 Greetings, we have set up the essential machinery to start working with second quantization techniques and these are very powerful techniques to deal with a many body system, whether it is a both system or a Fermi system. And today I will introduce how a many particle Hamiltonian is set up in the second quantization formulation and how the Schrodinger equation is set up in the second quantization formulation. So, essentially we have the fundamental commutation rules for the Boson operators. The statistics is now put into the creation and destruction operators and what their commutation properties are. In the first quantization formalism, we have the statistics in the symmetry of the wave function. So, as opposed to putting the statistics in the symmetry of the wave function, we now put it in the properties of the operators and how these operators commute and with regard to the creation operators B Dagger and the destruction operators B, you have got these fundamental commutation rules and these contain all information about the statistics of the many body system. So, these are the fundamental commutation relations for the Boson operators and let us consider just to illustrate how it works a simple Bosonic system excitations of the simple harmonic oscillator which you would have met even in your earlier quantum mechanics course, but this will be a very quick recapitulation of the excitations of the simple harmonic oscillator. So, you have the simple harmonic oscillator Hamiltonian and this has got a quadratic term in the position coordinate k is the so called spring constant omega is the natural frequency of the simple harmonic oscillator and this is your simple harmonic oscillator Hamiltonian in the first quantization notation. This is the usual one that we work with now what we do is to introduce creation and destruction operators. Destruction operators are also often referred to as annihilation operators it is the same thing and these operators are defined in terms of the operators x and p. So, these are linear super positions of x and p this is the coefficient of x and i times this one over root twice h cross m omega is the coefficient of p and when you construct this type of a summation you get the destruction operator B. If you take its adjoint you get B dagger so instead of the plus i over here you get the minus i. So, these are respectively the destruction and creation operators for this system. Now, if you now construct the operator B dagger B. So, B dagger is this operator it has got these two terms. So, these two terms come in the first beautiful bracket and then B has got these two terms which come in the second beautiful bracket. So, I have simply written B dagger B in terms of the position and momentum operators and now you have got two operators over here and two operators over here. So, you can carry out the operator algebra and look at the four terms that you will get out of this. So, you will get a term in x square from this one and this then you will get a term in x p this is x and this is p then you get a term in p and x and then finally, in p square and the appropriate coefficients have to be properly taken care of. So, all we have done is to write B dagger B and now if you look at this term over here you see that i over 2 h cross is common to both of these terms and essentially you have got the famous commutator between position and momentum. You get x p minus p x and you know what it is from the uncertainty principle. So, that is the famous x comma p commutator which you know is i h cross and you can write i h cross in place of this and you have got a term which is quadratic in x and a term which is quadratic in p and then this i h cross. So, this i together with this i will give you minus 1 this h cross will cancel the h cross in the denominator and you will get minus 1 over 2 from the last term. So, you have got the first quantization Hamiltonian which is quadratic in x and quadratic in p and if you write this Hamiltonian recognize these terms then essentially it is the Hamiltonian divided by h cross this h cross has cancelled. So, you get minus 1 over 2 and B cross B is nothing but the Hamiltonian divided by the energy h cross omega minus 1 over 2. So, you can now place the Hamiltonian on one side and everything else on the other and that will give you the expression for the Hamiltonian in terms of the destruction and creation operators. So, let us do that. So, this is the Hamiltonian in the first quantization notation you have got B dagger B equal to h over h cross omega minus 1 half which is what we just obtained and if you flip this equation you get h is equal to B dagger B plus 1 half h cross omega. Are you expect this to be you know that this half h cross omega will be the 0 point energy which is a well known result for the simple harmonic oscillator and typically the energy will be n plus half h cross omega. So, that will identify the operator B dagger B with the number operator. So, this is the Hamiltonian for the simple harmonic oscillator in the notation of second quantization destruction and creation operators. Now, these are the main results that we have got this is the Hamiltonian these are the destruction and creation operators in terms of position and momentum fundamental operators and you can now add and subtract these results and take care of the coefficients and you can write x and p in terms of B and B dagger. You have got these two linear equations B and B dagger in terms of x and p. So, you can flip them and get x and p in terms of B and B dagger. So, you get x in terms of B plus B dagger and p in terms of B minus B dagger weighted by appropriate coefficients. So, that is very easy to see. So, this is now your number operator B dagger B because this will give you n plus half h cross omega. This is your number operator and the fundamental commutation rule if you remind yourself of what it is. It is B B dagger minus B dagger B is equal to 1 which means that B dagger B is B B dagger minus 1. Now, we can play with these terms to get some very useful results. So, what we do is find the operator n B n is this number operator B dagger B. So, if you now find n B n is B dagger B. So, this is n B, but n we have already found over here is B B dagger minus 1 and now you can expand this. So, that will give you B B dagger B from the first term and minus B from the second which essentially gives you B times n minus 1. So, this is your operator n B. Now, if you operate by n B on an occupation number ket, n is an occupation number ket. It is a vector in the occupation number space. Then since n B is equal to B times n minus 1 you have effectively B times n minus 1 operating on n. This is of course, an Eigen state of the unit operator 1 and the occupation number ket is also an Eigen ket of the occupation number operator which is capital N. So, the Eigen value of n minus 1 operating on n will be little n minus 1 and this is then further to be operated upon by the operator little B which is the destruction operator. So, you get n B operating on n gives you B operating on n minus 1 times n, but n minus 1 is just a number. So, essentially you have got n minus 1 times B operating on n. Now, if you look at this last result you immediately recognize that the vector in the occupation number space you get from the result of the destruction operator B operating on n is also an Eigen value of the number operator. Essentially you have got an Eigen value equation. So, the destruction operator gives you an Eigen value equation and the Eigen value is n minus 1. Now, let us look at the other possibility say you have got again B dagger B as the number operator. We have already seen that B n is an Eigen vector of the number operator belonging to the Eigen value n minus 1. Now, let us look at the norm of this vector n is an occupation number vector. We presume that all of these are normalized you have got an orthonormal basis in the occupation number space and the norm of B will be B n projected on its adjoint vector which is this. Now, B dagger B is the number operator. So, n operating on n will give you the number n times the Eigen vector n whose norm is 1. So, the norm of B n will be equal to n. Now, if this relation is to be valid also in the context of the fact that these occupation number vectors are themselves normalized which means that the norm of the vector n is equal to 1 and the norm of the vector with n minus 1 excitations is also equal to 1 they are all normalized. So, if both of these are normalized what does it tell you about the result B n B n must be square root of n times n minus 1. B operating on n will give you n minus 1, but it will scale it by a factor root n because if you now construct the norm of this vector the root n and root n will give you the norm of B n which is equal to n. Now, I need you to remember this result because we are going to refer back to this in a short while. So, make sure that you take note of the fact that when the destruction operator operates on the occupation number ket for a Bose system this is valid we I just use the example of the simple harmonic oscillator, but then this is much more general than that you can take any Bose on excitation field and you have got similar results and then you get an occupation number vector with the number of excitations reduced by 1. So, from n you go to n minus 1, but then you get a little bit of scaling and the scaling is root of n this is the original number of excitations in the vector on which B operated. So, this factor is something that I need you to remember for our discussion which is to follow. Now, we had worked with the operator n B in the previous example. Now, let us work with n B dagger. Now, what is n B dagger? n is B dagger B. So, n B dagger is B dagger B B dagger which, but this operator multiplication is associative. So, I can think of this as B B dagger, but B B dagger is 1 plus B dagger B or it is equal to n plus 1, because B dagger B is the number operator. So, this operator is n plus 1. So, this is n B dagger becomes B dagger times 1 plus n or n plus 1. Now, this being the case let us now use the operator n B dagger and operate on an occupation number vector n, what do you get? Now, B dagger 1 plus n is n B dagger. So, this is n B dagger and this is effectively equal to B dagger n plus 1 and this is an Eigen state of n and also of the unit operator belonging to the Eigen value 1 plus n. So, the result is 1 plus n times B dagger n. So, now this is the primary result which we have used and if you now look at the norm of B dagger n, if you look at the norm of B dagger n, you find that this norm is equal to n plus 1. What does it tell you for the result B dagger operating on n, because again these vectors n and n plus 1 are independently normalized. So, that means that when B dagger operates on n, you get one extra excitation. So, the occupation number goes from n to n plus 1, but then the vector gets scaled by a factor root of n plus 1. It is exactly the same kind of analysis as we had in the previous case. So, you get root of n plus 1 and again I need you to take special note of this factor that you not only go from n to n plus 1, but in this case you get a scaling through root n plus 1. In the previous case, you got scaling through root of n when you operated by B. So, do make a mental note of these results and we will now consider the fermions. The focal interest in atomic physics is on fermions, because you are want to study the electronic structure of atoms and how the atom respond to collisions, electromagnetic radiation etcetera. So, a lot of work has to be done with fermion operators in atomic physics. So, the fermion anti-commutation rules, these are the fundamental fermion anti-commutation rules and here you have got a r a s dagger plus a s dagger a r equal to delta r s. The two creation operators always anti-commute and the two destruction operators fermions always be 0. And we worked with these operators in our previous class just to get a little handle on these operators and we discovered that why this is the number operator that it square is always equal to itself that the Eigen values for fermions will always be either 0 or 1 and we recognize these operators to be destruction and creation operators and the number operator. So, we acquainted ourselves with these properties in the previous class. Now, let us look at some of these results. So, when you have got a creation operator which operates on the vacuum state, you get a state with one particle. We will work with an ordered set and this I would like to emphasize this does not mean that here you are dealing with numbers which have got this inequality, but here these are quantum labels which are arranged in a particular sequence. So, that the label a 1 comes before a 2 and the label a 2 comes before label a 3, because these labels are not necessarily numbers, they are just quantum labels. They are in fact a set of 4 quantum labels and you order them in a particular manner and having fixed that order in occupation number state, an arbitrary occupation number state in which you have got n fermions can then be obtained by operating the creation operators for all of these states from a 1 up to a infinity either 0 or 1 time. If you operated 0 times, you have got an operator a 1 dagger to the power 0 that is a unit operator. So, it does not do anything to vacuum it leaves it alone, but if the index here is 1, then of course, it will create a particle in that state and that will be the occupation number in this occupation number space representation. So, the occupation numbers vectors can then be written as a result of the creation operators for each state operating n number of times on the vacuum. Now, this is your representation of an arbitrary occupation number state and because of completeness, you require all of these infinite single particle states out of which only a certain number of finite number depending on how many electrons you have in the system, those many states are occupied all the others are vacant. So, only those creation operators corresponding to those occupied states would have operated on vacuum giving you an occupation number state. So, this is your fermion occupation number state. Now, let us take an example, let us take 3 electron system in the lowest 3 fermion state. So, n 1 is equal to 1, n 1 is the lowest 1 corresponding to this state a 1 lowest in the sense, it is the first one in the sequence I am not referring to energies, I am not referring to angular momentum, I am just referring to a certain order in which the quantum numbers are written up. You have got infinite number of single particle states, I have written all of these single particle states which are coming from Eigen values of commuting operators, they are coming from measurements what are compatible measurements. Those measurements give you a set of quantum numbers, those are 4 of them are stacked together in one label which is a 1, similar you have another set of 4 labels in a 2 and these are now arranged in a certain sequence, one that sequence is it does not matter, I can write phi p before 1 s if I want it, it is just an ordered set of these quantum states. So, I have got n 1 is equal to 1, n 2 is equal to 1, n 3 is equal to 1 and all the other ones are 0, which means that you have got these 3 creation operators which operate on the vacuum to give you this state and the creation operators for all the other single particle states are missing. So, their powers in this expression would be 0, which would give you a unit operator. So, this is your 3 electron system for the first 3 lowest in the sense the first 3 states are occupied and not any of the other. So, this is your 3 particle 3 electron system, so you have got 1, n 1 is equal to 1 in state 1, the number of particles in state 2 is also equal to 1, the number of particles in state 3 is also equal to 1. Now, let us destroy an electron in the state 2, suppose you have some mechanism, you can shine light on it and extract one of the electrons out of it, you can do various things. So, you destroy one electron out of these 3, which one the one which was in state number 2. Now, this is the one that we have chosen for the purpose of this discussion to be destroyed. So, you have got the annihilation operator for the state 2, A 2 operating on this occupation number state. Now, this state itself is the result of A 1 dagger A 2 dagger A 3 dagger operating on vacuum, so now you have got A 2 operating on these. Now, what happens, now when A 2 operates on this, if you want to move A 2 to the right, does A 2 commutes with A 1 dagger, it anticommutes and because the states 2 and 1 are different, they have to be different, they may have the same energy, but they are different states, because we are not referring to an energy sequence or something, we are talking about a sequence of quantum numbers. So, this could be the lithium atom for example, so you have got 1 S 2 2 S 1, the energy of the 1 S up and the 1 S down is the same. So, the energy ordering is not of significance, but clearly 1 S up is different from 1 S down and one of them is what I have labeled as A 1 and the other as A 2. So, I know precisely which one is being destroyed, so now A 2 is certainly different from A 1, because the labels 2 and 1 are different and therefore, A 2 A 1 dagger will be equal to minus of A 1 dagger A 2, because this anticommutator is equal to delta R S. So, since R is not equal to S in this case, this is equal to 0 and A 2 A 1 dagger will be equal to minus of A 1 dagger A 2. Now, let us take this result to the next slide, which is here and now you want to move A 2 further to the right of A 2 dagger. Now, when you do that again you have this anticommutation relation, but this time A 2 A 2 dagger will be equal to 1 minus A 2 dagger A 2, because these are the same labels. So, this time you will get 1 minus A 2 dagger A 2 and now you will get two terms 1, which is this minus A 1 dagger 1 operating on A 3 dagger operating on vacuum and the second is minus A 1 dagger operating on this minus sign do not forget and then you have got A 2 dagger A 2 operating on A 3 dagger operating on the vacuum. So, you get two terms. So, these are the two terms first is minus A 1 dagger A 3 dagger operating on vacuum, second is now it comes with a plus sign, because of these two minus signs. So, you have got a plus sign here and you have got A 1 dagger A 2 dagger A 2 operating on A 3 dagger, which is operated on vacuum, which gives you one particle in the state 3. Now, your state vector look at the second term here, in the second vector you have got one particle in the state 3 and this fellow can only destroy a particle, if it existed in the state number 2. It cannot destroy a particle in state number 3. So, this term will vanish, because the operand of A 2 has got a particle, it has got an electron, but not in the state for which the destruction operator is A 2. A 2 can destroy a particle only in state 2, not in state 3. So, this second term vanishes and you are left with only the first term that when A 2, you try to destroy particle number 2 electron number 2 in a 3 electron system and you are destroying the one in the second state. The result is you are left with a state in which you have got particles, electrons in state number 1 and state number 3, but now you have picked up a minus sign. So, note that there is a minus sign which has resulted. So, far as the occupation number phase is concerned, it gets multiplied by minus 1. So, there is a minus sign which results in this process. So, now let us look at a general occupation number state and now we will take not just a 3 electron system, but an n electron system and some of these infinite states are occupied, some are not occupied, a total number of n are occupied whichever does not matter that depends on which of these powers of the creation operators are unity. Those whose powers are 0 will be missing. So, you may have a very complicated configuration, not necessarily in the lowest energy n states, but you have got n electrons in n single particle states, but these can be any combination of n from an infinite number of possibilities. Now, from these possibilities, you now annihilate the electron in the fermion state S. In the previous case, we destroyed an electron in the state number 2. Now, we are going to do it, we are going to destroy an electron out of these n electrons in a particular state which is state number S which is labeled by S. So, this is your result, A S can destroy an electron only if the occupation number of the S S state is equal to 1 not otherwise, because that is the only one it can destroy. So, N S will have to be equal to 1 and this is now your expression that A S now operates on this occupation number state and how do you get this occupation number state by operating by A 1 dagger n 1 times, where n 1 may be either 0 or 1 that depends on whether A 1 is occupied or not. And then it will have the creation operator for the S S state operating one time, because only then you will get N S equal to 1 not otherwise, if this was equal to 0 N S would be unoccupied. So, here N S is equal to 1 and then other ones it does not matter, because the only thing A S can kill is a state in which the S S single particle state is occupied. So, keep track of the fact that you are dealing with the destruction operator which can destroy a particle only if the S S state is occupied. So, here is your destruction operator and this occupation number state must have the creation operator raised once and not 0 times. Now, the question is what sign will we get, because when we destroyed the electron in state 2 in the 3 electron system you remember we got a minus sign yes well you have to define what that S is. So, we are going to count that. So, what is the sign that you should get. So, let us see how you get the correct sign, because you already know that if N S is equal to 0 you will get 0, because you will get the number 0 not the vacuum state. And if N S is equal to 1 that is the question that we have now raised as to what is the sign that we will get. So, now what you have to do is to move A S to the right now again we are going to make use of the anti commutation relations and these are the fundamental anti commutation rules. So, if A S to be moved to the right of a creation operator you can certainly do. So, when R is not equal to S, but when you do. So, you must get a minus sign. So, this is what happens that every time A S is moved to the right of a creation operator you get a minus sign every time. So, how many times do you have to move it to the right is the question that is what it boils down to that you have to move this A S operator to the right of this all the way up to here, but then again you have to also move it beyond this creation operator. So, let us first move it up to here. So, when you move it just behind this creation operator you have already every time you have moved it you will raise minus 1 to the power a sum of all the occupied states till this all the occupied states. If any of those is vacant it will not contribute because that a dagger to the power 0 is just the unit operator. So, it does not contribute. So, you have got minus 1 to the power n 1 plus n 2 and so on up to n s minus 1. And now if you want to move it further to the right you will get 1 minus a is dagger a s because these two labels are now the same. So, this label was different from all the previous labels, but it is exactly the same as this label. So, you get 1 minus a is dagger a s. So, now again you get two terms one coming from one and the other coming from this minus a is dagger a s. So, here minus 1 has to be raised a certain number of times and this number is given by the total occupancy of all the states till the s th state. So, it is really not s minus 1 you have to add the occupation numbers of those states. So, you need to add all the occupation numbers of states and these are n 1, n 2, n 3 and so on up to n s minus 1 and that is the phase that you pick up. And then you have got a s operating on the vector occupation number vector to its right, but the occupation number to the right occupation space vector to the right of this destruction operator certainly does not have an electron in the s th state. So, that will give you a 0 just the way a 2 operating on a 3 dagger vacuum gave you a 0. So, that will give you a 0 null vector and this is now your result. So, it is minus 1 to the power a certain sum which is not s, but it is n 1 plus n 2 plus everything up to n s minus 1. So, this is the phase that you get. Now, here n s must be equal to 1 whatever annihilation this is the result that you get that you certainly can annihilate you can destroy a particle. If it is occupied if it is 0 if it is not occupied the result is the number 0, if it is occupied you will get a destruction of an electron in that state giving you a new vector occupation number vector in which the occupancy has now dropped by 1. So, it becomes 0 likewise if you do the same exercise with creation operators you get a similar result. So, I leave that as an exercise it is a very similar kind of analysis and when the creation operator operates it cannot create an electron if the state is originally occupied. So, a dagger operating on a state in which is already occupied the first row will give you the number 0 this we discussed also in the previous class this is essentially the Pali exclusion principle. And then if the original state is unoccupied so, if n s were 0 then certainly the a s dagger can and does create an electron in the state giving you a state with increased occupation and then you will get n s equal to 1. So, you get a phase minus 1 to the power s where s is the sum of all the occupied states till the previous state. So, this is your result now if you now operate by a s dagger on a s then if n s is equal to 1 then you will get a phase minus 1 to this is what you get because first you operate by a s. So, you have destroyed this particle so, you get a result in which n s is equal to 0 and on this you will now create a s dagger, but once you do that again you will get a phase of minus 1 to the s. So, minus 1 to the 2 s no matter what it is will always be plus 1. So, this is the number operator which will leave this up vector in variant with n s equal to 1 if it were 0 it would give is 0. So, this is the Eigen value of the number operator it is either 1 or 0 we already know that. So, in this case it is equal to 1 and if it is 0 then you will get n s equal to 0 then you will get n s equal to 0 you will get the same state it will not be changed and essentially what you find is that n s the destruction operator would give you a 0 and you will get the Eigen value to be 0. So, essentially the a s dagger a s is the number operator whose Eigen values are either 1 or 0 as we have seen earlier. Now, I am going to rewrite this result in a slightly different form which is well adapted for the second quantization formulation it is the same result, but we are going to rewrite it in a slightly different form it is a essentially the same result which we have at the top, but now I have inserted a factor n s over here and this is strictly correct because if n s is 0 you get a 0 that here I can write these expressions with the occupation number n s over here and this works for both whether it is occupied or unoccupied. If it is unoccupied n s is 0 if it is occupied n s is equal to 1 the important thing is that here the occupation number becomes n s minus 1, but this is applicable only if n s were originally equal to 1 because there is nothing like a minus 1 occupancy. So, you do not go from 0 to minus 1 you can only go from 1 to 0 you can only go from 1 to 0 and not from 0 to minus 1 when you are operating by the destruction operator. So, this result which is written with n s can be equivalently written with a square root of n s is the same its numerical value is exactly the same, but the advantage here of writing it with the square root n s. So, this is the certain convention that you introduce because this makes it completely equivalent to the Bose case because in the Bose case you remember that you had when the destruction operator operated on an occupation number state you got another state with one Bose on less, but there was a scaling by a factor square root of n. When you created the scaling was square root of n plus 1 you remember that is the result which I had asked you to take note of this is where we use it that by writing this s square root of n s you can certainly write it as n s times this, but by writing it instead s square root of n s you have got a completely equivalent expression with the advantage that you have got a relation which is identical to what you have got for the Bose case except for the fact that you now have this phase factor minus 1 to the power s. So, this is an additional factor that you have to work with now likewise you remember that there is this phase factor minus 1 to the power s and now if you look at the expressions for the creation operators then you have got you can write these results as 1 minus n s because if n s were 0 you get 1 times this and the occupation occupancy goes up by 1 from 0 to 1. If n s were 1 then 1 minus 1 would give you 0 which is the fact that you cannot create a particle in an already occupied fermion state. So, this expression is completely equivalent to this, but again instead of writing this as 1 minus n s times this instead of this coefficient we can use the coefficient root of n s plus 1. So, notice that these two are completely equivalent because if n s is equal to 0 then you have got 0 plus 1 and you get a square root of 1. If n s is 1 you cannot create any particle any another fermion on that. So, these two expressions are completely equivalent and this makes it look just like the Bose case except for the phase factor which is minus 1 to the power s which you must always remember when you are working with fermions. So, now let us ask ourselves how to write the many electron Hamiltonian in the second quantization formulation. So, we have a many electron Hamiltonian in the first quantization formulation it is a sum of all these single particle operators and the two particle operators. This we have discussed at great length in the context of the Hartree-Fock formalism in the previous course. So, you have sufficient familiarity with this and I will use that in our discussion now. So, this is your many electron Hamiltonian which is a sum of the single particle and the two electron operators and from the previous class we wrote a general n electron Schrodinger equation. Here this is valid for those cases inclusive of correlations not just the uncorrelated system in the single particle approximation. Now, you understand what the correlations are what the single particle approximation is. The independent particle approximation is the Hartree-Fock's later determinant. When you have multi-configuration Hartree-Fock when you have a large number of slater determinants why large anything more than one anything more than one then you have got essentially an interaction between those two configurations which are represented by the two slater determinants. So, if you have a system of that kind then you have a correlated wave function which is again written in terms of product of single particle functions. So, this is the product of single particle functions which is all right because after all these electrons are fundamental elementary particles. So, which is why you write them as a product of single particle functions, but then you must have alternate possibilities because there is no guarantee that the electron at coordinate x 1 will be in state even prime or the electron at state x 2 will be in state e 2 prime they could be interchanged the electron at coordinate x 1 could be in the n th state and vice versa. These are all indistinguishable particles and you must therefore, consider all of those possibilities every time you carry out an interchange you pick up a minus one factor because of the anti symmetry of the wave function. But then there is not a single slater determinant to talk about. So, even prime and up to e n prime there is one coefficient corresponding to one slater determinant, but then you must sum over all of these possibilities. So, each even prime can go over the entire infinite set of possibilities. So, even prime will have access to infinite single particle states e 2 prime will also have an access to infinite single particle states and when you sum over all of them you have got the most general many electron wave function. Now, we worked with this in the previous class and we recognize that the symmetry of the wave function is built into the coefficient c and I am just going to remind you of some of the steps that we discussed already in the previous class, but I am not going to spend any time doing it which is why I have got this green arrow which is to tell me and to tell you that we are not going to spend any time on this, but you can refer back to the previous class and what we did was to take the Schrodinger equation multiply this by a particular set of product of a joint vectors. Having done this we integrated over all the coordinates from x 1 to x n we carried out integration over all of these. We did it in considerable detail in the previous class and then what we did was to exploit the orthogonality of these single particle states and that gave us contraction of these over the sums. This we did in some detail in the previous class, we exploited this orthogonality and as a result of the exploitation of this orthogonality then we plugged in the complete form of the Hamiltonian which is the sum of the kinetic energy terms and the potential energy terms or the single particle terms and the two electron terms. When we did this we separated the kinetic energy term and the potential energy term. So, all this we did in some detail in the previous class. So, I am taking you very quickly through those steps and then we carried out integration over all the independent degrees of freedom separately. So, the only thing which remains to be integrated are those terms such as this like the kinetic energy operator or the single particle operator which is sandwiched between E k and E k prime when the arguments of both of these are x k because the integration over x k is independent of that over any other coordinate all of these are independent degrees of freedom. So, we separated the integrations over various independent degrees of freedom from these the other ones we get orthogonality integrals. So, they give you either a 0 or a 1 and then when you sum over all of these even prime E 2 prime and so on you contract all of those summation. So, you left with only one sum in the kinetic energy term and with a double sum in the case of the potential energy term. So, this is where you get the contraction. So, you are left with a single summation over E k prime in the kinetic energy term and what happens in this term is that you have got this integral over x k which has to be carried out and this is the one which connects E k prime and E k and in the corresponding coefficient E k prime will appear once extra and E k will appear once less. Now, this is the important thing because we are working with this occupation number formalism and counting is important everything is everything hinges on the number of times the operators operate. So, counting is important E k prime appears once extra and E k appears once less in this term. Now, this is what happens in the kinetic energy term, in this business in the occupation number formalism it is important to count and you have already seen that in the kinetic energy term E k prime appears once extra E k appears once less, but then there is a summation over k going from 1 through n. The potential energy term E k prime and E l prime appear once extra and E k and E l appear once less, but then there is a summation over k and l each going from 1 through n. So, we will carry this information into our occupation number formalism in which we recognize we discussed this point in the previous class that the coefficients of which the arguments are these single particle states, which single particle states those which are occupied those are the ones which come in those coefficients. There is also a time dependence because of our single particle states are completely independent of time all the time dependence is in this coefficient c. So, there is a time dependence in the function f which is a function of all of these occupation number and these two coefficients are completely equivalent the one on the left hand side is immediately adaptable to the occupation number formalism. The one on the right hand side is the one that we use in the first quantization formalism, but then when you go over from the first quantization to the second quantization then the information which is contained in which of these coefficients which of these quantum states are occupied that information goes into these occupation numbers because if E 2 is occupied then N 2 is 1 if E 2 is not occupied then N 2 would be 0. So, there is a 1 to 1 correspondence which is why these two coefficients are completely equivalent. So, this is the information that we are going to carry and let us do that and subject to the consideration that E k prime appears once extra in the kinetic energy term and the potential energy term E k prime and E l prime appear once extra, but E k and E l appear once less. So, with reference to this if you now rewrite the same expression, but instead of the coefficient c we will now rewrite this expression for the coefficient f which is completely equivalent. So, the coefficient f is completely equivalent, but you must also carry this information about the occupancy. So, let us do that. So, to be able to do that I plug in the information about the occupation numbers explicitly over here and here you remember that we had these square root N plus 1 and square root N factors because if a term appears once less then you have got a destruction. If it appears once extra you have got a creation and you have to look at this matrix element of the single particle operator which is the kinetic energy operator over here, but it can also include a single particle potential energy term like what each electron experiences from the nuclear field. So, the z over r is also a single particle operator it is only the E square over r i j or r 1 2 which are the two center particles. So, you have square root of N and this will come into picture only if E k prime N E k prime this occupation number of the E k prime its state is equal to 1 because what this kinetic energy integral term is doing is you can think of this as if it is transferring a particle from E k prime state to the state E k and that would happen if and only if E k prime were occupied not otherwise. If E k prime were vacant to begin with there is no way this could be done which is why there is this chronicle delta which takes care of it likewise you have got a delta N E k comma 0 over here and this is scaled by the square root N plus 1 factor. Now, in the potential energy term you have got similar terms just like these two, but now you have got four of them it is exactly the same logic and here the potential energy integral is this E k E l this is the potential energy operator and on the right side you have got E k prime E l prime and now you instead of summing over k from 1 through l you now will sum over all the E k states because this was summation over all the occupied states. Here you will sum over all the single particle states, but whether or not they were occupied is taken care of by these square root of N and square root of N plus 1 and the chronicle delta. So, the same information is transferred, but we are now equipping ourselves with a reformulation of the many electron problem from first quantization formulation to the second quantization formulation. So, the occupation information is now contained in these factors, but now the summation there is a double summation this summation is over E k prime and this is a summation over E k. So, instead of k going from 1 through N you now sum over E k, but now the summation is over all the single particle states, but all of them were not involved to begin with, but so also now because now you have got this delta N E k 0 factor and here you have got the chronicle delta E k prime comma 1. So, that takes care of it. So, now you have plugged in the information about the occupations and now you have got a double summation over here and the quadrupole summation over here over these 4 quantum numbers. Each quantum number is a set of 4 quantum numbers, but here we write it as 4 labels and all the information which is there in the coefficient c is in the coefficient f as you know. So, now instead of these coefficients c with the information we now have about these occupation numbers, we can now replace these coefficients c by the coefficients f and go over to the occupation space vector. So, let us do that when we do that this E k prime which had appeared once more if you remember, but you are summing over E k prime and it could this excess E k prime could be anywhere from the first to the last, but remember that when you go to the occupation number space formalism you have to put it in exactly the same order in which you chose to identify your single particle states. So, we worked with an ordered set A 1, A 2, A 3 up to A infinity and you have to move this E k to its appropriate place and when you move it you must pick an appropriate phase which will be minus 1 to the power a certain number which will depend on how many times you have to move it to get it to its appropriate position. So, when you if E k prime were less than E k then you will have to pick up minus 1 to the power E k prime plus 1, because you will have to move it only beyond E k prime till you get to E k minus 1 to 1 preceding. So, that is the number of times you will pick minus 1 factor. So, you can add all of these powers of minus 1 and that is the phase you will have to plug in. So, now we will put this extra phase in our factor we have got the numbers already. Now, we have to put in this additional information about the phase and now with this additional phase information these numbers and the chronicle deltas you have got everything that you need to go over to the occupation number vectors. So, now we use this equivalence between the coefficient C and the coefficient F and with respect to the equivalence between the coefficient C to the equivalence between the coefficient C and the coefficient F you now have this partial time derivative of the function F which is the function of all these occupation numbers which is equal to h psi on the right side, but this is also now written in terms of the function F which is the function of all the occupation numbers. Here notice that the occupation number of E k prime is 1 less the occupation number of E k is 1 more in the kinetic energy term in the potential energy term it is occupation number of E k which is 1 more E k prime which is 1 less E l which is 1 more. So, it is n E l plus 1 and here E l prime it is 1 less and then you have got all the phases and all the occupation numbers here. Now, you have got everything we have used this equivalence and written the rate equation. So, this is the time evolution of a state vector, but now we are now able to carry over this discussion into the occupation number space. So, we can now do a little bit of simple manipulation of this phase factor because this is very simple because minus 1 to the power this phase can be written as minus 1 to the power S E k minus S E E k prime this is very easy to see. I will leave this as an exercise for you to figure out it is very straight forward and likewise you can work with the phases in the potential energy term also and write all of these phases here in terms of these minus 1 to the power S and minus 1 to the power minus S E k prime and using these phase factors you have got this term over here. So, I have got the same phases, but they have been written in a more compact form. Now, what is this? Now, look at this vector here here what is the information that you have got? You have got the function f which is the function of all these occupation numbers, how can you get this occupation numbers? You can get this if you destroy a particle in the state E k prime and if you create one in E k and then you also have these phases. Now, what you find is that that is exactly what you are doing that if you have a general occupation number vector n 1 n 2 all the way up to n infinity then only if E k prime is occupied which is why you have got E k prime comma 1 chronicle delta. This chronicle delta would be 0 if E k prime were not equal to 1 if the occupation number of E k prime were not equal to 1. So, this chronicle delta would give you a 0 if E k prime was unoccupied, but when it is occupied you can destroy a particle from E k prime and that would give you a new state in which the occupation number of E k prime would be 1 less which is n E k prime minus 1. So, this state is completely equivalent to this state on the right hand side and now you can write this result completely in terms of occupation number state vectors because of the kinetic energy term you have got a dagger a, but the destruction is in the state E k prime and the creation is in the state E k. This is exactly what gives you the correct phases and the correct square root signs over here. So, you have that for the kinetic energy term. So, for as a potential energy term is concerned it takes a little more time to see that, but it is based on exactly the same logic. There is nothing new in it absolutely no new logic in it that you have got the phases you have got these chronicle deltas which tells you whether you are going to get a 0 or 1 and you are going to get a 0 or 1 depending on whether you are trying to if you are trying to destroy a particle destroy an electron from a state which is unoccupied you will get a 0 because you cannot do that you can destroy an electron only that state is occupied. So, that information is sitting in these chronicle deltas and here you see that the occupation of E k has gone up by 1 occupation of E k prime has gone down by 1 occupation of E l has gone up by 1 occupation of E l prime has gone down by 1. So, now this result is completely equivalent to the operation of on and general occupation number state by these creation and destruction operators. There are two creation operators for E k and E l and there are two destruction operators E k prime and E l prime, but find you they must come in exactly the order in which you see them on the screen because they must satisfy the anti commutation rule. So, you cannot write them in any arbitrary order you must write them in exactly the in the order in which you see them here. So, what we can do is in place of what you have in these two square boxes you can write the right hand side and you have got for the kinetic energy term or right hand side which can replace this and in the potential energy term you have got this information on the right hand side in terms of the occupation number state vectors and the creation and destruction operators which can replace the corresponding term in the first quantized notation or what is some sort of a transition from the first quantization to the second quantization for management. So, using these two results you now have the Schrodinger equation the Schrodinger equation which is the time evolution of a state vector in the occupation number space is now given in terms of these are of course, integrals these are single center integrals integrals over a certain coordinate. Here you have got a two center integral and then you had a creation and a destruction, but you must sum over all of these states e k and e k prime both going from 0 not from 0 to infinity, but over all the infinite single particle states. All of them must be summed over so now we have the final expression for the Schrodinger equation in the second quantization formalism which is this, but mind you these operators must be written exactly in this order. So, I have brought that expression to the top of this slide here, but now for simplicity I will replace the labels e k, e l, e k prime and e l prime respectively by r s t u and mind you there has to be a one to one mapping do not mix them up because the ordering is very important. So, making the notation a little simple so that r s t u are effectively single particle quantum states right. If you just rewrite this expression with e k, e l, e k prime and e l prime written in terms of r s and t u then you have this expression here. I have only renamed this there is no new physics in it it is just a renominclature of the expression and this is what the Hamiltonian turns out to be. So, what is in the square bracket is the Hamiltonian i h cross del by del t of an occupation number state is now equal to h operating on this state in which this is now the Hamiltonian. So, now the Hamiltonian in the second quantization formalism is identified and it is now written in terms of these creation and destruction operators and of course, there are these integrals r t s and so on. So, this is your Hamiltonian in the second quantization formalism this is your Schrodinger equation in the second quantization formalism this is for a general n electron state. Now, we can write this as the Hamiltonian in a slightly different order because you can move a t to the right of a s dagger, but then you pick up a minus sign and you can move a t further to the right of u and again it will pick up a minus sign. So, it will become plus. So, the original ordering a r dagger a t a s dagger a u becomes a r dagger a s dagger a u a t. So, you can rewrite this Hamiltonian in terms of this a r dagger a s dagger a u a t instead of a r dagger a t a s dagger a u. This ordering is completely equivalent, but please note the fact that you can write it in some equivalent ordering, but not in arbitrary ordering. So, any equivalent ordering which duly respects the anti commutation rules of the Fermi on creation and destruction operators is acceptable, but not any arbitrary order. Note that here you have got t u in this two center integral over here you have got a u and a t. So, this order is different from this and you cannot mess up with that because this two center integral if you remember this is the we worked quite extensively with this in the discussion of the Hartree-Fock formalism in the previous course. This is an explicit expression for the two for the integral over q 1 and q 2 which are dummy variables which get integrated out and in this integration you have got on the right hand side phi t q 1 phi u q 2 which is telling you that this is the probability amplitude that particle at q 1 is in the state t and this is the probability amplitude that the particle at q 2 is in the state u. Now, you can interchange that, but if you do you must accommodate a minus sign which is what in the Hartree-Fock formalism gives you the Coulomb and the exchange integrals. In the Boson case it would not matter because a u a t come in the Fermi on case you have to be careful. So, I will take a break here and we will continue from here in the next class. If there is any question I will be happy to take, but in a sense you have the Schrodinger equation here in the second quantization formalism and this we are now going to use this in our subsequent applications of second quantization to deal with correlations in a many electron system. That is the main subject of this unit which is the second unit in this course that we will be working with the Schrodinger equation for a many electron system. This is inclusive of all the correlations now, all the correlations can be built in and now you must keep track of the order in which these operators come because the Fermion operators anticommute unlike the Boson operators. In the case of Boson it does not matter, in the case of Boson you also do not have those phase factors which you have in the case of fermions and then with respect to this we will now proceed with our discussion of correlations in the many electron system. So, we now have the Schrodinger equation in the second quantization formulation, we have got the Hamiltonian in the second quantization formulation. Question? On general question, yes, from where the Pauli's exclusion principle, from which basic principle, I mean historically Pauli's exclusion, spin studies theorem or came as an property of the anti-immunization model? No, the Pauli's exclusion principle came much ahead of the second quantization formulation. Historically, we have the second quantization formulation was developed in the late 40s, Dyson, Feynman, Wicks, many contributors to that. So, there are theorems named after Wicks, there are theorems named after Dyson, Feynman, all of them Pauli's exclusion principle came even before spin because Pauli recognized that if you start filling in electrons in like you begin with the hydrogen atom 1 s 1, in the helium atom you have 1 s 2 and then you go across in the periodic table, go to lithium 1 s 2 2 s 1. Now, if you start doing it over the entire periodic table, you cannot get the correct configurations unless you had some quantum number which had two values. This is what Pauli recognized even before spin was recognized, then subsequently when spin was recognized, which happened through experimental observations these were the interpretations of Ullinbeck and Goudsmit and they suggested that to understand Zeeman spectra, not just the Zeeman, but the family of Zeeman effect spectra including the Passchenbach effect and the anomalous Zeeman effect and everything, the entire range of Zeeman spectroscopy to understand that Ullinbeck and Goudsmit proposed that there has to be a half integer quantum number. Then it was recognized in the Dirac equation that there is an intrinsic angular momentum which is half for the electron. Then after Hartree's work it was recognized that electrons being identical particles, this symmetry of the function must accommodate the fact that these are half integer particles and this is the spin statistics here which Pauli formulated much later, it came much later because many electron systems were being studied. Hartree-Fock was in the 1928 and the Coulomb and the exchange integrals which were completely inspired by the fact that the wave function must change its sign when you interchange two particles. So, all this preceded, this is the formulation of quantum mechanics, this is not really new physics, but this is a new formulation which is very elegant, which is extremely convenient. So, for many electron atomic physics or molecular physics or even condensed matter physics, it provides great elegance and great convenience. In relativistic domain it becomes a necessity because you can actually create and destroy particles because energy and matter are convertible, but that requires energies which are more than the sum of the energy of a positron and an electron. So, only above 1.2 million electron volts will you have to work with that, but the kind of energies you work with an atomic physics or molecular physics, these are of the order of few electron volts, tens of electron volts, hundreds of electron volts, even thousands or even tens of thousands if you go to deep inner shell x rays and so on. We do not go to millions of electron, so in these processes in atomic physics you are really not considering, you are not working in the energy domain in which an electron positron would annihilate each other and you have got energy and you are not really carrying out creation and destruction in that sense, but what you are doing is you are considering configuration interaction once later determinant is not appropriate to describe an electron system. It gives you only one configuration, but there may be n number of configurations 2, 3, 4, may be hundreds and to be correct you really need to consider an infinite set. Now, that is what is being summed over here, because each of these states r and s, each r goes over every possible single particle state which is in the Eigen spectrum of the single particle Hamiltonian. So, you first stack those Eigen states, register them in a certain sequence which you call as a 1, a 2, a 3, a 4 and with reference to that ordering the rest of the formalism is developed. So, now we have got the Schrodinger equation in the second quantization formulation, we have got the Hamiltonian in the second quantization formulation and we are going to find it extremely useful to deal with many electron correlations, because the Hartree Fock takes into account only the exchange or the poly correlations, but not the Coulomb correlations. Now, our interest is in studying these Coulomb correlations in atomic physics, that is what this is about. Any other question? Thank you.