 linearly independent matrices can be built and you can classify them in different you know structures. So, you have got only one element of this kind you have got 4 matrices of this kind with mu going from 0, 1, 2, 3 you get 6 matrices of this kind you get 1 matrix of this kind and another 4 of this kind. So, you can you know place them in different sets or place them in 5 sets which are the conventional sets in which these are structured and the reason to do it because if you consider their transformation properties under Lorentz transformations then they have similar properties. This one in the first set you know it transforms as a scalar these transform as a vector these transform as a tensor this one as a pseudo scalar and this one as an extra vector. So, there are these are the reasons that they are put in different sets and I will not spend too much time on this I will proceed with the Dirac equation, but now we are going to invoke the electromagnetic potential that is of importance to us because we know our interest is in the hydrogen atom and the electron is in the presence of an electromagnetic potential right it meets the coulomb potential of the nucleus that is the electromagnetic potential right. So, we have to look not just for a relativistic quantum equation, but for a relativistic quantum equation which also has the electromagnetic potential. So, this is the electromagnetic potential now this is it has got the 4 components this is the electric scalar potential this is what you very often call as a magnetic vector potential and we make an ansatz for the Lagrangian for the system for the electron in the electromagnetic potential we make this ansatz that the Lagrangian will be given by this. And this is the point which I am sure has been emphasized a number of times in your introductory quantum mechanics class that whenever you set up the Hamiltonian you never write it as t plus v or anything like that the first thing to do is to set up the Lagrangian for the system then obtain the Lagrangian is always in terms of position and velocity right. And then from the Lagrangian you find the momentum the generalized momentum right and once you have it then you proceed to build the Hamiltonian and then you quantize it. So, we begin with the Lagrangian for the system we make an ansatz that this would be the Lagrangian we need to verify if it is the right Lagrangian it is not something that we are going to take for granted. So, we propose this Lagrangian and we ask if it satisfies the Lagrangian equation and if it does what kind of relationships come out of it because the Lagrangian Lagrangian equation must give you the equation of motion for the electron in the electromagnetic field. And we already know that do not we the equation of motion for an electron in an electromagnetic field mass times acceleration is equal to the force. And we know that the force is the Lorentz force which is f is equal to charge times what is it v which is or phi which is the scalar potential right plus the v cross beta the electric intensity e plus v cross b right e plus v cross b times the charge will give you the Lorentz force. So, does this Lagrangian give us the equation of motion is the question that we ask. So, we look at it set it up for each component. So, q are the three degrees of freedom for this Lagrangian and we set up the v square is the sum of these three components square right. I am using a simple Cartesian coordinate system it is very easy to use and it serves our purpose in this case. So, your m v square the kinetic energy part in the Lagrangian is given by this your Lagrangian which is t minus v gives you this minus q times phi. And then you have got this velocity term and the vector potential on the equation of motion that you expect is this. So, the what you do is find out what the momentum is. And the momentum of course, is the derivative of the Lagrangian with respect to the velocity momentum is not mass times velocity it is much more than that. The primary definition of momentum is the partial derivative of the Lagrangian with respect to the velocity. So, you take the derivative of the Lagrangian with respect to the velocity and you find from this term you get m into v x, but then this term also has got the velocity. So, the derivative of this term with respect to velocity will give you this q by c a x. So, your momentum now which is the generalized momentum not just the traditional mechanical momentum. So, this traditional mechanical momentum we can quantize using the gradient operator, but momentum itself will include this vector potential as well. So, the next thing you do in the Lagrangian's equation is to take the time derivative of the partial derivative of Lagrangian with respect to the velocity. So, that is what it is you take the time derivative of the right hand side. So, mass is a constant you get the derivative of velocity which is v dot which would be the classical acceleration. So, there is a dot on this v it is a tiny dot, but do not ignore it that is the time derivative of the velocity and you get the time derivative of the vector potential which could be 0 if the vector potential is not independent of time, but it would not be 0 in general. So, now you take the partial derivative of this vector potential with respect to time, but the dependence of the vector potential on time is through an explicit dependence of the component on time and an implicit dependence through the dependence of this a on the coordinates which in turn depend on time. This is the convective derivative like idea that you have used in fluid dynamics or in electromagnetic theory earlier. So, the same kind of you know reasoning is involved all you have to do is to recognize that the dependence on time is not just because of how a x depends explicitly on time, but also on how it depends implicitly on time via its dependence on the position r which in turn depends on time. So, you have a derivative with respect to x which in turn depends on time. So, this is d x by d t likewise you have got terms in y and z and then you have got the final term which comes from the explicit dependence of a x on time. So, you handle this derivative carefully and that gives you the derivative of the momentum which is akin to mass times acceleration, but not just the traditional Newtonian mass times acceleration, but it has got these terms coming from the vector potential as well. Then this must be equal to del l by del x and del l by del x is something that you can obtain from here because you find out which are the terms in this relationship which depend on x. So, here is one the phi the scalar potential depends on x. So, you get the derivative of phi with respect to x and you have got the vector potential a x which also depends on x because it depends on r likewise a y also depends on r. So, there is an x dependence of a x there is also an x dependence of a y. So, you get a term in del a x by del x times d x by d t and a term in del a x by del y times d y by d t and there is a term in z as well. So, you write all of these terms carefully make sure that you bring them to the next slide carefully and then you can insert them in the Lagrange's equation because now you have got both the left hand side of the Lagrange's equation and the right hand side. The left hand side is given by this and the right hand side which is del l by del x is given by this. So, you just put the two to be equal to each other done. What do you find simplify this a little bit notice that you have got the velocity terms over here likewise you have got d y by d t here and v y over here d z by d t here and v z over here. So, you can combine corresponding terms and you are left with a relationship for mass times acceleration which is the traditional Newtonian mass times acceleration which is what would go into the Lorentz force law and on the right hand side I have moved these terms to the right with appropriate signs combine them and I find combine the terms and velocity because I can combine the term in d y by d t here which is on the left with v y which is on the right and this will move to the right with the minus sign. So, I will have both of them with the minus sign now one of them with the minus sign this is with the minus sign and this is with the plus sign. So, I get these terms in which the v y and the v z terms are combined. Now, let us bring it up and we will make use of something else which we also know that the magnetic field is given by the curl of the vector potential and if you just write look for the x component of the velocity cross B term you find that you have identical terms which we have seen on the previous slide you see exactly the same slide. So, you can insert the x component of v cross p over there now this is very easy to see. So, I will not work out this determinant for you and you can see that with the recognition of B being the curl of A you find that the x component of v cross B is what gives you these terms which you have found in the equation of motion coming from the Lagrange's you insert the corresponding terms. So, this term in the Lagrange's equation is replaced by q over c times v cross B you have seen it for the x component you have got corresponding terms for the y and z component and you get exactly what you are looking for what you are hoping to find because now you find that the x component of the traditional mass times acceleration which is the Newtonian force is equal to q into e plus v cross B and we are using the Gaussian system of units. So, which is why I have got a 1 over c is taking along with the vector potential in s i you would not have the 1 over c, but in atomic physics it is more convenient to use the Gaussian system. So, everything hangs together and you get the Lorentz force relationship 1 over c times v cross B which gives us the confidence that the ansatz we made for the Lagrangian is the correct one and we can use the Lagrangian further. So, let us use this Lagrangian now we know that it is no longer just a postulate it has led to the correct equation of motion for the electron in the electromagnetic field. So, we will use this Lagrangian we obtain the momentum which we have done already and now we are ready to put the charge which is the electron charge which is minus e. So, this m v plus q over c a becomes m v minus e over c a and now you quantize this which is to replace all the momentum operators by the corresponding derivative operators along with the I h cross and so on. So, that is something that you know how to do go ahead and quantize it and what you have is you have the fourth component you have got the three traditional components of momentum and the fourth component p 0 of the generalized momentum four momentum is nothing but this gamma times m c as you can see and this is what leads you to the Dirac equation for the charge particle in the electromagnetic field. So, now you no longer have to ignore the electromagnetic field in fact that is what you are really interested in. So, let us collect all the terms and this is the four by four Dirac equation that you get along with the operators it has got the derivative operators it has got the magnetic vector potential which has come from the generalized momentum and you have all of these terms stacked together in a matrix equation but the wave function has got four components that is a new feature we did not have it in the Schrodinger equation. Now, you have a wave function which has got four components. So, this is a four by four matrix operator and at the right side is a four by one four rows and one column null matrix these are the four Dirac matrices and make sure that you use the indices carefully because of your signature of the g you have got this a 0 a 1 a 2 a 3 which translate to a 0 minus a 1 minus a 2 minus a 3. So, keep track of the sign carefully use the signature correctly and you can write these terms expand them in terms of the betas and the alphas which I have introduced write out the components of the vector potential explicitly and you get a relationship which you can also write in terms of the Cartesian components and this is your Dirac equation as it is refer to in the standard form. So, this is beta times alpha dot p this i h cross gradient gives you the momentum operator this plus sign goes over to the minus sign and this is what is refer to as the Dirac equation in the so called standard representation you can transform it and put it in different equivalent representations but this is the representation that I shall make use of. Now, we have used the momentum operator p but now we have the fourth component. So, it is called as the generalized energy momentum four vector it includes the magnetic vector potential and you can write the four vector function if you want to write it in a block diagonal form because you know that the four by four Dirac matrices can be written in these block diagonal form. So, you can write the wave function also in two blocks at top block made of two elements at the lower block made of two elements and you can write it as phi tilde chi tilde and alpha and beta matrices are the gamma matrices that we have defined earlier. Now, for a free electronic rest you can simplify it and it has got these solutions. So, before we consider the more general solutions let us look at the free electron solutions. So, you set the electromagnetic field equal to 0 the traditional mechanical momentum equal to 0 that is what gives you the particle at rest and this is a very simple first derivative equation involving only the beta operator and it has got four solutions not one psi 1 satisfies it psi 2 also satisfies it psi 3 and psi 4 also satisfy it. But here you have got minus m c square by h cross as you expect whereas over here you have plus m c square over h cross as you do not expect or you did not expect. These are the positive energy solutions these do not surprises the negative energy solutions do and one has to see where these are coming from and what do they have to what is their place in physics. So, you first of all you get a multi component wave function and whenever you have a multi component wave function that is a signature of a spin that the particle must have a spin. And all elementary particles which obey Fermi statistics have two components these are spin half particle this is the result that you get from quantum field theory electron is a fermion. So, it is a two component particle, but the Dirac equation has given us four components two more than what we want and we cannot avoid because it came out of whatever we did to get a relativistic quantum equation. And then these negative energy solutions which cannot be avoided in fact have a real place in physics and the history of the interpretation of the negative energy particles is a very fascinating one, but that goes beyond our domain of atomic physics you need to study this in particle physics, but this is what led Dirac to predict the positron the antiparticle. In fact his earlier prediction was that these were protons rather than positrons because antiparticles were not known Dirac had to postulate them invent them and then Carl Anderson actually found them the origin actually lies in this relationship that we began with because you have a quadratic energy term which will have two roots one with a plus sign and the other with a minus sign. So, the origins can be traced to that and these are what give you the antiparticles or antimatter. Now, where is all this antimatter? We see matter as if antimatter does not matter, but it does it comes out of the Dirac equation and where is all this antimatter? When I read this question when I was preparing this slide I first thought that I will ask this question to Cosmic Raj and ask him where the antimatter is because he knows the Cosmos. He rules the Cosmos, but he rules only the Cosmos and not the Anticosmos. So, he is going to tell us to go to the Anticosmic Raj, but that is a matter of particle physics and I will not discuss this. These are interesting questions now these are our conclusions that you have a four component function. Two components is what we need for the electron, the remaining two components admit the negative energy solutions they correspond to the antiparticles to the positron in the present case, but we need only two components and we need to find a mechanism to reduce our four component theory to a two component theory. It is not clear that it is possible, but we are going to attempt to do so there are two ways of doing it one is the poly reduction which I will discuss and the other which is the more correct one more appropriate one which is also what I will discuss and the folding with Eigen transformations are very important in this context and that will take a good bit of our time in the next few classes. So, today I will conclude the class over here I will be happy to take some questions. So, there are these two techniques of reducing the four component theory to a two component theory and I will discuss this in the next few classes next couple of classes, but for now if there are any questions I will be happy to take otherwise goodbye for now.