 We will start with the many electron problem or in general quantum many particle problem. So we have revised up to the basic quantum mechanics and the exactly solvable problems particle in a box, harmonic oscillator as well as hydrogen atom okay and we kind of said that if you go for more than one electron which are interacting then the problem is not exactly solvable and the solutions are usually done through some approximations and two important approximations that we mentioned a variation method and perturbation theory. We have done that in the previous class but we will again redo this variation and perturbation in this class as well but unlike the last class where I actually presented variation and perturbation back to back here I will present this as and when required. So first probably variation and then perturbation when is required we will also I will also tell you the symbols because at this point we have to be very consistent in the symbols. Symbols that we will use we will try to be consistent as much as possible there would be some points where there may be problem but we will introduce the symbols. So first of all we want to talk of many electron problem when we are saying many electron problem there by definition interacting many electron problems because if they are non-interacting of course the problem is nothing but sum of one particle problem as we noticed. So first is a trivial case of a many electron problem where the electrons or the particles are not interacting with each other. This is something that we have already discussed but this is also a part of many electron problem but they are non-interacting and hence their solutions are trivial. So what is non-interacting that your Hamiltonian for n particle problem is a sum of n 1 particle problem so h 1 plus h 2 plus h n. Let us say that your Hamiltonian is of this form in this case the results are very easy the total wave function is a product and we will later see actually this will be an anti-symmetry product the spin orbitals which are Eigen functions of the one particle operator h so it is very easy. So basically you have to solve only one particle problem. It may also happen that this h and this h may be different so it may be h 1, h 2, h 3 even in that case the problem is trivial. So you have to simply solve h 1, h 2 etc and then later on make an anti-symmetric product so we will see. So this is an easy thing and the energy or the Eigen value of the n particle problem or the n electron problem is simply a sum is a sum of n 1 particle. So basically solve this one particle problem you will get an Eigen function whose products is the total Eigen function of this Hamiltonian and this Eigen value sum is the total sum of this n electron problem. So this is a trivial problem so this is really not called a many many electron problem because this is actually one particle problem it is just that from the one particle problem you can construct the n particle problem by trivial product and sum. So when we will talk of many electron problem henceforth it essentially means that it is an interacting many electron problem and not non-interacting which means the Hamiltonian cannot be written in this form as a sum of n 1 particle Hamiltonian. So those are the problems which are of interest because they cannot be now trivially solved by solving the one particle problem like this problem can be solved trivially as long as you can solve the one particle problem you can solve the n particle problem. So let us say let us look at now the interacting problem so we actually discussed it last time so we have an interacting many particle problem where the Hamiltonian has a form of this kind it will still have a one particle term so I am just writing this as sum of i equal to 1 to n where h is again a one particle operator just like this so I am now writing as a sum plus it has an additional term which is usually for the Coulomb problem written as i less than j 1 by r ij where r ij is the distance between the electrons i and j the coordinates of the electron i and j. So r ij is nothing but the distance between r i and j this is an interacting problem because now this problem cannot be written as a sum of n one particle problem simply because as I told you this term cannot be written as a sum of one particle Hamiltonians the one by r ij term so essentially for two particle it will be h1 plus h2 plus 1 by r12 that 1 by r12 which is the Coulomb operator cannot be split as sum of two one particle operators hence these theorems this theorem cannot be used okay so you cannot simply get the eigenvalue and eigenfunction of this problem by solving one particle equations. So this is a difficult problem and the in principle the solution when I mean by solution is eigenvalue solution eigenvalue equation of this Hamiltonian in principle the solution is not the exact solution I should say the exact solution cannot be found. So any method that we will propagate in this course will be approximate in nature so that is the first very important point to realize the nothing that we will speak is exact what is important is however to understand how to get the exact solution or how to reach towards the exact solution which means how to improve the solutions one approximation to another approximation how the approximations is progressively improved I think that is very important to understand otherwise of course the theories that we will develop will have no meaning so that is what we will do before we do that let us first try to look at some important symmetry of this problem. So let us say that I write a two particle problem as an example as h of one plus h of two plus 1 by r12 so very simple two particle problem which can now be written as sum of two one particle Hamiltonians h of one and h of two plus 1 by r12 which means essentially only one term of this summation because there is only one term for two particle problem. So first to note that if I interchange the two particles the coordinates of the two particles so write this h of two one then it is identical to h of one two because r21 if I interchange one and two r21 is exactly same as r12 and of course h of one plus h of two will remain invariant just a question of writing h of one plus h of two or h of two plus h of one so the point is that the r12 and r21 each of them represents the distance between coordinates of one and coordinates of two and this distance is invariant whether I call it one to two or two to one. Hence the entire Hamiltonian is invariant if I interchange one and two this is usually called the permutation invariance so the Hamiltonian is permutation invariance. So this is an important point for any n particle problem whether they are non-interacting or interact of course if it is non-interacting this is trivial. If it is interacting also the interactions are of the kind which are basically permutation invariant. So if you recognize this then we can right away do some analysis based on these symmetry and come to some conclusion so that is what we will first do because that is a very general conclusion that we will have. So we will first try to analyze this problem that in case of permutation invariance what do you get? So let us try to understand the word permutation little bit more carefully. So again I will take an example of a two particle problem. So permutation is an operator let us say P of one two so permutation itself I define a new operator called permutation operator it is a two particle permutation operator P12 it can be P23, P34 does not matter that depends on where it is acting. So the if it acts on a function two particle function 12 then it generates a new function which is phi of 21 remember this is not a Hamiltonian it is acting on a function. So if it acts on a function phi 12 the definition that it generates another function where one is replaced by two, two is replaced by one so I hope it is clear what this function means so we can take some example of a function which is a function of coordinates of one and two which are actually I told you this coordinates are essentially space and spin coordinates that you should not forget four dimensional coordinates but you can take any other function. So if you have for example a two particle function sin of x1 into exponential minus x2 let us say I have a function f of x1 x2 so what will this permutation operator do? It will generate sin of x2 e to the minus x1 it will simply change x1 to x2 so this will now be actually a space spin coordinates in our case but whatever it is if it is space coordinate it will simply change space, spin coordinate it will simply change spin so it will simply make x1 to x2 x2 to x1 so this will generate another function now this function of course may not be equal to this function or may be equal that depends on the nature of the function if the nature of the function itself is such that it is permutation invariant then of course it will generate the same function. Remember this is a function permutation invariance that I talked about is for the operator now I am saying in any arbitrary function if an operator acts on that function it changes just like we defined for the harmonic oscillator parity operator where parity operator acts on a function it changes x2 minus x the coordinate for one particle problem very similar we are defining here they are not making it negative they are simply interchanging so this actually interchanges the coordinates of the two particles so that is why we call it permutation operator then we notice that for such a permutation operator if you have the product of the Hamiltonian n particle Hamiltonian or n electron Hamiltonian with h and p acting on an arbitrary function phi 1 2 so this is for any function phi I hope all of you understand this symbol it is an arbitrary function so any function phi if I allow this to act then let us see what happens is that first p 1 2 will act on phi 1 2 because p 1 2 can act on whatever is in the right of this so this will make it h of 1 2 phi of 2 1 right h of 1 2 will remain that is an operator and this axing on phi of 1 2 will make it phi of 2 1 by my definition correct is it okay then I allow the reverse to act p of 1 2 h of 1 2 phi of 1 2 so basically this operator is acting on phi of 1 2 or this operator is acting on phi of 1 so I am just analyzing these two when this operator acts on phi 1 2 note that the p 1 2 now acts on this combination of h 1 2 phi 1 2 whatever is on the right of p 1 2 so what will it generate it will generate now 1 will become 2 2 will become 1 so it will generate h of 2 1 and phi of 2 1 correct so both the coordinates of 1 and 2 sorry both the coordinates of 1 and 2 are interchanged in the Hamiltonian as well as on the phi okay so both of them will be interchanged and here only one of them has been interchanged the other has remained as it is however now we note because of the permutation invariant this h of 1 2 is nothing but h of 2 1 alright so this can be further written as h of 1 2 phi of 2 1 or this can be written as h of 2 1 phi of 2 1 the point to note is that both of them are identical okay is it clear so if I have h p acting on any function or p h acting on any function the result is identical okay because simply because of this permutation invariant these functions are different of course phi 1 2 is different from phi 2 1 because that is an arbitrary function this may or may not be same but the results is different simply because the results are the same simply because h of 1 2 is equal to h of 2 1 so with that we can now note that if this is true for any arbitrary function phi what does it mean that this operator is identical to this operator I hope all of you remember operator equality when a is equal to b if a acts on any function phi and gives you equal to b acting on any function phi I mean the same function of course okay for any arbitrary function a phi equal to b phi then a is equal to b remember operator equality is a very strict equality so if a acts on a function and b acts on the same function on a given function there equally does not mean that the operator is equal correct so the operator is equal only if a acts on a phi and b acts on the same phi for all arbitrary phi if this is equal then we say a equal to b correct so that is very important this again please understand this symbol arbitrary phi any phi but if it is true only for a set of phi then it is not necessarily true the same goes for the null operator so if for example if a phi is 0 for all arbitrary phi then a is a null operator null operator is 0 operator so there are certain rules so here also it is the same a minus b acting on phi is 0 so a minus b is a null operator which means a is equal to b correct so it comes exactly from the same so now we can argue from this that HP so we can now say that the operator HP is same as the operator PA which means that the two operators commute with each other and then the rest of the argument on symmetry will fall on the same lines that we did for the parity operator very similar argument so what is the next argument the argument is that instead of looking for the eigen function of the Hamiltonian we will now look at the eigen function of the permutation operator because they commute and the theorem of quantum mechanics says that if two operators commute they have common eigen functions they share simultaneously the same eigen functions hence instead of looking for the eigen function of the Hamiltonian we look for the eigen function of the permutation operator this is basically also the objective of the group theory that in group theory we try to identify symmetry operators which are commute to the Hamiltonian and then we analyze the eigen function of the Hamiltonian in terms of the eigen functions of the symmetry operators so then we get the symmetry properties of the wave function so exactly the same argument we will use here and the same thing we did for the parity operator so we will now look for the eigen function of the permutation operator and let us see how they will look like so that will give us a clue on the eigen function of the two particle wave function so let us now look at the eigen functions of the permutation operator so how will it look like so let us say that I write the eigen function of the permutation operator as P12 acting on psi 12 equal to lambda times psi 12 remember now this is not an arbitrary function the psi is not an arbitrary function this is an eigen function of the permutation operator so just note this okay this psi is an eigen function hence I am saying that the operator acting on psi of 12 gives you a number times psi of 12 so that is the eigen value equation right operator acting on a function gives you a number times a function this is the eigen value this function is the eigen function right so we know this so we want to analyze this the reason we are analyzing this that we know that this psi will also be an eigen function of the Hamiltonian because of the this commutation rule right so let us see how this looks like so to analyze this let us apply exactly like we did for the parity operator in the parity operator case apply this operator twice on this so let us analyze this equation P12 acting on psi 12 and another P12 acting from the left so what will it give if you look at the eigen value equation this gives you lambda times psi 12 right lambda is a number so it can be taken out so then P12 will again act on psi 12 to give you another lambda times psi 12 so the result will become lambda square psi 12 is it clear to everybody because the eigen value equation is repeated twice in fact all of you know that if a psi equal to lambda psi a to the power n psi equal to lambda to the power n psi okay so it any power of that operator also has the same eigen value eigen functions not eigen values are going to be the power eigen functions are same so you have a lambda square psi 12 on the other hand if I analyze the left hand side by definition by definition you remember if P12 acts on psi 12 they become psi 21 for any arbitrary function so it will be true for eigen function as well then P12 again acts on psi 21 it will bring it back to psi 12 right so if you analyze this then I get psi 12 is equal to lambda square psi 12 so this is so this is equal to lambda square psi 12 and the left hand side is equal to psi 12 itself so then up trivially I get a result that for all eigen functions of the permutation operator the eigen function is the eigen value square times the eigen function again and that gives me a handle to give me get the value of lambda because now I know that from here lambda square is equal to 1 and hence lambda is either plus or minus so I get the eigen value for this result then what do I do I go back to this equation again so I write P12 psi 12 and now I put the eigen value is equal to plus or minus psi 12 correct plus minus 1 means plus or minus psi 1 just multiply plus plus 1 minus 1 and then apply the definition again on the left hand side the definition so by the definition this is psi 21 equal to plus or minus psi 1 so we get an interesting result for the eigen function of the Hamiltonian or the eigen function of the permutation operator what is that interesting result that if I permute the coordinates of the two particles one and two the eigen function will either remain the same or chain sign so that is a result that we get in fact this result is generally true for not only electrons but in general for all quantum particles because remember we have not used anything about electrons in this definition for all quantum particles the Hamiltonian permutes to the permutation operator simply because the quantum particles are indistinguishable so this really comes from the fact that the quantum particles are indistinguishable so I cannot distinguish between one and two so that is the reason this permutation invariance come so this result is a very general result of quantum mechanics that all quantum particles follow or obey a rule that if I interchange the coordinates of the electrons or the particles they will be either positive for the particles or the negative in the sign so this is called the symmetric properties with respect to permutation for quantum particles so then without any further proof we mentioned that if it is symmetric if it is symmetric which means if it is plus then symmetric with respect to interchange of the coordinates then these are called the Bose particles which are called the bosons and they follow the very famous Bose Einstein statistics and those which are anti-symmetric which essentially means the negative sign here are the fermions which obey the Fermi Dirac statistics so there are two very important statistics in quantum mechanics so for our case so for electrons again we have mentioned this in the last class also the last course also the electrons are fermions they obey the Fermi Dirac statistics so for the moment we will not worry about for this this course we will not worry about the plus sign we will only worry about the minus sign so essentially it means that all the wave functions that we will construct for the Hamiltonian from now on must be anti-symmetric with respect to interchange of the coordinates of two particles note that this result although I derived only for two particles this result is general for any n particle so for any n particle function if I interchange the coordinates of two particles any two particles the wave function will be anti-symmetric with respect to the permutation of those two particles so every pair interchange should produce a negative sign that is important thing that we should remember so if you now look at our theorems even for non-interacting theorem I said that the wave function is a product of the one electron eigenfunctions so now we can see that a product is not acceptable unless it is anti-symmetric so that is the reason I wrote in bracket if you remember anti-symmetrize product so the product by itself is not acceptable unless it is anti-symmetric so all electron wave functions must be anti-symmetric even if they are non-interacting or interacting so any time we talk of a product remember that they have to be anti-symmetrize product many times is anti-symmetric is actually called anti-symmetrized so how do I make this product to do this first again let me go back to the exact case exact case is the non-interacting n particle problem or n electron problem so from now on we will only talk of n electron and that is the whole subject today that the symmetry of the many electron problem so let us now go back to the n electron non-interacting problem so let us again concentrate on a two particle problem because two particle problems are very instructive two particle non-interactive very instructive because they give you all the physics that is required and they are simple to analyze so my Hamiltonian now is just H of 1 plus H of 2 is very simple I do not have any interacting term so this is a two particle non-interacting Hamiltonian so I have H of 1 plus H of 2 then what do I say the eigenfunction of this is just a product of or anti-symmetrize product of the eigenfunction of these two so let us now write down the eigenfunction of this note that